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http://physics.stackexchange.com/questions/55147/em-waves-how-do-they-travel-for-billions-of-km-without-damping/55149
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# EM waves: How do they travel for billions of km without damping
If a star is 1 billion light years away, it means that the light we see from the star is emmitted billions of years ago.
How does this light not undergo a frequency change or get damped inspite of the collosal distance of travel? I am asking this as doppler effect is used to know star-distances and is one of the basis for big-bang theory
Also it travels in vacuum. According to Maxwell/Gauss, a charged particle is responsible for electric field whose change creates magnetic filed. Inturn, its change causes Electric field, a process that occurs infinitely. Even if we assume that the energy travel in vacuum as the energy produced from a charged particle elsewhere is used for propagation in vacuum, does the waves not damp significantly?
In the radio waves emitted/transmitted for earthly communications, power of source is directly proportional to the distance travelled by the wave. Is this not contradicting the fact that stars light reach us from the humongous distances?
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– Qmechanic♦ Feb 26 at 16:33
## 2 Answers
The light from distant galaxies does undergo a frequency change. It is red shifted, and the amount of red shifting is used to work how fast the distant galaxy is receding and therefore how far away it is. However this is not a damping effect. The light red shifts because the spacetime in between us and the distant galaxy is expanding, so although the light's energy is conserved it is spread over a larger distance.
You ask why light isn't damped, but why should it be? Since energy is conserved the light wave can't lose energy unless there is some mechanism to carry the energy away. For a light wave travelling in vacuum there is simply no mechanism by which it can lose energy, so it doesn't.
In your last paragraph I think you're getting mixed up with a different effect. When a terrestrial radio station broadcasts it sends the radio waves out as a half sphere (the half above the ground). As you get further away from the transmitter the field strength of the radio waves decreases as the inverse square of the distance because the energy of the transmitted wave is spread over a larger area. However the total energy of the light wave is conserved. This obviously happens with distant galaxies as well because the more distant a galaxy is the fainter it appears. This isn't due to damping, it's just the inverse square law dependance of the field strength.
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Thanks! So I understand that the brightness decreases NOT due to lower energy(amplitude) and frequency, but due to the amount/fraction of waves itself reaching us. – Aad Feb 26 at 8:50
Yes. The fraction that we see of the total energy emitted decreases as $1/r^2$. – John Rennie Feb 26 at 9:06
First, we don't observe individual stars that are 1 billion light years away (i.e. starlight emitted 1 billion years ago). Individual stars we observe are either supernovae which may be outside our galaxy but they are bright enough and visible for a short moment of time only; or stars in our galaxy, the Milky Way, which are at most 200,000 light years away from us.
The more distant objects we observe are whole galaxies – different from our galaxy. Quasars – visually "quasistellar objects" which are actually active galactic nuclei – are usually even further than that.
An object that is 1 billion light years away does get redshifted by Hubble's law. Hubble's constant is about 22 km/s per million of light years so 1,000 million light years gives them speeds of order 22,000 km/s, or 1/14 of the speed of light. This is surely a significant, observable redshift. The frequencies get lowered by the factor of $15/14$, too. It's not "quite" a coincidence that the number 14 is the age of the Universe in billions of years although you can't use this formula if you want a great accuracy.
Electromagnetic waves in the vacuum proceed indefinitely and they don't lose any energy whatsoever – except for the loss of the energy (and frequency) of photons described in the previous paragraph and (one more multiplicative factor of the same size, $14/15$) the decrease of the number of photons we collect each second.
The energy of the electromagnetic waves in the vacuum can't be lost due to the energy conservation: there's just nothing in the vacuum where the energy could go and it's easy to see that the precise, undamped waves are solutions to Maxwell's equations of electromagnetism. An even more obvious point is that the frequency can't change at all (except for the Doppler shift due to the relative speed, either due to Hubble's expansion or some extra velocity or due to different gravitational potentials in general relativity). Why it cannot change at all? Because if the electromagnetic waves are created by some process (acceleration of charges) that has some frequency, the "input" perturbations are periodic functions of time with the period $\Delta t = 2\pi/f$ which implies that all of their implications – such as waves measured a billion years ago – must be periodic with the same period, too. The frequency just can't get changed.
The total energy from a star or a similar localized source gets distributed to a surface $4\pi R^2$, the surface of the sphere, where $R$ is the distance. So indeed, how much light from a star we may see is decreasing as $1/R^2$. However, the angular (apparent) size of the star is also decreasing as $1/R^2$ which means that the density of energy (light) per unit solid angle is actually independent of the distance $R$. Stars that are further look "smaller" but the "intrinsic color" of the dot doesn't depend on the distance.
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Thanks a lot for the attention to detail. It really helps! – Aad Feb 26 at 8:51
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http://mathoverflow.net/questions/86498?sort=newest
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## Quantum coordinate ring at root of unity
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Noah Snyder gave a great answer to this question about different versions of a quantum group $U_q(\mathfrak g)$ when $q$ is a root of unity. I want to ask about forms of the deformed coordinate ring $\mathcal O_q(G)$ when $q$ is a root of unity.
Let's focus on $SL(2)$. Recall that the "small quantum group" (see Noah's answer) is obtained by dividing $U_q(\mathfrak s\mathfrak l_2)$ by the ideal generated by $E^e,F^e,K^e-1$ (which is central since $q^e=1$). The quotient is a finite-dimensional Hopf algebra $\overline{U_q}(\mathfrak s\mathfrak l_2)$. Since $\mathcal O_q(SL(2))$ and $U_q(\mathfrak s\mathfrak l_2)$ are in duality, one expects that there is a corresponding finite-dimensional subalgebra of $\mathcal O_q(SL(2))$ (the set of elements annihilated by $E^e,F^e,K^e-1$). What is it, and are there references about it? Is there a good reason why it is hard to work with compared to the Hopf algebra perspective?
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I think that if you take in $Q_q(G)$ elements of the form $X^n$ they form an ideal. Small check - if I take elements from the same column of $Q_q(GL)$ they are q-commuting variables xy=qyx (well-known), so x^n , y^n are central in column-subalgebra. However "duality O(G) and U(g)" is somewhat subtle - it requires completions - since U(g) - is "delta function at identity for O(G)"... These completions may not well respect q^n=1... So whether it is true that "(the set of elements annihilated by ..." - I am not sure – Alexander Chervov Jan 24 2012 at 5:26
By the way, what are finite-dim irreps of $Q_q(GL)$ for q^n=1 ? If we take elements from the same column - they are q-commuting xy=qyx so as a reperesentation we can take 'shift' and 'FT(shift)'. Can we say something interesting about FT (Fourier transform) from quantum groups point of view ? – Alexander Chervov Jan 24 2012 at 5:30
## 1 Answer
For what concerns De Concini-like integer form the Sl_2 case (and more) is treated in quite some detail in "Quantum function algebra at roots of 1" De Concini-Lyubashenko, Adv. Math. 108, 205-262 (1994). The powers of usual $a,b,c,d$ generators form a commutative Hopf subalgebra and the duality relation is explained in detail (Proposition 1.4 of ref. cit.)
A number of general algebraic properties are contained in two papers by Benjamin Enriquez
"Le centre des algèbres de coordonnèes des group quantiques aux racines $p^\alpha$-ièmes de l'unité" Bull. Soc. Math. Fr. 122, 443-485 (1994)
"Integrity, Integral closedness and finiteness over their centers of the coordinate algebras of quantum groups at $p^\nu$-th roots of unity" Ann. Sci. Math. Quebec 19, 21-47 (1995).
The multiparameter cas was considered by Costantini-Varagnolo "Multiparameter quantum function algebras at roots of 1" Math. Ann. 306, 759-780 (1996).
More recently, also,
Costantini "On the quantum function algebra at roots of 1" Comm. Algebra 32, 2377-2383 (2004).
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# Tagged Questions
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http://en.wikibooks.org/wiki/Physics_Study_Guide/Gravity
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# Physics Study Guide/Gravity
## Newtonian Gravity
Newtonian Gravity (simplified gravitation) is an apparent force (a.k.a. pseudoforce) that simulates the attraction of one mass to another mass. Unlike the three fundamental (real) forces of electromagnetism and the strong and weak nuclear forces, gravity is purely attractive. As a force it is measured in newtons. The distance between two objects is measured between their centers of mass.
$F = \frac{Gm_1m_2}{r^2}$
Gravitational force is equal to the product of the universal gravitational constant and the masses of the two objects, divided by the square of the distance between their centers of mass.
$g = \frac{Gm_1}{r^2}$
The value of the gravitational field which is equivalent to the acceleration due to gravity caused by an object at a point in space is equal to the first equation about gravitational force, with the effect of the second mass taken out.
$U = -\frac{GMm}{r}$
Gravitational potential energy of a body to infinity is equal to the universal gravitational constant times the mass of a body from which the gravitational field is being created times the mass of the body whose potential energy is being measured over the distance between the two centers of mass. Therefore, the difference in potential energy between two points is the difference of the potential energy from the position of the center of mass to infinity at both points. Near the earth's surface, this approximates:
$\Delta U_g = mgh\$
Potential energy due to gravity near the earth's surface is equal to the product of mass, acceleration due to gravity, and height (elevation) of the object.
If the potential energy from the body's center of mass to infinity is known, however, it is possible to calculate the escape velocity, or the velocity necessary to escape the gravitational field of an object. This can be derived based on utilizing the law of conservation of energy and the equation to calculate kinetic energy as follows:
$\boldsymbol{ke}_{initial} = \Delta U$ $\boldsymbol{ke}_{initial} = {U}_{infinity} - {U}_{initial}$ $\frac{1}{2}mv^2 = \frac{GMm}{r}$ ${v}_{esc} = \sqrt{\frac{2GM}{r}}$
## Variables
F: force (N) G: universal constant of gravitation, (6.67x10-11 N•m2/kg2) m1: mass of the first body m2: mass of the second body r: the distance between the point at which the force or field is being taken, and the center of mass of the first body g: acceleration due to gravity (on the earth’s surface, this is 9.8 m/s2) U: potential energy from the location of the center of mass to infinity (J) ΔUg: Change in potential energy (J) m and M: mass (kg) h: height of elevation (m) vesc: escape velocity (m/s)
## Definition of terms
Universal constant of gravitation (G): This is a constant that is the same everywhere in the known universe and can be used to calculate gravitational attraction and acceleration due to gravity. 6.67x10-11 N·m2/kg2 Mass one (m1): One of two masses that are experiencing a mutual gravitational attraction. We can use this for the mass of the Earth (1023 kg). Mass two (m2): One of two masses that are experiencing a mutual gravitational attraction. This symbol can represent the mass of an object on or close to earth. Units: kilograms (kg) Acceleration due to gravity (g): This is nearly constant near the earth's surface because the mass and radius of the earth are essentially constant. At extreme altitudes the value can vary slightly, but it varies more significantly with latitude. This is also equal to the value of the gravitational field caused by a body at a particular point in space (9.8 m/s2) Escape velocity (vesc): The velocity necessary to completely escape the gravitational effects of a body.
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http://mathoverflow.net/questions/19076/bringing-number-and-graph-theory-together-a-conjecture-on-prime-numbers/22862
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## Bringing Number and Graph Theory Together: A Conjecture on Prime Numbers
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Some MOers have been skeptic whether something like natural number graphs can be defined coherently such that every finite graph is isomorphic to such a graph. (See my previous questions [1], [2], [3], [4])
Without attempting to give a general definition of natural number graphs, I invite you to consider the following
DEFINITION
A natural number $d$ may be called demi-prime iff there is a prime number $p$ such that $d = (p+1)/2$. The demi-primes' distribution is exactly like the primes, only shrinked by the factor $2$:
$$2, 3, 4, 6, 7, 9, 10, 12, 15, 16, 19, 21, 22, 24, 27, 30, 31, 34, 36, 37, 40, 42, 45, 49, ...$$
Let D($k,n$) be the set which consists of the $k$-th up to the $(k+n-1)$-th demi-prime number.
After some - mildly exhaustive - calculations I feel quite confident to make the following
CONJECTURE
For every finite graph $G$ there is a $k$ and a bijection $d$ from the vertex set $V(G)$ to D($k,|G|$) such that $x,y$ are adjacent if and only if $d(x),d(y)$ are coprime.
I managed to show this rigorously for all graphs of order $n\leq$ 5 by brut force calculation, having to take into account all (demi-)primes $d$ up to the 1,265,487th one for graphs of order 5. For graphs of order 4, the first 1,233 primes did suffice, for graphs of order 3 the first 18 ones.
Looking at some generated statistics for $n \leq$ 9 reveals interesting facts(1)(2), correlations, and lack of correlations, and let it seem probable (at least to me) that the above conjecture also holds for graphs of order $n >$ 5.
Having boiled down my initial intuition to a concrete predicate, I would like to pose the following
QUESTION
Has anyone a clue how to prove or disprove the above conjecture?
My impression is that the question is about the randomness of prime numbers: Are they distributed and their corresponding demi-primes composed randomly enough to mimick – via D($k,n$) and coprimeness – all (random) graphs?
(1) E.g., there is one graph of order 5 - quite unimpressive in graph theoretic terms - that is very hard to find compared to all the others: it takes 1,265,487 primes to find this guy, opposed to only 21,239 primes for the second hardest one. (Lesson learned: Never stop searching too early!) It's – to whom it is of interest – $K_2 \cup K_3$:
````0 1 0 0 0
1 0 0 0 0
0 0 0 1 1
0 0 1 0 1
0 0 1 1 0
````
(2) Added: This table shows the position of the smallest prime (among all primes) needed to mimick the named graphs of order $n$. All values not shown are greater than $\approx 2,000,000$
````order | 3 4 5 6 7 8
-------------------------------------------------
empty | 14 45 89 89 89 3874
complete | 5 64 336 1040 10864 96515
path | 1 6 3063 21814
cycle | 5 112 21235 49957
````
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It is slightly amusing that you call your search up to 5 mildly exhaustive :P – Mariano Suárez-Alvarez Mar 23 2010 at 2:11
3
Exhaustive with regard to the 10 million primes I checked, not with regard to the order 5, of course. – Hans Stricker Mar 23 2010 at 7:17
GREAT QUESTION! Aside the intrinsic mathematical interest, it also has a foundational/metamathematical one (see my last question on Ackermann' s Yoga): basically what you seem to be after is an interpretation of (finite) graph theory inside arithmetics. Now, graphs are are the ground for a lot of other stuff, for instance categories, so... – Mirco Mannucci Jun 7 2011 at 14:06
## 3 Answers
Theorem: Schinzel's hypothesis H implies the conjecture.
Proof: Choose distinct primes $q_S > 100|G|$ indexed by the 2-element subsets $S$ of $G$. For each $i \in G$, let $Q_i$ be the set of $q_S$ for $S$ such that $i \in S$ and the edge $S$ is not part of $G$. Let $P_i$ be the product of the primes in $Q_i$. Let $P = 4 \prod_S q_S^2$.
By the Chinese remainder theorem, for each $i$ we can find a positive integer $a_i$ such that
$a_i \equiv 1 \bmod{\ell^2}$ for each prime $\ell \le 10|G|$,
$a_i \equiv q-1 \bmod{q^2}$ for each $q \in Q_i$, and
$a_i \equiv 1 \bmod{q_S}$ for each $q_S \notin Q_i$.
Moreover, we can choose the $a_i$ to be distinct. Let $J$ be the set of positive integers up to $\operatorname{max} a_i$, but excluding all of the $a$'s themselves (i.e., $J$ consists of the numbers in the gaps). For each $j \in J$ choose a prime $s_j$ much larger than all the $a_i$ and all the $q_S$.
Consider the linear polynomials $P n + a_i$ and $(P n + a_i + 1)/(2P_i)$ In $\mathbf{Z}[n]$. For each prime $\ell \le 10|G|$ and each $\ell$ of the form $q_S$, all these $2|G|$ polynomials are nonzero mod $\ell$ at $n=0$. For each other prime $\ell$, there exists $n$ such that all these polynomials are nonzero mod $\ell$, since $n$ needs to avoid no more than $2|G|$ residue classes mod $\ell$. Furthermore, we can impose the condition that $P n+j$ is divisible by $s_j^2$ for each $j \in J$, and still find $n$ as above. Therefore Schinzel's hypothesis H implies that there exist arbitrarily large positive integers $n$ such that the numbers $P n+a_i$ and $(P n + a_i + 1)/(2P_i)$ are all prime, and such that $P n+j$ is not prime for $j \in J$. This makes the numbers $p_i:=P n + a_i$ consecutive primes such that $(p_i+1)/2 = P_i r_i$ for some prime $r_i$. If $n$ is sufficiently large, then these primes $r_i$ are all distinct and larger than all of the $q_S$. So the greatest common factor of $(p_i+1)/2$ and $(p_j+1)/2$ for $i \ne j$ equals $1$ if there is an edge between $i$ and $j$, and `$q_{\{i,j\}}$` otherwise. $\square$
Remark: Given how little is known about consecutive primes, it seems unlikely that the conjecture can be proved unconditionally. But at least now we can be confident that it's true!
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Are you kidding (in your remark)? Or are you really confident? (To be honest: I guess not.) – Hans Stricker Mar 23 2010 at 7:06
7
@HS: I think Bjorn is serious. Schinzel's hypothesis is one of several "standard" conjectures in analytic number theory that most experts in the field strongly believe to be true. Indeed, lots of people believe it enough to prove theorems like "Assuming Schinzel's hypothesis to be true, it follows that..." as Bjorn has done here. – Pete L. Clark Mar 23 2010 at 13:46
1
@Pete, thanks for the clarification, now I understand the remark better. – Hans Stricker Mar 23 2010 at 13:58
1
That's good news! – Hans Stricker Mar 23 2010 at 13:59
1
Nice! Btw: strictly speaking, the $s_j$ should be distinct... – fherzig Apr 28 2010 at 23:27
show 1 more comment
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I don't really know the answer, but I suppose I would first start by trying to disprove the conjecture. After all, it has only been verified for graphs up to order 5. The obvious counterexamples I would check are large cliques and large anti-cliques.
So, do there exist arbitrary long sequences of consecutive demi-primes that are pairwise co-prime? What about arbitrary long sequences of consecutive demi-primes such that each pair has a common factor?
The number theorists can feel free to chime in here anytime.
If those don't work, then some other candidates for counterexamples would be large matchings or large cliques together with an isolated vertex.
Edit: I just read that it is strongly believed that there are arbitrarily long sequences of consecutive primes such that each prime is congruent to 3 (mod 4). If true, this would give a representation of arbitrarily large anti-cliques, since the corresponding sequence of demi-primes would all be even. Does anyone know if this has been proven?
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2
Yes, it has been proven - see MR1760689. – David Hansen Mar 23 2010 at 5:19
2
A little more bibliographic detail on the paper David Hansen references: D K L Shiu, Strings of congruent primes, J London Math Soc (2) 61 (2000) 359-373. According to the review, the author proves that for any relatively prime $a$ and $q$, there exist arbitrarily long lists of consecutive primes, each congruent to $a$ modulo $q$. – Gerry Myerson Mar 23 2010 at 6:12
What's the intuition behind your surmise that large cliques, matchings or cliques together with an isolated vertex are good candidates for counter-examples? – Hans Stricker Mar 23 2010 at 7:22
Thanks David and Gerry for the references. Quite interesting. – Tony Huynh Mar 23 2010 at 15:07
@Hans - I wouldn't say that they are good candidates for counterexamples, only that for large graphs they would be the first thing I would investigate. This is simply because it's hard to get to grips with the condition for other large graphs (even say a large tree). As I said, I had no intuition for the problem initially, it just looked easier to find a counterexample than to prove it so I took the lazy man's approach. Actually, after Gerry and David's references I started to believe that the conjecture is true, and after Bjorn's answer, I really believe it. Nice question. – Tony Huynh Mar 23 2010 at 15:15
On the general theme of Gödel codings of graphs, see my work on Riffs and Rotes, for example, here.
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http://math.stackexchange.com/questions/87326/what-would-this-curve-be-called?answertab=oldest
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# What would this curve be called?
In an resistor-capacitor circuit, the voltage stored is measured by taking the two values in Farads and Ohms - and the curve looks like this:
What would this curve be called, logarithmic? exponential (inverse?) I have not a clue in how to explain the nature of the growth to somebody without showing them the actual picture.
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## 3 Answers
It's in general impossible to say what a curve is just from looking at a picture of it, but your curve could be a logistic curve, with equation
$$y(t) = \frac{1 - e^{-t}}{1+e^{-t}}$$
and which looks like this.
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Thank you so much, although the average person may not know what one is - it fits perfectly for what I am trying to explain. To calculate voltage from time in my graph, you use a similar formula ($Vs * (1 - e^{-t/τ}$) - excellent! – Johnny Nero. Dec 1 '11 at 9:49
That is clearly an exponential curve:
$$i(t)=c-e^{a \cdot (b-t)}$$
for the graph to pass by $(0;0)$, you must have:
$$a \cdot b = \ln(c)$$
The precise values of $a$ and $b$ are very difficult to give, because I have not enough informations about the curve.
Example with $a=0.6$ and $b=5$ (and $c=e^3$)
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Thank you, even though this specific math diverges away from electronics, it and Wolfram make me want to mess with it more. – Johnny Nero. Dec 1 '11 at 10:02
No problem, my pleasure! – Oltarus Dec 1 '11 at 10:21
Since it is a resistor-capacitor circuit, this means that the differential equation describing your system must be of the following type:
$$R\frac{dQ}{dt}+\frac{Q}{C}=U \; ,$$
in which $R$ is the resistance, $C$ the capacity, $Q$ the charge stored in the capacitor at some time, $U$ the voltage applied on the system by some external source like a DC battery.
Solving this system gives
$$Q(t)=CU+(Q_0-CU)e^{-\frac{t}{RC}} \; ,$$
in which $Q_0$ is the charge stored at time $0$. If you need the voltage stored, just divide by $C$ to get:
$$V(t)=U+(V_0-U)e^{-\frac{t}{RC}} \; ,$$
with $V_0=Q_0/C$. The figure you show is giving the current flowing through the system, but that is related to the voltage by $V=IR$. Also the $V$ in the picture is what I called $U$.
As Oltarus already mentioned, this is an exponential curve, and the phenomenon it describes is often termed "exponential relaxation".
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1
This should be the accepted answer. – J. M. Dec 1 '11 at 11:21
Thank you Raskolnikov. I may be missing something simple, however if I apply the formula.. C=0.000330, R=4700, U=5, Q0 = 0?, it will equal ~0.00000000069510576900. – Johnny Nero. Dec 2 '11 at 6:24
First of all, you can't come out a number as a result since the formula is time dependent. So, unless you did specify a time, you did something wrong. Second, you also don't mention any units, so the numbers don't mean much. – Raskolnikov Dec 2 '11 at 9:09
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http://electronics.stackexchange.com/questions/29873/rc-rl-circuits-and-frequency-selection/29890
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# RC/RL circuits and frequency selection
We have a lab that wants us to simulate an RC and RL circuit, and then a CR and LR circuit (series). From reading it is my understanding that reversing order of components results in reversing the action (ie frequency selection curve). In other words
LR=RC CR=RL
Why is this?
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If you draw the circuit diagrams and think about the voltage divider rule and how the component impedance magnitude changes with frequency, you'll have your answer. – The Photon Apr 13 '12 at 20:39
Sorry, still not seeing it. If the resistor's frequency response is flat, how can its position affect the action? – Joe Stavitsky Apr 13 '12 at 20:53
When you say "RC" circuit I thought you meant a resistor and a capacitor. When you say "RL" I thought you meant a resistor and an inductor. If that's not what you meant, could you explain those terms? – The Photon Apr 13 '12 at 22:13
## 2 Answers
There is a lot you implied or are assuming that you didn't state. The order of two components in series only matters if you access the node between them. It doesn't matter either if they are in parallel. You really should be more careful in explaining what you are talking about than just "RC and RL circuit". There are many ways to hook up resistors, capacitors, and inductors.
You apparently mean the special case of a RC and RL circuit where the two components are used as a frequency-dependent voltage divider. In that case the order matters just like it would if two different resistors are used in a voltage divider. Clearly these two circuits are not the same:
If the 10 kΩ resistors were replaced by capacitors, it would be even more obviously not the same. For the purpose of analyzing R-C filters, you can think of the C as a resistor that goes down with frequency. If R1 were replaced with a capacitor, the voltage divider would attenuate less at high frequencies and more at low frequencies, which we call a high pass filter. At DC it would attenuate infinitely, meaning it blocks DC. Similarly, if R3 were replaced by a capacitor, the voltage divider would attenuate more at high frquencies and less at low frequencies, which we call a low pass filter. At DC the capacitor looks like a open circuit, so it wouldn't attenuate at all at DC.
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$V_{OUT} = V_{IN} \cdot \dfrac{Z_C}{Z_C + R}$
If the impedance of $C$ is much higher than $R$ this equation will approach 1. This is the case for low frequencies, for DC the impedance will be infinite and then $V_{OUT}$ = $V_{IN}$.
For higher frequencies the factor $R$ will become more prominent, and if $R$ >> $Z_C$ the output will approach zero.
Now replace the resistor by an inductor, and the capacitor by a resistor. Then
$V_{OUT} = V_{IN} \cdot \dfrac{R}{R + Z_L}$
If $R$ is much higher than the impedance of $L$ this equation will approach 1. This is the case for low frequencies, for DC the impedance will be zero and then $V_{OUT}$ = $V_{IN}$.
For higher frequencies the factor $Z_L$ becomes more prominent, and if $Z_L$ >> $R$ the output will approach zero.
Note how similar the paragraphs after the equations are!
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http://mathhelpforum.com/trigonometry/90670-solving-trig-equation.html
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# Thread:
1. ## Solving Trig Equation
Hi all,
$3\sin^2\theta + 7\cos\theta = 5$ solve for $0 <= \theta <= 2\pi$
The working I've done so far.
1. $3(1 - \cos^2\theta) + 7\cos\theta = 5$
2. $3 - 3\cos^2\theta + 7\cos\theta = 5$
3. $\cos\theta (7 - 3\cos\theta) = 2$
4. $\cos\theta = 2$ or $(7 - 3\cos\theta) = 2$
5. $7 - 3\cos\theta = 2$
6. $-3\cos\theta = -5$
7. $\cos\theta = \frac{3}{5}$
Therefore, $\theta = 53.13 deg$ or $\theta = 306.87 deg$
Which means that $\theta = \frac{53.13\pi}{180}$ or $\theta = \frac{306.87\pi}{180}$
I'm just not sure if step 4 is correct. Can I simply ignore one of the options because $\cos\theta = 2$ is invalid?
Cheers.
2. Step 4 is correct.
3. Thanks How about the rest of it lol.
4. I'm sorry; step 4 is incorrect!
How does step 4 follow from step 3.
3. $\cos\theta (7 - 3\cos\theta) = 2$
How did you get to step 4?
4. $\cos\theta = 2$ or $(7 - 3\cos\theta) = 2$
This conclusion can be made only if the RHS=0. I think you're missing something.
5. I was thinking along the lines of factorising.
I.e.
$(x^2 + 3x) = 0$
$x (x + 3) = 0$
$x = 0$ or $(x + 3) = 0$
$x = 0$ or $x = -3$
Edit - Posted before your addition, I can only do that when x = 0 ? Learnt something new (although I should have known that before I guess)
Any further idea's of where to look?
6. $3cos^2 \theta - 7cos \theta+2=0$
$(3cos^2 \theta-6cos \theta - cos \theta+2)=0$
$3cos \theta (cos\theta-2) - (cos \theta-2)=0$
$(cos \theta-2) (3cos \theta-1)=0$
$(cos \theta-2)=0$ or $(3cos \theta-1)=0$
7. Originally Posted by Peleus
Hi all,
$3\sin^2\theta + 7\cos\theta = 5$ solve for $0 <= \theta <= 2\pi$
The working I've done so far.
1. $3(1 - \cos^2\theta) + 7\cos\theta = 5$
2. $3 - 3\cos^2\theta + 7\cos\theta = 5$
3. $\cos\theta (7 - 3\cos\theta) = 2$
4. $\cos\theta = 2$ or $(7 - 3\cos\theta) = 2$
5. $7 - 3\cos\theta = 2$
6. $-3\cos\theta = -5$
7. $\cos\theta = \frac{3}{5}$
Therefore, $\theta = 53.13 deg$ or $\theta = 306.87 deg$
Which means that $\theta = \frac{53.13\pi}{180}$ or $\theta = \frac{306.87\pi}{180}$
I'm just not sure if step 4 is correct. Can I simply ignore one of the options because $\cos\theta = 2$ is invalid?
Cheers.
you can't do that on step three you need to do the following:
3-3cos^2x+7cosx=5
take everything to the right hand side
3cos^2x-7cosx+5-3=0
3cos^2x-7cosx+2=0
now you have to factorise
i would replace the cos with y first
3y^2-7y+2=0
(3y-1)(y-2)=0
so
3cosx=1
cosx=1/3
or cosx=2
x=error
so x=70.53 or 289.47 (2dp)
and i have used x instead of theta
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http://physics.stackexchange.com/questions/11975/what-potential-energy-functions-are-mostly-used-in-schrodinger-equation?answertab=oldest
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# What potential energy functions are mostly used in Schrodinger equation?
In the time-independent Schrodinger equation
$$\left(-\frac{\hbar^2}{2m}\Delta+V(\mathbf{r})\right)\psi(\mathbf{r})=E\psi(\mathbf{r})$$
What functions $V(\mathbf{r})$ are mostly used in research (e.g. Constant potential, Inverse Power-Law Potentials, Finite square well and Infinite square well)?
And who can give a brief history of different potential functions studied in Schrodinger equation?
And can anyone explain to me what is 'Hartree Potential'?
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1
You pretty much named them all (although one can of course study general properties of the equation for broader classes of potentials). So why are you asking if you know the answer? Also, what use is this information to you (because personally, I can't think of any use at all...)? – Marek Jul 7 '11 at 8:45
I need to write a paper about Schrodinger equation, and discuss some examples of the potential energy(as many as possible). – NGY Jul 7 '11 at 8:52
2
You forgot periodical functions in your list. They are heavily used in solid state physics, where they correspond to the potential seen by an electron in a metallic crystal. – Frédéric Grosshans Jul 7 '11 at 9:25
3
There is also the delta-function potential. It might be used to model point impurities in a metal. Potentials involving the spin components can also be included, like the Zeeman term or spin-orbit interactions. There are many other possibilities. But in most modern research, one-particle potentials won't do the job. The stuff you mention is mainly on the undergrad QM course level and therefore very far away from most research involving QM. What do you exactly mean by "research"? – Heidar Jul 7 '11 at 10:18
Thanks for your comments, Frédéric Grosshans and 4tnemele. My 'research' is same as what you mean(but as I know little about research in QM and other branches of physics, I could only mention those undergrad level). I just want to know potentials heavily used in research involving QM(or other branches of physics), and to summerize them. – NGY Jul 7 '11 at 10:27
## 2 Answers
The mostly used potentials are the parabolic potentials. You may have seen it in quantum harmonic oscillator with potential energy $\frac{1}{2}m\omega^2 x^2$.
The parabolic potential is of the form $\frac{1}{2}k x^2$ and corresponds to the force $F=-\frac{dV}{dx}=-kx$. It can be used as an approximation to potentials at local minima (stable equilibrium points), at which the second derivative is positive.
For example, suppose that you have a potential with local minimum at $x=0$ and $V''(0)>0$. Taking Taylor expansion of $V(x)$ near the $x=0$ you get:
$V(x)=V(0)+V'(0)x+\frac{1}{2}V''(0)x^2+\cdot\cdot\cdot$
You can choose $V(0)=0$ and of course $V'(0)=0$ (local minimum at $x=0$). So finally you get $V(x)=\frac{1}{2}V''(0)x^2=\frac{1}{2}kx^2$, where $k=V''(0)>0$, as a good approximation of the $V(x)$ near the $x=0$.
This parabolic approximation can be used for vibrations of diatomic or polyatomic molecules.
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Sometimes $V''(0)=0$ as well. Then harmonic approximation is no long valid. – C.R. Jul 7 '11 at 11:15
"Sometimes $V''(0)=0$" Can you give an example? And why it's not valid? – Andyk Jul 7 '11 at 14:29
A typical example is a quartic potential, $V(x) = a x^4.$ – Gerben Jul 8 '11 at 8:36
You can take the Taylor expansion of $V(x)=a x^4$ at x=0, I don't see any problem with that. Of course it will be just an approximation of V(x) at particular point. – Andyk Jul 8 '11 at 17:10
"You can take the Taylor expansion of $V(x)=ax^4$ at $x=0$" You can. But the $x^2$ term will have a co-efficient of zero, so the problem can not be approximated as harmonic. – dmckee♦ Jul 15 '11 at 14:50
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The most general widely studied examples of potentials are those that are derived from an exactly given ground state, and which have the property that they contain enough parameters to be closed under taking supersymmetric conjugates. These are called "shape invariant" potentials.
Given a real positive ground state, one can ask "Which potential has this ground state?" The answer is that if the ground state is $\exp(-W(x))$, and its energy is exactly zero, then the potential is:
$V(x) = {1\over 2} |\nabla W|^2 - {1\over 2} \nabla^2 W$
The conjugate potential is defined with a plus sign between the two terms instead of a minus sign. It is more usual to define W as the derivative of what I am calling W, but this convention is terrible in higher than one dimension.
The two potentials taken together define a supersymmetric quantum mechanics, as originally defined by Witten. The supersymmetric quantum mechanics in imaginary time is a stochastic Brownian process with a drift which is an analytic function of the position. The conjugate potentials correspond to reversing the direction of the drift, and their properties are similar because they are related by a stochastic version of time-reversibility.
If W(x) goes to plus infinity at infinity (so that it actually defines a normalizable ground state), then the ground state is unique, and the conjugate potential has the exact same spectrum as the original potential, except it omits the lowest energy state. This, plus the form of the supercharge, gives exact solutions of many classes of quantum potentials.
Here are some simple W's which correspond to usual elementary quantum mechanics examples:
• W(x) = |x|^2 is the Hamonic oscillator in any dimension
• W(x) = |x| is the delta function potential in 1d, and the Coulomb potential in 3d
• W(x) = log(|cos(x)|) this gives the infinite hard wall
http://arxiv.org/abs/hep-th/9405029 has a bunch of more interesting examples. Any quantum mechanical potential which has closed form energy states is in this class.
A completely diffeent class of widely studied potentials are random potentials, as studied by Halperin and others, to understand Anderson localization.
Later Edit: The original paper by Anderson which started the random potential field is http://prola.aps.org/abstract/PR/v109/i5/p1492_1 "Absence of diffusion in certain random lattices", and it's one of the great classics. The setup is a square lattice with a random potential at each site, an independent random number between -V and V. The continuum limit in one dimension, where the potential is a random gaussian at each point is analyzed by Halperin B. I. Halperin, Green ' s Functions for a Particle in a One-Dimensional Random Potential, Phys. Rev. 139 , A104 (1965). The field is enormous--- look up "localization" on google scholar. It includes "weak localization" effects, which were popular in the mid 90s because they imply that resistance can drop sharply in the presence of a magnetic field, because the perturbative precurser to the localization process is hindered.
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Hi Ron, can you give me some reference for random potentials? What theoretical results do we have for random potentials? – felix Aug 15 '11 at 0:03
I added some entry points. The basic theoretical results are that all the energy eigenstates in one and two dimensions with an arbitrarily small random potential decay exponentially at long distances. In three dimensions and higher, there is a phase transition as you vary the strength of the randomness between localized and extended wavefunctions. There is very little known rigorously, as far as I know, but a huge amount known by physics standards of certainty. I added a few very old literature pointers in the answer itself. – Ron Maimon Aug 15 '11 at 4:16
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http://complexzeta.wordpress.com/
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# An Idelic Life
Algebraic number theory and anything else I feel like telling the world about
## An Introduction to Tuning and Temperament
Friday, August 1, 2008 in music | 4 comments
In this post, I will explain some different tuning systems and their advantages and disadvantages. First, let’s see what the basic goals are of tuning systems.
1. There should be 12 notes per octave, in half-step intervals.
2. An octave (12 half steps) should be at a ratio of 2:1.
3. A fifth (7 half steps) should be at a ratio of 3:2.
We’ll look at some other goals later on, but let’s start with these. Since 7 and 12 are relatively prime, these goals are sufficient to generate an entire tuning system. However, this would require that 7 octaves be the same as 12 fifths, or $2^7=(3/2)^{12}$. Unfortunately, this is not the case; in fact, it’s not even particularly close. We call the discrepancy, $3^{12}/2^{19}$, a Pythagorean comma.
The first two principles really cannot be compromised. (Well, in some Indian, Arabic, and Turkish tuning systems, the first principle is modified. A common Turkish tuning system consists of 53 notes to an octave, for reasons we will see later on.) The first principle is essential because pianos and other keyboard instruments have 12 notes per octave, and Western music is always written with the 12 notes per octave model in mind. The second principle is essential since we want to treat two notes that differ by an octave as being the same note in many cases; therefore any fudging on goal 2 leads to unacceptable results.
The third principle, however, can be modified, if necessary (and, as we have just seen, it is necessary). The most naïve way of fudging is to make 11 out of 12 fifths be in a 3:2 ratio, and fit the last one to whatever is needed to make the cycle work out. This is known as Pythagorean tuning.
One immediate drawback of Pythagorean tuning is that it makes the so-called “wolf” fifth unusable beyond repair; it just sounds awful. But there are also more subtle problems with Pythagorean tuning. To see these, let’s see one other goal.
4. A major third (4 half steps) should be at a ratio of 5:4.
For simplicity, let’s ignore the problem of the wolf fifth for now. There are other more relevant problems to worry about already. Since $4=7\times 4-12\times 2$, a major third should be four fifths minus two octaves. In terms of the ratios, this means that it would be necessary that $5/4=(3/2)^4/2^2=81/64$. In other words, we’re off by a factor of $81/80$. While this ratio may seem rather small, it turns out to be a very serious problem with the Pythagorean tuning system.
Mathematically, the problem is that we’re trying to factor rational numbers in terms of only powers of two and powers of three. Therefore, as soon as we need to use any prime greater than 3 (in this case, 5), we end up in trouble. However, we don’t really have to factor all rational numbers in terms of powers of two and three; we just have to approximate these factorizations reasonably well. Still, our goals so far are quite mutually incompatible, so we have to sacrifice something, somewhere.
As we pointed out, 5:4 is a better major third than 81:64, so let’s try to construct a tuning system that takes that into account. For simplicity, we’ll assume that our scales always start with C, and we’ll construct ratios for intervals above C. A good possibility is the following:
C 1
C-sharp 16/15
D 9/8
E-flat 6/5
E 5/4
F 4/3
F-sharp 45/32
G 3/2
A-flat 8/5
A 5/3
B-flat 9/5
B 15/8
C 2
This list is generated by using the rules described so far and using the ratio with smallest denominator when there are multiple choices. This tuning system is known as (5-limit) Just intonation, the 5 meaning that we have allowed ourselves to use rational numbers whose factorizations contain primes up to 5.
Naturally, the Just intonation chart should be compared to the chart for Pythagorean tuning, which is as follows (assuming we place the wolf between the G-sharp and the E-flat, which is as good a place as any):
C 1
C-sharp 2187/2048
D 9/8
E-flat 32/27
E 81/64
F 4/3
F-sharp 729/512
G 3/2
G-sharp 6561/4096
A 27/16
B-flat 16/9
B 243/128
C 2
Note the much larger denominators needed in Pythagorean tuning.
So, I pulled these goals out of thin air at the beginning, so I really ought to explain why they are natural goals. For that, it is necessary to discuss the overtone series.
On a string instrument, when a note is sounded, all integer mutliple frequencies are also sounded simultaneously. However, the higher multiples are quieter than the lower ones. Therefore, it is beneficial to have a tuning system that emphasizes at least the first few overtones, while possibly discarding higher ones. The first overtone, at a ratio of 2, sounds one octave higher than the original note (called the fundamental). The second overtone, at a ratio of 3, sounds an octave and a fifth higher than the original note. Subtracting the octave (or, equivalently, dividing by 2) tells us that fifths should be in a ratio of 3:2 above the fundamental. Continuing on, the third overtone is at a ratio of 4, so it sounds two octaves above the fundamental. The next overtone, at a ratio of 5, sounds two octaves and a third above the fundamental.
We really want all these notes to be part of a scale, since we tend to play several notes together, and common overtones yield pleasant sounds. So that’s where these goals I mentioned above come from.
So, it would appear that Just intonation is a good tuning system. However, again there are serious drawbacks. For example, from D to A, which is a perfect fifth, we have a ratio of 40:27, rather than the desired 3:2. There are many other intervals that are similarly problematical, always differing by a factor of 81/80 from the desired ratio. The result is that many such intervals must be avoided in Just intonation. Therefore, Just intonation is only usable in music which specifically avoids certain intervals. So, it’s really not a very nice solution.
One really obvious solution to the entire problem of tuning is just to make all the half steps equal, in a ratio of $2^{1/12}$. In fact, this temperament, known as equal temperament, is what is generally used on pianos today. What is gained by using equal temperament is that there are no wolf tones, and all intervals are usable. However, $2^{1/12}$ is irrational, so we do not get any perfect overtones or intervals with perfect ratios, except the octaves. So, in using equal temperament, one sacrifices all the overtones (except octaves) in order to get a uniformly playable system.
To many people, the sacrifice of overtones is unacceptable. Therefore, we’ll ignore equal temperament from now on.
One attempt to resolve our problems is to use the idea of Pythagorean tuning, but instead of putting the wolf tone all in one place, we spread it out over the octave by putting a quarter of a Pythagorean comma in four different places. To do this, we make the fifths C-G, G-D, D-A, and B-F-sharp a quarter of a comma smaller than the usual 3/2 ratio. Here is that system, known as Werckmeister I:
C 1
C-sharp 256/243
D $64/81\times \sqrt{2}$
E-flat 32/27
E $256/243\times 2^{1/4}$
F 4/3
G $8/9\times 8^{1/4}$
G-sharp 128/81
A $1024/729\times 2^{1/4}$
B-flat 16/8
B $128/81\times 2^{1/4}$
C 2
(Werckmeister also devised several other temperaments, but this one is the most popular.) Of course, we’ve lost many of the perfect fifths here. However, they’re not that bad, and the thirds are also pretty good. (The ratio of the C-E here, $256/243\times 2^{1/4}$, and the desired C-E, 5:4, is about 1.002; that’s definitely acceptable. Of course, some thirds aren’t very good, but these ones are likely to be played less frequently in most “reasonable” keys. So Werckmeister I seems like a pretty good temperament.
I’ll post about some other temperaments (such as quarter comma and sixth comma meantone) another day, but this is a math blog, so I’d like to say something that has some semi-serious mathematical content. In particular, I mentioned earlier that 53 notes to an octave can lead to a very nice tuning system, so I’ll explain where I got that number. Ideally, we’d like some number of fifths to be equal to some other number of octaves. This would allow us to satisfy goals 2 and 3 perfectly. This means that we would like to solve $(3/2)^x=2^y$ as a Diophantine equation (meaning, look for integer solutions). Unfortunately, the only solution is $(0,0)$, which is completely useless. But all is not lost, as it is possible to approximate solutions to this equation. Really, we only require that $(3/2)^x\approx 2^y$. I prefer to get rid of the variable $y$ and look for approximate rational solutions to $(3/2)^x\approx 2$. In other words, I want good rational approximations of $\log_2(3/2)$.
We can easily find the best possible rational approximations by using continued fractions. The continued fraction coefficients begin with 0, 1, 1, 2, 2, 3, 1, 5, 2, 23, meaning that
$\log_2(3/2)=\frac{1}{1+}\frac{1}{1+}\frac{1}{2+}\frac{1}{2+}\frac{1}{3+}\frac{1}{1+}\frac{1}{5+}\frac{1}{2+}\frac{1}{23+\cdots}$.
(I can’t figure out how to get WordPress to allow nested fractions, unfortunately.) Cutting this off at various points gives the following sequence of rational approximations for $\log_2(3/2)$:
1, 1/2, 3/5, 7/12, 24/41, 31/53, 179/306, 207/353, and I don’t really want to work out the next one. In these fractions, the numerator represents the number of units that are to be in a fifth, and the denominator represents the number of units that are to be in an octave. So the usual system corresponds to the approximation 7/12. Good rational approximations are ones whose denominators are much smaller than the next term of the continued fraction approximation. Therefore, 7/12 is quite good, as is 31/53. Thus a 53-note scale is a very natural alternative to the standard 12-note scale.
## Galois Groups of Local Fields
Tuesday, July 29, 2008 in algebraic number theory | 2 comments
We begin with a few definitions.
Definition 1: A integral domain $R$ is called a Dedekind domain if it is noetherian, every nonzero prime ideal is maximal, and it is integrally closed in its field of fractions.
Definition 2: A ring $R$ is called a discrete valuation ring (DVR) if it is a principal ideal domain (PID) with exactly one nonzero prime ideal. (In other language, a DVR is a local PID of Krull dimension 0 or 1.)
One very important property of Dedekind domains is that ideals have unique factorizations as products of prime ideals. I used this property in the case of rings of integers in my last post to say that if $L/K$ is an extension of number fields with rings of integers $B/A$, so if $\mathfrak{p}\subset A$ is a prime ideal, then we can write $\mathfrak{p}B=\prod_{i=1}^g \mathfrak{P}_i^{e_i}$. But this result holds in more generality, for any Dedekind domain.
Also, it is very easy to check that a DVR is a Dedekind domain. But one very common occurrence of DVRs is as localizations of rings of integers. In particular, if $A$ is a Dedekind domain and $\mathfrak{p}$ is a prime ideal of $A$, then $A_{\mathfrak{p}}$ is a DVR.
One way to interpret a DVR is through the following filtration of ideals. Let $R$ be a DVR, and let $\mathfrak{p}$ be the unique nonzero prime ideal of $R$. Then every nonzero ideal of $R$ is of the form $\mathfrak{p}^n$ for some $n\ge 0$ (where by $\mathfrak{p}^0$ I mean $R$). Now, for any $x\in R\setminus\{0\}$, there is an integer $n$ so that $x\in\mathfrak{p}^n\setminus\mathfrak{p}^{n+1}$. We can now define a function $v:R\setminus\{0\}\to\mathbb{N}$ (where $\mathbb{N}$ includes 0 in this case) by $v(r)=n$ as above. We can extend our definition of $v$ to all of $R$ by setting $v(0)=+\infty$.
It is also possible to extend $v$ to the quotient field $K$ of $R$ by setting $v(x/y)=v(x)-v(y)$; it is easy to check that this is well-defined. Now, $v$ satisfies the following properties:
1) $v:K\setminus\{0\}\to\mathbb{Z}$ is a surjective homomorphism.
2) $v(x+y)\ge\min(v(x),v(y)$.
We call such a function $v$ a valuation of the field $K$.
Knowing $v$ is enough to reconstruct $R$, since $R=\{x\in K:v(x)\ge 0\}$. Furthermore, $\mathfrak{p}=\{x\in K:v(x)\ge 1\}$. We call $R$ the valuation ring of $K$.
Let’s look at a few places where DVRs arise naturally.
1) As we mentioned earlier, the localization of a Dedekind domain at a prime ideal is a DVR. So, for example, $\mathbb{Z}_{(p)}=\{x/y\in\mathbb{Q}:p\nmid y\}$ is a DVR if $p$ is a prime. The unique prime ideal is $p\mathbb{Z}_{(p)}$.
2) The ring $\mathbb{Z}_p$ of $p$-adic integers is a DVR with unique prime ideal $p\mathbb{Z}_p$. Also, finite extensions of the field $\mathbb{Q}_p$ of $p$-adic numbers inherit valuations from $\mathbb{Q}_p$, and so they contain DVRs as described above. In particular, if $K/\mathbb{Q}_p$ is a finite field extension, then the integral closure of $\mathbb{Z}_p$ in $K$ is a DVR.
Now, if $R$ is a DVR and $\mathfrak{p}$ is its prime ideal, then $R/\mathfrak{p}$ is a field. In the cases described above, this will always be a finite field; in what follows, we always assume that this field is finite. We call $R/\mathfrak{p}$ the residue field.
We can also put a topology on a valued field $K$ by letting the following sets be a basis for the topology: if $x\in K$ and $n\ge 0$ is an integer, then $\{y\in K:v(x-y)\ge n\}$ is an open set. These sets generate the topology. In what follows, we will assume that $K$ is complete as a topological space with this topology. Finite extensions of $\mathbb{Q}_p$ are complete with respect to this topology, so this will be our motivating example. The residue fields will also be finite.
Last post, I pointed out that if $L/K$ is a Galois extension of number fields, then $efg=[L:K]$. This holds more generally, however. If $L/K$ is a finite Galois field extension, and $A\subset K$ is a Dedekind domain so that $K$ is the quotient field of $A$, and $B$ is the integral closure of $A$ in $L$, then we still have $efg=[L:K]$.
We now interpret this in the case of $K$ a field complete with respect to a discrete valuation $v$, and $A$ the valuation ring of $K$. Let $L/K$ be a finite Galois extension, and let $B$ be the integral closure of $A$ in $L$, or, equivalently, the valuation ring of $L$. Then $L$ is also complete with respect to a discrete valuation $w$ that is very closely related to $v$, as we will see soon.
Let $\mathfrak{p}$ be the prime of $A$, and let $\mathfrak{P}$ be the prime of \$B\$. Since there is only one prime, $g=1$. Hence $\mathfrak{p}B=\mathfrak{P}^e$. Now, if $x\in K$, then $w(x)=ev(x)$, and if $x\in L$, then $fw(x)=v(N_{L/K}x)$. (But we won’t need these results in what follows, at least today.) The implication is the decomposition group of the extension $L/K$ is the entire Galois group.
We can put a filtration on the Galois group as follows: For $i\ge -1$, let $G_i=\{\sigma\in Gal(L/K):w(\sigma(x)-x)\ge i+1 \hbox{ for all } x\in L\}$. We call $G_i$ the $i^{\hbox{th}}$ ramification group of $L/K$. $G_{-1}=G$ is the entire Galois group (or the decomposition group; $G_0$ is the inertia group. Also, each $G_i$ is normal in $G$.
Now, I won’t prove it here, but it can be shown that if the residue field is finite of characteristic $p>0$ and $K$ is complete, then for each $i\ge 1$, $G_i/G_{i+1}$ is a direct product of copies of $\mathbb{Z}/p\mathbb{Z}$, and $G_0/G_1$ is a subgroup of the roots of unity of $B/\mathfrak{P}$ (and hence finite and cyclic of order prime to $p$). Hence, by basic group theory or otherwise, $G_0$ is a semidirect product of a normal Sylow $p$-subgroup and a cyclic group of order prime to $p$. In particular, $G_0$ is solvable. However, as shown in the last post, $G/G_0\cong Gal((B/\mathfrak{P})/(A/\mathfrak{p}))$ is cyclic since it is the Galois group of an extension of finite fields. Hence:
Theorem: $G$ is solvable.
## A Generalization of Dirichlet’s Theorem
Sunday, July 13, 2008 in Uncategorized | Tags: algebraic number theory | Leave a comment
The following is a well-known result:
Theorem: If $a$ and $m$ are integers, with $(a,m)=1$, then there are infinitely many primes congruent to $a\pmod m$.
It turns out that Dirichlet’s Theorem is actually a special case of Artin’s Reciprocity Law. So, we’ll discuss how this works.
Let $L/K$ be an extension of number fields. (That is, $L$ and $K$ are finite extensions of $\mathbb{Q}$.) Let $A$ and $B$ be the rings of integers of $K$ and $L$, respectively. (This means that $A$ and $B$ are the integral closures of $\mathbb{Q}$ in $K$ and $L$, respectively.) Now, let $\mathfrak{p}$ be a nonzero prime ideal in $A$. Then $\mathfrak{p}B=\prod_{i=1}^g \mathfrak{P}_i^{e_i}$ for some primes $\mathfrak{P}_i$ of $B$ and some positive integers $e_i$. If $e_i>1$, we say that $\mathfrak{P}_i$ is ramified over $\mathfrak{p}$. We call $e_i$ the ramification index. The primes $\mathfrak{P}_i$ are said to lie above $\mathfrak{p}$.
Since $A$ and $B$ are Dedekind domains, $\mathfrak{p}$ and $\mathfrak{P}_i$ are maximal ideals. Hence $k=A/\mathfrak{p}$ and $\ell=B/\mathfrak{P}_i$ are finite fields, and $\ell/k$ is a field extension. Let $f_i=[\ell:k]$. We call $f_i$ the residue degree.
It is not too difficult to show that if $n=[L:K]$, then $n=\sum_{i=1}^g e_if_i$. If $L/K$ is a Galois extension, then $e_i$ and $f_i$ are independent of $i$ (since the Galois group of $L$ over $K$ acts transitively on the $\mathfrak{P}_i$), so we can write $n=efg$.
Now, let’s define a few subgroups of $Gal(L/K)$. We’ll assume from now on that $L/K$ is a Galois extension. Furthermore, we fix a prime $\mathfrak{p}$ of $A$, and some prime $\mathfrak{P}$ of $B$ lying above $\mathfrak{p}$. Now, define $D_{\mathfrak{P}}=\{\sigma\in Gal(L/K):\sigma(\mathfrak{P})=\mathfrak{P}\}$. We call $D_{\mathfrak{P}}$ the decomposition group. We now have a homomorphism $\phi : D_{\mathfrak{P}}\to Gal(\ell/k)$. To define $\phi$, we note that an element of $D_{\mathfrak{P}}$ permutes cosets of $B/\mathfrak{P}$ and thus gives the desired homomorphism. Furthermore, this homomorphism is surjective. The kernel $T_{\mathfrak{P}}$ of $\phi$ is called the inertia group. Hence $D_{\mathfrak{P}}/T_{\mathfrak{P}}\cong Gal(\ell/k)$.
It is not hard to determine the sizes of $D_{\mathfrak{P}}$ and $T_{\mathfrak{P}}$ in terms of quantities we already understand: $|D_{\mathfrak{P}}|=ef$ and $|T_{\mathfrak{P}}|=e$. In particular, if $\mathfrak{p}$ is an unramified prime, then $D_{\mathfrak{P}}\cong Gal(\ell/k)$.
That’s particularly nice, because Galois groups of extensions of finite fields are always cyclic, generated by the Frobenius automorphism. Thus in the unramified case, $D_{\mathfrak{P}}$ is cyclic and generated by an automorphism $\sigma$ satisfying the congruence $\sigma(x)\cong x^{|k|} \pmod{\mathfrak{P}}$ for all $x\in B$. (We can extend $\sigma$ to all of $L$ by multiplicativity.) Furthermore, this element $\sigma$ is unique. The common notation for $\sigma$ is $(\mathfrak{P},L/K)$.
If $\mathfrak{P}$ and $\mathfrak{Q}$ are two primes lying above $\mathfrak{p}$, then there is some element $\sigma\in Gal(L/K)$ so that $\sigma(\mathfrak{P})=\mathfrak{Q}$. It is easy to verify that $(\mathfrak{Q},L/K)=\sigma(\mathfrak{P},L/K)\sigma^{-1}$. Therefore, if $Gal(L/K)$ is abelian, then $(\mathfrak{P}/L/K)$ depends only on $\mathfrak{p}$. In this case, we may write $(\mathfrak{p},L/K)$ for this element.
Finally, we’re ready to state (part of) the Artin reciprocity theorem.
Let $L/K$ be an abelian Galois extension of number fields, and let $\sigma\in Gal(L/K)$ be fixed. Then there are infinitely many primes $\mathfrak{p}$ of $A$ that are unramified and so that $\sigma=(\mathfrak{p},L/K)$.
(In fact, only finitely many primes ramify, since primes ramify if and only if they divide the discriminant, which can be easily verified. The other part of the statement is more interesting.)
Let’s look at one example. Take $K=\mathbb{Q}$ and $L=\mathbb{Q}(\zeta_n)$, where $\zeta_n=e^{2\pi i/n}$. Then $Gal(L/K)\cong(\mathbb{Z}/n\mathbb{Z})^\times$, where the isomorphism is as follows: if $(m,n)=1$, then there is an automorphism $\sigma_m\in Gal(L/K)$ defined by $\sigma_m(\zeta_n)=\zeta_n^m$. Now, if $(p,n)=1$, then $((p),L/K)=\sigma_p$. In particular, this case of the theorem is equivalent to Dirichlet’s Theorem on primes in arithmetic progressions.
It turns out that the theorem isn’t too much more general than this, since any abelian extension is contained in a cyclotomic extension (this is the Kronecker-Weber Theorem), and it’s not hard to see what happens to Frobenius elements when we pass to sub-extensions.
All this material can be found (with many more details included) in Serre’s Local Fields.
## Projective and Injective Modules
Tuesday, June 17, 2008 in homological algebra, ring theory | 2 comments
Let’s fix a ring $R$. A module (assumed to be a left module, I suppose, but it doesn’t really matter as long as we’re consistent) $M$ is said to be projective if the functor $Hom(M,-)$ is exact. (That is, if $0\to A\to B\to C\to 0$ is a short exact sequence of $R$-modules, then $0\to Hom(M,A)\to Hom(M,B)\to Hom(M,C)\to 0$ is also exact.) Dually, $M$ is said to be injective if the functor $Hom(-,M)$ is exact.
There are various equivalent conditions for projectives and injectives. One particularly useful result is that projective modules are exactly the direct summands of free modules. Another one is that injective modules satisfy a certain extension property: A module $J$ is injective if and only if for any map $\phi:A\to J$ and any injective map $f:A\to B$, there exists a map (not necessarily unique) $\theta:B\to J$ so that $\phi=\theta\circ f$.
Actually, we didn’t need to start with a ring $R$ at all; it would make just as much sense to allow $M$ to be an object in an arbitrary abelian category $\mathcal{C}$. We say that $\mathcal{C}$ has enough projectives if for every object $A$ of $\mathcal{C}$, there is an epimorphism (or, a surjective map, in the case of many interesting categories) $P\to A$, where $P$ is projective. Dually, $\mathcal{C}$ has enough injectives if for every object $A$ of $\mathcal{C}$, there is a monomorphism (or, an injective map, in the case of many interesting categories) $A\to J$, where $J$ is injective.
It is easy to see that the category of modules over a ring $R$ has enough projectives: if $A$ is an $R$-module, just take the free module on all the elements of $A$, and then quotient out by the submodule consisting of all relations in $A$. Hence $A$ is isomorphic to a quotient of a free (and hence projective) module.
It is also true that the category of modules over a ring $R$ has enough injectives, but this is a bit trickier to prove. To begin, we note that an arbitrary product of injective objects is injective. This follows from the isomorphism $Hom(A,\prod_{i\in I} B_i)\cong\prod_{i\in I} Hom(A,B_i)$. We also note (although I’m not going to prove it here) that in the category of abelian groups (or $\mathbb{Z}$-modules), injective modules are the same as divisible modules (i.e. modules $M$ so that the maps $m\mapsto nm$ for $n\neq 0$ are all surjective).
Let’s first show that the category of abelian groups has enough injectives. The abelian group that plays the most important role here is $\mathbb{Q}/\mathbb{Z}$. Let $A$ be any abelian group, and let $I(A)$ be the product of copies of $\mathbb{Q}/\mathbb{Z}$, indexed by the set $Hom(A,\mathbb{Q}/\mathbb{Z})$. Then $I(A)$ is injective, and there is a canonical map $e_A:A\to I(A)$. We now check that $e_A$ is actually an injective map. To do this, pick $0\neq a\in A$. It is easy to find some nontrivial map $a\mathbb{Z}\to\mathbb{Q}/\mathbb{Z}$. By the extension property for injective modules, this map extends to a map on all of $A$. This is enough to show that $e_A$ is an injective map.
Now let’s return to the category of modules over an arbitrary ring $R$. It can be shown that if $J$ is an injective abelian group, then $Hom(R,J)$ has the structure of an injective $R$-module. Now, let $M$ be an arbitrary $R$-module. Then let $I(M)$ be the product of copies of $I_0=Hom(R,\mathbb{Q}/\mathbb{Z})$, indexed by the set $Hom_R(M,I_0)$. Then, just as before, there is a canonical injective map $M\to I(M)$. This completes the proof that the category of $R$-modules has enough injectives.
## Martingale Monkeys
Friday, February 8, 2008 in probability theory | 8 comments
A friend told me about an interesting problem that can be solved easily using martingales. I’ll start with the problem and solution, and then I’ll talk a bit about martingales in general. (That’s all I can do; I don’t know much about them.)
Problem: A monkey is sitting at a typewriter, typing a letter (A-Z) independently and with uniform distribution each minute. What is the expected amount of time that passes before ABRACADABRA is spelled?
Solution: Suppose that, before every keystroke is made, a new monkey enters and wagers \$1 on the next keystroke being an A fairly (so that if the keystroke is indeed an A, then the payoff is \$26). If the keystroke is an A, the monkey stays and wagers everything (in this case \$26) on the next letter being B, and so on. If the monkey ever loses a wager, then it leaves. Now let’s analyze what happens when ABRACADABRA is finally spelled out. The monkey who kept making correct wagers all the way through won \$ $26^{11}$. But another monkey who got in on the second ABRA won \$ $26^4$, and a third monkey who got in on the final A got \$26. Hence the total payoff is \$ $26^{11}+26^4+26$. But all the wagers are fair, and the house gets \$1 on every turn from the new monkey, so the expected time before ABRACADABRA is spelled is $26^{11}+26^4+26$.
I’m trying to figure out intuitively why the expected time should be longer than $26^{11}$. Here’s my best understanding of it: if a lot of random letters are typed, then on average ABRACADABRA will be about as common as any other 11-letter string. However, they’ll be clumpier than some other strings, since we’ll see some instances of ABRACADABRACADABRA, which contain ABRACADABRA twice. But we can never clump, say, AAAAAAAAAAB more densely. Therefore, although the number of occurrences should be roughly equal in the limiting case, one can clump more easily than the other, so it should take more time to the first occurrence of ABRACADABRA than to the first occurrence of AAAAAAAAAAB.
In fact, we can make this clumping relationship even clearer with the following question: Suppose we have only two letters, A and B, and a monkey randomly types them. What is the probability that BA occurs before AA? The answer is $3/4$ for a rather simple reason: if the sequence does not begin with AA, then there will be a B before the first occurrence of A, so BA will occur before AA unless the sequence starts with AA, which happens only $1/4$ of the time.
Now on to martingales. A martingale is a sequence of random variables $X_1,X_2,X_3,\ldots$ so that the expectation $E(X_{n+1}\mid X_1,\ldots,X_n)=X_n$ (and $E(|X_n|)<\infty$). In this case, the random variables $X_n$ for each monkey are the payoffs after $n$ wagers by that particular monkey. If the $(n+1)^\text{th}$ wager is going to be made, then $X_n=26^n$, and $X_{n+1}$ is going to be $26^{n+1}$ with probability $1/26$ and 0 with probability $25/26$. Thus $E(X_{n+1}\mid X_1,\ldots,X_n)=26^{n+1}/26=26^n=X_n$. Hence this chain is a martingale.
## A Mathematician’s Take on Challah Braiding
Wednesday, September 12, 2007 in food, group theory | 3 comments
For many years, my mother has baked a challah nearly every Friday for Shabbat. Occasionally, however, she asks me to do some portion of the challah making, possibly including the braiding. For reasons we’ll see later, I don’t like her braiding algorithm. This post includes several algorithms, written in a way that someone who knows what a braid group is can understand. (I never had much success following those series of diagrams I sometimes see; I always wished someone would write out the braiding process in terms of generators of the braid group, so that’s what I’m going to do here after I give the preliminary definitions.)
Wikipedia’s page on braid groups has lots of interesting things, so I’ll only write a few essential points here, leaving the reader to explore Wikipedia at eir leisure. I’m finding it a bit tricky to give a good informal definition of braids, so I’ll just assume that my reader knows roughly what a braid is and skip to the formal definition.
The braid group on $n$ strands is the group $B_n=\langle a_1,\ldots,a_{n-1}\mid a_ia_{i+1}a_i=a_{i+1}a_ia_{i+1} \text{ for } 1\le i\le n-2, \ a_ia_j=a_ja_i \text{ for } |i-j|>1\rangle$. In terms of actually playing with braid strands, $a_i$ means interchanging strand $i$ with strand $i+1$ by putting strand $i$ over strand $i+1$. It is pretty simple to see that these generators do indeed induce all possible braids (although I haven’t yet said what a braid is), and that the relations ought to hold. Now, of course, a braid is an element of the braid group.
The braid groups become rather complicated quite quickly. While $B_0=B_1=0$ and $B_2=\mathbb{Z}$, already $B_3$ is nonabelian, and it’s isomorphic to the fundamental group of the complement of a trefoil knot in $\mathbb{R}^3$.
Note also that there is a natural homomorphism $\pi:B_n\to S_n$ that tells us where the strand that started in the $i^\text{th}$ position ends up.
Okay, now it’s time for some challah braiding algorithms. My mother’s usual challah has four strands on the bottom and three on the top. The algorithm for the top braid is pretty natural: $(a_1a_2^{-1})^n$, where $n$ is decided by the length of the dough ropes.
I’m more concerned about the element of $B_4$ used for the bottom braid. She uses $(a_1a_2^{-1}a_3^{-1}a_2)^n$. If $n=1$, we have $\pi(a_1a_2^{-1}a_3^{-1}a_2)=(142)(3)$ (in cycle notation). This is already bad news to me: one step of the algorithm produces a single fixed point! I think one step of the algorithm ought to give an $n$-cycle (here a 4-cycle) or else a pure braid (i.e. a braid in the kernel of $\pi$). But it gets worse: the strand that starts in position 3 has no undercrossings. So when we’re done, it sits on top of every other strand.
It turns out not to be so bad because the three-strand braid sits on top of the four-strand braid, so the central portion of the four-strand braid is not visible in the finished bread. But aesthetically (and mathematically), this feels like a serious flaw to me.
Fortunately, I found an alternate algorithm for four-strand braiding that lacks these flaws: $(a_2a_1a_3^{-1})^n$. If $n=1$, $\pi(a_2a_1a_3^{-1})=(1243)$, which is nice. Furthermore, every strand has both overcrossings and undercrossings. So this is my new preferred braid.
Sometimes, however, it is preferable to braid with six strands. There was an article in the newspaper that explained how to do it, but I was unable to follow it. Fortunately, I found a YouTube video that shows someone doing it (possibly the same way; I can’t tell). I was able to transcribe this method in terms of generators of the braid group. However, I’m not quite sure where it is supposed to end, so my braid may be slightly different from the one shown in the video. The braid is the video is $(e^{-1}d^{-1}c^{-1}b^{-1}a^{-1}(bcdeabd^{-1}c^{-1}b^{-1}a^{-1}e^{-1}d^{-1})^n$, except that it might stop somewhere in the middle of the $(\cdot)^n$. I don’t really want to calculate $\pi$ of this braid (computations like this have never been that easy for me), but I would guess that it is a 6-cycle if it stops at an appropriate moment. (Also, it’s not as complicated as the formula would make it seem, since there’s a lot of stuff like moving the strand on the right all the way over to the left, and it takes a lot of generators to express that, even though it’s not complicated when you’re actually braiding dough.)
## The Eilenberg Swindle
Tuesday, August 28, 2007 in K-theory, ring theory | 5 comments
Recall from the last post that if $R$ is a commutative ring, we define $K_0(R)$ to be the Grothendieck group of the isomorphism classes of finitely generated projective $R$-modules. It is natural to ask what happens if we replace finitely generated projective modules with countably generated projective modules. Let us write $\mathfrak{K}_0(R)$ for this group. It turns out that understanding $\mathfrak{K}_0(R)$ is extremely easy.
Theorem: For any commutative ring $R$, $\mathfrak{K}_0(R)=0$.
Proof: We have to show that if $A$, $B$, $C$, and $D$ are countably generated projective $R$-modules, there is some countably generated projective $R$-module $E$ so that $A\oplus D\oplus E\cong B\oplus C\oplus E$. Define $E=\bigoplus_{i=1}^\infty (A\oplus B\oplus C\oplus D)$. Hence $\mathfrak{K}_0(R)=0$.
A similar construction shows up in the theory of group rings. Here’s an exercise from T.Y. Lam’s Exercises in Classical Ring Theory:
Exercise 8.16: Let $G$ and $H$ be any two groups. Show that there is some ring $R$ so that $R[G]\cong R[H]$. (Here $R[G]$ is the ring of finite $R$-linear combinations of elements of $G$, and multiplication is defined by the group multiplication of $G$.)
Solution: Let $K=(G\times H)\times(G\times H)\times\cdots$, and set $R=\mathbb{Z}[K]$. Then $R[G]\cong R[H]$.
Lam makes the comment that, although consideration of the group rings $\mathbb{Z}[G]$ and $K[G]$ are very useful for determining properties of $G$ (for instance, the modules over these rings are the objects of study in group cohomology and representation theory, respectively), the group ring $R[G]$ for an arbitrary ring $R$ might not give us much information about $G$.
There’s an interesting article I found on more general Eilenberg swindles, but the authors don’t define progenerators, so I’ll include that here.
Let $R$ be a ring and $M$ a right $R$-module. Define $M^\ast=Hom_R(M,R)$ and $S=Hom_R(M,M)$. Then $M$ and $M^\ast$ are $S-R$ and $R-S$ bimodules, respectively. Furthermore, we can define multiplications $M^\ast M\subseteq R$ by $m^\ast m=m^\ast(m)$ and $MM\ast\subseteq S$ by $mm^\ast(m')=m(m^\ast m')$. We say that $M$ is a progenerator if $MM^\ast=S$ and $M^\ast M=R$.
## New objects from old using equivalence classes of pairs
Tuesday, August 14, 2007 in K-theory, ring theory | 1 comment
In many places in mathematics, we see some variant of the following simple construction. The first time we see it is in constructing the integers from the natural numbers:
Consider pairs $(a,b)$ of elements of $\mathbb{N}$ (which includes zero for our purposes, but it doesn’t really matter this time). We form equivalence classes out of these pairs by saying that $(a,b)\sim(c,d)$ if $a+d=b+c$. We can create a group structure on these pairs by setting $(a,b)+(c,d)=(a+c,b+d)$. The resulting group is isomorphic to $\mathbb{Z}$. So that’s how to construct the integers from the natural numbers.
We see a similar construction when we discuss localizations of rings. Let $R$ be a commutative ring and $S$ a multiplicative subset containing 1. (If $R$ is noncommutative, you can still localize provided that $S$ is an Ore set, but I don’t feel like going there now.) We now consider pairs $(r,s)\in R\times S$ under the equivalence relation $(r_1,s_1)\sim(r_2,s_2)$ if there is some $t\in S$ so that $tr_1s_2=tr_2s_1$. The set of equivalence classes has the structure of a ring, called the localization of $R$ at $S$, and denoted by $R_S$. This construction is generally seen with $S=R\setminus\mathfrak{p}$, where $\mathfrak{p}$ is a prime ideal of $R$. The resulting ring is then local (meaning that it has a unique maximal ideal, namely $\mathfrak{p}R_S$. (We generally write $R_{\mathfrak{p}}$ rather than $R_S$ in this situation.) Anyway, this construction is really useful because localizations at prime ideals are frequently principal ideal domains, and we know all sorts of interesting theorems about finitely generated modules over principal ideal domains. And then we can use some Hasse principle-type result to transfer our results back to the original ring.
Notice that I allowed the multiplicative set of localization to contain zero. However, in this case, the localization becomes the trivial ring (or not a ring, if you require that $0\neq 1$ in your definition of a ring, as many people do). More generally, allowing zero divisors in the multiplicative set causes various elements in $R$ to become zero in the localization.
A similar construction shows up in $K$-theory. Suppose $A$ is any commutative semigroup. We consider pairs $(a,b)\in A\times A$ under the equivalence relation $(a,b)\sim(c,d)$ if there is some $e\in A$ so that $a+d+e=b+c+e$. (This is necessary since we do not assume that $A$ satisfies the cancellation property.) The resulting equivalence classes form a group called the Grothendieck group of $A$ and denoted by $K_0(A)$.
The Grothendieck group satisfies the following universal property. Let $\phi$ be the map sending $a\in A$ to $(a+b,b)$ for any $b\in A$. (This is easily seen to be well-defined.) Now let $G$ be any abelian group and $\psi:A\to G$ any semigroup homomorphism. Then there is a (unique) map $\theta:K_0(A)\to G$ so that $\theta\phi=\psi$.
Grothendieck groups can be very helpful for studying rings. Let $R$ be a commutative ring, and let $A$ denote the semigroup of isomorphism classes of projective $R$ modules (under the operation of direct sum). Then $K_0(A)$ (or $K_0(R)$, as people often write) is an important object of study. If $R$ is a field, for instance, then $K_0(R)\cong\mathbb{Z}$. However, if $R$ is the ring of integers of a number field $K$, then $K_0(R)\cong\mathbb{Z}\oplus C(K)$, where $C(K)$ is the ideal class group.
Perhaps more interesting is Swan’s Theorem, which relates vector bundles over a compact topological space to the projective modules over its ring of continuous functions: they have isomorphic Grothendieck groups. But that’s probably the subject of another post, especially if I can manage to understand my notes from Max Karoubi’s lecture series in Edmonton.
## Tschirnhaus Transformations
Monday, August 13, 2007 in Galois theory | Leave a comment
This is based on a talk by Zinovy Reichstein from the PIMS Algebra Summer School in Edmonton.
The motivation comes from looking at ways to simplify polynomials. For example, if we start with a quadratic equation $x^2+ax+b$, we can remove the linear term by setting $y=x+\frac{a}{2}$; our equation then becomes $y^2+b'$.
We can do something similar with any degree polynomial. Consider the polynomial $x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots+a_n$. We may make the substitution $y=x+\frac{a_1}{n}$ to remove the $(n-1)$-degree term.
We can also make the coefficients of the linear and constant terms equal with the substitution $z=\frac{b_n}{b_{n-1}}y$ (where the $b$‘s are the coefficients of the polynomial expressed in terms of $y$).
Enough for motivation. Suppose $f(x)=x^n+a_1x^{n-1}+\cdots+a_n$ is a polynomial, and let $K=\mathbb{C}(a_1,\ldots,a_n)$. (So, in particular, $a_1,\ldots,a_n$ form a transcendence basis for $K$ over $\mathbb{C}$.) Let $L=K[x]/(f(x))$. A Tschirnhaus transformation is an element $y\in L$ so that $L=K(y)$.
Applying a Tschirnhaus transformation to a polynomial of degree $n$ gives us another polynomial of degree $n$ with different coefficients. They also allow us to simplify polynomial expressions in various senses. We will use the following two criteria of simplification:
1) A simplification involves making as many coefficients as possible 0.
2) A simplification involves making the transcendence degree of $\mathbb{C}(b_1,\ldots,b_n)$ over $\mathbb{C}$ as small as possible. (For a polynomial of degree $n$, we will write $d(n)$ for this number.)
Suppose $n=5$. Hermite showed that it is possible to make $b_1=b_3=0$ and $b_4=b_5$. Therefore $d(5)\le 2$. Klein showed that $d(5)$ is in fact equal to 2.
Now suppose $n=6$. Joubert showed that again we can make $b_1=b_3=0$ and $b_5=b_6$. Therefore $d(6)\le 3$.
It is unknown whether we can make $b_1=b_3=0$ when $n=7$. However, it is known that we cannot do so if $n$ is of the form $n=3^r+3^s$ for $r>s\ge 0$ or $n=3^r$ for $r\ge 0$. It is also known (Buhler and Reichstein) that $d(n)\ge\left\lfloor\frac{n}{2}\right\rfloor$.
## Dynamical Systems in Number Theory and Linear Algebra
Thursday, July 5, 2007 in algebraic number theory, dynamical systems, linear algebra | 2 comments
I have been reading Joseph Silverman’s new book on arithmetic dynamics lately. There’s a lot of really fascinating stuff in there, including a large number of potential research problems that are currently way beyond me, but I’ll continue thinking about them! Most interesting so far is the Uniform Boundedness Conjecture:
Let $d\ge 2$, $N\ge 1$, and $D\ge 1$ be integers. Then there exists a constant $C=C(d,N,D)$ such that for any number field $K$ with $D \ge [ K : \mathbb{Q} ]$ and any morphism $\phi:\mathbb{P}^N(K)\to\mathbb{P}^N(K)$ of degree $d$, the number of preperiodic points of $\phi$ in $\mathbb{P}^N(K)$ is at most $C$.
Not much is known about this conjectures; even the case $d=2$, $N=1$, and $D=1$ is open. It’s even open if we restrict to morphisms of the form $\phi_c(z)=z^2+c$. Bjorn Poonen has shown, however, that these maps have no rational periodic points of exact period 4 or 5; it is conjectured that they have no rational periodic points of exact period greater than 3.
However, there is a positive result of the above type that doesn’t depend that much on some of the above quantities:
Let $K$ be a number field and $\phi:\mathbb{P}^1\to\mathbb{P}^1$ be a rational map over $K$. Let $\mathfrak{p}$ and $\mathfrak{q}$ be prime ideals of $\mathfrak{o}_K$ so that $\phi$ has good reduction at $\mathfrak{p}$ and $\mathfrak{q}$ (meaning that when we reduce $\phi$ modulo $\mathfrak{p}$ and $\mathfrak{q}$, we end up with a map $\tilde\phi$ of the same degree as $\phi$) and whose residue characteristics are distinct. Then the period $n$ of any periodic point of $\phi$ in $\mathbb{P}^1(K)$ satisfies $n\le (N\mathfrak{p}^2-1)(N\mathfrak{q}^2-1)$, where $N$ denotes the (absolute) norm.
(See, for instance, my algebraic number theory notes for definitions of some of these terms.)
Anyway, that wasn’t really the point of this post, as you may have guessed from the title. I meant to talk about theorems that pretend not to be related to dynamical systems but actually are. First we need to discuss height functions a bit; there’s a lot more about them in Silverman’s book and in my elliptic curve notes.
We let $K$ be a number field and $M_K$ the set of standard absolute values on $K$ (These are the absolute values on $K$ whose restriction to $\mathbb{Q}$ is either the standard absolute value or one of the $p$-adic absolute values.) We write $n_v=[K_v:\mathbb{Q}_v]$ (where $F_v$ denotes the completion of $F$ with respect to the absolute value $v$). Suppose $P\in\mathbb{P}^N(K)$; we can then write $P=[x_0,\ldots,x_N]$ for some $x_0,\ldots,x_N\in K$. We then define the height of $P$ with respect to $K$ to be $H_K(P)=\prod_{v\in M_K} \max\{|x_0|_v,\ldots,|x_N|_v\}^{n_v}$. One can check that this is well-defined, and that if $L/K$ is a finite extension of number fields and $P\in K$, then $H_L(P)=H_K(P)^{[L:K]}$. Hence it is possible to define the absolute height of $P$ by $H(P)=H_K(P)^{1/[K:\mathbb{Q}]}$.
One of the key facts about heights is the following: If $B$ and $D$ are constants, then $\{P\in\mathbb{P}^N(\overline{\mathbb{Q}}):H(P)\le B \text{ and } [\mathbb{Q}(P):\mathbb{Q}]\le D\}$ is finite. A corollary is the following well-known result of Kronecker:
Let $\alpha\in\overline{\mathbb{Q}}$ be nonzero. Then $H(\alpha)=1$ if and only if $\alpha$ is a root of unity.
Proof: If $\alpha$ is a root of unity, then $H(\alpha)=1$ is clear. Now suppose that $H(\alpha)=1$. For any $\beta$ and $n$, we have $H(\beta^n)=H(\beta)^n$, so $H(\alpha^n)=H(\alpha)^n=1$, so $\{\alpha,\alpha^2,\alpha^3,\ldots\}$ is a set of bounded height and is therefore finite. Therefore there are integers $i>j>0$ such that $\alpha^i=\alpha^j$, so $\alpha$ is a root of unity.
And now for linear algebra. Sheldon Axler has a well-known book on linear algebra without determinants. He therefore uses dynamical systems to show the following familiar result:
Theorem: Every operator on a finite-dimensional nonzero complex vector space has an eigenvalue.
Proof: Let $V$ be such a vector space of dimension $n$. Let $T$ be an operator on $V$, and let $v\in V$ be nonzero. Then $\{v,Tv,T^2v,\ldots,T^nv\}$ cannot be a linearly independent set. Hence there exist $a_0,\ldots,a_n\in\mathbb{C}$, not all zero, so that $a_0v+a_1Tv+\cdots+a_nT^nv=0$. Suppose $m$ is maximal with respect to $a_m\neq 0$. Then $a_0v+a_1Tv+\cdots+a_mT^mv=0$. Since we’re working over $\mathbb{C}$, the polynomial $a_0+a_1z+\cdots+a_mz^m$ factors as $a_m(z-\lambda_1)\cdots(z-\lambda_m)$. We then have $0=a_0v+a_1Tv+\cdots+a_mT^mv=a_m(T-\lambda_1I)\cdots(T-\lambda_mI)v$. Hence some $T-\lambda_jI$ is not injective. This $\lambda_j$ is an eigenvalue for $T$.
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http://mathoverflow.net/questions/18797?sort=oldest
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## Contractible manifold with boundary - is it a disc?
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I'm sure this is standard but I don't know where to look. Let $M$ be a contractible compact smooth $n$-manifold with boundary. Does it have to be homeomorphic to $D^n$? What about diffeomorphic?
[UPDATE: the answer is well-known to be negative as many people kindly pointed out. But actually I assume more about the manifold, namely the following:]
There is a Riemannian metric on $M$ such that every two points are connected by a unique shortest path. So $M$ can be contracted to a point $p\in M$ by sending every point along a shortest path to $p$. These paths can bend along the boundary and can merge because of this. But they are relatively nice (namely $C^{1,1}$) curves and their first derivatives depend continuously on their endpoints. Given all this, can one conclude that $M$ is a disc?
ADDED: These curves are of course gradient curves of a function (the distance to $p$) which is $C^1$ and has no critical points in the interior of $M$, except at $p$.
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No. See the Wikipedia page on Mazur manifolds for examples and relevant history of your question. – Ryan Budney Mar 19 2010 at 22:42
I see. By the way, are there examples bounded by the standard sphere? – Sergei Ivanov Mar 19 2010 at 23:21
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If the boundary is a standard sphere then the contractible manifold has to be the standard disc. This is an h-cobordism theorem type argument -- puncture your contractible manifold, this gives an h-cobordism, etc. – Ryan Budney Mar 19 2010 at 23:25
## 3 Answers
Given a function $\psi:\mathbb R\to \mathbb R$, set $$\Psi=\psi\circ\mathrm{dist}_ {\partial M},\ \ \ \ \ f=\Psi\cdot(R-\mathrm{dist}_ p)$$ for some fixed $R>\mathrm{diam}\, M$.
Further, $$d\,f= (R-\mathrm{dist}_ p)\cdot d\,\Psi-\Psi\cdot d\,\mathrm{dist}_ p$$ Thus, we may choose smooth increasing $\psi$, such that $\psi(0)=0$ and it is constant outside of little nbhd of $0$ so that $\Psi$ is smooth. (It is possible since the function $\mathrm{dist}_ {\partial M}$ is smooth and has no critical points in a small neighborhood of $\partial M$.) Note that $d\,\Psi$ is positive muliple of $d\,\mathrm{dist}_ {\partial M}$. Thus $d_x\,f=0$ means that geodesic from $x$ to $p$ goes directly in the direction of minimizing geodesic from $x$ to $\partial M$, which can not happen.
Now we can apply Morse theory for $f$...
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Is $\Phi$ needed here? I don't see why. – Sergei Ivanov Mar 27 2010 at 18:37
ОЙ, right, I'll remove it :) – Anton Petrunin Mar 27 2010 at 19:08
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
If $M$ is contractible and the boundary of $M$ is simply-connected and $n\ge 6$ then $M$ is diffeomorphic to $D^n$. See Milnor's "Lectures on the h-cobordism theorem".
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Sergei, there are lots of compact contractible smooth manifolds; see e.g. my answer here.
I am a bit confused about what you say next. Are you claiming that any compact contractible manifold admits the metric as you describe?
You might be interested in a paper of Ancel-Guilbaut who put a negatively curved (in the comparison sense) metric on the interior of any compact contractible manifold; see also discussion of this paper on the bottom of page 4 of the paper by Alexander-Bishop here.
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No the metric is assumed to exist and this implies that the manifold is contractible. – Sergei Ivanov Mar 19 2010 at 22:47
Since you have a $C^1$ function with no critical points outside $p$, should not the standard arguments (like in the Soul Theorem) imply that the manifold is smoothly a disk. – Igor Belegradek Mar 19 2010 at 23:06
Formally no, because you can remove any subregion and the function is still there. The fact that the gradient never points outwards could help but I don't know how. – Sergei Ivanov Mar 19 2010 at 23:19
Am I correct that when you start with a small metric sphere around $p$ and flow it by the gradient flow, it stays a topological sphere for a while, and then it hits the boundary and possibly gets messed up, and stops beings a topological sphere, and this is what makes you unsure the manifold is a disk? – Igor Belegradek Mar 19 2010 at 23:43
Yes. And actually there is no gradient flow away from p, only towards p. – Sergei Ivanov Mar 20 2010 at 8:41
show 1 more comment
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http://math.stackexchange.com/questions/175287/area-arc-length-of-a-hyperbolic-segment
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Area/Arc Length of a Hyperbolic Segment
I'm given a hyperbolic segment, similar to the parabolic segment shown here: http://mathworld.wolfram.com/ParabolicSegment.html
I know the height of the segment ("h" in the wolfram article), and the length of the line segment joining the endpoints of the hyperbola ("2a" in the wolfram article).
Is it possible to find the area of the segment? Also, does there exist an approximation formula, or rapidly converging method to determine the approximate arc length of the given hyperbola?
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You mean you know the distance between the end points, and you want to know the area between the segment and the Y-axis? Or do you know the difference in X-coordinates of the end points, and you want to know the area of a region bounded (somehow) by the segment and the X-axis? – Beta Jul 26 '12 at 1:18
@Beta: sorry, width was a poor choice of words. yes I know the distance, and I want to find the area of the region bounded by the hyperbolic segment, and the line segment joining the endpoints. – Chris Dueck Jul 26 '12 at 1:20
My intuition tells me it is possible. Let's see... – Beta Jul 26 '12 at 1:24
Wait... Do you know the parameters of the hyperbola? Because if you don't, I think I can prove it's impossible. And by "height of the segment", do you mean Y-difference between the endpoints, or Y-value of one of, say, the upper one? – Beta Jul 26 '12 at 1:31
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@Beta, how can it be insoluble if the MathWorld article linked above gives the solution! I'm assuming, as does the article, that the figure is symmetric about the "$h$ axis", so to speak. – Rahul Narain Jul 26 '12 at 2:40
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2 Answers
It is not possible to find an answer just from the information supplied.
Consider the hyperbola with equation $$\frac{(y-h)^2}{d^2} -\frac{x^2}{c^2}=1.$$ One branch of this has shape roughly similar to the parabola illustrated in your picture. In particular, it has "height" $h$.
In order for the $x$-intercepts to be at $\pm a$ as in the picture, the relevant condition is $c\sqrt{h^2-d^2}=da$. There are infinitely many hyperbolas for specified $h$ and $a$. The areas are not all the same for these hyperbolas, and neither are the arclengths.
Once the hyperbola is completely specified, arclength, though somewhat unpleasant, can be handled by setting up the usual integral. It is one of the relatively rare cases where the integration can be carried out explicitly in terms of elementary functions.
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translate everything so that the vertices of the hyperbola are on the x axis.
find the partial integral of the positive part of the hyperbola from vertex to the positive intersection, and subtract the integral of the positive part of the line (from the x-intercept to the same intersection point).
do the same thing with the negative parts, and add the absolute value of both parts.
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Did you read the question? – Beta Jul 26 '12 at 1:33
@beta i get that a lot. how hard is it to find the functions of lines and parabolas of which you know multiple parameters? – user35945 Jul 26 '12 at 1:36
A hyperbola and parabola are two different things, bub. – J. M. Jul 26 '12 at 1:36
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http://mathhelpforum.com/geometry/207236-triangle-square.html
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# Thread:
1. ## A triangle in a square ?
Hi !
I've got some trouble to resolve this problem :
What is the most little surface area for a triangle containing a square of 1m² ?
It is admitted that a square' side is confunded with a triangle' side.
Thank you !
2. ## Re: A triangle in a square ?
Okay, set up a coordinate system so that your square has vertices at (0, 0), (1, 0), (0, 1), and (1, 1). The smallest area triangle that will fit around that is the right triangle having (0, 0) as the vertex of the right angle and hypotenuse passing through (1, 1). If we call the point at which the hypotenuse touches the x-axis (X, 0) and the point at which it touches the y axis (0, Y), then the area of the triangle is $\frac{1}{2}XY$ and, using "similar triangles", $\frac{Y}{X}= \frac{1}{X- 1}$ or $Y(X- 1)= XY- Y= X$.
So the problem becomes "minimize $\frac{1}{2}XY$ subject to the constraint $XY- Y= X$.
3. ## Re: A triangle in a square ?
Originally Posted by mathsforce
Hi !
I've got some trouble to resolve this problem :
What is the most little surface area for a triangle containing a square of 1m² ?
It is admitted that a square' side is confunded with a triangle' side.
Thank you !
Here comes a slightly different approach:
1. Draw a sketch (see attachment)
2. The grey right triangles and the blue right triangles are similar:
$\Delta(ADH) \simeq \Delta(HKC)$ ......... and ........... $\Delta(EBG) \simeq \Delta(KGC)$
3. Use proportions:
$\frac hs = \frac{k+x}{x}~\implies~x=\frac{k s}{h-s}$
$\frac hs = \frac{s-k+y}{y}~\implies~y=\frac{s(s-k)}{h-s}$
4. The area of the triangle $\Delta(ABC)$ is calculated by:
$a = \frac12 \cdot (x+s+y) \cdot h$
Replace the terms x and y by the terms of #2:
$a = \frac12 \cdot \left(\frac{k s}{h-s}+s+\frac{s(s-k)}{h-s}\right) \cdot h$
Simplify!
5. You have found out (I hope):
$a(h)=\frac s2 \cdot \frac{h^2}{h-s}$
Solve for h: $\frac{da}{dh}=0$
You should come out with $h = 2s$
6. Replace h by 2s in the equation at #5:
$a(2s)=\frac s2 \cdot \frac{4s^2}{2s-s}=\boxed{2s^2}$
Attached Thumbnails
4. ## Re: A triangle in a square ?
Thank you for your help but I don't know how to use similar triangles...
How can I do without using this ?
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http://math.stackexchange.com/questions/307893/what-characteristic-of-the-triangle-leads-the-the-existence-of-the-orthocenter
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# What characteristic of the triangle leads the the existence of the orthocenter
We all know that all three altitudes of a triangle meets in the orthocenter of the triangle. It's a quite classical problem and is proven. However, what I really wanna know is what characteristic of the triangle is the profound for this to happen? E.g: Is this because of the sum of 3 internal angles equals 180? In Non-Euclidean geometry, where sum of 3 internal angles is greater or smaller than 180 degree, does the 3 altitudes meets in a single point? Or is it because of another reason?
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Euler has quite a few things named after him, so there is no need to deprive Euclid of any honors! :D – Mariano Suárez-Alvarez♦ Feb 19 at 7:43
In the hyperbolic plane, the perpendicular bisectors of the sides of a triangle can be concurrent, have a common perpendicular or be asymptotic. In general, then, there is no orthocenter. – Mariano Suárez-Alvarez♦ Feb 19 at 7:53
+1 Damn intresting observation! – Arjang Feb 19 at 7:54
Have you heard, that the orthocenter is isogonal conjugate of circumcentre? It does not explain everything about the orthocenter, but for me it does explain a lot (isogonal conjugation contains traces of $z \mapsto z^{-1}$ transformation). – dtldarek Feb 19 at 8:11
Something else that bothers me more is that the existence of the Euler line: the orthocenter, centroid, and circumcenter of any triangle are always collinear. It sounds interesting but doesn't it seem a little bit coincident? I mean I know how to prove it but it doesn't sound natural to me. – Nhím Hổ Báo Feb 19 at 15:37
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## 3 Answers
It is perhaps interesting to note that the definition of altitudes is perfectly straightforward for simplexes in higher dimensions, but that already in dimension $3$ the altitudes of a general tetrahedron are not concurrent. For instance for the tetrahedron with vertices $(0,0,0)$, $(1,0,0)$, $(0,1,0)$, $(1,0,1)$, two of the altitudes meet in the origin and two others meet in $(1,0,0)$, but there are no other points of intersection; in general position none of the altitudes will intersect.
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The proofs I know all use Euklidean geometry (e.g. the orthocenter is the intersection of the middle orthogonals for a bigger triangle).
In synthetic geometry, one can consider translation planes with an orthogonality relation and the Fano axiom (diagonals of a nondegenerate parallelogram intersect), thus minimally allowing the proof above. One can show that this makes the geometry at least a Pappus plane.
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What is synthetic geometry? – Arjang Feb 20 at 2:57
Vladimir Arnol'd often emphasized that it was because of the Jacobi identity in Lie algebras. I think he might have meant the algebra so(3) represented as $R^3$ with vector cross-product as the multiplication, but am not sure, and it would be nice to see his remark explained.
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http://physics.stackexchange.com/questions/5317/dirac-equation-on-general-geometries
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# Dirac equation on general geometries?
I have a numerical method for computing solutions to the Dirac equation for a spin 1/2 particle constrained to an arbitrary surface and am interested in finding applications where the configuration space has a complicated geometry, i.e., not just $R^2$ or the sphere, but a more general curved surface possibly with special boundary conditions. On the sphere, for instance, one can take advantage of symmetry and simply use the spherical spinors, but for a more general configuration space (or one described only by measurements, for instance) it may not be possible to come up with a nice, closed-form solution. However, I am not a physicist and would like to better understand where (or indeed, if) such problems arise. Any pointers are greatly appreciated. Thanks!
Edit: Note that I already have a numerical method for solving the Dirac equation -- I am not looking for information on how to solve it. (And, interestingly enough, Clifford algebra / geometric algebra is already the starting point for what we do.) Also, I am specifically interested in the case of surfaces, i.e., 2-manifolds, so (many) applications in GR probably don't apply. Thanks!
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– user346 Feb 18 '11 at 7:47
## 4 Answers
The lagrangian for a massless Dirac fermion in a general background is given by:
$$L_D = i (\bar \psi e^{\mu}_I \gamma^I \mathcal{D}_{\mu} \psi + c.c.)$$
where $c.c.$ means complex conjugate. $\bar \psi = \gamma^0 \psi^+$, $e^{\mu}_I$ is the tetrad (or more generally n-bien) encoding the background metric via $g_{\mu\nu} = e^I_{\mu} e^J_{\mu} \eta_{IJ}$, where $\eta_{IJ}$ is the minkowski metric $diag(-1,1,1,1)$. $\mathcal{D}_{\mu}$ is the covariant derivative whose action on spinors is given by:
$$\mathcal{D}_{\mu} \psi = \partial_{\mu} \psi + g A_{\mu}^{IJ}\gamma_I \gamma_J \psi$$
Here $\gamma_I$ are the Dirac matrices and $\gamma_I\gamma_J$ are the generators of the Lorentz lie algebra. For spinors which transform under the action of a general group the second term in the above expression can be generalized to $A_{\mu}^I T_I \psi$ where $T_I$ are the lie-algebra generators of the relevant group.
The n-bien encodes the background metric. The gauge connection encodes the background curvature via $F_{\mu\nu}^{IJ} = \partial_{[\mu}A_{\nu]}^{IJ} + g [A_{\mu}, A_{\nu}]^{IJ}$. The $[..]$ in the first term indicates antisymmetrization over the contained indices. The second term is the commutator of the matrices which constitute the connection and is non-zero only for a non-abelian group. The complete Lagrangian including that for the background geometry and the Dirac fields is:
$$L_{GR + D} = \frac{1}{8\pi G} e \wedge e \wedge F + L_D$$
Of course, if the background geometry is or can be taken to be static then only the Dirac term matters. The corresponding equations of motion are easily determined via variation w.r.t $\psi$ and $\bar \psi$. The complete solution must take into account not only the bulk values of the n-bien and connection but also the boundary conditions imposed by the background geometry.
In the above, what is left unsaid is how the $e \wedge e \wedge F$ term corresponds to the Einstein-Hilbert lagrangian. This is essentially the connection formulation of GR for which an excellent pedagogical reference is the paper by Romano Geometrodynamics vs. Connection Dynamics.
Given all this formalism what is lacking is a concrete example. I have no doubt someone (@Lawrence ?) will soon supply one. But this should get you going. You can also go the Clifford/geometric algebra route @Carl mentioned and gain the benefits of a unified language at the expense of some time spent learning the framework. Hope this helps!
An explicit example of an exact solution can be found in this paper on Graphene Wormholes
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Very cool. I've seen something similar for C60, but in that case they just assume a spherical geometry (no fun). This looks like a nice lead. Thanks! – fuzzytron Feb 18 '11 at 21:51
– user346 Feb 19 '11 at 2:39
To write down Dirac's equation over curved spacetime, first express the geometry in terms of vierbeins and spin connections. Spinor bundles are vector bundles over spacetime transforming locally under the local Lorentz gauge group. Using the spin connection, we can write down covariant derivatives for sections of the spinor bundle. To get the Dirac operator, we contract the covariant derivative with the inverse vierbein and contract that with the gamma matrices.
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You might take a look at the version of general relativity (GR) done by the Cambridge Geometry Group. They translated GR into "geometric algebra", which amounts to the gamma matrices. So it's easy for them to do Dirac equation calculations on a black hole. Here's an example papers:
In P.G. Bergmann and V. de Sabbata eds, Advances in the Interplay Between Quantum and Gravity Physics, 251-283, Kluwer (2002), Anthony Lasenby and Chris Doran, Geometric Algebra, Dirac Wavefunctions and Black Holes
http://www.mrao.cam.ac.uk/~clifford/publications/abstracts/anl_erice_2001.html
Phys.Rev. D66 (2002) 024006, Chris Doran, Anthony Lasenby, Perturbation Theory Calculation of the Black Hole Elastic Scattering Cross Section http://arxiv.org/abs/gr-qc/0106039v1
Dirac equation for the Kerr metric (rotating black hole):
Phys.Rev. D61 (2000) 067503, Chris Doran, A new form of the Kerr solution
http://arxiv.org/abs/gr-qc/9910099v3
These, and papers they cite or are cited by, should be sufficient.
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I've never seen, and i think it would be fun to solve it inside the klein-bottle. Basically its a closed version of the Möebius strip (meaning that is also non-orientable); the boundary conditions are pretty simple
take a square, with vertices A,B,C,D and edges AB, BC,CD,DA. Now make the following identifications:
1) (A, AB, B) => (D, DC, C)
this gives you a periodic condition on x, making your square a cylinder
now if you would do (B, BC, C) => (A, AD, D) you would get back a torus $T^2$
if you do instead
2) (B , BC, C) => (D, DA, A) you will get a surface homeomorphic to the klein-bottle
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-1 nothing to do with the question. – Marek Feb 17 '11 at 7:02
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Sounds like fun, but if you're considering particles with a given chirality I think you're out of luck because you can't define Weyl spinors on a nonorientable surface. In other words, I believe it's impossible to consistently use each fiber of a nonorientable bundle as a representation space for the spin group (i.e., it has no spin structures). (Not taking any points off though!!) – fuzzytron Feb 18 '11 at 21:48
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http://math.stackexchange.com/questions/49990/the-p-adic-numbers-as-an-ordered-group
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# The p-adic numbers as an ordered group
So I understand that there is no order on the field of p-adic numbers $\mathbb{Q_p}$ that makes it into an ordered field (i.e.) compatible with both addition and multiplication.
Now, from the responses to a couple of my previous questions,
1. $\mathbb{Q_p}$ is a divisible abelian group under addition (being a field of characteristic $0$).
2. $\mathbb{Q_p}$ is torsion free.
3. It admits an order compatible with the group operation (addition), since every torsion free abelian group is orderable.
My question is, can I write an explicit ordering of $\mathbb{Q_p}$ compatible with the group operation? By "explicit", I mean an ordering in which, given two p-adic numbers, I can decide which is greater.
P.S.: I was not sure how to classify this problem, so please feel free to change the tags.
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Brittany, just a comment. The algebra tag is used for more elementary precalculus type algebra questions. You should instead use the abstract algebra tag. You can read that in the description of the algebra tag. – Adrián Barquero Jul 7 '11 at 0:55
@Adrian: Thanks. I actually did mean to say abstract algebra here, but I wrote algebra and chose the wrong item from the drop down menu. – B M Jul 7 '11 at 1:06
For what it's worth, G Rangan, On orderability of topologicaL groups, Internat J Math Math Sci 8 (1985) 747-754, proves that ${\bf Q}_p$ has no order compatible with both its topology and addition, although it has orders compatible with each separately. – Gerry Myerson Jul 7 '11 at 3:26
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– Gerry Myerson Jul 7 '11 at 3:48
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@Gerry: I see. However, we don't necessarily have to use that construction. It just shows that a compatible order exists. Perhaps there is a construction some other way, but as you said, seems unlikely. – B M Jul 7 '11 at 3:53
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## 2 Answers
EDIT: I thought it would be appropriate, given the perhaps unexpected descriptive set-theoretic nature of this answer, to give a foreword explaining the focus on the Baire property (BP for short). A subset of a topological space has the BP if it differs from an open set by a meager set (a set contained in the union of countably many nowhere dense sets). In the setting of a Polish space (such as $\mathbb{Z}_p$ or the real numbers), the BP sets contain the Borel sets (and continuous images of Borel sets) and satisfy some nice regularity properties. In this context BP sets should be considered analogous to measurable sets, with meager sets serving as analogs of the null sets. For example, the Polish space itself is not meager, and the Kuratowski-Ulam theorem asserts that a subset of the plane is meager if and only if only a meager set of vertical sections are nonmeager (this is basically a Fubini theorem).
One particularly noteworthy fact is that it's consistent with the axioms $\mathtt{ZF+DC}$ that every subset of a Polish space has the BP. Recall that $\mathtt{ZF}$ is the standard axiomatization of set theory without choice, and $\mathtt{DC}$ is the axiom of dependent choice. Intuitively, $\mathtt{DC}$ gives you the freedom to make a countable sequence of choices, where each choice is allowed to refer to properties of your previous choices (so they aren't "independent"). Dependent choice is enough to do almost all of common mathematics: you can perform most of analysis, carry out typical inductive constructions, Borel sets behave reasonably, the first uncountable cardinal is not a countable union of countable sets, etc. Some contexts in which $\mathtt{DC}$ doesn't bestow the full power of $\mathtt{AC}$ include performing wildly nonconstructive acts like building a Vitali set or choosing bases from huge vector spaces. So, I think that $\mathtt{ZF+DC}$ is a reasonable framework in which to carry out your request for an "explicit" linear order of $(\mathbb{Z}_p,+)$. Once we rule out the existence of such a linear order with the property of Baire, we're therefore forced to concede that no argument producing this order may be carried out in $\mathtt{ZF+DC}$, dashing our hopes of an explicit construction.
By the way, I focus on $(\mathbb{Z}_p, +)$ rather than $(\mathbb{Q}_p,+)$ simply for convenience. It should be clear that any order of the latter induces an order of the former, so if anything the problem is harder for $\mathtt{Z}_p$.
There is no linear order of the additive group $(\mathbb{Z}_2,+)$ of $2$-adic integers which has the property of Baire (with respect to the usual Polish topology). In particular, it is consistent with $\mathtt{ZF+DC}$ that no such order exists at all, so a large fragment of the axiom of choice is indeed required to build such an order. Analogous arguments will work for all $\mathbb{Z}_p$, with slightly more obnoxious notation.
From here on we will identify elements of $\mathbb{Z}_2$ with infinite binary strings, that is, elements of $2^\omega$ with the product topology. We define an equivalence relation $E_0$ on $2^\omega$ by setting two strings equivalent iff they differ in finitely many coordinates. This $E_0$ has a nice interpretation in $\mathbb{Z}_2$: $x$ and $y$ are $E_0$ related iff their difference is a (standard) integer. More precisely, $x E_0 y$ iff for some $n$, $x + 1 + 1 + \cdots + 1 (n \mbox{ times}) = y$ or vice-versa, where $1$ denotes the standard integer $1$ (i.e., the sequence $10000\ldots$).
(That last part isn't literally true, since the constant $1$ sequence plus $1$ equals the constant $0$ sequence. But it is true off of the eventually constant sequences, which is enough to make the below argument go through (since there are only countably many eventually constant sequences).)
We use without proof two standard facts about $E_0$:
1. If $A \subseteq 2^\omega$ has the Baire property (from now on abbreviated BP) and meets each $E_0$-class in at most one point, then $A$ is meager (this is essentially the Vitali argument);
2. If $A \subseteq 2^\omega$ has the BP and is $E_0$-invariant (i.e., $x \in A$ and $x E_0 y$ implies $y \in A$), then $A$ is either meager or comeager (this is a form of ergodicity) (*).
Now, given a putative order $<$ with the BP, we can partition $2^\omega$ into three $E_0$-invariant BP pieces:
• $X_- = \{x \in 2^\omega : \forall y (y E_0 x \implies y < 0)\}$;
• $X_+ = \{x \in 2^\omega : \forall y (y E_0 x \implies y > 0)\}$;
• $X_0 = 2^\omega \backslash (X_- \cup X_+)$
(here $0$ is the identity element of the group: the constant $0$ sequence). So $X_-$ is the union of the $E_0$-classes which are entirely negative, $X_+$ the union of those entirely positive, and $X_0$ the union of those which are sometimes positive and sometimes negative.
(Technically, these pieces might not have the BP, but by Kuratowski-Ulam there's some element in $2^\omega$ we can use in place of $0$ to make the pieces have the BP. For ease of notation, let's assume $0$ works.)
We first observe that $X_0$ is meager. Note that the set $\{x : 0 \leq x < 1\}$ (which is meager by Fact 1) hits each $X_0$ class in exactly one point, so $X_0$ is the union of countably many homeomorphic translations (namely, the standard integer shifts) of a meager set, thus is meager.
Now let $f: 2^\omega \to 2^\omega$ denote the bitflipping homeomorphism, so $f(01001110\ldots) = 10110001\ldots$. We note that $x \in A_-$ iff $f(x) \in A_+$, since $x + f(x) + 1 = 0$ for all $x$. This means that $A_-$ cannot be comeager, else $A_+ = f[A_-]$ would be a disjoint comeager set. But then by Fact 2, $A_-$ is meager, thus so is $A_+ = f[A_-]$, and consequently we've written $2^\omega$ as the union of three meager sets. So we've hit a contradiction.
(*) By request, here is a reference for Fact 2: Theorem 3.2 of G. Hjorth: Classification and Orbit Equivalence Relations, Mathematical Surveys and Monographs, 75, American Mathematical Society, Providence, RI, 2000. Although actually this theorem is overkill for this special case -- here's a sketch of a more elementary argument that works here.
Suppose that $B \subseteq 2^\omega$ is nonmeager; we want to show that $[B]_{E_0} = \{x: \exists y \in B\ (x E_0 y)\}$ is comeager. By localization, there's a basic open set $U$ such that $B \cap U$ is comeager in $U$. We can find a finite binary string $s$ let's say of length $n$ such that $U$ contains all elements of $2^\omega$ beginning with $s$. Now look at the $2^n$ homeomorphisms of $2^\omega$ which flip some subset of the first $n$ bits of a string and leave the rest unchanged. These maps send each $x$ to something $E_0$-related to $x$, so it follows that $[B]_{E_0}$ is comeager in the union of $U$'s images under these maps. But the union of these images is all of $2^\omega$!
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Thanks for your elaborate answer, though I don't fully understand it yet. Although, more importantly, I don't see why the order is required to satisfy the Baire property? I just need a linear order that is compatible with addition. – B M Jul 11 '11 at 14:54
There are two reasons I looked at orders with the Baire property. First, you ask for an "explicit" order, which suggests some sort of definability constraint. BP sets include Borel sets, analytic sets, and so on, so this argument rules out anything "explicit" in that sense. Second, and more important, it is consistent with the axioms of ZF that every subset of $2^\omega$ has the Baire property. So this argument shows that any argument that there is an order of the sort you want necessarily uses the axiom of choice. – ccc Jul 11 '11 at 15:34
Thanks for the explanation. So is it true that your proposition does not hold for the real numbers, since we can produce an order compatible with addition (and multiplication) explicitly? – B M Jul 11 '11 at 16:07
Yes, this very much uses the particular structure of addition in the $p$-adic integers. For instance, the subgroup generated by $1$ is dense in the $p$-adics whereas of course no cyclic subgroup is dense in the reals. – ccc Jul 11 '11 at 16:15
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I'm giving this +1 because it is the first answer which shows complete understanding of the question! (Whether or not it is actually correct is beyond my expertise -- I don't know any descriptive set theory, or whatever it is that is going into this answer. But at this point I consider that to be a bonus that others here will be qualified to judge.) I think it might be helpful to include a link to this "property of Baire" business. I conjecture that a lot of people interested in the question will not know what that is. The conjecture is true for me, at least. – Pete L. Clark Jul 11 '11 at 16:22
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$1$ is either positive or negative. Say it is positive (otherwise, reverse the order). Then $0<1<2<3<...$. But there is a subsequence of these that converges to $0$. Impossible.
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3
$@$GEdgar: I don't understand what you're trying to do here. As discussed above, since the additive group of $\mathbb{Q}_p$ is torsionfree, there does exist an ordering compatible with the group structure. The question is whether this can be made (in some precise sense) explicit. You seem to be trying to show that a putative ordering is in some sense not compatible with the topology...but that's not the question. – Pete L. Clark Jul 11 '11 at 4:46
@Pete: You are correct, this does not solve that problem. – GEdgar Jul 11 '11 at 13:11
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http://lukepalmer.wordpress.com/tag/math/page/3/
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# Luke Palmer
Functional programming and mathematical philosophy with musical interludes
# Domain Convergence
By on December 31, 2008 | 5 Comments
Okay, how is this for a definition of convergence:
An infinite sequence xs :: [()] converges if there exists an n such that for every m > n, xs !! m = xs !! n. In other words, this is a bunch of programs which either halt or don’t, and after some point, either they all halt or they all don’t.
Then an infinite sequence xs :: [a] converges if map f xs converges for every f :: a -> ().
Sound good?
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# Continuous Stream Fusion
By on December 29, 2008 | 8 Comments
I’ve been able to sketch-implement a version of FRP with a strange distinction: Env vs. Behavior. I’ll give you the intuition, then the rest of the article will develop the semantics more fully. I thought “backwards” to discover this (the word “discover” used under the assumption that it has some merit: questionable), from implementation, to concept, to semantics.
The semantics of Env a are plain ol’ Time -> a, but with a caveat: there is no reactivity allowed. I.e. an Env a yields an a only by looking at stuff “now”. In the implementation, this is just a subtype of IO. The semantics of Behavior are even less defined. I guess it’s semantics are Time -> Time -> a, so it is given a “start time” and returns a time function. It’s sortof like a relative behavior, but shifting it around does not exactly preserve behavior. The implementation is much like Reactive (Env a), where Reactive is from the popular reactive library. But through this exploration I am led to a very different implementation.
To formalize this silliness, first we need to add some composability and define these as transformations. Let’s just fix an e for the whole discussion, and then say the meaning of Env a is (a subtype of) (Time -> e) -> (Time -> a). Now it’s easy to say what we mean by not allowing reactivity: an Env a is a function f of the form f r = g . r for some g. So to determine the value of f at time t, you look at its environment at time t and do something to the result. No peeking at other points in its environment.
For Behavior, I want a scanl, but on continuous functions. You can think of this as a generalization of integration. For a continuous scan to be meaningful, I require its argument to be a continuous function f of type Real -> a -> a, such that f 0 = id. By continuity I mean Scott-continuity: treat Real as an infinite-precision computable real, and then any definable total function is continuous.
Then I can find the integral function ∫f on the interval [0,1] by computing the limit:
``` f 0.5 . f 0.5
f 0.25 . f 0.25 . f 0.25 . f 0.25
...
```
There is no guarantee that this will converge, or uniformly converge, or any of that good stuff. But we will cover our ears and pretend that it always does.
If f :: Real -> Bool -> Bool, once our precision goes past the modulus of uniform continuity of f (the precision of the information that f will never look at to determine what to do), then ∫f will be constant. So there is some notion of continuity preserved.1
Okay, now we can say what a Behavior is! The meaning of Behavior a is a transformation f :: (Time -> e) -> Time -> Time -> a of the form — hold your breath (I’ll explain) — f r t0 = i . ∫ (j r) (k (r t0)), for functions i,j,k, where j is an Env function with the scan restriction j 0 = id.
Let’s dissect: First, k (r t0) is the initial value when the Behavior is started, which depends on the environment exactly when the Behavior is started and no other time. j r is the integrand, which takes into account the “previous” value and the current environment (only current because j must be an Env function). Finally, the i . out front is so we can compute things more complex than their final type; i.e. a Behavior Bool could have more state under the hood than just a boolean.
This definition is very much like Haskell stream fusion; we take a recursive definition and “flatten it out” into a state and an iterator function. By imposing a simple constraint on the iterator function, we have retrieved continuity of the generated stream, and can also perform continuous stream fusion in the implementation!
I still haven’t figured out how to work events back into this model.
1 I’m committing a blatant foul here. Bool -> Bool is a discrete space, so any continuous function from Real to it has to be constant to begin with. So that paragraph was pretty much meaningless. I still liked it to get a computational handle on things.
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# Reactive spaces
By on December 20, 2008 | 1 Comment
My recent days have been spent staring at the ceiling, drawing abstract doodles on a whiteboard, or closing my eyes and watching higher-dimensional images fly through my consciousness. No, I haven’t been on drugs. I’m after a very specific piece of mathematics, to solve a specific problem. But I have no idea what that mathematics is.
I’m after the sameness between functions of time and functions that react to events. Here is my best attempt at putting my images into words:
I will call them (generalized) Reactives, to confuse classical FRP people :-). A Reactive is defined over an environment space, which is a set of environments (with some structure). Time is an environment space; it is uniform and boring, so all of its environments are identical or isomorphic. There is also an Event environment space, whose inhabitants are roughly Events from reactive (a set of separated occurrences).
A Reactive takes an environment and outputs values in terms of it somehow. Reactives have a notion of translation. Say you have a reactive over an event space which is “False until the next mouse click and then True”. Translating this switches which mouse click it is talking about, but not when the mouse clicks occur; so the transition point will always be exactly on an external mouse click. However, since time is a uniform space, translation of a reactive over time does correspond to simple translation, since there is no interesting structure to query.
I don’t know yet what exactly an environment is. I am trying to capture the fact that reactives over an event space can only switch on occurrences of events, whereas reactives over time correspond to continuous functions. If an event environment looks like an FRP Event, what does the time environment look like?
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# Compact data types
By on December 7, 2008 | 6 Comments
Join me, fellow readers, as I learn some topology by pretending I know something about it.
I recently uploaded Martin Escardó’s infinite search monad to hackage, under the name infinite-search. It is a delightful little thing to play with, and has inspired some deeper thoughts about the connection between computation and topology.
Quick refresher on topology: a topological space is a set X together with a collection of subsets of X, called the open subsets. The open subsets must include the empty set and the entirety of X itself, and must be closed under arbitrary unions and finite intersections.
So for example, you can generate the standard topology on the real line by taking all the open intervals (a,b) and closing them under arbitrary unions. So for example, the set $\cup \{ (a-\frac{1}{4},a+\frac{1}{4}) | a \in \mathbb{Z} \}$, which is ball of radius 1/2 around every integer, is open. No solitary points are open; every point must have an arbitrarily small cloud of nearby points.
As another example, there is a topology on the natural numbers called the discrete topology, in which you consider every singleton set to be an open set and take arbitrary unions over those. In this topology, every subset of the naturals is open.
A space is compact if “every open cover has a finite subcover”. That means for any collection S of open sets whose union is the entire space, there is a finite collection $S' \subseteq S$ whose union is also the entire space. Obviously all finite topologies are compact, because there are no infinite collections to begin with.
So the discrete topology on the naturals I talked about is not compact. Consider this cover: $\{ \{n\} | n \in \mathbb{N} \}$; just all the singleton open sets. This obviously has no finite subcollection which also unions to the entire set of naturals (if you leave out even one of these sets it fails to be a cover).
Okay, on to computation. Roughly what Martin Escardó’s work says is that compact spaces can be exhaustively searched. That is, if S is a compact space, then given a total function p :: S -> Bool, we can either compute an element of S satisfying p or compute that there is no such element. If we could figure out a way to form topological spaces on our data types, then we can know whether they are exhaustively searchable (and construct algorithms to search them).
A domain is the set of all elements of a data type, including partial ones (with ⊥s in them). So the domain for Bool is { ⊥, True, False }, and the domain for Either () is { ⊥, Left ⊥, Left (), Right ⊥, Right () }. To construct a topology (the Scott topology?) for the type, generate it by the sets of all compatible elements with some finite element of the domain. So the open sets for our Either () would be { { Left (), Right () }, { Left () }, { Right () } }. We also toss in the empty set, since it is the union of zero of these.
Earlier we saw that the naturals were not compact. This is a good sign, because if we could exhaustively search the naturals, then we could write an algorithm to eg. decide the twin-prime conjecture. However, I will now show that the lazy naturals are compact.
The lazy naturals are the data type:
```data Nat = Zero | Succ Nat
```
What is the set of elements of this domain, and what is the topology? The set of elements is just the regular naturals plus the special one let infinity = Succ infinity. The topology has all the regular finite naturals as singleton sets, just like the discrete topology, but it also has subsets generated by { ⊥, Succ ⊥, Succ (Succ ⊥), … }. That is, sets of all of them greater than some natural including infinity. The key is that the singleton { infinity } itself is not open, it only comes from one of these infinite sets.
Now let’s say you have a cover for Nat. It includes infinity, so it must include the a set { n | n > k for some k }. So we pick that one, and then we pick however many sets we need to cover all the naturals less than k; there will be at most k of them. So we have constructed a finite subcover.
So regular naturals are not compact, but Nat is. So regular naturals are not searchable, but Nat is. But we got Nat only by adding to the regular naturals; how did we make it searchable in the process?
There was a little snag in the definition of searchable above: we needed a total predicate. Many functions which we would not be able to search on the regular naturals would get into an infinite loop if we passed the function infinity, and that would mean the predicate is not total. Intuitively, that means we can only search on predicates which are bounded in the size of the natural; which stop looking at some point.
I conjecture that every ADT in Haskell (without strict fields) is compact. It is pretty easy to write combinators which construct Data.Searchable Sets for any ADT; whether they are proper compact Sets (i.e. the search function is in fact total) is another issue.
So that’s it. We can make a compact set of Naturals and exhaustively search them. For fun, here’s a little Haskell implementation demonstrating this:
```import Data.Searchable
data Nat = Zero | Succ Nat
nat :: Set Nat
nat = singleton Zero `union` fmap Succ nat
-- decidable equality on functions from Nat!
eqnatf :: (Eq a) => (Nat -> a) -> (Nat -> a) -> Bool
eqnatf f g = forevery nat $ \n -> f n == g n
```
Go ahead, write a Num instance for Nat and play around. Just remember that your predicates need to halt even when given infinity, otherwise they are not total. In particular, this means that you can’t use eqnatf on Nat -> Nat, because equality on Nat itself is not total (infinity == infinity loops).
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# The problem with Coq
By on September 7, 2008 | 7 Comments
I’ve written about an idea for a graphical interface to dependent types before. I just played with Coq a little bit, and it’s really fun. I spend 45 minutes proving that every number can be written as either 2n or 2n+1. A feeling of accomplishment washes over me as I finish proving the most substantial result in mathematics history!
Still, the fun of using this system only makes me want to explore this idea more! I like the presentation in the right panel, but I hate the interaction with it. I have to memorize the syntax for a bunch of tactics, typing sequential commands for something that are essentially non-sequential transformations. It’s hard to go back and forth between programming and proving, it’s hard to divert your attention elsewhere for a while. An interface where I could just right click on a hypothesis and see all the ways I could productively transform it would be great in itself.
Damnit! I want a good flexible gui library! I want reactive to be working!
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# All functions are continuous, always
By on August 11, 2008 | 4 Comments
Dan Piponi and Andrej Bauer have written about computable reals and their relationship to continuity. Those articles enlightened me, but only by way of example. Each of them constructed a representation for real numbers, and then showed that all computable functions are continuous on that representation.
Today, I will show that all functions are continuous on every representation of real numbers.
I’m going to use the following definition of analytic continuity:
f is continuous if for any chain of open sets x1 ⊇ x2 ⊇ …, ∩i f[xi] = f[∩i xi]
Where ∩ denotes the intersection of a set of sets, and f[x] denotes the image of f under x (the set {f(z) | z in x}).
This means that for a continuous function, the intersection of a bunch of images of that function is the same as the image of the intersection of the sets used to produce those images (whew!). It might take you a little while to convince yourself that this really means continuous in the same way as it is normally presented. The proof is left as an exercise to the reader (yeah, cop-out, I know).
I chose this definition because it is awfully similar to the definition of Scott continuity from domain theory, which all computable functions must have.
A monotone function f is scott-continuous if for any chain of values x1 ⊑ x2 ⊑ …, supi f(xi) = f(supi xi).
Where ⊑ is the “information” partial ordering, and sup is the supremum, or least upper bound, of a chain. Analogous to the last definition, this means that the supremum of the outputs of a function is the same as the function applied to the supremum of the inputs used to create those outputs.
What I will do to show that every computable function is continuous, no matter the representation of reals, is to show that there is a homomorphism (a straightforward mapping) from Scott continuity to analytic continuity.
But first I have to say what it means to be a real number. It turns out this is all we need:
```fromConvergents :: [Rational] -> Real
toConvergents :: Real -> [Rational]
```
These functions convert to and from infinite convergent streams of rationals. They don’t need to be inverses. The requirement we make is that the rational at position n needs to be within 2-n of the actual number represented (the same as Dan Piponi’s). But if a representation cannot do this, then I would say its approximation abilities are inadequate. To make sure it is well-behaved, toConvergents(fromConvergents(x)) must converge to the same thing as x.
Now we will make a homomorphism H from these Reals (lifted, so they can have bottoms in them) to sets (precisely, open intervals) of actual real numbers.
Range will be a function that maps lists to a center point and an error radius.
Range(⊥) = ⟨0,∞⟩
Range(r:⊥) = ⟨r,1⟩
Range(r:rs) = ⟨r’,e/2⟩ where ⟨r’,e⟩ = Range(rs)
And now H:
H(x) = (r-e,r+e) where ⟨r,e⟩ = Range(toConvergents(x))
H(⊑) = ⊇
H(sup) = ∩
H(f) = H ° f ° fromConvergents ° S, where S(x) gives a cofinal sequence of rational numbers less than x that satisfies the error bound requirement above.
The hard part of the proof is H ° f ° fromConvergents = H(f) ° H ° fromConvergents for fully defined inputs; in other words that the homomorphism does actually preserve the meaning of the function. It boils down to the fact that H ° fromConvergents ° S is the identity for fully defined inputs, since Range(x) is just {lim x} when x is fully defined. I expect some of the details to get a bit nasty though. Left as an exercise for a reader less lazy than the author.
And that’s it. It means when you interpret f as a function on real numbers (namely, H), it will always be continuous, so long as the computable real type you’re using has well-behaved toConvergents and fromConvergents functions.
Intuitively, H maps the set of convergents to the set of all possible numbers it could represent. So ⊥ gets mapped to the whole real line, 0:⊥ gets mapped to the interval (-1,1) (all the points within 1 of 0), etc. The analytical notion of continuity above can be generalized to any function on sets, rather than just f[] (the image function of f). This means we can define continuity (which is equivalent to computability) on, for example, functions from Real to Bool.
This was a fairly technical explanation, where I substituted mathematical reasoning for intuition. This is partially because I’m still trying to truly understand this idea myself. Soon I may post a more intuitive / visual explanation of the idea.
If you want to experiment more, answer: what does it mean for a function from Real -> Bool to be continuous?
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# The Curry-Howard isomorphism and the duality of → and ×
By on July 13, 2008 | 3 Comments
To people who are familiar with the Curry-Howard isomorphism, this post will probably be trivial and obvious. However, it just clicked for me, so I’m going to write about it. I had passively heard on #haskell that → and × (functions and tuples) were dual, so I played with types a bit, finding dual types, and it never really came together.
I’m not actually sure what I’m talking about here is exactly the CH isomorphism. I’ve been thinking about dependent types, and converting propositional formulae into that calculus. But if it’s not, it’s close.
Let’s use some sort of type bounded quantification in our logic. I was thinking about the following formula:
$\forall n \in \mathbb{N} \, \exists p \in \mathbb{N} \, (p > n \wedge \text{Prime}(p) \wedge \text{Prime}(p+2))$
The corresponding type is:
$(n : \mathbb{N}) \rightarrow [ (p : \mathbb{N}) \times (p > n) \times \text{Prime}\, p \times \text{Prime} (p+2) ]$
In other words, the proposition “for every natural, there is a twin prime greater than it” corresponds to the type “a function which takes a natural and returns a twin prime greater than it”. × corresponds to existentials, because it is the proof: it tells you which thing exists. The first element of the tuple is the thing, the second element is the proof that it has the desired property.
It’s interesting that ∀ corresponds to →, since logical implication also corresponds to →. The difference is that the former is a dependent arrow, the latter is an independent one. Beautifully in the same way ∃ corresponds to ×, and so does ∧.
Oh right, the duality. Knowing this, let’s take the above type and negate it to find its dual.
$(n : \mathbb{N}) \times [ (p : \mathbb{N}) \rightarrow (\neg (p > n) + \neg\text{Prime}\,p + \neg\text{Prime}(p + 2)) ]$
i didn’t bother expanding the encoding of not on the inner levels, because it doesn’t really make anything clearer.
The dual is a pair: a number n and a function which takes any number and returns one of three things: a proof that it is no greater than n, a proof that it is not prime, or a proof that the number two greater than it is not prime. Intuitively, n is a number above which there are no twin primes. If this pair exists, the twin prime conjecture is indeed false.
So yeah, that’s how → and × are dual. It’s pretty obvious now, actually, but it took a while to make sense.
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# Set Selectors
I am writing a poker game, and I got mildly annoyed when I went to write the hand classification functions. There was a disparity between the specification and implementation of poker hands; I had to come up with an algorithm to match each type. I didn’t like this, I want the code to match the specification more directly.
This is quite a geeky post. The types of poker hands are very unlikely to change, and the amount of time I’ve spent thinking about this problem already is many times that of solving it directly in the first place. I.e. it would be stupid for someone to pay me by the hour to solve it this way. Still, it gives rise to an interesting generalization that could be very useful.
I decided that the way I would like to specify these things is with “set selectors” over logical expressions. That is, given a finite set U, find a subset R of U such that some logical expression holds in R (i.e. all quantifiers are bounded on R).
This has a straightforward exponential time solution. I’m trying to do better.
I started by classifying logical expressions. In the following, let P(…) be quantifier-free.
• $\exists x P(x)$ is straightforward $O(n)$.
• More generally, $\exists x_1 \exists x_2 \ldots \exists x_k P(x_1, x_2, \ldots x_k)$ is straightforward $O(n^k)$.
• $\forall x P(x)$ is also $O(n)$ to find the largest solution (because the empty set would satisfy it, but that’s not very interesting).
• $\exists x \forall y P(x,y)$ has an obvious solution, same as $\exists x. P(x,x)$. There is no unique largest solution, but there is a unique largest for each x which can be found in $O(n^2)$ time. It’s unclear what the library should do in this scenario.
• $\forall x \forall y P(x,y)$ is called the Clique problem and is NP-complete! Damn!
But the most interesting one so far is the case: $\forall x \exists y P(x,y)$. It turns out that there is a unique largest solution for this, and here is an algorithm that finds it:
Given a finite set U, find the largest subset R such that $\forall x \! \in \! R \, \exists y \! \in \! R \, P(x,y)$.
Let $r_0 = U, r_{n+1} = \{ x \in r_n | \exists y\!\in\!r_n \, P(x,y) \}$. That is, iteratively remove x’s from r that don’t have corresponding y’s. Then define the result $R = \bigcap_i r_i$.
Lemma. There is a natural number $n_f$ such that $r_{n_f} = R$.
Proof. Follows from finite U and $r_{n+1} \subseteq r_n$.
Theorem. $\forall x \! \in \! R \, \exists y \! \in \! R \, P(x,y)$.
Proof. Given $x \in R = r_{n_f} = r_{n_f + 1}.$ Thus there exists $y \in r_{n_f} = R$ such that P(x,y), by the definition of $r_{n_f+1}$.
Theorem. If $R^\prime \subseteq U$ and $\forall x \! \in \! R^\prime \, \exists y \! \in \! R^\prime \, P(x,y)$, then $R^\prime \subseteq R$.
Proof. Pick the least n such that $R^\prime \not\subseteq r_n$. There is an $x \in R^\prime$ with $x \not\in r_n$. The only way that could have happened is if there were no y in rn-1 with P(x,y). But there is a y in R’ with P(x,y), so $R^\prime \not\subseteq r_{n-1}$, contradicting n’s minimality.
The time complexity of this algorithm can be made at least as good as $O(n^2 \log n)$, maybe better.
While that was interesting, it doesn’t really help in solving the general problem (which, I remind myself, is related to poker, where all quantifiers will be existential anyway!). The above algorithm generalizes to statements of the form $\exists w_1 \exists w_2 \ldots \exists w_j \forall x \exists y_1 \exists y_2 \ldots \exists y_k \, P(w_1,w_2,\ldots w_j,x,y_1,y_2, \ldots y_k)$. Each existential before the universal adds an order of magnitude, and I think each one after does too, but I haven’t thought it through.
In fact, I think that, because of the clique problem, any statement with two universal quantifiers will take exponential time, which means I’m “done” (in a disappointing sort of way).
Back to the real world, I don’t like that finding a flush in a hand will take $O(n^5)$ time (16,807 checks for a 7-card hand, yikes), when my hand-written algorithm could do it in $O(n)$. I’m still open to ideas for specifying poker hands without the use of set selectors. Any ideas?
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# What’s a natural transformation?
By on April 28, 2008 | 9 Comments
For some reason I’ve had a great amount of trouble learning category theory. I think a lot of that is that most of the literature sucks. But I don’t blame it, in fact I find it encouraging, because it indicates that there is a brain-rewiring involved: people who know category theory cannot teach it, because as a consequence of learning it, they are thinking in a fundamentally different way.
However, thanks to the Stanford Encyclopedia of Philosophy, a resource which consistently provides good intuition for mathematics when the literature is otherwise impenetrable, I think I finally get natural transformations.
I always heard that a natural transformation is “just like a polymorphic function”. I never really got that, especially since wikipedia said a natural transformation acted on functors but a function is a morphism (in Hask).
Let’s work in the context of an example. I’ll look at `concat :: [[a]] -> [a]` because it has only one argument and only one type variable, so that will simplify things for us. Let’s also recognize that there are two functors appearing here: `[]` on the right and `[]°[]` on the left.
Here’s the formal definition, which I wish to decode: A natural transformation η relates two functors (F and G) with identical source and target. It maps objects in the source category to morphisms in the target category. For every morphism $f: X \mapsto Y$ in the source category, $\eta_Y \circ F(f) = G(f) \circ \eta_X$.
So here F is `[]°[]` and G is `[]`. Our natural transformation `concat` must associate objects (types) to morphisms (functions) and satisfy the above equation. To expand that equation, let’s realize that `[](f)` (where [] is the functor) is `map f` and `[]°[](f)` is `map (map f)`. These are part of the definitions of these functors, and in Haskell correspond to the definition of `fmap`.
The equation is thus: for all functions `f :: x -> y`, `concaty . map (map f) = map f . concatx`. Holy smokes! That’s the free theorem for concat! (note: that paper was another that I never really grokked; maybe I’m grokking it now)
If I could go back to last week and help myself understand, I would have said: a polymorphic function in Haskell is a collection of morphisms in Hask, not a single one, and the equation above is what guarantees that it’s actually parametrically polymorphic (doesn’t care what type it is).
So the following pseudohaskell would not define a natural transformation:
```concat' :: forall a. [[a]] -> [a]
concat'
| a ~ Int = const []
| otherwise = concat
```
Because the equation fails to hold for, say, `show :: Int -> String`: `concat'String . map (map f) /= map f . concat'Int` because `concatInt = const []`. That makes sense, because `concat'` is not parametrically polymorphic.
A question I have is what a polymorphic function is when it’s acting on type constructors that aren’t functors. Is that just a “polymorphic function” and doesn’t necessarily have a categorical parallel?
I guess the weirdest thing about category theory is how natural trasformation was defined. I think I have an intuition for what it is now, but I would never fathom defining it that way. Thinking of objects only as their relationships to other objects is a mind bender.
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# Inane Patterns
I was just screwing around in ghci, playing with the pattern:
```> sum [1..1000]
500500
> sum [1..100]
5050
> sum [1..10]
55
> sum [1..1]
1
```
And it was unfortunate to see the pattern “break” at [1..1]. The pattern being, sum [1..10^n] = 5 followed by (n-1) zeros followed by 5 followed by (n-1) zeros.
But check this out:
``` 5050 = (5 * 10^1) * 10^2 + (5 * 10^1)
55 = (5 * 10^0) * 10^1 + (5 * 10^0)
```
Continuing the pattern…
``` (5 * 10^-1) * 10^0 + (5 * 10^-1) = 0.5 * 1 + 0.5 = 1
```
So 1 is 5 followed by -1 zeros (itself zero digits, 1 for the 5 and -1 for the zeros, which is why we multiply by 10^0), followed by 5 followed by -1 zeros, and the pattern didn’t break. Woah…
Yeah, not mathematically deep, just kinda funny.
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### Twitter
• RT @pigworker: Statutory "Ult" tweet. A tweet prefixed with "Ult" is a comment on the immediately preceding retweet. We all need something … 2 days ago
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http://mathhelpforum.com/number-theory/197072-p-adic-numbers.html
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# Thread:
1. ## P-adic Numbers
Hello, I am having difficultly getting started on these two questions. Some hints would be greatly appreciated.
1) Show that $\sqrt{2} \in \mathbb{Q}_7$ by definition (either the infinite series definition or the algebraic definition)
2) Show that a p-adic number $a = \sum_{v=-m}^\infty a_vp^v \in \mathbb{Q}_p$ is a rational number if and only if the sequence of digits is periodic.
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http://mathforum.org/mathimages/index.php/Vector
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# Vector
### From Math Images
Vectors are quantities that are specified by both a magnitude and direction. Perhaps the simplest vector is a Euclidean vector, represented by an arrow in Euclidean space. Refer to Image 1 below. This arrow has a length and points in some direction.
Image 1. Image of a vector
The length of a vector is called the magnitude. The magnitude is denoted by $\left| \vec A \right|$. The magnitude of a vector is a scalar quantity, a numerical value.
Image 1 is a vector. However, Image 1 is not very useful because it's unclear where it the vector is positioned. We need to introduce the Cartesian coordinate system (the x-y graph) to properly give the direction of a vector. Assigning x and y coordinates to a vector allows us to be more precise when we talk about the location of a vector. Refer to Modified Image 1 below. In this updated image, we now know the x and y coordinates. Our vector is 6 units in the x direction and 8 units in the y direction.
Image 2. Modified Image 1
Another way to locate a vector uses unit vectors. In the two-dimensional coordinate system, the vectors $\mathbf{i} = \left ( 1, 0 \right )$ and $\mathbf{j} = \left ( 0, 1 \right )$ are our unit vectors. It's clear that unit vectors have a length of one. In textbooks, unit vectors can be written in bold text or have a hat placed above the variables like so, $\hat{i}$.
Any vector may be written in terms of our unit vectors $\mathbf{i}$ and $\mathbf{j}$ through scalar multiplication and addition (this is discussed in the graphical introduction below). For now, imagine the unit vector is a rubber substance that can stretch or shrink. With this property, we change change the magnitude of the unit vector such that it can express any vector in the coordinate system.
For example, $\vec A = \left ( a_1, a_2 \right )$ can be written as the following:
$\left ( a_1, a_2 \right ) = \left ( a_1, 0 \right ) + \left ( 0, a_2 \right ) = a_1 \left ( 1, 0 \right ) + a_2 \left ( 0, 1 \right ) = a_1 \mathbf{i} + a_2 \mathbf{j}$.
We can write the vector in the Modified Image 1 as $\vec A = 6 \mathbf{i} + 8 \mathbf{j}$.
Image 2: Unit Vectors (i, j, k)
In order to give a vector's position using unit vectors, we write it as a combination of unit vectors that are placed along the the coordinate axes. Unit vectors correspond to the x-y-z coordinate system this is in three-dimensions. $\mathbf{i} = \left ( 1,0,0 \right )$ points along the x-axis $\mathbf{j} = \left ( 0,1,0 \right )$ points along the y-axis and $\mathbf{k} = \left ( 0,0,1 \right )$ points along the z-axis. Unit vectors (sometimes called the standard basis vectors) are used in physics, engineering and linear algebra.
The $\mathbf{ijk}$ notation is used to emphasize the "vector" nature of a vector while the coordinate notation is use to emphasize the "point" nature of a vector.
## Labeling Vectors
When we have a vector quantity we put an arrow on top of the labeling letter to remind us that it is a vector. It looks like so $\vec A$. One way of writing vectors is by components, like this: $\left ( a_1, a_2, a_3 \right )$. For example, suppose we want to write a specific vector in components, and we know the vector goes 3 units in the x direction, 2 units in the y direction and 0 units in the z direction. Then we can simply write: $\left ( 3, 2, 0 \right )$.
The components of the same vector can also be written as: $\begin{bmatrix} 3 \\ 2\\ 0\\ \end{bmatrix}$. This vector has an x-component of 3, a y-component of 2 and a z-component of 0. This is shown in the image on the below.
Vector (3, 2): Click to enlarge
With $\left ( a_1, a_2, a_3 \right )$ as any numerical values we can also write any vector in terms of standard unit vectors: $\vec A = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$. If we are given $\vec A = 3 \hat{i} + 3 \hat{j} + 2 \hat{k}$ we know that vector A is 3 units in the x and y direction and 2 units in the z direction. This vector could equivalently be written as $\left ( 3, 3, 2 \right )$ or $\begin{bmatrix} 3\\ 3\\ 2\\ \end{bmatrix}$. It is shown in the image below.
Vector (3,3,2): Click to enlarge
If a vector is still a bit abstract to you then think of a compass. The arrow has a certain length, this is our magnitude, and it points in any direction (north, south, east, and west).
When two vectors are added, the sum can be found by placing the two vectors 'head to tail' and finding the coordinate they reach. When a vector is multiplied by a scalar several things may occur:
• When multiplied by a number with absolute value greater than one, the vector stretches. (Figure 1)
• When multiplied by a number with absolute value less than one, the vector shrinks. (Figure 2)
• When multiplied by a negative number, the vector reverses direction. (Figure 3)
'Head to tail' addition of two vectors: (3, 2) + (1,-1) = (4, 1): Click to enlarge
Many mathematical and physical entities can be represented by vectors. For example, an object's velocity can be represented by a 'velocity vector', where each component represents the object's speed in a certain direction. Gravity can also be represented by a vector: the strength of gravity's pull corresponds to the magnitude of the vector, and the direction of the pull corresponds to the direction the vector points in. Vector Fields are another important application of vectors.
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|---------------------------------|--------------------------------|----------------------------------------------------------------|
| Figure 1: Stretching a vector A | Figure 2: Shrinking a vector A | Figure 3: Stretching and reversing the direction of a vector A |
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For vectors $\vec{A},\vec{B},$ and $\vec{C}$ and numbers $d$ and $e$,
The following equalities hold for all vectors and scalars:
• $\vec{A}+\vec{B} = \vec{B} + \vec{A}$ (Commutivity of vector addition)
• $\vec{A}+(\vec{B}+\vec{C}) = (\vec{A}+\vec{B})+\vec{C}$ (Associativity of vector addition)
• $d(\vec{A}+\vec{B}) = d\vec{A} + d\vec{B}$ (Distributivity of scalars)
• $d(e\vec{A}) = (de)\vec{A}$ (Associativity of scalar multiplication)
• the existence of a zero vector (additive identity), $\vec{0}$, such that $\vec{A} + \vec{0} = \vec{A}$
The standard algebraic representation of vectors is in terms of components, although not all vectors can be expressed this way:
$\vec{A} = (a_1, a_2, ... , a_k)$
$\vec{B} = (b_1, b_2, ... , b_k)$
Vectors in this form are added by components:
$\vec{A} + \vec{B} = (a_1, a_2, ... , a_k)+(b_1, b_2, ... , b_k)=(a_1+b_1, a_2+b_2, ... ,a_k + b_k)$
Multiplying a vector by a number, known as a scalar, means multiplying each component by that number:
$d\vec{A} = (da_1, da_2, ... , da_k)$
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http://physics.stackexchange.com/questions/29533/top-quark-and-z-w-bosons
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# top quark and Z,W bosons?
The masses of the Z and W particle sum almost exactly to the mass of the Top quark,within the errors:
Z + W = 80.385±0.015 + 91.1876±0.0021 = 171.57 ±0.0171 GeV
Top quark 172.9± 1.5 GeV
A: Is this one of those simple coincidences?
B: The Z,W particles are decays of the T?
C: Someone has a not too cranky idea connecting them?
EDIT: After consideration of dmckee and Lubos posts.
How about instead of a decay from a t quark, collide a $W^\pm$ and a $Z$ to produce a red top and anti-red bottom.
$$W^+ + Z^0 \to t(r) + \bar{b}(\bar{r})$$
this conserves charge, spin, color, confinement and energy - provided the excess energy of the bottom quark comes from the kinetic term of the collision. It immediately decays as lubos and dmckee pointed out in an early question
EDIT 2: Also note decay time of t-quark is $4.2\ 10^{-25} s$, nearly matching the W,Z decays times of $3.0\ 10^{-25} s$ , although I'm yet to find an uncertainty for these.
And with incredible hubris I'm calling this the Metzgeer Momentary Meson $t\bar{b}$
:) joke
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Interesting, I've never noticed this before. It sounds coincidental but I'll refrain from answering in favor of someone who may actually know about any such work that may have been done. – David Zaslavsky♦ Jun 5 '12 at 2:50
## 1 Answer
You run into several practical problems immediately.
• The total angular momentum of a system of a $W^\pm$ and a $Z$ is an integer, but the top has spin $\frac{1}{2}$.
• The charge of a system involving one $W$ and one $Z$ would be $\pm 1$. The top has charge $\frac{2}{3}$.
These two could be repaired by assuming that there is a spin $\frac{1}{2}$ charge $-\frac{1}{3}$ particle involved. However...
• The decay modes would generate some really funny implications. Start with the predominate mode: $t \to W^+ + b$ means that we're suggesting that a $Z$ is related to a $b$ (and a anti-down or anti-strange?) somehow.
Wait! what?
We started with one quark related to two weak vector bosons because the masses came out close and now we have some other quark related to one weak vector boson even though the masses are totally different. And even if we believe that what are we going to do about the origin of the lighter quark generation? Or are we to believe that only the top and bottom are related to the weak bosons?
Any way, as you may have guessed, I'm going with coincidence all the way. They have to happen sometimes.
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1
One also fails to conserve the color in the original decay - a confined colorful particle can't decay to colorless ones such as W,Z. ... I am utterly unimpressed by the "coincidence" that the sum of two particles' masses is a third particle mass within 1%. I wouldn't even call it a coincidence. It's just a description of one of the normal situations. – Luboš Motl Jun 5 '12 at 3:51
Wait up - there's another coincidence - the decay times of t,Z,W are all close to 10^-25 s, so the masses match up and the decay times. That hints at more than a coincidence. – metzgeer Jun 5 '12 at 9:50
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@metzgeer Ah...that is not a coincidence instead it is an example of "weak universailty". All three are a single weak vertex with roughly the same mass difference which means roughly the same phase space for the products, so roughly the same half-life follows. – dmckee♦ Jun 5 '12 at 16:33
Also, $m_{\rm Higgs} \approx \sqrt{m_Z m_t}$. Just a coincidence. There are lots of them if you look. They mean nothing. – Matt Reece Jun 6 '12 at 4:43
Alright, I can see now. Thanks all! – metzgeer Jun 6 '12 at 8:06
show 2 more comments
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http://mathoverflow.net/questions/110140?sort=newest
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## Isometric embedding of a Kaehler manifold as a special Lagrangian in a Calabi-Yau manifold
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Hallo,
I am reading the paper "Hyperkaehler structures on the total space of holomorphic cotangent bundles" by D.Kaledin and I am asking if it is possible to embedd a real-analytic Kähler manifold, isometrically, as a special Lagrangian in a Calabi-Yau manifold. Acctualy what I am looking for is the following: Start with a compact real-analytic Kähler manifold $(M, I, \omega)$ and in a neigbourhood of the zero-section in the cotangent bundle $T^{*}M$ there should exists a holomorphic $(n,0)-$form $\Omega$ (with respect to some complex structure on this neigbourhood) and a Kaehler form $\tilde{\omega}$ such that the forms $Im(\Omega)$ and $\tilde{\omega}$ vanishes when restricted to $M$ (the zero section) and $\tilde{\omega}^{2n} = C_{n} \Omega \wedge \bar{\Omega}$ for some constant $C_{n}$ that depends only on $n$. I know that one can do this. But I don't know some references where I can find a explanation of this. Is it sufficient just to read the paper of Kaledin or do I have also to switch to other references? By using Kaledin's paper what ingredients are necessary for a proof of this embedding problem? I am a beginner in Calabi-Yau manifolds and Hyperkaeler manifolds and I would be very thankfull if someone has the answers. I hope for a lot of replys and also hope that this question is not too trivial.
Best Regards, Pavel
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try looking at papers of Stenzel: math.osu.edu/~stenzel.3/research/publications/… – Spiro Karigiannis Oct 20 at 12:48
or by Calabi: archive.numdam.org/ARCHIVE/ASENS/… – Spiro Karigiannis Oct 20 at 12:49
Your question suggests that you are looking for an isometric embedding of the given Kähler manifold as a special Lagrangian in a Calabi-Yau manifold, but you don't mention this requirement in the text. I'll just point out that the induced metric on any special Lagrangian submanifold of a Calabi-Yau manifold is necessarily real-analytic, so it follows that it is not possible, in general to isometrically embed a given Kähler manifold as a special Lagrangian in some Calabi-Yau manifold. – Robert Bryant Oct 20 at 22:19
ok, I see. lets assume that the given Kaehler manifold is also real analytic. Is it then possible? how can one explain this? what are the ingredients in showing this? – Pavel Oct 21 at 6:01
I am not exactly sure what you mean by "explanation", but since you're asking for references, have a look at Birte Feix's thesis "Hyperkaehler metrics on cotangent bundles". There she constructs the HK metric in a different way. See also mathoverflow.net/questions/46752/… – Peter Dalakov Oct 21 at 14:30
show 2 more comments
## 2 Answers
Disclaimer: I am not sure what kind of "explanation" you are looking for. I would guess that you are after the observation (due to Hitchin), that complex Lagrangian submanifolds become special Lagrangian after rotating the complex structure.
Observation: Let $X$ be a hyperkaehler manifold. Let $\{I,J,K\}$ be a triple of complex structures, satisfying the quaternionic identities, and let $\{\omega_I,\omega_J,\omega_K\}$ be the respective Kaehler forms. Let $M\subset (X,I,\omega_I)$ be a complex-lagrangian submanifold for the complex-symplectic form $\omega^c= \omega_J+i\omega_K$. Then $M$ is a special lagrangian submanifod of $(X,J, \omega_J,\Omega = (\omega_K+i\omega_I)^{\dim_{\mathbb{C}} M})$.
(Actually, if $\dim_{\mathbb{C}} M$ is odd you must either take $i\Omega$ as your holomorphic volume form, or use the more relaxed definition of special Lagrangian. )
Here "complex-Lagrangian" means that $M\subset (X,I)$ is a complex submanifold and $\left. \omega^c\right|_M=0$.
So given a real-analytic Kaehler manifold, you embed it as the zero-section of the cotangent bundle, take the Kaledin-Feix metric on a (formal) tubular neighbourhood, and rotate the complex structure.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
How can you show then that, after a rotation, it satisfies the Calabi-Yau equation?
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You should not ask a new question as an answer to an existing question. You can either create a new question or post a comment. But it's not at all clear what you mean. A hyperKahler manifold is Calabi-Yau in an $S^2$ worth of ways. This is clear, because the triple $\omega_I$, $\omega_J$, and $\omega_K$ are all parallel with respect to the Calabi-Yau metric, so the $\Omega$ that Peter defines is parallel, thus the pair $(\omega_J, \Omega = (\omega_K + i \omega_I)^{\dim_{\mathbb C} M}$ is a Calabi-Yau structure. – Spiro Karigiannis Oct 23 at 18:19
Yes but does it follow then that $\omega_{J}^{n} = c_{n} \Omega \wedge \Omega$, where $c_{n}$ is a constant depending only on $n$, where $n = dim_{\mathbb{C}}M$? – Mina Oct 24 at 12:26
ok I will post it as a question :). – Mina Oct 24 at 16:46
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http://mathoverflow.net/revisions/14630/list
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## Return to Answer
1 [made Community Wiki]
I am Formalist. And from a formal point of view, GCH is independent from ZF(C) - nothing more. To me, sets are not really "existing" (have you ever seen an infinite ordinal flying around somewhere? well, i didnt.) - but assuming GCH could make mathematics easier - and that is what is important to me. Without GCH, there must be at least one set (and therefore infinitely many sets) with an undecidable cardinality.
On the other hand, with GCH, if you can prove $|A| > \aleph_i$ and $|A| \le \aleph_{i+1}$ then you are already done proving $|A| = \aleph_{i+1}$.
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http://math.stackexchange.com/questions/303834/common-knowledge-and-concept-of-coarsening-partition
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common knowledge and concept of coarsening partition
Here is a proof of the equivalence between my definition and Aumann's for "common knowledge". I'm assuming some familiarity with set partitions. Aumann's definition is in terms of the Kolmogorov model of probability. In particular, a proposition is identified with the set of possible worlds in which the proposition is true. Let P₁ be that partition of the possible worlds such that two worlds share the same block in P₁ if and only if I condition on the same body of knowledge when computing posterior probabilities in the two worlds. Let P₂ be the analogous partition of the possible worlds for you. For each world w, let P₁(w) denote the block in my partition containing w, and let P₂(w) be the block in your partition containing w. Let P denote the finest common coarsening of our respective partitions**, and let P(w) be the block of P containing w. (from http://lesswrong.com/r/discussion/lw/6je/an_explanation_of_aumanns_agreement_theorem/4hc5)
So, what exactly is common coarsening of partitions? And here what does it mean? Is it just union of two partition maps?
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2 Answers
Fix a base set $X$ and let $P$ and $Q$ be partitions of $X$.
$P$ is finer than $Q$ (or equivalently, $Q$ is coarser than $P$) if every set in $P$ is a subset of a set in $Q$. The partition where all the subsets have size 1 is the finest possible partition. The partition where the entire original set lies in only one set is the coarsest possible partition.
The finest common coarsening of $P$ and $Q$ is the finest partition $R$ such that $R$ is coarser than both $P$ and $Q$. (There is a dual notion, called the coarsest common refinement, which is the coarsest $R$ that is finer than both $P$ and $Q$.)
For example, if $$X = \{1,2,3,4,5,6,7\}$$ $$P = \{\{1,3,4\}, \{2,5\}, \{6,7\}\}$$ $$Q = \{\{1,2,5\},\{3,4\}, \{6\}, \{7\}\}$$ then the finest common coarsening of $P$ and $Q$ is $\{\{1,2,3,4,5\},\{6,7\}\}$ while the coarsest common refinement of $P$ and $Q$ is $\{\{1\}, \{2,5\}, \{3,4\}, \{6\}, \{7\}\}$.
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Ted's answer on common coarsening is correct. I'll try to brief Aumann's agreement theorem which is basically an application of his original formulation of common knowledge in terms of partitions(compared with David Lewis).
On one hand, Aumann(1976) assumed a common prior for all possible worlds, which assigns probabilities to all events conceivable. Moreover,common prior is known by two players, and the event they know the common prior is also known by them, and the event they know they know …… ad infinitum. On the other hand, two players' difference of knowledge are modeled in the same fashion as two people at different positions seeing the whole picture from afar, i.e. some states of world can not be distinguishable from others, while some others can. When one particular state of world, as a full specification of events of interest obtain, people receive some signal which can be utilized to define an equivalence relation on states of world. That's where partitions come from.
Noticeably, each players' partition are also common knowledge, even though it's not necessarily known by his opponent which atom that contains the true state. Events that are common knowledge are those contains at least one atom of common coarsening of their partitions. The point is, even though his opponent may not know which atom contains the true state, at least he know the candidate atoms are those that can derive the same posterior probability and those in the one atom of common coarsening of their partitions. In fact, the latter is a subset of the former, since the posterior is common knowledge as assumed. Thus posterior probabilities derived by two players' partition must be the same as that derived by common coarsening of their partitions, which shows that two players posterior probabilities must coincide.
Added:Why not have a look at Aumann's classic, and Geanakoplos' survay among others?
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http://mathoverflow.net/questions/61946/a-lower-bound-of-a-particular-convex-function/64279
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## A lower bound of a particular convex function
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Hello, I suspect this reduces to a homework problem, but I've been a bit hung up on it for the last few hours. I'm trying to minimize the (convex) function $f(x) = 1/x + ax + bx^2$ , where $x,a,b>0$. Specifically, I'm interested in the minimal objective function value as a function of $a$ and $b$. Since finding the minimizer $x^*$ is tricky (requires solving a cubic), I figured I'd try and find a lower bound using the following argument: if $b=0$, the minimizer is $x=1/\sqrt{a}$ and the minimal value is $2\sqrt{a}$. If $a=0$, the minimizer is $x=(2b)^{-1/3}$ and the minimal value is $\frac{3\cdot2^{1/3}}{2}b^{1/3}$. Therefore, one possible approximate solution is the convex combination
$(\frac{a}{a+b})\cdot2\sqrt{a} + (\frac{b}{a+b})\cdot\frac{3\cdot2^{1/3}}{2}b^{1/3}$.
Numerical simulations suggest that the above expression is a lower bound for the minimal value. Does this follow from some nice result about parameterized convex functions? It seems like it shouldn't be hard to prove. I guess in a nutshell I just want to prove that for all $x,a,b>0$ we have
$(\frac{a}{a+b})\cdot2\sqrt{a} + (\frac{b}{a+b})\cdot\frac{3\cdot2^{1/3}}{2}b^{1/3} \leq 1/x + ax + bx^2$. Thanks!
EDIT: It also appears that if I take the convex combination
$(\frac{a^{3/5}}{a^{3/5}+b^{2/5}})\cdot2\sqrt{a} + (\frac{b^{2/5}}{a^{3/5}+b^{2/5}})\cdot\frac{3\cdot2^{1/3}}{2}b^{1/3}$
then I get a tighter lower bound, and in fact the lower bound is within a factor of something like $3/2$ of the true minimal solution.
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can't you just split 1/x into the linear combination and use that. – Nishant Chandgotia Apr 16 2011 at 19:24
Sorry, I don't understand -- how does the "splitting" work? – Jennifer Gao Apr 17 2011 at 8:29
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but computing the roots of your cubic don't seem to be too much more complicated than computing the lower bounds? en.wikipedia.org/wiki/Cubic_function – S. Sra Apr 17 2011 at 8:50
@Suvrit: is it? The expression for the roots of a cubic looks pretty complicated to me, and the lower bound is just a pair of terms. – Jennifer Gao Apr 17 2011 at 20:40
## 2 Answers
The first inequality is true. Write $$f=\frac{a}{a+b}f_0+\frac{b}{a+b}f_1,$$ where $f_0$ and $f_1$ correspond to the case $b=0$ and $a=0$, respectively. You know that $f_0\ge2\sqrt a$ and $f_1\ge\frac{3\cdot2^{1/3}}{2}b^{1/3}$. This implies $$f\ge\frac{a}{a+b}2\sqrt a+\frac{b}{a+b}\frac{3\cdot2^{1/3}}{2}b^{1/3}.$$
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Is it true? We have $\frac{a}{a+b}f_0 + \frac{b}{a+b}f_1 = \frac{1}{x}+\frac{a^2}{a+b}x + \frac{b^2}{a+b}x^2$ – Maciej S. May 8 2011 at 9:10
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As Nishant Chandgotia sugessted: simply write $f(x) = \left(p\cdot \frac{1}{x} + ax\right) + \left((1-p)\frac{1}{x}+bx^2 \right)$ for some parametr $p\in[0,1]$.
For the first term, minimizer is equal to $p^{\frac{1}{2}}a^{-\frac{1}{2}}$ and the minimal value is $p^{\frac{1}{2}} 2a^{\frac{1}{2}}$. For the second therm, minimizer is equal to $(1-p)^{\frac{1}{3}}(2b)^{-\frac{1}{3}}$ and the minimal value is $(1-p)^{\frac{2}{3}} \cdot\frac{3\cdot 2^{\frac{1}{3}} }{2} b^{\frac{1}{3}}$
The best estimate is achieved when both minimizers are equal, which means, in therms of $p$, that
\begin{equation} \left(\frac{p}{a}\right)^{3} = \left( \frac{1-p}{2b}\right)^{2} \end{equation}
Note, that this equation has a solution in interval $0 < p < 1$ by Mean-value theorem, unfortunately not expressible in nice way.
Any way, we get for any $p\in[0,1]$ the following estimate:
\begin{equation} f(x) \geqslant p^{\frac{1}{2}} \cdot 2a^{\frac{1}{2}} + (1-p)^{\frac{2}{3}} \cdot\frac{3\cdot 2^{\frac{1}{3}} }{2} b^{\frac{1}{3}} \end{equation}
Finally, we check that $p^{\frac{1}{2}} + (1-p)^{\frac{2}{3}} \geqslant 1$. This inequality implies, that estimate reamins valid after using any convex combination instead of weights $p^{\frac{1}{2}}, (1-p)^{\frac{2}{3}}$, i.e.
\begin{equation} f(x) \geqslant \alpha \cdot 2a^{\frac{1}{2}} + (1-\alpha) \cdot\frac{3\cdot 2^{\frac{1}{3}} }{2} b^{\frac{1}{3}} \end{equation}
This can be seen immediately, however, by the inequality $f \geqslant \alpha f_0 + (1-\alpha) f_1$. My goal was to complete presented ideas and to show when exact optimum is attained.
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http://mathhelpforum.com/differential-geometry/74715-analysis-question.html
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# Thread:
1. ## analysis question
1. For some ε > 0 there exists an index N such that
if n > N then absolute value of an- a < ε
2.For each ε > 0 and each index N if n > N then absolutue value an-a <ε
3.There exists an index N such that for all ε > 0 if n > N then absolute value an-a < ε
4. For each ε > 0 and each index N if n > n then absolute value an-a<ε
Match from 1-4
a) The sequence {an} is bounded
b) The sequence {an} is a constant
c) All except finitely many terms of the sequence {an} are equal to constant a
d) Such a sequence does not exist
e) The assertion holds for every sequence
2. Originally Posted by wvlilgurl
1. For some ε > 0 there exists an index N such that
if n > N then absolute value of an- a < ε
2.For each ε > 0 and each index N if n > N then absolutue value an-a <ε
3.There exists an index N such that for all ε > 0 if n > N then absolute value an-a < ε
4. For each ε > 0 and each index N if n > N then absolute value an-a<ε
4. The sequence is constant since we may let $N=1$, then it asserts that for all $\varepsilon>0$ $:|a_n-a|<\varepsilon$, which implies that for all $n\ge1:\ a_n=a$.
CB
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http://physics.stackexchange.com/questions/28137/behavior-of-shock-waves-at-relativistic-speeds
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# Behavior of shock waves at relativistic speeds
Suppose I am in a spaceship traveling inertially at a velocity $v$ that is of the same order as $c$. As I pass by a metal bar that is oriented parallel to $v$, someone hits it with another metal bar, causing it to vibrate. The vibration waves travel down the bar with a velocity of $u$. Observing the bar from my spaceship, I see the waves travelling with velocity $u+v\over 1+uv$.
1. Suppose that the waves are longitudinal, so that $v$ and $u$ are parallel. The atoms in the bar are individually moving back and forth, with a maximum velocity $w$ which is quite different from $u$. $w$ and $u$ are related in a way that depends on the material properties of the bar; I think on its elasticity its density. By comparing $u$ and $w$ I can calculate these properties of the bar. If I observe the motion of the individual atoms, I will see them moving back and forth with a maximum velocity of $v+w\over 1+uv$ and by comparing this with the observed velocity of the travelling wave I will obtain a measurement of the stiffness and density of the bar. But this value will be different from the value that an observer would calculate at rest. What explains this discrepancy? Is it due to relativistic dilation of the bar's mass and volume, and hence its density? Is there some other effect?
2. Now consider transverse waves. Again $v$ and $u$ are parallel. But this time the individual atoms are moving back and forth perpendicular to $u$ so the observed dilation of their velocity $w$ is different than it was for longitudinal motion. If I do the same calculations as in (1) I should get different results for the physical properties of the bar: I see the transverse and longitudinal waves moving at the same speeds in the bar, but when I look closely at the motion of atoms, and try to calculate material properties of the bar I get two different sets of results, because the atomic motion causing the longitudinal wave is relativistically dilated to a different degree than is the atomic motion causing the transverse wave. What is going on here? What does it really look like?
Presumably there is no discrepancy and I see a consistent picture regardless of $v$. How do all these apparently conflicting measurements iron out in special relativity?
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## 2 Answers
I guess Ron is saying that by definition the bulk modulus is the value measured in frame at rest relative to the bar, and therefore it makes no sense to ask how the bulk modulus depends on speed. It's a bit like relativistic mass: few physicists assign any significance to relativistic mass - we just use rest mass, which is defined in the rest frame.
Having said this, I think your question is interesting, though I doubt it exposes any fundamental insights into relativity. You ask:
Is it due to relativistic dilation of the bar's mass and volume, and hence its density? Is there some other effect?
I started doing a few back of the envelope calculations and quickly concluded that the answer is "all of the above". However I also concluded that it would take more time than I have available in my coffee break to give you a detailed answer. The problem with SR is that it's easy to throw around ideas like velocity addition or time dilation, but doing this casually is to tread a dangerous path. To really work out what's going on you need to take your system, i.e. the bar, and apply the Lorentz transforms to calculate exactly how it looks in your frame. If you do this you'll find that the atomic motion, interatomic forces and interatomic spacing all change, so the bulk modulus and the density both change, and unsurprisingly the speed of sound changes as well. I did Google to see if anyone had done this calculation, but without any success.
Re your second point, the speed of shear and compression waves are usually different. In fact they are different in steel even without raising the spectre of relativistic motion. It's unsurprising that they would appear to be different when viewed from a relativistic frame.
I hope this helps; I feel it's a bit of a cop out since I haven't actually given you a straight answer. I think the answer would be a lot of work (if straightforward maths) and wouldn't reveal any fundamental insights. It's sort of the Physics equivalent of a crossword puzzle.
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Thanks for your answer and for the time you took to search Google and to do the envelope calculations. I did not expect it to expose any fundamental insights into relativity. – Mark Dominus May 15 '12 at 16:59
Can you recommend a textbook if I want to learn to do this kind of calculation myself? Assume that I have or can acquire the necessary mathematical background, and all I need is the physics-specific part of the instruction. – Mark Dominus May 25 '12 at 15:34
The calculations of the speed of sound (not the speed of a "shock wave", this is just sound you are describing) are done in the rest frame. They are dependent on the change in relative position of neighboring atoms. The analysis does not apply in a moving frame. In the moving frame, the speed of sound is the boost of the speed of sound in the stationary frame.
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I don't understand how this answers any of my questions. – Mark Dominus May 11 '12 at 0:42
You asked "why does the relation between the atomic velocity and the velocity of sound not seem to work in a moving frame". The answer is because it is derived in the rest frame, and the material picks out the rest frame. So why should you expect it to work? You didn't show a contradiction, you just said "how do you calculate the speed of sound in the moving frame"--- well, you go to the rest frame and find the speed of sound, then boost the speed of sound as you did. There is no question I can see. – Ron Maimon May 11 '12 at 0:45
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http://physics.stackexchange.com/questions/21955/ignorance-in-statistical-mechanics
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# Ignorance in statistical mechanics
Consider this penny on my desc. It is a particular piece of metal, well described by statistical mechanics, which assigns to it a state, namely the density matrix $\rho_0=\frac{1}{Z}e^{-\beta H}$ (in the simplest model). This is an operator in a space of functions depending on the coordinates of a huge number $N$ of particles.
The ignorance interpretation of statistical mechanics, the orthodoxy to which all introductions to statistical mechanics pay lipservice, claims that the density matrix is a description of ignorance, and that the true description should be one in terms of a wave function; any pure state consistent with the density matrix should produce the same macroscopic result.
Howewer, it would be very surprising if Nature would change its behavior depending on how much we ignore. Thus the talk about ignorance must have an objective formalizable basis independent of anyones particular ignorant behavior.
On the other hand, statistical mechanics always works exclusively with the density matrix (except in the very beginning where it is motivated). Nowhere (except there) one makes any use of the assumption that the density matrix expresses ignorance. Thus it seems to me that the whole concept of ignorance is spurious, a relic of the early days of statistical mechanics.
Thus I'd like to invite the defenders of orthodoxy to answer the following questions:
(i) Can the claim be checked experimentally that the density matrix (a canonical ensemble, say, which correctly describes a macroscopic system in equilibrium) describes ignorance? - If yes, how, and whose ignorance? - If not, why is this ignorance interpretation assumed though nothing at all depends on it?
(ii) In a though experiment, suppose Alice and Bob have different amounts of ignorance about a system. Thus Alice's knowledge amounts to a density matrix $\rho_A$, whereas Bob's knowledge amounts to a density matrix $\rho_B$. Given $\rho_A$ and $\rho_B$, how can one check in principle whether Bob's description is consistent with that of Alice?
(iii) How does one decide whether a pure state $\psi$ is adequately represented by a statistical mechanics state $\rho_0$? In terms of (ii), assume that Alice knows the true state of the system (according to the ignorance interpretation of statistical mechanics a pure state $\psi$, corresponding to $\rho_A=\psi\psi^*$), whereas Bob only knows the statistical mechanics description, $\rho_B=\rho_0$.
Presumably, there should be a kind of quantitative measure $M(\rho_A,\rho_B)\ge 0$ that vanishes when $\rho_A=\rho_B)$ and tells how compatible the two descriptions are. Otherwise, what can it mean that two descriptions are consistent? However, the mathematically natural candidate, the relative entropy (= Kullback-Leibler divergence) $M(\rho_A,\rho_B)$, the trace of $\rho_A\log\frac{\rho_A}{\rho_B}$, [edit: I corrected a sign mistake pointed out in the discussion below] does not work. Indeed, in the situation (iii), $M(\rho_A,\rho_B)$ equals the expectation of $\beta H+\log Z$ in the pure state; this is minimal in the ground state of the Hamiltonian. But this would say that the ground state would be most consistent with the density matrix of any temperature, an unacceptable condition.
Edit: After reading the paper http://bayes.wustl.edu/etj/articles/gibbs.paradox.pdf by E.T. Jaynes pointed to in the discussion below, I can make more precise the query in (iii): In the terminology of p.5 there, the density matrix $\rho_0$ represents a macrostate, while each wave function $\psi$ represents a microstate. The question is then: When may (or may not) a microstate $\psi$ be regarded as a macrostate $\rho_0$ without affecting the predictability of the macroscopic observations? In the above case, how do I compute the temperature of the macrostate corresponding to a particular microstate $\psi$ so that the macroscopic behavior is the same - if it is, and which criterion allows me to decide whether (given $\psi$) this approximation is reasonable?
An example where it is not reasonable to regard $\psi$ as a canonical ensemble is if $\psi$ represents a composite system made of two pieces of the penny at different temperature. Clearly no canonical ensemble can describe this situation macroscopically correct. Thus the criterion sought must be able to decide between a state representing such a composite system and the state of a penny of uniform temperature, and in the latter case, must give a recipe how to assign a temperature to $\psi$, namely the temperature that nature allows me to measure.
The temperature of my penny is determined by Nature, hence must be determined by a microstate that claims to be a complete description of the penny.
I have never seen a discussion of such an identification criterion, although they are essential if one want to give the idea - underlying the ignorance interpretation - that a completely specified quantum state must be a pure state.
Part of the discussion on this is now at: http://chat.stackexchange.com/rooms/2712/discussion-between-arnold-neumaier-and-nathaniel
Edit (March 11, 2012): I accepted Nathaniel's answer as satisfying under the given circumstances, though he forgot to mention a fouth possibility that I prefer; namely that the complete knowledge about a quantum system is in fact described by a density matrix, so that microstates are arbitrary density matrces and a macrostate is simply a density matrix of a special form by which an arbitrary microstate (density matrix) can be well approximated when only macroscopic consequences are of interest. These special density matrices have the form $\rho=e^{-S/k_B}$ with a simple operator $S$ - in the equilibrium case a linear combination of 1, $H$ (and vaiious number operators $N_j$ if conserved), defining the canonical or grand canonical ensemble. This is consistent with all of statistical mechanics, and has the advantage of simplicity and completeness, compared to the ignorance interpretation, which needs the additional qualitative concept of ignorance and with it all sorts of questions that are too imprecise or too difficult to answer.
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Is this not the same problem as the MaxEnt school "runs into" (scare quotes because they don't really) that physics seems to change depending on how much one chooses to ignore? The resolution there is that ultimately one is doing science, so one needs a condition like "this set of control variables is empirically sufficient to control the outputs". – genneth Mar 6 '12 at 15:15
– Arnold Neumaier Mar 6 '12 at 15:24
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@ArnoldNeumaier yes, but "everything that makes a difference to the system [as measured by macroscopic instruments]" != everything. MaxEnt is founded precisely on ignoring the microscopic details that do not make any difference to the macroscopic state, while not ignoring anything that does. Ignoring things that don't make any difference is good, because it means you don't have to calculate them! – Nathaniel Mar 6 '12 at 18:27
Arnold, perhaps this a minor point, but use of the canonical ensemble implies to me the penny is in thermal equilibrium with an environment. This would mean that the penny is entangled with the environment and therefore it would not be described by a pure state. Your questions do not seem to be as sharp if they are posed to the microcanonical ensemble. – BebopButUnsteady Mar 6 '12 at 19:19
@BebopButUnsteady: The penny is by assumption in thermal equilibrium, but need not be in equilibrium with the environment (e.g, if I just opened the window, thereby changing the environment.) - But any macroscopic body (not only a penny, and not only in a canonical ensemble, and even if far from equilibrium) is always entangled with its environment. The consequence is that no macroscopic object can be assigned a pure state, not even in principle. But this flatly contradicts the ignorance interpretation of statistical mechanics. Thus more things to defend for the upholders of orthodoxy! – Arnold Neumaier Mar 6 '12 at 19:36
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## 4 Answers
I wouldn't say the ignorance interpretation is a relic of the early days of statistical mechanics. It was first proposed by Edwin Jaynes in 1957 (see http://bayes.wustl.edu/etj/node1.html, papers 9 and 10, and also number 36 for a more detailed version of the argument) and proved controversial up until fairly recently. (Jaynes argued that the ignorance interpretation was implicit in the work of Gibbs, but Gibbs himself never spelt it out.) Until recently, most authors preferred an interpretation in which (for a classical system at least) the probabilities in statistical mechanics represented the fraction of time the system spends in each state, rather than the probability of it being in a particular state at the present time. This old interpretation makes it impossible to reason about transient behaviour using statistical mechanics, and this is ultimately what makes switching to the ignorance interpretation useful.
In response to your numbered points:
(i) I'll answer the "whose ignorance?" part first. The answer to this is "an experimenter with access to macroscopic measuring instruments that can measure, for example, pressure and temperature, but cannot determine the precise microscopic state of the system." If you knew precisely the underlying wavefunction of the system (together with the complete wavefunction of all the particles in the heat bath if there is one, along with the Hamiltonian for the combined system) then there would be no need to use statistical mechanics at all, because you could simply integrate the Schrödinger equation instead. The ignorance interpretation of statistical mechanics does not claim that Nature changes her behaviour depending on our ignorance; rather, it claims that statistical mechanics is a tool that is only useful in those cases where we have some ignorance about the underlying state or its time evolution. Given this, it doesn't really make sense to ask whether the ignorance interpretation can be confirmed experimentally.
(ii) I guess this depends on what you mean by "consistent with." If two people have different knowledge about a system then there's no reason in principle that they should agree on their predictions about its future behaviour. However, I can see one way in which to approach this question. I don't know how to express it in terms of density matrices (quantum mechanics isn't really my thing), so let's switch to a classical system. Alice and Bob both express their knowledge about the system as a probability density function over $x$, the set of possible states of the system (i.e. the vector of positions and velocities of each particle) at some particular time. Now, if there is no value of $x$ for which both Alice and Bob assign a positive probability density then they can be said to be inconsistent, since every state that Alice accepts the system might be in Bob says it is not, and vice versa. If any such value of $x$ does exist then Alice and Bob can both be "correct" in their state of knowledge if the system turns out to be in that particular state. I will continue this idea below.
(iii) Again I don't really know how to convert this into the density matrix formalism, but in the classical version of statistical mechanics, a macroscopic ensemble assigns a probability (or a probability density) to every possible microscopic state, and this is what you use to determine how heavily represented a particular microstate is in a given ensemble. In the density matrix formalism the pure states are analogous to the microscopic states in the classical one. I guess you have to do something with projection operators to get the probability of a particular pure state out of a density matrix (I did learn it once but it was too long ago), and I'm sure the principles are similar in both formalisms.
I agree that the measure you are looking for is $D_\textrm{KL}(A||B) = \sum_i p_A(i) \log \frac{p_A(i)}{p_B(i)}$. (I guess this is $\mathrm{tr}(\rho_A (\log \rho_A - \log \rho_B))$ in the density matrix case, which looks like what you wrote apart from a change of sign.) In the case where A is a pure state, this just gives $-\log p_B(i)$, the negative logarithm of the probability that Bob assigns to that particular pure state. In information theory terms, this can be interpreted as the "surprisal" of state $i$, i.e. the amount of information that must be supplied to Bob in order to convince him that state $i$ is indeed the correct one. If Bob considers state $i$ to be unlikely then he will be very surprised to discover it is the correct one.
If B assigns zero probability to state $i$ then the measure will diverge to infinity, meaning that Bob would take an infinite amount of convincing in order to accept something that he was absolutely certain was false. If A is a mixed state, this will happen as long as A assigns a positive probability to any state to which B assigns zero probability. If A and B are the same then this measure will be 0. Therefore the measure $D_\textrm{KL}(A||B)$ can be seen as a measure of how "incompatible" two states of knowledge are. Since the KL divergence is asymmetric I guess you also have to consider $D_\textrm{KL}(B||A)$, which is something like the degree of implausibility of B from A's perspective.
I'm aware that I've skipped over some things, as there was quite a lot to write and I don't have much time to do it. I'll be happy to expand it if any of it is unclear.
Edit (in reply to the edit at the end of the question): The answer to the question "When may (or may not) a microstate $\phi$ be regarded as a macrostate $\rho_0$ without affecting the predictability of the macroscopic observations?" is "basically never." I will address this is classical mechanics terms because it's easier for me to write in that language. Macrostates are probability distributions over microstates, so the only time a macrostate can behave in the same way as a microstate is if the macrostate happens to be a fully peaked probability distribution (with entropy 0, assigning $p=1$ to one microstate and $p=0$ to the rest), and to remain that way throughout the time evolution.
You write in a comment "if I have a definite penny on my desk with a definite temperature, how can it have several different pure states?" But (at least in Jaynes' version of the MaxEnt interpretation of statistical mechanics), the temperature is not a property of the microstate but of the macrostate. It is the partial differential of the entropy with respect to the internal energy. Essentially what you're doing is (1) finding the macrostate with the maximum (information) entropy compatible with the internal energy being equal to $U$, then (2) finding the macrostate with the maximum entropy compatible with the internal energy being equal to $U+dU$, then (3) taking the difference and dividing by $dU$. When you're talking about microstates instead of macrostates the entropy is always 0 (precisely because you have no ignorance) and so it makes no sense to do this.
Now you might want to say something like "but if my penny does have a definite pure state that I happen to be ignorant of, then surely it would behave in exactly the same way if I did know that pure state." This is true, but if you knew precisely the pure state then you would (in principle) no longer have any need to use temperature in your calculations, because you would (in principle) be able to calculate precisely the fluxes in and out of the penny, and hence you'd be able to give exact answers to the questions that statistical mechanics can only answer statistically.
Of course, you would only be able to calculate the penny's future behaviour over very short time scales, because the penny is in contact with your desk, whose precise quantum state you (presumably) do not know. You will therefore have to replace your pure-state-macrostate of the penny with a mixed one pretty rapidly. The fact that this happens is one reason why you can't in general simply replace the mixed state with a single "most representative" pure state and use the evolution of that pure state to predict the future evolution of the system.
Edit 2: the classical versus quantum cases. (This edit is the result of a long conversation with Arnold Neumaier in chat, linked in the question.)
In most of the above I've been talking about the classical case, in which a microstate is something like a big vector containing the positions and velocities of every particle, and a macrostate is simply a probability distribution over a set of possible microstates. Systems are conceived of as having a definite microstate, but the practicalities of macroscopic measurements mean that for all but the simplest systems we cannot know what it is, and hence we model it statistically.
In this classical case, Jaynes' arguments are (to my mind) pretty much unassailable: if we lived in a classical world, we would have no practical way to know precisely the position and velocity of every particle in a system like a penny on a desk, and so we would need some kind of calculus to allow us to make predictions about the system's behaviour in spite of our ignorance. When one examines what an optimal such calculus would look like, one arrives precisely at the mathematical framework of statistical mechanics (Boltzmann distributions and all the rest). By considering how one's ignorance about a system can change over time one arrives at results that (it seems to me at least) would be impossible to state, let alone derive, in the traditional frequentist interpretation. The fluctuation theorem is an example of such a result.
In a classical world there would be no reason in principle why we couldn't know the precise microstate of a penny (along with that of anything it's in contact with). The only reasons for not knowing it are practical ones. If we could overcome such issues then we could predict the microstate's time-evolution precisely. Such predictions could be made without reference to concepts such as entropy and temperature. In Jaynes' view at least, these are purely macroscopic concepts and don't strictly have meaning on the microscopic level. The temperature of your penny is determined both by Nature and by what you are able to measure about Nature (which depends on the equipment you have available). If you could measure the (classical) microstate in enough detail then you would be able to see which particles had the highest velocities and thus be able to extract work via a Maxwell's demon type of apparatus. Effectively you would be partitioning the penny into two subsystems, one containing the high-energy particles and one containing the lower-energy ones; these two systems would effectively have different temperatures.
My feeling is that all of this should carry over on to the quantum level without difficulty, and indeed Jaynes presented much of his work in terms of the density matrix rather than classical probability distributions. However there is a large and (I think it's fair to say) unresolved subtlety involved in the quantum case, which is the question of what really counts as a microstate for a quantum system.
One possibility is to say that the microstate of a quantum system is a pure state. This has a certain amount of appeal: pure states evolve deterministically like classical microstates, and the density matrix can be derived by considering probability distributions over pure states. However the problem with this is distinguishability: some information is lost when going from a probability distribution over pure states to a density matrix. For example, there is no experimentally distinguishable difference between the mixed states $\frac{1}{2}(\mid \uparrow \rangle \langle \uparrow \mid + \mid \downarrow \rangle \langle \downarrow \mid)$ and $\frac{1}{2}(\mid \leftarrow \rangle \langle \leftarrow \mid + \mid \rightarrow \rangle \langle \rightarrow \mid)$ for a spin-$\frac{1}{2}$ system. If one considers the microstate of a quantum system to be a pure state then one is committed to saying there is a difference between these two states, it's just that it's impossible to measure. This is a philosophically difficult position to maintain, as it's open to being attacked with Occam's razor.
However, this is not the only possibility. Another possibility is to say that even pure quantum states represent our ignorance about some underlying, deeper level of physical reality. If one is willing to sacrifice locality then one can arrive at such a view by interpreting quantum states in terms of a non-local hidden variable theory.
Another possibility is to say that the probabilities one obtains from the density matrix do not represent our ignorance about any underlying microstate at all, but instead they represent our ignorance about the results of future measurements we might make on the system.
I'm not sure which of these possibilities I prefer. The point is just that on the philosophical level the ignorance interpretation is trickier in the quantum case than in the classical one. But in practical terms it makes very little difference - the results derived from the much clearer classical case can almost always be re-stated in terms of the density matrix with very little modification.
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Thanks for the clarification on the origins. The problem with your answer to (iii) is that in the particular case mentioned in my edited statement on (iii), the ground state would be the most consistent pure state, irrespective of temperature. Thus the K/L measure doesn't allow me to assess whether treating the pure state $\psi$ as a canonical example (if I am only interested in macroscopic consequences) is or isn't acceptable. – Arnold Neumaier Mar 6 '12 at 17:33
The only lesson to draw from this is that it isn't always sensible to try and take a single "most representative" pure state from a probability distribution and expect it to have similar properties. If you're interested in macroscopic properties you should be calculating expectations. If there is a pure state whose properties (or at least the ones you're interested in) behave similarly to expectations calculated from the density matrix then you'd be justified in what you're triyng to do. I agree that the KL measure by itself doesn't tell you this, of course. – Nathaniel Mar 6 '12 at 18:09
But if I have a definite penny on my desk with a definite temperature, how can it have several different pure states? Either this penny has a particular wave function $\psi$ which gives its complete quantum mechanical description (even though we are never able to say which one it is), then this state must somehow have an associated temperature , since Nature knows this temperature, and the description is complete. - Or such a unique $\psi$ doesn't exist, in which case the concept of microstates breaks down, and there is only the density matrix to describe the system. – Arnold Neumaier Mar 6 '12 at 18:15
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In Jaynes' view, the macrostate is a probability distribution over the microstates, and the temperature is a property of the macrostate, not the microstate. $T=\partial S/\partial U$, where $S$ is the entropy of the macrostate. If we completely knew the microstate we would be talking about a probability distribution where one state has $p=1$ and the rest 0. There would be no entropy, and hence no temperature. – Nathaniel Mar 6 '12 at 18:30
In ignorance terms, $\partial S/\partial U$ means something like "if I added a little bit more energy to this penny, how much more ignorance would I then have about its microstate?" I will update my answer to make some of this clearer. – Nathaniel Mar 6 '12 at 18:32
I'll complete @Natahniel's answer with the fact that 'knowledge' can have physical implication linked with the behaviour of nature. The problem goes back to Maxwell's demon, who converts his knowledge of the system into work. Recent works (like arXiv:0908.0424 The work value of information) shows that the information theoretical entropies defining the knowledge of the system is connected to the work which is extractable in the same way than the physical entropies are.
To sum al this into a few words, "Nature [does not] change its behaviour depending on how much we ignore", but "how much we ignore" changes the amount of work we can extract fro Nature.
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– Nathaniel Mar 6 '12 at 16:05
Thanks for the reference. – Frédéric Grosshans Mar 6 '12 at 16:56
– Arnold Neumaier Mar 6 '12 at 17:28
@ArnoldNeumaier : Thanks for the reference. I've just read the chapter 10.1. For me (but I'm biased towards information theory), the choice of a description level is precisely what is related to the physicist's knowledge. But I agree that it is a (useful) philosophical debate, and the whole question is linked to the study of the model choice itself. – Frédéric Grosshans Mar 7 '12 at 18:21
By the way, the paper linked to in my answer is not directly related to Gibbs paradox, but is a computation of the work which can (probabilistically) be extracted from a system on which we have a partial knowledge (quantified by Shannon/Smooth-Rényi entropies) – Frédéric Grosshans Mar 7 '12 at 18:21
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When it comes to discussion of these matters, I make a following comment witch starts with the citation fom Landau-Lifshitz, book 5, chapter 5:
The averaging by means of the statisitcal matrix ... has a twofold nature. It comprises both the averaging due to the probalistic nature of the quantum description (even when as complete as possible) and the statistical averaging necessiated by the incompleteness of our information concerning the object considered.... It must be borne in mind, however, that these constituents cannot be separated; the whole averaging procedure is carried out as a single operation, and cannot be represented as the result of succesive averagings, one purely quantum-mechanical and the other purely statistical.
... and the following ...
It must be emphasized that the averaging over various $\psi$ states, which we have used in order to illustrate the transition from a complete to an incomplete quantum-mechanical description has only a very formal significance. In particular, it would be quite incorrect to suppose that the description by means of the density matrix signifies that the subsystem can be in various $\psi$ states with various probabilities and that the average is over these probabilities. Such a treatment would be in conflict with the basic principles of quantum mechanics.
So we have two statements:
Statement A: You cannot "untie" quantum mechanical and statistical uncertainty in density matrix.
(It is just a restatement of the citations above.)
Statement B: Quantum mechanical uncertainty cannot be expressed in terms of mere "ignorance" about a system.
(I'm sure that this is self-evident from all that we know about quantum mechanics.)
Finally:
Therefore: Uncertainty in density matrix cannot be expressed in terms of mere "ignorance" about a system.
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2
The conclusion does not follow from the premises. I could just as easily say "1. quantum and statistical uncertainty cannot be untied in the density matrix formalism. 2. the uncertainty in a density matrix cannot be expressed as mere 'quantum' uncertainty (otherwise it would be a pure state). Therefore, 3. uncertainty in the density matrix cannot be expressed in terms of mere 'quantum' uncertainty." A much more reasonable conclusion is that some of the uncertainty in the density matrix is quantum and some is statistical; it's just impossible to untie them. – Nathaniel Mar 6 '12 at 16:16
@Nathaniel I agree with your statement 3 and see no problem with it. It doesn't contradict anything. And also it doesn't in any way refute my statement. While the "much more reasonable conclusion" is just restatement of statement 1. – Kostya Mar 6 '12 at 16:25
@Nathaniel: Why should your point 2 in your comment be true? Surely a density matrix is a quantum object and expresses quanutm uncertainty. The success of statistical mechanics together with the fact that you cannot untie the information in a density matrix rather suggests that the density matrix is the irreducible and objective quantum information, and the pure state is only a very special, rarely realized case. – Arnold Neumaier Mar 6 '12 at 17:18
@Kostya, sorry - in that case I misunderstood - I interpreted you as saying that none of the uncertainty in the density matrix can be expressed in terms of ignorance. If you were only saying that some of it can't then no problem. (Though having said that, for someone who supports a non-local hidden variable interpretation, it can all be expressed as ignorance. Some people might find that more palatable than abandoning locality; I'm not sure whether I do or not.) – Nathaniel Mar 6 '12 at 17:52
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@ArnoldNeumaier consider a machine that mechanically flips a coin, then based on the result prepares an electron in one pure state (call it $|A\rangle$) or another ($|B\rangle$). To model the state of an electron from this machine you would use the density matrix $\frac{1}{2}\left( |A\rangle\langle A| + |B \rangle \langle B| \right)$. Surely this represents both the quantum uncertainty inherent in the pure states and your classical uncertainty about the outcome of a (hidden) coin flip. So at least in some situations some of the density matrix's uncertainty is ignorance. – Nathaniel Mar 6 '12 at 18:01
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The ignorance interpretation of the density matrix was introduced by vonNeuman in close analogy to the ignorance interpretation in earlier classical statistical mechanics, where probabilities were associated to ignorance. But in quantum theory probabilities are intrinsic not related to our ignorance.
Dirac and Landau introduced the quantum mechanical interpretation of the density matrix as the more general description of the quantum state of a system. Feynman reasoned that the usual wavefunction theory works only when one considers isolated systems and ignores the rest of the universe.
Prigogine and other members of the Brussels school have shown that wavefunction theory only applies to stable quantum systems but not to unstable ones, which require density matrices outside the Hilbert space.
In this modern perspective, the old supposition that the quantum state is given by some $|\Psi \rangle$ is merely a supposition based in ignorance and approximations on the underlying micro state given by a density matrix. Regarding your three questions:
(i) No it can't, even if we ignore the recent results on unstable quantum systems and focus on simpler systems. The old ignorance interpretation is a non-scientific hypothesis, because it first assumes the existence of a pure state and next claims that this hypothetical pure state cannot be measured. This is not different from hidden variables approaches to QM or from parodies of religion based in the famous invisible pink unicorn.
(ii) $Tr\{ O \rho_A \} = Tr\{ O \rho_B \}$ for any observable $O$, if both descriptions are mutually consistent, which implies either $\rho_A = \rho_B$ or that some of them has redundant information. See (iii) for some detail.
(iii) The problem in quantum statistical mechanics is that two completely different interpretations of the density matrix are usually confused in the literature. There are two kind of contractions of the description of an atomic-molecular system: exact and inexact. Suppose that the vector $(\mathbf{n})$ of variables describes the state of a given system. We can split this into two sets $(\mathbf{n}_1,\mathbf{n}_2)$; suppose now that we contract the description using only $(\mathbf{n}_1)$ and ignoring the rest. If the dynamical description of the system is unchanged by using this contracted description $(\mathbf{n}_1)$ instead of the whole $(\mathbf{n})$ then the contraction is exact and $(\mathbf{n}_2)$ are redundant variables; otherwise $(\mathbf{n}_1)$ only provides an approximated description of the system. The redundant variables denote relaxed variables in the scale of time chosen to study the system. Let me a simple example.
Consider a simple nonequilibrium gas (constant composition) described by a $\rho(t_0)$. Writing down the equation of motion, we can check that there exists a hierarchy of time scales $t_1 < t_2 < t_3 ...$ for which different sets of variables relax, achieve equilibrium values, and no more participate in the dynamical description. For instance, there exists a scale $t_C$, which is roughly of the order of duration of a collision so that for $t \gg t_C$, the binary correlations $g_2$ take an equilibrium value and the corresponding integral, in the exact equation of motion, vanishes (doing unneeded its computation). There exists a scale $t_R \gg t_C$, which is roughly of the order of relaxation time so that for $t \gg t_R$, the deviations of $\rho$ from its equilibrium value vanish and the corresponding relaxation kernel in the exact equation of motion vanishes (doing unneeded its computation). This hierarchy of contractions explains why an equilibrium gas systems can be described by a highly contracted description: e.g. using only pressure and temperature at equilibrium. The existence of contracted descriptions is not related to ignorance but to the dynamical survival of the more 'robust' mode of evolution for the characteristic times scales.
In fact, the old ignorance interpretation of statistical mechanics does not explain why we can use $p$ and $T$, and ignore the rest of variables, for an equilibrium gas at equilibrium, but we cannot use the same contracted description for the same gas in a turbulent regime. Evidently it has nothing to see with ignorance and/or the ability to measure.
For instance, we can measure fields $p(x,t)$, $n(x,t)$, and $T(x,t)$ for both equilibrium and nonequilibrium regimes. However, those fields are completely redundant for an equilibrium gas, whereas the same fields describe the gas for not too far nonequilibrium regimes, and miserably fail for far-from-equilibrium turbulent regimes. The modern interpretation explains why. For turbulent regimes the fast variables did not relax still and you need to use a broad set of variables to describe the system (precisely extended thermodynamics adds an extended set of variables to the above fields for describing far from equilibrium regimes). For linear non-equilibrium regimes the fast variables did relax and can be ignored, providing a contracted description where you only need $p(x,t)$, $n(x,t)$, and $T(x,t)$ to describe the system. Finally, at equilibrium, all the variables did relax and $p(x,t)$, $n(x,t)$, and $T(x,t)$ is contracted again to $p$ and $T$.
Regarding your (ii) Alice could use $p(x,t)$, $n(x,t)$, and $T(x,t)$ whereas Bob use only $p$ and $T$ for a gas at equilibrium and both would agree on the description of the same system (except Alice would describe the system redundantly with lots of non-useful information).
P. S: I would add that the old ignorance interpretation generates many problems and paradoxes both in quantum and classical contexts.
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von Neumann carefully distinguished between the intrinsic, and essentially new, quantum probabilities which inhere even to a pure state, and the ignorance-based probabilities which were in analogy with those in Classical Stat Mech, which are added on top of the QM probabilities, to produce the density matrix. He was also aware, and commented on it, that what I have carelessly called «added on top» was also something new to Quantum Stat Mech, it was not exactly like the way ignorance probabilities were constructed in Classical Stat Mech, and this point is all that Landau is getting at. – joseph f. johnson Feb 12 at 17:34
Landau never expresses himself clearly on foundational matters, the way von Neumann always did. Yet Landau was incomparably the greater physicist, I would in fact deny that von Neumann was a physicist at all, not even a mathematical physicist. He was superb at foundational considerations in maths, and logic, but had no physical intuition at all. You are also wrong about Dirac: Dirac had not the slightest intention to make the density matrix a description of the quantum state of a system. It was, for him, a description of the mixed quantum state in analogy to classical mixed state. – joseph f. johnson Feb 12 at 17:38
@josephf.johnson Von Neumann introduced the statistical interpretation for mixed states in 1927. This would not be confused with the statistical interpretation of the wavefunction (pure states) due to Born (1925). The quantum mechanical interpretation of the density matrix was first introduced by Landau in 1927 and latter (1930,1931) by Dirac who also discussed the statistical interpretation and even normalized each $\rho$ in a different way. – juanrga Feb 15 at 20:06
@josephf.johnson No. It is not required to introduce "ignorance-based probabilities [...] to produce the density matrix." There is no ignorance for pure density matrices $\rho^2 =1$, neither there is ignorance for a mixture $\rho^2 \neq 1$ in the Landau/Dirac approach. The ignorance interpretation is exclusive to the von Neumann's statistical approach. The quantum mechanical interpretation is the basis for the Brussels School approach to LPSs and foundational issues of QM. – juanrga Feb 15 at 20:13
– joseph f. johnson Feb 16 at 2:25
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http://mathhelpforum.com/calculus/105957-finding-limit.html
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# Thread:
1. ## finding a limit
hi guys...
Im having difficulty finding this limit and i wanted to know if anyone can help me out:
lim e^(x)/ (1+(e^(2x))^(3/2))
x -> infiniti
I tried using L'Hospital's rule but that only made it more difficult to see. I feel like there's an easy trick that im just forgetting. If anyone could help me out that would be great. Thanks!
2. Originally Posted by collegestudent321
hi guys...
Im having difficulty finding this limit and i wanted to know if anyone can help me out:
lim e^(x)/ (1+(e^(2x))^(3/2))
x -> infiniti
I tried using L'Hospital's rule but that only made it more difficult to see. I feel like there's an easy trick that im just forgetting. If anyone could help me out that would be great. Thanks!
Not difficult
$\mathop {\lim }\limits_{x \to \infty } \frac{{{e^x}}}<br /> {{{{\left( {1 + {e^{2x}}} \right)}^{3/2}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{e^x}}}<br /> {{{e^{3x}}{{\left( {\frac{1}<br /> {{{e^{2x}}}} + 1} \right)}^{3/2}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{1}<br /> {{{e^{2x}}{{\left( {\frac{1}<br /> {{{e^{2x}}}} + 1} \right)}^{3/2}}}} = 0.$
3. oh wow, ok i see. haha, that was easy. Thank you!!!
4. no wait... why is it e^(3x)????
5. Originally Posted by collegestudent321
no wait... why is it e^(3x)????
Because
$\begin{gathered}<br /> {\left( {1 + {e^{2x}}} \right)^{3/2}} = {\left( {\sqrt {1 + {e^{2x}}} } \right)^3} = {\left( {{e^x}\sqrt {\frac{1}<br /> {{{e^{2x}}}} + 1} } \right)^3} = \hfill \\<br /> = {e^{3x}}{\left( {\sqrt {\frac{1}<br /> {{{e^{2x}}}} + 1} } \right)^3} = {e^{3x}}{\left( {\frac{1}<br /> {{{e^{2x}}}} + 1} \right)^{3/2}}. \hfill \\ <br /> \end{gathered}$
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http://quant.stackexchange.com/questions/1710/proof-that-you-cannot-beat-a-random-walk/1714
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# Proof that you cannot beat a random walk
There is much speculation to what degree financial series are random (and what kind of randomness prevails).
I want to turn the question on its head and ask:
Is there a mathematical proof that whatever trading strategy you use you cannot beat a random walk (that is the expected value will always be 0 assuming no drift)?
(I found this blog post where the author used the so called "75% rule" to purportedly beat a random walk but I think he got the distinction between prices and returns wrong. This method would only work if you had a range of allowed prices (e.g. a mean reverting series). See e.g. here for a discussion.)
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You are not going to be able to create a "mathematical proof" without mathematical definitions of the processes followed by the financial series. Obviously an OU process in price space would be exploitable. You need to exclude that from your process definitions to have a hope of generating a proof. – Brian B Aug 18 '11 at 13:34
@Brian: Do you know of any literature that does exactly that? I mean it seems intuitively the case but we all know intuition is not always the best judge when it comes to math... – vonjd Aug 18 '11 at 16:31
## 5 Answers
I can help you beat random walk 'in the way you want', i.e. the expected value $E[\$]$will always be positive even assuming no drift. However, I have to warn people that$E[\$] > 0$ is NOT really an adequate condition for 'beating' in reality (at least to myself).
Let's define some mathematical notations for derivation, and rephrase (simplify) vonjd's question without losing generality. Assume a trader plays a fair game, and his surplus $X(0), X(1), X(2), ... X(t)$ is a martingale.
Q: Can the trader find a stopping time $s$ such that $E[X($s$)] > X(0)$?
• A proof supporting Bootvis' answer, for comparison, consider a normal trading strategy that bets evenly. Then,
$$\begin{align*}E[X(s)] &= E[ E[X(s)|X(s-1), X(s-2),..., X(0)] ] \\ &= E[X(s-1)] = E[X(s-2)] = ... = E[X(0)] = X(0).\end{align*}$$
• Now, consider a 'double-betting' strategy. We keep doubling your losing trade until first win. Let's set the initial surplus, $X(0) = 0$ for simplicity.
Accordingly, $X(k) = X(k-1) + G(k)$, where $G(k)=\pm 2^{k}$ with probability $1/2$. Note that we get the power $(k)$ of $2$ in $G(k)$ because of 'double-betting'. Our market is still random walk.
This strategy is designed stop at a time $s = min{k}$ s.t. $G(k) > 0$ (Note that $Prob{s=infinity} = 0$)
Compute $E[X(s)]$ by conditioning on s:
$$\begin{align*} E[X(s)] &= E[E[X(s)|s]] = \sum_{k=1}^{\infty} E[X(s)|s=k] * Prob{s=k} \\ &= \sum_{k=1}^{...} (-1-2-4-8...-2^{(k-1)} + 2^{k}) * (1/2)^{k} \\ &= \sum_{k=1}^{...} 1 * (1/2)^k = 1 > 0 = X(0)\end{align*}$$
Conclusion
A trader can make $E[X(s)]>0$ for random walk using the double-betting strategy. We proved that you can beat random walk in your definition of 'beating', i.e. expected value > 0.
This is actually a simplified proof supporting Akshay's answer. Whatever it's called: volatility pumping, Kelly strategy, optimal growth portfolio, and etc. These ideas simply ask one more question: why double? Is there an optimal betting ratio because of ... (various reasons and assumptions)?
• WARNING: Yes, the expected value is indeed positive, and it might be an adequate proof for people who believe winning strategy is all about searching for $E[X(s)]>0$. Unfortunately, this is NOT adequate in reality, at least to myself. You have been warned.
• A $E[X(s)]>0$ strategy is guaranteed to make you real fortune if and only if we have 'unlimited amount of capital'. For details (long story), see wiki: Martingale betting system.
• You might ask what should we do if we only have limited capital? The Kelly criteria actually kind of offers the effect of the double-betting strategy for limited capital. For example, if you have a very weak trading signal (close to random walk in which there is no signal at all), the Kelly criteria will recommend you to bet something like \$1 (initially) for \$1M capital, and increase/decrease your position by certain % when you lose/win. Yeah, \$1M indeed looks like unlimited capital to \$1.
• (From comment) There is no contradiction to the common sense that 'pure independence = zero E[PnL]'. $E[] > 0$ in my example and vonjd'd Parrondo's paradox are indeed exploited from sort of dependency. While the Parrondo's paradox exploits the dependency between two losing games, mine is exploiting the dependency from my losing trades (which is less obvious). But warn again: This is at the cost of ruin risk! Though Kelly and vol-pump strategies eliminate ruin risk, they still suffer from trending risk.
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Thank you - how does this square with "constant rebalancing"? Is this the same as "volatility pumping"? Could you show a more direct connection between your first proof and the other concepts that seem to use some kind of "Parrondo's paradox"? Thank you again. – vonjd Aug 22 '11 at 7:43
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Thank you for accepting the answer, vonjd. I also like the analogy you found in Parrondo paradox. Though not sure how to answer your questions? here is my attempt: First, all mentioned strategies propose to add(cut) risk exposure when lose(win). For example, if you long stock and you lose/gain 1 dollar when market moves, kelly or vol-pump will require you buy/sell in order to maintain constant betting ratio. This makes volatility your friend (pumping). But trend is your enemy! In this case, is random walk more like your enemy or your friend? – 楊祝昇 Aug 24 '11 at 8:29
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Second, there is no contradiction to the common sense that 'pure independence = zero E[PnL]'. E[] > 0 in my example and your Parrondo's paradox is indeed exploited from sort of dependency. While Parrondo exploits the dependency between two losing games, mine is exploiting the dependency on my losing trades (which is less obvious). But (warn again), this is at cost of ruin risk! Note that Kelly/Vol-pump eliminate ruin risk, but still suffer tail risk. Conclusion? Find dependency had better, create it if you must. – 楊祝昇 Aug 24 '11 at 9:13
there is at least one hole in this answer, how do you know that financial valuations are really dependent on the past, not on the last stage (Markov assumption) or perhaps something else? I agree that you can beat RW with arbitrary assumptions but are you sure about your premise here about the historical dependency? – hhh Dec 8 '11 at 9:09
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＠hhh, Nobody here tried to beat real financial data. Vonjd was asking whether we can/can't beat the 'random walk', not the 'real world', in the way he wants. We are indeed aware of the difference between RW and real world, but thanks for reminding us. :) – 楊祝昇 Feb 18 '12 at 4:37
If the price of every asset follows an independent random walk without drift then every position has an expected return of zero. So, in expectation, there is no combination of positions that has an expectation different from zero.
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Thank you. I think this is an important part of a proof but not the full story. E.g. a mean reverting process can also have an expected value of 0, yet you can exploit the underlying mathematical structure (see the paper above). A random walk also has some mathematical structure, the problem is to show that there is no way to exploit this. – vonjd Aug 18 '11 at 10:38
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@vonjd, I think Bootvis's proof already handles that objection. Presumably, a mean-reverting process has an increased chance of returning to the starting point (to 0), compared to a rnadom walk. Thus a mean-reverting process differs from a random walk in the following respect: fix a point in time where the value is positive; then the conditional expectation of the future returns for a random walk is zero, but negative for a mean-reverting process. Bootvis's proof relies upon the fact that, for a random walk, conditional expectation is always 0. – D.W. Aug 18 '11 at 19:58
@D.W.: Good point: Both expectation and conditional expectation are 0 with a random walk. Thank you. – vonjd Aug 18 '11 at 20:12
@Bootvis - this argument does not take into account dynamic replicability of a position. I can have a risk-free bond and a stock - both having iid returns/price processes. I can create an option to exploit the volatility characteristics of the processes. See my answer below. – Akshay Aug 18 '11 at 21:06
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@akshay I think vonjd's original question was only regarding the trading of a single asset. – Tal Fishman Aug 19 '11 at 13:41
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That post has been up since March. Either he hasn't figured it out, or he's trying to get people to click through to the book.
In the following statement, isn't he implying that "rw" is a return (as in....random walk)?
rw <- rnorm(100)
In the following statements, isn't he calling a "trade" the DIFFERENCE IN RETURNS? Isn't that meaningless?
if(rw[i] < m) trade[i] <- (rw[i+1]-rw[i])
if(rw[i] > m) trade[i] <- (rw[i]-rw[i+1])
From there on, isn't the whole thing a waste of time? Likewise, when I opened that .pdf file, the first thing I saw was Sornette's name. There's no need to read further.
As far as "proofs" go, how are you going to agree on the properties of the market? If you can't, then the idea of a "proof" goes poof.
-
I am not so much trying to find a 1-to-1 correspondence to the statistical properties of the market. It is more a purely intellectual endavour to find out what underlying structure in a random process is exploitable (e.g. mean reverting random walk) and what is not. There is always some structure but it won't be always exploitable. I started with the well known (i.i.d.) random walk, but this can be broadened later on. – vonjd Aug 18 '11 at 16:11
...and to add to your point: What is wrong with Sornette? At least he is bringing some new ideas into the field. And in this article he shows that you cannot exploit the 75%-rule. – vonjd Aug 18 '11 at 16:40
@vonjd, Actually, it is a 1-to-1 property issue. Just try to get two people to agree on the properties. Also, it is one thing to explain what happened in the market, it is totally different to predict the same thing. And, we can't even explain it. The "exploited issues" are typically something different. If I'm a market maker, I have a lot of exploitable advantages over my prey. If I'm a hedge fund manager, I really can't lose. 2 and 20 means I'll get at least 2% of your money. As far as Sornette goes, just read 4 or 5 of his articles (or his book). You'll understand. – bill_080 Aug 18 '11 at 17:30
– bill_080 Aug 18 '11 at 17:42
Here is a more or less formal proof of the fact that "the system can't be beaten". The argument works whenever the underlying process is a martingale. In particular, it is valid for a random walk without drift.
Let $S=\{S_n\}$ be a discrete-time martingale which represents a series of games played at times $n=1,2,...$. Assume that $S_0=0$ (no game at time $n=0$). Let $\mathcal A=\{\mathcal A_n\}$ be a filtration of $\sigma$-algebras $\mathcal A_0\subset \mathcal A_1\subset\ ...$, such that the process $S$ is adapted with respect to $\mathcal A$, i.e. $S_n$ is $\mathcal A_n$-measurable for each $n$. Intuitively speaking, this implies that $\mathcal A_n$ contains all information about the outcomes of the game after the first $n$ rounds, and the value of $S_n$ is known to the player at time $n$.
We may think of the difference $S_n-S_{n-1}$ as the net winnings per unit stake in game $n$. Since $S$ is a martingale, $$\mathbb E(S_n-S_{n-1}|\mathcal A_{n-1})=0,\qquad n=1,2,...$$ Now let $C=\{C_n\}$ be a previsible bounded process, i.e. $C_n$ is $\mathcal A_{n-1}$-measurable and $$\sup|C_n(\omega)|\leq K$$ for each $n=1,2,...$ and some constant $K$. $C_n$ represents the player's stake in game $n$. Obviously, the value of $C_n$ must be determined based on the history up to time $n-1$ (the player has no information about the value of $S_n$ before the n-th round is played). Thus the total winnings up to time $n$ are $$X_0=0,\qquad X_n=\sum\limits_{k=1}^{n}C_k(S_k-S_{k-1}),\quad n\geq1.$$ Since the process $C$ is previsible and bounded, using the standard properties of the conditional mean, we have that
$$\mathbb E(X_n-X_{n-1}|\mathcal A_{n-1})=\mathbb E(C_n(S_n-S_{n-1})|\mathcal A_{n-1})= C_n\mathbb E(S_n-S_{n-1}|\mathcal A_{n-1})=0$$ for each $n=1,2,...$. In other words, $X=\{X_n\}$ is an adapted integrable process such that $$\mathbb E(X_n|\mathcal A_{n-1})=X_{n-1}$$ for all $n\geq 1$, i.e. it's a martingale itself. It immediately follows that $\mathbb E(X_n|\mathcal A_{m})=X_m$ when $m<n$ and that the unconditional mean $\mathbb E(X_n)=\mathbb E(X_0)=0$ does not depend on $n$.
"Probability with Martingales" by Williams is a good and fairly standard reference for this stuff.
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Thank you. What I learned by reading the other answers is that a random walk CAN in fact be beaten. You can call the mechanisms Shannon's demon, volatility pumping, growth-optimal portfolio or Kelly criterion. How would one resolve this contradiction?!? Thank you again. – vonjd Aug 23 '11 at 15:21
I think you mean to say any self-financing trading strategy would give you zero returns - which it will - the principle of no-arbitrage ensures that.
For example, say you have a stock whose returns follow a random walk pattern with expected return equal to zero. The only way this can happen is (a) the stock stays fixed at a value K or (b) the stock has an equal probability of being +x% or -x% tomorrow or (c) the stock can go +x% with probability p1 and -y% with probability p2 with p1*x + p2*y = 0.
Now, for all of these cases, I can have an options-based long straddle centered around K. With non-zero volatility (stochastic as well as determinate), it is far more likely that my straddle will earn me a positive return than not. The down-side is - other traders also realize this fact and so I can't have that straddle for free (and thus earning a clean expected non-zero return).
So it is not the iid process by itself which is non-exploitable (sure it is, as in the strategy above if arbitrage exists or, even simpler, in the case of a "buy-low, sell-high" strategy which can earn a positive return from an iid). The "mathematical structure"which can be exploited is known as the process volatility. It is no-arbitrage which disallows an iid process from being exploited for returns in excess of those commensurate with its volatility - not any property of the process itself.
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– vonjd Aug 19 '11 at 7:58
2
– vonjd Aug 19 '11 at 8:47
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http://quant.stackexchange.com/questions/834/how-to-conduct-monte-carlo-simulations-to-test-validity-of-black-scholes-for-a-s/836
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# How to conduct Monte Carlo simulations to test validity of Black Scholes for a specific option?
In reference to the original Black Scholes model, what approach is best to test the model in a rigorous way? Is there a standard approach that can accomplish this in a reasonable amount of time?
Details I require:
• number of trials,
• which software to use, formulas etc.
• any other information that I should be aware of
* This should be able to be done on a laptop with a Core i5 processor with a graphics card.
-
the original model - you mean the one based on log-normal distribution (3-line formula...) ? \\ Also what kind of validity are you looking for ? That it matches market price - it does if you calibrate it with implied vol... \\ And what is the Monte-Carlo for ? if you assume the same dynamics as BS - there is convergence of MC simulated option price... showed in every textbook – Vytautas Mar 28 '11 at 11:28
– Vytautas Mar 28 '11 at 11:35
I know it converges, but how do I do it for a specific option? What software should I use? I am looking to start with the original Black Scholes and make a few modifications, that's why I asked for a standard approach. – Beatrice Mar 28 '11 at 12:10
– richardh♦ Mar 30 '11 at 13:18
## 5 Answers
1. I recommend to use MATLAB / Excel for simplicity - depends which one do you already know.
2. Write down the SDE for geometic brownian motion (to simulate stock price over time) on paper, as quant_dev mentioned.
1. Discretize it using i.e. forward Euler discretization (see Wikipedia), code up a MC simulation to simulate it for the time period you want to price your options.
2. Don't forget to use the risk-free dynamics in the SDE, otherwise you wont converge to BS price.
3. Code the $f(S_T)$ payoff function for your option payoff.
4. Calculate the expected (average over simulations), discounted payoff.
With 10 000 simulations, or even 100 000, there should be a decent convergence of your simulation (error at $10^{-4}$) - your CPU should handle this in a few mins max.
-
## Did you find this question interesting? Try our newsletter
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Write out your model as an SDE, simulate it and compare the result with an analytical solution (if you've got one).
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On the software end, if you want something quick/dirty I would personally go with Matlab/R/python however if you want something a bit more rigorous (e.g. payoff classes, "better" SDEs) something OO like C++ would really be the route to take.
The basic is fairly simple here's a quick sample of what it should look like:
````double variance = vol*vol*expire;
double rootVariance = sqrt(variance);
double halfVar = -0.5*variance;
double SpotPlusOne = s*exp(r*expire+halfVar);
double Spot;
double runningSum=0;
for (unsigned long i=0; i < NumOfPaths; i++)
{
double SN = SNByBoxMuller();
Spot = SpotPlusOne*exp(rootVariance*SN);
double PayOff = Spot – strike;
PayOff = PayOff >0 ? PayOff : 0;
runningSum += PayOff;
}
double mean = runningSum / NumOfPaths;
mean*=exp(-r*expire);
return mean;
````
The SNByBoxMuller() is just the standard way of generating a random number from a standard normal distribution from Box Muller.
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I think what you really need to do is test the results of the analytic/simulated solution against the actual AND future historical price. So you need to get historical data for the specific option. That, to me, is much more interesting since it will tell you if Black-Scholes is working vs. the reality. Of course you will need a large number of historical data points to tackle this.
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It's not that easy, because the simulation will be performed in risk-neutral measure, and historical data give you information about the physical (a.k.a. real-world measure). You can use them to verify your models, but it's not something a beginner will be able to do. – quant_dev Mar 28 '11 at 21:10
Good point. However if the OP is versed in Monte Carlo simulation, she is not a beginner, and she can make any assumptions she wishes regarding IV, interest rate etc. and let the model run. The only real constraint is time. – Ralph Winters Mar 31 '11 at 18:36
It seems like everyone here is MC but you can use PDE methods as well.
Anyway there is two things that you can usually check, the Price and ... the Hedge (or replication price).
Let's look at the first case:
• If you have closed-form formulas (as is usually the case in the BS "fantas(ma)tic-wishfull thinking"-setting), then that's all you need. If not, then you are not comfortable with your math (or your model but this is another issue).
• If you don't have such an analytical solution at hand, then usually MC comes in naturally (as every one suggests here) but you could use PDE methods aswell (after all it was the original methods for derivation of the BS Call/Put options prices). And you have plenty of books and nice articles that will tell you how to proceed in both cases (especially in BS settings). An easy check that I recommand, is to compare the "closed-form formulas" vs "MC (and/or PDE)" in the vanilla cases. Moreover those methods provide a good introdution to the replication prices that you might be willing to check in the second case.
Note that by using finite difference (or element) methods for the PDE you get an error and that when using discretization sheme for SDE you get at the of the day "random variable" for your P&L. There it is really a matter of taste in my opinion both methods have pros and cons.
For the second case, that I called replication price then it is usually provided in a (at least in principle) straigthforward manner of the methods you used for the PDE and/or SDE discretization.
Still regarding the replication prices, the very recent article of Wilmott and Ahmad "Which Free Lunch Would You Like Today, Sir?: Delta Hedging, Volatility Arbitrage and Optimal Portfolios" is really illuminating in many ways and stays in the BS setting you want to stay within I think you should read it.
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http://programmingpraxis.com/2011/06/24/thank-god-its-friday/?like=1&source=post_flair&_wpnonce=0df65f7478
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# Programming Praxis
A collection of etudes, updated weekly, for the education and enjoyment of the savvy programmer
## Thank God It’s Friday!
### June 24, 2011
We have previously given two algorithms to calculate the day of the week, one in the Standard Prelude and one in the exercise on Zeller’s Congruence. In today’s exercise we give three more algorithms to calculate the day of the week.
Our first method is due to Carl Gauss, and is based on moving January and February to the end of the preceding year, then fitting a straight line through the number of days in each month. Gauss gives the formula $w = \left( d + \lfloor 2.6 m - 0.2 \rfloor + y + \lfloor \frac{y}{4} \rfloor + \lfloor \frac{c}{4} \rfloor - 2c \right) \pmod{7}$ where Y is the input year, except that it is reduced by 1 in January and February, d is the day of the month, m is the number of the month, with 1 for March through 12 for February, y is the last two digits of Y, c is the first two digits of Y, and w is the day of the week, with 0 for Sunday through 6 for Saturday. For instance, June 24, 2011 is calculated as 24 + floor(2.6×4−0.2) + 11 + floor(11÷4) + floor(20÷4) – 2×20, modulo 7, which is 5 for Friday.
Our second method is due to Tomohiko Sakamoto, who gives a table of offsets for the days of each month from the day at the start of the year: t = {0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4}. Then Sakamoto subtracts 1 from the input year in January and February and calculates the day of the week with the formula $\left( y + \lfloor \frac{y}{4} \rfloor - \lfloor \frac{y}{100} \rfloor + \lfloor \frac{y}{400} \rfloor + t[m-1] + d \right) \pmod{7}$. For instance, June 24, 2011 is calculated as 2011 + floor(2011÷4) − floor(2011÷100) + floor(2011÷400) + t[6−1] + 24 = 2011 + 502 – 20 + 5 + 3 + 24 = 2525, modulo 7, which is 5 for Friday.
Our third method is due to John Horton Conway, and is intended for mental calculation. Conway’s method is based on calculating the anchor day for the requested century, the doomsday for the requested year, and interpolating from various repetitions of the doomsday through the year.
The anchor day is calculated as $\left( 5c + \lfloor \frac{c-1}{4} \rfloor \right) \pmod{7} + Thursday$, where c is the century; note that century years, such as 2000, are part of the succeeding century, so c=21 for the year 2000. For example, the anchor day for the 21st century is Tuesday, calculated as 5×21 + floor(20÷4), modulo 7, plus Thursday. Anchor days repeat every four centuries, in the cycle Friday, Wednesday, Tuesday, Sunday starting from the year 1800.
The doomsday is calculated by dividing the last two digits of the year by 12 to calculate the quotient and remainder. Then the doomsday is the quotient, plus the remainder, plus the quotient of the remainder divided by 4, plus the anchor day for the century. For example, the doomsday for 2011 is Monday, calculated as 0 + 11 + floor(11÷4) = 13, which is 6 modulo 7, plus the anchor day Tuesday.
Once the doomsday is known, the day of the week is calculated by locating the nearest doomsday in each month, which can be memorized in the following manner: For April, June, August, October, and December, the doomsday is the month number: 4/4, 6/6, 8/8, 10/10, and 12/12. For May, July, September, and November, the doomsday can be calculated by the ditty “I worked 9 to 5 at 7-11″ which gives 5/9, 7/11, 9/5, and 11/7. The last day of February is a doomsday, whether a common year or a leap year, and this gives an easy way to calculate the day of the week for March, where doomsday is 3/0 (the “zeroth” day of March). All that’s left is January, for which the doomsday is 1/10 in common years and 1/11 in leap years.
Thus, the day of the week for June 24, 2011 is calculated as 24−6=18 ≡ 4 (mod 7) days past the doomsday, which gives an answer of Friday. Conway claims to be able to calculate any day of the week in two seconds, though I confess I have not been able to make the required calculations reliably except by using pencil and paper to assist.
Your task is to write programs to calculate the day of the week using the three functions described above; for Conway’s algorithm, you should calculated the doomsday for the year rather than the day of the week for the date. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
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### 5 Responses to “Thank God It’s Friday!”
1. June 24, 2011 at 9:45 AM
[...] today’s Programming Praxis exercise, our goal is to implement three functions related to dates: two ways to [...]
2. Remco Niemeijer said
June 24, 2011 at 9:46 AM
My Haskell solution (see http://bonsaicode.wordpress.com/2011/06/24/programming-praxis-thank-god-it%E2%80%99s-friday/ for a version with comments):
```data Weekday = Sun | Mon | Tue | Wed | Thu | Fri | Sat deriving (Enum, Eq, Show)
gauss :: Int -> Int -> Int -> Weekday
gauss y m d = toEnum $ mod (d + floor (2.6 * fromIntegral
(mod (m - 2) 12) - 0.2) + y' + div y' 4 + div c 4 - 2*c) 7 where
(c,y') = divMod (if m < 3 then y - 1 else y) 100
sakamoto :: Int -> Int -> Int -> Weekday
sakamoto y m d = toEnum $ mod (y + div y 4 - div y 100 +
div y 400 + [0,3,2,5,0,3,5,1,4,6,2,4] !! (m - 1) + d) 7
conway :: Int -> Weekday
conway y = toEnum $ mod (q + r + div r 4 + 5*(c+1) + div c 4 + 4) 7
where (c, (q,r)) = (div y 100, divMod (mod y 100) 12)
```
3. Graham said
June 25, 2011 at 11:57 PM
My Python submission (basically just a translation of the Scheme solution):
```#!/usr/bin/env python
from fractions import Fraction
DAYS = {0: "Sunday", 1: "Monday", 2: "Tuesday", 3: "Wednesday", 4: "Thursday",
5: "Friday", 6: "Saturday"}
def gauss(year, month, day):
yr = (year - 1) if (month < 3) else year
m = (month + 10) if (month < 3) else (month - 2)
c, y = divmod(yr, 100)
return DAYS[sum((day, int((Fraction('13/5') * m) - Fraction('1/5')), y,
(y / 4), (c / 4), - 2 * c)) % 7]
def sakamoto(year, month, day):
t = [0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4]
y = (year - 1) if (month < 3) else year
return DAYS[sum((day, y, (y / 4), - (y / 100), (y / 400), t[month - 1]))
% 7]
def doomsday(year):
return DAYS[sum((2, year, year / 4, - (year / 100), year / 400)) % 7]
if __name__ == "__main__":
print gauss(2011, 6, 24)
print sakamoto(2011, 6, 24)
print doomsday(2011)
```
4. Mike said
June 28, 2011 at 4:56 AM
Python version
Tried all the formulas from Wikipedia for conway’s method. And went all the way to calculate the day of the week.
```dow = "Sun Mon Tue Wed Thu Fri Sat".split()
def leap(year):
return bool(not year % 400 or not year % 4 and year % 100)
def gauss(year, month, day):
c, y = divmod(year if month > 2 else (year - 1), 100)
m = (month - 3)%12 + 1
return dow[(day + int(2.6*m - 0.2) + y + y//4 + c//4 - 2*c) % 7]
def sakamoto(year, month, day):
t = [0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4]
y = year if month > 2 else (year - 1)
return dow[(day + y + y//4 - y//100 + y//400 + t[month - 1]) % 7]
def conway(year, month, day):
c, y = divmod(year, 100)
# original formula
#c += 1
#anchor = (5*c + (c-1)//4 + 4)%7
# modified formula-don't add 1 to first 2 digits of year.
anchor = (5*c + c//4 + 2)%7
# original conway formula
#doomsday = (anchor + y//12 + y%12 + y%12//4)%7
# alternative from wikipedia. I think this is easiest to do in head.
doomsday = (anchor + y + y//4)%7
# odd+11 rule from wikipedia
#y = (y+11) if y&1 else y
#y = y//2
#y = (y+11) if y&1 else y
#y = 7 - y%7
#doomsday = (anchor + y)%7
#computer formula from wikipedia
#doomsday = (2 + y + y//4 - y//100 + y//400)%7
nearest = (10, 28, 0, 4, 9, 6, 11, 8, 5, 10, 7, 12)[month - 1]
if month < 3 and leap(year):
nearest += 1
return dow[(doomsday + day - nearest) % 7]
```
5. razvan said
July 5, 2011 at 3:56 PM
Here’s my solution in Java:
```package dayofweek;
public class DayOfWeek {
private static final int[] SAKAMOTO_TABLE = { 0, 3, 2, 5, 0, 3, 5, 1, 4, 6,
2, 4 };
private static void checkDate(int d, int m, int y) {
if (m < 1 || m > 12)
throw new NumberFormatException("Bad month");
if (d < 0 || d > 31)
throw new NumberFormatException("Bad day");
if (m == 2
&& (d > 29 || d > 28
&& ((y % 4 == 0 && y % 100 != 0) || y % 400 == 0)))
throw new NumberFormatException("Bad day");
if (y < 0)
throw new NumberFormatException("Bad year");
}
public static int gauss(int d, int m, int y) {
checkDate(d, m, y);
if (m >= 3 && m <= 12)
m -= 2;
else if (m < 3 && m > 0) {
m += 10;
y--;
}
int c = y / 100;
y %= 100;
return (d + (int) Math.floor(2.6 * (double) m - 0.2) + y + y / 4 + c
/ 4 - 2 * c) % 7;
}
public static int sakamoto(int d, int m, int y) {
checkDate(d, m, y);
if (m < 3 && m > 0)
y--;
return (y + y / 4 - y / 100 + y / 400 + SAKAMOTO_TABLE[m - 1] + d) % 7;
}
public static int conway(int d, int m, int y) {
checkDate(d, m, y);
int c = y / 100 + 1;
int anchor = ((5 * c + (c - 1) / 4) % 7 + 4) % 7;
int yy = y % 100;
int quo = yy / 12, rem = yy % 12;
int doomsday = ((quo + rem + rem / 4) % 7 + anchor) % 7;
int doom = 0;
if (m == 4 || m == 6 || m == 8 || m == 10 || m == 12)
doom = m;
else if (m == 2) {
if ((y % 4 == 0 && y % 100 != 0) || y % 400 == 0)
doom = 29;
else
doom = 28;
} else if (m == 3)
doom = 0;
else if (m == 3) {
if ((y % 4 == 0 && y % 100 != 0) || y % 400 == 0)
doom = 11;
else
doom = 10;
} else
switch (m) {
case 5:
doom = 9;
break;
case 7:
doom = 11;
break;
case 9:
doom = 5;
break;
case 11:
doom = 7;
break;
}
if (d > doom)
return ((d - doom) % 7 + doomsday) % 7;
else
return ((doomsday - (doom - d) % 7) + 7) % 7;
}
public static String intToDate(int d) {
switch (d) {
case 0:
return "Sunday";
case 1:
return "Monday";
case 2:
return "Tuesday";
case 3:
return "Wednesday";
case 4:
return "Thursday";
case 5:
return "Friday";
case 6:
return "Saturday";
}
throw new NumberFormatException("Bad day of week");
}
public static void main(String[] args) {
System.out.println(intToDate(gauss(24, 6, 2011)));
System.out.println(intToDate(sakamoto(24, 6, 2011)));
System.out.println(intToDate(conway(11, 5, 2010)));
}
}
```
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http://mathoverflow.net/questions/108055/function-with-all-but-mixed-second-partial-derivatives-twice-differentiable
|
## Function with all but mixed second partial derivatives twice differentiable?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $f(x,y)$ be a a real valued function on an open subset of $\mathbf{R}^2$ with continuous partial derivatives $\frac{\partial^2 f}{\partial x^2}$ and $\frac{\partial^2}{\partial y^2}$. Is $f$ twice differentiable?
-
## 2 Answers
Here is an explicit counterexample in $B_1$, `$$f(x,y) = x\, y \, \log(-\log(x^2+y^2))$$`
Both $\partial_{xx} f$ and $\partial_{yy} f$ are continuous but $f$ is not twice differentiable at the origin.
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Upon a quick check it looks just fine. Nice idea to take the double logarithm of the troublesome term in the usual counterexample for the insufficiency of single partial differentiability to ensure the total one. My curiosity came from the observation that over the p-adic numbers this criterion oddly seems sufficient. Thank you! – Tiffy Nov 12 at 19:32
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For very general open sets, this is not true.
However, assume true if the boundary is sufficiently smooth and the boundary conditions are nice enough (say, Dirichlet). In that case, your question boils down to asking whether a continuous function whose Laplacian is continuous, too, is also in C^2. This is surely true if instead of continuity you assume Hölder continuity: i.e., if your aim is to pass from $\Delta u\in C^{0,\alpha}$ to $u\in C^{2,\alpha}$: this is exactly the assertion of Schauder's estimates; or else if you content yourself with weak differentiability instead of classical one: in that case, see e.g. §6.3.2 in Evans' Partial Differential Equations.
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The are well known counterexamples of functions for which $\Delta f$ is continuous, but f is not $C^2$. The question asks what happens under the stronger condition that both $f_{xx}$ and $f_{yy}$ are continuous. – Michael Renardy Sep 25 at 20:50
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http://mathoverflow.net/questions/5443?sort=votes
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## Wick rotation in mathematics
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In physics, esp. quantum field theory, Wick rotation (i.e. putting $t \mapsto i\tau$, imaginary time) is often used to simplify calculations, make things convergent or make connections between different models (e.g. quantum and statistical mechanics). Does the Wick rotation trick occur anywhere in mathematics not related to QFT (analysis, PDE etc.)?
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## 4 Answers
A general form of the Wick's rotation is the "Weyl's unitary trick". This construction allows to relate group actions of noncompact forms of a complex Lie group to those of the compact one by changing the signature of the Cartan-Killing form . Although, the representations of the compact and noncompact forms are different, the unitary trick introduces relations among their invariants and between the transition functions, hence the use in quantum field theory. Also, it introduces relations between their homogeneous spaces (see the example above of the sphere and the hyperboloid).
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Yes. It's called analytic continuation.
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Analytic continuation is much more general than Wick rotation. I was asking about the specific instances of $t \mapsto it$ substitution. – Marcin Kotowski Nov 14 2009 at 8:25
One instance of (something similar to) that is the construction of models of hyperbolic geometry using a sphere of radius $i$.
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I don't understand what you mean by this. – Qiaochu Yuan Nov 14 2009 at 4:25
I would bet that the reference is to the following fact: consider Minkowski space R^{n,1} with the Lorentzian metric x_1^2 + ... x_n^2 - t^2. The "sphere of radius -1" is the hyperboloid |x|^2 = t^2 + 1. The restriction of the Lorentzian metric to this hyperboloid is positive definite, and in fact has constant sectional curvature -1. Thus this hyperboloid gives a model of hyperbolic n-space H^n. [This exhibits the identification Isom^+(H^n) = SO^+(n,1).] But I cannot exactly see how to interpret the answer above in these terms. – Tom Church Nov 14 2009 at 7:22
Typo: The "sphere of radius -1" is the hyperboloid |x|^2 = t^2 - 1. – Tom Church Nov 14 2009 at 7:23
Tom's correct, up to the following: the radius is $i$, not $-1$: the $-1$ is the square of the radius. This model of hyperbolic geometry is very very old, and nicely explains why hyperbolic trigonometry and spherical trogonometry are so similar. It is discussed, for example, in Kolmogorov's and Yushkievich´s book on 19th Centruty Geometry and Function theory. The idea is due to Lambert aroung the 1770s. – Mariano Suárez-Alvarez Nov 14 2009 at 17:55
Well, since you are looking at it like a "trick", I think that it may be suitable to mention that it can be used to relate the trigonometric functions with the hyperbolic ones (and also to solve somewhat related integrals).
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http://physics.stackexchange.com/questions/34217/why-do-people-categorically-dismiss-some-simple-quantum-models/34402
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# Why do people categorically dismiss some simple quantum models?
Deterministic models. Clarification of the question:
The problem with these blogs is that people are inclined to start yelling at each other (I admit, I got infected and it's difficult not to raise one's electronic voice). I want to ask my question without an entourage of polemics.
My recent papers were greeted with scepticism. I've no problem with that. What disturbes me is the general reaction that they are "wrong". my question is summarised as follows:
Did any of these people actually read the work and can anyone tell me where a mistake was made?
Now the details. I can't help being disgusted by the "many world" interpretation, or the Bohm - deBroglie "pilot waves", and even the idea that the quantum world must be non-local is difficult to buy. I want to know what is really going on, and in order to try to get some ideas, I construct some models with various degrees of sophistication. These models are of course "wrong" in the sense that they do not describe the real world, they do not generate the Standard Model, but one can imagine starting from such simple models and adding more and more complicated details to make them look more realistic, in various stages.
Of course I know what the difficulties are when one tries to underpin qm with determinism. Simple probabilistic theories fail in an essential way. One or several of the usual assumptions made in such a deterministic theory will probably have to be abandoned; I am fully aware of that. On the other hand, our world seems to be extremely logical and natural.
Therefore, I decided to start my investigation at the other end. Make assumptions that later surely will have to be amended; make some simple models, compare these with what we know about the real world, and then modify the assumptions any way we like.
The no-go theorems tell us that a simple cellular automaton model is not likely to work. One way I tried to "amend" them, was to introduce information loss. At first sight this would carry me even further away from qm, but if you look a little more closely, you find that one still can introduce a Hilbert space, but it becomes much smaller and it may become holographic, which is something we may actually want. If you then realize that information loss makes any mapping from the deterministic model to qm states fundamentally non-local - while the physics itself stays local - then maybe the idea beomes more attractive.
Now the problem with this is that again one makes too big assumptions, and the math is quite complicated and unattractive. So I went back to a reversible, local, deterministic automaton and asked: to what extent does this resemble qm, and where does it go wrong? With the idea in mind that we will alter the assumptions, maybe add information loss, put in an expanding universe, but all that comes later; first I want to know what goes wrong.
And here is the surprise: in a sense, nothing goes wrong. All you have to assume is that we use quantum states, even if the evolution laws themselves are deterministic. So the probability distributions are given by quantum amplitudes. The point is that, when describing the mapping between the deterministic system and the quantum system, there is a lot of freedom. If you look at any one periodic mode of the deterministic system, you can define a common contribution to the energy for all states in this mode, and this introduces a large number of arbitrary constants, so we are given much freedom.
Using this freedom I end up with quite a few models that I happen to find interesting. Starting with deterministic systems I end up with quantum systems. I mean real quantum systems, not any of those ugly concoctions. On the other hand, they are still a long way off from the Standard Model, or even anything else that shows decent, interacting particles.
Except string theory. Is the model I constructed a counter example, showing that what everyone tells me about fundamental qm being incompatible with determinism, is wrong? No, I don't believe that; the idea was that, somewhere, I will have to modify my assumptions, but maybe the usual assumptions made in the no-go theorems, will have to be looked at as well.
I personally think people are too quick in rejecting "superdeterminism". I do reject "conspiracy", but that might not be the same thing. Superdeterminism simply states that you can't "change your mind" (about which component of a spin to measure), by "free will", without also having a modification of the deterministic modes of your world in the distant past. It's obviously true in a deterministic world, and maybe this is an essential fact that has to be taken into account. It does not imply "conspiracy".
Does someone have a good, or better, idea about this approach, wthout name calling? Why are some of you so strongly opinionated that it is "wrong"? Am I stepping on someone's religeous feelings? I hope not.
References:
Relating the quantum mechanics of discrete systems to standard canonical quantum mechanics, arXiv:1204.4926 ;
Duality between a deterministic cellular automaton and a bosonic quantum field theory in 1+1 dimensions, arXiv.org/abs/1205.4107 ;
Discreteness and Determinism in Superstrings, arXiv:1207.3612[hep-th].
Further reactions on the answers given
(Writing this as "comment" failed, then writing this as "answer" generated objections. I'll try to erase the "answer" that I should not have put there...):
First: thank you for the elaborate answers.
I realise that my question raises philosophical issues; these are interesting and important, but not my main concern. I want to know why I find no technical problem while constructing my model. I am flattered by the impression that my theories were so "easy" to construct. Indeed, I made my presentation as transparent as possible, but it wasn't easy. There are many dead alleys, and not all models work equally well. For instance, the harmonic oscillator can be mapped onto a simple periodic automaton, but then one does hit upon technicalities: The hamiltonian of a periodic system seems to be unbounded above and below, while the harmonic oscillator has a ground state. The time-reversible cellular automaton (CA) that consists of two steps A and B, where both A and B can be written as the exponent of physically reasonable hamiltonians, itself is much more difficult to express as a hamiltonian theory, because the BCH series does not converge. Also, explicit 3+1 dimensional QFT models resisted my attempts to rewrite them as cellular automata. This is why I was surprised that the superstring works so nicely, it seems, but even here, to achieve this, quite a few tricks had to be invented.
@Ron Maimon. I here repeat what I said in a comment, just because there the 600 character limit distorted my text too much. You gave a good exposition of the problem in earlier contributions: in a CA the "ontic" wave function of the universe can only be in specific modes of the CA. This means that the universe can only be in states $\psi_1,\ \psi_2,\ ...$ that have the property $\langle\psi_i\,|\,\psi_j\rangle=\delta_{ij}$, whereas the quantum world that we would like to describe, allows for many more states that are not at all orthonormal to each other. How could these states ever arise? I summarise, with apologies for the repetition:
• we usually think that Hilbert space is separable, that is, inside every infinitesimal volume element of this world there is a Hilbert space, and the entire Hibert space is the product of all these.
• normally, we assume that any of the states in this joint Hilbert space may represent an "ontic" state of the Universe.
• I think this might not be true. The ontic states of the universe may form a much smaller class of states $\psi_i$ ; in terms of CA states, they must form an orthonormal set. In terms of "Standard Model" (SM) states, this orthonormal set is NOT separable, and this is why, locally, we think we have not only the basis elements but also all superpositions. The orthonormal set is then easy to map back onto the CA states.
I don't think we have to talk about a non-denumerable number of states, but the number of CA states is extremely large. In short: the mathematical system allows us to choose: take all CA states, then the orthonormal set is large enough to describe all possible universes, or choose the much smaller set of SM states, then you also need many superimposed states to describe the universe. The transition from one description to the other is natural and smooth in the mathematical sense.
I suspect that, this way, one can see how a description that is not quantum mechanical at the CA level (admitting only "classical" probabilities), can "gradually" force us into accepting quantum amplitudes when turning to larger distance scales, and limiting ourselves to much lower energy levels only. You see, in words, all of this might sound crooky and vague, but in my models I think I am forced to think this way, simply by looking at the expressions: In terms of the SM states, I could easily decide to accept all quantum amplitudes, but when turning to the CA basis, I discover that superpositions are superfluous; they can be replaced by classical probabilities without changing any of the physics, because in the CA, the phase factors in the superpositions will never become observable.
@Ron I understand that what you are trying to do is something else. It is not clear to me whether you want to interpret $\delta\rho$ as a wave function (I am not worried about the absence of i, as long as the minus sign is allowed). My theory is much more direct; I use the original "quantum" description with only conventional wave functions and conventional probabilities.
( new since Sunday Aug. 20: )
There is a problem with my argument (I correct some statements I had put here earlier). I have to work with two kinds of states: 1: the template states, used whever you do quantum mechanics, these allow for any kinds of superposition; and 2: the ontic states, the set of states that form the basis of the CA. The ontic states $|n\rangle$ are all orthonormal: $\langle n|m\rangle=\delta_{nm}$, so no superpositions are allowed for them (unless you want to construct a template state of course). One can then ask the question: how can it be that we (think we) see superimposed states in experiments? Aren't experiments only seeing ontic states?
My answer has always been: who cares about that problem? Just use the rules of QM. Use the templates to do any calculation you like, compute your state $|\psi\rangle$, and then note that the CA probabilities, $\rho_n=|\langle n|\psi\rangle|^2$, evolve exactly as probabilities are supposed to do.
That works, but it leaves the question unanswered, and for some reason, my friends on this discussion page get upset by that.
So I started thinking about it. I concluded that the template states can be used to describe the ontic states, but this means that, somewhere along the line, they have to be reduced to an orthonormal set. How does this happen? In particular, how can it be that experiments strongly suggest that superpositions play extremely important roles, while according to my theory, somehow, these are plutoed by saying that they aren't ontic?
Looking at the math expressions, I now tend to think that orthonormality is restored by "superdeterminism, compined with vacuum fluctuations. The thing we call vacuum state, $|\emptyset\rangle$, is not an ontological state, but a superposition of many, perhaps all, CA states. The phases can be chosen to be anything, but it makes sense to choose them to be +1 for the vacuum. This is actually a nice way to define phases: all other phases you might introduce for non-vacuum states now have a definite meaning.
The states we normally consider in an experiment are usually orthogonal to the vacuum. If we say that we can do experiments with two states, A and B, that are not orthonormal to each other, this means that these are template states; it is easy to construct such states and to calculate how they evolve. However, it is safe to assume that, actually, the ontological states $|n\rangle$ with non-vanishing inner product with A, must be different from the states $|m\rangle$ that occur in B, so that, in spite of the template, $\langle A|B\rangle=0$. This is because the universe never repeats itself exactly. My physical interpretation of this is "superdeterminism": If, in an EPR or Bell experiment, Alice (or Bob) changes her (his) mind about what to measure, she (he) works with states $m$ which all differ from all states $n$ used previously. In the template states, all one has to do is assume at least one change in one of the physical states somewhere else in the universe. The contradiction then disappears.
The role of vacuum fluctuations is also unavoidable when considering the decay of an unstable particle.
I think there's no problem with the above arguments, but some people find it difficult to accept that the working of their minds may have any effect at all on vacuum fluctuations, or the converse, that vacuum fluctuations might affect their minds. The "free will" of an observer is at risk; people won't like that.
But most disturbingly, this argument would imply that what my friends have been teaching at Harvard and other places, for many decades as we are told, is actually incorrect. I want to stay modest; I find this disturbing.
A revised version of my latest paper was now sent to the ArXiv (will probably be available from Monday or Tuesday). Thanks to you all. My conclusion did not change, but I now have more precise arguments concerning Bell's inequalities and what vacuum fluctuations can do to them.
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Dear @QuestionAnswers, you're misinterpreting what I say: I don't think that these questions are illegal to ask. They're legal to be asked, they were asked about 90 years ago and they were answered 85 years ago. It's silly, and not illegal, to ask them again in 2012 because physics has known the answer for quite some time. It's a pretty long time. 85 years after physicists determined that heliocentrism was more right than geocentrism, it was generally viewed as silly to question heliocentrism again. Learning in the modern age should be faster but it's apparently not. – Luboš Motl Aug 16 '12 at 6:10
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@LubošMotl: Yes, of course you can't simulate quantum computation. The thing is, these types of quantum computing states are incredibly entangled, and very hard to realize without decoherence spoiling them, so much so that we haven't realized any such states experimentally. The question is whether you can simulate workaday QM, lots of decoherence, no quantum computer around, by a linearly scaling computer. You could say "do collapse", but that's harder than it looks computationally. In a discrete QM analog, you get automatic collapse, and you can always do monte-carlo. – Ron Maimon Aug 16 '12 at 8:58
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@LubošMotl: You are repeating ridiculous propaganda as if it were fact, this is just deplorable fear-mongering. Zeilinger's "classes of models" is a very myopic class that doesn't include mine or any other reasonable nonlocal model. Locality and Lorentz invariance are unrelated, the naive arguments nonwithstanding. Here is a Lorentz invariant nonlocal action: $\int {\phi(x)\phi(y) \over ((x-y)^2 + 1)^{.73}} d^4x d^4y$, there are lots of others. Locality is absent in string theory, there are no local fields, and it is completely absent in AdS/CFT bulk, where the whole spacetime is emergent. – Ron Maimon Aug 16 '12 at 17:09
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... regarding "realism", what is disproved is local realism, with Bell's inequality, and "small realism" that reproduces exact QM, using Shor's algorithm. All other constraints are weaker. Bohm's theory shows you can't disprove (exponentially large) realism, because it works and it is realistic, so there is no general no-go. But a modern "realist" is not looking for Bohm's theory, but a theory that fails to reproduce QM in the highly entangled domain, and therefore gives actual different predictions. Such a theory is what t'Hooft is after, and it makes sense to look for this. – Ron Maimon Aug 16 '12 at 17:15
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... even if it is physically wrong, if mathematically right, it is a prescription for reducing highly entangled states. Of course entanglement is the norm! But we usually call it "collapse", not entanglement, and it usually only goes one-way--- reducing a quantum system in complexity. The delicate cases of quantum computation require fine-tuning to make the entanglement go back and forth, not getting collapse, but nontrivial computation. Most cases, you can approximate entanglement as collapse. A "realist approximation" to quantum mechanics is useful for automatic collapse simulation. – Ron Maimon Aug 16 '12 at 17:19
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## 7 Answers
I can tell you why I don't believe in it. I think my reasons are different from most physicists' reasons, however.
Regular quantum mechanics implies the existence of quantum computation. If you believe in the difficulty of factoring (and a number of other classical problems), then a deterministic underpinning for quantum mechanics would seem to imply one of the following.
• There is a classical polynomial-time algorithm for factoring and other problems which can be solved on a quantum computer.
• The deterministic underpinnings of quantum mechanics require $2^n$ resources for a system of size $O(n)$.
• Quantum computation doesn't actually work in practice.
None of these seem at all likely to me. For the first, it is quite conceivable that there is a polynomial-time algorithm for factoring, but quantum computation can solve lots of similar periodicity problems, and you can argue that there can't be a single algorithm that solves all of them on a classical computer, so you would have to have different classical algorithms for each classical problem that a quantum computer can solve by period finding.
For the second, deterministic underpinnings of quantum mechanics that require $2^n$ resources for a system of size $O(n)$ are really unsatisfactory (but maybe quite possible ... after all, the theory that the universe is a simulation on a classical computer falls in this class of theories, and while truly unsatisfactory, can't be ruled out by this argument).
For the third, I haven't seen any reasonable way to how you could make quantum computation impossible while still maintaining consistency with current experimental results.
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@Jim: A deterministic theory acting in a $2^n$-dimensional Hilbert space would indeed meet my criteria for a reasonable thing to think about. The problem is that it seems hard to reconcile a deterministic theory that increases with the size of the system with the criterion of locality. But this seems to be exactly what 't Hooft hopes to happen ... the equations for the deterministic theory itself aren't local, but they somehow translate into local physics. – Peter Shor Aug 17 '12 at 17:07
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A question: If you view things holographically, do you need to look at the entire universe to get a complete description of local dynamics? If it does, it's not very useful for doing calculations. And if it doesn't, it's hard for me to understand where the power of quantum computation is coming from. – Peter Shor Aug 18 '12 at 16:26
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Finally a meaningful and valid answer, Prof @PeterShor. Ron: there don't exist any "approximately quantum" theories. Theories are either classical, quantum, or logically inconsistent. However, I agree with your comment that holography doesn't cause any problem for quantum computing because for the relevant states - which are far from forming black holes etc. - the laws for the qubits in the quantum computer are as local in the right variables as they've always been. Prof Shor, if you don't try to make the density of degrees of freedom too high (grav. collapse!), things work just like before. – Luboš Motl Aug 20 '12 at 6:15
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@Lubos: unitarity is a very good condition, but saying that it is required for consistency requires a very strange definition of consistency. It does seem to be required to make thermodynamics work, but if you look at topological quantum error correction a la Kitaev, there are good reasons for believing that there could be laws of physics in which unitarity breaks down at the Planck scale but is preserved at macroscopic scales. – Peter Shor Aug 21 '12 at 14:34
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@Jon: I haven't read the papers, but I've looked at their abstracts. Both say that some quantum systems start behaving classically as their size grows large. But it doesn't follow logically that all quantum systems behave classically when their size is very large, which (unless I misunderstand) is what you're claiming. Both Aram Harrow (your commenter) and I are fairly famous researchers in the field of quantum computing, as you would have discovered if you Googled, and one of us is definitely not impersonating the other. – Peter Shor Aug 26 '12 at 19:06
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[ text I had put here is moved to the original question, but I prefer not to erase the comments that were posted here. ]
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Dear Prof. 't Hooft, the most obvious reason for downvoting this answer is that it doesn't answer the original question. The second reason is that the question and the answer were posted by the same user which is... strange. The third reason is that the content is entirely invalid. Superpositions are never "superfluous" in any quantum mechanical theory. They must always be allowed, the coefficients must always be allowed to be complex, and the phases always matter. – Luboš Motl Aug 17 '12 at 14:56
You would see that if you looked at least at one particular system, e.g. the ammonia molecule motls.blogspot.cz/2012/08/two-state-systems-masers.html?m=1 , and figured out what can be measured and in what state the molecule is actually finding itself at various times. The complex probability amplitudes always matter - after all, the rate of change of the phase as a function of one observable tells us about the value of the complementary observable, and so on. For an actual answer to your question, see e.g. motls.blogspot.cz/2012/08/… – Luboš Motl Aug 17 '12 at 14:58
@ Lubos Motl, you still didn't get it: the ammonia molecule is in the SM, and there superpositions are meaningful. But if you use the CA as a basis, superposition of two basis elements acts as a classical composition with classical probabilities. "Superdeterminism" here boils down to saying that 2 ammonia states that are not orthogonal only become "ontic" if you make them orthogonal by including other quantum states elsewhere, and making those orthogonal. It looks inelegant but I do think that this actually happens. – G. 't Hooft Aug 17 '12 at 20:39
@G.'tHooft would you mind also transferring the relevant content of the comments to your question? It's much better for the site if this sort of information appears in questions or answers rather than (just) in comments. – David Zaslavsky♦ Aug 22 '12 at 20:50
This could have been a comment, but as it actually anwers the question asked in the title, I'll post it as such:
As far as I can tell there's no rational reason to dismiss these models out of hand - it's just that quantum mechanics (QM) has set the bar awfully high: So far, there's no experimental evidence that QM is wrong, and no one has come up with a viable alternative.
Ultimately, your theory needs to reproduce all experimentally verified predictions of QM (or rather may only deviate within the experimental precision). However, there's of course no need to reproduce arbitrary predictions - in fact, if you did, you'd end up with a re-formulation - ie a new interpretation - of ordinary QM. If your model tells us large-scale quantum computation is impossible, then it's up to the experimentalists to prove you wrong.
Any objections beyond that are just psychology at work: It takes quite some effort for most people to convince themselves that QM is a valid description of the world we live in, and once such a believe is ingrained, it easily becomes dogma.
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There are two questions here: Why criticize your models? And are there better ideas? I will try to answer the second question in a separate answer. Here I only give some comments of a general nature to adress the first question.
I personally agree with you, and I think most people who care about this stuff do too, that it is disconcerting to have a theory in which the information produced by observations is not contained in the theory itself, but is produced out of thin-air by an act of measurement. The natural idea is that when we see a bit of information produced through an act of observation, then the value of this bit was somehow contained in the complete description of nature independent of the act of observation. This was Einstein's reality principle, and I agree that it is preferred for a theory to obey it.
When a theory doesn't obey the reality principle, one has to note that macroscopic reality does obey it, and find the bits in the macroscopic world by a philosophically contorted roundabout exercize in mysticism. But since physics is empirical, and positivism is fruitful, I take the point of view that any framework that explains the results of observations must ultimately be philosophically ok, even if it requires contortions, and even if it is not correct! So Newton's mechanics, even though it is wrong, is not necessarily empirically disproved given only observations of human beings and so on, so it must not be philosophically incompatible with free will. Similarly, quantum mechanics might be wrong, but we have no empirical data that shows it is wrong, so it should be philosopically consistent to say QM is all there is. This means QM should describe observers too, and if there is no mathematical contradiction with this view, there should not be a philosophical contradiction either, even if there is a contradiction with experiment. This is the many-worlds philosophy, and it's the self-consistent answer if quantum mechanics is correct. It might be annoying, but I don't think it is too annoying--- one should just learn to live with many-worlds as a fine philosophical position.
But it is wrong to just say "many-worlds" at this point, because the quantum description has not been tested in the realm where the many-worlds have a real logical-positivist manifestation--- most obviously when doing factoring of enormously large numbers using a quantum computer. Until we do this, it is definitely conceivable that nature is only very closely approximately quantum for small systems of a few particles, in the cases we tested the theory already, and is just not quantum for highly entangled many particle systems.
Even if the world turns out to be really quantum, and a quantum computer factors numbers all the time, finding a deterministic substructure is useful for giving a computationally tractable small truncation of quantum mechanics in cases which are not a quantum computer, and it is possible that this truncation can be useful for quantum simulations. This is so needed that I think finding a substructure to quantum mechanics is a central important problem, personally, regardless of whether it turns out to be right. For this reason, I devoted much time to understanding your approach.
The problem with your construction is that it works too well, it is too easy to transform a quantum system into a beable basis, so that the global wavefunction is evolving in a way that is deterministic using the global Hamiltonian. Since you introduce the Hilbert space early and use it to do the transformation of basis to the internal states of the automaton, there is no obvious barrier to transforming a quantum computer into a beable basis, nor is there any barrier to violating Bell's inequality locally. These do not suggest that the no-go theorems are flawed, rather they suggest that the transformation to a beable basis with a permutation Hamiltonian does not produce a true classical system.
The precise way in which I believe that this system fails to be classical is in state-preparation on the interior. The process of state-preparation involves a measurement, entangling some interior subsystem with a macroscopic subsystem, and then a reduction of the macroscopic system according to Born's rule, leaving a pure quantum state of the interior subsystem. In your paper on Born's rule, you suggested how reduction should happen in a CA system, but your precise models don't really respect this intuition, in that the measurement of intermediate states always produces one of the eigenstates of the observable in the interior, no matter how complicated the observable and how entangled its eigenstates. This is what allows you to reproduce quantum mechanics on the interior subsystems, I am somewhat certain that this does not keep the state unsuperposed in the beable basis. Because these internal reductions don't respect the probability structure, you are really doing quantum mechanics not CA, and this is the only reason that you have such an easy time sidestepping the no-gos.
The fact that you sidestep the no-gos with no difficulty suggests strongly that your construction is leaving the space of allowed classical probability distributions on the CA somehow. The only place where this can happen is during internal state-preparation, during measurements of internal operators. This is how you prepare Bell states or quantum computers, after all. These interior operations must be producing states (after projection) which cannot be interpreted as classical probability states of the automaton, although the Hamiltonian evolution never does this. This is not a proof, but I would bet lots of money (if I had any). I asked for a proof here: In t'Hooft beable models, do measurements keep states classical?
This is part I of the answer, I post it separately, so that people who agree with this part don't have to upvote the second part, which is devoted to a different approach to getting quantum mechanics out of automata, to answer the second question.
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Foundational discussions are indeed somewhat like discussions about religious convictions, as one cannot prove or disprove assumptions and approaches at the foundational level.
Moreover, it is in the nature of discussions on the internet that one is likely to get responses mainly from those who either disagree strongly (the case here) or who can add something constructive (difficult to do in very recent research). I think this fully explains the responses that you get.
I myself read superficially through one of your articles on this and found it not promising enough to spend more time on the technical issues.
However, I agree that both many-worlds and pilot waves are unacceptable physical explanations of quantum physics, and I am working on an alternative interpretation.
In my view, particle nonlocality is explained by negating particles any ontological existence. Existent are quantum fields, and on the quantum field level, everything is local. Nonlocal features appear only when one is imposing on the fields a particle interpretation, which, while valid under the usual asssumptions of geometric optics, fails drastically art higher resolution. Thus nothing needs to be explained in the region of failure. Just as the local Maxwell equations for a classsical electromagnetic field explain single photon nonlocality (double slit experiments), and the stochastic Maxwell equations explain everything about single photons (see http://www.mat.univie.ac.at/~neum/ms/optslides.pdf), so local QFT explains general particle nonlocality.
My thermal interpretation of quantum mechanics (see the section http://www.mat.univie.ac.at/~neum/physfaq/topics/found0.html from my theoretical physics FAQ at http://www.mat.univie.ac.at/~neum/physfaq/physics-faq.html, and Chapter 10 of my book http://lanl.arxiv.org/abs/0810.1019) gives a view of physics consistent with actual experimental practice and without any of the strangeness introduced by the usual interpretations. I believe this interpretation to be satisfactory in all respects, though it requires more time and effort to analyse the standard conundrums along these lines, with a clear statistical mechanics derivation to support my so far mainly qualitative arguments.
In presenting my foundational views in online discussions, I had similar difficulties as you; see, e.g., the PhysicsForums thread ''What does the probabilistic interpretation of QM claim?'' http://www.physicsforums.com/showthread.php?t=480072
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The problem is that your interpretation is standard QM, and whether you take fields or particles as the basic variables, you still have the exponential explosion. The field wavefunction is just as horrendous as particle wavefunction. The model t'Hooft would like (and I would like too), would not require exponential resources for a large grid. You can't just rejigger the philosophy to make this happen, by changing the "ontic" status of objects in the theory (honestly, a positivist doesn't even care about the ontic status). – Ron Maimon Aug 18 '12 at 2:32
@RonMaimon: Interpretations of quantum physics have nothing per se to do with computational complexity, but with making commong sense out of the standard quantum phenomena. – Arnold Neumaier Aug 26 '12 at 9:54
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Yes, ordinarily this is true, and this is why this question is not ordinary. The point of 'tHooft's line of investigation is to formulate a new theory (not quantum mechanics) which is nevertheless approximately quantum, but which does not suffer from the exponential explosion of computatinal complexity that plagues ordinary QM, and therefore is experimentally different from ordinary QM for the case of a quantum computer. The goal of such a theory is to not be experimentally different in cases where you would disagree with present experiments. This is a tough thing to do. – Ron Maimon Aug 26 '12 at 10:00
@RonMaimon: While this may be the case in his model, it is not his expressed goal. Instead, he wrote above: ''I can't help being disgusted by the "many world" interpretation, or the Bohm - deBroglie "pilot waves", and even the idea that the quantum world must be non-local is difficult to buy. I want to know what is really going on, and in order to try to get some ideas, I construct some models with various degrees of sophistication.'' – Arnold Neumaier Aug 26 '12 at 10:04
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Yes, I know, the stated motivations invite a lot of philosophical answers, but this is really not the motivation. I know this from understanding the holographic principle, reading the papers, and from talking to him briefly once ten years ago. The motivation is to make small hidden variables, meaning a number of bits which makes classical computation of reasonable size given holographic constraints, and which reproduces QM approximately. It is not to reproduce QM exactly. But he makes a mistake in his papers, and gets too good a QM, so he thinks it reproduces it locally and almost exactly. – Ron Maimon Aug 26 '12 at 10:09
This question tries to reproduce quantum mechanics from classical automata with a probabilistically unknown state.
### Probability distributions on Automata states
Start with a classical CA and a probability distribution on the CA. To keep things general, I allow the CA to have some non-determinstic evolution, but only stochastic probability, no quantum evolution, and it's not necessary, you can always put the probability in the initial conditions, with no stochasticity at intermediate times, it's just an option.
The first point about these stochastic systems is detailed here: Consequences of the new theorem in QM? (in the section on duck feet). If the flow of probability is always between states where the probability is only infinitesimally different from te stationary distribution, then the classical flow conserves entropy, and is reversable, even if it is probabilistic and diffusive. This is the central motivation for the construction, and one should review how the particle diffusing the in the heat exchanger diffuser bounces back and forth reversibly from room to room, in a linear way described by an operator with a mostly complex eigenvalue, even though it is at all times only diffusing between different allowed regions.
Consider a classical probability distribution on a CA, $\rho(B)$ where B is the state of all the bits comprising the automaton, then
$$I = - \sum_B \rho(B) \log(\rho(B))$$
is the information contained in fully knowing the automaton state, above that provided by the distribution. If you make a perturbation to first order, changing $\rho$ to $\rho+\delta\rho$, you find
$$I = - \sum_B (\rho(B)+\delta\rho(B)) \log(\rho(B) + \delta\rho(B))$$
When $\rho$ is uniform, the first order correction vanishes since the sum of $\delta\rho$ is zero, and the second order correction gives a quadratic metric structure on $\delta\rho$.
$$I = - \sum_B \delta\rho(B)^2$$
This is what I identify as a pre-quantum structure on the space of perturbations to the uniform distribution. The reason it is so symmetrical (like a sphere, not like a simplex) is because the perturbation is small. The reversibility is required by conservation of entropy, and conservation of entropy requires that all the transformations on $\delta\rho$ are orthogonal.
The picture to zeroeth order is that nearly every state is equally likely, but some states are slightly more likely than others, and the information revealed by experiments is only producing a slight bias for some states rather than others. These slight biases then are more symmetric than the underlying space of probabilities on automata states, because these distributions never deviate enough from uniformity to see the corners of the probability space simplex. The corners are the states where the automata bits are known for certain, and if you are always far from these, you can find a symmetric and reversible probabilistic dynamics.
Here is the central problem with this approach--- it is impossible for an information containing perturbation $\delta \rho$ to be everywhere small. The reason is that an everywhere small $\delta \rho$ necessarily produces a state which is almost indistinguishable from the uniform state, and which therefore makes a perturbation which corresponds to you having learned much less than even 1 bit of information. For example, if you have N bit automaton, and you make a distribution where the probability of every bit value is between ${1\over 2} - \epsilon$ and ${1\over 2} + \epsilon$, you get an information content bounded above by a small multiple of $\epsilon$ bits.
The reason is that learning even one bit of information about an automaton state roughly cuts down the number of states you can occupy by a factor of 2. This means that the true probability distribution must be significantly small on at least half the configurations, and it cannot be a small perturbation. This means that the information expansion breaks down, and this is where I was stuck for a long time
### Locally small perturbations
The reason the notion of "small perturbation" is failing is because a small perturbation, as in the duck-feet example, is not globally small, it only has the property that the ratio of the probabilities between two nearby states is small. If the states are made by independently varying lots of bits, there are many states with the same ratio of probability.
The fix might as well be the following easy trick: just raise everything to the M-th power. If you have a system with states indexed by i an integer in the range 1,2,...,N, and a perturbation
$$(\rho_i + \delta\rho_i)$$
You can take the M-th tensor power of $\rho$, to produce a product distribution on the tensor space with M indices $i_1,i_2,...,i_M$. This product distribution is defined by the condition that changing every value i from one value to another produces the same ratio change in probability.
Now it is allowed for $\delta\rho$ to be small even when the information in $\delta\rho$ is not, because the M-th power isn't small at all. In fact, in this system, because it's a tensor product, if you know that the information content of $\rho+\delta\rho$ is I bits overall, then you learn that
$$M \sum_B \delta\rho^2 = 1$$
In other words, the finite information perturbations to the stationary distribution on a system with M-copies forms a (real, not complex) Hilbert space, ever more perfectly as M goes to infinity. If the dynamics is duck-feet, meaning that the entropy is conserved with the small perturbation, then the time evolution of $\delta\rho$ is necessarily an orthogonal transformation, no matter what the underlying stochastic or deterministic evolution law is.
The basic idea is that you can make a quantum mechanics emerge from stochastic evolution of systems with many identical copies, under the condition that the copies interact symmetrically with each other, so that you don't know which copy is which.
To see how the inner product comes out, you consider the mutual information, which tells you how independent two different distributions are. To lowest order, this is found by taking the information in $\delta\rho_1$ and $\delta\rho_2$ and subtracting the information in $\delta\rho_1$ and $\delta\rho_2$ separately. Since these are the norms, you find
$$I_{12} = ||\delta\rho_1 + \delta\rho_2 ||^2 - || \delta\rho_1 ||^2 - ||\delta\rho_2||^2 = \langle\rho_1 , \rho_2 \rangle$$
So that if you have two distributions, they share states to the degree that their inner product is nonzero.
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– user1247 Aug 19 '12 at 19:43
– Ron Maimon Aug 19 '12 at 20:30
There's no doubt it's possible to reproduce quantum integrable models rather efficiently and simply using classical systems. And of all integrable systems, harmonic oscillators are one of the simplest. The real challenge is to reproduce quantum nonintegrable systems. Can you reproduce quantum chaos? Can you reproduce quantum nonintegrable spin models over a 1d spatial lattice? Trying perturbation theory from an integrable models runs into the problem that the number of feynman diagrams grows exponentially with the number of loops.
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Its just the start in the future he may be able to. – Asphir Dom Aug 18 '12 at 10:35
There is doubt! It is not trivial to reproduce simple quantum mechanics from cellular automata, and I don't think 'tHooft does it (although I think he came very close to doing so, and intuitively he is spot on). – Ron Maimon Aug 19 '12 at 21:35
@Scary Monster: The claim is that any CA can be cast in the language of QM, although in most cases the QM models you get will be uninteresting; there will be states, and they will obey Schroedinger equations. Now many CA models are computationally universal, so certainly not integrable, and therefore the asociated QM theory is also expected to be non-trivial. But of course the math is much harder; it's much more instructive to search for cases where you can do (perturbative) calculations. – G. 't Hooft Aug 20 '12 at 15:57
## protected by David Zaslavsky♦Aug 26 '12 at 1:06
This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.
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http://physics.stackexchange.com/questions/46908/liquid-with-freezing-point-above-0-celsius-that-could-be-use-at-ice-rinks?answertab=oldest
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# Liquid with freezing point above 0 Celsius that could be use at ice rinks
Is there a liquid that could be used to fill an ice rink (non-explosive, non-poisonous, etc), and have the freezing point above 0 Celsius?
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Iron- it has a freezing point well above zero degrees. Seriously though, water is used because it is plentiful, freezing point is at a useful temperature, plumbing is already understood and simple at this temperature and it is largely safe. – Rory Alsop Dec 15 '12 at 13:37
Thank you for your comment,if such a liquid exist will help me spare a lot of electricity used to freeze the water. – Muresan Dec 16 '12 at 5:27
## 1 Answer
The answer to this question is "probably not". The reason for this is quite interesting.
Ice skates have such low friction because a layer of water forms in between the ice and the blades. In order for this to happen, you need a substance that will turn from solid to liquid when it's compressed, which (according to thermodynamics) is the same thing as having a liquid that expands when it freezes. If you don't use a substance with this property then your skates are just resting on a solid surface and friction will prevent you from going anywhere.
But water is quite unusual in having this property. There are other substances that do it, but not many. The most comprehensive list I can find includes only water, silicon, gallium, antimony, bismuth and acetic acid. The metals can all be ruled out because you'd have to heat the floor up to close to their melting point. Acetic acid comes close: its melting point is $16$-$17^\circ C$, so you could in principle skate on a floor made of the stuff at only a little below room temperature. But unfortunately pure acetic acid is corrosive and has a pungent smell (it's what you can smell in vinegar, but this would be the purified version, so much more intense) so it wouldn't be suitable for a public place.
Maybe there is another molecule that has the desirable properties, but it seems kind of unlikely, because these substances are easy to spot - the solid phase floats on top of the liquid one - so if there was another one it would probably already be known.
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Thak you very much Nathaniel, perhaps a solutions, a mix of more than one liquid will have such freezing point.For me it's important to find such a liquid or mixture to spare the electricity used to freeze the ice rink. – Muresan Dec 16 '12 at 5:30
@Muresan the problem is that all the known substances with the expanding-on-freezing property are metals, apart from acetic acid. The list I linked to says some of bismuth's alloys also expand, so maybe you could lower its freezing point by mixing it with mercury, while still keeping the expansion property (this seems unlikely to me but I don't know enough chemistry to say for sure) - but mercury releases toxic vapour, so it wouldn't be suitable for a public place. – Nathaniel Dec 16 '12 at 9:00
@Nathaniel: "you need a substance that will turn from solid to liquid when it's compressed, which (according to thermodynamics) is the same thing as having a liquid that expands when it freezes" could you point me toward some reading material on that ? – josinalvo Dec 16 '12 at 15:26
– Nathaniel Dec 17 '12 at 1:56
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http://www.cfd-online.com/W/index.php?title=Dimensional_analysis_(introductory)&oldid=4954
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Home > Wiki > Dimensional analysis (introductory)
# Dimensional analysis (introductory)
### From CFD-Wiki
Revision as of 22:15, 17 December 2005 by Stefan (Talk | contribs)
```Every relation between n physical variables can be written in a an equivalent nondimensional form,
and the number of the dimensionless variables of this new relation is always smaller than the original one.
```
### What's the point?
• Reduction of the number of variables
Engineers often have to make experiments, because the governing equations of fluid flow are impossible to solve analytically in many important applications. The outcome of an experiment depends on the physical properties of the fluid, such as viscosity, density ond so on. To fully understand how a variation of these properties will effect the outcome, one might think that many experiments with different values for each of these parameters would have to be made. Let's say, for example, we want to measure the drag on a smooth sphere in an incompressible fluid, e.g. in a large and deep river in which the fluid flows over the sphere with a certain velocity. The drag will depend on the diameter of the sphere, the velocity of the fluid, the density of the fluid and the its viscosity. If we hold three parameters fixed and see how the result changes for 10 different values of the fourth parameter (e.g. velocity) , we would conclude in what way an alteration of velocity affects the drag, given the certain values for the rest of the variables. But what if we changed the diameter of the sphere to a different value now? Would we have to do ten more experiments for finding the drag in function of the velocity with the new sphere? Considering the time we have to spend for conducting a single experiment, and the number of possible combinations of the four parameters (10^4 = 10000), we would maybe regret our choice of becoming experimental physicists,-). The good thing about dimensional analysis is that it shows us that we don't need to try out every possible combination of parameters.
• Transforming data obtained from models to the prototype
Dimensional anlysis tells us how to build a model so that we can convert the data from the experiment with the model to the actual prototype.
### How do we obtain the dimensionless Parameters?
Every relation between n dimensional variables can be expressed in dimensionless form:
$f\left(x_{1},x_{2}\ldots x_{n}\right)$
can be expressed in dimensionless form:
$g\left(\pi_{1},\pi_{2}\ldots \pi_{k}\right)$
Let's consider the example from above:
We will state that that the drag force is a function of diameter, velocity, density and viscosity:
$F=f\left(D,U_{\infty},\rho, \mu \right)$
We have five variables: $n=5$.
• Diameter of the sphere: $D$
• Velocity of the river: $U_{\infty}$
• Density of the Medium: $\rho$
• Viscosity of the fluid: $\mu$
Now we find the corresponding dimensions to each variable:
$\begin{Bmatrix} F & D & U_{\infty} & \rho& \mu \\ MLT^{-2} & L & LT^{-1} & ML^{-3} & ML^{-1}T^{-1}\end{Bmatrix}$
Where $M$ is mass, $L$ is length, $T$ is time and $\Theta$ is temperature (in this problem, we don't consider temperature.
As mentioned before, the advantage of the dimensionless form is that the number of variables is always less. The Pi-Theorem show us how many variables will be left:
$k=n-rank$
where $rank$ is the rank of the Matrix $c_{ij}$:
M L T $\Theta$
$x_{1}$ $c_{11}$ $c_{12}$ $c_{13}$ $c_{14}$
$x_{2}$ $c_{21}$ $c_{22}$ $c_{23}$ $c_{24}$
...
$x_{n}$ $c_{n1}$ $c_{n2}$ $c_{n3}$ $c_{n4}$
In our case, this is:
M L T $\Theta$
$F$ $1$ $1$ $-1$ $0$
$D$ $0$ $1$ $0$ $0$
$U_{\infty}$ $0$ $1$ $-1$ $0$
$\rho$ $1$ $-3$ $0$ $0$
$\mu$ $1$ $-1$ $-1$ $0$
The rank of this matrix is three, so $k=2$.
This means that we will have 2 dimensionless Parameters (also called "pi products") in our dimensionless relation. We now have to choose 5-2=3 variables that will appear (with a certain exponent which can sometimes be 0) in both of the pi products. The important thing is that these variables thrown together contain all of the basic dimensions used (MLT) and that we cannot build a dimesionless parameter out of them. Let's see what choices we can make:
• $FLU_{\infty}$
They contain obviously all parameters, but only $F$ contains $M$, so
we cannot multiply them to get a dimensionless parameter. Thus, we could choose them to appear in all
pi products.
• $LU_{\infty}\rho$
That works too.
• Other combinations work as well.
The other two variables left will appear each in a different pi product. Since we are interested in expressing the drag force, we choose $F$ to be one of the two variables that will only appear in one of the pi products. In all the books I consulted, the other variable that appears only in one of the pi products is the viscosity, $\mu$. That means that usually engineers choose $LU\rho$ as the three variables that appear in each pi product. I don't know why this is a good choice, do you? My guess is that if we make this choice, then one of the pi products is (as you will see) $\frac{\rho U_{\infty}L}{\mu}$, which is called Reynolds number. This dimensionless pi product is very important, and it also appears in the dimensionless form of the Navier-Stokes equation. So we take $LU_{\infty}\rho$ and multiplicate them with $F$ to obtain the first pi product, and then we take $LU_{\infty}\rho$ and build the second pi product together with $\mu$.
Let's see how the pi products look like: We know that the first one will be made up of $FLU_{\infty}\rho$ and that it needs to be dimensionless, so we have to find the correct exponents for the variables:
$\pi_{1} = FL^{a}U_{\infty}^{b}\rho^{c} =(MLT^{-2})(L)^{a}(LT^{-1})^{b}(ML^{-3})^{c}$
$a+b-3c+1 =$ 0
$c+1 =$ 0
$-b-2=$ 0
Note that if we did everything right, then this system has exactly one solution:
$a=-2 \qquad b=-2 \qquad c=-1$
So
$\pi_{1}=\frac{F}{\rho U_{\infty}^{2}L^{2}}$
If we do the same to get the other pi product:
$\pi_{2}=\frac{\rho U_{\infty} L}{\mu}$
We finally obtained the dimensionless form of the relationship:
$\frac{F}{\rho U_{\infty}^{2}L^{2}}=h\left(\frac{\rho U_{\infty} L}{\mu}\right)$
Imagine we want to make our experiment with water (there are obvious reasons for that: it's cheap and doesn't harm our health), but we are interested in the drag force caused by a much less viscous $\left(\mu_{o}\right)$fluid with the same density at a certain velocity $u_{o}$: We could still make the experiment with water, but we would have to increase either the diameter of the sphere or the velocity of the "river" (or the density, but that's complicated or impossible). Doing this, (in this case we chose velocity as an example) we obtain the same Reynolds number.
$\frac{\rho U_{o} L}{\mu_{o}}=\frac{\rho U_{w} L}{\mu_{w}}$
The Pi Theorem guarantees us that
$\frac{F_{o}}{\rho U_{o}^{2}L^{2}}=\frac{F_{w}}{\rho U_{w}^{2}L^{2}}$
and so
$\frac{F_{o}}{F_{w}}=\left(\frac{U_{o}}{U_{w}}\right)^{2}$
Since we know the values of the two velocities, we can make our experiment with water, obtain $F_{w}$ and convert to $F_{o}$ with the relationship.
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http://mathforum.org/mathimages/index.php?title=Parametric_Equations&oldid=17009
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# Parametric Equations
### From Math Images
Revision as of 20:27, 25 May 2011 by Ifayanj (Talk | contribs)
Butterfly Curve
Field: Algebra
Image Created By: Direct Imaging
Website: [1]
Butterfly Curve
The Butterfly Curve is one of many beautiful images generated using parametric equations.
# Basic Description
Parametric Equations can be used to define complicated functions and figures in simpler terms, using one or more additional independent variables, known as parameters . In particular, Parametric Equations can be used to define and easily generate geometric figures, including(but not limited to) conic sections and spheres.
We often graph functions by letting one coordinate be dependent on another. For example, graphing the function $f(x) = y = x^2$ has y values that depend upon x values. However, some complicated functions are best described by having the coordinates be described using an equation of a separate independent variable, known as a parameter. Changing the value of the parameter then changes the value of each coordinate variable in the equation. We choose a range of values for the parameter, and the values that our function takes on as the parameter varies traces out a curve, known as a parametrized curve. Parametrization is the process of finding a parametrized version of a function.
### Parametrized Circle
One curve that can be easily parametrized is a circle of radius one:
We use the variable t as our parameter, and x and y as our normal Cartesian coordinates.
We now let $x = cos(t)$ and $y = sin(t)$, and let t take on all values from $0$ to $2\pi$.
When $t=0$, the coordinate $(1,0)$ is hit. As t increases, a circle is traced out as x initially decreases, since it is equal to the cosine of t, and y initially increases, since it is equal to the sine of t. The circle continues to be traced until t reaches $2\pi$, which gives the coordinate $(1,0)$ once again.
It is also useful to write parametrized curves in vector notation, using a coordinate vector: $\begin{bmatrix} x \\ y\\ \end{bmatrix}= \begin{bmatrix} cos(t) \\ sin(t)\\ \end{bmatrix}$
The butterfly curve in this page's main image uses more complicated parametric equations as shown below.
# A More Mathematical Explanation
Note: understanding of this explanation requires: *Linear Algebra
[Click to view A More Mathematical Explanation]
[[Image:Animated_construction_of_butterfly_curve.gif|thumb|right|500px|Parametric construction of the [...]
[Click to hide A More Mathematical Explanation]
Parametric construction of the butterfly curve
Sometimes curves which would be very difficult or even impossible to graph in terms of elementary functions of x and y can be graphed using a parameter. One example is the butterfly curve, as shown in this page's main image.
This curve uses the following parametrization:
$\begin{bmatrix} x \\ y\\ \end{bmatrix}= \begin{bmatrix} \sin(t) \left(e^{\cos(t)} - 2\cos(4t) - \sin^5\left({t \over 12}\right)\right) \\ \cos(t) \left(e^{\cos(t)} - 2\cos(4t) - \sin^5\left({t \over 12}\right)\right)\\ \end{bmatrix}$
### Parametrized Surfaces
The surface of a sphere can be graphed using two parameters.
In the above cases only one independent variable was used, creating a parametrized curve. We can use more than one independent variable to create other graphs, including graphs of surfaces. For example, using parameters s and t, the surface of a sphere can be parametrized as follows: $\begin{bmatrix} x \\ y\\ z\\ \end{bmatrix}= \begin{bmatrix} sin(t)cos(s) \\ sin(t)sin(s) \\cos(t) \end{bmatrix}$
### Parametrized Manifolds
While two parameters are sufficient to parametrize a surface, objects of more than two dimensions, such as a three dimensional solid, will require more than two parameters. These objects, generally called manifolds, may live in higher than three dimensions and can have more than two parameters, so cannot always be visualized. Nevertheless they can be analyzed using the methods of vector calculus and differential geometry.
### Parametric Equation Explorer
This applet is intended to help with understanding how changing an alpha value changes the plot of a parametric equation. See the in-applet help for instructions.
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# Related Links
### Additional Resources
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Butterfly Curve
Field: Calculus
Image Created By: Direct Imaging
Website: [2]
Butterfly Curve
The Butterfly Curve is one of many beautiful images generated using parametric equations.
# Basic Description
We often graph functions by letting one coordinate be dependent on another. For example, graphing the function $f(x) = y = x^2$ has y values that depend upon x values. However, some complicated functions are best described by having the coordinates be described using an equation of a separate independent variable, known as parameters. Changing the value of the parameter then changes the value of each coordinate variable in the equation. We choose a range of values for the parameter, and the values that our function takes on as the parameter varies traces out a curve, known as a parametrized curve. Parametrization is the process of finding a parametrized version of a function.
### Parametrized Circle
One curve that can be easily parametrized is a circle of radius one:
We use the variable t as our parameter, and x and y as our normal Cartesian coordinates.
We now let $x = cos(t)$ and $y = sin(t)$, and let t take on all values from $0$ to $2\pi$.
When $t=0$, the coordinate $(1,0)$ is hit. As t increases, a circle is traced out as x initially decreases, since it is equal to the cosine of t, and y initially increases, since it is equal to the sine of t. The circle continues to be traced until t reaches $2\pi$, which gives the coordinate $(1,0)$ once again.
It is also useful to write parametrized curves in vector notation, using a coordinate vector: $\begin{bmatrix} x \\ y\\ \end{bmatrix}= \begin{bmatrix} cos(t) \\ sin(t)\\ \end{bmatrix}$
The butterfly curve in this page's main image uses more complicated parametric equations as shown below.
# A More Mathematical Explanation
Note: understanding of this explanation requires: *Linear Algebra
[Click to view A More Mathematical Explanation]
[[Image:Animated_construction_of_butterfly_curve.gif|thumb|right|500px|Parametric construction of the [...]
[Click to hide A More Mathematical Explanation]
Parametric construction of the butterfly curve
Sometimes curves which would be very difficult or even impossible to graph in terms of elementary functions of x and y can be graphed using a parameter. One example is the butterfly curve, as shown in this page's main image.
This curve uses the following parametrization:
$\begin{bmatrix} x \\ y\\ \end{bmatrix}= \begin{bmatrix} \sin(t) \left(e^{\cos(t)} - 2\cos(4t) - \sin^5\left({t \over 12}\right)\right) \\ \cos(t) \left(e^{\cos(t)} - 2\cos(4t) - \sin^5\left({t \over 12}\right)\right)\\ \end{bmatrix}$
### Parametrized Surfaces
The surface of a sphere can be graphed using two parameters.
In the above cases only one independent variable was used, creating a parametrized curve. We can use more than one independent variable to create other graphs, including graphs of surfaces. For example, using parameters s and t, the surface of a sphere can be parametrized as follows: $\begin{bmatrix} x \\ y\\ z\\ \end{bmatrix}= \begin{bmatrix} sin(t)cos(s) \\ sin(t)sin(s) \\cos(t) \end{bmatrix}$
### Parametrized Manifolds
While two parameters are sufficient to parametrize a surface, objects of more than two dimensions, such as a three dimensional solid, will require more than two parameters. These objects, generally called manifolds, may live in higher than three dimensions and can have more than two parameters, so cannot always be visualized. Nevertheless they can be analyzed using the methods of vector calculus and differential geometry.
### Parametric Equation Explorer
This applet is intended to help with understanding how changing an alpha value changes the plot of a parametric equation. See the in-applet help for instructions.
If you can see this message, you do not have the Java software required to view the applet.
# Teaching Materials
There are currently no teaching materials for this page. Add teaching materials.
Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.
Hyperboloid
Mobius Strip
Torus
Vector Fields
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http://unapologetic.wordpress.com/2011/02/14/intertwinors-from-semistandard-tableaux-span-part-3/?like=1&_wpnonce=68d26d17fa
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# The Unapologetic Mathematician
## Intertwinors from Semistandard Tableaux Span, part 3
Now we are ready to finish our proof that the intertwinors $\bar{\theta}_T:S^\lambda\to M^\mu$ coming from semistandard generalized tableaux $T$ span the space of all intertwinors between these modules.
As usual, pick any intertwinor $\theta:S^\lambda\to M^\mu$ and write
$\displaystyle\theta(e_t)=\sum\limits_Tc_TT$
Now define the set $L_\theta$ to consist of those semistandard generalized tableaux $S$ so that $[S]\trianglelefteq[T]$ for some $T$ appearing in this sum with a nonzero coefficient. This is called the “lower order ideal” generated by the $T$ in the sum. We will prove our assertion by induction on the size of this order ideal.
If $L_\theta$ is empty, then $\theta$ must be the zero map. Indeed, our lemmas showed that if $\theta$ is not the zero map, then at least one semistandard $T$ shows up in the above sum, and this $T$ would itself belong to $L_\theta$. And of course the zero map is contained in any span.
Now, if $L_\theta$ is not empty, then there is at least some semistandard $T$ with $c_T\neq0$ in the sum. Our lemmas even show that we can pick one so that $[T]$ is maximal among all the tableaux in the sum. Let’s do that and define a new intertwinor:
$\displaystyle \theta' = \theta - c_T\bar{\theta}_T$
I say that $L_{\theta'}$ is $L_\theta$ with $T$ removed.
Every $S$ appearing in $\bar{\theta}_T(e_t)$ has $[S]\trianglelefteq[T]$, since if $T$ is semistandard then $[T]$ is the largest column equivalence class in $\theta_T(\{t\})$. Thus $L_{\theta'}$ must be a subset of $L_\theta$ since we can’t be introducing any new nonzero coefficients.
Our lemmas show that if $[S]=[T]$, then $c_S$ must appear with the same coefficient in both $\theta(e_t)$ and $c_T\bar{\theta}_T(e_t)$. That is, they must be cancelled off by the subtraction. Since $T$ is maximal there’s nothing above it that might keep it inside the ideal, and so $T\notin L_{\theta'}$.
So by induction we conclude that $\theta'$ is contained within the span of the $\bar{\theta}_T$ generated by semistandard tableaux, and thus $\theta$ must be as well.
## 3 Comments »
1. You know what? This makes no sense to me. I mean, it shouldn’t, as I don’t have the prerequisite knowledge to formulate an understanding…
… but isn’t that a HUGE issue with mathematics? Why can’t this information be explained in laymen’s terms? Isn’t that the point of this blog?
Why do mathematicians constantly shroud their art form?
Comment by Haakon | February 15, 2011 | Reply
2. Haakon, have you tried tracing back the links to the more foundational material?
Comment by | February 15, 2011 | Reply
3. [...] Now we’ve finished our proof that the intertwinors coming from semistandard tableauxspan the space of all [...]
Pingback by | February 17, 2011 | Reply
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://mathhelpforum.com/math-topics/39387-problem-numbers.html
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# Thread:
1. ## problem with numbers
For which numbers N, is N*-2 greater than -2? In other words, for which numbers N, is N*-2>-2?
Investigate theis question as follow:
A) Without using a calculator, multiply -2 by the given numbers. Show your work. In each case, determine wheather the resulting product is greater than -2 or not.
3.15, ,1.01, 0.85, 0.002, -0.3, -4.2
4 3/5, 1 2/3, 9/10, 3/5, -7/8, -1 1/8
B) Based on your answers in part a and on th emeaning and rules of multiplication, describe the collection of all numbers N, for which N8-2 is greater than -2.
Thank you
2. Well, nobody seems to want to respond to this one, so here goes.
In order for $N\cdot-2>-2$,
$N<1$
For instance; if $N=\frac{1}{2}$, then $\frac{1}{2}\cdot-2=-1$ which is greater than -2.
If $N=-2$, then $-2\cdot-2=2$, which is greater than -2.
So, there you go.
I assume you can multiply -2 times all your sample numbers and compare the products to see if any are greater than -2.
You'll find that 0.85, 0.002, -0.3, -4.2, $\frac{9}{10} , \frac{3}{5} , \frac{-7}{8} and -1\frac{1}{8} \$, all conform to the fact that N < 1 is the set of numbers that satisfy your original statement.
3. Thank you so much!
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http://en.m.wikibooks.org/wiki/Calculus_Optimization_Methods/Constrained_Optimization
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Calculus Optimization Methods/Constrained Optimization
Presentation
There are two main ways to present a subset of $\mathbf{R}^n$ – explicitly and implicitly.
An explicit presentation expresses a set as the image of a function, $X \to \mathbf{R}^n,$ most often the image of a function from another Euclidean space or cube, $\mathbf{R}^k \to \mathbf{R}^n$ or $[0,1]^k \to \mathbf{R}^n.$ This function can be interpreted as a vector-valued function with k inputs and n outputs.
An implicit presentation expresses a set as the pre-image of a function or functions, $\mathbf{R}^n \to Y,$ most often the pre-image of a function to another Euclidean space $\mathbf{R}^n \to \mathbf{R}^k,$ which can be interpreted as k separate real-valued functions (each with n inputs), or as a single vector-valued function in n inputs. Implicit presentations generally present subsets as level sets or sublevel sets: via equalities or inequalities that elements of a subset must satisfy.
Briefly, explicit presentations list points, while implicit presentations test points: in an explicit presentation, as one ranges over the input, one obtains all the output, while in an implicit presentation, one can test possible points to see if they fall in the set: if the satisfy the constraints or not.
One can sometimes convert between these presentations, but in general this can be quite difficult.
Example: circle and disk
Circle:
• $x^2+y^2=1$ is an implicit presentation.
• $(\cos \theta, \sin \theta)$ for $\theta \in [0,1] \times [0,2\pi]$ is an explicit presentation.
• $\left(\frac{1-t^2}{1+t^2},\frac{t}{1+t^2}\right)$ for $t \in \mathbf{R}$ is an explicit presentation.
Disk:
• $x^2+y^2 \leq 1$ is an implicit presentation
• $(r \cos \theta, r \sin \theta)$ for $(r,\theta) \in [0,1] \times [0,2\pi]$ is an explicit presentation.
• $\left(u\frac{1-t^2}{1+t^2},u\frac{t}{1+t^2}\right)$ for $t \in \mathbf{R}, u \in [0,1]$ is an explicit presentation.
Different presentations may be more or less useful.
Caveats: Imperfect parametrizations
Note that explicit presentations sometimes hit the same point more than once (in the first presentation of the circle, $\theta = 0$ and $\theta = 2\pi$ map to the same point, and in the presentation of the disk, all points with $r=0$ map to the same point, the origin, $(0,0)$), or miss a point (in the second presentation, the point $(-1,0)$ is not hit by any finite t – if one extends the domain to include $\infty,$ then it can be hit by some input). This is often simply a minor technicality to deal with, such as by excluding $\theta = 2\pi$ or including $t=\infty,$ or by checking points that are not hit or hit multiply separately and with care. The underlying mathematical issue is that the space being parametrized may not be homeomorphic or diffeomorphic to the space being used to parameterize: one says that there is a "topological obstruction" to giving a parametrization without these imperfections.
One need not in general concern oneself with the details of the topology, and topology is a major field of mathematics, but, as with the boundedness principle and the maximum principle, topological theories underlie much of why calculus optimization methods work.
One topological, or more precisely geometric, observation that is worth mentioning, however, is that in many applications, the subset being considered is a convex set – it bulges out, not in, and, further, is connected (in one piece) and does not have holes in the middle. In these settings, the shape is topologically a disk or in higher dimensions a ball, and one can thus talk about the interior of a shape (the open n-ball), and a single boundary, which will topologically be a simple sphere (the ($n-1$)-sphere) rather than something more complicated. Thus there is little loss in generality of bearing the disk in mind as the archetypal example of a subset in thinking about these problems.
Explicit: parametrization
A set may be given explicitly by a parametrization, such as by a parametric equation in 1 parameter for a curve, or a parametric surface (parametric equation in 2 parameters).
Implicit: constraints
• Equality constraints
• Inequality constraints
↑Jump back a section
Explicit solution
↑Jump back a section
Implicit solution
↑Jump back a section
Discussion
When finding the extreme values of $f(x_1,x_2,\cdots, x_n)$ subject to a constraint $g(x_1,x_2,\cdots, x_n)=k$, the stationary points found above will not work. This new problem can be thought of as finding extreme values of $f(x_1,x_2,\dots, x_n)$ when the point $(x_1,x_2,\dots, x_n)$ is restricted to lie on the surface $g(x_1,x_2,\dots, x_n)=k$. The value of $f(x_1,x_2,\dots, x_n)$ is maximized(minimized) when the surfaces touch each other,i.e , they have a common tangent line. This means that the surfaces gradient vectors at that point are parallel, hence,
$\nabla f(x_1,x_2,\cdots, x_n) = \lambda \nabla g(x_1,x_2,\cdots, x_n) .$
The number $\lambda$ in the equation is known as the Lagrange multiplier.
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http://mathoverflow.net/questions/119899?sort=newest
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## Mirror symmetry for hyperkahler manifold
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Hi there,
I have some questions about the mirror symmetry of hyperkahler manifold and K3 surface.
The well-known result said: the mirror symmetry for K3 surface is just given by its hyperkahler rotation.
1) In what sense, the rotation gives the mirror map?
2) Does this means:
if we start from $(M,\omega_I,I)$, here the k3 surface $M$ has a special lagrangian fibration structure with respect to $\omega_I$ and $I$, and also a special lagrangian section. Denote its SYZ mirror by $(M^{mirror},J_{mirror})$. Then exist a (fiberwised) diffeomorphism $\phi: M \rightarrow M^{mirror}$, s.t. $\phi^{*} (J_{mirror}) = K$?
(Here $I,K$ are the standard base of the $S^2$-family of compatible complex structure of the hyperkahler metric on $M$.)
Thanks!
-
I recommend you read section 7 of this paper of Gross arxiv.org/pdf/math/9809072.pdf, section 1 of Gross-Wilson arxiv.org/pdf/math/0008018v3.pdf, and this paper of Dolgachev arxiv.org/pdf/alg-geom/9502005v2.pdf – YangMills Jan 26 at 2:10
arxiv.org/abs/hep-th/9512195 Mirror Symmetry for hyperkaehler manifolds Misha Verbitsky We prove the Mirror Conjecture for Calabi-Yau manifolds equipped with a holomorphic symplectic form. Such manifolda are also known as complex manifolds of hyperkaehler type. We obtain that a complex manifold of hyperkaehler type is Mirror dual to itself. The Mirror Conjecture is stated (following Kontsevich, ICM talk) as the equivalence of certain algebraic structures related to variations of Hodge structures. We compute th – Alexander Chervov Jan 26 at 13:11
thank you so much! i have not expect such direct approach.. the kahler form gives the deformation of the complex structure directly through Tian-Todorov coordinates. inspiring! – Jay Feb 6 at 18:48
## 1 Answer
Thanks, YangMills, for the references to my papers. I want to elaborate, because I disagree with the statement that mirror symmetry is given by hyperkahler rotation. It may be the case for certain choices of K3, but I think this happens by accident and that it's not a useful principle. Here is how I view mirror symmetry for K3 surfaces. Choose a rank 2 sublattice of the K3 lattice generated by $E$ and $F$ with $E^2=F^2=0, E.F=1$. Consider a K3 surface $X$ with a holomorphic $2$-form with $E.\Omega\not=0$. We can assume after rescaling $\Omega$ that $E.\Omega=1$, and then write $\Omega=F+\check B+i\check\omega \mod E$ for some classes $\check B,\check\omega$ in $E^{\perp}/E$. The K3 surface will be equipped also with a Kaehler form $\omega$ and a B-field $B$, which we write as $B+i\omega$. We choose this data in $E^{\perp}/E \otimes {\mathbb C}$, although the class of $\omega$ is determined in $E^{\perp}$ by its image in $E^{\perp}/E$ by the fact that $\omega\wedge \Omega$ must be zero. Then the mirror $\check X$ is taken to have holomorphic form $\check\Omega=F+B+i\omega\mod E$ and complexified Kaehler class $\check B+i\check\omega$.
Note that there is no particular reason to expect this new K3 surface to be a hyperkaehler rotation, as the mirror complex structure depends on $B$, which gives far too many parameters worth of choices: there is only a two-dimensional family of hyperkaehler rotation of $X$.
Note that we can hyperkaehler rotate $X$ so that special Lagrangians become holomorphic. The new holomorphic form is $\check\omega + i \omega \mod E$. If we multiply this form by $i$, we get $-\omega +i\check\omega\mod E$. A change of Kaehler form followed by another hyperkaehler rotation will give the mirror for certain choices of $B$-field, but note this involves two hyperkaehler rotations with respect to different metrics.
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http://math.stackexchange.com/questions/69373/expected-number-of-rolls-until-a-number-appears-k-times?answertab=active
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# Expected number of rolls until a number appears $k$ times
Let $N$ be the number of rolls until the same number appears $k$ consecutive times. Show the expected value $E[N]=\dfrac{6^k-1}{5}$. I've tried conditioning this on the first occurrence of the expected number, but I'm having a hard time generalizing further than 2 consecutive times. I think I need to use the conditional expectation formula, $E[N]=E[E[N|Y]]$ where $Y$ is another random variable which I've previously taken to be the first appearance of the number.
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What is the probability that the second roll yields a different number than the first one ? This should help you compute the probability that no number appears two consecutive times up until the $N^{\textrm{th}}$ roll. – Joel Cohen Oct 2 '11 at 23:19
## 2 Answers
You can prove this by induction. Let $a_k$ be the expected number of rolls until the same number appears $k$ consecutive times. Clearly $a_1=1=(6^1-1)/5$. Now assume $a_{k-1}=(6^{k-1}-1)/5$.
To get $k$ consecutive rolls with the same number, you first need to get $k-1$, and this is expected to take $a_{k-1}$ rolls. On the next roll, you have a $1$ in $6$ chance to finish and a $5$ in $6$ chance to go back to square one. Thus
$$a_k=a_{k-1}+\frac16\cdot1+\frac56\cdot a_k\;,$$
or
$$a_k=6a_{k-1}+1\;.$$
Substituting $a_{k-1}=(6^{k-1}-1)/5$ yields $a_k=(6^k-1)/5$.
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Well, for k=1, the value is trivially 1.
For k=2, you can find the probability of your event happenning in 2,3,4,.... throws.
The probability of N=2 , is the probability of the first two throws having the same outcome. So the expected number of throws needed for N=2 is:
P(Outcome aa).2+ [P(Outcome aba)+ P(Outcome baa)].3+....+[P( same number appears twice after j rolls].j+......
But, notice that every six rolls, you necessarily get at least one repetition.
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While the last sentence is correct, it is not applicable as OP asked about repetitions on successive rolls. – Ross Millikan Oct 2 '11 at 23:49
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http://mathoverflow.net/questions/24835/any-paradoxical-theorems-arising-from-large-cardinal-axioms/24851
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## Any paradoxical theorems arising from large cardinal axioms?
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If we accept the axiom of choice, we take the responsibility of living in a world in which, e.g., a ball in Euclidean 3-space is equiscindable to two isometric copies of itself (Banach-Tarski). So we have a very natural and seemingly harmless axiom (that most matehmaticians are probably willing to accept and use) on one hand, and a quite paradoxical and counterintuitive phenomenon arising from accepting that axiom on the other.
I would like to know if the assumpion of some large cardinal axiom is known to produce some sort of paradoxical phenomena in the "everyday world", as it happens in the case of the axiom of choice.
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## 3 Answers
In fact, the opposite is true. If suitable large cardinals exist, then all projective sets of reals (i.e. the definable sets that you are likely to come across in real life) are non-pathological. More precisely, they are measurable, have the Baire property, are either countable or have a perfect subset, etc.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
(This was a reply to John that grew beyond the alloted limit.)
large cardinals have consequences that were initially surprising to set theorists.
But, John, this is simply because our intuitions about large cardinals took a bit to form, just as one would expect with any relatively recent theory. Now we understand significantly better how far from $L$ the universe ought to be in the presence of large cardinals, and find the puzzlement at Scott's result curious. We know, for example, that much much much less than a measurable already contradicts $V=L$.
Similarly, when our understanding of large cardinals was relatively poor (early 80s), it was the common belief that for any large cardinal there was a nice inner model with a $\Delta^1_3$ well-ordering of ${\mathbb R}$, and it was by refuting this that we eventually arrived at Woodin cardinals and our current view of the set theoretic universe, where (as Simon mentioned) large cardinals make such well-orderings impossible, since all projective sets are measurable.
(This significant shift in understanding started with the Martin's Maximum paper by Foreman-Magidor-Shelah, Annals of Mathematics, 127 (1988), 1-47, and is very nicely explained in 'Iteration trees' by Martin-Steel, Journal of the American Mathematical Society, 7(1):1–73, 1994.)
This also led, by the way, to the proof that large cardinals imply determinacy in nice inner models, again shifting the prior poorly developed intuitions that had us expect much larger consistency strength for determinacy than it turned out to be the case.
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Andres, welcome to MO! – Joel David Hamkins May 16 2010 at 2:11
Andres, I admit that today's set theorists understand large cardinals much better than their predecessors. I just want to highlight how much our understanding has changed (and therefore, how much it could change in the future). – John Stillwell May 16 2010 at 2:52
2
Hi Joel! Oh, John, I agree. Determinacy itself is a good example of the same phenomenon. From being a somewhat ad hoc theory about infinite games, it has gone to become a systematic way of studying and establishing regularity properties. – Andres Caicedo May 16 2010 at 14:30
I believe that, back when supercompact cardinals were a new invention, Solovay mentioned in print that they might imply L[R]-determinacy (which turned out to be true, even with weaker large cardinals). Nevertheless, I believe Andres's description of the situation later, in the early 80's, is accurate. Determinacy had, in the meantime, started to look much stronger. – Andreas Blass Jul 3 2010 at 20:36
I broadly agree that large cardinals don't have paradoxical consequences, especially at the "everyday world" level. However, large cardinals have consequences that were initially surprising to set theorists. I think it was a surprise when Scott proved ( in 1961) that measurable cardinals imply the existence of non-constructible sets, and Gödel was astounded that the first measurable cardinal has to be larger that the first inaccessible.
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# The Purplemath Forums
Helping students gain understanding and self-confidence in algebra.
## find equation for line passing through pts of intersection
Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
2 posts • Page 1 of 1
### find equation for line passing through pts of intersection
by testing on Sun Feb 22, 2009 12:37 am
I have to find an equation for the line that passes through the points of intersection of the circles $x^2+y^2=25$ and $x^2-3x+y^2+y=30$. Can I just subtract and get $-3x+y=5$?
testing
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by stapel_eliz on Sun Feb 22, 2009 1:56 pm
You'd need first to find the slope of that line, in order to answer the question asked. To check your answer, go the long way 'round:
Solve the line equation you got for "y=", and plug this into either of the original equations in place of "y". (I'd use the x2 + y2 = 25: it's simpler.) Solve the resulting quadratic for two values of x, and then back-solve for the corresponding values of y. This will give you the two points of intersection, which you can then plug into the slope formula.
And you can always check your answer by graphing. You can either confirm that the two points are on each circle, or else you can just graph the two circles and the line, and make sure they overlap at the right spots. If you're plugging into a graphing calculator, you'll need to do the circles in two halves, since of course their equations aren't functions.
. . . . .$\mbox{Y1}\, =\, -0.5\, -\, \sqrt{32.5\, -\, \left(\mbox{X}\, -\, 1.5\right)^2$
. . . . .$\mbox{Y2}\, =\, -0.5\, +\, \sqrt{32.5\, -\, \left(\mbox{X}\, -\, 1.5\right)^2$
. . . . .$\mbox{Y3}\,=\, -\sqrt{25\, -\, \mbox{X}^2}$
. . . . .$\mbox{Y4}\,=\, \sqrt{25\, -\, \mbox{X}^2}$
. . . . .$\mbox{Y5}\, =\, 3\mbox{X}\, +\, 5$
Have fun!
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http://physics.stackexchange.com/questions/14128/when-is-many-body-perturbation-theory-valid
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# When is many-body perturbation theory valid?
I'm calculating expectation values (thermal, time-independent) using many-body perturbation theory, but I'm unsure how to work out what values the parameter I'm expanding the perturbation series in can take.
I read that it's when the matrix elements, $\langle i | H_{pert} | j \rangle$ where $H_{pert}$ is the perturbative term of the Hamiltonian and $| i \rangle$ and $| j \rangle$ are the eigenvectors of the unperturbed Hamiltonian, are much smaller than the energy difference between $i$ and $j$. But I don't really understand what that means, or how it helps me to calculate what values my perturbation parameter can take - Is there a method I can use to figure it out?
Edit:
As requested, to make this concrete, I have a one dimensional fermionic Hubbard model with Hamiltonian
\begin{equation} H= -t \sum_{\langle l,m \rangle} (c^\dagger_l c_m + h.c.) + U \sum_l (n_{l\uparrow} - 1/2)(n_{l\downarrow}-1/2) \end{equation}
I have a special case where I know that $U$ is very small and I want to use many-body perturbation theory to see its effects on correlation functions. I calculate the correlation functions using the functional integral method (i.e. calculating a functional partition function). For this case, how would I go about finding out how small $U$ has to be in order for perturbation theory to be valid?
Secondly, (if this should be a separate question, please let me know!) if instead I have a random $U$, dependent on its position in the lattice,
\begin{equation} H= -t \sum_{\langle l,m \rangle} (c^\dagger_l c_m + h.c.) + \sum_l U_l (n_{l\uparrow} - 1/2)(n_{l\downarrow}-1/2) \end{equation}
I can then use a similar functional integral technique, but take an average over the functional partition function (e.g. over a Gaussian distribution). This average removes the $U_l$ and leaves $\Delta$, the variance of the distribution we've averaged over. In this case, it is $\Delta$ which the perturbation series is expanded in. How would I go about finding how small $\Delta$ has to be for the perturbation series to be valid?
I don't want an answer that's true for any system, I just want to understand how to go about finding it for any system. So if anyone knows of another system where it is shown how small the expansion term has to be, please let me know.
Thanks.
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Is `many-body` important in your question? For me it looks like careful reading about perturbation theory would help. – Misha Aug 30 '11 at 17:37
@misha Yes many-body is important. I calculate the expectation values from a functional partition function. I've read about this kind of perturbation theory in lots of places, but none that I've found focus on this. If you know of any, that'd be great! – Calvin Aug 30 '11 at 19:48
You should say that you are asking for a perturbative treatment of a spatially variable hopping in a Hubbard model, this makes the problem concrete. There is no general answer to the question, because it depends sensitively on things that are impossible to state in terms of crude matrix element bounds. – Ron Maimon Aug 31 '11 at 2:57
@Calvin Trying to make the question general you lost some important details. No, I do not know many of Hubbard models. However, your idea that perturbation theory there should be different from perturbation theory elsewhere seems unnatural. Thus I assume that your question is not about perturbation theory but about how to apply perturbation to the particular problem which you (unfortunately) did not formulate in the question. – Misha Aug 31 '11 at 4:27
@Misha I've updated my question. The reason I made it general is that I want to know know to find out when the perturbation is valid for any model, not just for this one. By that I don't mean something that's true for any system; I mean that I would like to understand how to go about finding out for a system. If you know of any examples (not just with the Hubbard model), please point me in the right direction. Thanks. – Calvin Aug 31 '11 at 12:42
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## 2 Answers
Your "perturbation parameter" should just be something that sets the scale of $H_{pert}$ -- that is, the size of the matrix elements you specified compared to the bare energies of the unperturbed problem, which you know the exact solution of. To see why this is you just need to look at the explicit second order contribution from (nondegenerate!) perturbation theory, say for the ground state: $$E_0^{(2)}=-\sum_{j \neq 0}\frac{\langle 0 | H_{pert }| j \rangle \langle j | H_{pert }| 0 \rangle}{E_j - E_0}$$
You can read off that this is only going to be a small correction (and it has to be a small correction if we want to truncate the expansion here) if you can justify those matrix elements being smaller than the energy difference between the states. If you can't do that (say, you have no control over the strength of the perturbation) then you can't trust the perturbation expansion.
Edit 1: When we derive the perturbation expansion, the usual recipe is to do something along the lines $H=H_0+\alpha H_{pert}$, as you say, then expand in alpha -- but we must keep in mind that this guy $\alpha$ is a bookkeeping device that lets us make sure we group all the terms correctly by their order in alpha! After we're done bookkeeping and grouping terms, $\alpha \rightarrow 1$ always. It is not part of the physics and is not ours to play with. We expanded in it as a formal device. The real small parameter is going to be the ratio of the overall energy scale of $H_{pert}$ to $H_0$.
Let me give an example. The fermion Hubbard model has two terms -- (1) an easily diagonalized local interaction with a Coulomb energy scale called $U$; (2) a nearest neighbor hopping term that cannot be diagonalized in the same basis, with an energy scale (~the bandwidth) called $t$. Whether we can trust perturbation theory starting from this basis [that is, taking the hopping as a perturbation] depends only on the dimensionless scale $t/U$, and not the bookkeeping parameter we used to derive the formal terms of the expansion.
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Yes, by perturbation parameter I mean $\alpha$ with the term $\alpha H_{pert}$ in the Hamiltonian. But I consider the thermodynamic limit, so there's no way I could calculate whether it is small for all possible $i$ and $j$ anyway. Is there another way to determine how small $\alpha$ has to be (or have I misunderstood)? – Calvin Aug 30 '11 at 14:34
Hm. It might help me understand your question if you tell me the explicit model you're using. I'm going to make an edit to my answer, and you tell me if I'm completely missing the point. :) – wsc Aug 30 '11 at 15:16
For me, the parameter $\alpha$ is the $t$ in the fermion Hubbard model, and I want to expand in $\alpha$. The Hubbard model is fine as an example, but I can edit question to include more specifics of my own model if you like? – Calvin Aug 30 '11 at 15:43
"the overall energy scale of $H_{\rm{pert}$" may involve the density of states and/or temperature of the system. – Slaviks Aug 30 '11 at 15:53
@wsc Actually, more specifically, if I have a Hubbard model, but with $t$ dependent on where it is in the system (i.e. it depends on $l$, the position). I then average over $n$ replicas of the partition function to remove the $t_l$, and then I'm left with a parameter $\Delta$ which is the variance of the distribution of $t_l$. It is $\Delta$ I then expand in. This is a different question, but I'd like to know the answer to both this and the case without averaging using $t$. – Calvin Aug 30 '11 at 16:07
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To assess whether perturbation theory is applicable in a particular case, you can apply variational perturbation theory:
Incorporate some parameters into your free comparison theory, with corresponding counterterms in the interaction, and do the perturbation theory as a function of these parameters. Typically, the best parameter choice is the one where the responses are least dependent on small changes of the parameters, and how much the responses change give you an idea of how much accuracy you can expect of your calculation.
In some cases, the improvement is drastic; see, e.g., http://en.wikipedia.org/wiki/Variational_perturbation_theory
The renormalizations in quantum field theory are a particular instantiation of variational perturbation theory where doing the variation is essential for getting finite results. See my paper ''Renormalization without infinities - a tutorial'' http://www.mat.univie.ac.at/~neum/ms/ren.pdf
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http://en.wikipedia.org/wiki/Finite_element_analysis
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# Finite element method
(Redirected from Finite element analysis)
Differential equations
Scope
Classification
Operations
Attributes of variables
Relation to processes
• Difference (discrete analogue)
• stochastic
• Delay
Solutions
Solution topics
Mathematicians
In mathematics, finite element method (FEM) is a numerical technique for finding approximate solutions to boundary value problems. It uses variational methods (the Calculus of variations) to minimize an error function and produce a stable solution. Analogous to the idea that connecting many tiny straight lines can approximate a larger circle, FEM encompasses all the methods for connecting many simple element equations over many small subdomains, named finite elements, to approximate a more complex equation over a larger domain.
## History
While it is difficult to quote a date of the invention of the finite element method, the method originated from the need to solve complex elasticity and structural analysis problems in civil and aeronautical engineering. Its development can be traced back to the work by A. Hrennikoff and R. Courant. Although the approaches used by these pioneers are different, they share one essential characteristic: mesh discretization of a continuous domain into a set of discrete sub-domains, usually called elements.
Hrennikoff's work discretizes the domain by using a lattice analogy, while Courant's approach divides the domain into finite triangular subregions to solve second order elliptic partial differential equations (PDEs) that arise from the problem of torsion of a cylinder. Courant's contribution was evolutionary, drawing on a large body of earlier results for PDEs developed by Rayleigh, Ritz, and Galerkin.
The finite element method obtained its real impetus in the 1960s and 70s by the developments of J.H. Argyris and co-workers at the University of Stuttgart, R.W. Clough and co-workers at UC Berkeley, O.C. Zienkiewicz and co-workers at the University of Swansea, and Richard Gallagher [1] and co-workers at Cornell University. Further impetus was provided in these years by available open source finite element software programs. NASA sponsored the original version of NASTRAN, and UC Berkeley made the finite element program SAP IV [2] widely available. A rigorous mathematical basis to the finite element method was provided in 1973 with the publication by Strang and Fix.[3] The method has since been generalized for the numerical modeling of physical systems in a wide variety of engineering disciplines, e.g., electromagnetism, heat transfer, and fluid dynamics; see O.C. Zienkiewicz, R.L.Taylor, and J.Z. Zhu,[4] and K.J. Bathe.[5]
## Technical discussion
### General principles
The subdivision of a whole domain into simpler parts has several advantages:[6]
• Accurate representation of complex geometry
• Inclusion of dissimilar material properties
• Easy representation of the total solution
• Capture of local effects.
A typical work out of the method involves (1) dividing the domain of the problem into a collection of subdomains, with each subdomain represented by a set of element equations to the original problem, followed by (2) systematically recombining all sets of element equations into a global system of equations for the final calculation. The global system of equations has known solution techniques, and can be calculated from the initial values of the original problem to obtain a numerical answer.
A feature of FEM is that it is numerically stable, meaning that errors in the input and intermediate calculations do not accumulate and cause the resulting output to be meaningless. In the first step above, the element equations are simple equations that locally approximates the original complex equations to be studied, where the original equations are often partial differential equations (PDE). To explain the approximation in this process, FEM is commonly introduced as a special case of Galerkin method. The process, in mathematics language, is to construct an integral of the inner product of the residual and the weight functions and set the integral to zero. In simple terms, it is a procedure that minimizes the error of approximation by fitting trial functions into the PDE. The residual is the error caused by the trial functions, and the weight functions are polynomial approximation functions that project the residual. The process eliminates all the spatial derivatives from the PDE, thus approximating the PDE locally with
• a set of algebraic equations for steady state problems,
• a set of ordinary differential equations for transient problems.
These equation sets are the element equations. They are linear if the underlying PDE is linear, and vice versa. Algebraic equation sets that arise in the steady state problems are solved using numerical linear algebra methods, while ordinary differential equation sets that arise in the transient problems are solved by numerically integration using standard techniques such as Euler's method or the Runge-Kutta method.
In the second step above, a global system of equations is generated from the element equations through a transformation of coordinates from the subdomains' local nodes to the domain's global nodes. This spatial transformation includes appropriate orientation adjustments as applied in relation to the reference coordinate system. The process is often carried out by FEM software using coordinates data generated from the subdomains.
FEM is best understood from its practical application, known as finite element analysis (FEA). FEA as applied in engineering is a computational tool for performing engineering analysis. It includes the use of mesh generation techniques for dividing a complex problem into small elements, as well as the use of software program coded with FEM algorithm. In applying FEA, the complex problem is usually a physical system with the underlying physics such as the Euler-Bernoulli beam equation, the heat equation, or the Navier-Stokes equations expressed in either PDE or integral equations, while the divided small elements of the complex problem represent different areas in the physical system.
FEA is a good choice for analyzing problems over complicated domains (like cars and oil pipelines), when the domain changes (as during a solid state reaction with a moving boundary), when the desired precision varies over the entire domain, or when the solution lacks smoothness. For instance, in a frontal crash simulation it is possible to increase prediction accuracy in "important" areas like the front of the car and reduce it in its rear (thus reducing cost of the simulation). Another example would be in numerical weather prediction, where it is more important to have accurate predictions over developing highly nonlinear phenomena (such as tropical cyclones in the atmosphere, or eddies in the ocean) rather than relatively calm areas.
FEM mesh created by an analyst prior to finding a solution to a magnetic problem using FEM software. Colours indicate that the analyst has set material properties for each zone, in this case a conducting wire coil in orange; a ferromagnetic component (perhaps iron) in light blue; and air in grey. Although the geometry may seem simple, it would be very challenging to calculate the magnetic field for this setup without FEM software, using equations alone. FEM solution to the problem at left, involving a cylindrically shaped magnetic shield. The ferromagnetic cylindrical part is shielding the area inside the cylinder by diverting the magnetic field created by the coil (rectangular area on the right). The color represents the amplitude of the magnetic flux density, as indicated by the scale in the inset legend, red being high amplitude. The area inside the cylinder is low amplitude (dark blue, with widely spaced lines of magnetic flux), which suggests that the shield is performing as it was designed to.
### Illustrative problems P1 and P2
We will illustrate the finite element method using two sample problems from which the general method can be extrapolated. It is assumed that the reader is familiar with calculus and linear algebra.
P1 is a one-dimensional problem
$\mbox{ P1 }:\begin{cases} u''(x)=f(x) \mbox{ in } (0,1), \\ u(0)=u(1)=0, \end{cases}$
where $f$ is given, $u$ is an unknown function of $x$, and $u''$ is the second derivative of $u$ with respect to $x$.
P2 is a two-dimensional problem (Dirichlet problem)
$\mbox{P2 }:\begin{cases} u_{xx}(x,y)+u_{yy}(x,y)=f(x,y) & \mbox{ in } \Omega, \\ u=0 & \mbox{ on } \partial \Omega, \end{cases}$
where $\Omega$ is a connected open region in the $(x,y)$ plane whose boundary $\partial \Omega$ is "nice" (e.g., a smooth manifold or a polygon), and $u_{xx}$ and $u_{yy}$ denote the second derivatives with respect to $x$ and $y$, respectively.
The problem P1 can be solved "directly" by computing antiderivatives. However, this method of solving the boundary value problem works only when there is one spatial dimension and does not generalize to higher-dimensional problems or to problems like $u+u''=f$. For this reason, we will develop the finite element method for P1 and outline its generalization to P2.
Our explanation will proceed in two steps, which mirror two essential steps one must take to solve a boundary value problem (BVP) using the FEM.
• In the first step, one rephrases the original BVP in its weak form. Little to no computation is usually required for this step. The transformation is done by hand on paper.
• The second step is the discretization, where the weak form is discretized in a finite dimensional space.
After this second step, we have concrete formulae for a large but finite dimensional linear problem whose solution will approximately solve the original BVP. This finite dimensional problem is then implemented on a computer.
### Weak formulation
The first step is to convert P1 and P2 into their equivalent weak formulations.
#### The weak form of P1
If $u$ solves P1, then for any smooth function $v$ that satisfies the displacement boundary conditions, i.e. $v=0$ at $x=0$ and $x=1$, we have
(1) $\int_0^1 f(x)v(x) \, dx = \int_0^1 u''(x)v(x) \, dx.$
Conversely, if $u$ with $u(0)=u(1)=0$ satisfies (1) for every smooth function $v(x)$ then one may show that this $u$ will solve P1. The proof is easier for twice continuously differentiable $u$ (mean value theorem), but may be proved in a distributional sense as well.
By using integration by parts on the right-hand-side of (1), we obtain
(2)$\begin{align} \int_0^1 f(x)v(x) \, dx & = \int_0^1 u''(x)v(x) \, dx \\ & = u'(x)v(x)|_0^1-\int_0^1 u'(x)v'(x) \, dx \\ & = -\int_0^1 u'(x)v'(x) \, dx = -\phi (u,v), \end{align}$
where we have used the assumption that $v(0)=v(1)=0$.
#### The weak form of P2
If we integrate by parts using a form of Green's identities, we see that if $u$ solves P2, then for any $v$,
$\int_{\Omega} fv\,ds = -\int_{\Omega} \nabla u \cdot \nabla v \, ds = -\phi(u,v),$
where $\nabla$ denotes the gradient and $\cdot$ denotes the dot product in the two-dimensional plane. Once more $\,\!\phi$ can be turned into an inner product on a suitable space $H_0^1(\Omega)$ of "once differentiable" functions of $\Omega$ that are zero on $\partial \Omega$. We have also assumed that $v \in H_0^1(\Omega)$ (see Sobolev spaces). Existence and uniqueness of the solution can also be shown.
#### A proof outline of existence and uniqueness of the solution
We can loosely think of $H_0^1(0,1)$ to be the absolutely continuous functions of $(0,1)$ that are $0$ at $x=0$ and $x=1$ (see Sobolev spaces). Such functions are (weakly) "once differentiable" and it turns out that the symmetric bilinear map $\!\,\phi$ then defines an inner product which turns $H_0^1(0,1)$ into a Hilbert space (a detailed proof is nontrivial). On the other hand, the left-hand-side $\int_0^1 f(x)v(x)dx$ is also an inner product, this time on the Lp space $L^2(0,1)$. An application of the Riesz representation theorem for Hilbert spaces shows that there is a unique $u$ solving (2) and therefore P1. This solution is a-priori only a member of $H_0^1(0,1)$, but using elliptic regularity, will be smooth if $f$ is.
## Discretization
A function in $H_0^1,$ with zero values at the endpoints (blue), and a piecewise linear approximation (red).
P1 and P2 are ready to be discretized which leads to a common sub-problem (3). The basic idea is to replace the infinite dimensional linear problem:
Find $u \in H_0^1$ such that
$\forall v \in H_0^1, \; -\phi(u,v)=\int fv$
with a finite dimensional version:
(3) Find $u \in V$ such that
$\forall v \in V, \; -\phi(u,v)=\int fv$
where $V$ is a finite dimensional subspace of $H_0^1$. There are many possible choices for $V$ (one possibility leads to the spectral method). However, for the finite element method we take $V$ to be a space of piecewise polynomial functions.
### For problem P1
We take the interval $(0,1)$, choose $n$ values of $x$ with $0=x_0<x_1<...<x_n<x_{n+1}=1$ and we define $V$ by:
$V=\{v:[0,1] \rightarrow \Bbb R\;: v\mbox{ is continuous, }v|_{[x_k,x_{k+1}]} \mbox{ is linear for } k=0,...,n \mbox{, and } v(0)=v(1)=0 \}$
where we define $x_0=0$ and $x_{n+1}=1$. Observe that functions in $V$ are not differentiable according to the elementary definition of calculus. Indeed, if $v \in V$ then the derivative is typically not defined at any $x=x_k$, $k=1,...,n$. However, the derivative exists at every other value of $x$ and one can use this derivative for the purpose of integration by parts.
A piecewise linear function in two dimensions.
### For problem P2
We need $V$ to be a set of functions of $\Omega$. In the figure on the right, we have illustrated a triangulation of a 15 sided polygonal region $\Omega$ in the plane (below), and a piecewise linear function (above, in color) of this polygon which is linear on each triangle of the triangulation; the space $V$ would consist of functions that are linear on each triangle of the chosen triangulation.
One often reads $V_h$ instead of $V$ in the literature. The reason is that one hopes that as the underlying triangular grid becomes finer and finer, the solution of the discrete problem (3) will in some sense converge to the solution of the original boundary value problem P2. The triangulation is then indexed by a real valued parameter $h>0$ which one takes to be very small. This parameter will be related to the size of the largest or average triangle in the triangulation. As we refine the triangulation, the space of piecewise linear functions $V$ must also change with $h$, hence the notation $V_h$. Since we do not perform such an analysis, we will not use this notation.
### Choosing a basis
Basis functions vk (blue) and a linear combination of them, which is piecewise linear (red).
To complete the discretization, we must select a basis of $V$. In the one-dimensional case, for each control point $x_k$ we will choose the piecewise linear function $v_k$ in $V$ whose value is $1$ at $x_k$ and zero at every $x_j,\;j \neq k$, i.e.,
$v_{k}(x)=\begin{cases} {x-x_{k-1} \over x_k\,-x_{k-1}} & \mbox{ if } x \in [x_{k-1},x_k], \\ {x_{k+1}\,-x \over x_{k+1}\,-x_k} & \mbox{ if } x \in [x_k,x_{k+1}], \\ 0 & \mbox{ otherwise},\end{cases}$
for $k=1,...,n$; this basis is a shifted and scaled tent function. For the two-dimensional case, we choose again one basis function $v_k$ per vertex $x_k$ of the triangulation of the planar region $\Omega$. The function $v_k$ is the unique function of $V$ whose value is $1$ at $x_k$ and zero at every $x_j,\;j \neq k$.
Depending on the author, the word "element" in "finite element method" refers either to the triangles in the domain, the piecewise linear basis function, or both. So for instance, an author interested in curved domains might replace the triangles with curved primitives, and so might describe the elements as being curvilinear. On the other hand, some authors replace "piecewise linear" by "piecewise quadratic" or even "piecewise polynomial". The author might then say "higher order element" instead of "higher degree polynomial". Finite element method is not restricted to triangles (or tetrahedra in 3-d, or higher order simplexes in multidimensional spaces), but can be defined on quadrilateral subdomains (hexahedra, prisms, or pyramids in 3-d, and so on). Higher order shapes (curvilinear elements) can be defined with polynomial and even non-polynomial shapes (e.g. ellipse or circle).
Examples of methods that use higher degree piecewise polynomial basis functions are the hp-FEM and spectral FEM.
More advanced implementations (adaptive finite element methods) utilize a method to assess the quality of the results (based on error estimation theory) and modify the mesh during the solution aiming to achieve approximate solution within some bounds from the 'exact' solution of the continuum problem. Mesh adaptivity may utilize various techniques, the most popular are:
• moving nodes (r-adaptivity)
• refining (and unrefining) elements (h-adaptivity)
• changing order of base functions (p-adaptivity)
• combinations of the above (hp-adaptivity).
### Small support of the basis
Solving the two-dimensional problem $u_{xx}+u_{yy}=-4$ in the disk centered at the origin and radius 1, with zero boundary conditions.
(a) The triangulation.
(b) The sparse matrix L of the discretized linear system.
(c) The computed solution, $u(x, y)=1-x^2-y^2.$
The primary advantage of this choice of basis is that the inner products
$\langle v_j,v_k \rangle=\int_0^1 v_j v_k\,dx$
and
$\phi(v_j,v_k)=\int_0^1 v_j' v_k'\,dx$
will be zero for almost all $j,k$. (The matrix containing $\langle v_j,v_k \rangle$ in the $(j,k)$ location is known as the Gramian matrix.) In the one dimensional case, the support of $v_k$ is the interval $[x_{k-1},x_{k+1}]$. Hence, the integrands of $\langle v_j,v_k \rangle$ and $\phi(v_j,v_k)$ are identically zero whenever $|j-k|>1$.
Similarly, in the planar case, if $x_j$ and $x_k$ do not share an edge of the triangulation, then the integrals
$\int_{\Omega} v_j v_k\,ds$
and
$\int_{\Omega} \nabla v_j \cdot \nabla v_k\,ds$
are both zero.
### Matrix form of the problem
If we write $u(x)=\sum_{k=1}^n u_k v_k(x)$ and $f(x)=\sum_{k=1}^n f_k v_k(x)$ then problem (3), taking $v(x)=v_j(x)$ for $j=1,...,n$, becomes
$-\sum_{k=1}^n u_k \phi (v_k,v_j) = \sum_{k=1}^n f_k \int v_k v_j dx$ for $j=1,...,n.$ (4)
If we denote by $\mathbf{u}$ and $\mathbf{f}$ the column vectors $(u_1,...,u_n)^t$ and $(f_1,...,f_n)^t$, and if we let
$L=(L_{ij})$
and
$M=(M_{ij})$
be matrices whose entries are
$L_{ij}=\phi (v_i,v_j)$
and
$M_{ij}=\int v_i v_j dx$
then we may rephrase (4) as
$-L \mathbf{u} = M \mathbf{f}.$ (5)
It is not necessary to assume $f(x)=\sum_{k=1}^n f_k v_k(x)$. For a general function $f(x)$, problem (3) with $v(x)=v_j(x)$ for $j=1,...,n$ becomes actually simpler, since no matrix $M$ is used,
$-L \mathbf{u} = \mathbf{b}$, (6)
where $\mathbf{b}=(b_1,...,b_n)^t$ and $b_j=\int f v_j dx$ for $j=1,...,n$.
As we have discussed before, most of the entries of $L$ and $M$ are zero because the basis functions $v_k$ have small support. So we now have to solve a linear system in the unknown $\mathbf{u}$ where most of the entries of the matrix $L$, which we need to invert, are zero.
Such matrices are known as sparse matrices, and there are efficient solvers for such problems (much more efficient than actually inverting the matrix.) In addition, $L$ is symmetric and positive definite, so a technique such as the conjugate gradient method is favored. For problems that are not too large, sparse LU decompositions and Cholesky decompositions still work well. For instance, MATLAB's backslash operator (which uses sparse LU, sparse Cholesky, and other factorization methods) can be sufficient for meshes with a hundred thousand vertices.
The matrix $L$ is usually referred to as the stiffness matrix, while the matrix $M$ is dubbed the mass matrix.
### General form of the finite element method
In general, the finite element method is characterized by the following process.
• One chooses a grid for $\Omega$. In the preceding treatment, the grid consisted of triangles, but one can also use squares or curvilinear polygons.
• Then, one chooses basis functions. In our discussion, we used piecewise linear basis functions, but it is also common to use piecewise polynomial basis functions.
A separate consideration is the smoothness of the basis functions. For second order elliptic boundary value problems, piecewise polynomial basis function that are merely continuous suffice (i.e., the derivatives are discontinuous.) For higher order partial differential equations, one must use smoother basis functions. For instance, for a fourth order problem such as $u_{xxxx}+u_{yyyy}=f$, one may use piecewise quadratic basis functions that are $C^1$.
Another consideration is the relation of the finite dimensional space $V$ to its infinite dimensional counterpart, in the examples above $H_0^1$. A conforming element method is one in which the space $V$ is a subspace of the element space for the continuous problem. The example above is such a method. If this condition is not satisfied, we obtain a nonconforming element method, an example of which is the space of piecewise linear functions over the mesh which are continuous at each edge midpoint. Since these functions are in general discontinuous along the edges, this finite dimensional space is not a subspace of the original $H_0^1$.
Typically, one has an algorithm for taking a given mesh and subdividing it. If the main method for increasing precision is to subdivide the mesh, one has an h-method (h is customarily the diameter of the largest element in the mesh.) In this manner, if one shows that the error with a grid $h$ is bounded above by $Ch^p$, for some $C<\infty$ and $p>0$, then one has an order p method. Under certain hypotheses (for instance, if the domain is convex), a piecewise polynomial of order $d$ method will have an error of order $p=d+1$.
If instead of making h smaller, one increases the degree of the polynomials used in the basis function, one has a p-method. If one combines these two refinement types, one obtains an hp-method (hp-FEM). In the hp-FEM, the polynomial degrees can vary from element to element. High order methods with large uniform p are called spectral finite element methods (SFEM). These are not to be confused with spectral methods.
For vector partial differential equations, the basis functions may take values in $\mathbb{R}^n$.
## Various types of finite element methods
### AEM
The Applied Element Method, or AEM combines features of both FEM and Discrete element method, or (DEM).
Main article: Applied element method
### Generalized finite element method
The Generalized Finite Element Method (GFEM) uses local spaces consisting of functions, not necessarily polynomials, that reflect the available information on the unknown solution and thus ensure good local approximation. Then a partition of unity is used to “bond” these spaces together to form the approximating subspace. The effectiveness of GFEM has been shown when applied to problems with domains having complicated boundaries, problems with micro-scales, and problems with boundary layers.[7]
### hp-FEM
The hp-FEM combines adaptively, elements with variable size h and polynomial degree p in order to achieve exceptionally fast, exponential convergence rates.[8]
### hpk-FEM
The hpk-FEM combines adaptively, elements with variable size h, polynomial degree of the local approximations p and global differentiability of the local approximations (k-1) in order to achieve best convergence rates.
### XFEM
Main article: Extended finite element method
### S-FEM
Main article: Smoothed finite element method
### Spectral methods
Main article: Spectral method
### Discontinuous Galerkin methods
Main article: Discontinuous Galerkin method
### Finite element limit analysis
Main article: Finite element limit analysis
### Stretched grid method
Main article: Stretched grid method
## Comparison to the finite difference method
This section does not cite any references or sources. Please help improve this section by adding citations to reliable sources. Unsourced material may be challenged and removed. (November 2010)
The finite difference method (FDM) is an alternative way of approximating solutions of PDEs. The differences between FEM and FDM are:
• The most attractive feature of the FEM is its ability to handle complicated geometries (and boundaries) with relative ease. While FDM in its basic form is restricted to handle rectangular shapes and simple alterations thereof, the handling of geometries in FEM is theoretically straightforward.
• The most attractive feature of finite differences is that it can be very easy to implement.
• There are several ways one could consider the FDM a special case of the FEM approach. E.g., first order FEM is identical to FDM for Poisson's equation, if the problem is discretized by a regular rectangular mesh with each rectangle divided into two triangles.
• There are reasons to consider the mathematical foundation of the finite element approximation more sound, for instance, because the quality of the approximation between grid points is poor in FDM.
• The quality of a FEM approximation is often higher than in the corresponding FDM approach, but this is extremely problem-dependent and several examples to the contrary can be provided.
Generally, FEM is the method of choice in all types of analysis in structural mechanics (i.e. solving for deformation and stresses in solid bodies or dynamics of structures) while computational fluid dynamics (CFD) tends to use FDM or other methods like finite volume method (FVM). CFD problems usually require discretization of the problem into a large number of cells/gridpoints (millions and more), therefore cost of the solution favors simpler, lower order approximation within each cell. This is especially true for 'external flow' problems, like air flow around the car or airplane, or weather simulation.
## Application
Visualization of how a car deforms in an asymmetrical crash using finite element analysis.[1]
A variety of specializations under the umbrella of the mechanical engineering discipline (such as aeronautical, biomechanical, and automotive industries) commonly use integrated FEM in design and development of their products. Several modern FEM packages include specific components such as thermal, electromagnetic, fluid, and structural working environments. In a structural simulation, FEM helps tremendously in producing stiffness and strength visualizations and also in minimizing weight, materials, and costs.
FEM allows detailed visualization of where structures bend or twist, and indicates the distribution of stresses and displacements. FEM software provides a wide range of simulation options for controlling the complexity of both modeling and analysis of a system. Similarly, the desired level of accuracy required and associated computational time requirements can be managed simultaneously to address most engineering applications. FEM allows entire designs to be constructed, refined, and optimized before the design is manufactured.
This powerful design tool has significantly improved both the standard of engineering designs and the methodology of the design process in many industrial applications.[9] The introduction of FEM has substantially decreased the time to take products from concept to the production line.[9] It is primarily through improved initial prototype designs using FEM that testing and development have been accelerated.[10] In summary, benefits of FEM include increased accuracy, enhanced design and better insight into critical design parameters, virtual prototyping, fewer hardware prototypes, a faster and less expensive design cycle, increased productivity, and increased revenue.[9]
FEA has also been proposed to use in stochastic modelling, for numerically solving probability models. See the references list.[11][12]
## References
1.
2. Strang, Gilbert; Fix, George (1973). An Analysis of The Finite Element Method. Prentice Hall. ISBN 0-13-032946-0.
3. Zienkiewicz, O.C.; Taylor, R.L.; Zhu, J.Z. (2005). The Finite Element Method: Its Basis and Fundamentals (Sixth ed.). Butterworth-Heinemann. ISBN 0750663200.
4. Bathe, K.J. (2006). Finite Element Procedures. Cambridge, MA: Klaus-Jürgen Bathe. ISBN 097900490X.
5. Reddy, J.N. (2005). An Introduction to the Finite Element Method (Third ed.). McGraw-Hill. ISBN 9780071267618.
6. Babuška, Ivo; Banerjee, Uday; Osborn, John E. (June 2004). "Generalized Finite Element Methods: Main Ideas, Results, and Perspective". International Journal of Computational Methods 1 (1): 67–103. doi:10.1142/S0219876204000083.
7. P. Solin, K. Segeth, I. Dolezel: Higher-Order Finite Element Methods, Chapman & Hall/CRC Press, 2003
8. ^ a b c Hastings, J. K., Juds, M. A., Brauer, J. R., Accuracy and Economy of Finite Element Magnetic Analysis, 33rd Annual National Relay Conference, April 1985.
9. McLaren-Mercedes (2006). "Vodafone McLaren-Mercedes: Feature - Stress to impress". Archived from the original on 2006-10-30. Retrieved 2006-10-03.
10. "Methods with high accuracy for finite element probability computing" by Peng Long, Wang Jinliang and Zhu Qiding, in Journal of Computational and Applied Mathematics 59 (1995) 181-189
11. Achintya Haldar and Sankaran mahadan: "Reliability Assessment Using Stochastic Finite Element Analysis", John Wiley & sons.
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http://math.stackexchange.com/questions/124102/what-does-tarski-mean-by-a-tautological-operation-on-a-boolean-algebra?answertab=active
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# What does Tarski mean by a “tautological operation” on a Boolean algebra?
I am reading Part II of Chin and Tarski's "Distributive and Modular Laws in the Arithmetic of Relation Algebras". In the beginning of section 4, the authors say "In general, if $\odot$ is a binary operation which is commutative, associative, and tautological, then both the left and right distributive laws for $\odot$ under $\odot$ are identically satisfied". My question is: what does "tautological" mean in this context?
We are working within a fixed Boolean algebra $B$. $\odot$ is some binary operation on the universe of $B$. He is saying that if $\odot$ is commutative, associative, and "tautological", then $\odot$ is left and right distributive over $\odot$, i.e., for all $x$, $y$, and $z$ in the universe of $B$, $x\odot (y\odot z)=(x\odot y)\odot (x\odot z)$ and $(y\odot z)\odot x=(y\odot x)\odot (z\odot x)$. I can't find the word "tautological" (in reference to a binary relation) used anywhere else.
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Do they use the term anywhere else? Can you give us any more of the text, or context here? – Doug Spoonwood Mar 27 '12 at 1:31
Could it possibly mean idempotent? – Miha Habič Mar 27 '12 at 5:44
## 1 Answer
Yes, it means idempotence (for binary operations). A binary operation $\odot$ is tautological iff $\forall x\ [x \odot x = x]$. Perhaps the name comes from the fact that conjunction and disjunction satisfy this tautologically. I think this is old terminology, but I personally find it somewhat appealing to reserve idempotence for unary operations since the idea is somewhat different. Anyway, here's an example, where it's called the law of tautology (pg. 211):
Frink, Orrin. New Algebras of Logic. The American Mathematical Monthly , Vol. 45, No. 4 (Apr., 1938), pp. 210-219 http://www.jstor.org/stable/2302605
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Thank you, Rachel :). – MathMastersStudent Mar 28 '12 at 2:11
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http://mathoverflow.net/questions/45586?sort=newest
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## The diameter of the Erdös component of the collaboration graph
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This site claims that the diameter of the Erdös component of the collaboration graph in 2004 was 23. What is it now? Is it increasing or decreasing with time? Recall that the vertices of the collaboration graph are mathematicians and two vertices are connected if the mathematicians co-author a paper. MathSci allows one to find the collaborative distance between any two mathematicians. So in principle one can find the diameter just by using MathSci. In general (and more seriously) is there a mathematical theory which describes the growth of "real life" networks like the collaborative graph?
Update 1 A process that I had in mind is something like this. At every step one of the following things can happen.
1. A new vertex $a$ is born with some probability $p$. Usually that vertex is a "student" of some other vertex $x$. We can connect $a$ and $x$ by an edge.
2. A randomly chosen vertex $y$ gets connected with a randomly chosen vertex $z$. But the probability for choosing $z$ is not uniform. Those vertices that are closer to $y$ have more chances getting connected to $y$.
3. A randomly chosen vertex "dies" (with some probability $q$) meaning it does not participate in new edges any longer.
Update 2. Many thanks to Balazs and Joseph for their answers. But the first question still remains: what is the diameter of the Erdös component now?
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en.wikipedia.org/wiki/Preferential_attachment – Steve Huntsman Nov 10 2010 at 19:04
@Steve: Thank you! I meant something like that, but it is not quite the same as the collaborative graph. I explain what kind of process I mean in the Update to the question, because there is not enough space in a comment. – Mark Sapir Nov 10 2010 at 19:10
## 3 Answers
There is a very large literature on this, written by people doing "network science". One of the names you might want to look up is that of Mark Newman, see for example his papers The structure of scientific collaboration networks or Who is the best connected scientist? A study of scientific coauthorship networks as well as Clustering and preferential attachment in growing networks. Another important player in this field is Albert-Laszlo Barabasi, whose group had a model (for the web graph) which is somewhat similar to what you suggest, Emergence of scaling in random networks. Some of this work is reviewed nicely in Statistical Mechanics of Complex Networks. The model by Barabasi et al is studied in mathematical terms by Bollobas-Riordan-Spencer-Tusnady's The degree sequence of a scale-free random graph process.
To answer your specific question on the diameter, I would expect it to be quite small ("six degress of separation"); whether it decreases in time must depend on the relative rate of birth of new vertices versus new edges (collaborations). If you take a finite graph and add completely random links, then already a few links lead to a diameter which is logarithmic in the size of the network. I think this remains the case also if you weigh your edge creation mechanism by the distance of the vertices; I once played with a model where the only new edges were created between 2-step neighbours, which still lead to small diameter.
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Thanks! This might be exactly what I wanted. – Mark Sapir Nov 10 2010 at 23:25
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
In the long run, the diameter will grow. Let k be an upper bound to the number of years between a mathematician's first and last publication. Then nk years after Erdos' last publication, all new nodes in the Erdos component will have Erdos number at least n, so the diameter will be at least n.
(I am assuming here that there will be new nodes, which is not true for very large values of n. Also, I am assuming that k is constant, but only to keep the estimate "nk" simple.)
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Of course, for my question to be non-trivial, one should consider the max. distance between "live" vertices only. – Mark Sapir May 21 2011 at 20:28
The paper "Some analyses of Erdös collaboration graph" by Vladimir Batagelj and Andrej Mrvar (Social Networks Volume 22, Issue 2, May 2000, Pages 173-186) is filled with fascinating data and analysis as of 2000. (Because this is a decade old, it does not answer your question.) Here is their figure of "cliques of the main core," with many recognizable names (if you can read them at this resolution!).
I see many (63) later papers that cite this one, but none that directly update it for this specific Erdös graph.
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@Joseph: Thanks! I did not know about this article. Do they consider the diameter also? I wonder if it was smaller or larger than 23 in 2000. I did not find it in the paper yet. – Mark Sapir Nov 10 2010 at 20:37
@Mark: There were 17 connected components (after removing the Erdös-root!). One of them was large: 6045 authors. The diameter of that component was 12. They do not seem to report the diameter of the connected graph, but it might be implied by data in their tables. – Joseph O'Rourke Nov 10 2010 at 20:46
@Joseph: Thanks for sending me the paper by email. I thought that 12 in the paper was the diameter of the Erdös component. In principle, it could be that for appropriately chosen parameters in "my" model (described in the question) one can get similar results. Then we would be able to predict the future structure of the mathematics community. – Mark Sapir Nov 10 2010 at 22:57
@Mark, p.176: "The diameter of the large component in graph $E'$ is 12 with three diametric pairs of vertices..." Not that it matters. Re predicting the future of the math community: I assume you are kidding! Regardless, collaboration may be stronger in Erdös-fields than in other areas of mathematics. – Joseph O'Rourke Nov 10 2010 at 23:47
Laszlo Lovasz (in the clique above) just received the Kyoto Prize! – Joseph O'Rourke Nov 11 2010 at 20:24
show 3 more comments
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http://www.physicsforums.com/showthread.php?p=3792360
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Physics Forums
## Parallel transport
Assuming that we are working with an infinitessimally small region of a manifold so that we can consider only first order effects, does parallel transport in the absence of torsion necessarily "close the quadrilateral"? What I mean is, if I have two vectors (very small vectors) V and U, and I parallel transported V along U, and vice versa, will the resulting 4 vectors (2 original, 2 parallel transported) form a closed quadrilateral?
I'm trying to get an intuition for torsion, and it seems to me that torsion would be the quantity which prevented the closing of the quadrilateral. I just wanted to confirm this, it seems right to me, but I can't be sure.
PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
Recognitions: Science Advisor What you have is correct as long as the vector fields U and V commute. For non-commuting vector fields on flat space, you would have a pentagon, where the last piece is given by the commutator of the vector fields. So in curved space, torsion measures the failure of this pentagon to close. Or thought of another way, it measures the failure of quadrilaterals to close, after taking into account the fact that vector fields might not commute. There are many ways to write the torsion tensor in various notation systems, but I think the easiest to interpret is $$T(X,Y) = \nabla_X Y - \nabla_Y X - [X,Y]$$
I did try to take account of the lie bracket. Since the Lie bracket operation is valid with no mention of the connection, I would think that it was always the "closer of quadrilaterals". The 5 sides of the pentagon are the vector fields U, V, evaluated at the base point, U, V evaluated at the "tips" of each other, and [U,V] closing the quadrilateral. Is this not always valid? I haven't mentioned anything about parallel transport at this point, so how could a definition of parallel transport affect this picture? When you have parallel transport with no torsion, is it not required that U, V, and the parallel transported U and V (as distinct from the U and V which naturally reside "at each others' tips") close the quadrilateral? This discussion would be much easier if I could draw this out haha.
## Parallel transport
I made a picture to show what I was talking about.
This is my conception of what things look like (infinitessimally of course) in the absence of torsion. I used deltas instead of nablas to denote the covariant derivative because I can't find the nabla symbol in word. I have also suppressed the subscripts. The outer quadrilateral doesn't close, and requires [U,V] to close it, but the inner one does.
One can easily see from that picture then that $\nabla_v u-\nabla_u v=[u,v]$ which implies no torsion. I seem to have derived a "proof", so to speak, from this picture, that it is always the case that the inside quadrilateral will always close in the absence of torsion. In this "proof", though, I just wantonly add and subtract vectors like I was in Euclidean space, so I wasn't sure that it's valid even in the infinitesimal case since I haven't checked how errors scale.
My conception of torsion, then, is that it will open up that inner quadrilateral, leaving the outer pentagon untouched. This will obviously make the above equality that I just wrote not true.
Is this a valid picture?
Recognitions: Science Advisor I think that makes sense. There is another way to think of torsion that makes it clearer why it is called "torsion": First off, any two neighboring tangent spaces at points $x$ and $x + dx$ are related by some matrix in $GL(n)$. Therefore, a connection is really just an object that tells you, for each possible displacement $dx$, what matrix to use in order to "glue" these tangent spaces together. Since the neighboring points are infinitesimally close together, the matrix in $GL(n)$ must be infinitesimally close to the identity; hence the connection is a 1-form that takes values in the Lie algebra $\mathfrak{gl}(n)$. Then we say that $GL(n)$ is the "structure group" of the manifold. There are some various properties you might ask of a manifold that correspond to "reduction of the structure group" to some subgroup of $GL(n)$. For example, if the manifold is orientable, then its structure group must be in $SL(n)$. We could also ask that the connection is "metric compatible", meaning that it preserves the lengths of vectors. In this case, the structure group will be $O(n)$, or $SO(n)$ if the manifold is also orientable. In any case, the Lie algebras of $O(n)$ and $SO(n)$ are the same. Suppose the connection is metric-compatible, such that all parallel transports correspond to some $SO(n)$ rotation. Now let us travel along a geodesic with tangent vector X. Hence $$\nabla_X X = 0$$ As you may notice, the geodesic equation doesn't care about the torsion. Since X appears twice, only the symmetric part of the connection participates, and the torsion is purely the antisymmetric part. Hence if two connections $\nabla$ and $\nabla'$ differ only by torsion, then they have the same geodesics. But what happens to the rest of the tangent space as it is parallel transported along a geodesic? The parallel transport is in $SO(n)$, and the geodesic equation tells us that along this particular path, one vector (namely X) must stay fixed. Hence the rest of the tangent space must rotate under $SO(n-1)$, which is the subgroup of $SO(n)$ that leaves X fixed. So, the rest of the tangent space "twists around" the geodesic. So then this is what torsion means. Connections differing only by torsion must have the same geodesics, and if we travel along a geodesic, the only degrees of freedom left are those that twist around the geodesic. Hence torsion corresponds to the amount of twisting around a geodesic, as we parallel transport along the geodesic. If I have time later, I'll see if I can put this into symbols, so you can see exactly how torsion mathematically relates to twisting around geodesics. Or maybe you can give it a shot.
By "twist around", you mean like rotation in the plane(s) perpendicular to the plane that includes the X vector and a small piece of the curve that is at the point of interest (this plane is sometimes called the osculating plane right)?
Recognitions: Science Advisor No. X is the only thing that has to remain fixed, by the geodesic equation. No vectors orthogonal to X need to remain fixed. For example, in 5 dimensions, X can be fixed, and ALL vectors perpendicular to X can be rotating under SO(4).
Ah, so if we reduce the SO(n) possible rotations to SO(n-1), we actually reduce the possible rotations by several planes right. For example, SO(4) has 6 possible planes of rotation, and SO(3) only has 3. I'm just trying to figure out which planes of rotations are "forbidden". Let's work in 3-D for now because that's easy to visualize. If I don't travel along geodesics, my tangent vectors can rotate in any of the 3 planes. If I do travel along a geodesic in the x-direction, i.e. imagine my geodesic coincides with the x-coordinate axis for some stretch, which plane can I still rotate in? SO(2) implies I can only rotate in 1 plane right. By symmetry this would have to be the y-z plane right? This is the so-called bi-normal plane right (binormal to the osculating plane). Tryin to visualize this hehe.
Ah visualizing this in the shower made me get it. I can rotate in any plane perpendicular to the tangent vector to the geodesic (which is what you said lol, but now I got it). This seems so obvious now that I "saw" the picture in my head. Thanks! In my picture above, Torsion would correspond to my parallel translated vectors possibly twisting (rotating) into or out of the page right. =D It may be a little bit before I can put it into mathematical language though...hehe. Follow up question. MTW states that torsion violates the equivalence principle, how so? If torsion keeps all geodesics the same, I don't see how it would violate the equivalence principle.
Recognitions: Science Advisor The twist rotations will always be in the subspace orthogonal to the geodesic. So whatever collection of planes make up that subspace. The planes that are forbidden are ANY planes that contain the tangent vector to the geodesic.
Yes, I got that now, thanks for clearing that up for me. =]
Recognitions:
Science Advisor
Quote by Matterwave Follow up question. MTW states that torsion violates the equivalence principle, how so? If torsion keeps all geodesics the same, I don't see how it would violate the equivalence principle.
Depends exactly how you define the "equivalence principle". Point particles with no other features will still follow the same geodesics. But the torsion affects the motion of spinning particles...it will contribute to their precession.
Could I use this to then see which situation I'm in: free falling in gravitational field, or free free floating in empty space? Would spinning particles precess in two different ways? Does this mean that Einstein-Cartan theory violates the equivalence principle? (At least the Einstein or Strong versions, perhaps not the weak version)? I believe it is only the weak equivalence principle that has been tested to like 10^-10 precision right.
Recognitions: Science Advisor You'll have to tell me what definitions you're using. People use the words "equivalence principle" to mean a lot of different things.
Basically, for the weak equivalence principle, I just mean m_i=m_g, i.e. objects with same mass fall at the same rate. For the strong equivalence principle, I mean, that there are not experiments which can locally tell if a frame is falling freely in a gravitational field, or floating freely in space.
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http://mathoverflow.net/questions/17461?sort=newest
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Conductor of monomial forms with trivial nebentypus
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Is it true that the conductor of a holomoprhic or a Maass cusp form with trivial nebentypus corresponding to a two-dimensional dihedral representation (over $\mathbb{Q}$ )is non-square-free?
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The conductor could be 1, right? Isn't that squarefree? Are you talking about 2-dimensional representations? Of an arbitrary number field? – Kevin Buzzard Mar 8 2010 at 12:03
A Dirichlet character is monomial and can have prime conductor. I think your question is too terse/ambiguous currently. – Kevin Buzzard Mar 8 2010 at 13:14
There's a weight 1 cuspidal modular form of level 23, whose associated Galois representation is induced from an unramified character of Q(sqrt(-23)). OK so I really give in now ;-) – Kevin Buzzard Mar 8 2010 at 13:28
Sorry. I was too careless in phrasing the question. There are certainly many dihedral forms of prime conductors. I forgot to add the condition that I believe should give the desired conclusion. I have modified the question. – Idoneal Mar 8 2010 at 17:02
Is this a p-adic or a complex representation? I am envisaging the 2-adic Tate module of X_0(32), which is "dihedral" for some values of dihedral, and has "conductor 1" for some values of conductor. But apart from what, read what Emerton said about the Steinberg: it's the only smooth irred rep of GL_2(Q_p) of conductor p and with unramified central character, and it can't show up for global reasons, so if you have trivial det you can't have p dividing the conductor eactly once in a dihedral setting. – Kevin Buzzard Mar 8 2010 at 20:51
2 Answers
Thinking about Idoneal's question to Emerton about what is being used about the Steinberg, it's not just standard facts about the Steinberg one needs via this approach, but also local-global compatibility. The standard facts about the Steinberg are also easily proved if one uses local Langlands (i.e. works on the Galois side). This made me realise that in fact one can pull off the entire argument on the Galois side! Let's assume you're talking about complex 2-dimensional representations of the Galois group. Emerton has already observed that they must be ramified at some prime $p$, so it suffices to prove that $p^2$ divides the conductor. But now say
`$\rho:Gal(\overline{\mathbf{Q}}_p/\mathbf{Q}_p)\to GL_2(\mathbf{C})$`
has trivial determinant and conductor $p$. One instantly gets a contradiction: the inertial invariants can't be 0-dimensional because already this would mean the conductor is at least $p^2$, and they can't be 1-dimensional because if $i\in I_p$ has one eigenvalue 1 then the other eigenvalue must be 1 too, and $\rho(i)$ is diagonalisable and hence trivial, so the inertial invariants must be 2-dimensional but now the conductor must be 1. This answers the question completely without recourse to the smooth representation theory side of things and is surely the easiest approach to the question.
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Yes, the Galois side is a slightly more familiar territory for me and this proof is quite clear. Thanks. I shall appreciate very much, nevertheless, if you (or anyone) can suggest some references for the representation theory facts used in the earlier arguments. I looked up Bump's text book but that seems not to be adequate. – Idoneal Mar 9 2010 at 11:51
In some sense I'm not sure that the representation theory facts shed any more light on the situation---they're just a translation of the Galois comments above to the automorphic side. For example, the statement that the Steinberg is (up to unram twist) the only smooth admiss rep of GL_2(Q_p) with unramified central char and conductor p is just a trivial consequence of the local Langlands conjectures and a calculation on the Galois side very analogous to the above (but with Weil-Deligne groups instead of Galois groups, so you're allowed a monodromy operator). Read any intro to LL for GL_n/Q_p! – Kevin Buzzard Mar 9 2010 at 12:40
You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Just to augment Kevin's series of comments: I think that the conductor of the induction of some character $\chi$ over a quadratic field to $\mathbb Q$ would normally equal $D N(C)^2$, where $D$ is the discriminant of the quadratic field, $C$ is the condutor of the character (an ideal in the quadratic field) and $N$ is the norm from the quadratic field to $\mathbb Q$. E.g. in Kevin's $23$ example, one inducing a character of conductor 1 from $\mathbb Q(\sqrt{-23})$, so the conductor is $23$. [Added in response to an edit in the question: This form has nebentypus equal to the Legendre symbol mod 23.]
In the Maass case one should be able to do something similar, by e.g. choosing a prime $p \equiv 1 mod 4$ such that $\mathbb Q(\sqrt{p})$ has non-trivial class group, and then inducing a non-trivial character of conductor 1. [Added in response to an edit in the question: Such examples will have nebentypus equal to the Legendre symbol mod p, I think.]
[Added in response to an edit in the question:] Based on the formula above for the conductor, I think that to have square free conductor one will need to induce a character with trivial conductor, i.e. coming from the (strict) class group. I think that such an induction will always have non-trivial nebentypus, though. (The key point being that if $H/{\mathbb Q}$ is the stict Hilbert class field of the real quadratic field, then this is a generalized dihedral extension.)
Another argument, pointed out me by a colleague, is that if the conductor is square fee and the nebentypus is trivial, then all the local factors of the automorphic representation at primes in the conductor are Steinberg, which is not possible for the induction of a character.
[One more remark:] It seems to me that if we replace $\mathbb Q$ by some well-chosen number field $F$, then it will be possible to find an unramified quadratic extension $E$ of $F$ such that $E$ in turn admits a degree $4$ extension $K$, everywhere unramified, so that $K$ over $F$ (a degree 8 extension) is Galois with the quaternion group as Galois group (as opposed to a dihedral group). I think if we then take the corresponding order 4 ideal class character of $E$ and induce it to $F$, we get a monomial representation of $F$ with trivial determinant whose conductor is equal to one (and in particular, is square-free).
In other words, one is a little bit "lucky" in the $\mathbb Q$-case that Hilbert class fields of real quadratic fields are dihedral over $\mathbb Q$.
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I think this answers my question although I don't understand it fully due to my ignorance about Steinberg representations. Can someone point out a reference for the fact being used about Steinberg representation? As I understand, the crucial fact is the uniqueness of Steinberg representation that Buzzard has mentioned in his last comment. – Idoneal Mar 9 2010 at 6:16
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http://mathhelpforum.com/advanced-statistics/48581-urn-problem-conditional-probability.html
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# Thread:
1. ## Urn problem - conditional probability
Consider three urns, A, B, and C. Urn A contains 6 white balls and 4 black balls, urn B contains 2 white balls and 6 black balls, and urn C is empty. Two balls are drawn at random from each one of the urn A and urn B, and are placed into urn C. (4 balls are now in urn C.) Then a ball is drawn at random from urn C. What is the probability the ball drawn from urn C is black?
I've figured out that P(K\A) = 2/3 and P(K\B) = 5/6. At least I think so. But now I don't know what to do with that. (My backslashes are for vertical lines I don't know how to type that.)
2. Hello, ban26ana!
I wouldn't call this "conditional probability" . . .
There are three urns, A, B, and C.
Urn A contains 6 white balls and 4 black balls,
urn B contains 2 white balls and 6 black balls,
and urn C is empty.
Two balls are drawn at random from each of urn A and urn B,
and are placed into urn C.
Then a ball is drawn at random from urn C.
What is the probability the ball drawn from urn C is black?
This takes a lot of preliminary work . . . bear with me.
Urn A
There are: . ${10\choose2} = 45$ outcomes.
$\begin{array}{cccccccc}<br /> WW\!: & {6\choose2} = 15\text{ ways} & P_A(WW) &=& \frac{15}{45} &=& \frac{5}{15} \\ \\[-3mm]<br /> WB\!: & 6\cdot 4 = 24\text{ ways} & P_A(WB) &= & \frac{24}{45} &=& \frac{8}{15} \\ \\[-3mm]<br /> BB\!: & {4\choose2} = 6\text{ ways} & P_A(BB) & = & \frac{6}{45} &=& \frac{2}{15}\end{array}$
Urn B
There are: . ${8\choose2} = 28$ outcomes.
$\begin{array}{cccccccc}<br /> WW\!: & {2\choose2} = 1\text{ way} & P_B(WW) &=& \frac{1}{28} \\ \\[-3mm]<br /> WB\!: & 2\cdot6 = 12\text{ ways} & P_B(WB) &=& \frac{12}{28} \\ \\[-3mm]<br /> BB\!: & {6\choose2} = 15\text{ ways} & P_B(BB) &=& \frac{15}{28}<br /> \end{array}$
Urn C
$WWWW\!:\;\;\begin{Bmatrix}<br /> P_A(WW)\cdot P_B(WW) &=&\frac{5}{15}\cdot\frac{1}{28} &=&\frac{5}{420} \end{Bmatrix} \qquad P(W,W,W,W) \:=\:\frac{5}{420}$
$WWWB\!:\;\;\begin{Bmatrix}<br /> P_A(WW)\cdot P_B(WB) & =&\frac{5}{15}\cdot\frac{12}{28} &=& \frac{60}{420} \\ \\[-3mm]<br /> P_A(WB)\cdot P_B(WW) &=& \frac{8}{15}\cdot\frac{1}{28} &=& \frac{8}{420} \end{Bmatrix} \qquad P(W,W,W,B) \:=\:\frac{68}{420}$
$WWBB\!:\;\;\begin{Bmatrix}<br /> P_A(WW)\cdot P_B(BB) &=& \frac{5}{15}\cdot \frac{15}{28} &=& \frac{75}{420} \\ \\[-3mm]<br /> P_A(BB)\cdot P_B(WW) &=& \frac{2}{15}\cdot\frac{1}{28} &=& \frac{2}{420} \\ \\[-3mm]<br /> P_A(WB)\cdot P_B(WB) &=& \frac{8}{15}\cdot\frac{12}{28} &=& \frac{96}{420}\end{Bmatrix} \qquad P(W,W,B,B) \:=\:\frac{173}{420}$
$WBBB\!:\;\;\begin{Bmatrix}<br /> P_A(WB)\cdot P_B(BB) &=& \frac{8}{15}\cdot\frac{15}{28} &=& \frac{120}{420} \\ \\[-3mm]<br /> P_A(BB)\cdot P_B(WB) &=& \frac{2}{15}\cdot\frac{12}{28} &=&\frac{24}{420} \end{Bmatrix} \qquad P(W,B,B,B) \:=\:\frac{144}{420}$
$BBBB\!:\;\;\begin{Bmatrix}<br /> P_A(BB)\cdot P_B(BB) &=& \frac{2}{15}\cdot\frac{15}{28} &=& \frac{30}{420}\end{Bmatrix} \qquad P(B,B,B,B) \:=\:\frac{30}{420}$
Now in each of these five cases, what is the probability of drawing a black ball?
$\begin{array}{cccccccc}<br /> P(WWWW) \:=\: \frac{5}{420}, & P(B) \:=\: 0 && P(WWWW \wedge B) &=& \frac{5}{420}\cdot0 &=& 0 \\ \\[-3mm]<br /> P(WWWB) \:=\:\frac{68}{420}, &P(B) \:=\:\frac{1}{4} && P(WWWB \wedge B) &=& \frac{68}{420}\cdot\frac{1}{4} &=& \frac{68}{1680} \\ \\[-3mm]<br /> P(WWBB) \:=\:\frac{173}{420}, & P(B) \:=\:\frac{2}{4} && P(WWBB \wedge B) &=&\frac{173}{420}\cdot\frac{2}{4} &=& \frac{346}{1680} \\ \\[-3mm]\end{array}$
$\begin{array}{cccccccc}<br /> P(WBBB) \:=\:\frac{144}{420}, & \;\;\;P(B) \:=\:\frac{3}{4} && P(WBBB \wedge B) &\;\;=&\frac{144}{420}\cdot\frac{3}{4} &=& \frac{432}{1680} \\ \\[-3mm]<br /> P(BBBB) \:=\:\frac{30}{420}, & \;\;\;P(B) \:=\:\frac{4}{4} && P(BBBB \wedge B) &\;\;=&\frac{30}{420}\cdot\frac{4}{4} &=& \frac{120}{1680}\end{array}$
Therefore: . $P(B) \;=\;\frac{68}{1680} + \frac{346}{1680} + \frac{432}{1680} + \frac{120}{1680} \;=\;\frac{966}{1680} \;=\;\boxed{\frac{161}{280}}$
I need a nap . . .
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3. Thank you very much. Really, thank you. When solving a problem like this, what if it is impossible to list all the possible outcomes? What if there were 30 outcomes for each step. You obviously wouldn't list them all. Is there a way to solve it for that, or are you stuck doing this huge amount of work for every problem.
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http://www.blkmage.net/tag/an-application-of-this-math-is-other-math/
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# Suugaku Girl supplementary handout for chapter 1: Sequences
Posted on August 21, 2011 by
That really isn't enough terms to identify as Fibonacci
Suugaku Girl is a manga about people who really like math. The great thing about it is that it contains a substantial amount of math. This is also great because it will give me plenty to blog about.
The first chapter of Suugaku Girl is about the main character’s initial meeting with Milka, our socially retarded genius meganekko. She just starts spouting out numbers and, for whatever reason, he feels compelled to guess what comes next. And that is apparently the beginning of this love story.
In the chapter, we’re introduced to a few sequences, some of which are famous and some of which you might not have considered a sequence. It’ll probably help to understand what a sequence is beyond just being a bunch of numbers in a prescribed order.
Formally, we define a sequence as a function $a:S\to R$. The set $S$ is basically our index and is $\{1,2,…,n\}$ if $a$ is a finite sequence or the set of natural numbers $\mathbb{N}$ if we’re dealing with an infinite sequence. Normally, functions are written as $a(n)$, but, as was alluded to before, we refer to terms by index as in $a_n$.
However, $R$ can be any set. In the case of Suugaku Girl, it seems to be sticking with integers, but we can have sequences of bitstrings, vectors, complex numbers, functions, or whatever. What this also means is that a sequence doesn’t necessarily need to have a “pattern”, but can really be any ordered list of numbers (or functions or vectors or…).
Milka also brings up the idea of infinite sequences. A lot of the time, people will try to “solve” a sequence by completing it when they’re given only the first few terms. But, like Milka suggests, what they’re doing in that case is assuming that the rest of the sequence goes on as initially implied. Really, we can define any sequence we like with any behaviour we like. Again, remember that a sequence can be anything we want it to be. In fact, the sequence that Milka defines using the digits of π is kind of like that in that it’s completely arbitrary and doesn’t really have the kind of sequence definitions we’re used to seeing.
For finite sequences, we can just list all of the terms of our sequence like $(1,1,1,1,1,1000000,1)$. Obviously, we can’t do that for an infinite sequence. Luckily, a lot of the time we define sequences that have some sort of useful pattern that we can represent in a succinct way. Sometimes, like the digits of π sequence, this is harder.
We can try to formally define all of the sequences that were given in the manga. For instance, the Fibonacci sequence $(a_n)_{n=1}^\infty$ is commonly defined as $a_n = a_{n-2} + a_{n-1}$, but we have to give the first two terms $a_1 = 1, a_2 = 1$. The second sequence (which we’ll call $(b_n)_{n=1}^\infty$) takes a bit more work to define. We’ll need to define $p_n$ to be the $n$th prime number and then we can define $b_n = p_n \times p_{n+1}$. The third sequence $(c_n)_{n=1}^\infty$ is easy, it’s just $c_n = n^n$. And we can formalize the last one, which I’ll call $(d_n)_{n=1}^\infty$, just like the first two with a few more words. We let $\pi = q_1q_2q_3\cdots$ be the decimal expansion of π. Then $d_n = 2\cdot q_n$.
So that’s all fine, but what exactly are sequences used for? I’m pretty sure everyone’s learned about arithmetic and geometric sequences in grade school. Obviously, we can study sequences and their behaviour on their own. We can talk about whether they increase or decrease or how fast they grow or whether they converge. Apart from that, I don’t remember seeing sequences used for something besides sequences until analysis.
Analysis is basically the field of pure math that formalizes the concepts that we’re introduced to in calculus and generalizes them to spaces. Limits are a fundamental idea in calculus and analysis and these are defined by how a sequence converges. And this is where those weird sequences of vectors or functions comes into play, since we can talk about the convergence of a sequence of vectors or a sequence of functions in these other spaces that we want to do analysis in.
That’s probably the easiest example of an “application” of sequences. For myself, over the last few months I’ve read about automatic sequences, which are sequences that can be generated by a deterministic finite automaton. This gives us a way to relate automata theory to number theory and algebra. For instance, once we have k-automatic sequences, we can talk about k-regular sequences and come up with power series with respect to certain rings and fields and bla bla bla.
If you ever want to find out what crazy sequence you might have a part of, check out the Online Encyclopedia of Integer Sequences.
Posted in Anime | |
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http://diracseashore.wordpress.com/2008/11/05/first-quantization-first-pass/
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# Shores of the Dirac Sea
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## First quantization, first pass
November 5, 2008 by Moshe
Suppose you want to solve a linear partial differential equation of the form $O \psi(x) = j(x)$, which determines some quantity $\psi(x)$ in terms of its source $j(x)$. Here x could stand for possibly many variables, and the differential operator $O$ can be pretty much anything. This is a very general type of problem, not even specific to physics. An example in physics could be the Klein-Gordon equation, or with some more bells and whistles the Maxwell equation, which determines the electric and magnetic fields.
Let us replace this problem with the following equivalent one. If we find a function $\psi(x,s)$ such that:
$\frac{\partial \psi}{\partial s} +O \psi =0$
with the initial condition $\psi(x, s=0) = j(x)$, and assuming the regularity condition $\psi (x,s= \infty) \rightarrow 0$, then it is easy to see that the function $\psi(x) = \int_0^\infty \psi(x,s) ds$ satisfies the original equation we set out to solve.
Now, this new equation for $\psi(x,s)$ looks kind of familiar, if we are willing to overlook a few details. If we wish, we can think about $\psi(x,s)$ as a time dependent wave function, with the parameter s playing the role of time. The equation for $\psi(x,s)$ could then be interpreted as a Schrödinger equation, with the original operator $O$ playing the role of the Hamiltonian. We are ignoring a few issues to do with convergence, analytic continuation, and the related fact that the Schrödinger equation is complex, and the one we are discussing is not. Never mind, these are subtleties which need to be considered at a later stage.
The point is that we can now use any technique we learned in quantum mechanics to solve the original equation – path integral, canonical quantization, you name it. We can talk about the states $|x\rangle$ and the Hilbert space they form, Fourier transform to get another basis for that Hilbert space, even discuss “time” evolution (that is, the dependence of various states on the auxiliary parameter s). We can get the state $\psi(x,s)$ by summing over all paths of a “particle” with an appropriate worldline action and boundary conditions. Depending on the problem, we may be interested in various (differential) operators acting on $\psi(x,s)$, and they of course do not commute, resulting in uncertainty relations. You get the picture.
This technique is sometimes called first quantization, or Schwinger proper time method, or heat Kernel expansion. Whatever you call it, it has a priori nothing to do with quantum mechanics, there are no probabilities, Planck constant or any wavefunctions in any real sense. At this point we may be discussing the financial markets, population dynamics of bacteria, or simply classical field theory.
In the second pass, we can apply this idea to linear fields, generating solutions to various linear differential equations. Some of those equations are Lorentz invariant (Klein-Gordon, Dirac, Maxwell equations), but they have nothing to do with quantum mechanics, despite the original way they were referred to as “relativistic wave equations”. Once we add spin to the game, we start having the fascinating structures of (worldline) fermions and supersymmetry (not to be confused with spacetime fermions and supersymmetry), and we are also in a good shape to make the leap from classical field theory to classical string theory. Maybe I’ll get to that sometime…
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Posted in high energy physics, Physics, quantum fields, Quantum Gravity, relativity | 6 Comments
### 6 Responses
1. There are a few more properties that make the heat-kernel so attractive: Often, you are interested in the special case where j is the delta function, in that case psi is called the Green function of O or the propagator for O. Of course, delta is not really a function but more singular, something mathematicians call a distribution. Similarly, in that case psi is also a distribution.
However, for any s>0, everything is smooth enough and psi(x,s) is actually an ordinary function of x.
Furthermore, if you have some sort of local symmetry like in gauge theory or in gravity, it turns out you can find gauge invariant expressions for psi and thus you do not have to use gauge dependent stuff like potentials.
Finally, it turns out, the one loop effective action can be expressed as integral ds/s psi(0,s). In this expression, all UV divergences show up only as divergences of the indefinite integral for s->0. The singularity can then very elegantly computed in a gauge invariant way using some recursive relation for psi.
2. As far as I understand, there are two approaches to QFT:
a) we can take a free relativistic particle, write the action for it and quantize it. That’s first quantization and that is how string is quantized (starting from the Nambu-Goto action).
b) we can start with a field from the very beginnin (second quantization).
Nice thing about (b) is that I can understand easily the effect of interactions, i.e., treat processes involving several particles, such as particle production in a strong external field, etc.
How does one deal with multiparticle processes in the first quantization?
Cheers
3. Dmitry, when you first-quantize the relativistic particle, you generate solutions to the classical Klein-Gordon equation. With worldline fermions you could do the same for the Dirac equation, and presumably also for the Maxwell and linearized Einstein equations (though I would be interested in seeing a reference to the latter, since I am curious how gauge invariance is dealt with). In all those cases this is “fake quantization”, you solve some auxiliary Schrodinger equation for the purpose of finding classical solutions.
Interactions in this formalism come from including explicitly vertices, then building tree graphs from propagators and those vertices. Quantum corrections come when including loops, exactly as in string perturbation theory.
The main advantage for “second” (i.e. field) quantization (annoying counting scheme for quantizations notwithstanding) is that non-perturbative physics is accessible. Unfortunately this is not yet fully accessible in string theory.
4. on November 8, 2008 at 8:09 pm Giotis
Hi Moshe
BTW what is the current status of the second quantization of the string (i.e. string field theory)?
It still attracts the attention of the string community? It is argued that is the only hope for a consistent non-perturbative formulation of a background independent string theory. I have the impression though that after ADS/CFT, the interest for this field is declining. Am i right?
BR
5. Hi Giotis, there is a constant interest in open string field theory, which in my eyes is similar in many ways to second quantized gauge theory (on a fixed manifold). As far as I know, not much is going on in the closed string field theory front, but maybe I am wrong.
My personal opinion, for what its worth, is that this is the wrong way to go, the approach is too tied to perturbation theory (and as a result, among other things, it is not background independent). This opinion only gets stronger with each example of fully non-perturbative (and background independent) formulation of string theory, like ads/cft, matrix theory, etc. etc., in all of them closed SFT seems to miss the interesting physics.
6. I hope I haven’t missed all the interesting discussion here, or maybe I should wait for the “second pass”.
In any case, the degrees of freedom in matrix models (probably “old” matrix models) allow for the description of a variable number of fundamental perturbation objects, already at the classical level (in the form of block-diagonal sectors). Is this an example of a system with particle-like excitations that doesn’t follow the first/second quantization pattern, or does the fact that the system have a classical description in terms of infinite-dimensional matrices hint at a hidden first quantization?
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http://math.stackexchange.com/questions/276685/does-a-graph-with-0-vertices-count-as-simple
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# Does a graph with $0$ vertices count as simple?
Does a graph with $0$ vertices count as a simple graph?
Or does a simple graph need to have a non-empty vertex set?
Thanks!
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That would really depended on the definition you're using on your course, book or whatever. Sometimes authors don't even consider the "empty graph" to be a graph, in that case it can't be a simple graph. – Git Gud Jan 13 at 1:55
## 2 Answers
It is typical to refer to a graph with no vertices as the null graph. Since it has no loops and no parallel edges (indeed, it has no edges at all), it is simple.
That said, if your present work finds you writing "Such and such is true for all simple graphs except the null graph", then it could be a good idea to announce at the beginning of your document that you will not consider the null graph to be simple.
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The term empty graph is also very often applied to graphs with no edges and a positive number number of vertices. – Chris Godsil Jan 14 at 19:31
@ChrisGodsil Thank you for the remark. It seems the more common term for a zero-vertex graph is "null graph". I've corrected this in my answer. – Austin Mohr Jan 14 at 19:51
By wikipedia:
Simple graph
As opposed to a multigraph, a simple graph is an undirected graph that has no loops (edges connected at both ends to the same vertex) and no more than one edge between any two different vertices. In a simple graph the edges of the graph form a set (rather than a multiset) and each edge is a distinct pair of vertices. In a simple graph with n vertices every vertex has a degree that is less than n (the converse, however, is not true — there exist non-simple graphs with n vertices in which every vertex has a degree smaller than n).
It satisfies all conditions, so, it should be.
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http://mathforum.org/mathimages/index.php?title=Henon_Attractor&diff=32961&oldid=6905
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# Henon Attractor
### From Math Images
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| - | {{Image Description | + | {{Image Description Ready |
| | |ImageName=Henon Attractor | | |ImageName=Henon Attractor |
| - | |Image=Henon.jpg | + | |Image=HenonMain.jpg |
| - | |ImageIntro=This image is a variation of the Henon Attractor (named after astronomer Michel Henon), which is a fractal in the division of the chaotic strange attractor. The Henon Attractor emerged from Henon's attempt to model the orbits of celestial objects. | + | |ImageIntro=This image is a Henon Attractor (named after astronomer and mathematician Michel Henon), which is a fractal in the division of the chaotic strange attractor. |
| | |ImageDescElem= | | |ImageDescElem= |
| | | + | The Henon Attractor is a special kind of fractal that belongs in a group called [[Strange Attractors]], and can be modeled by two general equations. The Henon Attractor is created by applying this system of equations to a starting value over and over again and graphing each result. |
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| - | <div style="width:265px; height:225px; float:left;">{{#eqt: HenonAnimation.avi|250|left}}</div> | | |
| - | The Henon Attractor is a special kind of fractal that belongs in a group called [[Strange Attractors]]. | | |
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| | | + | ===Making the Henon Attractor=== |
| | | + | [[Image:HenonAnimation.gif|thumb|250px|left]] |
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| - | A characteristic of this strange fractal is that it is drawn irregularly. The Henon Attractor is described by two equations. Let us say that we take a starting value (x,y) and apply the equations to the starting values and then the resulting outcome over and over (a process called iteration). If we plot every outcome from this iteration one at a time, we would observe that the points jump from one random location to another within the image. If you take a look at the animation, you can see the irregularity of a number of plotted points. Eventually, the individual points become so numerous that they appear to form lines and an image emerges. | + | Say we took a single starting point (x,y) and plotted it on a graph. Then, we applied the <balloon title="One equation for the x-value and one equation for the y-value.">two</balloon> Henon Attractor equations to the initial point and emerged with a new point that we graphed. Next, we took this new point and again applied the two equations to it and graphed the next new point. If we continued to apply the two equations to each new point in a process called '''iteration''' and plotted every outcome from this iteration, we would create a Henon Attractor. Click here to learn more about [[Iterated Functions|iterated functions]]. |
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| - | This image results from an [[Iterated Functions|iterated function]], meaning that the equations that describe it can be applied to itself an infinite amount of times. In fact, if you magnify this image, you would find that the lines (really many, many points) that appear to be single lines on the larger image are actually sets or bundles of lines, who, if magnified closer, are bundles of lines and so on. This property is called '''self-similarity''', which means that even as you look closer and closer into the image, it continues to look the same. In other words, the larger view of the image is similar to a magnified part of the image. | + | Furthermore, if we plotted each outcome one at a time, we would observe that the points jump from one random location to another within the image. If you take a look at the animation, you can see the irregularity of the plotted points. Eventually, the individual points become so numerous that they appear to form lines and an image emerges. |
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| - | <gallery caption="Magnification of the Henon Attractor" widths="220px" heights="150px" perrow="4"> | + | ===Magnification of the Henon Attractor=== |
| | | + | <gallery caption="" widths="220px" heights="150px" perrow="4"> |
| | Image:HenonMag1.png|1X | | Image:HenonMag1.png|1X |
| | Image:HenonMag2.png|8X | | Image:HenonMag2.png|8X |
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| | Image:HenonMag4.png|512X | | Image:HenonMag4.png|512X |
| | </gallery> | | </gallery> |
| | | + | |
| | | + | |
| | | + | If you magnify this image, you would find that the lines (really many, many points) that appear to be single lines on the larger image are actually sets or bundles of lines, that, if magnified closer, are bundles of lines and so on. This property is called '''self-similarity''', which means that even as you look closer and closer into the image, it continues to look the same. In other words, the larger view of the image is similar to a magnified part of the image. |
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| | | + | |
| | | + | ===History of the Henon Attractor=== |
| | | + | Michel Henon was a French mathematician and astronomer who developed the Henon Attractor in the 1970s. At that time, Henon was interested in dynamical systems and especially the complicated orbits of celestial objects. The Henon Attractor emerged from Henon's attempt to model the chaotic orbits of celestial objects (like stars) in the mist of a gravitational force. |
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| | |MiddleSchool=Yes | | |MiddleSchool=Yes |
| | |HighSchool=Yes | | |HighSchool=Yes |
| - | |ImageDesc= The image featured at the top of this page is a modified version of the Henon Attractor that will be described below. | + | |ImageDesc= |
| - | | + | |
| | ==Fractal Properties== | | ==Fractal Properties== |
| - | |1=[[Image:HenonZoomAnimation.gif|150px]] | | |
| - | |2=[[Image:HenonZoomAnimation.gif|right|thumb|Zooming in on the Henon Attractor]] | | |
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| - | The shape of the Henon Attractor is often described as a smooth fractal in one direction and as a [[Cantor Set]] in another direction. If we zoom into the Henon Attractor near the doubled-tip of the fractal (as seen in the animation), we can see that the points form layers of lines that appear to resemble the Canter Set. If we follow the attractor backwards from the doubled-tip, we can see that the fractal is more smooth and contains less bundles of lines. | + | The Henon Attractor is often described as being similar to the [[Cantor Set]]. Let us zoom into the Henon Attractor near the doubled-tip of the fractal (as seen in the animation). We can see that as we continue to magnify the lines that form the structure of the Henon Attractor, these lines become layers of increasingly deteriorating lines that appear to resemble the Canter Set. |
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| | The [[Fractal Dimension]] of the Henon Attractor is not calculable using a single <balloon title = "load:equation"> equation</balloon><span id="equation" style="display:none"><math>D = \frac{log(n)}{log(e)}</math></span>, but it is estimated to be about 1.261. | | The [[Fractal Dimension]] of the Henon Attractor is not calculable using a single <balloon title = "load:equation"> equation</balloon><span id="equation" style="display:none"><math>D = \frac{log(n)}{log(e)}</math></span>, but it is estimated to be about 1.261. |
| - | }} | | |
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| | ==Chaotic System== | | ==Chaotic System== |
| - | |1=[[Image:Henon2.jpg|150px]] | | |
| - | |2=[[Image:Henon2.jpg|thumb|200px|left|Original Henon Attractor , a = 1.4, b = 0.3]] | | |
| - | The Henon system can be described as [[Chaos|chaotic]] and random. However, the system does have structure in that its points settle very close to an underlying pattern called a '''chaotic attractor''' or '''basin of attraction'''. The Henon Attractor can be described by the following equations: | | |
| | | | |
| | | + | [[Image:Henon2.jpg|right|thumb|Original Henon Attractor, a = 1.4, b = 0.3]] |
| | | + | The Henon system can be described as [[Chaos|chaotic]] and random. However, the system does have structure in that its points settle very close to an underlying pattern called a <balloon title="In a chaotic system like this fractal, over time plotted points are evolving towards (or being 'attracted' to) a region that is called the chaotic attractor.">chaotic attractor</balloon>. The basic Henon Attractor can be described by the equations, where <math>x_n</math> is the x-value at the nth iteration. |
| | | | |
| - | <math>x_{n+1} = y_n + 1 - ax^2_n</math> | + | ::<math>x_{n+1} = y_n + 1 - ax^2_n</math> |
| | | | |
| | | + | ::<math>y_{n+1} = bx_n\,</math> |
| | | | |
| - | <math>y_{n+1} = bx_n\,</math> | + | Astronomer Michel Henon created the original Henon Attractor using the values ''a'' = 1.4 and ''b'' = 0.3 and starting point (1,1). These are also the values used by the artist to create the featured image at the top of the page. However, by changing the values of ''a'' and ''b'', we can obtain Henon Attractors that look slightly different. |
| | | | |
| | | | |
| - | The Henon Attractor uses the values a = 1.4 and b = 0.3 and begin with a starting point (1,1). | + | ==Changing "a" and "b"== |
| - | }} | + | Although the original Henon Attractor uses ''a'' = 1.4 and ''b'' = 0.3, we can alter those values slightly to produce various-looking Henon Attractors. However, the values of ''a'' and ''b'' are limited to a small range of values, outside of which the fractal ceases to resemble the Henon Attractor. |
| | | + | |
| | | + | Here are some more examples of Henon Attractors with different ''a'' and ''b'' values. |
| | | + | |
| | | + | <gallery widths="250px" heights="250px" perrow="3" align ="center"> |
| | | + | Image:Henon.jpg|a = 1; b = 0.542 |
| | | + | Image:Henon_a1.2_b0.3.png|a = 1.2; b = 0.3 |
| | | + | Image:Henon_a1.3_b0.3.png|a = 1.3; b = 0.3 (points need to be enlarged) |
| | | + | Image:Henon_a1.4_b0.3.png|a = 1.4; b = 0.3, original Henon Attractor |
| | | + | Image:Henon_a1.5_b0.2.png|a = 1.5; b = 0.2 |
| | | + | Image:Henon_a1.4_b0.1.png|a = 1.4; b = 0.1 |
| | | + | </gallery> |
| | | | |
| | | | |
| | ==Fixed Points== | | ==Fixed Points== |
| - | {{Switch|link1=Show|link2=Hide | + | Looking at the <balloon title="load:eqns">system of equations</balloon> <span id="eqns" style="display:none"><math>x_{n+1} = y_n + 1 - ax^2_n</math> and <math>y_{n+1} = bx_n\,</math></span> that describe the fractal, the Henon Attractor uses only two variables (x and y) that are evaluated into themselves. This results in two equilibrium or fixed points for the attractor. Fixed points are such that if the two Henon Attractor equations are applied to the fixed points, the resulting points would be the same fixed points. In algebraic terms: |
| - | |1=[[Image:HenonFixedPoints1.png|180px]] | + | |
| - | |2=[[Image:HenonFixedPoints1.png|thumb|300px|right|Original Henon Attractor with fixed points]] | + | |
| - | Looking at the system of equations that describe the fractal, the Henon Attractor uses only two variables (x and y) that are evaluated into themselves. This results in two equilibrium or fixed points for the attractor. Fixed points are such that if the system of equations are applied to the fixed points, the resulting output would be the same fixed points. Therefore, if the system ever plotted onto the fixed points, the fractal would become stagnant. | + | |
| | | | |
| | | | |
| - | The two fixed points of the Henon Attractor must satisfy <math>x_{n+1} = x\,</math> and <math>y_{n+1} = y\,</math>. | + | :<math>x_{n+1} = x_n\,</math> and <math>y_{n+1} = y_n\,</math> |
| | | | |
| - | Using the Henon Attractor's system of equations, the fixed points are (0.6314 , 0.1894) and (-1.1314 , -0.3394). | + | |
| | | + | :where <math>x_n</math> is the x-value at the nth iteration and <math>x_{n+1}</math> is the x-value at the next iteration. |
| | | + | |
| | | + | |
| | | + | Therefore, if the system ever plotted onto the fixed points, the fractal would become stagnant. By solving the Henon Attractor's system of equations with a = 1.4 and b = 0.3, we can find that the fixed points for the original Henon Attractor are (0.6314 , 0.1894) and (-1.1314 , -0.3394). |
| | | | |
| | {{HideThis|1=Solving the System of Equations|2= | | {{HideThis|1=Solving the System of Equations|2= |
| Line 75: | | Line 91: | |
| | | | |
| | | | |
| - | If <math>x_{n+1} = x\,</math> and <math>y_{n+1} = y\,</math> then | + | Since <math>x_{n+1} = x_n\,</math> and <math>y_{n+1} = y_n\,</math>, we can simplify the equations and refer to the variables as just <math>x</math> and <math>y</math>, respectively |
| | | | |
| | :<math>x = y + 1 - ax^2</math> | | :<math>x = y + 1 - ax^2</math> |
| Line 81: | | Line 97: | |
| | :<math>y = bx\,</math> | | :<math>y = bx\,</math> |
| | | | |
| | | + | |
| | | + | By substituting the value of <math>y</math> defined by the second equation into the <math>y</math> in the first equation, we get |
| | | | |
| | :<math>x = bx + 1 - ax^2</math> | | :<math>x = bx + 1 - ax^2</math> |
| Line 86: | | Line 104: | |
| | | | |
| | Using the <balloon title = "load:quadeqn"> quadratic equation </balloon> | | Using the <balloon title = "load:quadeqn"> quadratic equation </balloon> |
| - | <span id="quadeqn" style="display:none"> <math>x_{1,2} = \frac{-b \pm sqrt{b^2 - 4ac}}{2a}</math> </span> | + | <span id="quadeqn" style="display:none"> <math>x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}</math> </span> |
| | | | |
| - | :<math>x_{1,2} = \frac{-(b-1) \pm \sqrt{(b-1)^2 - 4(-a)(1)}}{2(-a)}</math> | + | :<math>x = \frac{-(b-1) \pm \sqrt{(b-1)^2 - 4(-a)(1)}}{2(-a)}</math> |
| | | | |
| | | | |
| - | :<math>x_{1,2} = \frac{-(b-1) \pm \sqrt{(b-1)^2 + 4a}}{-2a}</math> | + | :<math>x = \frac{-(b-1) \pm \sqrt{(b-1)^2 + 4a}}{-2a}</math> |
| | | | |
| | | | |
| | Using a = 1.4, b = 0.3: | | Using a = 1.4, b = 0.3: |
| | | | |
| - | :<math>x_{1,2} = 0.6314,-1.1314 \, </math> | + | :<math>x = 0.6314, -1.1314 \, </math> |
| | | | |
| | | | |
| | Using y = bx: | | Using y = bx: |
| | | | |
| - | :<math>y_{1,2} = 0.1894,-0.3394 \, </math> | + | :<math>y = 0.1894, -0.3394 \, </math> |
| | }} | | }} |
| | | | |
| - | | + | There are two types of fixed points, '''stable''' and '''unstable'''. The first fixed point (0.6314, 0.1894), labeled "1" on the image, is located within the attractor itself and is stable. This means that if a point is plotted close to the fixed point, the next iterated points will remain close to the fixed point. The second fixed point (-1.1314 , -0.3394), labeled "2", is considered unstable, and it is located outside of the bounds of the attractor. An unstable fixed point is such that if the system gets close to the fixed point, the next iterated points rapidly move away from the fixed point. |
| - | There are two types of fixed points, '''stable''' and '''unstable'''. The first fixed point (0.6314, 0.1894), labeled "1" on the image, is located within the bounds of the attractor and is unstable. This means that if the system gets close to the point, it will exponentially move away from the fixed point to continue chaotically. The second fixed point, labeled "2", is considered stable, and it is located outside of the bounds of the attractor. | + | |
| - | }} | + | |
| | |other=Algebra | | |other=Algebra |
| - | |AuthorName=SiMet | + | |AuthorName=Piecewise Affine Dynamics |
| - | |AuthorDesc=The images created by this author were found on the author's (username SiMet) Picasa Web Album under the category "Computer Art". | + | |AuthorDesc=Piecewise Affine Dynamics is a wiki site that was created by a group of French mathematicians that is dedicated to providing information about "dynamic systems defined by piecewise affine transformations". |
| - | |SiteName=SiMet's Gallery (Picasa) | + | |SiteName=Lozi Maps |
| - | |SiteURL=http://picasaweb.google.com/lh/photo/6iojFK86zEuRY3Zi5DBq-w | + | |SiteURL=http://padyn.wikidot.com/lozi-maps |
| | |Field=Dynamic Systems | | |Field=Dynamic Systems |
| | |Field2=Fractals | | |Field2=Fractals |
| - | |ImageRelates= The artistic image of the Henon Attractor that is the subject of this page instead uses the values a = 1 and b = 0.542. | | |
| | |References= | | |References= |
| - | Glenn Elert, [http://hypertextbook.com/chaos/21.shtml The Chaos Hypertextbook] | + | :*Glenn Elert, [http://hypertextbook.com/chaos/21.shtml The Chaos Hypertextbook] |
| - | Heinz-Otto Peitgen, Hartmut Jürgens, Dietmar Saupe, '''Chaos and fractals''' | + | :*Heinz-Otto Peitgen, Hartmut Jürgens, Dietmar Saupe, '''Chaos and fractals''' |
| | | + | |
| | | + | :*Bill Casselman, [http://www.ams.org/featurecolumn/archive/henon.html Simple Chaos-The Hénon Map] |
| | | + | |
| | | + | :*www.ibiblio.org [http://www.ibiblio.org/e-notes/Chaos/strange.htm Henon Strange Attractors] |
| | | | |
| - | Bill Casselman, [http://www.ams.org/featurecolumn/archive/henon.html Simple Chaos-The Hénon Map] | + | :*Michele Henon, [http://www.exploratorium.edu/turbulent/CompLexicon/henon.html Michele Henon] |
| | | + | |ToDo= |
| | | + | A better, less vague description of how sections of the Henon Attractor resembles the Cantor Set |
| | | + | Also, the description of the Henon Attractor can be expanded to include a discussion about the fractal's "basin of attraction". For more information, click [http://www.ams.org/featurecolumn/archive/henon.html here]. |
| | | | |
| - | www.ibiblio.org [http://www.ibiblio.org/e-notes/Chaos/strange.htm Henon Strange Attractors] | | |
| - | |InProgress=Yes | | |
| - | |HideMME=No | | |
| | }} | | }} |
## Current revision
Henon Attractor
Fields: Dynamic Systems and Fractals
Image Created By: Piecewise Affine Dynamics
Website: Lozi Maps
Henon Attractor
This image is a Henon Attractor (named after astronomer and mathematician Michel Henon), which is a fractal in the division of the chaotic strange attractor.
# Basic Description
The Henon Attractor is a special kind of fractal that belongs in a group called Strange Attractors, and can be modeled by two general equations. The Henon Attractor is created by applying this system of equations to a starting value over and over again and graphing each result.
### Making the Henon Attractor
Say we took a single starting point (x,y) and plotted it on a graph. Then, we applied the two Henon Attractor equations to the initial point and emerged with a new point that we graphed. Next, we took this new point and again applied the two equations to it and graphed the next new point. If we continued to apply the two equations to each new point in a process called iteration and plotted every outcome from this iteration, we would create a Henon Attractor. Click here to learn more about iterated functions.
Furthermore, if we plotted each outcome one at a time, we would observe that the points jump from one random location to another within the image. If you take a look at the animation, you can see the irregularity of the plotted points. Eventually, the individual points become so numerous that they appear to form lines and an image emerges.
### Magnification of the Henon Attractor
1X 8X 64X 512X
If you magnify this image, you would find that the lines (really many, many points) that appear to be single lines on the larger image are actually sets or bundles of lines, that, if magnified closer, are bundles of lines and so on. This property is called self-similarity, which means that even as you look closer and closer into the image, it continues to look the same. In other words, the larger view of the image is similar to a magnified part of the image.
### History of the Henon Attractor
Michel Henon was a French mathematician and astronomer who developed the Henon Attractor in the 1970s. At that time, Henon was interested in dynamical systems and especially the complicated orbits of celestial objects. The Henon Attractor emerged from Henon's attempt to model the chaotic orbits of celestial objects (like stars) in the mist of a gravitational force.
# A More Mathematical Explanation
Note: understanding of this explanation requires: *Algebra
[Click to view A More Mathematical Explanation]
## Fractal Properties
The Henon Attractor is often described as being similar to the Cantor Set [...]
[Click to hide A More Mathematical Explanation]
## Fractal Properties
The Henon Attractor is often described as being similar to the Cantor Set. Let us zoom into the Henon Attractor near the doubled-tip of the fractal (as seen in the animation). We can see that as we continue to magnify the lines that form the structure of the Henon Attractor, these lines become layers of increasingly deteriorating lines that appear to resemble the Canter Set.
The Fractal Dimension of the Henon Attractor is not calculable using a single equation$D = \frac{log(n)}{log(e)}$, but it is estimated to be about 1.261.
## Chaotic System
Original Henon Attractor, a = 1.4, b = 0.3
The Henon system can be described as chaotic and random. However, the system does have structure in that its points settle very close to an underlying pattern called a chaotic attractor. The basic Henon Attractor can be described by the equations, where $x_n$ is the x-value at the nth iteration.
$x_{n+1} = y_n + 1 - ax^2_n$
$y_{n+1} = bx_n\,$
Astronomer Michel Henon created the original Henon Attractor using the values a = 1.4 and b = 0.3 and starting point (1,1). These are also the values used by the artist to create the featured image at the top of the page. However, by changing the values of a and b, we can obtain Henon Attractors that look slightly different.
## Changing "a" and "b"
Although the original Henon Attractor uses a = 1.4 and b = 0.3, we can alter those values slightly to produce various-looking Henon Attractors. However, the values of a and b are limited to a small range of values, outside of which the fractal ceases to resemble the Henon Attractor.
Here are some more examples of Henon Attractors with different a and b values.
| | | |
|--------------------------------------------|------------------|-----------------------------------------------|
| a = 1; b = 0.542 | a = 1.2; b = 0.3 | a = 1.3; b = 0.3 (points need to be enlarged) |
| a = 1.4; b = 0.3, original Henon Attractor | a = 1.5; b = 0.2 | a = 1.4; b = 0.1 |
## Fixed Points
Looking at the system of equations $x_{n+1} = y_n + 1 - ax^2_n$ and $y_{n+1} = bx_n\,$ that describe the fractal, the Henon Attractor uses only two variables (x and y) that are evaluated into themselves. This results in two equilibrium or fixed points for the attractor. Fixed points are such that if the two Henon Attractor equations are applied to the fixed points, the resulting points would be the same fixed points. In algebraic terms:
$x_{n+1} = x_n\,$ and $y_{n+1} = y_n\,$
where $x_n$ is the x-value at the nth iteration and $x_{n+1}$ is the x-value at the next iteration.
Therefore, if the system ever plotted onto the fixed points, the fractal would become stagnant. By solving the Henon Attractor's system of equations with a = 1.4 and b = 0.3, we can find that the fixed points for the original Henon Attractor are (0.6314 , 0.1894) and (-1.1314 , -0.3394).
[Show Solving the System of Equations][Hide Solving the System of Equations]
To solve the system of equations:
$x_{n+1} = y_n + 1 - ax^2_n$
$y_{n+1} = bx_n\,$
Since $x_{n+1} = x_n\,$ and $y_{n+1} = y_n\,$, we can simplify the equations and refer to the variables as just $x$ and $y$, respectively
$x = y + 1 - ax^2$
$y = bx\,$
By substituting the value of $y$ defined by the second equation into the $y$ in the first equation, we get
$x = bx + 1 - ax^2$
Using the quadratic equation $x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$
$x = \frac{-(b-1) \pm \sqrt{(b-1)^2 - 4(-a)(1)}}{2(-a)}$
$x = \frac{-(b-1) \pm \sqrt{(b-1)^2 + 4a}}{-2a}$
Using a = 1.4, b = 0.3:
$x = 0.6314, -1.1314 \,$
Using y = bx:
$y = 0.1894, -0.3394 \,$
There are two types of fixed points, stable and unstable. The first fixed point (0.6314, 0.1894), labeled "1" on the image, is located within the attractor itself and is stable. This means that if a point is plotted close to the fixed point, the next iterated points will remain close to the fixed point. The second fixed point (-1.1314 , -0.3394), labeled "2", is considered unstable, and it is located outside of the bounds of the attractor. An unstable fixed point is such that if the system gets close to the fixed point, the next iterated points rapidly move away from the fixed point.
# Teaching Materials
There are currently no teaching materials for this page. Add teaching materials.
# About the Creator of this Image
Piecewise Affine Dynamics is a wiki site that was created by a group of French mathematicians that is dedicated to providing information about "dynamic systems defined by piecewise affine transformations".
# References
• Glenn Elert, The Chaos Hypertextbook
• Heinz-Otto Peitgen, Hartmut Jürgens, Dietmar Saupe, Chaos and fractals
• Bill Casselman, Simple Chaos-The Hénon Map
• www.ibiblio.org Henon Strange Attractors
• Michele Henon, Michele Henon
# Future Directions for this Page
A better, less vague description of how sections of the Henon Attractor resembles the Cantor Set Also, the description of the Henon Attractor can be expanded to include a discussion about the fractal's "basin of attraction". For more information, click here.
Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.
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http://www.physicsforums.com/showthread.php?t=591659
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Physics Forums
## Equivalence of definitions for regular representations
There seem to be two definitions for a regular representation of a group, with respect to a field k. In particular, one definition is that the regular representation is just left multiplication on the group algebra kG, while the other is defined on the set of all functions $f: G \to k$. I do not see why these are equivalent, and would appreciate any advice as to why this is the case.
PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
Recognitions: Homework Help Science Advisor (I'm assuming G is a finite group.) The element ##\sum_{g \in G} c_g g## in kG can be thought of as the function ##G \to k## defined by ##g \mapsto c_g##. Conversely, a function ##f \colon G \to k## gives rise to the element ##\sum_g f(g) g \in kG##. From this it's easy to see that the two vector spaces kG and {functions ##G \to K##} are isomorphic; in fact the map ##\sum_g c_g g \mapsto (g \mapsto c_g)## is an isomorphism. Now all you have to do is check that this isomorphism respects the G-action. You've already indicated that the action on kG is given by left multiplication. The action of G on a function ##f \colon G \to k## is defined by ##(h \cdot f)(g) = f(h^{-1}g)## (for ##h \in G##). Now note that $$h \sum_g c_g g = \sum_g c_g hg = \sum_{h^{-1}g} c_{h^{-1}g} g.$$ This shows that the isomorphism is G-linear.
Excellent, thank you.
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http://crypto.stackexchange.com/questions/5234/is-there-a-practical-zero-knowledge-proof-for-this-special-discrete-log-equation
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# Is there a practical zero-knowledge proof for this special discrete log equation?
We have a multiplicative cyclic group $G$ with generators $g$ and $h$, as in El Gamal. Assume $G$ is a subgroup of $(\mathbb{Z}/n\mathbb{Z})^*$. There are two parties, Alice and Bob:
• Alice knows: $g$, $h$, $x_1$, $x_2$ and $(a,b,c)$,
• Bob knows: $g$, $h$ and $(a,b,c)$.
Can Alice prove to Bob in zero knowledge that she knows $x_1$ and $x_2$ such that $(a,b,c) = (g^{x_1} , x_2·(h^{x_1}), h^{x_2})$?
This proof must be practical and non-interactive.
-
Do you have a specific setup in mind? $\:$ – Ricky Demer Oct 31 '12 at 19:38
Could you post that commitment format you're proposing? Maybe I can change it and still keep the remaining desired properties of the system. Thanks. – SDL Oct 31 '12 at 20:12
1
@SDL, I was thinking of $\textrm{commit}(x_2) = (g^{x_3}, x_2 h^{x_3})$, but I just now realized this might be problematic: this is binding for everyone, but not concealing against the person who holds the private key (the person who knows the discrete log of $h$ to base $g$ can infer what value was committed to without permission), which I suspect might not meet your needs. So, I withdraw my previous comment. Sorry for my error. – D.W. Oct 31 '12 at 20:27
1
wait a minute: $\;\;$ How do you make sense of $\: x_2 \cdot \left(h^{x_1}\right) \:$ in a cyclic group? $\hspace{1.05 in}$ – Ricky Demer Oct 31 '12 at 20:37
ElGamal is defined over a multiplicative cyclic group of order q. – SDL Nov 1 '12 at 13:32
show 4 more comments
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http://mathoverflow.net/questions/90544/is-there-a-homological-way-to-compute-quiver-presentations
|
Is there a homological way to compute quiver presentations?
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I have recently been studying with colleagues the representation theory of certain finite monoids that come up in probability theory and combinatorics, see Ken Brown's beautiful survey here.
These monoid have a split basic algebra over any field which is directed quasi-hereditary (standard modules are projective/costandards are simple). Thus they have acyclic quivers and can be written as the quotient of the path algebra of their quiver by an admissible ideal.
We can compute $\mathrm{Ext}^n(S,S')$ between simple modules $S,S'$ and so can get the quiver by taking dimensions when $n=1$. We can also get the number of relations from $n=2$.
Question. Is there some homological way to get at a quiver presentation of a split basic algebra with an acyclic quiver? You can assume the field is algebraically closed if it helps.
We do have explicit complete sets of primitive idempotents for these algebras, but they are sufficiently complicated that we don't even know how to recover the description of the quiver using the primitive idempotents.
The only cases where quiver presentations are known are for oriented matroids (in particular hyperplane arrangements) and of course the hereditary case.
The situation is so embarrassing that we have global dimension 2 examples with a single relation and we can't find the relation.
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I'm afraid I can't help with the question, but out of curiosity, are these monoids certain types of band? – Yemon Choi Mar 8 2012 at 5:10
They are bands. – Benjamin Steinberg Mar 8 2012 at 12:39
Can you do the Yoneda product of these Ext groups? The quiver presentation of the algebra in particular contains the information about products of $Ext^1$-groups. – Dag Oskar Madsen Mar 16 2012 at 13:20
I don't know how to compute the Yoneda product. How does it relate to the quiver presentation? – Benjamin Steinberg Apr 6 2012 at 0:27
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http://mathoverflow.net/questions/10860/why-no-abelian-varieties-over-z/10867
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## Why no abelian varieties over Z?
### Motivation
I learned about this question from a wonderful article Rational points on curves by Henri Darmon. He gives a list of statements (some are theorems, some conjectures) of the form
• the set `$\{$` objects $\dots$ over field $K$ with good reduction everywhere except set $S$ `$\}$` is finite/empty
One interesting thing he mentions is about abelian schemes in the most natural case $K = \mathbb Q$, $S$ empty. I think according to the definition we have a trivial example of relative dimension 0.
### Question
Why is the set of non-trivial abelian schemes over $\mathop{\text{Spec}}\mathbb Z$ empty?
### Reference
This is proven in Il n'y a pas de variété abélienne sur Z by Fontaine, but I'm asking because: (1) Springer requires subscription, (2) there could be new ideas after 25 years, (3) the text is French and could be hard to read (4) this knowledge is worth disseminating.
-
8
It was Shafarevich who at the 1962 ICM asked whether this set was empty, so maybe he should get the mathoverflow "reputation". :) According to Parshin, Abrashkin proved the result independently of Fontaine at about the same time: see the addendum at the end of Fontaine's paper, and see Abrashkin, V. A. Galois modules of group schemes of period $p$ over the ring of Witt vectors. (Russian) Izv. Akad. Nauk SSSR Ser. Mat. 51 (1987), no. 4, 691--736, 910; translation in Math. USSR-Izv. 31 (1988), no. 1, 1--46. – Bjorn Poonen Jan 6 2010 at 4:42
I found a way to "donate reputation to Shafarevich": I can post a question with bounty "on his behalf". Would you like to try giving a question? – Ilya Nikokoshev Jan 6 2010 at 12:33
@Bjorn. Then how did it so happen that this theorem is known in the name of Fontaine alone? Perhaps, in the context of the "big picture" of Fontaine theory? – Anweshi Jan 10 2010 at 15:27
@Anweshi: I'm not so sure that this is true. Careful people often cite it as "proved independently by Fontaine and Abrashkin". Here's a conjecture as to why Fontaine gets more hits and Abrashkin though: when Faltings proved Mordell there was a book written containing a lot of the ideas of the proof and accessible to grad students called "Arithmetic Geometry" by Cornell-Silverman, and in one of the articles there, much read by grad students presumably, they say it's "a very recent theorem of Fontaine". – Kevin Buzzard Feb 10 2010 at 7:20
On the other hand, Russian translation of Lang's Diophantine Geometry contains a very good survey of Zarkhin and Parshin about finiteness problems in diophantine geometry, including a sketch of Faltings' proof, and they give attribution to Abrashkin and Fontaine (independently). That's where I read it first. So it depends on the country. – Victor Protsak Jun 9 2010 at 0:49
## 5 Answers
It's a result related in spirit to Minkowski's theorem that $\mathbb Q$ admits no non-trivial unramified extensions. If $A$ is an abelian variety over $\mathbb Q$ with everywhere good reduction, then for any integer $n$ the $n$-torsion scheme $A[n]$ is a finite flat group scheme over $\mathbb Z$. Although this group scheme will be ramified at primes $p$ dividing $n$, Fontaine's theory shows that the ramification is of a rather mild type: so mild, that a non-trivial such family of $A[n]$ can't exist.
In the last 25 years, there has been much research on related questions, including by Brumer--Kramer, Schoof, and F. Calegari, among others. (One particularly interesting recent variation is a joint paper of F. Calegari and Dunfield in which they use related ideas to construct a tower of closed hyperbolic 3-manifolds that are rational homology spheres, but whose injectivity radii grow without bound.)
EDIT: I should add that the case of elliptic curves is older, due to Tate I believe, and uses a different argument: he considers the equation computing the discriminant of a cubic polynomial $f(x)$ (corresponding to the elliptic curve $y^2 = f(x)$) and shows that this solution equation has no integral solutions giving a discriminant of $\pm 1$.
This direction of argument generalizes in different ways, but is related to a result of Shafarevic (I think) proving that there are only finitely many elliptic curves with good reduction outside a finite set of primes. (A result which was generalized by Faltings to abelian varieties as part of his proof of Mordell's conjecture.)
Finally, one could add that in Faltings's argument, he also relied crucially on ramification results for $p$-divisible groups, due also to Tate, I think, results which Fontaine's theory generalizes. So one sees that the study of ramification of finite flat groups schemes and $p$-divisible groups (and more generally Fontaine's $p$-adic Hodge theory) plays a crucial role in these sorts of Diophantine questions. A colleague describes it as the black magic'' that makes all Diophantine arguments (including Wiles' proof of FLT as well) work.
P.S. It might be useful to give a toy illustrative example of how finite flat group schemes give rise to mildly ramified extensions: consider all the quadratic extensions of $\mathbb Q$ ramified only at $2$: they are ${\mathbb Q}(\sqrt{-1}),$ ${\mathbb Q}(\sqrt{2})$, and ${\mathbb Q}(\sqrt{-2})$, with discriminants $-4$, $8$, and $-8$ respectively. Thus ${\mathbb Q}(\sqrt{-1})$ is the least ramified, and not coincidentally, it is the splitting field of the finite flat group scheme $\mu_4$ of 4th roots of unity.
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Re: generalized by Faltings to abelian varieties. That's one more reason I ask: Darmon seems to call this a Shafarevich Conjecture (page 16) rather then a result. – Ilya Nikokoshev Jan 5 2010 at 23:33
2
From a number theorist's point of view, $p$-adic Hodge theory is one of the key ingredients in the theory of motives, so these arguments are motivic, in a certain sense. (Perhaps one can say that $p$-adic Hodge theory encodes arithmetic properties of motives in a way analogously to the way that Hodge theory encodes geometric and analytic properties.) – Emerton Jan 6 2010 at 0:23
1
One might say that the elliptic curve case follows from nonexistence of weight 2 level 1 cusp forms + modularity, but there may be some hidden circular reasoning (and it seems like a backwards way to look at things). – S. Carnahan♦ Jan 6 2010 at 5:13
4
I don't think there is any hidden circular reasoning there, and in fact is a perfectly good way to think about it. Actually, Tate's result served as an early consistency check for the modularity conjecture. (Just as his result on the non-existence of continuous representations $\rho:G_{\mathbb Q} \to GL_2(\bar{F}_2)$ unramified outside 2 served as an early consistency check on Serre's conjecture.) – Emerton Jan 6 2010 at 5:17
1
I should also add, regarding my paranthetical remark, that Tate's non-existence result for mod 2 2-dim'l Galois reps. unram. outside 2 is a crucial ingredient in the proof of Serre's conjecture by Khare and Wintenberger, so logically that situation is quite different from the elliptic curve case, where Tate's non-existence result plays no role in the proof of modularity. – Emerton Jan 6 2010 at 6:02
show 3 more comments
### Comments by Anweshi
The essential point is what Emerton mentioned, ie the analogy with Minkowski's theorem on number fields with ramification. The basic principle is that "arithmetic is geometry". Number rings are in a sense zero dimensional objects, elliptic curves one dimensional objects and abelian varieties correspond to higher dimensions. So we have Minkowski's theorem. And we ask, can we extend it to higher dimensions? Tate, after setting up the theory correctly as in his famous survey article on the arithmetic of elliptic curves, proved it rather trivially for elliptic curves(as Emerton mentions). Now the task is for abelian varieties.
Fontaine comes along, and proves that it is indeed the case. But the proof turns out to be much more complicated than expected. He built a whole lot of "Fontaine theory" around this. It goes into $p$-adic Hodge theory, $p$-adic Galois representations etc. He worked on it for some 15 years in isolation, it is said. The first major success of his theory was this theorem, and later it gained popularity. Now it is a major stream of research in arithmetic geometry.
References:
• Neukirch, Algebraic number theory, for the general philosophy that "arithmetic is geometry".
• Notes of Robert Coleman's course on Fontaine's theory of the mysterious functor
• The Bourbaki expose of Bearnadette Perrin-Riou. Fonctions L p-adiques des représentations p-adiques, Astérisque 229, (1995).
• Tate, The Arithmetic of Elliptic Curves, Survey Article, Inventiones.
It could be also worthwhile to have a look at the articles on finite flat group schemes in the volume Arithmetic Geometry of Cornell and Silverman, and in the volume Modular forms and Fermat's Last Theorem by Cornell, Silverman and Stevens. This is all intimately connected with them, as Emerton mentions. In fact, you can find a particular viewpoint by Fontaine on Finite Flat group Schemes.
There could be also be a simpler motivic explanation of this, without getting into the intricacies of Fontaine theory. The reason I think so is the following. I have heard the answer that there is no elliptic curve over $F_1$ because from the zeta functions the motives turn out to be mixed Tate. But, on the other hand, my own "proof" of this fact was that if there were an elliptic curve or abelian variety over $F_1$, it would be extensible to $Spec\ Z$ and there by Fontaine's theorem the only abelian scheme is the trivial one. Ever since I have wondered, whether it is possible to substitute Fontaine's theory arguments with motivic ones.
Emerton clarified to me in this connection: From a number theorist's point of view, p-adic Hodge theory is one of the key ingredients in the theory of motives, so these arguments are motivic, in a certain sense. (Perhaps one can say that p-adic Hodge theory encodes arithmetic properties of motives in a way analogously to the way that Hodge theory encodes geometric and analytic properties.)
Thus, by Emerton's answer, Fontaine theory seems to be thus a deeper part of motives. However, this "no abelian variety over Z" theorem of Fontaine was the first major application of Fontaine's theory. I imagined, if any results of Fontaine's theory were to be replaced by usual motivic arguments, then this ought to be the first candidate.
Before stopping, I must mention the intimate connection all this has with Iwasawa theory. Fontaine's theory is very much tangled with it, as could be seen in the expose of Perrin-Riou. However the more knowledgeable people should clarify on this.
This might be an apt place to mention the conference in honor of Fontaine. He is about to retire, after his great achievements.
### Comment by Ilya
I think this should be indeed related to motives. (update: I think others provided some good references.)
### Comments by Emerton
(1) There were earlier applications of Fontaine's results on finite flat group schemes; e.g. they played a role in Mazur's proof of boundedness of torsion of elliptic curves over $\mathbb Q$. I say this just to emphasize that Fontaine's theory didn't really develop in isolation. His theory is deep and technical, and it took people time to absorb it. But the theory of finite flat group schemes and $p$-divisible groups has a long history intertwined with arithmetic: there are results going back to Oda, Raynaud, and Tate; Fontaine generalized these; they were used by Mazur in his work, and by Faltings; Fontaine generalized further to $p$-adic Hodge theory (a theory whose existence was in part conjectured earlier by Grothendieck, motivated by, among other things, the work of Tate); ... . One shouldn't think of these ideas as being esoteric (despite the black magic'' label); they are and always have been at the forefront of the interaction between geometry and arithmetic, in one guise or another. (As another illustration, Fontaine's theory also closely ties in with earlier themes in the work of Dwork.)
(2) I'm not sure that there is any particular kind of usual motivic argument. The phrase motive conjures up a lot of different images in different peoples minds, but one way to think of what motivic means is that it is the study of geometry via structures on cohomology. From this point of view, $p$-adic Hodge theory is certainly a natural and important tool.
Here are some papers that give illustrations of $p$-adic Hodge theoretic reasonsing in what might be regarded as a motivic context:
Grothendieck, Un theoreme sur les homomorphismes de schemas abeliens, a wonderful paper. Although the results are essentially recovered and generalized by Delignes work in his Hodge II paper, it gives a fantastic illustration of how $p$-adic Hodge theoretic methods can be used to deduce geometric theorems.
Kisin and Wortmann, A note on Artin motives
Kisin and Lehrer, Eigenvalues of Frobenius and Hodge numbers
James Borger, Lambda-rings and the field with one element
These three are chosen to illustrate how $p$-adic Hodge theory arguments can be used to make geometric/motivic deductions. The paper of Borger is also an attempt in part to provide foundations for the theory of schemes over the field of one element, and illustrates how $p$-adic Hodge theory plays a serious role in their study.
Maulik, Poonen, Voisin, Neron-Severi groups under specialization, a terrific paper, which illustrates the possibility of using either $p$-adic Hodge-theoretic arguments or classical Hodge-theoretic arguments to make geometric deductions. (This is the same kind of complementarity as in Grothendieck's paper above compared to Deligne's Hodge II.)
### Comments by Anweshi.
@Emerton, or anybody else: If there is something which does not make sense in my foray into "motivic" pictures, or something else which does not make sense, please feel free to erase and edit in whatever way you wish.
### a further question by Thomas:
The great references given above let me ask about the current status of the many conjectures and open questions in Illusie's survey, e.g. finiteness theorems, crystalline coefficients, geometric semistability,... ?
• ilya's comment: I think it would be very useful if somebody posted a question along the lines of what Thomas suggests, especially filling in some background from Illusie's paper (I would do it, but I don't have the paper itself).
** Anweshi's comment:** Fontaine's theory uses a great deal of crystalline cohomology. For instance please see Robert Coleman's notes referred above.
-
Fontaine's initial result was proven using his results on Hodge-Tate decomposition which refine Tate's theorem. But later on (around 1990, Schémas propres et lisses sur $\mathbf Z$), he revisited his theorem using Faltings's theorem (that the $p$-adic cohomology of a proper variety with good reduction mod $p$ is crystalline) and showed that proper smooth schemes over $\mathbf Z$ have a fairly special cohomology in degree $\leq 3$ (the Hodge numbers $h^{i,j}$ are zero for $i\neq j$ and $i+j\leq 3$).
As for Abelian varieties, this theorem has also been proved independently by Abrashkin.
-
Here's another motivation for believing that there is no abelian variety over Z: the existence of such a beast would imply that the "motivic" expectations concerning L-functions are false. More precisely, it is not very difficult to show (Mestre does this in a paper in Compositio in 1986 using explicit formulas) that if an abelian variety over the rationals of dimension g at least 1 has the property that its L-function is entire with the expected functional equation, then its conductor is at least 10^{g}, and in particular is >1.
(There's a very short proof of a weaker fact, sufficient to "imply" the theorem of Fontaine and Abrashkin, in Th. 5.51 of my book with Iwaniec, though the printed version has an unfortunate mistake, and -- I must apologize here for not checking the history at the time -- we did not mention Abrashkin...)
-
Here is a nice script of Schoof: http://www.cems.uvm.edu/~voight/notes/274-Schoof.pdf
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http://mathhelpforum.com/trigonometry/64022-trigonometry-help.html
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# Thread:
1. ## Trigonometry help?
How would you find the length of RS? What do you have to find first? Can someone please explain this to me step by step? Help is appreciated!
Same for this one, how would you find the length of TU? Where would you start and what do you have to do?
2. Problem #1
$\sin$ = $\frac{opposite}{hyp}$
If you use the angle with 32 deg you would have Opposite = 25 and
Hyp = x
when you get that you can use a $^2$ + b $^2$ = c $^2$
A = 25
B = RS
C = the hyp you found earlier
Problem #2
Im not sure on the second one because I don't know if you can use S $\frac{o}{h}$ C $\frac{A}{H}$ T $\frac{O}{A}$ or not without a Right angle its been too long for me if i have time ill figure out how to do it and help you out.
3. Originally Posted by tjsfury
Problem #1
$\sin$ = $\frac{opposite}{hyp}$
If you use the angle with 32 deg you would have Opposite = 25 and
Hyp = x
when you get that you can use a $^2$ + b $^2$ = c $^2$
A = 25
B = RS
C = the hyp you found earlier
Problem #2 ---- give me a second to look at it
I don't think the opposite side of the triangle (if you're looking through at 32 degrees) would be 25 though. The base of the triangles aren't the same length, but together they add up to 50 meters.
Edited:
One of triangle's bases has a bigger length than the other (for the first question)
4. Yea i agree let me get a pencil out and think about it im bored enough to help lol
5. Originally Posted by tjsfury
Yea i agree let me get a pencil out and think about it im bored enough to help lol
Haha, alright. I appreciate you helping me =)
6. ok i figured it out, i think..... 32+26 = 58 (1 angle of big triangle)
180 - (32 + 90) = 58 (angle of top side) 2 angles are the same 2 opposite sides are the same in length
.... been a long time all angles of a triangle added together = 180 right?
7. Originally Posted by 4AM
How would you find the length of RS? What do you have to find first? Can someone please explain this to me step by step? Help is appreciated!
Same for this one, how would you find the length of TU? Where would you start and what do you have to do?
The 1st triangle is an iscolses triangle. The top left angle is 58 degrees and the bottom left is 64 degrees, and the middle right is 58 degrees
So $\cos {26} = \frac{RS}{50}$
8. ok yea i looked it up all angles = 180
So your bottom leg SQ = 50m
Because of the equal angles then you can use..... Sin with 26 knowing that H = 50
9. Originally Posted by 11rdc11
The 1st triangle is an iscolses triangle. The top left angle is 58 degrees and the bottom left is 64 degrees, and the middle right is 58 degrees
Lol i figured that out right before you posted it :] so how about his second one?
for some reason i remember something about having a line like a Z shape inner and outter angles being equal....
in his problem would the bottom right angle be 38.8? Probably not
10. Originally Posted by tjsfury
ok yea i looked it up all angles = 180
So your bottom leg SQ = 50m
Because of the equal angles then you can use..... Sin with 26 knowing that H = 50
U mean cos not sin right
11. Originally Posted by 11rdc11
U mean cos not sin right
Yea i do I was looking at the 90 degree mark and didn't look at the angles (if you look i said that H=50.... yea that side isn't H its A) what about his #2? any ideas
12. Originally Posted by tjsfury
Yea i do I was looking at the 90 degree mark and didn't look at the angles (if you look i said that H=50.... yea that side isn't H its A) what about his #2? any ideas
Yea you have to find all the angles and then use the law of sines
$\frac{\sin{15}}{12.5} = \frac{\sin{38.8}}{TV}$
13. Ha I was right about the 38.8 angle.... Z angles (alternate angles) are equal.... shouldn't be too hard knowing that
Angles
14. From there either use law of sines or sin or cos to TU
15. Originally Posted by 11rdc11
The 1st triangle is an iscolses triangle. The top left angle is 58 degrees and the bottom left is 64 degrees, and the middle right is 58 degrees
So $\cos {26} = \frac{RS}{50}$
Wait, I don't get why we have to use cos?
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http://cs.stackexchange.com/questions/tagged/quantum-computing
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http://mathhelpforum.com/number-theory/137271-irreducible-polynomial.html
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# Thread:
1. ## irreducible polynomial
Hi guys,
I was wondering how I can prove that $x^4+x+2$ in the field $\mathbb{Z}_3[x]$ is irreducible?
I know how to do it if the degree is 3 or less, but here I have no idea..
Thanks in advance,
Orbis
2. Originally Posted by Orbis
Hi guys,
I was wondering how I can prove that $x^4+x+2$ in the field $\mathbb{Z}_3[x]$ is irreducible?
I know how to do it if the degree is 3 or less, but here I have no idea..
Thanks in advance,
Orbis
It's easy to see there are no linear factors.
As for quadratic terms, WLOG its ok to assume the two constant terms are $2$ and $1$, since $-2=1$ and $-1=2$.
So we have $x^4+x+2=(x^2+ax+2)(x^2+bx+1) = x^4+(a+b)x^3+abx^2+(a+b)x+2$
The $x^3$ term tells us that $a+b\equiv0\bmod{3}$.
But the $x$ term tells us that $a+b\equiv1\bmod{3}$.
This is impossible, hence $x^4+x+2$ is irreducible in $\mathbb{Z}/3\mathbb{Z}$.
3. Ah, so simple! I was thinking along the lines of proving that $<x^4+x+2>$ was a maximal ideal, but that wasn't getting me anywhere.
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http://mathhelpforum.com/advanced-applied-math/153257-potential-energy-problem-print.html
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# Potential energy problem
Printable View
• August 10th 2010, 08:13 AM
FGT12
Potential energy problem
The potential energy function of a particle of mass m is $V=\frac{-1}{2}c(x^2-a^2)^2$, where a,c > 0. Sketch V as a function of x and describe the possible types of motion in the three cases (a) E > 0, (b) $E < \frac{-1}{2}ca^4$ and (c) $\frac{-1}{2}ca^4< E < 0$
I can sketch the graph which gives you an upside-down parabola, but i cannot see how to determine the motion, i think it have something to do with T+V=E (conservation of energy)
• August 10th 2010, 08:25 AM
Ackbeet
Think of the potential energy function as rising and falling ground (with gravitational potential energy equal to mgh, this analogy is actually fairly precise!). Then, the energy E you can plot on the same graph as the potential energy, only this time it's just a horizontal line. The intersections of V with the horizontal line E represent the turning points. For example: suppose you had a hill V=x, and you had a ball rolling in from the negative x direction. If the energy E were equal to 0, it would intersect V right at the origin, right? Well, that's where the ball would approach, slowly, and then when it hit the origin, it would start rolling back out to negative infinity, and never come back. That's the sort of analysis required in this problem. Make sense?
Another example: the simple harmonic oscillator (pendulum, spring, RLC circuit, etc.). This time, the potential looks like V=x^2. It's a valley. Suppose your energy E = 1. (Note that E must always be greater than V_min; that's a law of physics.) Then the intersection of E with V occurs at +1 and -1. Therefore, the ball is going to go to +1, reverse direction, go to -1, reverse direction, go to +1, etc. It will oscillate between the two turning points. Make sense?
Final example: V = e^(-x^2). This is the bell curve. The peak of it is at the origin. If 0<E<1, then the ball could come in from either positive or negative infinity, hit the turning point (intersection) on the corresponding side of the origin, and then bounce right back out to infinity. But, if the ball has energy greater than 1, it will be able to get over the hill. Then it will go over the hill, lose some kinetic energy in the process, and regain it as it goes back downhill and out to the opposite infinity from which it came.
Those are some ideas. Does this help?
All times are GMT -8. The time now is 11:18 AM.
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http://mathoverflow.net/questions/95514?sort=oldest
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Incidence Correspondence
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
A useful tool in Algebraic Geometry is the incidence correspondence. Loosely speaking, it is a set of the form $$\{(p,X): p \text{ a fixed dimension subscheme of } Y \text{ and } X \text{ a specific type of subscheme} \}.$$ For example one could consider the incidence correspondence of lines in $\mathbb{P}^2$ with a point on them, or cubics in $\mathbb{P}^3$ with a line on them.
It is not too hard to see that in each of the previous cases, the resulting scheme is a variety by writing down explicit equations in coordinates. In the first case the variety lives in $\mathbb{P}^2\times \mathbb{P}^2$ and with a bit of work it is not too hard to show it is a projective bundle over $\mathbb{P}^2.$ The second case is a subvariety of $\mathbb{P}^{19}\times G(1,3)$.
My question is the following. Is there a way to get both of the previous examples in a more natural way then explicitly writing down equations? Should I even expect there to be one? Generally when working with incidence correspondences one is interested in properties such as smoothness and irreducibility and most authors I have seen conclude these from the equations. Since the first case above does end up being a projective bundle, I would really hope for there to be a natural way to construct it.
Thanks.
-
1 Answer
There's several issues here.
(1) Is a given incidence correspondence actually a closed variety?
(2) What are explicit equations for the correspondence in the product of the relevant spaces?
(3) What are geometric properties of the incidence correspondence?
Most often, questions (1) and (3) are studied and little attention is paid to (2).
For (1), things can be generalized a fair bit. For instance, suppose $X$ is a variety, $H$ is an ample divisor, and $P$, $Q$ are two Hilbert polynomials with respect to the ample divisor $X$. Then there are (projective) Hilbert schemes $Hilb_P(X)$ and $Hilb_Q(X)$ parameterizing closed subschemes of $X$ with Hilbert polynomials $P$ and $Q$. Then there are a couple different natural incidence correspondences, for example
$\{(Z,Z'):Z\subset Z'\} \subset Hilb_P(X)\times Hilb_Q(X)$
$\{(Z,Z'):Z\cap Z'\neq \emptyset\}\subset Hilb_P(X)\times Hilb_Q(X),$
and it is easy to verify that these conditions are closed (although the correct scheme structure may be less clear). One can instead restrict attention to a closed subvariety of the Hilbert scheme (so as to not use all the components of the Hilbert scheme, for instance, in case the geometric objects you care about are not entirely determined by their Hilbert polynomials). It is also easy to generalize to cases with more factors. These types of constructions mean that arguments for the closedness of incidence correspondences are almost never written down, as anything reasonable that you can write down will be closed so long as the families of objects under consideration are themselves closed.
In practice, (2) is rarely of any theoretical interest, unless these are very special varieties. Perhaps there is a large algebraic group acting and the ideal can be studied via representation theory, or perhaps the variety has very small dimension or is otherwise very simple, in which case some information might be learned from the ideal.
Regarding (3), geometric properties of an incidence correspondence are only very rarely (roughly in the same cases as discussed for (2)) determined by studying the defining equations. Most often, the reason we study an incidence correspondence $\Sigma \subset X\times Y$ is because we have some question about one of the projection maps, say $\Sigma \to X$; for instance, we may wonder if it is dominant. We then ideally answer this by studying the other projection, which ideally is easier to understand. Likewise, properties like dimension and irreducibility are hopefully easily understood by studying the projections, and smoothness can sometimes be analyzed as well (although smoothness is often not so important in basic applications).
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Thanks. That was an helpful explanation. – Lalit Jain May 2 2012 at 18:25
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http://unapologetic.wordpress.com/2011/07/23/de-rham-cohomology-is-functorial/?like=1&_wpnonce=67418a1937
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# The Unapologetic Mathematician
## De Rham Cohomology is Functorial
It turns out that the de Rham cohomology spaces are all contravariant functors on the category of smooth manifolds. We’ve even seen how it acts on smooth maps. All we really need to do is check that it plays nice with compositions.
So let’s say we have smooth maps $f:M_1\to M_2$ and $g:M_2\to M_3$, which give rise to pullbacks $f^*:\Omega(M_2)\to\Omega(M_1)$ and $g^*:\Omega(M_3)\to\Omega(M_2)$. All we really have to do is show that $g^*\circ f^*=(f\circ g)^*$, because we already know that passing from chain maps to the induced maps on homology is functorial.
As usual, we calculate:
$\displaystyle\begin{aligned}\left[\left[\left[g^*\circ f^*\right](\omega)\right](p)\right](v_1,\dots,v_k)&=\left[\left[g^*(f^*\omega)\right](p)\right](v_1,\dots,v_k)\\&=\left[\left[f^*\omega\right](g(p))\right](g_*v_1,\dots,g_*v_k)\\&=\left[\omega(f(g(p)))\right](f_*g_*v_1,\dots,f_*g_*v_k)\\&=\left[\omega(\left[f\circ g\right](p))\right]((f\circ g)_*v_1,\dots,(f\circ g)_*v_k)\\&=\left[\left[(f\circ g)^*\omega\right](p)\right](v_1,\dots,v_k)\end{aligned}$
as asserted. And so we get maps $f^*=H^k(f):H^k(M_2)\to H^k(M_1)$ and $g^*=H^k(f):H^k(M_3)\to H^k(M_2)$ which compose appropriately: $H^k(g)\circ H^k(f)\to H^k(f\circ g)$.
### Like this:
Posted by John Armstrong | Differential Topology, Topology
## 1 Comment »
1. [...] We know that a map induces a chain map , which induces a map on the de Rham cohomology. This is what we mean when we say that de Rham cohomology is functorial. [...]
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://stats.stackexchange.com/questions/33065/proportional-hazard-ratio-formula-in-r
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# Proportional hazard ratio formula in R
I'm new proportional hazard ratio models and new-ish to R, so I suspect this is a basic question:
The scenario is modeling the speed of an effect through hazard ratios. People are joining an organization, and most are invited by someone else within the organization*. The speed measured is the time between when the invitation is made and the person joins the organization, called the `registration_interval`. The way that seems evident to model this in R is like so:
````analysis<-coxph(Surv(registration_interval) ~ Factor1 + Factor2 + Factor3 ...)
````
However, there's also the prospect that people are faster/slower to join the organization as the date of their invitation moves along (ex. people invited in 2012 might join faster than people joining in 2009), and this could interact with the various factors. What is a sensible way to incorporate this `date` information?
*We only have data for those that do ultimately join the organization. We consider our model as simply modeling "joiners", and not right-censoring everyone whose invitation-registration intervals are infinitely long.
Edit for clarification: We know when people are invited, so on any given date there are only a certain number of people invited who could join. If everyone were invited all at once the proper model setup would just be:
````analysis<-coxph(Surv(registration_interval) ~ date + Factor1 + Factor2 + Factor3 ...)
````
But since the `registration_interval` is determined by two dates, there is some interaction between that and `date`. The hazard function would, for example, look quite different if there is a span of time in which nobody is invited, versus a span of time in which thousands of people are invited. Perhaps this is all packaged within `Surv`, and I'll look into it.
Edit regarding putting both `date` and `registration_interval` or join time information in Surv: Nope. That appears to be a red herring.
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## 1 Answer
If `date` is the date people are invited, I would write a model just like you suggest in your second formula:
````coxph(Surv(registration_interval) ~ date + Factor1 + Factor2 + Factor3 ...)
````
When in doubt, it is always useful to write the full model.
$$\lambda(t|X) = \lambda_0(t) \cdot \exp(\beta_{date} X_{date} + \beta_1 X_1 + \beta_2 X_2 + ...)$$
In essence, this means that for the same values of `Factor1, Factor2, ...` (the variables $X_1, X_2, ...$) the hazard (the probability of answering the invitation at a particular instant) is multiplied by a factor $\exp(\beta_{date}X_{date})$. Note that if `date` increases linearly with time, the effect on the hazard will increase or decrese exponentially, so you might want to use `log(date)`, or another function of `date` instead.
I think that if everybody was invited at the same date, you would not use this model, because $X_{date}$ would be the same for every subject and would carry no information whatsoever (it would be absorbed in $\lambda_0(t)$ and could be removed from the model).
I also think that the hazard function does not change if there is a span of time in which nobody is invited, versus a span of time in which thousands of people are invited. What changes is the values of $X_{date}$, which is a property of individuals, not of the model.
Regarding the possibilities of interactions with the other variables, you can add product terms to the model as shown below.
````coxph(Surv(registration_interval) ~ date + Factor1 + Factor2 + Factor3 +
date:Factor1 + date:Factor2 + date:Factor3)
````
In summary, I would try following the approach that you suggest yourself.
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http://math.stackexchange.com/questions/227365/is-this-function-constructed-using-ac-necessarily-discontinuous-everywhere?answertab=votes
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# Is this function constructed using AC necessarily discontinuous everywhere?
Assume AC. Let $x_\alpha$ be a well-ordering of $\mathbb{R}$. For all $\alpha < \mathfrak{c}$, let $F(x_\alpha) = x_{\alpha+1}$.
Can it be proven that $F$ is discontinuous everywhere?
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Is $F$ intended to be defined on $x_\alpha$ where $\alpha \geq c$? If so, how is it defined? – Logan Nov 2 '12 at 6:39
I guess I'm confused. Unless I misunderstand, as phrased there is no guarantee that $F$ is defined on all of $R$. Equipollent is a cardinal relation, so even though $c$ is equipollent to $R$ it might not be order-isomorphic to the well ordering on $R$. There are many ordinals equipollent to $R$ and by choosing the smallest you make it likely that many elements of $R$ won't be included in the definition of $F$. – Logan Nov 2 '12 at 6:46
1
Dan, I feel that the answer is yes or it depends on CH in some way. I have no idea what the argument should be, though. Interesting question, by the way (+1). – Asaf Karagila Nov 2 '12 at 7:16
3
For some reason noone has pointed this out yet: $F$ is not a bijection. Its range misses $x_0$ and all $x_\alpha$ with $\alpha$ limit ordinal. – Stefan Geschke Nov 2 '12 at 7:43
1
@Dan Brumleve: Yes, sure, the question is completely valid. I just wanted to clear up this misconception. – Stefan Geschke Nov 2 '12 at 7:49
show 7 more comments
## 1 Answer
$F$ can be continuous at some points, if the well-ordering is defined in the right way. For example, choose your well ordering so that each real number in $(0,1)$ is a unique limit ordinal in $c$. That is, if $T:\mathbb{R} \rightarrow c$ maps each element $x \in \mathbb{R}$ to the element $\alpha$ of $c$ such that $x_\alpha=x$, then we want $x \in (0,1) \Rightarrow \alpha$ is a limit ordinal or 0.
There are enough limit ordinals to accomplish this, because the cardinality of $c$ is the cardinality of $\mathbb{N} \times L(c)$ where $L(c)$ is (the set of all limit ordinals in $c$) $\cup$ 0. So the number of limit ordinals must have the same cardinality as $c$, and the same cardinality as (0,1).
Further define $T$ so that $T(x) = T(x-2)+1$ for all $x \in (2,3)$. At this point $T$ is still injective, because $T(x-2)+1$ is not a limit ordinal.
Now extend $T$ to the rest of $\mathbb{R}$ where it hasn't already been defined in such a way that it is bijective.
With $x_\alpha$ defined in terms of this $T$, F will be continuous on $(0,1)$. In fact, it will be identically equal to $f(x)=x+2$ on $(0,1)$.
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Points whose index is not a successor ordinals are not in the image of $F$. So if you make $(0,1)$ into limit ordinals you just ensured $F(x)\notin(0,1)$ for all $x\in\mathbb R$. – Asaf Karagila Nov 2 '12 at 7:57
@AsafKaragila I wasn't trying to put $F(x) \in (0,1)$. I was trying to get $F$ to map $(0,1)$ onto (2,3) continuously. $T$ is a bijection sending (0,1) to limits and (2,3) to successors of limits, and $F$ is a continuous function R->R sending (0,1) to (2,3). – Logan Nov 2 '12 at 8:05
I think this works fine for the domain $\mathbb{R}^+$. That is good enough for me, but I am going to work through it a little more before accepting. – Dan Brumleve Nov 3 '12 at 17:01
The domain can be $\mathbb{R}$ if we map $[-1,1)$ to $L(\frak c)$ instead, counting down from $[-1,0)$ and counting up from $[0, 1)$. In that case $f(x) = x+1$ when $x \ge 0$ and $f(x) = x-1$ when $x \lt 0$ and it has a single discontinuity at $x = 0$. – Dan Brumleve Nov 3 '12 at 18:16
as described here, the domain can be literally anything that contains (0,1)$\cup$(2,3) and has the cardinality of the continuum. – Logan Nov 3 '12 at 20:35
show 3 more comments
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http://physics.stackexchange.com/questions/tagged/mechanics+angular-momentum
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# Tagged Questions
1answer
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### Spin angular momentum of a system of particles : Is there any energy associated with it?
Consider a system of point particles , where the mass of particle $i$ is $μ_i$ and its position vector is $\vec{r}_i$. Let $\vec{r}_\text{cm}$ is the position vector of the center of mass of the ...
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### Angular Momentum and Force
I'm stuck on number 5. The answers to the first 4 are correct, but I dont know how to set up number 5. Any idea that I would have would require me having some kind of time information, but thats not ...
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### Angular Momentum and Average Torque
Refer to number 6. This is the one I'm stuck on. So angular momentum is conserved right, so initial angular momentum is equal to final angular momentum. Initial is 7.87 so final must be 7.87, right? ...
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http://math.stackexchange.com/questions/282405/what-does-inf-int-mean?answertab=votes
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# What does $\inf\int$ mean?
What does notation
$$\inf\int f(x) \,\mathrm{d}x$$
stand for? I noticed it in a question on this site.
Sorry for such a basic question, but i can't find a reasonable keyword to look this up anywhere.
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– Jeremy Jan 20 at 1:43
2
The notation, exactly as put, does not make sense. I hope that it was not used. – André Nicolas Jan 20 at 1:44
@André - I have linked the usage that I saw. If that is correct and my usage is wrong, please specify. I just wanted to write some more simple/general example instead of that double integral. – Juris Jan 20 at 1:51
@AndréNicolas In the original question, the infimum was taken over all functions in a given space. In that context, the usage made sense. – Ayman Hourieh Jan 20 at 1:51
## 3 Answers
It means infimum, which is a generalization of the notion of minimum. For example, $f(x)=1/x$ doesn't have a minimum on $x\in[1,\infty)$, but its infimum is $0$.
In the context in which it was used,
$$\inf \iint\limits_{x^2+y^2\leqslant1}\left(\frac{\partial u}{\partial x}\right)^2+\left(\frac{\partial u}{\partial y}\right)^2\mathrm dx\,\mathrm dy$$ for $C^\infty$ functions $u$ that... [more conditions on $u$]
it means the tightest bound on how small the value of the integral can get for any such function $u$.
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$$\inf \int f(x)dx$$ is the largest real number that is less than or equal to $$\int f(x) dx$$ inf stands for infimum. See more information here:
http://en.wikipedia.org/wiki/Infimum
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To write $\displaystyle\inf\int f(x)\,dx$ without the words that came after that fails to convey what was said. It says "for $C^\infty$ functions that vanish at $0$ and [. . . . .]", and the expression inside the integral has "$u$" in it. In other words, it identifies a particular set. An infimum is an infimum of a set.
The infimum of a set with a smallest member is the smallest member. Thus the infimum of the set of all nonnegative numbers is $0$. The infimum of some sets with no smallest member exists. For example, there is no smallest positive number, and the infimum of the set of all positive numbers is $0$.
The infimum of a set is the greatest lower bound of the set. $0$ is a lower bound of the set of all positive numbers because $0$ is less than or equal to every positive number. No number bigger than $0$ is a lower bound of the set of all positive numbers. So $0$ is the greatest among all lower bounds.
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http://chronicle.com/blognetwork/castingoutnines/2007/10/16/fear-courage-and-place-in-problem-solving/
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Previous: Next:
# Fear, courage, and place in problem solving
October 16, 2007, 9:32 pm
By Robert Talbert
Sorry for the slowdown in posting. It’s been tremendously busy here lately with hosting our annual high school math competition this past weekend and then digging out from midterms.
Today in Modern Algebra, we continued working on proving a theorem that says that if $$a$$ is a group element and the order of $$a$$ is $$n$$, then $$a^i = a^j$$ if and only if $$i \equiv j \ \mathrm{mod} \ n$$. In fact, this was the third day we’d spent on this theorem. So far, we had written down the hypothesis and several equivalent forms of the conclusion and I had asked the students what they should do next. Silence. More silence. Finally, I told them to pair off, and please exit the room. Find a quiet spot somewhere else in the building and tell me where you’ll be. Work on the proof for ten minutes and then come back.
As I wandered around from pair to pair I was very surprised to find animated conversations taking place about the proof. It wasn’t because of the time constraint — they’d been at this for three days now. For whatever reason, they were suddenly into it. One pair was practically arguing with each other over the right approach to take. By the end of the 10 minutes, two of the groups had come up with novel and mathematically watertight arguments. Between the two, and with a little bit of patching and a lemma that needs to be proven still, they generated the proof.
One student made the remark that she had been thinking of these ideas all along, but she didn’t feel like it was OK to say anything. This is a very verbal, conversational class done in Moore method style, so I can only interpret that comment to mean that she didn’t feel free enough, or bold enough, to say what she was thinking. The right proof was just bottled up in her mind all this time.
There’s something about our physical surroundings which figures in significantly to our effectiveness as problem solvers. Getting out of the classroom, for this one student at least, was tantamount to giving her permission to have the correct thoughts she was already having and to express them in a proof. I think our problem solving skills are highly inhibited by fear — fear that we will be wrong. And it takes a tremendous amount of confidence and/or courage as a problem solver to overcome that fear.
When you feel that fear in a classroom, it becomes compounded by the dread of looking like an idiot. Changing the surroundings — making things a little less cozy, a little more unusual and uncertain — doesn’t seem to make the fear go away as much as it helps us feel like that fear is perfectly normal and manageable, if not less fearful.
This entry was posted in Abstract algebra, Critical thinking, Education, Higher ed, Math, Problem Solving, Teaching and tagged abstract algebra, Education, Problem Solving, proof, Teaching, theorem proving. Bookmark the permalink.
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### Search Casting Out Nines
• I am a mathematician and educator with interests in cryptology, computer science, and STEM education. I am affiliated with the Mathematics Department at Grand Valley State University in Allendale, Michigan. The views here are my own and are not necessarily shared by GVSU.
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http://physics.stackexchange.com/questions/2787/how-to-correctly-show-units-with-a-vector
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# How to correctly show units with a vector
Supposed I have an position vector $$\vec{r}=\begin{pmatrix} 10.0 & -30.0 & 25.0\end{pmatrix}$$ expressed in $\mathrm{millimeters}$.
What is the correct notation to display $\vec{r}$
1. $\begin{pmatrix} 10.0 & -30.0 & 25.0\end{pmatrix}\text{ mm}$
2. $\begin{pmatrix} 10.0\text{ mm} & -30.0\text{ mm} & 25.0\text{ mm}\end{pmatrix}$
3. $\begin{pmatrix} 10.0 & -30.0 & 25.0\end{pmatrix}\text{ in }\mathrm{mm}$
If the answer is 2. then why add all those redundunt units when all elements of a vector have to be of the same unit. If you have a long list of values then usualy you present this a table with the units in the header (and not on each element). What if the units are complex (with powers and fraction), do we really have to write them out for each element?
How would you consicely write out a vector while describing the units those values are in also at the same time?
PS. I did not post this in the `Math SE` because they have never heard of units :-) and only physics deals with real situations.
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I think the correct notation is the second because the dimension of each component is a length. – Andrea Amoretti Jan 13 '11 at 18:43
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From a typesetting point of view I think you want $123.45\text{ mm}$ rather than $123.45\text{mm}$. To get this in LaTeX consider `123.45\text{ mm}`. This applies to your form #1 as well: it should be $(10.0, -30.0, 25.0)\text{ mm}$. You'll also note that I like commas in vector expansions performed inline (but not in matrices in general). – dmckee♦ Jan 13 '11 at 18:50
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If you have a units package like `siunitx` available in LaTeX you can write `\SI{123.45}{mm}` and it will add the space automatically. If there's no units package available, I usually use `123.45\,\mathrm{mm}` although I guess it doesn't really matter whether you use that or dmckee's method (which is a little shorter) - nobody here is going to complain about having the wrong size space in your measurements! – David Zaslavsky♦ Jan 13 '11 at 19:02
## 4 Answers
I would say 1. and 2. are correct. In the first you are multiplying by the scalar 1 mm.
Also elements of vectors dont need to have the same units.
Just consider the 4-vector (3 sec, 1 µm, 82 ly, 5.6 parsec)
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1
Although technically you're right about units, in that it is possible to perform well-defined operations on vectors in which the components have different units, it's a major pain in the butt when you're working with linear transformations. It also requires you to introduce a non-identity metric to take the inner product. So in practice usually we normalize the units so that they're the same on all components. – David Zaslavsky♦ Jan 13 '11 at 19:00
The 4-vector above is a block-vector, with 1 time + 3 position quantities. Those do not behave like regular vectors because you cannot get a magnitude out of them without a non-scalar metric (as mentioned elsewhere correctly). In a pure sense a vector has to have the same units, although when it comes to SI prefixes it get complicated like (1 μm,1 km,1 ly) ?? – ja72 Jan 13 '11 at 19:26
@jalexiou: then you should make that clear. I thought $c = 1$ is implied ;-) – Marek Jan 13 '11 at 22:42
@jalexiou: sorry, I didn't notice you are not an author of this answer. In that case, I don't know why you think $c = 1$ is not implied :-) – Marek Jan 13 '11 at 22:44
I write it as 10i + 20j + 5k and set the scale somewhere before the calculations. Of course, you could add it after the vector, but that just doesn't look neat and tidy.
Or you could write the unit vector on the side and multiply the magnitude (with correct units) into it. That could also clarify things. I guess it just depends on how you grasp things and what's clearer to you. As long as you keep others in the loop then it should be okay.
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I have to ask something though, what's the shortcode for writing i-cap in latex? – Anna Jan 13 '11 at 19:12
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I think what you're looking for is \hat{\imath}. (Looks like: $\hat{\imath}$) The "\imath" is an undotted i, so you get the hat without the dot. (There's also a \jmath: $\jmath$.) – Matt Reece Jan 13 '11 at 21:16
Both 1 and 2 are perfectly fine notation. As someone who teaches introductory physics a lot, I would use either in class and accept either as correct from a student. #3 expresses the meaning perfectly clearly but is not the usual form of expression. I would certainly have no problem if a student wrote that in my class, for instance.
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The correct notation is that of multiplying the whole thing by a unit (as a multiplication constant):
`(a,b,c)*mm`
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So how would you show (1 mile, 1 mm, 1 angstrom) in the same vector? I wonder... Maybe (1,0,0)*mile. – ja72 Jan 15 '11 at 0:46
think of units as of multiplication by unknown constant. You can put it inside, outside, whereever you want. – Pavel Radzivilovsky Jan 16 '11 at 23:20
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http://math.stackexchange.com/questions/195833/searching-for-a-multiplicative-function-to-separate-complex-numbers
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# Searching for a multiplicative function to separate complex numbers
Let $g$ be a complex number, where $a$, $b$, $c$, $d$ are real numbers, and $i = \sqrt{-1}$.
$g = \frac{{(a + bi)\exp (ci)}}{{{\rm{abs}}(a + bi)}}$
Since the absolute value (i.e. modulus) of a product of complex numbers is the product of the absolute values:
${\rm{abs((a + b}}i{\rm{)(c + d}}i{\rm{)) = abs(a + b}}i{\rm{)abs(c + d}}i{\rm{)}}$
In addition,
$\exp (ix) = \cos (x) + {\rm{ }}i\sin (x)$
${\rm{abs}}(\exp (ix)) = 1$
So from the above,
${\rm{abs(g) = 1}}$
I am searching for a Multiplicative function $f(g)$ such that:
$f(g) \neq 1$
$f(g) = f\left( {\frac{{(a + bi)}}{{{\rm{abs}}(a + bi)}}} \right)f\left( {\exp (ci)} \right) \ne 1$
Ideally,
$f\left( {\frac{{(a + bi)}}{{{\rm{abs}}(a + bi)}}} \right) = 1$
or
$f\left( {\exp (ci)} \right) = 1$
but
$f\left( {\frac{{(a + bi)}}{{{\rm{abs}}(a + bi)}}} \right) \ne f\left( {\exp (ci)} \right)$
Does such a function exist? Why or why not? What if $a,b,c$ are not known, and only $g$ is known?
Note that it appears that the complex logarithm (Wikipedia) of $g$ will always have a real part equal to zero, so that $\log(g)$ has only an imaginary part:
$\log (g) = \log \left( {\frac{{a + bi}}{{{\rm{abs(}}a + bi{\rm{)}}}}} \right) + \log (\exp (ci))$
$\log (g) = \log ({\rm{abs(}}a + bi)) + \arg (a + bi)i - \log ({\rm{abs}}(a + bi)) + \log (\exp (ci))$
$\log (g) = \arg (a + bi)i + \log (\exp (ci))$
If there is a logarithm to another base $k$, such that ${\log _k}\left( {\frac{{a + bi}}{{{\rm{abs(}}a + bi{\rm{)}}}}} \right)$ is real and ${\log _k}\left( {\exp (ci)} \right)$ is complex (or vice versa), then it might be possible to separate ${\frac{{a + bi}}{{{\rm{abs(}}a + bi{\rm{)}}}}}$ from ${\exp (ci)}$.
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If you assume $g\neq 1,$ then complex conjugation, or even identity, works. – Andrew Sep 14 '12 at 18:35
I think he actually wants a group homomorphism $f: U(1) \to \mathbb{C}^*$ so that $Im(f) \nsubseteq U(1)$. – N. S. Sep 14 '12 at 18:38
N.S. Yes, N.S., I think that is a very elegant way of saying it, and Andrew's comment is correct as well. Ideally I would like $f\left( {\frac{{(a + bi)}}{{{\rm{abs}}(a + bi)}}} \right) = 1$ or $f\left( {\exp (ci)} \right) = 1$ – Nicholas Kinar Sep 14 '12 at 18:49
Can such a group homomorphism be constructed? Is there a good book that can be used as a reference? – Nicholas Kinar Sep 14 '12 at 19:42
## 1 Answer
If you want this to hold all the time (i.e. for any $a,b,c$ excluding $a=b=0$), then your second point has to be relaxed. $\frac{a+bi}{|a+bi|}$ and $e^{ci}$ are both points on the unit circle. So if one of those values maps to $1$, the other can have its constants chosen so that it is the same value, and so is also mapped to one. (e.g, with $a=0,b=1$ we choose $c=\pi/2$). So if the product can never map to $1$, then nothing can map to $1$. At the moment I'm unclear as to why something such as $f(z)=2|z|$ wouldn't suit your purposes, since that seems a bit simple for the complexity of the question.
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Thanks for your response. Ideally, I would like $f\left( {\frac{{(a + bi)}}{{{\rm{abs}}(a + bi)}}} \right) = 1$ or $f\left( {\exp (ci)} \right) = 1$, if possible. – Nicholas Kinar Sep 14 '12 at 18:52
That's the thing: if $f(e^{ic})$=1, I can pick $a,b$ such that $\frac{a+bi}{|a+bi|}=e^{ic}$, so that the whole product will be $1$. If anything ever maps to one, then a product exists that will as well. So if the product can never map to one, then nothing can. – Robert Mastragostino Sep 14 '12 at 18:59
Thanks Robert. Okay, but what if I cannot pick $a,b$, and these are "given"? Is there still a transformation $f(x)$ with the properties that I am searching for? – Nicholas Kinar Sep 14 '12 at 19:09
@NicholasKinar if they're given, you can pick $c$ such that they're equal and you get the same problem. If you can't pick anything, then it's trivial because you're only dealing with a domain of three numbers: just pick $f(\frac{a+bi}{|a+bi|})=m, f(e^{ic})=n$ and define $f(\frac{(a+bi)e^{ic}}{|a+bi|})=mn$. – Robert Mastragostino Sep 14 '12 at 19:43
If I'm given $g = \frac{{(a + bi)\exp (ci)}}{{{\rm{abs}}(a + bi)}}$ without knowing $a,b,c$, can I still find an $f(x)$ such that $f(g) = f\left( {\frac{{(a + bi)}}{{{\rm{abs}}(a + bi)}}} \right)$ or $f(g) = f\left( {\exp (ci)} \right).$ Thanks again, Robert. – Nicholas Kinar Sep 14 '12 at 19:50
show 3 more comments
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http://mathoverflow.net/questions/107664/intersection-of-hilbert-class-fields-of-imaginary-quadratic-fields/107672
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## Intersection of Hilbert class fields of imaginary quadratic fields
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
In this question http://mathoverflow.net/questions/76628/hilbert-class-field-of-quadratic-fields it is mentioned that if $d\equiv 1 \mod 4$ then the Hilbert class field of $\mathbb{Q}(\sqrt{-d})$ contains $\mathbb{Q}(i,\sqrt{d})$.
Could someone point me to where the intersections of Hilbert class fields of imaginary quadratic fields is discussed in more detail?
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This is similar to a CM version of this question: mathoverflow.net/questions/74906/… – Dror Speiser Sep 20 at 18:56
@Dror: I think in the question you link to, as JSE comments, it is really a group theoretic question about $SL_2(\mathbb{Z}_p)\times SL_2(\mathbb{Z}_p)$. For an elliptic curve with CM by $\mathcal{O}$, a CM version of the other question would be talking about the extra structure on the Tate module as a rank 1 $\mathcal{O} \otimes \mathbb{Z}_p$-module. I see this question having more to do with the action of $Pic(\mathcal{O})$ on the Hilbert class field. I'm struggling to see the whole picture at the moment with CM curves, could you please expand a little as to the similarities you see? – Adam Harris Sep 20 at 21:58
Leaving the solution to the other question aside, torsion points of a CM elliptic curve generate the abelian extensions of the hilbert class field of the relevant quadratic field. So, this question is one level before the CM version of the linked question: here you want an intersection before taking torsion points, but just the defining field of the elliptic curve. So it's not exactly the CM version, but it's related. It's just a comment, and I'm not sure how to pursue it. (Also, I did not realise before that you asked the other question as well.) – Dror Speiser Sep 21 at 0:23
## 2 Answers
The generalization of the phenomena you see is genus theory. If $K = \mathbf{Q}(\sqrt{-d})$ and $H = K(j_d)$ then $H$ contains the Genus field $G$.
If $d = \prod_{i=1}^n p_i$ is squarefree (and odd for convenience's sake) then $G = K(\sqrt{p_i^*})$ where $p_i^* = (-1)^{(p_i-1)/2}p_i$. In particular $p_i^* = p_i$ if and only if $p_i \equiv 1\bmod 4$. Therefore if $d \equiv 1\bmod 4$ then $d = \prod_i p_i^*$. Therefore if $d \equiv 1\bmod 4$ then $G$ and thus $H$ contains $K(\sqrt d) = K(i)$.
In general, if $d_1$ and $d_2$ have lots of prime factors in common, then the genus fields of their corresponding imaginary quadratic fields will also have large intersections. Outside of the genus field, I don't know of any studies into the intersections of Hilbert Class Fields.
The standard reference for this is Cox's wonderful book "Primes of the form $x^2 + ny^2$."
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@stankewicz: Thanks! - I think you must mean $p_i^*=(-1)^{(p-1)/2}p_i$ though? – Adam Harris Sep 20 at 14:06
Yes, I always do that for some reason. Thank you very much. – stankewicz Sep 20 at 14:08
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
It is not very clear to me what you mean by "intersection of Hilbert Class Fields [...] is discussed". The theory of Complex Multiplication (see, for instance, Serre's short note in Cassels and Frohlich's Algebraic Number Theory, or Silverman's Advanced Topics in the Arithmetic of Elliptic Curves, or directly the bible from Shimura, Introduction to the Arithmetic Theory of Automorphic Functions) tells you that there is an explicit way, given an imaginary quadratic field $K=\mathbb{Q}(\sqrt{d})$ to construct not only its Hilbert class field, but all ray class fields of different conductors. For instance, for the Hilbert class field, you can first create the elliptic curve $\mathbb{C}/\mathcal{O}_K$: it is an elliptic curve with complex multiplication by $\mathcal{O}_K$. Then, the theory will tell you that the smallest extension of $\mathbb{Q}$ containing $\sqrt{d}$ over which this curve is defined is the Hilbert Class Field of $K$. Just to be convinced that what I say is believable (beside being true, for which you might look at the references), observe that there is an elliptic curve with CM by $\mathbb{Z}[i]$ which admits a Weierstrass equation $$y^2=x^3+x\;.$$ This is defined over $\mathbb{Q}$, so the smallest field of definition which contains $\sqrt{-1}$ is $\mathbb{Q}(i)$, which is indeed its own Hilbert class field.
In more concrete terms, if you are given a squarefree $d<0$ then the Hilbert class field of $K=\mathbb{Q}(\sqrt{d})$ is$K(j(\sqrt{d})$, where $j$ is the modular function $$j(q)=\frac{1}{q}+744+196884q+\dots$$ whose definition you'll find in the above references, or at http://en.wikipedia.org/wiki/J-invariant . All the fun is in proving that $j(\sqrt{d})$ is actually an algebraic number (it is even an algebraic integer!); then, of course, one proves the abelian+unramified property of the (now, finite!) extension $K(j(\sqrt{d}))/K$. Quiteremarkably, it is an algebraic number ''because" its conjugates are precisely the $j$-invariants of the elliptic curves $\mathbb{C}/\mathfrak{a}_i$ for $\mathfrak{a}_i$ running through a set of integral representatives of the ideal class. This, already, shows that the degree of the extension coincides with the class number (and rapidly leads to it being at least Galois).
In your case, life is easier: you only want to be sure that if $d\equiv 1\pmod{4}$ is negative and squarefree, then $K(i)/K$ is unramified (being abelian, this would force it to lie inside the Hilbert class field). For this, it is enough to observe that $K(i)/\mathbb{Q}$ is a biquadratic extension with Galois group $(\mathbb{Z}/2)^2$ and has therefore three quadratic subfields: $\mathbb{Q}(i),K$ and $F=\mathbb{Q}(\sqrt{-d})$. Now pick a prime $\ell$ dividing $d$: it is necessarily odd. Its ramification degree is $2$ both in $K$ and in $F$, while it is unramified in $\mathbb{Q}(i)$. Therefore it needs be unramified in $K(i)/K$, since ramification degrees are multiplicative in towers of extensions. Similarly, $2$ is unramified in $F/\mathbb{Q}$ and cannot ramify in $K(i)/K$. For what concerns the infinite primes, observe that $K$ is already totally complex, so no ramification can occur. Done!
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@Filippo: Thanks! I was thinking about intersections in general though, for example does a similar situation occur for $d \equiv 3 \mod 4$? Is there a standard reference for the statement for $d \equiv 1 \mod 4$? – Adam Harris Sep 20 at 11:37
2
The situation in the case $d\equiv 3\pmod{4}$ is slightly different because the extension $K(i)/K$ needs be ramified at $2$ for the same reason as before. It is then a little bit hard to understand what happens, but if you restrict to $\ell$ being a prime $\equiv 3\pmod{4}$ then $2$ cannot divide the class number of $K$ (see Theorem 10.4 (b) in Washington's Introduction to cyclotomic fields). I know of no standard references, and I would call this kind of results "small/clever/ingenous/trivial trick" according to your taste...Otherwise, Stankewicz' answer correctly pointing to genus theory. – Filippo Alberto Edoardo Sep 20 at 12:56
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http://gilkalai.wordpress.com/2011/07/10/a-couple-updates-on-the-advances-in-combinatorics-updates/?like=1&source=post_flair&_wpnonce=eb9f8a7f00
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Gil Kalai’s blog
## A Couple Updates on the Advances-in-Combinatorics Updates
Posted on July 10, 2011 by
In a recent post I mentioned quite a few remarkable recent developments in combinatorics. Let me mention a couple more.
## Independent sets in regular graphs
A challenging conjecture by Noga Alon and Jeff Kahn in graph theory was about the number of independent sets in regular graphs. The conjecture asserts that the number of independent sets in an N-vertex d-regular graph is at most $(2^{d+1}-1)^{N/2d}$. Alon proved in 1991 a weaker form of the conjecture that was posed by Granville. Kahn proved in 2001 the conjecture for bipartite graphs. In 2010, Yufei Zhao, an undergraduate student, proved the conjecture. (The number of independent sets in a regular graph, Combinatorics, Probability and Computing 19 (2010), 315-320. A link to the arxived paper.) The proof is based on a surprising reduction to the bipartite case. Zhao’s result came as a surprise to several experts in the field who were working on this problem.
## Around Roth’s theorem
The second development is about Roth’s theorem that we often discuss (E.g., here, and here and here).
Suppose that $R_n$ is a subset of $\{1,2,\dots, n \}$ of maximum cardinality not containing an arithmetic progression of length 3. Let $g(n)=n/|R_n|$. The gap between the upper and lower bounds for g(n) is exponential. Can we look behind the horizon and make an educated guess about the true behavior of g(n)?
A recent paper entitled “Roth’s theorem in many variables” by Tomasz Schoen and Ilya D. Shkredov contains a remarkable piece of evidence which is strong enough for me to update my beliefs on the problem.
Schoen and Shkredov proved that if a subset A of {1, 2,…, N} has no nontrivial solution to the equation $x_1+x_2+x_3+x_4+x_5=5y$ then the cardinality of A is at most $N e^{-c(log N)^{1/7-t}}$, where $t>0$ is an arbitrary number, and c>0 is an absolute constant. In view of the well-known Behrend construction their estimate is close to best possible.
Roth’s theorem is about the equation $x_1+x_2=2y$. I used to think that much better bounds than Behrend are quite possible. (But I was aware that most experts have the opposite view.) Schoen and Shkredov’s result is a strong piece of evidence that the truth is near the Behrend’s bound. Two comments: First, if there are conceptual differences between the case of 6 variables and the case of 3 variables I will be happy to know about them. Second, additional interesting examples of sets without 3-term AP are still very welcome.
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### 4 Responses to A Couple Updates on the Advances-in-Combinatorics Updates
1. Terence Tao says:
Well, on a technical level the key improvement is that iterated sumsets such as A+A+A+A are much “smoother” than single sum sets A+A, in that the former are known to contain large structured sets (e.g. generalised arithmetic progressions or Bohr sets), whereas the latter can still have a lot of “holes”. Consider for instance the case when A is a subset of Z/NZ of the form $B \backslash -B$, where B is a random set of density 1/2. Then A has density 1/4 but A+A omits 0; nevertheless, A+A+A (and larger sumsets) are almost certain to fill up all of Z/NZ.
The equation $x_1+\ldots+x_5=5x_6$ is basically the assertion that the dilate $5 \cdot A$ cannot “hide” in the holes of the five-fold sumset $A+A+A+A+A$ (modulo a little technicality involving the trivial solutions). This will be a lot harder to be possible if, say, $A+A+A+A$ does not have many holes. In contrast, the set $A \hat{+} A$ (sums of distinct elements of A) has a lot more holes, and it is a priori conceivable that $2 \cdot A$ manages to completely miss this set.
Nevertheless, this is indeed a very nice psychological and conceptual breakthrough. The main tool is Sander’s near-solution to the polynomial Freiman-Ruzsa conjecture. It does raise the hope that further progress on the PFR conjecture will be useful in getting better bounds on other additive combinatorics problems.
• Gil Kalai says:
Dear Terry, this is very interesting. Probably it is very hard “to look behind the horizon” and all we can say is that Behrend’s bound would be tight unless we can eventually come up with a new type of construction showing otherwise. I think that for the cap set problem (say over $F_p^n$ it is not known that you can prove an upper bound of $(p-t)^n$ (for t>0) even if you consider one equaltion with many variables as Schoen and Shkredov do.
2. Yufei Zhao says:
I feel very honored to see my result featured on your blog. Thanks!
GK: My pleasure, Yufei
3. harrison says:
While I don’t mean to downplay Yufei’s result, it’s worth noting that we still can’t prove the analogous conjecture for other homomorphisms, in even as simple a case as 3-colorings. (Galvin and Tetali proved it for bipartite graphs using a method similar to Kahn’s, but Yufei’s reduction trick doesn’t work — although he does have, I believe, a preprint extending his method as far as it’ll go, it doesn’t work for colorings.)
Again, I don’t mean to insult anyone with that statement (although maybe I’d be insulted most of all, as I spent last summer doing an REU in which I tried to extend Yufei’s trick to 3-colorings) — it just seems to me to be a classic example of a problem which sounds like it should be much easier than it actually is. It’s definitely very interesting stuff, though!
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http://crypto.stackexchange.com/questions/5173/is-there-an-algorithm-for-factoring-n-which-is-just-as-simple-as-this-one-but?answertab=oldest
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# Is there an algorithm for factoring N, which is just as simple as this one, but faster?
I found a simple algorithm for factoring semiprime numbers, you can read about it in Factoring Semiprimes and Possible Implications for RSA.
It basically works like this:
• You reverse the digits in $N$, let's call this value $Ň$
• You pick an integer, $k$ (normally 1, but can be other values)
• A variable, $Δ$, can be any integer value between the square root of $N$ and the square root of $Ň$
You then calculate four values, if any return a value other than 1, then you have one of your prime factors:
• $gcd[N, (k × Ň) + Δ]$
• $gcd[N, (k × Ň) - Δ]$
• $gcd[N, (Δ × Ň) + k]$
• $gcd[N, (Δ × Ň) - k]$
Using this algorithm and the GMP library, it would factor 18014417929109603 in 3 seconds.
I was wondering if there were any other algorithms which were faster than this, but just as easy to implement? I know the GNFS is the fastest, but it is also incredibly hard to implement.
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2
Did you look into the naive quadratic sieve? It's reasonably sane to implement and kind of fun. There's also a good open-source implementation you can look into, MSieve. On a related note, the pdf feels and reads like snake oil... I whipped up a Python implementation and while 18014417929109603 factored in 5 seconds, 31226716938897156373 at least takes longer than 10 minutes. – Thomas Oct 28 '12 at 0:00
Huh, that paper's author doesn't even know that an equation should have an $=$ sign in it. Also, there is no reasoning that the decimal system is in any way preferable for calculating the digit-reversal $Ň$ (why not use binary, ternary, 13-ary, ... system)? – Paŭlo Ebermann♦ Oct 28 '12 at 0:43
I've looked into the quadratic sieve, but I'm not a mathematician and find it hard to understand. It looks very promising if I could get it to work though. – Sam Kennedy Oct 28 '12 at 1:13
## 2 Answers
Pure nonsense. For choosing the random $\Delta$ between $\sqrt{\min(N, Ň)}$ and $\sqrt{\max(N, Ň)}$ there are too many possibilities for it to work. For example whenever the first and last digits of $N$ differ, you get something like $\frac{1}{10} \cdot \sqrt N$ possibilities (the exact formula doesn't matter).
So you can replace the first formula $gcd[N, (k × Ň) + Δ]$ with $k=1$ by this one: $gcd[N, \Delta]$. Try all $\Delta$ between $1$ and $\sqrt N$, the number of possibilities is within some small factor.
Compared to the division, computing $gcd(N, x)$ has the advantage that at each attempt you try all factors of $x$. This is good in case $N$ has some small factors, which in case of RSA is surely not true.
I see no advantage of the linked algorithm to trial division. Trying all primes up to $\sqrt N$ is probably faster, terminates also when $N$ is prime, and has a bounded time.
Summary: pure nonsense.
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Are either of these better alternatives? coolissues.com/mathematics/Goldbach/goldbach.htm mirlabs.info/cddump/data/4437a108.pdf – Sam Kennedy Oct 28 '12 at 2:05
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This doesn't really answer the title question though, even if it refutes the proposed algorithm. – Thomas Oct 28 '12 at 2:27
– maaartinus Oct 28 '12 at 2:27
@SamKennedy The second document is full of grammatical errors and references the original one, which pretty much seals its fate as crackpot material. The first one claims to solve three open mathematical questions (goldbach conjecture, twin primes conjecture, and the ubiquitous factoring problem) through a single quadratic equation.. not buying it. – Thomas Oct 28 '12 at 2:33
@Thomas: Doesn't it? I'm basically saying that the choice of Ň gains nothing. So yes, there are better algorithms out there, and for number as small as the given one it's easy to implement. – maaartinus Oct 28 '12 at 2:34
show 1 more comment
Yes there are several algorithms, as simple or even simpler as the proposed algorithm, that are expected to factor $N$ much faster. The simplest and oldest is the Sieve of Eratosthenes (which works for any $N$); there's also Fermat's factoring method, preferably with the easy sieve improvement to recognize squares efficiently (which works when $N$ is the product of two primes $P$ and $Q$ with $|P-Q|$ of the same order of magnitude as $\min(P,Q)$, which seems the aim of the article). Arguably, the more efficient Pollard's Rho is on the other side of the simplicity threshold.
The algorithm proposed in the question and article essentially attempts to factor a semi-prime $N=P\cdot Q$ by repeatedly picking a pseudo-random integer $R$ smaller than a few times $N$, and computing $G=\mathtt{GCD}(N,R)$, until that is not $1$. The last $G$ is generally either $P$ or $Q$ (the only other possibility, $G=N$, is rare).
Odds of success at each $\mathtt{GCD}$ test is about $1/P+1/Q$ (thus less than $N^{1/2}/2$ steps are expected). For a 1024-bit RSA key odds of success per $\mathtt{GCD}$ may be $2^{-510}\ll10^{-153}$. The distributed attack envisioned at the end of the article expects to make $10^{14}$ such $\mathtt{GCD}$ tests, and thus has odds less than $10^{-139}$ to succeed. Hélas, I have no reason to doubt that the article was accepted in an IEEE-endorsed conference despite absence of practical interest and discussion about its (in-)efficiency.
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I quickly wrote a program using Pollard's Rho algorithm, it's much faster than than the original (factored 31226716938897156373 in 0.055 seconds). I have some questions about Pollard's Rho algorithm now, should I ask here or start a new question with the relevant topic? – Sam Kennedy Oct 28 '12 at 16:29
@Sam Kennedy: Congratulations for implementing Pollard's Rho (I suggest a new question if you want info about that). Logical next steps may be Pollard's p-1 and improvements; then Quadratic Sieve and variations (MPQS, SIQS), where earlier work is reused to some degree; or Pollard's p+1 then ECM. – fgrieu Oct 29 '12 at 9:02
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http://math.stackexchange.com/questions/60165/stuck-with-ieee-754-format-binary-to-decimal-conversion
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# Stuck with IEEE-754 format (binary to decimal conversion)
I am stuck with a problem,
Now here is the formula applied to a question,
The red colored bits show the exponent(e) while, the yellow bits are fraction (f). Now I am confused with how did the author convert the fraction (f) into 1-2^(-23). I understand that he has converted from binary to decimal to get this value but how do I convert such a big number from binary to decimal?
EDIT:
One trick that my teacher told me was to convert from binary to hexadecimal and then convert the value to decimal easily. But doing that has given me a value `8388607` which is not equal to `1 - 2 ^(-23)`
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## 2 Answers
The conversion from $23$ bits, all $1$ to $1-2^{23}$ comes from the fact that summing $\sum_{i=1}^{23}2^{-i}=1-2^{-23}$. The final conversion is from $-2^{128}$ (the $2^{104}$ is negligible) to $-3.4E38$, which is done using the base $10$ logarithm of $2$. $\log_{10}2^{128}=128\log_{10}2\approx 128\cdot.30103\approx 38.532$, so $2^{128} \approx 3.4E38$
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I think that knowledge about U2 (a.k.a ZU2) binary coding and number conversions might be helpful :
http://en.wikipedia.org/wiki/Two%27s_complement
For exercises with IEEE-754 you might like this on-line calc:
http://www.binaryconvert.com/
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http://mathoverflow.net/questions/88741/methods-for-determining-domains-of-influence/89107
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## Methods for determining domains of influence
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Given a hyperbolic PDE, the domain of influence of a spacetime point $x$, say $I_x$ though $x$ could be replaced by any set, can be defined in two ways. Lets call one of them geometric ($I_x^G$) and the other analytical ($I_x^A$). In Lorentzian geometry, the geometric domain of influence consists of the interior of the cone of null geodesics emanating from $x$ (let me not bother about whether to include the boundary in the definition of $I_x^G$ or not). In general, a similar definition can be given using characteristic cones instead of null cones. The analytical domain of influence can be defined as the set of all spacetime points $y$ such that for every neighborhood neighborhood $O$ of $x$ there exist two solutions $u_1$ and $u_2$ satisfying the condition $u_1(x')=u_2(x')$, for all $x'$ on a Cauchy surface passing through $x$ except for $x'\in O$, and also the condition $u_1(y)\ne u_2(y)$. The latter one is the definition used in Lax's book on Hyperbolic PDEs.
Similar defintions can be given for the geometric and analytical domain of dependence, say $D_S^G$ and $D_S^A$. Such a definition should capture the desired equality $D_K = I_{S\setminus K}$, for a Cauchy surface $S$ and $K\subset S$ (once again, being sloppy with boundaries). I know that energy methods can be used to establish that $D^G_K \subseteq D^A_K$ (the analytical domain of dependence is at least as large as the geometric one). Hence, by duality, the same methods establish $I_K^A \subseteq I_K^G$ (that the geometric domain of influence is at least as large as the analytical one).
My question is about the reverse inclusion, $I_K^G \subseteq I_K^A$ or by duality $D_K^A \subseteq D_K^G$. Maybe it's too much to ask for the analytical and geometric definitions to coincide. But when they do, what methods are used to establish that? When they don't what methods can identify the obstruction? Except briefly in Lax's book, I don't know what references discuss this problem explicitly, so those would also be appreciated!
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Just added a bounty to raise the question's profile. – Igor Khavkine Feb 21 2012 at 13:41
Your definition of analytic domain of dependence needs to be more carefully re-written. I think instead of $\exists O$ you need $\forall O$: otherwise take any Cauchy surface $\Sigma$ through $x$ and any $y\in \Sigma$: you can certainly choose neighbourhood $O$ of $x$ composed of $O_x\cup O_y$, where $O_y$ is a neighbourhood of $y$. Then by your definition $y$ is in the domain of influence, which is nonsensical. – Willie Wong Feb 24 2012 at 16:15
Willie, you're right. I want the analytical domain of influence to be as small as possible, so I should ask for the relevant conditions to be satisfied for arbitrarily small neighborhoods $O$ of $x$. I'll edit the question accordingly. – Igor Khavkine Feb 28 2012 at 10:51
## 2 Answers
I highly doubt the result you actually asked for is true.
Consider the linear wave equation on $(1+3)$ Minkowski space. The analytic domain of influence of a point $x$ as Lax defined it, which morally says that $y$ is in the analytic domain only if one can find perturbations in arbitrary small neighborhoods of $x$ that change $y$ (if I interpret your question statement correctly), actually consists of only the null cone emanating from $x$ and nothing more, since strong Huygen's principle holds.
The same is true for the linear wave equation on $(1+(2k+1))$ Minkowski spaces.
The opposite conclusion can be drawn on $(1+2k)$ dimensional Minkowski spaces, where the Green's function have support inside the cone.
For more general situations, you may want to consult the classical result of Atiyah-Bott-Garding on the existence of Petrowsky lacunae. For any linear hyperbolic equation that admits a lacuna, the analytic domain of influence cannot cover the entirety of the geometric one.
But for the result that you seem to actually want, where you should replace the analytic domain of dependence by a suitable "convex" envelope of it, I don't know if such a result is proven anywhere, but my guess is that, at least for the "local" version one can approach it using some sort of geometric optics construction.
For possible references (I haven't actually finished reading either, so they may not contain what you want), maybe you want to look at Michael Beals' book on propagation of singularities (sorry, the title escapes me at the moment) or Rauch's notes on Hyperbolic PDEs and Geometric Optics which I think you can find floating around on the internet.
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Willie, that's for the answer! I had the feeling that the equality of the geometric and analytical domains of influence/dependence was probably too good to be true. I'm basically happy to learn of the relevant PDE lingo that can be seen as addressing the difference between the two. From your answer and from looking through the literature, I see that the main relevant key words are indeed lacunae and geometric optics. I'll accept this answer. And thanks for the references. – Igor Khavkine Feb 28 2012 at 11:00
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I'm not quite sure if this is really the situation you are interested in, but in the book of Bär, Ginoux, and Pfäffle: Wave Equations on Lorentzian Manifolds and Quantization. ESI Lectures in Mathematics and Physics, European Mathematical Society, 2007, they discuss in quite some detail the Cauchy problem for hyperbolic linear wave equations on, and that is the catch, globally hyperbolic spacetimes. I guess that one should require something like that since otherwise you can at best hope for some local statements. But in their situation, I'm pretty sure to remember correctly, they have statements like the one you are looking for (don't they?). In any case, this is maybe a too special situation for you, but the book is nevertheless very nice. Unlike many other texts on hyperbolic PDE, it emphasizes the geometry very much.
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Hi, Stefan. Thanks for the tip! I know that book and also think highly of it. I know for sure that they have a discussion of the domain of dependence theorem of the kind I described above ($D^G\subseteq D^A$). I'll take a look again to see if they talk about inclusion in the other direction or something related. – Igor Khavkine Feb 21 2012 at 23:45
Hi Igor, maybe you're right and it is only about this inclusion. So the point would be to construct a particular wave equation which extremizes this support of its solutions. So I fear that my reference does not answer your question :( sorry – Stefan Waldmann Feb 22 2012 at 8:37
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http://math.stackexchange.com/questions/198646/length-and-area-in-hyperbolic-geometry
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# Length and area in hyperbolic geometry
I am reading a book about modern geometries by Michael Henle. He gives formulas for length of a curve and area of a region (in upper half plane model: $l(\gamma)=\int _a^b \frac{|z'(t)|}{y(t)}dt, A(R)=\iint_{R} \frac{dxdy}{y^{2}})$ as definitions. I am pretty sure these aren't Ad Hoc definitions though, but derived from something else. Am I right?
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– Will Jagy Sep 18 '12 at 18:44
So in the wikipedia article they start with a metric. Can one start, as the book did, with transformations, instead of with a metric (define the geometry to be the upper half plane where two sets are congruent if one set can be obtained by applying tranforms of the form (az+b)/(cz+d) on the other) and derive the metric from that? (Sound improbable.) – tsho Sep 18 '12 at 19:03
I don't particularly know. Give it a try. – Will Jagy Sep 18 '12 at 19:05
## 1 Answer
The invariance under transformations of the form $(az+b)/(cz+d)$ determines the metric up to a constant scalar factor. Think of a metric as a prescription of length of each tangent vector to the plane (the length may depend on the point at which the vector is attached to the plane). To normalize the metric, let's say that the vector $v_0=\langle 1,0\rangle$ attached at the point $i=(0,1)$ has unit length. The transformation $f(z)=(az+b)/(cz+d)$ (where we may assume $ad-bc=1$ for simplicity) has derivative $f'(z)=1/(cz+d)^2$. Thus, it maps $v_0$ to the vector $1/(ci+d)^2$ attached to the plane at $w=f(i)=(az+b)/(cz+d)$. It follows that this vector also has invariant length $1$. Since the Euclidean length of this vector is $1/(c^2+d^2)$, the invariant metric scales it by $c^2+d^2$.
Next observe that $$\mathrm{Im} w= \frac{(ai+b)(-ci+d)}{c^2+d^2}=\frac{1}{c^2+d^2}$$ Hence, the scaling factor at $w$ is $1/\mathrm{Im}\, w$. Note that this factor applies to vectors in all directions, because by varying $a,b,c,d$ we can map $v_0$ into a vector of any direction at any point.
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http://mathoverflow.net/questions/78772/hyperplane-sections-of-isolated-hypersurface-singularities/78776
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## hyperplane sections of isolated hypersurface singularities.
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Given an isolated singularity $p$ in a hypersurface $Y$ of dimension $n$ (let say a surface in $\mathbb{P}^3$). I can intersect $Y$ with a hyperplane $H$ passing through $p$ such that it induces a singularity in a lower dimension. For example, if we start with a surface $Y$, we obtain a plane singularity on $H$.
I want to use the singularities in the hyperplanes for classifying(?) the original singularity in the hypersurface. Using homeomorphism for defining an equivalence relation. There is a finite number of singularities types that I can obtained in the hyperplanes, and one of those types will be "generic".
I want to believe that those hyperplane singularities + some "vicinity" data is enough for characterizing the original singularity. However, I cannot find related work, or theorems to start with. I appreciate any help or references.
Thanks
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## 1 Answer
This is a good strategy and it does work for many types of singularities. In fact, many times you don't even need "vicinity" data if your singularity is isolated. (In other words, being isolated is the "vicinity" data). If you do not assume that the singularities are isolated, then you need to assume something about $X\setminus H$.
Typical theorem
Let $X\subseteq \mathbb P^N$ be a quasi-projective variety and let $H\subseteq X$ be a hyperplane section and assume that $H$ only has singularities of type $\mathfrak T$. Then $X$ only has singularities of type $\mathfrak T$ along $H$. In particular, if in addition $X\setminus H$ only has singularities of type $\mathfrak T$, then so does $X$.
Obviously, whether or not the above theorem is in fact true depends on your choice of $\mathfrak T$. It is known in many cases and conjectured in others.
If singularities of type $\mathfrak T$ satisfy the general inverse of the above, that is if the
General hyperplane section theorem
Let $X\subseteq \mathbb P^N$ be a quasi-projective variety and let $H\subseteq X$ be a general hyperplane section and assume that $X$ only has singularities of type $\mathfrak T$. Then $H$ only has singularities of type $\mathfrak T$.
holds, then the two theorems together imply that small deformations of varieties with singularities of type $\mathfrak T$ still have singularities of type $\mathfrak T$.
In other words, a possible way to find the statement you are looking for is to look for papers that claim that certain singularity types are invariant under small deformations.
Here is a list of singularities for which both of these theorems hold:
• smooth (Sketch of proof): being smooth means that the local ring is regular. If a ring mod a regular element is regular, then so is the ring.
• Cohen-Macaulay : basically by the definition of CM
• Gorenstein : Gor = CM + $\omega$ is a line bundle. CM follows from above and line bundle from the fact that if the restriction of a coherent sheaf to a hypersurface is a line bundle, then so is the original sheaf.
• rational : This is a result of Elkik (Inventiones, cca. 1978)
• klt,dlt,lc,etc : this is essentially inversion of adjunction
• Du Bois : this is a result of Kovács-Schwede (available on arXiv.org)
Remark
In fact, there is a tendency toward $X$ having better singularities than $H$ does. Namely it is sometimes true that (for example)
If $H$ only has singularities of type $\mathfrak T$ and $X\setminus H$ is smooth, then $X$ only has singularities of type $\mathfrak T^+$ for some class of singularities that is milder than $\mathfrak T$.
An example for this is Karl Schwede's theorem (Thm.5.1 in A simple characterization of Du Bois singularities, Compositio Math. 143 (2007) 813–828) that says exactly this with $\mathfrak T$ being "Du Bois" and $\mathfrak T^+$ being "rational.
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I wanted to add a more elementary example of the phenomenon in the last remark. The statement there is also true if we take $\mathfrak T$ to be "reduced" and $\mathfrak T^+$ to be "normal." – Dustin Cartwright Oct 23 2011 at 20:58
That's right. And more generally it works with $\mathfrak T$ being $S_n$ and/or $R_k$ and $\mathfrak T^+$ being $S_{n+1}$ and/or $R_{k+1}$. – Sándor Kovács Oct 23 2011 at 21:30
Thanks a lot for the answers and references. I was also wondering in the line of Arnolds's classification of singularities. Indeed, the Arnold's list, plus C.T.C Wall's work provide a complete classification up to modality ~3, and multiplcity ~6, but higher multiplicity singularities classification seems incomplete. However, I acknowledge those singularities are uglier than usual, and not that often exposed. Any case, thanks. – pmath Oct 29 2011 at 0:57
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|
http://openwetware.org/index.php?title=User:Pranav_Rathi/Notebook/OT/2011/03/01/Laser_Shutter_.2&diff=cur&oldid=657445
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# User:Pranav Rathi/Notebook/OT/2011/03/01/Laser Shutter .2
### From OpenWetWare
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| | ==Motivation== | | ==Motivation== |
| - | Motivation behind designing the shutter is speed, accuracy and variability (time). In our optical tweezers we need to center the trap over the tethered bead, to get the geometry right (because this affects the force measurement) and make our feedback program run. To center the tether we need to turn the laser intensity (trap) on/off quickly for various time intervals. To do this we used to use AOM (acoustooptic module), because it’s extremely quick (nano second on/off time). But AOM has some inherent oscillation problem which makes it use as a switch, unsuccessful. So I needed some-thing which can replace this AOM function. | + | Motivation behind designing this shutter is speed, accuracy and variability (activation-time). In our optical tweezers we need to center the trap over the tethered bead, to get the geometry right (this affects the force measurement) and make our feedback program run. To center the tether we need to turn the laser intensity (trap) on/off quickly for various time intervals. To do this we used to use AOM (acoustooptic module), because it’s extremely quick (nano second on/off time). But AOM has some inherent oscillation problem which makes it use as a switch, unsuccessful. So I needed something else, (a shutter) which can replace this AOM function. |
| | | | |
| - | This shutter does it exact. It is fast (activation time) with an opening time of 4 and closing time of 2 μsec. The shutter runs on a +5 volts voltage. It moves to on-position with an active voltage (controlled by the toggle foot switch) working against a restoring-spring, it remains on when the voltage is applied and turns to off when the voltage is removed. This gives a freedom of choice for the time shutter is active and very simple electronics controls the speed (activation time) | + | This shutter does it exact and it is fast (activation-time) with an opening time of 4 and closing time of 2 μsec. The shutter runs on a +5 volts voltage. It moves to an on-position with an active voltage (controlled by the toggle foot switch) working against a restore-spring and remains on when the voltage is applied and turns to off when the voltage is removed. This gives a freedom of choice for the duration shutter is active with very simple electronics used to control the speed (activation time.) |
| | | | |
| - | In design the laser passes through an aperture, and the only moving part is the cylinder. No gears, and no electronics in the actual shutter, makes this design very stable and accurate, even under the heat produced by the laser beam. | + | In the design the laser passes through an aperture on the only moving part a cylinder. No gears and no electronics in the design make it very simple, stable and accurate, even under the heat produced by a laser beam. The shutter needs no special power supply and can be run on a cell phone charge with an output of roughly 300mA/5V. |
| - | This shutter needs no special power supply, it can be run through a cell phone charge with an output of roughly 300mA/5V. | + | |
| - | Cost and construction time is also important. With this design, a shutter can be prepared under \$40 with 10 hours of construction time (10 x 25 (hourly wage of a technician)=\$250+40=\$290). Still better than many commercially available shutter systems with same performance. | + | Cost and construction time is also important. With this design, a shutter can be prepared under \$40 with 10 hours of construction time (10 x 25 (hourly wage of a technician) =\$250+40=\$290), still better than many commercially available shutter systems with same performance. |
| | | | |
| | == Design & Construction== | | == Design & Construction== |
| Line 42: | | Line 42: | |
| | | | |
| | === Construction=== | | === Construction=== |
| - | The buildup is divided is the following categories. In the start make sure that you have a good set of screw driver, players, wire cutter & stripper and solder. | + | The buildup is divided is the following categories. In the start make sure that you have a good set of screw driver, pliers, wire cutter, stripper and solder. |
| | ==== Shutter==== | | ==== Shutter==== |
| - | I will start with the choice of the motor. A 12V DC motor is a good choice because it can provide a wide rang of torques. The motor and the spring works against each other so it important to have the right set. So choose a 12V DC motor with a shaft length of at least 15mm. Mount the motor on the cage plate as shown in the picture, you will need 4/40 screws to tight it. once this is done, unscrew and take the motor out. | + | I will start with the choice of the motor. A 12V DC motor is a good choice because it can provide a wide range of torques. The motor and the spring works against each other so it important to have the right set. So choose a 12V DC motor with a shaft length of at least 15mm. Mount the motor on the cage plate or platform as shown in the picture, you will need 4/40 screws to tight it (in present setup). Once this is done, unscrew and take the motor out. |
| | | + | |
| | | + | Now next task is to choose the right spring. The spring should be half the length of the shaft and the shaft should easily fit through the spring. I want to keep the spring-choice procedure really simple and easy. To choose the spring first need to know the torque of the motor. This is really simple; it can be mathematically calculated if the voltage and the current are known. But I like the experimental way; the setup is shown in the slide2. In the slide I have a DC motor with a leverage pivoted at the shaft and a free weight hanging with a string on the other side. In this, all I am doing is balancing the weigh at that point by giving the motor just enough power. |
| | | + | So the torque is equal to the weight at the distance from the shaft: |
| | | | |
| - | Now next task is to choose the right spring. The spring should be half the length of the shaft and the shaft should easily fit through the spring. I want to keep all this really simple, so there is a very easy way to choose the right spring. To choose the spring first need to know the torque of the motor. This is really simple, it can be mathematically calculate if the voltage and the current is known. But I like the experimental way; the setup is shown in the slide2. In the slide I have a DC motor with a leverage pivoted at the shaft and a free weight hanging with a string on the other side. In this, all I am doing is balancing the weigh at that point by giving the motor just enough power. | | |
| - | so the torque is equal to the weight at the distance from the shaft: | | |
| | *:: <math> \mathbf{\tau}=r \times F = r \times m .g </math> | | *:: <math> \mathbf{\tau}=r \times F = r \times m .g </math> |
| | | | |
| - | Once this is know we can choose the spring with less torque than this. The restoring torque applied by the spring can also be measure in the same fashion (as motor) as shown in slide1, with hanging known weight. The spring stiffness and the torque can be related through: | + | Once this is know we can choose the spring with less torque. The restoring torque applied by the spring can also be measure in the same fashion (as motor) as shown in slide1, with hanging known weight. The spring stiffness and the torque can be related through: |
| | | + | |
| | *:: <math> \mathbf{\tau}=2{\pi}n.K.r^2 </math> | | *:: <math> \mathbf{\tau}=2{\pi}n.K.r^2 </math> |
| | | | |
| - | Where K is the stiffness, r distance from the shaft center and n number of times spring twisted one complete 360 degrees circle before it was hacked. Since the torque is n dependent, it is a matter of great convenience, because now it is really easy to pick the right spring. All you have to do, is twist the required number of times before hacked to get the right restoring torque. | + | Where K is the stiffness, r distance from the shaft center and n number of times spring twisted one complete 360 degrees circle before it was hacked. Since the torque is n dependent, it is a matter of great convenience, because now it is really easy to pick the right spring. All you have to do is twist the required number of times before hacked to get the right restoring torque. |
| | | + | |
| | | + | There is even easier way to deal with mambo-jumbo, get a motor get a spring and put them together. |
| | | + | |
| | | + | Now we are ready to put the shutter together. Start with the motor; most of the motors have little holes for the screws to hold slide3. Choose a screw which can fit into that but still sticking out few millimeters slide4. Next is the rotation stage which joins the cylinder to the motor and holds the spring assembly. I choose the wood for this function, because it’s easy to machine and a good thermal-insulator slide5. I drilled two hole on both sides; to fit the shaft and 1/4 screw on either sides. To drill for 1/4 I used 15/64 size drill-bit slide6, 7, 8 & 9. Now the motor can be attached to one side of the stage and cylinder to other slide10 & 11. |
| | | | |
| - | Now we are ready to put the the shutter together. Start with the motor, most of the motors have little holes for the screws to hold slide3. Choose a screw which can fit into that but still sticking out few millimeters slide4. Next is the rotation stage which joins the cylinder to the motor and holds the spring assembly. I choose the wood for this function, because its easy to machine and a good thermal-insulator slide5. I drilled two hole on both sides; to fit the shaft and 1/4 screw on either sides. To drill for 1/4 I used 15/64 size drill-bit slide6,7,8 & 9. Now the motor can be attached to one side of the stage and cylinder to other slide10 & 11. | | |
| | | | |
| - | Now we can machine the spring assembly. Slide 12 shows the parts of spring assembly. We need two metal strips of .75 and 1 cm long with a hole in the end for the screws. This strips with the screw on the motor will define the boundaries of the movement. Cylinder will rotate end to end between these strips. To screw the strips directly on the stage we need to drill two holes for the screws. The holes are 135 degree apart over the face of the stage as shown in the slide13. We also drill a small hole for the spring attachment near the edge. | + | Now we can machine the spring assembly. Slide 12 shows the parts of spring assembly. We need two metal strips of .75 and 1 cm long with a hole in the end for the screws. These strips with the screw on the motor will define the boundaries of the movement. Cylinder will rotate end to end between these strips. To screw the strips directly on the stage we need to drill two holes for the screws. The holes are 135 degree apart over the face of the stage as shown in the slide13. We also drill a small hole for the spring attachment near the edge. |
| | | | |
| - | Now all parts are ready to get together slide 14. This shutter is little different of what is shown in the video but follows the same concept. Put the shorter end of the spring in the hole on the stage like slide 15. Now the motor shaft will go through the spring into the stage. The other side of the spring is hold against the screw. You can twist the required number of turns before it holds to get the right restoring torque. Now the spring wants to move the stage clockwise but the strip against the screw does not let it (this is position 1). As the voltage is applied the motor turns counterclockwise working against the spring until the second strip stops it. When the voltage is applied the stage remains in the second position. As soon as the voltage is turned down the spring restores the stage into the previous position. These two positions can be chosen for the shutter to be on/off slide 16,17 & 18. | + | Now all parts are ready to get together slide 14. This shutter is little different of what is shown in the video/pictures but follows the same concept. Put the shorter end of the spring in the hole on the stage like slide 15. Now the motor shaft will go through the spring into the stage. The other side of the spring is hold against the screw. You can twist the required number of turns before it holds to get the right restoring torque. Now the spring wants to move the stage clockwise but the strip against the screw does not let it (this is position 1). As the voltage is applied the motor turns counterclockwise working against the spring until the second strip stops it. When the voltage is applied the stage remains in the second position. As soon as the voltage is turned down the spring restores the stage into the previous position. These two positions can be chosen for the shutter to be on/off slide 16, 17 & 18. |
| | | | |
| | A fully assembled shutter is ready. Now the next task is to prepare a control box. | | A fully assembled shutter is ready. Now the next task is to prepare a control box. |
## Motivation
Motivation behind designing this shutter is speed, accuracy and variability (activation-time). In our optical tweezers we need to center the trap over the tethered bead, to get the geometry right (this affects the force measurement) and make our feedback program run. To center the tether we need to turn the laser intensity (trap) on/off quickly for various time intervals. To do this we used to use AOM (acoustooptic module), because it’s extremely quick (nano second on/off time). But AOM has some inherent oscillation problem which makes it use as a switch, unsuccessful. So I needed something else, (a shutter) which can replace this AOM function.
This shutter does it exact and it is fast (activation-time) with an opening time of 4 and closing time of 2 μsec. The shutter runs on a +5 volts voltage. It moves to an on-position with an active voltage (controlled by the toggle foot switch) working against a restore-spring and remains on when the voltage is applied and turns to off when the voltage is removed. This gives a freedom of choice for the duration shutter is active with very simple electronics used to control the speed (activation time.)
In the design the laser passes through an aperture on the only moving part a cylinder. No gears and no electronics in the design make it very simple, stable and accurate, even under the heat produced by a laser beam. The shutter needs no special power supply and can be run on a cell phone charge with an output of roughly 300mA/5V.
Cost and construction time is also important. With this design, a shutter can be prepared under \$40 with 10 hours of construction time (10 x 25 (hourly wage of a technician) =\$250+40=\$290), still better than many commercially available shutter systems with same performance.
## Design & Construction
### Components
There are three major parts of the system.
• Shutter.
• Control box.
• Power supply.
The components used:
#### Shutter
• 12V DC motor
• Wood rotation stage
• Spring with torque of .1 N m
• Rubber padding
• Pillar Post Extension, Length=1" from Thorlabs (shutter cylinder)
• 30mm Cage Plate Optic Mount from Thorlabs
• Post-holder, base-plate ext...
#### Control Box
• 1 power jack M&F
• 1 1/4" mono Panel-Mount Audio Jack M&F
• 1 Foot Paddle
• 1 100Ω pot with, 1 220Ω resistor
• 1 on/off toggle switch
• 1 LED
• 1 box enclosure
• some connection wires, solder gun and solder wire
#### Power Supply
Any power-supply which can provide 300mA at 5V and above. The motor torque is power dependent and the shutter speed is resorting spring's stiffness (torque) dependent. So choose the spring carefully before decide on the power supply. I would recommend a variable power-supply which can be bought easily from any where.
### Construction
The buildup is divided is the following categories. In the start make sure that you have a good set of screw driver, pliers, wire cutter, stripper and solder.
#### Shutter
I will start with the choice of the motor. A 12V DC motor is a good choice because it can provide a wide range of torques. The motor and the spring works against each other so it important to have the right set. So choose a 12V DC motor with a shaft length of at least 15mm. Mount the motor on the cage plate or platform as shown in the picture, you will need 4/40 screws to tight it (in present setup). Once this is done, unscrew and take the motor out.
Now next task is to choose the right spring. The spring should be half the length of the shaft and the shaft should easily fit through the spring. I want to keep the spring-choice procedure really simple and easy. To choose the spring first need to know the torque of the motor. This is really simple; it can be mathematically calculated if the voltage and the current are known. But I like the experimental way; the setup is shown in the slide2. In the slide I have a DC motor with a leverage pivoted at the shaft and a free weight hanging with a string on the other side. In this, all I am doing is balancing the weigh at that point by giving the motor just enough power. So the torque is equal to the weight at the distance from the shaft:
• $\mathbf{\tau}=r \times F = r \times m .g$
Once this is know we can choose the spring with less torque. The restoring torque applied by the spring can also be measure in the same fashion (as motor) as shown in slide1, with hanging known weight. The spring stiffness and the torque can be related through:
• $\mathbf{\tau}=2{\pi}n.K.r^2$
Where K is the stiffness, r distance from the shaft center and n number of times spring twisted one complete 360 degrees circle before it was hacked. Since the torque is n dependent, it is a matter of great convenience, because now it is really easy to pick the right spring. All you have to do is twist the required number of times before hacked to get the right restoring torque.
There is even easier way to deal with mambo-jumbo, get a motor get a spring and put them together.
Now we are ready to put the shutter together. Start with the motor; most of the motors have little holes for the screws to hold slide3. Choose a screw which can fit into that but still sticking out few millimeters slide4. Next is the rotation stage which joins the cylinder to the motor and holds the spring assembly. I choose the wood for this function, because it’s easy to machine and a good thermal-insulator slide5. I drilled two hole on both sides; to fit the shaft and 1/4 screw on either sides. To drill for 1/4 I used 15/64 size drill-bit slide6, 7, 8 & 9. Now the motor can be attached to one side of the stage and cylinder to other slide10 & 11.
Now we can machine the spring assembly. Slide 12 shows the parts of spring assembly. We need two metal strips of .75 and 1 cm long with a hole in the end for the screws. These strips with the screw on the motor will define the boundaries of the movement. Cylinder will rotate end to end between these strips. To screw the strips directly on the stage we need to drill two holes for the screws. The holes are 135 degree apart over the face of the stage as shown in the slide13. We also drill a small hole for the spring attachment near the edge.
Now all parts are ready to get together slide 14. This shutter is little different of what is shown in the video/pictures but follows the same concept. Put the shorter end of the spring in the hole on the stage like slide 15. Now the motor shaft will go through the spring into the stage. The other side of the spring is hold against the screw. You can twist the required number of turns before it holds to get the right restoring torque. Now the spring wants to move the stage clockwise but the strip against the screw does not let it (this is position 1). As the voltage is applied the motor turns counterclockwise working against the spring until the second strip stops it. When the voltage is applied the stage remains in the second position. As soon as the voltage is turned down the spring restores the stage into the previous position. These two positions can be chosen for the shutter to be on/off slide 16, 17 & 18.
A fully assembled shutter is ready. Now the next task is to prepare a control box.
#### Control Box
I used a 2X3 inch aluminium box to house the electronics slide20. The electronics is really simple. The circuit is shown in the slide21. The foot switch is used to connect the motor ground to the power-supply hence activates the shutter. Since the shutter works on active voltage, the voltage is always on, when the foot paddle is active. So it is very important to give the right voltage to the motor to avoid damage to the motor over the long duration of active voltage. To do this i use a 100 ohm POT. POT let me choose the right voltage and current. This can be done once the shutter is ready.
Now connect the shutter to the shutter input and foot-paddle to its input.
#### Power Supply
The good thing about the system is that it can use any power supply which can give enough power to operate the motor. The POT inside the control box let choose the appropriate voltage and current and hence makes easy to choose a power-supply.
## Comments
Motor shutter buildup
View more presentations from pranavrathi.
Full Shutter System
Shutter
Control box
Shutter different view
Control box different view
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|
http://stats.stackexchange.com/questions/32839/will-logistic-regression-fit-this-situation
|
# Will logistic regression fit this situation?
I have a response variable with 2 categories and $500$ predictor variables. The $500$ coefficient $a_1, a_2, a_3, \ldots, a_{500}$ ranges $(-1, 1)$. Positive $a_i$ indicates category A; negative $a_i$ indicates category B. The larger the coefficient, the stronger it indicates its correspondent category. I get the coefficient from a researcher, who score each coefficient from -1 to 1 based on the importance and influence in classification.
To classify an object that have attributes $x_1, x_2, x_3,\ldots,x_{500}$, I am thinking of using logistic regression. But I do not know how to deal with continuous data (the range of the coefficient is continuous from -1 to 1). Is logistic regression viable?
If not, will someone help with other methods and post your code? I prefer using R.
-
Are you saying that you have a response variable with 500 categories, or a response variable with 2 categories and 500 predictor variables? Also, if you haven't fit a model yet, where did the coefficients come from? You ask about continuous data, are these response variables or covariates? – gung Jul 23 '12 at 15:48
What is your dependent variable? A number ranging from $(-1,1)$? – Macro Jul 23 '12 at 15:48
Whether logistic regression is viable depends (in large part) on the nature of the dependent variable. It sounds like you have a dichotomous DV, which would be right for logistic regression. There may be other problems - 500 predictors is very rarely going to be sensible. – Peter Flom Jul 23 '12 at 15:57
By "rarely going to be sensible," do you mean the predictors are too many? For each object, about 10 to 20 xi are larger than 0, the others = 0. – juju Jul 23 '12 at 16:01
@Macro I have revised the question. Thanks in advance. – juju Jul 23 '12 at 16:09
show 3 more comments
default
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http://mathoverflow.net/questions/120137/how-to-define-laplacian-on-l-2/120145
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## How to define Laplacian on $L_2$
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
This might be a dumb question, but I thought the Laplacian (classical) is defined for $C^2$ functions. How do we extend that to be a self-adjoint operator on all of $L_2$? Is it the so called Friedrich's extension theorem? Is the extension explicit? How would I, for example, compute the Laplacian of |x|
Thanks
-
## 3 Answers
(1) Let me answer first to the last question: $\Delta \vert x\vert$ is homogeneous of degree $-1$ and radial. On $\mathbb R^d$ ($d\ge 2$) it is $$(\partial_r^2+\frac{d-1}{r}\partial_r)(r)=\frac{d-1}{\vert x\vert}$$ which is an $L^1_{loc}$ function.
(2) Now you can define $\Delta$ on Distributions $T$ on some open set $\Omega$ of $\mathbb R^d$ with $$\langle\Delta T,\phi\rangle_{\mathscr D'(\Omega), \mathscr D(\Omega)}= \langle T,\Delta \phi\rangle_{\mathscr D'(\Omega), \mathscr D(\Omega)}.$$ A particular case with $T$ in $L^2$ is given in the previous answer.
(3) You can also consider $\Delta$ as an unbounded operator on $L^2(\mathbb R^d)$ with domain
$D(\Delta)$={$u\in L^2(\mathbb R^d), \Delta u\in L^2(\mathbb R^d)$}
where the term $\Delta u$ is taken in the distribution sense as in (2). The point is to prove that the operator $(-\Delta)$ is non-negative selfadjoint, which means that it is symmetric nonnegative and that the domain of the adjoint is the same as the domain of $-\Delta$. Thanks to Friedrichs extension theorem, since $-\Delta$ is nonnegative, there is no other selfadjoint extension.
-
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I can think of two differing viewpoints, and though many would prefer the latter due to its simplicity inherited from the Hilbert structure, I prefer the former for its connections to Sobolev spaces and PDE.
I. The Laplacian can be defined in the sense of distributions, since the Laplacian of an $L^2$ function is a distribution. In particular, for $u \in L^2(\Omega)$ define
$<\Delta u, \phi> := \int_\Omega u \Delta \phi\;dx$
for $\phi \in C^\infty_c(\Omega)$.
This would then say that in one dimension, for example, $\Delta |x| = 2\delta_0$, where I have used $\delta_0$ to denote the Dirac mass at zero, a distribution/measure.
II. For $u \in L^2(0,1)$, $sin(n\pi x)$ and $cos(n\pi x)$ form a basis and we can write
$u(x)= b_0 + \sum_n a_n sin(n\pi x) + b_n cos(n\pi x)$,
and we have $\sum_n a_n^2+b_n^2 <\infty$
Then we can define, formally, $\Delta u := \sum_n (n\pi)^2(-a_n sin(n\pi x) - b_n cos(n\pi x))$. In general, $\Delta u$ will not make sense pointwise, unless we know that $\sum_n n^4(a_n^2+b_n^2) <\infty$ (but as above, we can write $<\Delta u, \phi> = \int u \Delta \phi\;dx = \sum_n (n\pi)^2 (-a_n a_n^\prime-b_nb_n^\prime)$
where $a_n^\prime$ and $b_n^\prime$ are the Fourier coefficients of $\phi$. Now this makes sense for any $u \in L^2(0,1)$ if $\phi$ satisfies $\sum_n n^4((a_n^\prime)^2+(b^\prime_n)^2)<\infty$.
Of course, II can be done in higher dimensions - I have only chosen one dimension to illustrate with a simple example the basis functions. I do comment that I is more general, since it does not require the Hilbert structure.
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Let me comment a bit on Daniel Spector's answer. Yet two further possible alternative approaches are based on the notion of weak derivative and derivative almost everywhere. For the sake of simplicity, I only discuss the case of functions defined over $(0,1)$.
1) An $L^2(0,1)$ function $f:(0,1)\to \mathbb R$ is said to be weakly differentiable if there exists a $g\in L^2(0,1)$ such that $$\int_0^1 f(x)h'(x)dx=-\int_0^1 g(x)h(x)dx\qquad \hbox{for all }h\in C^1[0,1],$$ and in this case $g$ is uniquely defined and called the weak derivative of $f$. So far so good. Now, you can define in the very same way the second (weak) derivative $f''$ of $f$ and define the (weak) Laplacian as the mapping $f\mapsto f''$.
2) If a function is Lipschitz continuous, then by Rademacher's theorem it is differentiable almost everywhere. Hence, a Lipschitz continuous functions that is differentiable with a Lipschitz continuous derivative is twice differentiable a.e. It is then natural to associate with each function its second derivative. This is possible at least a.e., giving rise to another notion of Laplacian that is well-behaved if one wants to solve integrated versions of elliptic problems.
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Part 1) is equivalent to the assumption $u \in H^2$ for which we know we can give meaning to the Laplacian in precisely the sense you describe. Part 2) is the assumption $u \in W^{2,\infty}$, which is a stronger hypothesis even than $H^2$. $H^1$ is enough to consider weak solutions to Laplace's equation, though defining the Laplacian on $H^1$ is another matter. – Daniel Spector Jan 29 at 8:19
of course. well, clearly i was trying to avoid to use explicitly sobolev space theory, which i was assuming not to be known by the OP. in fact, another possibility would be to study everything in $W^{2,1}$, if one already knows what an absolutely continuous function is, and in which sense one can think of it as differentiable. – Delio Mugnolo Jan 29 at 14:18
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http://physics.stackexchange.com/questions/31293/practical-matter-of-the-higgs-mechanism/31297
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# Practical matter of the Higgs-Mechanism
My maybe very naive question is, of what practical importance will the discovery of the Higgs-Mechanism be for our technological advance in the near future?
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2
In the near future, likely none. Observing it at all took the most expensive and complicated machine ever built. – Jerry Schirmer Jul 4 '12 at 18:06
You mean technological advance as in: (1) what can we do know that we have the detector technology what else can we do with it? Or (2) what can we do theoretically now that we are pretty sure the Higgs exists? – DJBunk Jul 4 '12 at 18:12
@DJBunk sorry for being unclear, I mean the second case. – bamboon Jul 4 '12 at 18:16
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What? You haven't started getting the adds for the Mr. Higgs home collider yet? – dmckee♦ Jul 4 '12 at 18:30
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@bamboon: The real insights will come once we find the rest of the stuff in the Higgs sector. It's not just the Higgs--- there should be other stuff there, either supersymmetry or some internal structure to the Higgs, making it a light composite of heavier things (the second possibility seems excluded considering the low mass and decay properties of the Higgs). Depending on the stability and SM charges of the new particles, maybe there are astounding applications. If they all are as unstable as their mass, or if the only stable one is dark matter, it will have no technological applications. – Ron Maimon Jul 5 '12 at 14:20
show 2 more comments
## 2 Answers
I don't think that's a naive question at all, as there's quite a lot to it.
The immediate answer is "no difference". If you, like many of us, read science fiction as an impressionable youngster you're probably vaguely disappointed that warp engines and interdimensional drives haven't been invented yet, and you probably harbour hopes that some new discovery could still make them possible.
The problem is that the Standard Model is an extremely accurate description of the world at low energies. There are undoubtably new discoveries to come, but by definition they'll be at high energies that we don't see in the world around us. This probably means the technologies we'll invent have to be based on Standard Model physics. Any new discoverties aren't likely to have a big impact on technology. I say "probably" because I'm keenly aware of Arthur C. Clarke's first law and it would be a brave man to claim discoveries like the Higgs will never affect everyday technology.
However, discovering the Higgs has required enormous advances in areas like computing, and this is likely to have a big effect on all of us in the near future, especially if cloud computing takes off. After all, remember that the World Wide Web was invented at Cern.
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People tend to talk about the role of the Higg's in particle mass, but this is in some sense a peripheral issue to something much deeper. This deeper issue is that without the Higgs, W boson scattering violates unitarity above $\sim 1000 \text{ GeV}$, or equivalently the Standard Model minus the Higgs is non-renormalizable above $\sim 1000 \text{ GeV}$.
Now to break down some of these buzzwords: This deeper issue I speak of is due to the fact that when we do calculations in particle physics we are using quantum mechanics which is an intrinsically probabilistic theory. In such a theory we don't have classical notions of trajectory etc, and all we can calculate is the probability that some process is going to occur. For example, a typically thing we might be interested in is the probability that some particle will be scattering off at some angle with some energy. Note that this is not a deficiency of the theory that we are limited to probabilistic predictions since as far as we know there is no meta theory that will ever be able to use to predict how a single given particle will behave. Now, since we are dealing with probabilities we should never calculate a probablity that is more than %100 or else the theory we are dealing with ceases to make sense. This is exactly what I mean by ' violates unitarity' - the Standard Model minus the Higgs makes no sense above $\sim 1000 \text{ GeV}$ as it starts giving nonsensical probabilities in this regime. With the Higgs however it is a logically complete theory to much higher energies (perhaps arbitrarily large ) in that the probabilities always come out bounded by 100%. It may not (probably won't) be the correct theory that describes our world up to arbitrarily large energies, but it is at least mathematically consistent, which is more than you can say for the Standard model without the Higgs.
So to summarize and answer your question: now that we know (pretty sure) there is a Higgs we can make predictions using the Standard Model to higher energies than if we didn't see a Higgs. Any deviations between experiment and theory will now be a sign of new physics instead of an artifact of our theory logically breaking down.
If you want a deeper explanation in either laymans terms or physics terms let me know.
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http://math.stackexchange.com/questions/216992/paraboloidal-coordinates
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# Paraboloidal Coordinates
How would one find the transforms for paraboloidal coordinate systems. ie) I want to find $x,y$, and $z$ in terms of other variables so that I can use the Jacobian to find the differential volume.
The paraboloid in question is $z = 16 - x^2 - y^2$
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## 1 Answer
If you use cylindrical coordinates \begin{align} x &= r \cos \theta \\ y &= r \sin \theta \\ z &= z \end{align} then $$z = 16 - r^2$$ and $$\frac{D(x,y,z)}{D(r,\theta,z)} = r.$$ which leads to a pretty easy volume calculation (if top and bottom of paraboloid are simple enough).
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Is that $\frac{D(x,y,z)}{D(r,θ,z)} = r$ notation referring to the Jacobian? – Cactus BAMF Oct 19 '12 at 18:09
@CactusBAMF It's the determinant of the Jacobian matrix. Indeed $$dx\,dy\,dz = \left|\frac{D(x,y,z)}{D(r,\theta,z)}\right|\,dr\,d\theta\,dz$$ Sorry for the obscure notation. That's how I learned calculus, the Courant style. – Pragabhava Oct 19 '12 at 18:24
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http://mathhelpforum.com/trigonometry/126181-solved-proving-sin-confirmation-needed-print.html
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# [SOLVED] proving sin, and confirmation needed.
Printable View
• January 29th 2010, 01:46 PM
MathBlaster47
[SOLVED] proving sin, and confirmation needed.
I have a weird proof(?) that I can't figure out how to work.
It goes as follows:
Show that $\sin \theta=\cos\theta\tan\theta$
I also have a few true/false questions I wanted to make sure I have right:
1: $\sin\theta=\cos(90-\theta)$ my answer: False
2: $\tan\theta=-\cot(90-\theta)$ my answer: False
3: $\sin\theta=\frac{1}{\csc\theta}$ my answer: True
Many thanks for all of your help so far!
• January 29th 2010, 02:03 PM
e^(i*pi)
Quote:
Originally Posted by MathBlaster47
I have a weird proof(?) that I can't figure out how to work.
It goes as follows:
Show that $\sin \theta=\cos\theta\tan\theta$
I also have a few true/false questions I wanted to make sure I have right:
1: $\sin\theta=\cos(90-\theta)$ my answer: False
2: $\tan\theta=-\cot(90-\theta)$ my answer: False
3: $\sin\theta=\frac{1}{\csc\theta}$ my answer: True
Many thanks for all of your help so far!
I'd use the basic definitions of the trig ratios:
$\frac{\text {opp}}{\text {hyp}} = \frac{\text {adj}{\text {hyp}} \times \frac{\text {opp}}{\text {adj}}$
The RHS cancels to give the LHS proving the identity.
$cos(90-\theta) = cos90cos\theta + sin90sin\theta$
This identity is true
• January 29th 2010, 02:10 PM
Archie Meade
Quote:
Originally Posted by MathBlaster47
I have a weird proof(?) that I can't figure out how to work.
It goes as follows:
Show that $\sin \theta=\cos\theta\tan\theta$
I also have a few true/false questions I wanted to make sure I have right:
1: $\sin\theta=\cos(90-\theta)$ my answer: False
2: $\tan\theta=-\cot(90-\theta)$ my answer: False
3: $\sin\theta=\frac{1}{\csc\theta}$ my answer: True
Many thanks for all of your help so far!
Hi Mathblaster,
Write $tan\theta$ using $Sin\theta$ and $Cos\theta$
$Cos\theta\ tan\theta=Cos\theta\ \frac{Sin\theta}{Cos\theta}$
For the next one, draw a right-angled triangle with $\theta$ in one corner.
The acute angle in the other corner is $90-\theta$
Now label the sides "a", "b", "c" and discover whether your statement is true or false.
For the next, write $tan\theta$ and $cot(90-\theta)$ using $Sin\theta$ and $Cos\theta$
and $Sin(90-\theta)$ and $Cos(90-\theta)$
You were correct that time but were you guessing?
Your final answer is also correct.
• January 29th 2010, 03:08 PM
MathBlaster47
Quote:
Originally Posted by Archie Meade
Quote:
Originally Posted by MathBlaster47
I have a weird proof(?) that I can't figure out how to work.
It goes as follows:
Show that $\sin \theta=\cos\theta\tan\theta$
I also have a few true/false questions I wanted to make sure I have right:
1: $\sin\theta=\cos(90-\theta)$ my answer: False
2: $\tan\theta=-\cot(90-\theta)$ my answer: False
3: $\sin\theta=\frac{1}{\csc\theta}$ my answer: True
Many thanks for all of your help so far!
Hi Mathblaster,
Write $tan\theta$ using $Sin\theta$ and $Cos\theta$
$Cos\theta\ tan\theta=Cos\theta\ \frac{Sin\theta}{Cos\theta}$
For the next one, draw a right-angled triangle with $\theta$ in one corner.
The acute angle in the other corner is $90-\theta$
Now label the sides "a", "b", "c" and discover whether your statement is true or false.
For the next, write $tan\theta$ and $cot(90-\theta)$ using $Sin\theta$ and $Cos\theta$
and $Sin(90-\theta)$ and $Cos(90-\theta)$
You were correct that time but were you guessing?
Your final answer is also correct.
Thanks Archie!
What I did for all of the true/false questions was:
I took an arbitrary value for $\theta$ (10) evaluated for the measure of given function, then I worked the other side of the equation in terms of the given function.
Let me make sure I have what you said straight:All of the true/false questions are correct, and all I have to do for the identity is rewrite the equation as: $Cos\theta\ tan\theta=Cos\theta\ \frac{Sin\theta}{Cos\theta}$?
• January 29th 2010, 03:18 PM
Archie Meade
Yes,
because tan is sine divided by cosine.
You could also do what $e^{i\pi}$ recommended,
which is to use the identity for Cos(A-B) and also Sin(A-B) for the tan and cotan, working from there.
In using an arbitrary value, you are proving the equation for that value only,
not in general.
You should go the extra mile and either draw the right-angled triangle
or use identities as $e^{i\pi}$ recommended.
Your original answers to numbers 2. and 3. were correct,
however I'd recommend you work on them a bit more.
I have to go to sleep but I will continue with this tomorrow,
in the meantime try.
• January 29th 2010, 03:47 PM
MathBlaster47
Quote:
Originally Posted by Archie Meade
Yes,
because tan is sine divided by cosine.
You could also do what $e^{i\pi}$ recommended,
which is to use the identity for Cos(A-B) and also Sin(A-B) for the tan and cotan, working from there.
In using an arbitrary value, you are proving the equation for that value only,
not in general.
You should go the extra mile and either draw the right-angled triangle
or use identities as $e^{i\pi}$ recommended.
Your original answers to numbers 2. and 3. were correct,
however I'd recommend you work on them a bit more.
I have to go to sleep but I will continue with this tomorrow,
in the meantime try.
Thanks so much Archie!I'll be sure to do that!
• January 30th 2010, 04:24 AM
Archie Meade
1 Attachment(s)
Hi MathBlaster47,
attached is a sketch to help work through those examples...
We use a right-angled triangle for these (one corner is 90 degrees)
Since
$Sin\theta=\frac{b}{c},\ Cos\theta=\frac{a}{c}\ and\ Tan\theta=\frac{b}{a}$
then
$Cos\theta\ Tan\theta=\frac{a}{c}\ \frac{b}{a}=\frac{a}{a}\ \frac{b}{c}=\frac{b}{c}=Sin\theta$
or...
$Tan\theta=\frac{Sin\theta}{Cos\theta}$
$Cos\theta\ Tan\theta=Cos\theta\frac{Sin\theta}{Cos\theta}=\fr ac{Cos\theta\ Sin\theta}{Cos\theta}=Sin\theta$
True/false 1.
On the attachment,
$Sin\theta=\frac{b}{c}$
the side marked "b" is opposite the top corner angle and the side marked "a" is adjacent to it, while "c" is still the hypotenuse...
$Cos(90-\theta)=\frac{b}{c},\ so\ Sin\theta=Cos(90-\theta)$
Alternatively, you can a the trigonometric identity for this...
$Cos(X-Y)=CosXCosY+SinXSinY$
$X=90,\ Y=\theta$
$Cos90^o=0\ and\ Sin90^o=1$
$Cos(90^o)Cos\theta+Sin(90^o)Sin\theta=(0)Cos\theta +(1)Sin\theta=Sin\theta$
True\false 2.
$Tan\theta=\frac{a}{b},\ Tan(90^o-\theta)=\frac{b}{a}$
$Cot(90^o-\theta)=\frac{1}{Tan(90^o)-\theta}=\frac{a}{b}$
$Cot(90^0-\theta)=Tan\theta$
hence
$Tan\theta\ \ne\ -Cot(90^o-\theta)$
True/false 3.
$Cosec\theta=\frac{1}{Sin\theta}$
so
$Sin\theta=\frac{1}{Cosec\theta}$
• February 1st 2010, 02:35 PM
MathBlaster47
AHHH!
Thank you!!
I made the mistake of not remembering that those fancy words (Sine, Cosine, Tangent, and so on) stand for the relationship between the sides of a triangle.
So correct me if I'm wrong but....had I remembered that, all I would have had to do to answer the question is multiply the fractional form of the function(s) and multiply or divide them as prescribed in the question?
• February 1st 2010, 02:59 PM
Archie Meade
Quote:
Originally Posted by MathBlaster47
AHHH!
Thank you!!
I made the mistake of not remembering that those fancy words (Sine, Cosine, Tangent, and so on) stand for the relationship between the sides of a triangle.
More specifically, the ratios of the sides of a right-angled triangle, to put it simply, though the idea is extended to cover all angles.
You will see Sine and Cosine being used for non-right-angled triangles, for example in the Sine Rule and Cosine Rule, but they are not defined by the ratio of the sides of those triangles.
So correct me if I'm wrong but....had I remembered that, all I would have had to do to answer the question is multiply the fractional form of the function(s) and multiply or divide them as prescribed in the question? Yes
Or, use the "trigonometric identities", best to know both ways.
• February 2nd 2010, 08:45 AM
MathBlaster47
Quote:
Originally Posted by MathBlaster47
AHHH!
Thank you!!
I made the mistake of not remembering that those fancy words (Sine, Cosine, Tangent, and so on) stand for the relationship between the sides of a triangle.
More specifically, the ratios of the sides of a right-angled triangle, to put it simply, though the idea is extended to cover all angles.
You will see Sine and Cosine being used for non-right-angled triangles, for example in the Sine Rule and Cosine Rule, but they are not defined by the ratio of the sides of those triangles.
So correct me if I'm wrong but....had I remembered that, all I would have had to do to answer the question is multiply the fractional form of the function(s) and multiply or divide them as prescribed in the question? Yes
This brings me to another question: What would I have done if the question had said "assume that the triangle is non-right angled" or something to that effect?
• February 2nd 2010, 11:08 AM
Archie Meade
All the questions you were given initially allow you to use the definitions of Sine, Cosine, Tangent, Cosec, Sec and Cotan as per the right-angled triangle.
If you were given a diagram of a non-right-angled triangle or asked to model a real life situation in terms of angles and lengths and there was no way to find a right-angle,
then you would need to examine the clues for alternative ways,
such as "The Law of Sines" or "The Law of Cosines", and other ways.
Even though Sine and Cosine is still used in these, the ratio of the sides of non-right-angled triangles are not SinA or CosA or TanA etc...
You just need to practice and become familiar with the differences.
You need visual comprehension of all that.
• February 2nd 2010, 11:20 AM
MathBlaster47
Ah....ok so practice, practice, practice is the best way to do it......
Makes sense, it depends on the question.
Thank you very, very much for your help Archie! (Rock)(Rock)
All times are GMT -8. The time now is 02:56 AM.
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http://math.stackexchange.com/questions/146776/a-problem-about-a-sequence-and-prime-factorization
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# A problem about a sequence and prime factorization
A long time ago I solved the following theorem
Let $p_1,p_2,\ldots,p_k$ be distinct primes. Let $\{a_i\}^\infty_{i=1}$ be the increasing sequence of positive integers whose prime factorization contains only these primes (not necesarily all). Show that $\forall c>0\exists n\in \mathbb{N}:a_{n+1}-a_n>c$
Solution. Let $m$ be an integer such that $p^m_i>c$ for all $i=1,2,\ldots,k$. Let $a_n=(a_1a_2\ldots a_k)^m$. As $a_{n+1}>a_n$, there exist a prime $p_j$ such that $p_j^m|a_{n+1}$. Then $p^m|a_{n+1}-a_n$ from which $a_{n+1}-a_n>c$.
Then a friend told me that this stronger version is also true.
Let $p_1,p_2,\ldots,p_k$ be distinct primes. Let $\{a_i\}^\infty_{i=1}$ be the increasing sequence of positive integers whose prime factorization contains only these primes (not necesarily all). Show that $\forall c>0\exists n_c\in \mathbb{N}:n>n_c\implies a_{n+1}- a_n>c$
I've tried to solve it but I couldn't. I'm 90% sure that the stronger version is also true. I'm interested in elementary solutions but a complex one is also welcome.
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## 2 Answers
This is a result of Thue. For much stronger information, see the 1973 paper "On integers with many small prime factors" (Tijdeman, Compositio Mathematica).
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This is not quite an answer but possibly a push in the right direction. I remembered the following beautiful idea to show that the reciprocal sum of $a_i$ must be finite:
Recall by Euler's product formula that $\prod_{i=1}^\infty \left(1-\frac{1}{q_i}\right)^{-1}=1+1/2+1/3+\ldots=\infty$ where $q_i$ is the $i'th$ prime. Notice that for your problem you have just $p_1,\ldots,p_k$. Then,
$S:=\prod_{i=1}^k\left(1-\frac{1}{p_i}\right)^{-1}$ is the sum of the reciprocals of all numbers whose factors are $p_1,\ldots,p_k$. Notice that unlike the full Euler product, $S<\infty$. Then
$\sum_{i=1}^\infty \frac{1}{a_i}\leq \prod_{i=1}^k\left(1-\frac{1}{p_i}\right)^{-1} <\infty$
Hence the sum of reciprocals of $a_i$ converges. So by the ratio test, we must have that $\lim_{n\rightarrow\infty} \frac{a_{n}}{a_{n+1}}\leq 1$. In fact this result is inherently obvious since $a_n\leq a_{n+1}$ in the first place but, the point is thinking about it in terms of a sum may be useful.
Now, suppose there exists $c\geq 1$ with $a_{n+1}-a_n>c$ for all $n$. Then, rearranging this and using the fact that $a_n$ is strictly increasing gives
$1>\frac{a_n}{a_{n+1}}>1-\frac{c}{a_{n+1}}$
The problem is that we would like to conclude with a contradiction that $\lim_{n\rightarrow\infty} \frac{a_n}{a_{n+1}}>1$ but, we actually get $\lim_{n\rightarrow\infty} \frac{a_n}{a_{n+1}}=1$ and the equality at 1 gives us an indeterminate result from the ratio test. Perhaps someone more clever than me can show that $\lim_{n\rightarrow\infty} a_{n}/a_{n+1}<1$ which would imply the stronger result by taking the limit along a subsequence $n_i$ for which $a_{n_{i+1}}-a_{n_i}> c$. My initial hope was to try and show the sum must diverge if the limit is 1.
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http://mathoverflow.net/questions/34027/uncertainty-on-sum-of-values-sampled-from-unknown-probability-distribution-functi
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Uncertainty on sum of values sampled from unknown probability distribution function
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I am dealing with a Monte-Carlo simulation which provides me with a value $x$ as a result. I can run the simulation $n$ times, which gives me $n$ values. I don't know in advance what the probability distribution function of the output values is. What I want to be able to do is to place an uncertainty on the sum or the mean of these values, but I need the uncertainty to be sensible even for small values of $n$.
For example, if I have $n=2$, and the two resulting values of $x$ are very similar (say 9.9 and 10.1), then intuitively, I know that the mean is still not necessarily accurate, because there are only two samples (whereas a standard error in the mean would give me a small uncertainty, because it uses the standard deviation).
Is there a formal way to deal with this kind of situation? In a way, I'm looking for something analogous to Poisson statistics, which allows one to place an uncertainty even on a single value, but in a more generalized way.
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If you have any guess on the probability distribution of the result, you can either: (1) find the best parameters that describe your distribution, and estimate the uncertainty from there; or (2) use some prior on the probability distribution and proceed Bayesically. The simplest example is assuming that everything is normal, estimating the standard deviation, and concluding the approximate distribution of the mean. – Yuval Filmus Jul 31 2010 at 23:58
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Could you provide more information? As you have described the problem, the simulation could even fail to terminate for some inputs, which makes problems of bounding uncertainty really hard. – András Salamon Aug 1 2010 at 2:37
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http://unapologetic.wordpress.com/2010/10/01/endomorphism-and-commutant-algebras/?like=1&_wpnonce=ab93ff8ff3
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The Unapologetic Mathematician
Endomorphism and Commutant Algebras
We will find it useful in our study of $G$-modules to study not only the morphisms between them, but the structures that they form.
A couple days ago we mentioned the vector space $\hom_G(V,W)$. Today, we specialize to the case $V=W$, where we use the usual alternate name. We write $\mathrm{End}_G(V)=\hom_G(V,V)$ and call it the “endomorphism algebra” of $V$. Not only is it a vector space of $G$-morphisms, but it has a multiplication from the fact that the source and target of each one are the same and so we can compose them.
We also have an analogous concept for matrix representations. Given a matrix representation $X$, a $G$-morphism from $X$ to $X$ is given by a linear map $T$ so that $TX(g)=X(g)T$ for all $g\in G$. That is, $T$ must commute with each of the matrices $X(g)$. And so we call the algebra of such matrices the “commutant algebra” of $X$, and write it $\mathrm{Com}_G(X)$. This is the matrix analogue of the endomorphism algebra because if we get $X$ by starting with a $G$-module $(V,\rho)$, picking a basis for $V$, and writing down $X(g)$ as the matrix of $\rho(g)$ corresponding to this basis, then we find that $\mathrm{Com}_G(X)\cong\mathrm{End}_G(V)$.
Let’s start our considerations by letting $X=Y$ by any matrix irrep, and let’s calculate its commutant algebra. By definition for any $T\in\mathrm{Com}_G(X)$ we have $TX(g)=X(g)T$ for all $g\in G$. We can subtract $cIX$ from both sides of this equation to find
$\displaystyle(T-cI)X(g)=TX(g)-cX(g)=X(g)T-X(g)c=X(g)(T-cI)$
where $I$ is the identity matrix. The matrix $T-cI$ commutes with $X(g)$ for every complex scalar $c$, and so Schur’s lemma will apply to all of them.
Since $\mathbb{C}$ is algebraically closed, we must be able to find an eigenvalue $\lambda$. Letting $c$ be this eigenvalue, we see that $T-\lambda I$ commutes with $X(g)$ for all $g\in G$, and so Schur’s lemma tells us that either it’s invertible or the zero matrix. But since $\lambda$ is an eigenvalue the matrix $T-\lambda I$ can’t possibly be invertible, and so we must have $T-\lambda I=0$, and $T=\lambda I$.
Therefore the only matrices that commute with all the $X(g)$ in a matrix irrep of $G$ are scalar multiples of the identity matrix. And since the product of two such matrices is just the product of their scalars, we find that $\mathrm{Com}_G(X)=\mathbb{C}$ as a complex algebra.
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Posted by John Armstrong | Algebra, Group theory
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1. [...] Commutant Algebras We want to calculate commutant algebras of matrix representations. We already know that if is an irrep, then , and we’ll move on [...]
Pingback by | October 4, 2010 | Reply
2. [...] Commutant Algebras We continue yesterday’s discussion of commutant algebras. But today, let’s consider the direct sum of a bunch of copies of the same [...]
Pingback by | October 5, 2010 | Reply
3. [...] we can describe the most general commutant algebras. Maschke’s theorem tells us that any matrix representation can be decomposed as the direct [...]
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4. [...] want to calculate the centers of commutant algebras. We will have use of the two easily-established [...]
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5. [...] approach to generalizing last week’s efforts. We’re not just going to deal with endomorphism algebras, but with all the [...]
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6. [...] morphism between two irreducible modules is either or an isomorphism. And, as we’ve seen in other examples involving linear transformations, all automorphisms of an irreducible module are scalars times the [...]
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About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://math.stackexchange.com/questions/301196/proving-big-theta-if-and-only-if-big-o-and-big-omega/301218
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# Proving Big-$\Theta$ if and only if Big-$O$ and Big-$\Omega$
Given the definitions of Big-$O$ and Big-$\Omega$, I'd like to prove that $f(n) = \Theta(g(n))$ if and only if $f(n) = O(g(n))$ and $f(n) = \Omega(g(n))$.
Here's what I've come up with, but I'm not sure it's a rigorous proof, or even correct. I'm looking for comments on both the structure and style of my proof as well as correctness.
Proof by contradiction.
Suppose $f(n) = \Theta(g(n))$ and either $f(n) \ne O(g(n))$ or $f(n) \ne \Omega(g(n))$.
By the definition of $\Theta$,
$0 \le c_1 g(n) \le f(n) \le c_2 g(n)$
For some positive constants $c_1$, $c_2$, and $n > n_0$.
But it was given that $f(n) \ne O(g(n))$, which means there exist no positive constants $c$, $n_0$ such that for all $n>n_0$,
$0 \le f(n) \le c \cdot g(n)$
Which is a contradiction. $\blacksquare$
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But it wasn't given that $f(n) \ne O(g(n))$, but rather that $f(n) \ne O(g(n)) \vee f(n) \ne \Omega(n)$ – anorton Feb 12 at 14:02
Could I revise my supposition to be "Suppose $f(n)=\Theta(g(n))$ and $f(n) \ne O(g(n))$"? Is that enough to contradict "if and only if"? – proc-self-maps Feb 12 at 14:11
## 1 Answer
To prove an if and only if statement, you must prove that $p \to q$ and $q\to p$. Applied to your statement, you must prove that: $$f(n) = \Theta(g(n)) \implies f(n) = \mathcal{O}(g(n)) \wedge f(n)=\Omega(g(n)) \tag{1}$$ ...and... $$f(n) = \mathcal{O}(g(n)) \wedge f(n)=\Omega(g(n))\implies f(n) = \Theta(g(n)) \tag{2}$$
Your proof only deals with the second case, and is therefore incomplete.
Also, I must ask, why did you go with a proof by contradiction for part 1? It appears that a direct proof would suffice:
Let $f$ and $g$ be functions, such that $f(n) = \Theta(g(n))$.
By definition of $\Theta$, there exist positive constants $k_1$ and $k_2$ such that, for sufficiently large $n$: $$k_1\cdot |g(n)| \le |f(n)| \le k_2\cdot |g(n)|$$
Thus, for sufficiently large $n$: $$|f(n)| \le k_2|g(n)|$$ Therefore $f(n) = \mathcal{O}(g(n))$ by definition.
And, for sufficiently large $n$: $$k_1|g(n)| \le |f(n)|$$ Therefore $f(n) = \Omega(g(n))$ by definition. $$\blacksquare$$
Still, part 2 needs to be proven for it to be an if and only if.
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Exactly the kind of feedback I was looking for! Thanks. As for why I chose contradiction, I am unpracticed and a novice with proofs; proofs by contradiction are all I remember how to do from my undergraduate years. – proc-self-maps Feb 12 at 14:40
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http://ardwulfslair.wordpress.com/2012/10/12/exploring-levels-again-the-numbers/?like=1&source=post_flair&_wpnonce=5d5a17dad5
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# Exploring Levels Again: The Numbers
Posted by on October 12, 2012
Getting back to the leveling discussion and the idea for an RPG without a level cap, I’m putting some numbers together just to see how they feel and to get an idea for the scope of a level range using WoW-style advancement. Let us make the following stipulations:
• We have a leveling system with no true cap.
• Characters and mobs both have levels.
• The XP needed to advance to a new level is calculated as a number of mob kills of a level equal to that of the character. This is how WoW does it, by the way, although of course in practice you have other factors such as rest XP, XP from quests and so on that will need to eventually be figured in.
• Mobs not at the character’s level will of course yield a different amount of XP, but that’s not important for the purposes of this particular calculation.
• The power curve, i. e. just how much a level means in mechanical effect, is not considered here. In a non-levelcapped system I suggest that the power curve be pretty gradual.
Okay, so given those things, we have a basic leveling calculation:
$x=16c^2e^r$
Where x is the number of XP needed to get to the next level, c is the character’s current level, e is the mathematical constant e (equal to approximately 2.718) and r is a scaling factor equal to the character’s level divided by 10 and rounded to one decimal place. Using this forumla some scaling could be done either by moving the constant up or down or adding some number (probably less than 1) to r.
Truncating the actual XP numbers and plugging in some sample figures, wee see that a level 1 character needs 70 XP to reach level 2, while a level 11 character needs 21,890 XP to reach level 12, or 73,790 accumulated total XP. This works out to over a thousand mobs killed. I’ve calculated this up to level 100 (not a cap, just as high as I have taken the numbers) and it seems likely that very, very few players would surpass level 40, which requires you to have gained the equivalent of over 100K worth of at-level mob kills.
Now, as I said before, you would almost certainly have quests as a means to get XP, and those would yield as much XP as a batch of mobs. Plus there might be XP gain rate boosts of some kind, or XP from crafting, exploration, gathering and so forth. So it might not be as slow in practice as these numbers suggest, but hitting level 20 would be something. If the power curve is indeed gradual and as steep as I suggest, you could tailor quests to target a range of, say, seven levels, and have a lot of overlap within the level ranges in which players will tend to cluster, therefore reducing the need for the same kinds of content in multiple level ranges.
On the other hand, many players under such a scheme might feel, say, in the 20s or even the teens that it takes forever to level. So maybe there is some kind of sub-level advancement a la DDO or Vanguard, where you get your benefits for the next level during that level as opposed to only when you hit it. If the net power gain per level in +5%, for example (it’s probably impossible to quantify this so precisely, but there ought to be some kind of target to shoot at,) you might get +1% at 20% of the next level, another +1% at 40% and so on. You might also want level-independent intangibles (reputation might be one example) that players could go for as measures of accomplishment.
Would crafting be dependent on this overall character level? I’m not sure, but my inclination is to say that it would… but there’s also my other notion that level is supposed to be the result of power advancement rather than the cause. In an arrangement like that, a more complicated progression formula would be in order, probably one based on skill ratings. But I’m not sure you’d need to implement both ideas, actually.
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This entry was posted in Design and tagged World of Warcraft, XP to Level. Bookmark the permalink.
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# Ardwulf’s Twitverse
• Played Neverwinter (PC) in the last 24 hours. raptr.com/Ardwulf?type=t… 1 day ago
• I felt bad about my Minecraft world until Mrs Ardwulf showed me her progress in Candy Crush. 1 day ago
• Played Neverwinter (PC) in the last 24 hours. raptr.com/Ardwulf?type=t… 2 days ago
• The Wargaming Urge nblo.gs/LgN4Q 2 days ago
• Played Neverwinter (PC) in the last 24 hours. raptr.com/Ardwulf?type=t… 3 days ago
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http://math.stackexchange.com/questions/179041/characterization-of-rational-normal-curve
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Characterization of Rational Normal Curve
Given a curve $C\subset P^n$ in projective space such that any $n+1$ points on $C$ are linearly independent. I've heard from multiple people that this implies that $C$ is a rational normal curve, some people said that one needs that $C$ is rational for that.
I would like to know what precisely is true and how I can prove it, google is not my friend here...
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This is certainly not true for $n=2$: any two points on an any plane curve are linearly independent. – Georges Elencwajg Aug 5 '12 at 8:50
Sorry- I of course mean "any n+1 points" – Miklos Aug 5 '12 at 9:05
Hint: project from a point on the curve and proceed by induction. – M P Aug 5 '12 at 9:41
2 Answers
Let me elaborate on my hint, since there was some skepticism! ;)
Let $p$ be a point on $C$ and let $C' \subset \mathbb{P}^{n-1}$ denote the projection of $C$ from $p$. Think of $C'$ as a curve contained in a hyperplane in $\mathbb{P}^n$. Suppose that there are $n$ points of $C'$ not in general position; this means that there is a hyperplane of $\mathbb{P}^{n-1}$ containing them all. Then the hyperplane of $\mathbb{P}^n$ containing those $n$ points of $C'$ and the point $p$ is a hyperplane intersecting $C$ at $n+1$ points, contrary to the independence condition. Therefore, also the curve $C'$ has the independence property.
If we are very formal, then we reduce to the case $n=1$, where the assertion is a tautology. Otherwise, we stop at the case $n=2$ and the assumption of independence implies that no line in $\mathbb{P}^2$ intersects the curve in 3 points: $C$ is a conic.
Finally, to show that $C$ is a rational normal curve, you would need to specify what is the definition of rational normal curve you use: some people say it is a smooth connected curve of degree $n$ in $\mathbb{P}^n$, others say that it is the image of $\mathbb{P}^1$ under the complete linear system $\mathcal{O}(d)$, still others might even define it by the property you mention. What is your definition?
Comment. The argument given here is not especially different from the one given by Georges Elencwajg, except that he did not use the more geometric terminology: for instance, his parameterization is the result of projecting the curve away from the $n-1$ points $q_1,\ldots,q_{n-1}$.
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When I say "rational normal curve", i mean projectively equivalent to $[X_0^n: X_0 ^{n-1} X_1^1:...:X_1^n]$. I think this is some really old use of the word "normal" that doesn't have much to do with the scheme-theoretic word normal. It's clear why it is smooth and rational, just "rational normal" is not clear to me. – Miklos Aug 5 '12 at 15:01
Here is then a quick, although not entirely elementary, way to show what you want. Once you know that the curve is smooth and rational, observe that the linear system used to embed it is the complete linear system determined by $\mathcal{O}(n)$: it has degree $n$ and if it were not complete, then there would be some linear dependence among some $n+1$ points on the curve. – M P Aug 5 '12 at 15:17
You don't need to know in advance that $C$ is rational: here is a proof that an irreducible curve $C\subset \mathbb P^n$ is normal and rational as soon as all its $(n+1)$-tuples of points are linearly independent.
Step 1: The curve $C$ has degree exactly $n$
Choose $n$ points $p_1,...,p_n$ on your curve. They lie on a unique hyperplane and this hyperplane cannot contain $C$ because of the hypothesis, hence the hyperplane contains $n$ points of $C$ (perhaps with multiplicity $\gt 1$) and so $deg(C)\geq n$.
Conversely, a generic hyperplane will cut $C$ tranversally and contain at most $n$ points of $C$ (by linear independence again) so that $deg(C)\leq n$.
These two inequalities show that $deg(C)=n$
Step 2: The curve $C$ is normal (=non-singular)
If it had a singular point $s$, the hyperplane through $s$ and $n-1$ smooth other points would cut it in a divisor of degree at least $n+1$, which is impossible for a curve of degree $n$.
Step 3: The curve $C$ is rational
Choose $n-1$ points $q_1,...,q_{n-1}$ on your curve and let $P\cong \mathbb P^{n-2}$ be the projective linear space they span.
Consider the pencil $\Pi(\lambda) \;(\lambda \in \mathbb P^1)$ of hyperplanes $\Pi(\lambda) \subset \mathbb P^n$ through $P$.
Since $C$ has degree $n$, each $\Pi(\lambda)$ will cut the curve $C$ in a new point point $q_n(\lambda)$ and we obtain the required rational parametrization $\mathbb P^1\xrightarrow {\cong } C:\lambda \mapsto q_n(\lambda)$
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http://mathhelpforum.com/algebra/211039-determine-quadratic-function-whose-graph-given-print.html
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# Determine the quadratic function whose graph is given
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• January 8th 2013, 10:11 PM
alex95
Determine the quadratic function whose graph is given
Hi All,
I would really appreciate some help/hints for the following problem I am stuck on.
Determine the quadratic function whose graph is given:
The function has given points at (-1,5), (3,5), and y-intercept at (0,-1)
I know that the axis of symmetry (h) is at 1 because it is the mid-point between -1 and 3. If I plug that into the function a(x-h)^2 + k, I still have two unknowns, a and k
a(x-1)^2+k
Can someone please explain to me how to solve this. Thank you very much in advance.
Alex
• January 8th 2013, 10:21 PM
MarkFL
Re: Determine the quadratic function whose graph is given
Let:
$f(x)=ax^2+bx+c$ and use the given points to state:
$f(-1)=5$
$f(0)=-1$
$f(3)=5$
The second equation tells you $c=-1$, and this along with the first and third equations will give you two equations and two unknowns.
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http://www.physicsforums.com/showpost.php?p=4223313&postcount=1
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View Single Post
## Acceleration of acceleration?
Hi everyone.
If the position of an object changes in time, the object has velocity.
If the velocity of an object changes in time, the object is accelerating (decelerating)
If the acceleration of an object changes in time, could we hypothetically have acceleration of acceleration.
I have this scenario in my head:
An asteroid is at rest, far away from earth. The earth starts dragging the asteroid towards it. The asteroid will accelerate towards earth ever so slightly (since gravity depends on distance)
The closer it gets to earth, not only will it go faster, but it will also accelerate faster.
So we will have the change of acceleration in time:
$a\prime = \lim_{\Delta t\to\ 0}\frac{\Delta a}{\Delta t}$
So acceleration of acceleration would be in $\frac{m}{s^3}$ or rather $\frac{\frac{m}{s^2}}{s}$ if it appeals more.
Would this be useful? We could calculate the exact acceleration in any given moment as opposed to having the average acceleration.
But that's beside the point since we could always calculate the exact acceleration if we know how far it is from a planet.
But if we needed to know the exact speed of the asteroid after let's say 20 hours, we would get an incorrect answer if we treated the acceleration as if it were constant.
So $a_1 = a_0 \pm a\prime t$
so if $a_0 = 0$ then $a_1 = a\prime t$
and if $v_1 = v_0 \pm at$
and if $v_0 = 0$ then $v_1 = at$
=> $v_1 = a\prime t^2$ if the object is starting to move from rest
Does any of this make sense?
Waiting for someone to point out a flaw in this.
PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus
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http://mathhelpforum.com/calculus/209917-proof-following-equation-has-only-one-real-root-print.html
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# Proof: the following equation has only one real root
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• December 16th 2012, 03:32 AM
wilhelm
**solved** Proof: the following equation has only one real root
Hi. Could somebody show me how to prove that $ae^x=1+x+\frac{x^2}{2}$ has only one real root for $a>0$ ? All I know so far is that this equation has at least one root because $1+x+\frac{x^2}{2}>0$ for all real x.
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Already solved
• December 16th 2012, 06:54 AM
jakncoke
Re: Proof: the following equation has only one real root
I got a solution that i thought was cool so im posting anyways. Now if $ae^x = 1 + x + \frac{x^2}{2}$. then $y = ae^x$ and $y = 1 + x + \frac{x^2}{2}$ should be solutions to the differential eqn $y' - y = 0$ but for $y = 1 + x + \frac{x^2}{2}$, we get $y ' - y = \frac{x^2}{2} =0$ implies $x = 0$ so to have a solution to $ae^x = 1 + x + \frac{x^2}{2}$, x = 0 on the right side, so you get $ae^x = 1$ which for a > 1, has only 1 real root, namely $x = ln(\frac{1}{a})$
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http://stochastix.wordpress.com/2012/11/25/generalizing-herons-method/
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# Rod Carvalho
## Generalizing Heron’s method
Suppose we are given a positive rational number $y$. We would like to compute an approximation of the $n$-th root of $y$, which we denote by $\sqrt[n]{y}$. If $n = 2$, we can use Heron’s method to approximate $\sqrt{y}$, as we saw a few days ago. But, what if we have that $n > 2$? Is it possible to generalize Heron’s method?
Indeed, it is. Let us introduce a function $f_{n,y} : \mathbb{R} \to \mathbb{R}$, defined by $f_{n,y} (x) = x^n - y$, where $n \geq 2$ and $y > 0$. By construction, the positive zero of function $f_{n,y}$ is $\sqrt[n]{y}$. Do note that if $n$ is odd, then $-\sqrt[n]{y}$ is not a zero of $f_{n,y}$. Recall that Newton’s method uses the recurrence relation $x_{k+1} = g_{n,y} ( x_k )$, where function $g_{n,y} : \mathbb{R} \to \mathbb{R}$ is defined by
$g_{n,y} (x) = x - \displaystyle\frac{f_{n,y} (x)}{f_{n,y}' (x)}$
where $f_{n,y}'$ is the first derivative of $f_{n,y}$. Hence, we obtain
$g_{n,y} (x) = x - \displaystyle\frac{f_{n,y} (x)}{f_{n,y}' (x)} = x - \displaystyle\left(\frac{x^n - y}{n x^{n-1}}\right) = \displaystyle\frac{1}{n} \left( (n-1) x + \frac{y}{x^{n-1}}\right)$
Thus, we have the generalized Heron’s method
$x_{k+1} = \displaystyle\frac{1}{n} \left( (n-1) x_k + \frac{y}{x_k^{n-1}}\right)$
If we make $n = 2$, we obtain the standard Heron’s method, as we expected. The following Haskell script implements this generalized Heron’s method using arbitrary-precision rational numbers (included in the Data.Ratio library) to represent $y$ and the sequence of approximations $(x_k)_{k \in \mathbb{N}_0}$:
```import Data.Ratio
-- define generalized Heron map
g :: Integer -> Rational -> Rational -> Rational
g n y x = (1 % n) * ((fromIntegral (n-1) * x) + (y / (x^(n-1))))
-- initial approximation
x0 :: Rational
x0 = 1 % 1
-- list of approximations of the n-th root of y
roots :: Integer -> Rational -> [Rational]
roots n y | n > 1 = iterate (g n y) x0
| n <= 1 = error "Invalid value of n!!"```
We load this script into GHCi. Let us now perform some numerical experiments.
__________
Experiment #1
We would like to compute a rational approximation of $\sqrt{2}$. We have $n = 2$ and $y = 2$. Here is a very brief GHCi session:
```*Main Data.Ratio> take 6 (roots 2 2)
[1 % 1,3 % 2,17 % 12,577 % 408,665857 % 470832,
886731088897 % 627013566048]```
We obtain the following rational approximation after five iterations
$x_5 = \displaystyle\frac{886731088897}{627013566048} \approx \sqrt{2}$
which is the same approximation I obtained twice before.
__________
Experiment #2
We would like to compute a rational approximation of $\sqrt{5}$. We have $n = 2$ and $y = 5$. Here is a very brief GHCi session:
```*Main Data.Ratio> take 7 (roots 2 5)
[1 % 1,3 % 1,7 % 3,47 % 21,2207 % 987,4870847 % 2178309,
23725150497407 % 10610209857723]```
We obtain the following rational approximation after six iterations
$x_6 = \displaystyle\frac{23725150497407}{10610209857723} \approx \sqrt{5}$
How good is this approximation? Let us check:
```*Main Data.Ratio> (23725150497407 % 10610209857723)^2 - 5
4 % 112576553224922323902744729
*Main Data.Ratio> 4 / 112576553224922323902744729
3.5531377408652764e-26```
Not bad! We could now obtain a rational approximation of the famous golden ratio $\varphi := \frac{1 + \sqrt{5}}{2}$, as follows:
```*Main Data.Ratio> let xs = roots 2 5
*Main Data.Ratio> let phis = map (/2) (map (+1) xs)
*Main Data.Ratio> take 7 phis
[1 % 1,2 % 1,5 % 3,34 % 21,1597 % 987,3524578 % 2178309,
17167680177565 % 10610209857723]```
We obtain the following rational approximation after six iterations
$\tilde{\varphi}_6 = \displaystyle\frac{17167680177565}{10610209857723} \approx \varphi$
How good is this approximation? Let us recall that the golden ratio is one of the solutions of the equation $x^2 - x - 1 = 0$. Hence, we can check how close to zero $\tilde{\varphi}_6^2 - \tilde{\varphi}_6 - 1$ is:
```*Main Data.Ratio> let phi6 = 17167680177565 % 10610209857723
*Main Data.Ratio> phi6^2 - phi6 - 1
1 % 112576553224922323902744729
*Main Data.Ratio> 1 / 112576553224922323902744729
8.882844352163191e-27```
I would say that is fairly close.
__________
Experiment #3
We now would like to compute a rational approximation of $\sqrt[3]{10}$. We have $n = 3$ and $y = 10$. Here is a very brief GHCi session:
```*Main Data.Ratio> take 6 (roots 3 10)
[1 % 1,4 % 1,23 % 8,4909 % 2116,55223315303 % 25495981298,
83759169926117983945469262167029 % 38876457805393768546966848104041]```
We obtain the following rational approximation after five iterations
$x_5 = \displaystyle\frac{83759169926117983945469262167029}{38876457805393768546966848104041} \approx \sqrt[3]{10}$
Is this approximation any good? Let us check:
```*Main Data.Ratio> let x5 = (roots 3 10) !! 5
*Main Data.Ratio> x5^3
5876207108075025254603649658428809744645894398375582704
87986732898174481543266076676033790365389 % 58757060813
2677736272189042625165443146909261971794220978628909441
43919908404893015241356540921
*Main Data.Ratio> 5876207108075025254603649658428809744
6458943983755827048798673289817448154326607667603379036
5389 / 587570608132677736272189042625165443146909261971
79422097862890944143919908404893015241356540921
10.000852709004354```
This is somewhat disappointing. The rational approximation $x_5$ is a ratio of two enormously long integers, but we do not have such a small error. Note that $x_5^3 - 10 \approx 10^{-3}$. We can do some quick error analysis. Let $f_{3,10} (x) = x^3 - 10$. The 1st order Taylor approximation of $f_{3,10}$ around $x_5$ evaluated at $x = \sqrt[3]{10}$ is
$f_{3,10} (\sqrt[3]{10}) \approx f_{3,10} (x_5) + f_{3,10}' (x_5) (\sqrt[3]{10} - x_5)$
and, since $f_{3,10} (\sqrt[3]{10}) = 0$, we obtain the estimate of the error
$x_5 - \sqrt[3]{10} \approx \displaystyle\frac{f_{3,10} (x_5)}{f_{3,10}' (x_5)}$
In other words, the output deviation $f_{3,10} (x_5) - f_{3,10} (\sqrt[3]{10})$ should be divided by the slope $f_{3,10}' (x_5)$ to obtain an estimate of the magnitude of the approximation error $x_5 - \sqrt[3]{10}$. Let us use GHCi yet once again:
```*Main Data.Ratio> let error = (x5^3 - 10) / (3 * x5^2)
*Main Data.Ratio> error
1670089160826306272530773923851043922672595525468316978
5941152245094153072382174540074985393 % 272741620880842
8590518540423731512512814773109630109912907276686930689
12248623218095571481624481
*Main Data.Ratio> 1670089160826306272530773923851043922
6725955254683169785941152245094153072382174540074985393
/ 2727416208808428590518540423731512512814773109630109
91290727668693068912248623218095571481624481
6.123338108179484e-5```
This is still quite a huge error, I would say.
__________
Bonus Experiment
Let us go crazy and attempt to compute a rational approximation of $\sqrt[100]{100}$. We have $n = 100$ and $y = 100$. We expect the rational numbers to quickly become ratios of astronomically long integers. Therefore, we will not be showing any rational approximations on GHCi. Here is something scary:
```*Main Data.Ratio> let x3 = (roots 100 100) !! 3
*Main Data.Ratio> let error = (x3^100 - 100) / (100 * x3^99)
*Main Data.Ratio> let error_float = fromIntegral (numerator error)
/ fromIntegral (denominator error)
*Main Data.Ratio> error_float```
After some 20 minutes without an output, I decided to abort this experiment. If you want to show `x3`, you can, but you will see thousands of digits. Thousands! You have been warned.
### Like this:
Tags: Arbitrary-Precision Arithmetic, Data.Ratio, Golden Ratio, Haskell, Heron’s Method, Newton’s Method, Numerical Methods, Rational Approximations, Rational Arithmetic
This entry was posted on November 25, 2012 at 10:53 and is filed under Haskell, Numerical Methods. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.
### 4 Responses to “Generalizing Heron’s method”
1. Simon Read Says:
November 29, 2012 at 13:54 | Reply
“rational approximation of \sqrt[3]{10}.
a very brief GHCi session:
*Main Data.Ratio> take 6 (roots 3 10)
1 % 1…
Note that x_5^3 – 10 \approx 10^{-3}.”
I see the problem: you started with 1/1 so used the first three iterations getting roughly near the root. If you start with (for example) 5/2, you’ll be much closer so allowing Newton-Raphson to do its best work. You’ll still have a lot of digits, though!
However, you can then use Haskell to get a rational approximation to your (super-multi-digit) approximation, so you might get something of similar accuracy but with fewer digits.
2. John Baez Says:
November 30, 2012 at 13:45 | Reply
Are 17167680177565 and 10610209857723 Fibonacci numbers? The best rational approximations to the golden ratio are ratios of Fibonacci numbers.
• Rod Carvalho Says:
November 30, 2012 at 23:06 | Reply
Prof. Baez,
Thanks for the insightful comment. I had not thought of using the Fibonacci sequence to compute the golden ratio, though now that you mention it, it seems obvious.
Indeed, those are Fibonacci numbers! Here’s the proof:
```Prelude> let fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
Prelude> take 20 fibs
[1,1,2,3,5,8,13,21,34,55,89,144,233,
377,610,987,1597,2584,4181,6765]
Prelude> 17167680177565 `elem` fibs
True
Prelude> 10610209857723 `elem` fibs
True
```
3. Simon Read Says:
December 15, 2012 at 04:35 | Reply
Surely they are fibonacci numbers. If you use rational arithmetic with Newton-Raphson to solve the polynomial $x^2-x-1=0$, then you get a selected tour of fractions using fibonacci numbers, *PROVIDED* you start with one pair in the first place, like $8/5$ or $5/3$.
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http://math.stackexchange.com/questions/81620/higher-ext-groups-of-skyscraper-sheaf
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Higher Ext groups of skyscraper sheaf
I would like to understand the calculation of higher Ext groups of a skyscraper sheaf $\mathcal{O}_p$ at $p$.
The calculation I have seen does this using a Koszul resolution. It starts out like this
"Assuming X is affine, local coordinates near p define a section s of \mathcal{O}^n (n = dim X) vanishing transversely at p. "
I know that for such a section we get a koszul resolution of $\mathcal{O}_p$ that will lead to the Ext groups. However, I fail to understand the quotation. More specifically: how is this section obtained and why is $p$ the zero locus?
Thanks in advance for your answers.
Carsten
ps: the calculation is from http://math.mit.edu/~auroux/18.969-S09/mirrorsymm-lect16.pdf
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1 Answer
If $p\in X=Spec(A)$ is a regular point corresponding to the maximal ideal ${\frak m}\subset A$, you may assume (by definition of "regular point") that there exist functions $s_1,s_2,...,s_n\in A$ such that their germs $s_{1,p},s_{2,p},...,s_{n,p} \in \mathcal O_{X,p}=A_{\frak m}$ generate $\frak m$ .
The required section of $\mathcal O_{X}^n$ is then $(s_1,s_2,...,s_n)\in \Gamma(X,\mathcal O_{X}^n)=\Gamma(X,\mathcal O_{X})^n=A^n$.
"Transversally" refers to the fact that our sections locally generate the maximal ideal.
In the more classical language of complex manifolds, you would take a point $p$ of a holomorphic manifold and an open neighbourhood $X$ of $p$ on which you have local coordinates $z_1,z_2,...,z_n$ with $z_i(p)=0$.
Consider the section $(z_1,z_2,...,z_n)$ of the trivial bundle $X\times \mathbb C^n$.
The zero set $z_1=z_2=...=z_n=0$ of that section is exactly $\lbrace p\rbrace$ and moreover the differentials $d_pz_1, d_pz_2,...,d_pz_n$ are linearly independant linear forms on $T_p(X)$ : this is the differential-geometric interpretation of transversality.
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Great, thanks a lot. – Carsten Nov 13 '11 at 14:15
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http://math.stackexchange.com/questions/321621/how-can-this-trigonometric-inequality-related-to-a-limit-be-proved
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# How can this trigonometric inequality related to a limit be proved?
I want to prove that $\;\;\displaystyle \left|\frac{\sin x-x}{x^{2}}\right|\leq\frac{4(\pi/2-1)}{\pi^{2}}\;\;$ for all $x$ such that $x\in\left[0,\pi/2\right]$.
If you look at the graph of the expression on the left, it is clearly (appearing to be) monotonically increasing, so the maximum value of the left hand side is the output of the expression when $x=\pi/2$.
Would like a proof that does not involve calculus since this inequality is being used to prove $\displaystyle\lim_{x\rightarrow0}\frac{\sin x}{x}=1$.
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1
Minor detail: you want $x$ in $(0,\pi/2]$. – julien Mar 5 at 17:24
1
But that inequality is quite stronger than $\lim\frac{\sin x}x=1$ (i.e. the limit follows immediately from the iniequality and just an arbitrary bound would be enough). So, what can we use about $\sin$? How is it defined? Purely geometrically? – Hagen von Eitzen Mar 5 at 17:24
## 2 Answers
You may already know that $\sin x < x< \tan x$ for $0<x<\frac\pi 2$. Then for such $x$ we have $$0< \frac{x-\sin x}{x^2}< \frac{\tan x-\sin x}{x^2}=\frac{(1-\cos x)\tan x}{x^2}=\frac{\sin^2x\tan x}{x^2(1+\cos x)}< \tan x.$$ This is not the bound you were asking for, but it is weaker only for big $x$, hence is more than enough to show $\lim_{x\to 0}\frac{\sin x}{x}=1$. (In fact, using $\cos x\to 1$, we obtain $\sin x=x+O(x^3)$)
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Actually we can expect this function
$$f(x)=\dfrac{x -\sin x }{x^2}$$
as an increasing function.
$f'(x) = \dfrac{2\sin x-x\cos x-x}{x^3}$
we just have to prove
$g(x) = 2\sin x-x\cos x-x\ge 0$ , for $x\in[0,\pi/2]$
since $g(0) = 0$ , we just prove $g(x)$ is also increasing on $[0,\pi/2]$.
$g'(x) = \cos x+x\sin x -1$, with $g'(0) = 0$
and $g''(x) = x\cos x\ge 0$ on the interval $[0,\pi/2]$, thus $g'(x)$ is non-decreasing, thus $g'(x)\ge 0$, which means $g(x)$ is non-decreasing, thus $g(x)\ge 0.$
Therefore,
$f'(x)\ge 0$, $f$ is non-decreasing, thus $f(\dfrac{\pi}{2})$ will be the maximum.
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http://mathoverflow.net/questions/120522?sort=votes
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## How to identify bridge nodes between nearly connected graph components in partitioned adjacency matrices?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I have adjacency matrices which have nearly connected components. That is partitions with a dense number of edges between nodes in the same group and few edges acting as bridges between these groups. I have clustered them so that they form a band matrix.
What ways exist to identity these nodes linking groups? These nodes that act as bridges between these nearly connected components.
It appears that each method I have seen has some element of it being a heuristic in some way.
-
Do you mean nearly complete components. How big are they relative to the vertex set and how dense? Imagine a square grid with $n^2$ points and $2n$ lines. Make it into a graph where each vertex is on $2n-2$ edges. If you order the matrix by rows then you have a block matrix with each block complete and a vertex having only one bridge to each other block. But you could organize by columns and swap roles of bridge and non-bridge. – Aaron Meyerowitz Feb 1 at 16:43
## 2 Answers
As i commented, you need to be more specific about the details. Here is an idea though which might work well if within groups there are lots (or at least a reasonable number of) triangles but no triangles involve bridges: Take the adjacency matrix $A.$ and compare it to $A^2.$ If the $u,v$ position is positive in both then the edge $uv$ is definitely not a bridge. So temporarily consider just these edges. They will split the graph into disjoint connected components which one hopes will be your blocks.
This will not work perfectly if some bridges are in triangles. It also won't work if the groups are complete bipartite graphs because there are then no triangle within a block. However you could look at powers of $A+I$ and the larger numbers should tend to be for vertices in the same group. Perhaps compute $M=(A+I)^k$ for $k=3$ or $4$, pick some cutoff value $v$ (maybe the median of the entries of $M$) and replace each entry $m_{ij}$ by $\lfloor\frac{m_{i,j}}{v}\rfloor$ (the integer quotient). Then multiply that by $A+I.$ The exact $k$ and $v$ might have to be tuned to your matrix. But a program could vary them until the final matrix has a sharp dichotomy of values indicating vertices in the same and different blocks.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I think that in this very special case you can get the answer just by computing a minimum cut. If the two subgraphs are really dense and the bridges between them are few, then the minimum cut will be just the set of bridges.
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