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http://en.wikipedia.org/wiki/Earth_radius
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# Earth radius
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Earth radius is the distance from Earth's center to its surface, about 6,371 kilometres (3,959 mi). This length is also used as a unit of distance, especially in astronomy and geology, where it is usually denoted by $R_\oplus$.
This article deals primarily with spherical and ellipsoidal models of the Earth. See Figure of the Earth for a more complete discussion of models. The Earth is only approximately spherical, so no single value serves as its natural radius. Distances from points on the surface to the center range from 6,353 km to 6,384 km (3,947–3,968 mi). Several different ways of modeling the Earth as a sphere each yield a mean radius of 6,371 kilometres (3,959 mi).
While "radius" normally is a characteristic of perfect spheres, the term as used in this article more generally means the distance from some "center" of the Earth to a point on the surface or on an idealized surface that models the Earth. It can also mean some kind of average of such distances, or of the radius of a sphere whose curvature matches the curvature of the ellipsoidal model of the Earth at a given point.
The first scientific estimation of the radius of the earth was given by Eratosthenes about 240 BC. Estimates of the accuracy of Eratosthenes’s measurement range from within 2% to within 15%.
## Introduction
### Radius and models of the earth
Earth's rotation, internal density variations, and external tidal forces cause it to deviate systematically from a perfect sphere.[a] Local topography increases the variance, resulting in a surface of unlimited complexity. Our descriptions of the Earth's surface must be simpler than reality in order to be tractable. Hence we create models to approximate the Earth's surface, generally relying on the simplest model that suits the need.
Each of the models in common use come with some notion of "radius". Strictly speaking, spheres are the only solids to have radii, but looser uses of the term "radius" are common in many fields, including those dealing with models of the Earth. Viewing models of the Earth from less to more approximate:
• The real surface of the Earth;
• The geoid, defined by mean sea level at each point on the real surface;[b]
• An ellipsoid: geocentric to model the entire earth, or else geodetic for regional work;[c]
• A sphere.
In the case of the geoid and ellipsoids, the fixed distance from any point on the model to the specified center is called "a radius of the Earth" or "the radius of the Earth at that point".[d] It is also common to refer to any mean radius of a spherical model as "the radius of the earth". On the Earth's real surface, on other hand, it is uncommon to refer to a "radius", since there is no practical need. Rather, elevation above or below sea level is useful.
Regardless of model, any radius falls between the polar minimum of about 6,357 km and the equatorial maximum of about 6,378 km (≈3,950 – 3,963 mi). Hence the Earth deviates from a perfect sphere by only a third of a percent, sufficiently close to treat it as a sphere in many contexts and justifying the term "the radius of the Earth". While specific values differ, the concepts in this article generalize to any major planet.
### Physics of Earth's deformation
Rotation of a planet causes it to approximate an oblate ellipsoid/spheroid with a bulge at the equator and flattening at the North and South Poles, so that the equatorial radius $a$ is larger than the polar radius $b$ by approximately $a q$ where the oblateness constant $q$ is
$q=\frac{a^3 \omega^2}{GM}\,\!$
where $\omega$ is the angular frequency, $G$ is the gravitational constant, and $M$ is the mass of the planet.[e] For the Earth q−1 ≈ 289, which is close to the measured inverse flattening f−1 ≈ 298.257. Additionally, the bulge at the equator shows slow variations. The bulge had been declining, but since 1998 the bulge has increased, possibly due to redistribution of ocean mass via currents.[2]
The variation in density and crustal thickness causes gravity to vary on the surface, so that the mean sea level will differ from the ellipsoid. This difference is the geoid height, positive above or outside the ellipsoid, negative below or inside. The geoid height variation is under 110 m on Earth. The geoid height can change abruptly due to earthquakes (such as the Sumatra-Andaman earthquake) or reduction in ice masses (such as Greenland).[3]
Not all deformations originate within the Earth. The gravity of the Moon and Sun cause the Earth's surface at a given point to undulate by tenths of meters over a nearly 12 hour period (see Earth tide).
### Radius and local conditions
Biruni's (973–1048) method for calculation of Earth's radius improved accuracy.
Given local and transient influences on surface height, the values defined below are based on a "general purpose" model, refined as globally precisely as possible within 5 m of reference ellipsoid height, and to within 100 m of mean sea level (neglecting geoid height).
Additionally, the radius can be estimated from the curvature of the Earth at a point. Like a torus the curvature at a point will be largest (tightest) in one direction (North-South on Earth) and smallest (flattest) perpendicularly (East-West). The corresponding radius of curvature depends on location and direction of measurement from that point. A consequence is that a distance to the true horizon at the equator is slightly shorter in the north/south direction than in the east-west direction.
In summary, local variations in terrain prevent the definition of a single absolutely "precise" radius. One can only adopt an idealized model. Since the estimate by Eratosthenes, many models have been created. Historically these models were based on regional topography, giving the best reference ellipsoid for the area under survey. As satellite remote sensing and especially the Global Positioning System rose in importance, true global models were developed which, while not as accurate for regional work, best approximate the earth as a whole.
## Fixed radii
The following radii are fixed and do not include a variable location dependence. They are derived from the WGS-84 ellipsoid.[4]
The value for the equatorial radius is defined to the nearest 0.1 meter in WGS-84. The value for the polar radius in this section has been rounded to the nearest 0.1 meter, which is expected to be adequate for most uses. Please refer to the WGS-84 ellipsoid if a more precise value for its polar radius is needed.
The radii in this section are for an idealized surface. Even the idealized radii have an uncertainty of ± 2 meters.[5] The discrepancy between the ellipsoid radius and the radius to a physical location may be significant. When identifying the position of an observable location, the use of more precise values for WGS-84 radii may not yield a corresponding improvement in accuracy.
The symbol given for the named radius is used in the formulae found in this article.
### Equatorial radius
The Earth's equatorial radius $a$, or semi-major axis, is the distance from its center to the equator and equals 6,378.1370 kilometres (3,963.1906 mi). The equatorial radius is often used to compare Earth with other planets.
### Polar radius
The Earth's polar radius $b$, or semi-minor axis, is the distance from its center to the North and South Poles, and equals 6,356.7523 kilometres (3,949.9028 mi).
## Radii with location dependence
### Notable radii
• Maximum: The summit of Chimborazo is 6,384.4 km (3,968 mi) from the Earth's center.
• Minimum: The floor of the Arctic Ocean is ≈6,352.8 kilometres (3,947.4 mi) from the Earth's center.[6]
### Radius at a given geodetic latitude
The distance from the Earth's center to a point on the spheroid surface at geodetic latitude $\varphi\,\!$ is:
$R=R(\varphi)=\sqrt{\frac{(a^2\cos\varphi)^2+(b^2\sin\varphi)^2}{(a\cos\varphi)^2+(b\sin\varphi)^2}}\,\!$
where $a$ and $b$ are the equatorial radius and the polar radius, respectively.
### Radius of curvature
These are based on an oblate ellipsoid.
Eratosthenes used two points, one almost exactly north of the other. The points are separated by distance $D$, and the vertical directions at the two points are known to differ by angle of $\theta$, in radians. A formula used in Eratosthenes' method is
$R= \frac{D}{\theta}\,\!$
which gives an estimate of radius based on the north-south curvature of the Earth.
#### Meridional
In particular the Earth's radius of curvature in the (north-south) meridian at $\varphi\,\!$ is:
$M=M(\varphi)=\frac{(ab)^2}{((a\cos\varphi)^2+(b\sin\varphi)^2)^{3/2}}\,\!$
#### Normal
If one point had appeared due east of the other, one finds the approximate curvature in east-west direction.[f]
This radius of curvature in the prime vertical, which is perpendicular, or normal, to M at geodetic latitude $\varphi\,\!$ is:[g]
$N=N(\varphi)=\frac{a^2}{\sqrt{(a\cos\varphi)^2+(b\sin\varphi)^2}}\,\!$
Note that N=R at the equator:
At geodetic latitude 48.46791 degrees (e.g., Lèves, Alsace, France), the radius R is 20000/π ≈ 6,366.197, namely the radius of a perfect sphere for which the meridian arc length from the equator to the North Pole is exactly 10000 km, the originally proposed definition of the meter.
The Earth's mean radius of curvature (averaging over all directions) at latitude $\varphi\,\!$ is:
$R_a=\sqrt{MN}=\frac{a^2b}{(a\cos\varphi)^2+(b\sin\varphi)^2}\,\!$
The Earth's radius of curvature along a course at geodetic bearing (measured clockwise from north) $\alpha\,\!$, at $\varphi\,\!$ is derived from Euler's curvature formula as follows:
$R_c=\frac{{}_{1}}{\frac{\cos^2\alpha}{M}+\frac{\sin^2\alpha}{N}}\,\!$
The Earth's meridional radius of curvature at the equator equals the meridian's semi-latus rectum:
$\frac{b^2}{a}\,\!$ = 6,335.437 km
The Earth's polar radius of curvature is:
$\frac{a^2}{b}\,\!$ = 6,399.592 km
## Mean radii
The Earth can be modeled as a sphere in many ways. This section describes the common ways. The various radii derived here use the notation and dimensions noted above for the Earth as derived from the WGS-84 ellipsoid;[4] namely,
$\textstyle a =$ Equatorial radius (6,378.1370 km)
$\textstyle b =$ Polar radius (6,356.7523 km)
A sphere being a gross approximation of the spheroid, which itself is an approximation of the geoid, units are given here in kilometers rather than the millimeter resolution appropriate for geodesy.
### Mean radius
The International Union of Geodesy and Geophysics (IUGG) defines the mean radius (denoted $R_1$) to be[7]
$R_1 = \frac{2a+b}{3}\,\!$
For Earth, the mean radius is 6,371.009 kilometres (Bad rounding here4,000 mi).[8]
### Authalic radius
Earth's authalic ("equal area") radius is the radius of a hypothetical perfect sphere which has the same surface area as the reference ellipsoid. The IUGG denotes the authalic radius as $R_2$.[7]
A closed-form solution exists for a spheroid:[9]
$R_2=\sqrt{\frac{a^2+\frac{ab^2}{\sqrt{a^2-b^2}}\ln{\left(\frac{a+\sqrt{a^2-b^2}}b\right)}}{2}}=\sqrt{\frac{a^2}2+\frac{b^2}2\frac{\tanh^{-1}e}e} =\sqrt{\frac{A}{4\pi}}\,\!$
where $e^2=(a^2-b^2)/a^2$ and $A$ is the surface area of the spheroid.
For Earth, the authalic radius is 6,371.0072 kilometres (Bad rounding here4,000 mi).[8]
### Volumetric radius
Another spherical model is defined by the volumetric radius, which is the radius of a sphere of volume equal to the ellipsoid. The IUGG denotes the volumetric radius as $R_3$.[7]
$R_3=\sqrt[3]{a^2b}\,\!$
For Earth, the volumetric radius equals 6,371.0008 kilometres (Bad rounding here4,000 mi).[8]
### Meridional Earth radius
Another mean radius is the rectifying radius, giving a sphere with circumference equal to the perimeter of the ellipse described by any polar cross section of the ellipsoid. This requires an elliptic integral to find, given the polar and equatorial radii:
$M_r=\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}\sqrt{{a^2}\cos^2\varphi + {b^2} \sin^2\varphi}\,d\varphi$.
The rectifying radius is equivalent to the meridional mean, which is defined as the average value of M:[9]
$M_r=\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}\!M(\varphi)\,d\varphi\!$
For integration limits of [0…π/2], the integrals for rectifying radius and mean radius evaluate to the same result, which, for Earth, amounts to 6,367.4491 kilometres (Bad rounding here4,000 mi).
The meridional mean is well approximated by the semicubic mean of the two axes:
$M_r\approx\left[\frac{a^{3/2}+b^{3/2}}{2}\right]^{2/3}\,$
yielding, again, 6,367.4491 km; or less accurately by the quadratic mean of the two axes:
$M_r\approx\sqrt{\frac{a^2+b^2}{2}}\,\!$;
about 6,367.454 km; or even just the mean of the two axes:
$M_r\approx\frac{a+b}{2}\,\!$;
about 6,367.445 kilometres (Bad rounding here4,000 mi).
## Notes
1. For details see Figure of the Earth, Geoid, and Earth tide.
2. In a geocentric ellipsoid, the center of the ellipsoid coincides with some computed center of the earth, and best models the earth as a whole. Geodetic ellipsoids are better suited to regional idiosyncrasies of the geoid. A partial surface of an ellipsoid gets fitted to the region, in which case the center and orientation of the ellipsoid generally do not coincide with the earth's center of mass or axis of rotation.
3. The value of the radius is completely dependent upon the latitude in the case of an ellipsoid model, and nearly so on the geoid.
4. This follows from the International Astronomical Union definition rule (2): a planet assumes a shape due to hydrostatic equilibrium where gravity and centrifugal forces are nearly balanced.
5. East-west directions can be misleading. Point B which appears due East from A will be closer to the equator than A. Thus the curvature found this way is smaller than the curvature of a circle of constant latitude, except at the equator. West can exchanged for east in this discussion.
6. N is defined as the radius of curvature in the plane which is normal to both the surface of the ellipsoid at, and the meridian passing through, the specific point of interest.
## References
1. ^ a b
2. ^ a b c Moritz, H. (1980). Geodetic Reference System 1980, by resolution of the XVII General Assembly of the IUGG in Canberra.
3. ^ a b c Moritz, H. (March 2000). "Geodetic Reference System 1980". Journal of Geodesy 74 (1): 128–133. Bibcode:2000JGeod..74..128.. doi:10.1007/s001900050278.
4. ^ a b Snyder, J.P. (1987). Map Projections – A Working Manual (US Geological Survey Professional Paper 1395) p. 16–17. Washington D.C: United States Government Printing Office.
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http://mathhelpforum.com/differential-geometry/161809-proof-involving-closures-functions-metric-spaces.html
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# Thread:
1. ## Proof involving closures of functions in metric spaces
If f is a continuous mapping of a metric space $X$ into a metric space $Y$, prove that $f(\overline{E}) \subset \overline{f(E)}$ for every set $E \subset X$. Show by an example that $f(\overline{E})$ can be a proper subset of $\overline{f(E)}$.
2. That's a lovely little problem but what have you done on it? There are, typically, many different ways of proving things like this, depending on exactly what definitions and basic ideas you use. Without seeing what you know about this situation, we don't know which to recommend for you.
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http://math.stackexchange.com/questions/239491/time-to-find-first-copy-of-a-duplicate?answertab=active
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# Time to find first copy of a duplicate
Consider a random process which samples uniformly with replacement from [n]. The expected time to find a duplicate is a constant factor times $\sqrt{n}$. This is a version of the famous Birthday Problem. How can one find the expected time to find the first copy of the first duplicate? This will obviously occur some time before. Also, what is the distribution of this time?
If you fix the position of the later copy of the duplicate then the earlier copy seems to occur uniformly before it but I am not sure how helpful that is to answer the questions.
By time I simply mean that each sample takes one unit of time so the time is just the number of samples at that point. This is not a question about algorithms or computation.
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Your comment at the end is quite helpful, at least for the mean. – André Nicolas Nov 17 '12 at 22:10
## 1 Answer
Call $2\leqslant D_n\leqslant n+1$ the time of the first duplicate and $1\leqslant C_n\leqslant D_n-1$ the time of the first copy of the first duplicate.
As you noted, for each $2\leqslant k\leqslant n+1$, conditionally on the event $[D_n=k]$, $C_n$ is uniformly distributed on $\{1,2,\ldots,k-1\}$ hence $\mathbb E(C_n\mid D_n=k)=\frac12k$. Thus, $\mathbb E(C_n)=\frac12\mathbb E(D_n)$ and each asymptotics on $\mathbb E(D_n)$ when $n\to\infty$ translates into an asymptotics on $\mathbb E(C_n)$.
Likewise, for every $1\leqslant i\leqslant n$, decomposing the event $[C_n=i]$ into its intersections with the events $[D_n=k]$ for $i+1\leqslant k\leqslant n+1$, one gets $$\mathbb P(C_n=i)=\sum_{k=i+1}^{n+1}\frac{\mathbb P(D_n=k)}{k-1}.$$ Recall finally that, for every $2\leqslant k\leqslant n+1$, $$\mathbb P(D_n\geqslant k)=\prod_{\ell=1}^{k-2}\frac{n-\ell}n,$$ hence, for every $1\leqslant i\leqslant n$, $$\mathbb P(C_n=i)=\frac1n\sum_{k=i}^{n}\prod_{\ell=1}^{k-1}\frac{n-\ell}n=\frac{n!}{n^{n+1}}\sum_{k=0}^{n-i}\frac{n^k}{k!}.$$ Edit: Let $N_n$ denote a Poisson random variable with parameter $n$, then $$\mathbb P(C_n=i)=\frac{n!\mathrm e^n}{n^{n+1}}\mathbb P(N_n\leqslant n-i).$$ Asymptotics follow from this identity since the prefactor $n!\mathrm e^n/n^{n+1}$ is equivalent to $\sqrt{2\pi/n}$ when $n\to\infty$, and $(N_n-n)/\sqrt{n}$ converges in distribution to a standard normal random variable. Thus, if $i_n/\sqrt{n}\to x$, $$\mathbb P(C_n=i_n)\sim\sqrt{\frac{2\pi}n}\Phi(-x)=\frac1{\sqrt{n}}\int_x^{+\infty}\mathrm e^{-s^2/2}\mathrm ds.$$ Summing these yields $$\mathbb P(C_n\geqslant i_n)\sim\int_x^{+\infty}(s-x)\mathrm e^{-s^2/2}\mathrm ds=\int_0^{+\infty}s\mathrm e^{-(s+x)^2/2}\mathrm ds.$$ In other words, $C_n/\sqrt{n}$ converges in distribution to a random variable $C$ whose probability density function $f_C$ is defined, for every $x\geqslant0$, by $$f_C(x)=\int_x^{+\infty}\mathrm e^{-s^2/2}\mathrm ds.$$
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That last equality is not correct. In fact $\mathbb{E}(D_n) = 1 + \sum_{k=1}^n \tfrac{1}{k}$ if I recall correctly. – WimC Nov 18 '12 at 9:33
@WimC Where do you see a formula for $\mathbb E(D_n)$ in my post? – Did Nov 18 '12 at 9:35
Nowhere, I just put two remarks in a single comment. :-) – WimC Nov 18 '12 at 9:35
@WimC Care to elaborate about the first part of your comment, then? – Did Nov 18 '12 at 9:40
Take $k=3$. I get $\mathbb{P}(D_n \geq 3) = 1-1/n$ right? – WimC Nov 18 '12 at 9:43
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http://physics.stackexchange.com/questions/33766/the-energy-level-of-electrons-in-an-atom-depend-on-l?answertab=oldest
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# The energy level of electrons in an atom depend on $l$?
According to quantum mechanics, energy level of electrons in an atom only depend on principal quantum number, $n$, and more accurately also depend on $j$.
However, according to the below picture,
it suggests that the orbit that has same $j$ and same $n$ does not have the same energy level, and this seems to be accurate, as I learned chemistry :)
So, what am I missing in my comprehension of quantum mechanics?
(I know that magnetic field somehow affects situation, but this case does not have any external magnetic field that influences the case, and internal magnetic field has already been counted in)
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The nucleus may have a magnetic field, generating the fine- and hyperfine-structure. I'll leave it to someone who remembers this stuff to write a real answer, but those are the search terms you want. – dmckee♦ Aug 9 '12 at 1:52
## 2 Answers
In hydrogen atom, this is called Lamb shift, an effect of QED, specifically vacuum polarization. This is outside the realm of quantum mechanics.
But judging from the enormous energy difference between $3p_{1/2}$ and $3s_{1/2}$ I guess this isn't a hydrogen atom. In that case, most likely the interaction between electrons is the cause of the difference.
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According to quantum mechanics, energy level of electrons in an atom only depend on principal quantum number, $n$.
This is only true for Hydrogen, and only for the gross structure. Based on your picture, it looks like you're talking about Sodium.
Disclaimer: this is a really long answer.
In non-Hydrogenic systems, (that is, multi-electron atoms), you're not just considering interaction between one electron and the nucleus - you also need to account for the repulsion between the other electrons. The full Hamiltonian of an $N$-electron system (neglecting further corrections like fine structure, which is responsible for the splitting between the $^3P_{3/2}$ and $^3P_{1/2}$ levels in your diagram) is
$$H = \sum\limits_{i=1}^N \left( \frac{-\hbar^2}{2m_e} \nabla_i^2 - \frac{Z e^2}{4\pi\epsilon_0 r_i} \right) + \sum\limits_{j>i} \frac{e^2}{4\pi\epsilon_0 r_{ij}}.$$
The first term in this expression is the kinetic energy of each electron, the second is the potential energy of each electron due to the nucleus, and the third is the electron-electron interaction. So we can write the Hamiltonian in short-hand as $H = K + V_{nuclear} + V_{electronic}$.
In order to do perturbation theory with this Hamiltonian, we wish to write it in the form $H = H_0 + H_1$, where $H_1 \ll H_0$. A first guess would be to use the (separable) first two terms as $H_0$, but the electron-electron interaction is too big to be treated as a perturbation. There are around the same number of electrons as there are protons in the nucleus $(N \approx Z)$, and the distance scales are around the same $(r_{ij} \approx r_i)$, so the magnitude of the electron-electron repulsion is around the same as the nuclear attraction. So we can't use perturbation theory with the Hamiltonian in its current form.
We make progress through the observation that closed shells of electrons are spherically symmetric. This means that, even though the electronic potential is large, most of it is spherically symmetric. We make the central field approximation and absorb as much as possible of the electron-electron term into the spherically symmetric term:
$$H = H_0 + H_{re}$$ where $$H_0 = \sum\limits_{i=1}^N \left( \frac{-\hbar^2}{2m_e} \nabla_i^2 + U(r_i) \right)$$ $$H_{re} = \sum\limits_{i=1}^N \left( - \frac{Z e^2}{4\pi\epsilon_0 r_i} - U(r_i) \right) + \sum\limits_{j>i} \frac{e^2}{4\pi\epsilon_0 r_{ij}}.$$
Here $U(r)$ is known as the central potential, and $H_{re}$ is the residual electrostatic interaction. All that has happened in this step is that we have added-and-subtracted $U(r_i)$ to the right hand side, but with a careful choice of $U(r)$, we can ensure $H_{re} \ll H_0$. Hereafter we will talk exclusively about the central potential - $H_{re}$ is interesting, but not relevant to what we're talking about right now.
The key to understanding the lifting of the $l$-degeneracy is understanding the form of the central potential. We can use computers to deduce a precise numerical form for $U(r)$, but it's not as important as a basic physical understanding of its behaviour.
Consider screening of a single valence electron in an atom with closed inner shells of electrons. Since the inner shells are spherically symmetric, the valence electron experiences an entirely spherically symmetric potential $(H_{re}=0)$. When the valence electron is far from the nucleus, Gauss's law tells us that the charge of the inner electrons 'cancels out' most of the nuclear charge, and the potential is a Coulomb potential due to a charge $+e$ (since there are $Z-1$ electrons doing the screening). Conversely, when the valence electron is near the nucleus, it is 'inside' the other electrons and sees the Coulomb potential of the full nuclear charge $+Ze$.
This argument tells us that far from the nucleus, the potential tends towards a $-1/r$ proportionality. Near the nucleus, the potential tends towards a $-Z/r$ proportionality. We may write it in terms of an effective nuclear charge $Z_{eff}(r)$: $$U(r) = \frac{-Z_{eff}(r) e^2}{4\pi\epsilon_0 r},$$ where $Z_{eff}(r)$ tends to 1 at large $r$, and $Z$ at small $r$.
Please excuse my poorly drawn and photographed sketches of $U(r)$ and $Z_{eff}(r)$:
So the potential felt by the valence electrons departs from the $1/r$ dependency seen in Hydrogen. This is what lifts the degeneracy between states of different $l$. The potential near the nucleus is 'lower down' than in a Hydrogenic $1/r$ system, so states that are localised near the nucleus are lower in energy than they would be in Hydrogen. Since the states of lower angular momentum (like $s$-states) are nearer the nucleus, these are lower in energy. There is a small subtlety here: in Hydrogen, this effect is exactly cancelled by the kinetic energy of the higher-AM state, but this doesn't apply in non-$1/r$ potentials.
Tl,dr: atoms with more than one electron are not degenerate in $l$, due to the departure of the central potential from the Hydrogenic $1/r$ potential.
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http://math.stackexchange.com/questions/215412/given-a-parallelepiped-how-do-i-find-the-determinant-given-vertices
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# Given a parallelepiped, how do I find the determinant given vertices?
Here are the given vertices of a given parallelepiped...
$(-1, 0, 0), (0, 4, 0), (-3, -5, 2), (-2, 2, -1)$
I know that first, we should translate all to the origin...
$(0, 0, 0), (1, 4, 0), (-2, -5, 2), (-1, 2, -1)$
Is the matrix to evaluate the determinant by $\begin{bmatrix}1 & 4 & 0\\-2 & -5 & 2\\-1 & 2 & -1\end{bmatrix}$ or is this incorrect?
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It looks right. – Aleks Vlasev Oct 17 '12 at 5:18
Yes it is. The volume is 15. – copper.hat Oct 17 '12 at 5:25
Great! Thank you – LearningPython Oct 17 '12 at 5:57
## 1 Answer
There are two approaches: either you transform one point to the origin, or you add a $1$ to every vector:
$$\begin{vmatrix} -1 & 0 & -3 & -2\\ 0 & 4 & -5 & 2 \\ 0 & 0 & 2 & -1 \\ 1 & 1 & 1 & 1 \end{vmatrix} = -\begin{vmatrix} 1 & -2 & -1 \\ 4 & -5 & 2 \\ 0 & 2 & -1 \end{vmatrix} = 15$$
So the two solutions differ in their sign. Which doesn't matter if you only care about absolute values, otherwise you should make sure to use one way exclusively. Of course, transforming a matrix doesn't change its determinant, so you might as well write the vectors as rows, the way you do in your question.
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http://math.stackexchange.com/questions/201881/optimal-strategy-for-a-2-player-game
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# Optimal Strategy for a 2 player game
• 2 players A & B are playing a game involving a number n
• Player A makes the first move & both players play alternately.
• In each move the player takes the number n,chooses a number i such that 2^i < n and replaces n with k = n - 2^i iff the number of 1s in the binary representation of k is greater than or equal to the number of 1s in the binary representation of n
• Game ends when no player can make a move, ie there does not exist such an i
For example:
````n = 13 = b1101
````
Only possible i=1
````k = n - 2^i = 11 = b1011
````
Again,only possible i = 2
````k = n - 2^i = 7 = b111
````
Since Player A cant make any more moves, Player B wins
I've deduced that at any step,we can only choose an i,such that there is a 0 at the corresponding position in the binary representation of n.
For Example: if n=1010010,then i can only be {0,2,3,5}.
But I cant move any further.A minimax algorithm isnt exactly striking me.I would appreciate any help.Thanks in advance
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## 1 Answer
It’s a disguised version of Nim. Each block of consecutive $0$’s in the binary representation of $n$ is a pile. For example, from $10000_2$ you can subtract $8$ to get $1000_2$; $4$ to get $1100_2$; $2$ to get $111_2$; or $1$ to get $1111_2$. These moves correspond to reducing a pile of $4$ stones to $3$, $2$, $1$, or $0$ stones, respectively. I’ll leave it to you to fill in the details. You probably already know the strategy for Nim.
(This is a cute variant; I’d not seen it before.)
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This doesn't seem right, since the removal of zeroes from one block will create new zeroes in the next block to the left. The value of $1010_2$, for instance, is $*$, not $0$, even there are two equal "piles" of zeroes in that number. – Théophile Sep 24 '12 at 23:50
@Théophile: You’re right: I was concentrating so much on the case of adjacent $1$’s that I forgot that the piles get amalgamated in the basic case! – Brian M. Scott Sep 25 '12 at 0:03
I’ll revise this when I get back. – Brian M. Scott Sep 25 '12 at 0:10
It seems that every position has value $0$, $*$, or $*2$. – Théophile Sep 25 '12 at 0:31
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http://mathoverflow.net/questions/75231/longest-element-of-a-weyl-group/75239
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## Longest element of a Weyl group
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $G$ an algebraic (reductive) group. $T$ a maximal torus, $B$ a Borel subgroup containing $T$, and $w_0$ the longest element of the Weyl group.
I'm looking for a reference explaining why when you conjugate $B$ by $w_0$, the result is the opposite Borel subgroup $B^-$.
Is there a proof involving roots of $G$ relative to $T$ ?
I've found a proof in a book of M. Geck, but this proof doesn't involve roots at all, but only the fact that $B^-$ is uniquely defined by the relation $B \cap B^- = T$.
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## 3 Answers
The proof depends on how you're setting things up. In my opinion the cleanest approach is the Lie algebraic one, and it goes as follows. Your Borel subalgebra `$\mathfrak b$` determines a choice of simple roots `$\Delta$` and consequently a choice of positive roots `$\Phi^+$`: `$\mathfrak b = \mathfrak t \oplus \bigoplus_{\alpha \in \Phi^+} \mathfrak g_\alpha$`. The action of `$w \in W$` takes `$\mathfrak b$` to `$\mathfrak b_w = \mathfrak t \oplus \bigoplus_{\alpha \in \Phi^+} \mathfrak g_{w\alpha}$`. With respect to the length function defined using `$\Delta$`, the longest element `$w_0$` of `$W$` takes `$\Phi^+$` to `$-\Phi^+$`. It follows that `$b_{w_0}$` is the Borel subalgebra opposite to `$\mathfrak b$`.
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Thank you ! I was trying to do the same thing but on the group-level, meaning that I was trying to look at the $U_\alpha$ (whose Lie algebras are the $\mathfrak{g}_\alpha$). However, I was trying to explicitely determine the image $w_0(\alpha)$ and I couldn't figure it out... I'll look at the global image of the positive roots then. – th.ng Sep 12 2011 at 19:21
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@Faisal: What you outline over an algebraically closed field of characteristic 0 for the Lie algebra is the right blueprint for passing to the algebraic (or Lie) group using the classical dictionary. But in prime characteristic the Lie algebra becomes too cumbersome to work with, as Chevalley found, even though his end results are remarkably close to the classical ones. – Jim Humphreys Sep 13 2011 at 22:45
@Jim: My outline does betray the fact that I'm much more accustomed to working over an alg. closed field of char. 0. However, I did check my references to make sure what I wrote still held true in the case of positive characteristic. Did I overlook something? I would appreciate any comments on the matter, as I am ignorant of this aspect of the theory. In any case, as you mention in your answer, the outline does rely on some nontrivial structure theory, but that is to be expected. – Faisal Sep 13 2011 at 23:04
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In prime characteristic, your description of the Lie algebra of a reductive group depends on knowing virtually all of the Borel-Chevalley structure theory for the groups. But the bigger complication is using information about subalgebras of the Lie algebra to get back to the group. There is no machine that does this directly, only a lot of partial results which often depend on the prime or Lie type. Chevalley abandoned the Lie algebra approach to get uniform results including classification of semisimple groups. But finding the root system (etc.) in the group is not easy. – Jim Humphreys Sep 14 2011 at 13:25
### 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.
The role of the longest element in `$W$` emerges only gradually in the Chevalley structure theory. This is developed similarly but in slightly different styles in the three books with the same title Linear Algebraic Groups (by Borel, Springer, and me). It's important to understand that the desired statement about opposite Borel subgroups of a reductive group depends on a long series of steps culminating in the Bruhat decomposition. In my book, the later steps occupy Sections 26-28. There's already a proof in 26.2 of the existence of a (necessarily unique) Borel subgroup intersecting a given Borel subgroup `$B$` in a specified maximal torus `$T$` of `$B$`. Theorem 26.3(b) almost gives the answer you want, but before enough details of the structure theory are in place. The underlying strategy is to fix `$B$` and the flag variety `$G/B$`, then see that the fixed points of `$T$` on the latter are in natural bijection with both the elements of `$W$` and the Borels containing `$T$`. (I don't think this got articulated explicitly enough, however.)
Only in my Section 27 is the root system explored, followed in Section 28 by the full details of the Bruhat decomposition. There is still some work to be done on the internal structure of `$W$` (including the length function and longest element) in relation to the root system. In my treatment this gets folded into the more general study of `$BN$`-pairs (Tits systems), which axiomatize the Bruhat decomposition efficiently. Along the way it turns out for instance that `$W$` has a natural structure of Coxeter group.
Unfortunately, the precise statement you want to see a proof of is left somewhat implicit in all these textbook treatments. But this is due partly to the fact that so much heavy theory has to be developed systematically before that kind of statement becomes obvious. It's hard to take any real shortcuts, but also unnecessary to get into the theory of buildings and such.
EDIT. The basic outline here looks deceptively simple: Start with the fixed data `$B$` and `$T$` and Weyl group `$W$`. Then `$W$` has a unique longest element `$w_0$`, which has order 2 and interchanges positive and negative roots. Here the positive roots are defined by the choice of `$B$`, so that `$B= T U$` with `$U$` the product in any order of 1-dimensional root subgroups corresponding to positive roots. Then the Borel subgroup `$B^- = w_0 B w_0$` contains `$T$` and has the form `$B^- =T U^-$` with `$U^-$` the product of root subgroups for negative roots.
But to work this out for a reductive group over an algebraically closed field of arbitrary characteristic requires virtually all of the Borel-Chevalley structure theory through the Bruhat decomposition. For instance, it's highly nontrivial to find the root subgroups and define the root system intrinsically for the group, as well as to work out basic facts about the Weyl group `$N_G(T)/T$` and its length function.
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Indeed, I worked with those 3 references and especially with yours, and I was a wondering if I missed something about this longest element or not. Though you are right, with 26.3(b) I thought I was really next to a proof, but still, as I commented above, one has to examine precisely the action of $w_0$, and the fact that $w_0$ send $\phi^+$ on $-\phi^+$ does not seem obvious to me, without using the action on Weyl chambers. – th.ng Sep 12 2011 at 20:21
One way to show this is by using BN-pairs and the associated (spherical) building (see section 6.2 the book "Buildings: Theory and Applications" by Abramenko and Brown, especially section 6.2.6).
The argument is basically the following. The chambers of the associated building correspond to the left cosets of $B$. The parabolic associated to a chamber $gB$ is then given by the conjugate of $B$ by $g$. The Weyl distance between two chambers associated to $gB$ and $hB$ is given by the element of the unique element $w$ in the Weyl group such that $Bg^{-1}hB$. So for your situation this yields that the chamber associated to $B$ is opposite to the chamber associated to $w_0B$.
I hope this helps. While typing this I also realized the answer may depend on which way you define the opposite Borel subgroup.
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Thank you, indeed, I've seen a proof that a Weyl chamber is sent to the opposite under the action of the longest element using combinatorics, but that wasn't really my setting... I should have been more precise. – th.ng Sep 12 2011 at 19:16
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http://mathoverflow.net/questions/94387/non-archimedean-non-standard-models-for-r/94394
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## Non-Archimedean non-standard models for R
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $\langle \mathbb{R}, 0, 1, +, \cdot, <\rangle$ be the standard model for $R$, and let $S$ be a countable model of $R$ (satisfying all true first-order statements in $R$). Is it true that the set $1,1+1,1+1+1,\ldots$ is bounded in $S$? My intuition says "no", but I am yet to find a counter example. I read something about rational functions, but I cannot verify it is, indeed, a non-standard model of R.
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3
Why does a routine application of compactness not yield the model you want? – Juris Steprans Apr 18 2012 at 11:42
@Juris Steprans: I'm sorry, I can't see what you're implying. Could you be more precise, please? – Dave Apr 18 2012 at 11:51
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Dave, due to the high intersection of users between the MO and MSE communities it is considered impolite to post a question on both sites simultaneously. Please remember that for future reference. (Cross posted on math.SE math.stackexchange.com/q/133418/622) – Asaf Karagila Apr 18 2012 at 15:50
## 1 Answer
If $S$ is the set of real algebraic numbers then $1, 1+1, 1+1+1, \dots$ is unbounded in $S$. On the other hand, by compactness of first order logic (as Juris points out), there are models $S$ for which $1, 1+1, 1+1+1, \dots$ is bounded.
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Thanks for your answer. I'm afraid I don't understand how compactness comes in here. I'm guessing I should add the statements $\exists{x} x>1$, $\exists{x} x>1+1$ and so on to my set, and show that there are models that satisfy that set. But I think it only gives me that there are greater elements than any element of the type $1+1+\ldots+1$; I need one element that "covers them all". – Dave Apr 18 2012 at 12:13
No, that doesn´t work. You first need to add a constant to the language. – Ramiro de la Vega Apr 18 2012 at 12:15
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Yes, indeed. I think that a good solution for that would include adding another constant c to the language, then the theorems c>1, c>1+1 and so on, find a model and "forget" c (this is a model of the reduced language). Thanks! – Dave Apr 18 2012 at 12:19
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http://physics.stackexchange.com/questions/44316/thermal-expansion-is-an-expression-of-which-conservation-laws
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# Thermal expansion is an expression of which conservation laws?
Many objects get larger as they heat up and contract as they cool down. Which conservation laws are applied to describe this phenomenon? How do they interact with each other to produce this effect?
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## 2 Answers
Imagine a 1-D axis with a point that moves between two walls and bounces off them. With each collision it imparts momentum to the wall. Now imagine the walls will only accept a certain amount of momentum per unit time. If the point moves faster (increasing temperature) the walls will have no choice but to move further apart.
What I just described is effectively a constant pressure process of an ideal gas. For now, you can conceptually think of the "walls" as either the boundary of the gas, or neighboring molecules. The point on the number line (roughly) follows the ideal gas law.
$$PV=nRT$$
This is one of, though not the only, state equation for a substance. The equation, as written above, has 4 independent variables (whereas $R$ is a universal constant). In order to reduce this, we will eliminate the macroscopic physical size, $V$ from the equation by division. It is then reformulated as a true state equation.
$$\ P = \rho R_{\rm specific}T$$
Setting aside information about what type of gas you're dealing with, this equation has 3 unknowns. If you specify 2 properties, the last property is then set by physics. A state equation can be generally written as:
$${\ f(p,V,T) = 0}.$$
Expansion and compression due to changes in temperature is a subset of this relationship. Thus, your question is answered by the physics that give rise to the the relationship.
To the extent that we speak of the pressure of the wall, we are doing force balance which reflects Newton's third law and balance between the impulse of the molecules and the force on the walls is ultimately a consequence of $F=m a$ (Newton's second law).
For what you're asking about (solid thermal expansion), $P$ is the least important of the 3 variables. The state equation is mostly insensitive to pressure for solid materials, and the interplay between $T$ and $\rho$ forms the concept of thermal expansion. The other hairy detail of solids is that they're not free to move like an ideal gas, they're held together like springs. Indeed, it is difficult to convince oneself of solid or liquid thermal expansion (actually, sometimes the opposite happens in nature). If I have an object on a spring, then even if it moves with a greater energy/amplitude/average speed/temperature, its average position would remain in the same (equilibrium) position. Thus, we have to consider how chemical bonds are non-ideal springs.
To make this argument, the Lennard-Jones potential profile is sufficient. In this illustration I will illustrate the basic physical mechanism of thermal expansion of liquids and solids.
I hope it's apparent that if the potential was a perfectly symmetric "bowl", the material WOULD NOT expand at all. After all, water and ice contract with increasing temperature over a certain range.
It's hard to quote a specific physical law, but from what I've argued here, I would say that the expansion of gases with temperature is largely explained by conservation of momentum and force distribution laws, while the expansion of solids is explained by the conservation of energy specific to real-world chemical potential profiles. Of course, the real state equation of real substances is probably somewhat of a combination of these two.
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They are not necessarily related to conservation laws (or they are, but indirectly in the sense that energy is overall conserved, for instance when it flows from a hot to a cold source). The reason that an an objects tends to expand when heat flows into it is due to the fact that heat makes their molecules to move faster, so they put more pressure on the container when they collide with it, and make it expand (like a gas within a balloon, unless the container is rigid. The opposite happens when a gas is cooled
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http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Heat_pump
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# All Science Fair Projects
## Science Fair Project Encyclopedia for Schools!
Search Browse Forum Coach Links Editor Help Tell-a-Friend Encyclopedia Dictionary
# Science Fair Project Encyclopedia
For information on any area of science that interests you,
enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.).
Or else, you can start by choosing any of the categories below.
# Heat pump
A heat pump is a machine which moves heat from a low temperature reservoir to a higher temperature reservoir under supply of work.
Common examples are:
Heat pumps are realized through several physical effects, but they are classified depending on their applications (driving energy, source and sink of heat, or a heat pump which is basically a refrigeration machine). Refrigerators, air conditioners and heating systems are all common applications of heat pumps.
An easy way to imagine how a heat pump works is to imagine the heat in a given space - say the volume of a football (a soccerball for those in the USA). The air within the volume of the football has say 100 units of heat. This air is then compressed to the size of ping pong ball (table tennis ball for affectionados); it still contains the same 100 units of heat, but the heat is much more concentrated and thus the average heat per volume unit is much higher. The ping pong volume of heat is then moved from the heat source area to the target area that has a lower per volume concentration of heat. Since the heat of the ping pong ball volume is now a higher concentration than the surrounding heat, the heat is given off until the ping pong ball volume heat reaches the same concentration of heat as the surrounding area. The ping pong ball volume is then moved outside the target area back to the heat source area and allowed to expand. In expanding the heat per unit volume of the football is now much lower than the source and the football volume absorbs heat from the surrounding area. The process then repeats.
When comparing the performance of heat pumps, it is best to avoid the word "efficiency", as it has many different meanings. The term coefficient of performance or COP is used to describe the ratio of heat output to electrical power consumption. A typical heat pump has a COP of about three, whereas a typical electric heater has a COP of one.
The COP of a heatpump is restricted by the second law of thermodynamics.
$COP_{\mathrm{heating}} = \frac{\Delta Q_{\mathrm{hot}}}{\Delta A} \leq \frac{T_{\mathrm{hot}}}{T_{\mathrm{hot}}-T_{\mathrm{cool}}} = \frac{1}{\eta_{\mathrm{carnotcycle}}}$
$COP_{\mathrm{cooling}} = \frac{\Delta Q_{\mathrm{cool}}}{\Delta A} \leq \frac{T_{\mathrm{cool}}}{T_{\mathrm{hot}}-T_{\mathrm{cool}}}$
All temperatures T are measured in kelvins.
## See also
03-10-2013 05:06:04
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http://math.stackexchange.com/questions/135405/module-m-im-of-finite-length-implies-ring-a-i-of-finite-length/139240
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# Module M/IM of finite length $\implies$ Ring A/I of finite length
This question is due to a proof in an algebra book (on the topic of dimension theory) which I don't fully understand (specifically, the proof of Thm 6.9b) in Kommutative Algebra by Ischebeck). It may have a very simple answer which I currently don't see.
Let A be a semilocal, noetherian ring, $M$ a finite A-module with $\mathrm{Ann}_A(M)=0$ and $a_1,\ldots,a_s\in \mathrm{Jac}(A)$ with $l_A(M/(a_1,\ldots,a_s)M)\lt\infty.$ Why is $l_A(A/(a_1,\ldots,a_s))\lt\infty$ ?
(where $\mathrm{Ann}_A(M)=\{x\in A\mid xM=0\}$, $\mathrm{Jac}(A)=\bigcap\limits_{\text{m is maximal ideal in } A} m$, $l_X(Y)=$ length of $Y$ as an $X$-module)
My ideas: I know that from $l_A(M/(a_1,\ldots,a_s)M)\lt \infty$ it follows that the ring $A/\mathrm{Ann}_A(M/(a_1,\ldots,a_s)M)$ is of finite length, because it can be embedded (as an $A$-module) in $M/(a_1,\ldots,a_s)M.$ But I don't know if/why $\mathrm{Ann}_A(M/(a_1,\ldots,a_s)M)=(a_1,\ldots,a_s)$.
Also I don't know if $A$ being semilocal or $a_1,\ldots,a_s\in \mathrm{Jac}(A)$ instead of $a_1,\ldots,a_s\in A$ is of any relevance to the question.
As further information, the final goal is to show that the degree of the polynomial p(n), which for large n gives the length $l_{A/Jac(A)}(M/Jac(A)^nM)$, is lesser or equal the minimum number s for which $l(M/(a_1,..,a_s)M)<∞$ as above.
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It is always useful to mention the title and author of a book one talks about—saying «an algebra book» is too vague! – Mariano Suárez-Alvarez♦ Apr 22 '12 at 23:08
The proof is the one of Thm 6.9b) in Kommutative Algebra by Ischebeck, a German textbook which probably noone here knows. :) The goal is to show that the degree of the polynomial p(n), which for large n gives the length $l_{A/Jac(A)}(M/Jac(A)^nM)$, is lesser or equal the minimum number s for which $l(M/(a_1,..,a_s)M)<\infty$ as above – juffo Apr 23 '12 at 1:09
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It is best if you edit the body of the question and add all that information there. – Mariano Suárez-Alvarez♦ Apr 23 '12 at 1:15
## 3 Answers
Suppose that $A$ is a noetherian commutative ring acting faithfully on a module $M$, and let $I$ be an ideal of $A$. It is not true in general that $A/I$ acts faithfully on $M/IM$, but one can show (e.g. using the Artin--Rees lemma) that the map $A/I \to End(M/IM)$ has nilpotent kernel.
In particular, if $M/IM$ has finite length, then $End(M/IM)$ also has finite length, and hence $A/I$ modulo a nilpotent ideal has finite length. This last statement in turn suffices to imply that $A/I$ itself is of finite length. (I didn't think about whether there is a more direct proof in your particular situation.)
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(I fixed a typo) – Mariano Suárez-Alvarez♦ Apr 23 '12 at 2:27
Thank you for your answer. Could you elaborate a bit on how to show that the kernel is nilpotent, i.e. on which ideal and submodule to apply Artin-Rees? – juffo Apr 23 '12 at 19:31
$\mathrm{Supp}(M/IM)=V(I+\mathrm{Ann}M)=V(I)=\mathrm{Supp}(R/I)$. Since $M/IM$ is of finite length, $\mathrm{Supp}(M/IM)$ consists of maximal ideals, so the same is true for $\mathrm{Supp}(R/I)$. Thus $R/I$ is of finite length. The second equality above follows because $\mathrm{Ann}M=0$.
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This looks like a perfect answer ! – QiL'8 Jan 3 at 11:37
The answer Matt E gave is absolutely right.
I gave the proof that the kernel is nilpotent in other way.
Since $M$ is finite over $A$, say $M$ is generated by $m_1,\ldots,m_n$, then we have an injective map
$$A\to \bigoplus_{i=1}^nM,a\mapsto (am_1,\ldots,am_n)$$
So we have a map $A/I\to \bigoplus_{i=1}^nM/IM$. Let the image of $a$ in $A/I$ be in the kernel, i.e., $aM\subset IM$, so by Nakayama's lemma, there is an $i\in I$ such that $(a^n+i)M=0$, because $M$ is a faithful $A$-module, we obtain that $a^n+i=0$, hence $a^n\in I$. Because $A/I$ is Noetherian, the kernel (denoted by $K$) is finitely generated, thus $K$ is nilpotent, i.e., there exists an $l$ such that $K^l=0$.
Okay, replacing $A/I$ by $A$, we have the conditions: $A$ is a Noetherian semilocal ring, $A/K$ is of finite length and $K$ is nilpotent. We want to show $A$ itself is of finite length.
The condition $A/K$ is of finite length implies that $A/K$ is an Artinian ring. Since $K$ is nilpotent, this implies that $A$ itself is an Artinian ring (dim=0,and Noetherian)! Hence $A$ is of finite length.
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http://mathhelpforum.com/differential-equations/166817-pde-u_-xxy-1-a.html
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# Thread:
1. ## PDE u_{xxy}=1
Find the general solution of $u_{xxy}=1$
I know how to find the general solution of $u_{xx}=1$
Since this is $u_{xxy}$, does this mean integrate with respect to y, than x, and x again?
Thanks.
Would this be the general solution:
$\displaystyle \int u_{xxy}dy=\int dy\Rightarrow u_{xx}(x,y)=y+f(x)\Rightarrow\int u_{xx}(x,y)dx=\int (y+f(x))dx$
$\displaystyle\Rightarrow u_{x}(x,y)=yx+f(x)+h(y)$
$\displaystyle\Rightarrow \int u_{x}(x,y)dx=\int (yx+f(x)+h(y))dx$
$\displaystyle\Rightarrow u(x,y)=\frac{yx^2}{2}+f(x)+xh(y)+g(y)$
2. This seems correct. Differentiating back again for your found solution: $\frac{\partial}{\partial y}\frac{\partial}{\partial x}\frac{\partial}{\partial x}u(x,y)$ yields 1 as well.
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http://crypto.stackexchange.com/questions/5889/calculating-rsa-private-exponent-when-given-public-exponent-and-the-modulus-fact
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# Calculating RSA private exponent when given public exponent and the modulus factors using extended euclid
When given p = 5, q = 11, N = 55 and e = 17, I'm trying to compute the private key d.
I can calculate phi(N) = 40, but my lecturer then says to use the extended Euclidean algorithm to compute d. That's where I get stuck.
Here's my work so far:
First I use the euclid algorithm to calculate :
40 = 2(17) + 6
17 = 2(6) + 5
6 = 1(5) + 1 = gcd
So I know the gcd is 1.
6 = 40-2(17)
5 = 17-2(6)
1=6-1(5)
1 = 6-1(17-2(6))
3(6) = 1 (17)
3(40 - 2(17)) - 1(17)
3(40) - 3(17)
I know the answer is 33, but I have no idea how to get there using the extended Euclidean algorithm. I can't figure out why I'm getting 3(40) - 3(17) when I know the answer should contain 33?
Any help is much appreciated!
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– CodesInChaos Jan 3 at 13:30
Unfortunately it has to be computed by hand... I've edited the question to included more of my work. – DougalMaguire Jan 3 at 13:32
## 1 Answer
The extended Euclidean algorithm is essentially the Euclidean algorithm (for GCD's) ran backwards.
Your goal is to find $d$ such that $ed \equiv 1 \pmod{\varphi{(n)}}$.
Recall the EED calculates $x$ and $y$ such that $ax + by = \gcd{(a, b)}$. Now let $a = e$, $b = \varphi{(n)}$, and thus $\gcd{(e, \varphi{(n)})} = 1$ by definition (they need to be coprime for the inverse to exist). Then you have:
$$ex + \varphi{(n)} y = 1$$
Take this modulo $\varphi{(n)}$, and you get:
$$ex \equiv 1 \pmod{\varphi{(n)}}$$
And it's easy to see that in this case, $x = d$. The value of $y$ does not actually matter, since it will get eliminated modulo $\varphi{(n)}$ regardless of its value. The EED will give you that value, but you can safely discard it.
Now, we have $e = 17$ and $\varphi{(n)} = 40$. Write our main equation:
$$17x + 40y = 1$$
We need to solve this for $x$. So apply the ordinary Euclidean algorithm:
$$40 = 2 \times 17 + 6$$
$$17 = 2 \times 6 + 5$$
$$6 = 1 \times 5 + 1$$
Write that last one as:
$$6 - 1 \times 5 = 1$$
Now substitute the second equation into $5$:
$$6 - 1 \times (17 - 2 \times 6) = 1$$
Now substitute the first equation into $6$:
$$(40 - 2 \times 17) - 1 \times (17 - 2 \times (40 - 2 \times 17)) = 1$$
Note this is a linear combination of $17$ and $40$, after simplifying you get:
$$(-7) \times 17 + 3 \times 40 = 1$$
We conclude $d = -7$, which is in fact $33$ modulo $40$ (since $-7 + 40 = 33$).
As you can see, the basic idea is to use the successive remainders of the GCD calculation to substitute the initial integers back into the final equation (the one which equals $1$) which gives the desired linear combination.
As for your error, it seems you just made a calculation error here:
3(40 - 2(17)) - 1(17)
which incorrectly became:
3(40) - 3(17)
It seems you forgot the factor of 3 for the left 17, the correct result would be:
3(40 - 2(17)) - 1(17) = 3 * 40 - 3 * 2 * 17 - 1 * 17 = 3 * 40 + (-7) * 17
Which is the -7 expected.
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Wow, that explains it perfectly. Thanks a lot! – DougalMaguire Jan 3 at 14:54
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http://physics.stackexchange.com/questions/6898/giant-string-in-space/6900
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# Giant string in space
I saw an interview with Ed Witten, where he said one way to confirm string-theory is to observe a giant string floating in space, left over from the Big-Bang.
How does one observe such a string, does it have thickness? Is this string any different then the strings that (hypothetically) make up elementary particles?
What is required to produce such a large string, why is this string so much bigger then others?
What happends if one tries to interact with this string, can one stretch/cut it etc?
Can we have a stable mega-string on earth, and can it be used for anything?
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## 3 Answers
The things in question are called "cosmic strings." They are predicted as a byproduct of some but by no means all theories of the early Universe. There's no experimental evidence that they actually exist, although I suppose the possibility hasn't been ruled out.
• How would we observe this?
A long cosmic string would have a couple of observable signatures. The main effect is a sort of gravitational lensing. If a cosmic string lies directly between you and a background object, you will see two copies of the object, one from light that passed around the string to the left, and one from light that went to the right.
To go into a bit more detail, a cosmic string produces a "conical singularity" in the space around it. This means that the circumference of a circle surrounding the string will be slightly less than $2\pi$ times the radius. The fractional deficit is independent of radius. It's as if someone had removed a triangular wedge of space, and then "glued together" the faces adjoining the removed bit. To see why this is called a conical singularity, imagine taking a piece of paper, cutting out a triangular wedge (i.e., removing all points with $0<\phi<\alpha$ for some constant $\alpha$), and then gluing together the edges $\phi=0$ and $\phi=\alpha$. The paper will now have the shape of a cone; an ant walking around the singularity will cover a circumference $C=(2\pi-\alpha) r$.
Light travels in straight lines (geodesics) in this conical space, but two such straight lines can pass around opposite sides and meet again. (Draw it on your conical paper if you don't see this.)
So one way to look for cosmic strings is to look for pairs of identical images of something.
If the cosmic string happens to be moving transverse to your line of sight, then there's another observational signature. Background light that reaches you from the "downstream" side will be blueshifted compared to background light reaching you from the upstream side. In effect, the conical geometry gives a kick to photons coming from one side relative to the other. (That's actually a kind of imprecise way to put it: this is a purely geometrical effect, not an actual force.)
• Does it have a thickness?
Cosmic strings are predicted to have some nonzero thickness, but very small. I don't remember the details.
• Are they the same as the strings that (are theorized to) make up elementary particles?
Cosmic strings are not the same as the strings in string theory. There are connections between the two theories, which I don't understand at all. I understand cosmic strings pretty well and string theory not at all. I hope someone else will talk about the connections.
• What is required to produce such a large string?
Cosmic strings are predicted to result in some theories of the early Universe. They are "topological defects" resulting from a phase transition. During the phase transition, some field tries to "relax" into a lower-energy configuration. There are multiple different versions of the lower-energy state, and the field chooses different versions in different parts of space. A cosmic string results from different regions making incompatible choices, so that the whole system can't relax to the same state.
If you simulate such a phase transition, in which different points "choose" randomly which state to relax into and then the whole system tries to settle down into a low-energy final configuration, you find that many such strings result. There are large numbers of small loops and smaller numbers of big strings. The very long ones that Witten is talking about are predicted to be rare, but there should be a small number in our observable Universe (if a string-producing phase transition actually occurred).
• What happens when you interact with such a string?
The strings are under high tension and have a high energy density confined to a very small cross-sectional area. It'd be hard to cut one, without supplying enough energy to "undo" the phase transition. When two strings (or two sections of the same string) meet, they can "reconnect," essentially swapping with each other at the intersection point. A loop of string can "pinch off" smaller loops via this process: imagine the string twisting around into a figure-8 shape, and then splitting into two smaller loops.
There's a lot more to the dynamics of these strings. They can lose energy by emitting gravitational waves, for example.
• Can we have one on Earth?
In principle we could probably produce some small ones, if we could get the energy density in some region up high enough to undo the phase transition, but nobody's done anything remotely like this. If we did have a big one, I don't know how easy it would be to control it.
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This is the conversion of a superstring into a cosmic string. A cosmic string is a one dimensional region where the vacuum state is that of the unbroken Lagrangian. It is similar to the physics of supercooled water that can be colder than the freezing point. In the phase transition of a gas or liquid to solid there can be boundary zones or grains where the gas or liquid state persists.
The connection to string theory is with F-theory, which takes one into 12-dimensions, or with D-strings that are related to F-strings by S-duality between strong and weak interactions. One of the corner stones of string theory is the Nambu-Goto action, which is a starting point for the string action determined by the area of the string world sheet. The one spatial quantum dimension theory also describes the cosmic string. If a superstring is drawn into a very large filament so that its states remain invariant under the symmetries of the Lagrangian the string can be converted into this "defect" that is a cosmic string. The string has a huge tension, related to the string parameter $\alpha'$, and if this string is stretched to enormous lengths this can have a large gravitation. The curvature of space can be thought of as an orbit around the string, but where the area enclosed by the loop is the disk with a wedge cut out of it. This deficit angle $\theta$ then defines a curvature bounded within that loop as $R~\sim~\theta/2\pi r^2$, where $r$ is the radius of the loop.
The universe might have these floating around out there, sort of menacing cosmic bullwhips of sorts, that could have an imprint on the CMB. It would be bad new if one of these cut through the Earth, for it would gravitationally pancake the Earth some. As I recall the Earth would be squashed inwards at 4km/sec as it passed through. We would get a global quake many orders of magnitude worse than the recent Sendai quake.
It is hard to know how stable these are, and it could be that interactions with them might break the string up into small strings near the string length. The deficit angle above would have a gravitational lensing effect, and if one passed between the Earth and a distant object it might be detected that way.
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This paper: http://arxiv.org/PS_cache/astro-ph/pdf/0302/0302547v1.pdf was a speculative look at a possible double galaxy image. They considered that it had been mirrored down the centre by a cosmic string to repeat the galaxy. It turned out to be untrue however when looked at in more detail by hubble.
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http://math.stackexchange.com/questions/216792/bivariate-lognormal-distribution
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# Bivariate Lognormal Distribution
Is there a pdf for bivariate lognormal distribution? Suppose there are 2 random variable $X_1$ and $X_2$ with standard normal distribution, how would the pdf of a bivariate lognormal distribution look like?
Thanks!
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1
If $X_1$, $X_2$ are normally distributed, why would you expect their joint distribution to be a bivariate lognormal distribution? – fgp Oct 19 '12 at 9:50
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http://mathoverflow.net/revisions/53623/list
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## Return to Answer
2 added 322 characters in body
The problem is nontrivial already in the finite dimensional case $E= \mathbb R^d$, $F=\mathbb R$. The space $C^{\omega}(\Omega)$ of real-valued real analytic functions on the open bounded set $\Omega\subset \mathbb R^d$ does not have any obvious or natural metric which would make it a Fréchet space.
The good news is that there is a "canonical" topology which renders $C^{\omega}(\Omega)$ as a complete (reflexive nuclear separable) space. In fact, it is natural to endow $C^{\omega}(\Omega)$ with either an inductive limit or a projective limit topology but these two are equivalent on $C^{\omega}(\Omega)$ as was shown by Martineau in 1966.
For practical purposes, the topology can be described following the suggestion of Piero D'Ancona in his comment above. Let $\{U_j\}_{j\in\mathbb N}$ be a monotonically decreasing sequence of open sets of $\mathbb C^d$ such that $\Omega=\bigcap U_j$. Let $\{h_j\}_{j\in\mathbb N}$ be a sequence of bounded holomorphic functions $h_j:U_j\to\mathbb C$ such that $h_j|_{U_k}=h_k$ for $k\geq j$. Then a subbase element of the topology on $C^{\omega}(\Omega)$ has the form $$\mathcal V_{j, K}=\left\{f\mbox{ is real analytic on }\Omega:\ \sup_{x\in K} \left|\partial^{\alpha} f\right|\leq C_j[\delta_j(K)]^{-|\alpha|}\ \mbox{ for every }\alpha\in\mathbb N^{d}_{0}\right\},$$ where the set $K\subset\Omega$ is compact, $\delta_j(K)=\mbox{dist}\{K,\partial U_{j+1}\}$ and $C_j$ is a constant which depends on the supremum of $h_j$ on $U_{j+1}$.
A sketch of the construction in the finite dimensional setting can be found, for instance, in A Primer of Real Analytic Functions by Krantz and Parks. Hopefully, it generalizes to the case of Banach spaces in a straightforward way.
[EDIT. Concerning your specific question whether the limit of a sequence of real analytic functions is itself an analytic function. Let $f\in C^\infty(\mathbb T)$ be a periodic smooth but non-analytic function. Then the partial Fourier sums $S_N f$ converge to $f$ in the uniform metric with all their derivatives.]
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The problem is nontrivial already in the finite dimensional case $E= \mathbb R^d$, $F=\mathbb R$. The space $C^{\omega}(\Omega)$ of real-valued real analytic functions on the open bounded set $\Omega\subset \mathbb R^d$ does not have any obvious or natural metric which would make it a Fréchet space.
The good news is that there is a "canonical" topology which renders $C^{\omega}(\Omega)$ as a complete (reflexive nuclear separable) space. In fact, it is natural to endow $C^{\omega}(\Omega)$ with either an inductive limit or a projective limit topology but these two are equivalent on $C^{\omega}(\Omega)$ as was shown by Martineau in 1966.
For practical purposes, the topology can be described following the suggestion of Piero D'Ancona in his comment above. Let $\{U_j\}_{j\in\mathbb N}$ be a monotonically decreasing sequence of open sets of $\mathbb C^d$ such that $\Omega=\bigcap U_j$. Let $\{h_j\}_{j\in\mathbb N}$ be a sequence of bounded holomorphic functions $h_j:U_j\to\mathbb C$ such that $h_j|_{U_k}=h_k$ for $k\geq j$. Then a subbase element of the topology on $C^{\omega}(\Omega)$ has the form $$\mathcal V_{j, K}=\left\{f\mbox{ is real analytic on }\Omega:\ \sup_{x\in K} \left|\partial^{\alpha} f\right|\leq C_j[\delta_j(K)]^{-|\alpha|}\ \mbox{ for every }\alpha\in\mathbb N^{d}_{0}\right\},$$ where the set $K\subset\Omega$ is compact, $\delta_j(K)=\mbox{dist}\{K,\partial U_{j+1}\}$ and $C_j$ is a constant which depends on the supremum of $h_j$ on $U_{j+1}$.
A sketch of the construction in the finite dimensional setting can be found, for instance, in A Primer of Real Analytic Functions by Krantz and Parks. Hopefully, it generalizes to the case of Banach spaces in a straightforward way.
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http://mathoverflow.net/questions/56186/the-vanishing-of-the-2nd-plurigenus-of-a-sextic-threefold/56196
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## The vanishing of the 2nd plurigenus of a sextic threefold
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I'm reading a recent preprint by Beauville on the nonrationality of a specific sextic threefold $X$ which is a complete intersection of a quadric and a cubic in $\mathbb P^5$. At some point he uses that $H^0(X,\Omega^2)=0$, and I was having trouble figuring out why that was. It occurred to me that it might use the Hodge decomposition, its symmetry, and two applications of the Lefshetz hyperplane theorem, but I couldn't get it to work. Anyone know a quick answer?
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Here's the argument I was thinking of which uses Lefschetz, which I should seen immediately: By Lefschetz $H^{p,q}(\mathbb P^5)$ is isomorphic to $H^{p,q}(C)$ for $p+q<4$, where $C$ is the cubic hypersurface. Applying this again by cutting with the quadric $Q$ to get our $X$, we see that $H^{p,q}(X)$ is isomorphic to $H^{p,q}(C)$ for $p+q<3$. Since the $h^{p,q}(\mathbb P^n)=0$ for $p\neq q$, this gives our result. I suppose since one proves Lefschetz with Kodaira vanishing, it's all the same really, but this is what I had in mind originally. – HNuer Feb 21 2011 at 20:36
Of course this is another possibility. The point is that this is a more general property of Fano manifolds... Of course all this needs $X$ to be smooth... – diverietti Feb 21 2011 at 21:55
## 2 Answers
What you are asking for is not the second plurigenus: the second plurigenus is $h^0(X,2K_X)$. So do you need the vanishing of the second plurigenus or of the space of global holomorphic two forms?
If you need just the second plurigenus, then this is very easy since by adjunction $K_X\simeq\mathcal O_X(−1)$ and so $h^0(X,2K_X)=h^0(X,\mathcal O_X(−2))=0$ since $\mathcal O_X(−2)$ is negative.
On the other hand, if you need the vanishing of global holomorphic two-forms, this is quite easy, too. Just observe that since $-K_X\simeq \mathcal O_X(1)$, then $-K_X$ is positive. So, you get by Kodaira's vanishing $$H^q(X,K_X-K_X)=H^q(X,\mathcal O_X)=0,\quad q\ge 1.$$ But now, by Dolbeault's isomorphism, $H^q(X,\mathcal O_X)\simeq H^{0,q}(X,\mathbb C)$. By the Hodge symmetry $h^{0,2}(X,\mathbb C)=h^{2,0}(X,\mathbb C)=h^0(X,\Omega_X^2)$, where the last equality is again thanks to the Dolbeault isomorphism, and you are done.
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Do you need any explanation about how to use adjunction formula to find the expression of the canonical bundle in term of the tautological bundle of the projective space? – diverietti Feb 21 2011 at 19:31
No, I understand why $K_X=\mathcal O_X(-1)$. – HNuer Feb 21 2011 at 20:27
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$h^{q,0}(X) = h^{0,q}(X) = 0$ for any Fano $X$ and $q > 0$, because $-K_X$ ample implies $$H^q(X, \Omega^0) = H^q(X, K_X \otimes (-K_X)) = 0$$ by Kodaira vanishing.
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http://math.stackexchange.com/questions/237147/showing-a-function-has-only-one-point-of-continuity
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# Showing a function has only one point of continuity.
Let
$$f(x) = \begin{cases}\;\;\, x\;\;,\;\text{ if } x \in \mathbb{Q}\\ -x\;\;,\; \text{ if } x \in \mathbb{R}\setminus \mathbb{Q} \end{cases}$$
(i) Determine the point or points of continuity of $f$. (ii) Show that the point of points of continuity of $f$ are the only points.
Clearly its continuous at $0$. I'm not sure how to prove it is continuous at $0$, but I know how to prove that it has no other points of continuity.
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## 2 Answers
Let $\,\{x_n\}_{n\in\Bbb N}\,$ be a real sequence s.t. $\,x_n\xrightarrow [n\to\infty]{} 0\,$ (and thus also $\,-x_n\xrightarrow [n\to\infty]{} 0\,$) .
It's easy to see that we have
$$f(x_n)=\pm x_n\xrightarrow [n\to\infty]{}0=f(0)$$
And $\,f\,$ is thus continuous at $\,x=0\,$ as the above's true for any real sequence converging to zero.
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Let $\epsilon >0$. Then for $\delta = \epsilon$ if $|x|<\delta \Rightarrow |f(x)-f(0)|=|f(x)|=|x|<\delta=\epsilon \Rightarrow$
$f$ is continuous at $0$ .
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http://mathoverflow.net/questions/74816?sort=oldest
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## Is there a standard name for the intersection of all maximal linearly independent subsets of a given set in a vector space?
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The title more or less says it all.... Let $V$ be a vector space (over your favorite field; $V$ not necessarily finite dimensional), and let $S$ be a subset of $V$. A maximal linearly independent subset of $S$ is exactly that: a subset of $S$ that is linearly independent yet not properly contained in any other linearly independent subset of $S$. (Equivalently, it is a basis for the subspace of $V$ that is spanned by $S$.)
Let $T$ be the intersection of all maximal linearly independent subsets of $S$. This $T$ might be as large as $S$, when $S$ itself is linearly independent. Alternatively, $T$ might be empty: if ${v_1,v_2,v_3}$ is a basis for $V$, then both examples $S = {}${$v_1,2v_1$} and $S = {}${$v_1,v_2,v_3,v_1+v_2+v_3$} have corresponding $T=\emptyset$. There are plenty of intermediate cases as well: in the same notation, if $S = {}${$v_1,v_2,v_3,v_2+v_3$} then $T={}${$v_1$}.
Does this object $T$ have a standard name?
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## 1 Answer
For a matroid the elements that are contained in every basis are called coloops, dual to the notion of a loop, which is an element not contained in any basis. Since you are interested in linearly independent sets perhaps adopting the language of matroids is not such a bad idea.
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http://mathoverflow.net/questions/21418/monoidal-closed-structures-on-the-category-bicategories-with-strict-functors/22373
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Monoidal closed structure(s) on the category “bicategories, with strict functors”?
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I'm working with globular operadic higher categories (with the Batanin/Leinster definitions) and ending up working a lot in the categories $P$-$\mathrm{Alg}$ of algebras for some globular operad $P$, together with (literal, strict, algebraic) morphisms between them. A familiar case of these is $\textrm{Bicat}_\textit{str}$: the category of bicategories and strict functors between them.
It would be very handy if there were some kind of mapping space constructions in these categories --- that is, some reasonable monoidal closed structures on them. Does anyone know what's been shown, either to exist or not to, either in the general case, or (more likely) for $\textrm{Bicat}_\textit{str}$?
(The most obvious specific candidate, of course, is Cartesian closure. However, even $\textrm{Bicat}_\textit{str}$ fails to be Cartesian closed: chasing through the Yoneda argument that shows what a Cartesian closure would have to look like if it did exist leads one to a counterexample showing that the product doesn't preserve pushouts. The next best hope would presumably be some sort of Gray tensor product; this is where I've not yet been able to find anything further on the closure question.)
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1 Answer
You may know this, but there are people thinking about this sort of question at least from the monoidal viewpoint. For instance, there is this paper, which shows that if you can defined what you want to mean by "a category enriched in P-Alg," then you can automatically recover from that a corresponding monoidal structure on P-Alg. In particular, this can produce the Gray tensor product from the notion of Gray-category. Although in general what you get is only a lax monoidal structure, and I don't think they have (yet) asked when it will be closed.
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Thanks — yes, I have looked at that (though not in detail), but as you say it doesn't (iirc) start looking into the question of closure; and unfortunately, the mapping spaces are what I really want... – Peter LeFanu Lumsdaine Apr 29 2010 at 17:31
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http://mathhelpforum.com/algebra/48431-help-ratios-ack.html
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# Thread:
1. ## HELP!!! ratios! ack!!
I reallly wanna go to bed but I need to get my homework done and I'm absolutely stuck on this one.
"The measures of two angles are in the ratio 5:3. The measure of the larger angle is 30 greater than half the difference of the angles. Find the measure of each angle."
2. Originally Posted by A dumb person
I reallly wanna go to bed but I need to get my homework done and I'm absolutely stuck on this one.
"The measures of two angles are in the ratio 5:3. The measure of the larger angle is 30 greater than half the difference of the angles. Find the measure of each angle."
Let $\alpha$ represent the smaller angle, and let $\beta$ represent the larger angle.
Thus, we see that $\frac{\beta}{\alpha}=\frac{5}{3}$
Now, the larger angle is 30 more than half of the difference of the angles. This means that $\beta=30+\tfrac{1}{2}(\beta-\alpha)$
We can now solve this system of equations:
$\left\{\begin{aligned}30+\tfrac{1}{2}(\beta-\alpha)&=\beta\\\tfrac{3}{5}\beta&=\alpha\end{alig ned}\right.$
Try to take it from here. It shouldn't be that bad.
I hope this makes sense!
--Chris
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http://physics.stackexchange.com/questions/236/how-does-positronium-exist/29291
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# How does Positronium exist?
I've just recently heard of Positronium, an "element" with interesting properties formed by an electron and positron, and I was shocked to hear that physicists were actually working with this element, even if for a very short lifetime. I was always under the impression that matter and antimatter annihilated when they came even remotely close to each other, which is apparently not the case.
How do these two particles combine to form an element if they're oppositely charged and roughly the same mass? What kind of interactions could possibly take place before they're pulled together and annihilated?
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The annihilation is relativistic, and happens in a Compton wavelength, while the orbit is nonrelativistic and 1/alpha times bigger. – Ron Maimon May 31 '12 at 19:45
## 5 Answers
As you've noticed, it's not automatically true that a particle and its antiparticle will annihilate each other when they get close to each other. In fact, no interaction between particles is really certain to happen. Quantum mechanics (and at a higher level, quantum field theory) tells you that all these interactions happen with certain probabilities. So for instance, when a particle and its antiparticle come into close proximity, there is only a chance that they will interact within any given amount of time.
However, the longer the particles remain together, the greater the probability that they will interact and annihilate each other. This is responsible for the 142 ns lifetime of positronium as reported in the Wikipedia article: the probability of annihilation increases with time in such a way that the average lifetime of an "atom" of positronium is 142 ns.
As Cedric said, as long as the positron and electron don't annihilate each other (and remember, there is only a limited chance of that happening in any given time), they can interact in much the same way as any other charged particles, such as the proton and electron. Being bound together by the electromagnetic interaction, as in a hydrogen atom or a positronium "atom," is just one example.
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Not only that, but I see from Google that there is a research field on Rhyberg or multiply excited states of Positronium, also magnetized ones, which should have much longer lifetimes. If anyone knows more, please chime in. – sigoldberg1 Nov 4 '10 at 22:58
– sigoldberg1 Nov 7 '10 at 0:13
Wouldn't that be a "half-life" or 142 ns? Not a "lifetime". – endolith Nov 13 '10 at 19:11
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@endolith: the lifetime of a particle is usually taken to be the decay time constant, which is the time $\tau$ such that the fraction of particles remaining undecayed after time $t$ is $\exp(-t/\tau)$. It's proportional to the half-life. (Some physicists, usually of the nuclear persuasion, do use "lifetime" to mean the half-life, or so I've heard) – David Zaslavsky♦ Nov 13 '10 at 22:11
Just to add. Not only does the positronium exist, it can also interact with matter and allows you to do some interesting physics. For example, in a recent paper S. Mariazzi, P. Bettotti, R.S. Brusa, Positronium Cooling and Emission in Vacuum from Nanochannels at Cryogenic Temperature, Phys. Rev. Lett. 104, 243401 (2010) positronium created by deposition of positrons on a nanostructured surface was cooled down by collision with walls of nanochannels and thermalized(!) at about 150K. Here is a citation from the abstract of that paper:
High formation yield and a meaningful cooled fraction of positronium below room temperature were obtained by implanting positrons in a silicon target in which well-controlled oxidized nanochannels (5–8 nm in diameter) perpendicular to the surface were produced. We show that by implanting positrons at 7 keV in the target held at 150 K, about 27% of positrons form positronium that escapes into the vacuum. Around 9% of the escaped positronium is cooled by collision with the walls of nanochannels and is emitted with a Maxwellian beam at 150 K.
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+1 for quoting Phys. Rev – Cedric H. Nov 6 '10 at 11:46
That's because they are oppositely charged that they can form a bound state: even classically you can understand that: oppositely charged charges attract each other.
While it is true that a particle and its antiparticle can annihilate each other, they first have to interact.
Positronium is a purely electromagnetic bound state: the positron and the electron will form a bound state by electromagnetic interaction (no strong interaction as they are leptons, and the weak interaction does not play a role to form the bound state).
They have the same mass, but it is not a real problem.
Quantum mechanically this problem is treated exactly the same way as the textbook example of the hydrogen atom. You first separate the centre of mass from the problem, but here as they have the same mass this cannot be neglected in the final result.
Then you calculate the interaction of one particle with the centre of mass (in the case of the H atom, this is unambiguously the interaction of the electron with the proton, but here it is one of the two lepton with the centre of mass which is in the middle).
I should also be noted that even if the bound state is stable from that point of view, the positronium will eventually annihilate because the two wave function will overlap and thus these two anti-particles can interact and annihilate.
Positronium can be formed in a variety of ways, one example, where you can create positronium in your bathroom is to have an element which is $\beta^+$ unstable. After this decay, a positron is emitted. It can then interact with the very large number of electron present in the matter and they can form a bound state: the positronium.
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For a slightly different spin on more or less the same thing that's already been said: The idea that an electron coming sort of close to a positron will inevitably annihilate with is comes from the same sort of misconception that leads people to think that objects coming sort of close to a black hole will inevitably fall in. Neither one is inevitable, because the effect of both forces is just to pull the objects toward one another, which may or may not lead to a collision that consumes them, depending on the details of the motion of the particles before they begin interacting.
You might be able to claim that a positron and an electron that start off perfectly at rest would annihilate without ever forming a bound state, but that's a completely unrealistic situation for a bunch of reasons, including the uncertainty principle. If they start out far apart from each other with some initial velocity, though, their fate will depend on the exact arrangement of the initial conditions. Forming a stable bound state most likely requires a third particle, as well, to conserve energy and momentum.
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Depending on the relative spin-orientation of electron and positron, the positronium may have two spin states: if the electron and positron have anti-parallel spins (+1/2 and -1/2), the positronium will be of spin-singlet state with intrinsic lifetime 0.125ns and decays via two-photon annihilation. On the otherhand, if electron and positron have parallel spins, the positronium will be of spin-triplet state with intrinsic lifetime of 142ns and annihilates in three-photon mode in vacuum. However, in presence of material, the positronium has finite probability to exchange its own electron with the opposite-spin electron from the surroundings, as a result the triplet-positronium may decay in a singlet-mode (two-photon) faster than 142 ns. This mode of annihilation is called pick-off decay, which shows its importance in the appliaction in calculating the void size in any porous material.
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http://mathoverflow.net/questions/42234/rugged-manifold/104175
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## Rugged manifold
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It is well known that any compact smooth $m$-manifold can be obtained from $m$-ball by gluing some points on the boundary.
Is it still true for topological manifold?
Comments:
• To proof the smooth case, fix a Riemannian metric and consider exponential map up to cutlocus.
• The question was asked by D. Burago. I made a bet that a complete answer will be given in an hour — please help :)
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If your manifold has a topological handle decomposition you'd get what you want by finding a maximal tree in the dual 1-skeleton. My understanding is topological handles were worked out by Kirby and Siebenmann but I've never really understood the details or even the precise statements of the theorems. But that's where I'd start looking. – Ryan Budney Oct 15 2010 at 0:49
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To add on what Ryan said: Kirby-Siebenmann proved that every closed topological manifold of dimension > 4 has a CW-structure, so one can get a manifold by identifying boundary points of finitely many disks. Whether one disk suffices, I do not know – Igor Belegradek Oct 15 2010 at 1:15
## 2 Answers
The answer is yes -- Morton Brown's mapping theorem says that for every closed (connected) topological $n$-manifold $M$ there is a continuous map $f$ from the $n$-cube $I^n$ onto $M$ which is injective on the interior of the cube (for manifolds with boundary, see Remark 1 below). This was proved in early sixties and can be found in M.Brown,"A mapping theorem for untriangulated manifolds," Topology of 3-manifolds, pp.92-94, M.K.Fort, Jr.(Editor), Prentice-Hall, Englewood Cliffs, N.J.,1962. MR0158374
Main idea of the proof is simple: "Use local PL structures to expand a small $n$-cell in $M$ gradually, until it becomes the whole manifold."
This can be realized the "infinite composition" of engulfings of finitely many points at a time. This argument is not too difficult, so I will try to sketch it.
First consider a closed $n$-cell $C$ in $M$ and a finite set $X={x_1,\ldots,x_k}$ of points of $M$ disjoint from $C$ which we want to engulf. Assume that each $x_i$ lies in some open $n$-cell $U_i$ which intersects $C$. For each $i$, fix a PL structure on $U_i$, and join $x_i$ and some point $y_i\in \partial C$ by a PL arc $\alpha_i\subset U_i$ relative to this structure. If we assume $\dim M\geq 3$ (this is not a restriction because the theorem is clear for $\dim M=1$, and easily follows from the surface classification for $\dim M=2$), we may require that $\alpha_i$'s should be disjoint. The regular neighborhood $Q_i$ of $\alpha_i$ within $U_i$ is a closed $n$-cell, and $Q_i$'s can be made disjoint. Let $h$ be a homeomorphism of $M$ pushing $y_i$ towards $x_i$ within $Q_i$ for each $i$ which is identity outside $Q_i$'s. Then $X\subset h(C)$, and we can apply this process again to $h(C)$ and another finite set $X'\subset M\setminus h(C)$. Repeating this process we obtain a sequence of engulfing homeomorphisms $h_1, h_2,\ldots$.
We can arrage that the uniform limit $f\colon M\to M$ of the composition of engulfing homeomorphisms $f_n=h_n\circ\cdots\circ h_1$ exists and $f(C)=M$. If we choose sufficiently small $Q_i$'s in each stage, one can make sure that $f$ is injective on the interior of $C$ (indeed, we can arrange that for each interior point $x$ there is $n$ such that $f_n(x)=f(x)$).
Remark 1: In Brown's paper, one can find a corollary that "If $M$ is a compact connected manifold with nonempty boundary $B$, then there is a surjection $f\colon B\times [0,1]\to M$ that restricts to the identity on $B\times 0$ and is injective on $B\times[0,1)$". Notice also that the above theorem can be applied to the closed manifold $B$. Then it follows that any compact topological $n$-manifold (possibly with boundary) can be obtained by identifying some points in the boundary of the $n$-ball.
Remark 2: Berlanga's theorem extended the Brown's theorem to noncompact manifolds: This theorem states that for every (connected) $\sigma$-compact $n$-manifold $M$, there is a nice kind of surjection $I^n\to \overline{M}$ similar to Brown's map, where $\overline{M}$ denotes the end compactification of $M$.
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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.
To expand on Ryan's comment: A handle decomposition of a manifold is a restricted type of quotient of a disjoint union of handles. An $n$-dimensional $k$-handle is by definition the manifold $B^k \times B^{n-k}$, with the decoration that $(\partial B^k) \times B^{n-k}$ is its lower boundary and $B^k \times (\partial B^{n-k})$ is its upper boundary. The allowed quotient is obtained by induction. The $k$-skeleton $M_k$ is the union of all of the $\le k$-handles. To make $M_{k+1}$, you identify part of $\partial M_k$ with all of the lower boundary of the $k+1$-handles using a closed embedding of the latter into the former. That is a complete definition, although maybe more conditions can also be added.
In addition to upper and lower boundary, the handles have upper and lower 1-dimensional radial foliations. The upper radial foliation is singular at the core of the handle and the lower radial foliation is singular at the co-core. There is a lemma that a handle decomposition can be extended radially. The lemma allows you to convert a handle decomposition of $M$ into a special kind of atlas of charts of $M$. It also allows you to convert the handle decomposition into a CW complex using the cores of the handles, or a dual CW complex using the co-cores.
Moreover, you can always simplify a handle decomposition of a closed $n$-manifold so that there is only one $0$-handle and only one $n$-handle. This answers Burago's question if $M$ has a handle decomposition.
It is easy to make a handle decomposition of any smooth or PL manifold (in the latter case, say, any triangulated PL manifold) where everything as described holds. For the topological case, it was proven by Kirby and Siebenmann that every $n$-manifold with $n \ge 5$ has a handle decomposition. Also every $n$-manifold with $n \le 3$ has a unique PL and a unique smooth structure. It was proven by Freedman that a 4-manifold with no PL structure (equivalently, no smooth structure) does NOT have a handle decomposition. So, the answer to Burago's question is yes for all closed manifolds except possibly for non-smoothable 4-manifolds. I suspect that the answer is still yes for all 4-manifolds.
Topological manifolds are a difficult theory and I am hand-waving a bit with this answer. But I think that it is a correct description of topological handle decompositions.
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http://math.stackexchange.com/questions/33522/coin-betting-expectation/33531
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# Coin Betting Expectation
Suppose I have a biased coin with probability of heads p, and tails q=(1-p). It is then used in a game which lasts at most N tosses, and start with a stake of £1. Each time the coin is tails my money is doubled. The first time it comes down heads my money is reduced to £1, and the second time it comes down heads, I lose all my money. The game ends after N tosses, or after the second head. What is the expectation of my money at the end of the game?
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– Undercover Mathematician Apr 17 '11 at 21:01
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@Undercover Mathematician In the gambler's ruin problem, there is no limit on the number of coin tosses. So this problem is not quite the same. – Byron Schmuland Apr 17 '11 at 21:13
Ah, you're right. Thanks! – Undercover Mathematician Apr 17 '11 at 21:26
## 2 Answers
With probability $(1-p)^N$ you get no heads, and end up with $2^N$.
For $k=1,..,N$, with probability $p(1-p)^{N-1}$ you get a single "heads" on the $k$th toss. Then you end up with a value of $2^{N-k}$.
Any other result you end up with zero.
So the expected value is:
$$(1-p)^N2^N + p(1-p)^{N-1}(1+2+2^2+...+2^{N-1}) =$$ $$(1-p)^{N-1}[(1-p)2^N+p(2^N-1)]=$$ $$(1-p)^{N-1}(2^N-p)$$
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Let $m$ be the amount of money you have before your first toss. (The problem specified $m=1$, but I did not notice. So we might as well look at this slightly more general situation.) If you are very lucky, all tails, probability $q^N$, you end up with $m2^N$.
Or else there could be $1$ head and the rest tails. The probability this happens at any particular one of the $N$ tosses is $pq^{N-1}$. Suppose it happens at the last toss. Then you end up with $1$. If it happens at the next to last toss, you end up with $2$. If it happens on the toss before that, you end up with $4$, and so on. Finally, if head happens on the very first toss, you end up with $2^{N-1}$. (Doesn't seem reasonable if $m=1$: the toss was bad but the casino let you keep your $1$ pound!) So the expectation is $$m2^Nq^N+ (1+2+\cdots +2^{N-1})pq^{N-1}$$ The expression can be simplified by noting that $1+2+\cdots+2^{N-1}=2^N-1$.
Now put $m=1$ for the actual problem as asked.
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http://math.stackexchange.com/questions/243352/why-is-an-associative-semisimple-algebra-the-direct-sum-of-minimal-left-ideals?answertab=oldest
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# Why is an associative semisimple algebra the direct sum of minimal left ideals?
I'm reading Clausen, Fast Fourier Transforms, and he states (p. 34):
Let A be an associative algebra.
A is semisimple $\Leftrightarrow$ A is a direct sum of minimal left ideals.
Can anyone explain why?
Followup question: since A is semisimple, its left A-modules are completely reducible, i.e. every left A-module is the direct sum of simple left A-modules. Is there an easy correspondence between the simple A-modules and the left ideals of A?
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– YACP Nov 23 '12 at 23:46
## 1 Answer
The equivalence between these conditions in an associative ring $R$ are covered in almost every book with noncommutative rings:
1. Every right ideal of $R$ is a direct summand. (Same for left ideals)
2. $R$ is a direct sum of simple right ideals. (Same for left ideals)
3. Every $R$ module is a direct sum of simple $R$ modules.
As for the followup question (usually it's better to post these separately, but this one is easy enough to cover here) the simple right $R$ modules will all be represented by the simple right ideals. There may be duplicate copies though!
Here is how to see it. If $M$ is a simple right $R$ module, then there is a homomorphism $R\rightarrow M$ that is onto. Thus $M\cong R/T$ where $T$ is a maximal right ideal. But since $R$ is semisimple $R\cong T\oplus N$, and you can verify that $N$ is a minimal right ideal. Finally, $M\cong N$.
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http://math.stackexchange.com/questions/204472/question-about-a-proof-of-ansel-stricker-in-a-paper/205043
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# Question about a proof of Ansel-Stricker in a paper
I was working through a paper by M. De Donno, which proves the Ansel-Stricker lemma in a different way. The paper can be found here. I've chosen this paper instead of the original one by Ansel-Stricker, simply because my French is to bad. Reading this paper I have some question concerning the proof of Theorem 1.
1. Why is it needed that $\sum_n P(\tau_n< T)<\infty$. This looks like Borel-Cantelli, but I do not see where we use this.
2. How do I derive the bound for $(\Delta X^n_{\sigma_m})^-$?
3. Why is $M^n_{t\wedge \sigma_m}\ge \theta_m-1-(m-\theta_m)$ and why is this positive? Or why do we need this calculation?
hulik
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@ hulik : by the way the point (ii) of th 1 uses that $\eta_k$ converges stationarily to $T$. What does stationarily means precisely in this context ? – TheBridge Oct 1 '12 at 12:44
@TheBridge that's a good question! I was also unsure about that, since I've never heard this expression. I think it should mean stationary. If I look at Corollary 2 below of theorem 1, I guess it means monotone increasing – hulik Oct 1 '12 at 12:57
@ hulik : Iguess you are right, by the way I think I have figured out your point 1. – TheBridge Oct 1 '12 at 15:38
## 2 Answers
Suppose you have a sequence of martingales converging uniformly in probability to some process, $X$ say. The first theorem of the paper gives you a couple of conditions to check to ensure that the limit $X$ is a local martingale.
Understanding the theorem statement
1. What does stationarily mean? This word describes a sequence which is eventually stationary. That is, there is some random variable $N$ such that, for $n\geq N$, $\eta_n=T$.
2. What is the "right" definition of a local martingale indexed by a compact set of times? It seems as though it should be similar to the familiar definition, except that the reducing sequence $(\sigma_n)_{n\in\mathbb{N}}$ should converge stationarily to T, as this gives the useful property that $$\lim_{n\to\infty}X^{\sigma_n}_T = X_T.$$ This is a property that the familiar local martingales have, which doesn't necessarily hold if you drop the stationarity requirement, since $X$ isn't necessarily continuous at $T$.
Understanding the theorem proof
Here we define a sequence of stopping times $(\tau_n)_{n\in\mathbb{N}}$ is such a way that the system is well-bahaved up to time $\tau_n$, for each $n$, but so that we still have $\tau_n\uparrow T$ - note here that $X$ is cadlag and thus bounded on compacts.
I think the author passes to a subsequence here as then that sequence $(\tau_{n_k})_{k\in\mathbb{N}}$ converges stationarily to $T$, and thus, since $\eta$ converges stationarily to $T$, $\sigma$ does also, which would fit with the above definition of a local martingale.
$$(\Delta X_{\tau_n})^- = (X_{\sigma_n}-X_{\sigma_n-})^-\leq n - \theta_n,$$ since $\sigma_n\leq \tau_n$, and $\sigma_n\leq \eta_n$.
Now note that it's not necessarily true that $$M^n_{t\wedge\sigma_m}\geq X_{t\wedge\sigma_m} -1.$$
Why? Take, for example $t=T$, and suppose $X$ jumps far from $M^n$ at time $\sigma_m$. For this reason, we need to take into account a possible jump of $X$ at time $t\wedge \sigma_m$, which we've just shown is bounded by $m-\theta_m$. This gives the required inequality.
We can now apply Fatou's lemma as follows: $$\liminf_n ~~\mathbb{E}[ M^n_{t\wedge\sigma_m} - (2\theta_m-m-1)]\geq \mathbb{E} [X_{t\wedge\sigma_m} - (2\theta_m-m-1)],$$
and we can cancel $2\theta_m-m-1$ from both sides since it is integrable. Alternatively, as TheBridge mentions, we can learn this as an extension of Fatou's lemma.
I hope that helps!
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@ Ben Derrett : In my answer I have interpreted $()^-$ as the negative part of the classical decoposition of a function $f$ as $f(x)=f(x)^+-f(x)^-$. I think that you interpret it the other way around, but from point (iii) in th 1 in the article I think that my interpretation makes more sense, don't you think ? – TheBridge Oct 2 '12 at 14:24
@TheBridge Thanks for drawing my attention to that - it's now fixed. I haven't received a down vote before, so I'd be very curious to know if anything else can be improved? – Ben Derrett Oct 2 '12 at 15:37
@ Ben Derrett : Actually I did the downvote, and I feel a bit sorry that I can't take it back. It was an epidermic reaction of mine due to the downvote that I received (just after your answer was posted) and that I interpreted coming from you. I found that a bit harsh as there were no comments explaining what I did wrong. I'll edit my answer accordingly to your comment when I find time. Best regards – TheBridge Oct 2 '12 at 20:11
@ Ben Derret: You can apply directly Fatou's Lemma extension that I found on wiki's page to $M^n_{t\wedge \sigma_n}$ to get to the result (you are basically redemonstrating it, though it is a quick thing to do). Best regards. – TheBridge Oct 2 '12 at 20:32
@TheBridge: No worries. – Ben Derrett Oct 3 '12 at 11:50
show 2 more comments
Hi for the first part of your 3rd question unless mistaken, I think that you have (using the elements in the proof of the paper):
$M^n_{\sigma_m\wedge t}\geq X_{\sigma_m\wedge t}-1 -(\Delta X_{\sigma_m\wedge t})^-\geq \theta_m -1 - (m-\theta_m)=2.\theta_m - m -1$
The need to substract the negative part of the jump of $X$ at $\sigma_n$ comes from $\tau_n$'s definition which implies that it is possible that $X$ jumps beyond $M^n_{\sigma_m\wedge t}+1$ at $\sigma_n$ (as pointed out by Ben Derrett).
For the second part, I think that it need not be positive, but the thing is that it is still integrable and I think that it is this hypothesis that is used here in a slight extension of Fatou's lemma which the authors abusively used without mentionning (check wiki's page of Fatou's Lemma).
Edit: Regarding 1 - The more I look at it, the more I think that the place where implicitly the Borel-Cantelli's lemma is used is when the authors define $\sigma_n$ where an $inf_{m>n}\tau_m$ is involved. So the fact that $\sigma_n$ is inferior to $T$ a.s. is coming from the fact that $inf_{m>n}\tau_m<T$ for some n>0 from Borel-Cantelli's Lemma as $P(limsup_n E_n)=0$ where $E_n=\{\tau_n<T\}$. Otherwise said it means that almost surely for a fixed $\omega$ there exists a $p>n$ such that $inf_{m>n}\tau_m(\omega)=\tau_p(\omega)$.
Edit two: Now point 2.Please check it carefully.
Suppose that we are on the event that $X$ has a strictly negative jump at $\sigma_n$, then we have $X_{\sigma_n-}<n$ (due to $\tau_n$'s definition and to the fact that $\sigma_n\leq \tau_n$), and on the other hand $X_{\sigma_n}>\theta_n$ by point (ii).So :
$\Delta X_{\sigma_n}^-=X_{\sigma_n-}-X_{\sigma_n}\leq n-\theta_n$
The case $\Delta X_{\sigma_n}^-=0$ doesn't matter.
Best regards
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1) Unfortunately, your first inequality doesn't hold quite as stated (hence the necessity of the $m-\theta_m$ term), since there may be a jump in $X$ at time $\sigma_m\wedge t$. 2) We define $\Delta X_{\sigma_n} = X_{\sigma_n}-X_{\sigma_n-}$. Hope that helps. :) – Ben Derrett Oct 2 '12 at 15:50
@ Ben Derrett : Hi you are right about 1) I misunderstood $\tau_n$ definition, I edit the first part of my answer accordingly. – TheBridge Oct 2 '12 at 20:16
Thank you so much for your help! I'm sorry for the late response, I had a busy week. Back to my question: There still some small points which are not entirely clear to me. regarding 1): What do you mean by $\sigma_n$ is inferior to $T$? 2) Why do I have $X_{\sigma_n-}< n$ From the definition of $\tau_n$ I have $X_{\tau_n}\ge n$ by right continuity, or am I wrong? 3) Sorry I really don't get the first inequality: $M_{\sigma_m\wedge t}^n\ge X_{\sigma_m\wedge t}-1-(\Delta X_{\sigma_m\wedge t})^-$. – hulik Oct 8 '12 at 13:32
Actually I even do not understand why $M^n_t\ge X_t-1$ for $n\ge m$ and $t<\sigma_m$ – hulik Oct 8 '12 at 13:35
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http://mathhelpforum.com/calculus/83425-lim_-x-rightarrow-0-sin-frac-1-x.html
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# Thread:
1. ## \lim_{x\rightarrow 0} sin (\frac{1}{x})
Show that $\lim_{x\rightarrow 0} sin (\frac{1}{x})$ doesn't exist by using the squeeze theorem.
Thanks in advance.
2. Originally Posted by Biscaim
Show that $\lim_{x\rightarrow 0} sin (\frac{1}{x})$ doesn't exist by using the squeeze theorem.
Thanks in advance.
I don't see how the squeeze theorem will help but here is a method.
By using the sequental char of limits all we need to do is find two different sequences that converge to 0 that have different function limits.
$x_n=\frac{1}{n\pi}$
$y_n=\frac{1}{(\frac{\pi}{2}+2n\pi)}$
note that both of these go to zero and n goes to infinity, but
$\sin\left(\frac{1}{x_n}\right)=\sin(n\pi)=0$ and
$\sin\left( \frac{1}{y_n}\right)=\sin\left( \frac{\pi}{2}+2\pi\right)=1$
There for the limit does not exists at 0
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http://mathhelpforum.com/number-theory/28477-simple-proof.html
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# Thread:
1. ## simple proof
How would you prove this, I'm stuck. Thanks
Let A and B be sets. Then A union B = A if and only if B is subset of A.
2. First, you can see from the definition of a subset and from the definition of the union of two sets that for any sets A and B, $A\subseteq{A\cup{B}}$ and $B\subseteq{A\cup{B}}$
Now, to prove our "if and only if" statement, we break it up into the two direction if-then statements:
1. If $A\cup{B}=A$ then $B\subseteq{A}$.
As noted above, $B\subseteq{A\cup{B}}$ for any sets A and B. Thus, if $A\cup{B}=A$, then $B\subseteq{A}$.
2. If $B\subseteq{A}$ then $A\cup{B}=A$.
Assume opposite: we have $B\subseteq{A}$, $A\cup{B}\ne{A}$. As $A\subseteq{A\cup{B}}$, if $A\cup{B}\ne{A}$, then there must be at least one element x in $A\cup{B}$ not in A. By definition of the union of sets, every element of $A\cup{B}$ must be an element of A or of B (or both). Thus if $x\in{A\cup{B}}$ and $x\notin{A}$, then $x\in{B}$. Thus there must be at least one element in B not in A. But this contradicts $B\subseteq{A}$. Thus, we show that if $B\subseteq{A}$ then $A\cup{B}=A$.
With (1) and (2), we have proved that $A\cup{B}=A$ if and only if $B\subseteq{A}$.
-Kevin C.
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http://www.physicsforums.com/showthread.php?t=285480
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Physics Forums
## Trig question
$$| a + bi | = \sqrt{(a)^2+(b)^2}$$
$$\theta = tan^{-1}(\frac{b}{a})$$
$$|(a + bi)^n| = (\sqrt{(a)^2+(b)^2})^n$$
$$\theta = n(tan^{-1}(\frac{b}{a}))$$
$$(a+bi) = (\sqrt{(a)^2+(b)^2})^n(cos(n(tan^{-1}(\frac{b}{a})) + isin(n(tan^{-1}(\frac{b}{a}))))$$
Why is $$\theta = n(tan^{-1}(\frac{b}{a})$$ ? For example when I have
$$5 + i$$
$$\theta = tan^{-1}(\frac{1}{5}) = 0.1974...$$
$$(5+i)^2 = 25 + 10i -1$$
$$(5+i)^2 = 24 + 10i$$
$$\alpha = tan^{-1}(\frac{10}{24}) = 0.3948...$$
$$2(0.1974) =0.3948$$
Why is it that the exponent of the vector can be used to get the angle of the resultant by simply multiplying it with the tan function? Also in the first part of that:
$$(\sqrt{(a)^2+(b)^2})^n(cos(n(tan^{-1}(\frac{b}{a})) + isin(n(tan^{-1}(\frac{b}{a}))))$$
Just to make sure, it is only
$$isin(n(tan^{-1}(\frac{b}{a}))))$$
and not simply
$$(\sqrt{(a)^2+(b)^2})^n(cos(n(tan^{-1}(\frac{b}{a})) + sin(n(tan^{-1}(\frac{b}{a}))))$$
because one of the components is complex?
Recognitions: Gold Member Science Advisor Staff Emeritus you are talk about putting complex numbers in "polar form". If you represent a+ bi as the point (a, b) in Cartesian coordinates and the polar form is $(r, \theta)$ (That is, distance from (0,0) to (a,b) is r and the line from (0,0) to (a,b) is $\theta$, it is easy to see that $a= rcos(\theta)$ and $b= rsin(\theta)$ so that $a+ bi= r (cos(\theta)+ i sin(\theta))$. Then [/itex](a+bi)^2= r^2(cos^2(\theta)-sin^2(\theta)+ i(2sin(\theta)cos(\theta)))[/itex] and recognise that $cos^2(\theta)- sin^2(\theta)= cos(2\theta)$ and [/itex]2cos(\theta)sin(\theta)= sin(2\theta)[/itex]. The more general formula follows from the identites for $sin(\theta+ \phi)$ and $cos(\theta+ \phi)$. Even simpler is to note that $cos(\theta)+ i sin(\theta)= e^{i\theta}$ so that $(cos(\theta)+ i sin(\theta)^n= (e^{i\theta})^n= e^{ni\theta}$.
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http://mathoverflow.net/questions/83251?sort=votes
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## Do Shintani zeta functions satisfy a functional equation?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Probably my questions are known or evident to the experts but I'm a bit puzzled. First of all there seem to be two kinds of zeta functions that go under the name of Shintani zeta functions.
First, there are zeta functions $\zeta^{SS}$ associated with so called prehomogenous vector spaces going back to important work by Sato and Shintani (see the original article or this book by Yukie) and then, second, zeta functions $\zeta^S$ that appeared in Shintani's work on special values of Dedekind zeta functions of totally real number fields at negative integers (see Shintani's article or Neukirch's book for example).
1) I'm mainly interested in the question if it is known (or expected) if the latter zeta functions $\zeta^S$ satisfy functional equations. (From what I understand the $\zeta^{SS}$ satisfy functional equations or are expected to satisfy in case it is not proven).
Let me just note that one can write Shintani zeta functions in the following form $$\Gamma(s)^n \zeta^S(s,z,x) = \int_0^\infty \cdots\int_0^\infty \sum_{z_1,\dots , z_n=0}^\infty e^{-\sum_{i=1}^n t_i L_i(z+x)}(t_1\cdots t_n)^{s-1} dt_1\cdots dt_n$$ where the $L_i(x)$ are linear forms, i.e. essentially we could say that we're looking at multivariable theta-like functions and Mellin transforms thereof. So the question can be rephrased in asking whether these theta-like functions occurring in the above integral satisfy a functional equation/theta inversion formula. (Note that these theta-like functions do in general not come from symplectic structures, i.e. they are not related to abelian varieties (at least as far as I see)).
2) But next to this question I'm also extremely interested in the relationship of these two kinds of zeta functions. In which cases do the two constructions agree? Is there anything known?
Thank you very much in advance!
EDIT: OK, so I could speak a bit with one of the absolute authorities in this field and I learned, that
1) one shouldn't expect functional equations for single functions $\zeta^S$ but rather for certain finite linear combinations and
2) one shouldn't expect relations between the two notions of "zeta" functions.
This doesn't destroy the applications I had in mind with my question but I have to rethink the question and will try to give a better and less naive version of it soon. Thank you so far very much for your helpful comments!
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3
Regarding "there seem to be two kinds of zeta functions that go under the name of Shintani zeta functions": I can definitively confirm this. I know about the prehomogeneous ones, but unfortunately not about the ones you asked about. AFAIK the only connection is that Shintani worked on both of them. But if there is some further connection I would be very interested to hear about it! – Frank Thorne Dec 12 2011 at 19:43
1
@Bora: My problem is that the expression $\prod_{i=1}^n L_i(x)^{-s}$ is not theta-like. To me a theta-like function is an infinite series such as $\sum_{n\in\mathbb{Z}} e^{-n^2 x}$ whose Mellin-transform is a zeta-like function. – GH Dec 12 2011 at 20:01
1
@Bora and GH : probably one needs to take an infinite sum (over $x$) of the expression Bora has written. One can also introduce denominators in the integrand. – François Brunault Dec 12 2011 at 20:01
1
In case it is not clear: the integral kernel is akin to one used by Riemann in one of his proofs for plain-old-zeta. Not the theta series, but $\sum e^{-nx} = 1/(e^x-1)$. The "cone" here is just the positive reals. For other number fields, the cone is a product of positive reals, but the rational structure is more complicated. So, in principle, it is related to the bigger-group Shintani zetas. I think Shintani did a few things about their special values, and I recall Satake wrote a paper about this. – paul garrett Dec 12 2011 at 20:28
1
I don't know if this will answer your question but Colmez gave a talk recently in Banff about Shintani's method. The notes (by Matt Greenberg) are available here : temple.birs.ca/~11w5125/colmez.pdf What I remember is that Colmez managed to express special values at negative integers without using the functional equation. – François Brunault Dec 13 2011 at 0:07
show 10 more comments
## 1 Answer
there was a talk recently in RIMS Kyota titled something like "the functional equation of the Shintani zeta function". You might try looking that up.
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Searching gives me this: math.stanford.edu/~fthorne/…. Who was the talk by that you are thinking of? – David Roberts Aug 27 at 7:15
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http://math.stackexchange.com/questions/133030/finding-a-point-above-the-line-in-o-log-n/133038
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# Finding a point above the line in $O(\log n)$
I am trying to solve the following problem. So far with no success.
Let $S$ be a set of $n$ points in the plane. Preprocess $S$ so that, given a (non-vertical) line $l$, one can determine whether there is any point of $S$ above $l$ in time $O(\log n)$. Few details: preprocess $S$ without knowing line $l$ in advance. Preprocessing doesn't have any special requirements.
According to big-$O$ requirement $O(\log n)$ the determination should be implemented similarly to binary search.
I have considered few options, $y$ - coordinate, slopes, but in any case it seems not relevant.
If you have any idea, I will appreciate sharing it with us.
Thanks!
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This seems strange, consider this problem in one less dimension - i.e given a set of points in $\mathbb{R}$ determine in $O(logn)$ if there is a point in this set which is the same as the given point. this of course means searching the set which can not be done without some assumption such as $S$ being sorted – Belgi Apr 17 '12 at 17:56
@HenningMakholm: My thought (a poor one) was that some point of the line will have lower $y$ coordinate than some points of $S$ and I would just answer yes. That is not usually the interpretation. – Ross Millikan Apr 17 '12 at 18:15
## 3 Answers
Find those among the given points that are the on the the upper boundary of the convex hull of $S$, and sort them by they $x$ coordinate. (Any point that is not on this upper boundary can be ignored). Then you should be able to do a binary search along these lines:
Given $l$, check it against the middle point. If the middle point is above $l$, you're done. Otherwise compare the slope of $l$ with the average of the slopes of the two segments of the upper boundary that adjoin the point you just compared to. If the slope of $l$ is greater, then any point of $S$ above $l$ must be somewhere to the left of the middle point; if the slope if $l$ is smaller, then any point above $l$ must be to the right of the middle point.
Now you have reduced the number of points to consider to half of the original size. Proceed by induction.
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Henning Makholm, Thank you very much for the answer, how it can be extended to the segment instead of line. Let's say now instead of line $l$ we have a segment, of course x coordinates of point must lie between x coordinates of the endpoints of e. – com May 25 '12 at 5:14
Find the Convex Hull in O(nlogn), and for every point on it remember the slope of the segment between it and the next on the convex hull. then, (they are already sorted), given any line l with a slope m, binary search the slope in the slope array. then you only have to check the point before and after m in the array and determine if they're above the line. i'm pretty sure you can't do any better than O(nlogn), although i have no proof of a lower bound.
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What are you allowed to do during preprocessing, and how quickly must it be done?
For example, you could compute the residual from each point to the line in $O(n)$ time, and then sort them in $O(nlog(n))$ time, in which case, determining whether or not there is a point of $S$ above the line $l$ is merely an $O(1)$ process - look at the largest residual and see if it is positive or not. Must the preprocessing be done in $O(logn)$ time as well?
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For the question to make sense, I think the preprocessing has to be done without knowing $l$. – Henning Makholm Apr 17 '12 at 17:58
yes, on the step of preprocessing we don't know what's line $l$ – com Apr 17 '12 at 18:01
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http://math.stackexchange.com/questions/271233/why-isnt-the-family-of-semi-algebras-aka-semi-rings-of-sets-closed-under-inte?answertab=votes
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# Why isn't the family of semi-algebras (aka semi-rings) of sets closed under intersection?
Any type of algebraic structure on subsets of $S$ that is defined purely in terms of closure properties will be preserved under intersection. Examples are σ-algebras, π-systems, λ-systems, or monotone classes of subsets.
...
Note however, this does not apply to semi-algebras, because the semi-algebras is not defined purely in terms of closure properties (the condition on $A^c$ is not a closure property).
...
$S$ is said to be a semi-algebra if it is closed under intersection and if complements can be written as finite, disjoint unions:
• If $A,B∈S$ then $A∩B∈S$.
• If $A∈S$ then there exists a finite, disjoint collection $\{B_i:i∈I\}⊆S$ such that $A^c=⋃_{i∈I} B_i$.
In "the condition on $A^c$ is not a closure property",
• what does "the condition on a set operation such as taking complement is not a closure property" mean?
• What is the meaning of "closure properties"?
How do you see the family of semi-algebras (aka semi-rings) of sets isn't closed under intersection?
2. Michael Greinecker also commented: The family of semi-rings on a set are not closed under intersections.
BTW, if I am correct, the concept of a semi-algebra of sets is the same as semi-ring of sets in Wikipedia.
Thanks and regards!
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The family of all bounded intervals is a semi-ring on $\mathbb{R}$ but not a semi-algebra (since the complement is unbounded and hence not a finite union of bounded sets). – Michael Greinecker Jan 6 at 0:47
## 1 Answer
Closure properties can be formulated in terms of concepts from universal algebra. Let $X$ be the underlying set (in our examples, $X$ is a famly of sets itself). Let $I$ be an index set, $(\kappa_i)_{i\in I}$ be a family of cardinal numbers and $(f_i)_{i\in I}$ a family of function satisfying $f_i:X^{\kappa_i}\to X$ for all $i$. We say that $C\subseteq X$ is closed under $(f_i)_{i\in I}$ if we have for all $i\in I$ that $f_i(x)\in C$ for all $x\in C^{\kappa_i}$. One can show that the family of sets closed under $(f_i)_{i\in I}$ forms a Moore collection.
Let's look an an example: Let $U$ be a set and $X\subseteq 2^U$. We let $I=\{s,c,u\}$, $\kappa_s=0$, $\kappa_c=1$, and $\kappa_u=\omega$. We identify constants and nullary functions, so we can let $f_s=U$. We let $f_c(A)=A^C$ for all $A\in X$, and we let $f_u(A_0,A_1,\ldots)=\bigcup_n A_n$. That $X$ is closed under these three functions means simply that it contains $X$, is closed under complements and countable unions- it is a $\sigma$-algebra.
Now, one cannot write down semi-algebras this way, since there is no unique decomposition of the complement into disjoint sets. If $\mathcal{S}$ is a semi-algebra and $A\in\mathcal{S}$, then there exists a number $n$ and sets $B_1,\ldots,B_n\in\mathcal{S}$ that are disjoint and such that $A_c=B_1\cup\ldots\cup B_n$. Now if there exists a unique such family and if this family only depended on $A$, we could write down this property as closure under some functions in the following way: We let $f_{c_1}=B_1,\ldots, f_{c_n}=B_n$, and for $m>n$ we let $f_{c_m}=f_{c_n}$. We use the last condition because we have no a priori bound on how many sets are needed. But these sets are not a function of $A$, so this property can not be viewed as a closure property.
Here is an explicit example (taken from Alprantis & Border) that shows that the intersection of sem-algebras might fail to be a semi-algebra: Let $X=\{0,1,2\}$, $\mathcal{S}_1=\big\{\emptyset, X,\{0\},\{1\},\{2\}\big\}$, $\mathcal{S}_2=\big\{\emptyset, X,\{0\},\{1,2\}\big\}$, and $A=\{0\}$. We have $\mathcal{S}_1\cap\mathcal{S}_2=\big\{X,\emptyset,\{0\}\big\}$, and $A^C=\{0\}^C=\{1,2\}$ is not the disjoint union of elements of this intersection.
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Thanks! (1) How is $C\subseteq X$ closed under $(f_i: X^{\kappa_i} \to X)_{i \in I}$ related to the example for semi-algebras? (2) For semi-algebras, why do "there do no unique decomposition of the complement into disjoint sets (even then, one would need a hack)" and "a function can have only one value" matter? What is "a function"? – Tim Jan 6 at 14:33
@Tim I've edited it and hope have clarified these issues. – Michael Greinecker Jan 6 at 22:19
Thanks! I edited some places I thought were typos. Feel free to edit again if I am wrong. I was wondering if the definition of a set operation having closure property comes from some references? – Tim Jan 7 at 14:02
@Tim All things closure operations I learned from the Handbook of Analysis and its Foundations by E. Schechter. – Michael Greinecker Jan 7 at 14:13
Thanks! Besides the two books, what other books in analysis do you recommend? – Tim Jan 11 at 21:36
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http://mathoverflow.net/questions/21028/shape-of-long-sequences-in-c-1/21084
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## Shape of long sequences in C(ω_1)
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Apologies for the vague title - I couldn't come up with a single sentence that summarised this problem well. If you can, please edit or suggest a better one!
This question is also rather specific and contains lots of annoying technical detail. I must admit to not really expecting an answer unless there's an obvious solution I'm missing (which is very possible - I feel like any solution is either going to be obvious or very deep), but some pointers in plausible sounding directions would be greatly appreciated. I suspect the answer will depend on the combinatorics of $\omega_1$, which I know relatively little about.
Let $V$ be a normed space. For $A \subseteq V$, define `$r(A) = \inf \{ r : \exists V, A \subseteq B(v, r) \}$`. Define a bad sequence in $V$ to be a sequence ${ v_\alpha : \alpha < \omega_1 }$ with the properties that:
`$\forall \beta, r(\{ v_\alpha : \alpha < \beta \}) \leq 1$`
`$\inf_\beta r(\{ v_\alpha : \alpha \geq \beta \}) > 1$`
An example of a space with a bad sequence is $c_0(\omega_1)$ (the set of all bounded real-valued sequences of length $\omega_1$ such that `$\{ \alpha : |x_\alpha| > 0 \}$` is countable). The sequence `$2 * 1_{\{\alpha\}}$` is bad. The radius of any tail is $2$ because the center must be eventually 0. The radius of the initial segments is $\leq 1$ because the segment up to $\alpha$ is contained in the closed ball of radius 1 around `$1_{[0, \alpha]}$`, which is in $c_0(\omega_1)$ because $\alpha < \omega_1$.
I have two (three depending on how you count it) major examples of spaces which have no bad sequences:
• Any separable space: you can choose centers to lie in the countable dense set, so one center must work as a radius for the initial segment for unboundedly many and thus for all $\alpha$.
• Any space which has what I'm imaginatively calling the chain-radius condition: The union of a chain of sets of radius $\leq r$ has radius $\leq r$. This includes:
• Any reflexive space: If $U_\alpha$ forms a chain, the sets `$F_\alpha = \bigcap_{v \in U_\alpha} \overline{B}(v, r + \epsilon)$` form non-empty closed and bounded convex sets with the finite intersection property, so compactness in the weak topology implies they have non-empty intersection. Any element of the intersection contains the union of the chain in `$\overline{B}(c, r + \epsilon)$`
• any space with the property that `$\textrm{diam}(A) = 2 r(A)$` (in particular the $l^\infty$ space on any set) because it's clear that unions of chains of diameter $\leq 2r$ have diameter $\leq 2r$.
So... that's all the backstory for this question. Given that, my actual question is very simple: Does $C(\omega_1)$ contain a bad sequence?
I feel like the answer "must" be no. In particular note that the projection of any sequence onto the first $\alpha$ entries is not bad (because it's a sequence in a separable space) and that if you drop the restriction for continuity the answer is immediately yes. So it sits right between two classes of examples where there are no bad sequences, and I feel that one really should be able to take advantage of that. But on the other hand, functions in $C(\omega_1)$ are eventually constant, so maybe you can take advantage of that to construct some sets with arbitrary bad tails.
For bonus kudos, I'd love to know for what compact Hausdorff spaces $K$, $C(K)$ contains a bad sequence.
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1
Can you explain why your example sequence is bad? Isn't it contained in the closed unit ball centered at 0? – Sergei Ivanov Apr 11 2010 at 19:43
1
Apologies, I actually meant twice that sequence. The sequence as previously mentioned actually has each of the initial segments having radius $\leq \frac{1}{2}$ and the radius of the tails is 1. I've fixed the post now. – David R. MacIver Apr 11 2010 at 19:50
In response to your comment below about citing MO, you may find the discussion at meta.mathoverflow.net/discussion/64 useful. – Joel David Hamkins Apr 13 2010 at 18:17
## 3 Answers
I claim that there are no bad sequences in C(ω1).
Suppose to the contrary that xα is bad. For any countable ordinal β, there is rβ in C(ω1) such that the distance between rβ and xα for α < β is at most 1. For any countable ordinal β and any positive rational number ε, there is a smaller ordinal γ < β such that all rβ(α) are within ε of rβ(β) for α in (γ,β]. For fixed ε, this is a regressive function on the countable ordinals. Thus, by Fodor's Lemma, there is a stationary and hence unbounded set of ordinals on which the function has constant value, which we may call γε. Since there are only countably many ε, we may find a countable ordinal γ above all γε. This ordinal has the property that for all ordinals β above γ, we have rβ(α) = rβ(β) for all α in the interval (γ,β], since the values are within every ε of each other. That is, every rβ function is constant from the same fixed γ up to β.
Let Cβ be the closed interval of values s such that the constant sequence s of length β lies within 1 of all xη(α) for all η ≤ β and all γ < α ≤ β. These are nested and not empty, since rβ(β) is in Cβ. By compactness, there is a value s in all Cβ. Thus, the number s is within xη(α) for all η and all α above γ.
Thus, we may form the desired sequence r by finding a center that works for the sequences up to stage γ, using the separability idea in your question, augmented with the constant value s at the stages above γ up to ω1. That is, we solve the problem separately on the first γ many coordinates, and then append the constant s sequence up to ω1. This sequence is continuous, and it lies within 1 of every xη, as desired.
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Thanks. I'm pretty sure I should have spotted something like that (it looks very similar to the proof which got me to this problem in the first place), but I'm not sure I would have! What's the etiquette/convention for citing a mathoverflow answer in a paper? – David R. MacIver Apr 12 2010 at 7:44
Oh, by the way, it doesn't actually damage the proof, but your interval is slightly wrong. I think you actually mean $(\gamma, \beta]$. When $\beta$ is a successor ordinal you can't guarantee that there are any smaller $\gamma$ with value close to it. – David R. MacIver Apr 12 2010 at 7:49
Another wrong but fixable detail: I don't believe it works to use $r_\gamma$ up to and including stage $\gamma$. The problem is that later elements of the sequence may add variation before $\gamma$ even while being well behaved after it. However, $C([0, \gamma])$ is separable, so has no bad sequences, so we can still find a center that works for all of $[0, \gamma]$ and then use the constant center after that as per your argument. – David R. MacIver Apr 12 2010 at 8:14
Actually... I'm starting to have serious doubts about the core idea of this proof. It doesn't seem to use anything about the $r_\alpha$ other than that they're continuous, and it's certainly not the case that every $\omega_1$ sequence of continuous functions is mutually constant after some point. Why, for example, would this not work with the following: Let $x_\alpha = 1_{[0, \alpha]}$ and let $r_\alpha = x_\alpha$. Obviously this isn't a bad sequence, but there's no common point above which the $r_\alpha$ are all constant. – David R. MacIver Apr 12 2010 at 8:21
David, my proof wasn't claiming that the r_beta are simultaneously constant from gamma out to omega_1, but only that r_beta is constant on the interval [gamma,beta]. In the example of your comment, gamma=0 works for this, since every r_alpha is constant from [0,alpha] in your example. And the sequence I build would have s=1. So, the sequence provided by my solution has value 1 from gamma=0 onwards, which works for this example. – Joel David Hamkins Apr 12 2010 at 12:30
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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 can partially answer the second question. If $X$ is a compact Hausdorff space whose topology has a countable base at every point [Edit: $\omega_1$ has this property but not compact], then there are no bad sequences. Moreover the following holds:
If $F\subset C(X)$ is a family such that every countable subfamily has radius $\le 1$, then $r(F)\le 1$.
Define a function $S=S_F:X\to\widehat{\mathbb R}=[-\infty,+\infty]$ (the "essential supremum" of $F$) as follows: $S(x)$ is the maximum $t$ such that for every neighborhood $U$ of $x$ one has `$\sup\{f(y):f\in F,y\in U\}\ge t$`. Observe that $S$ is upper semi-continuous: for every $t\in\widehat{\mathbb R}$, the set `$\{x\in X:S(x)\ge t\}$` is closed.
Define the essential infimum $I_F$ similarly, this function is lower semi-continuous.
For every $x\in X$ there is a countable family $G\subset F$ such that $S_G(x)=S_F(x)$ and $I_G(x)=I_F(x)$. Indeed, using countable base at $x$, one can realize $S(x)$ by a sequence $x_i:i\in\mathbb N$ converging to $x$ and functions $f_i\in F$ such that $f_i(x_i)\to S(x)$.
It follows that $S(x)\le I(x)+2$ for all $x\in X$. Indeed, take $G$ as above, it is contained in a $(1+\epsilon)$-ball centered at some $f\in C(X)$, then $S_G(x)\le f(x)+1+\epsilon$ and $I_G(x)\ge f(x)-1-\epsilon$.
Fix $\epsilon>0$ and let us prove that $F$ is contained in a $(1+\epsilon)$-ball. For $x\in X$, define $C_x=\frac12(S(x)+I(x))$. Note that $S(x)\le C_x+1$ and $I(x)\ge C_x-1$. By semi-continuity, there is a neighborhood $U_x$ of $x$ such that `$S(y)<C_x+1+\epsilon$` and `$I(y)>C_x-1-\epsilon$` for all $y\in U_x$. Choose a finite subcovering `$V_i=U_{x_i}:i\le N$`.
On each neighborhood $V_i$ we have a constant function $f_i:=C_{x_i}$ which works as a center within this neighborhood. It suffices to construct a function $g\in C(X)$ such that for every $x\in X$, $g(x)$ is between the minimum and maximum of these partially defined constant functions at $x$. This is easy to do by induction in the number of sets in the covering. Suppose we have already defined $g=g_{n-1}$ that works on `$\bigcup_{i<n} V_i$`. By Urysohn's lemma there is $\phi:X\to[0,1]$ such that $\phi=0$ on $X\setminus V_n$ and $\phi=1$ on $X\setminus\bigcup_{i\ne n} V_i$. Then $g_n:=\phi f_n+(1-\phi)g_{n-1}$ works on $\bigcup_{i\le n}V_i$.
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Believe it or not, my answer was not inspired by this, even though they are quite similar in essence. It occurred to me while I was on the train and I wrote it up before I saw this one. – David R. MacIver Apr 12 2010 at 9:38
Sorry, that sounded a bit ungrateful. Thanks for the answer! – David R. MacIver Apr 12 2010 at 9:44
For some reason, I cannot edit my answer. To clarify, it does not work for $\omega_1$: it has countable base at every point but it is not compact. – Sergei Ivanov Apr 12 2010 at 12:05
Hm. I was about to say that that wasn't a problem because $C(\omega_1 + 1) = C(\omega_1)$, but of course that's compact but doesn't have a countable base at every point. Vexing. – David R. MacIver Apr 12 2010 at 12:15
I tricked this stupid software to let me edit the post. By the way, you don't really need compactness, only normality and that every countable covering has a finite subcovering. But I don't know whether $\omega$ is normal or not. – Sergei Ivanov Apr 12 2010 at 12:25
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Edit: This proof is wrong, but preserving for posterity.
Oh! In a very wizard of oz manner, the answer was within my power all along (it's within $\epsilon$ of a proof I'd already written for something else).
Here's a cute little proof that C(K) has the property that diam(A) = 2 * r(A), and thus has the chain-radius condition and thus has no bad sequences.
Let $A \subseteq C(K)$ be non-empty. Define
`$ g(x) = \sup_{f \in A} f(x)$`
`$ h(x) = \inf_{f \in A} f(x)$`
`$g$` is upper semicontinuous: `$g(x) > a$` iff there exists $f \in A$ such that $f(x) > a$. Similarly `$h$` is lower semicontinuous.
Further, `$g(x) - h(x) \leq \textrm{diam}(A)$`.
Therefore `$g(x) - \frac{1}{2}\textrm{diam}(A) \leq h(x) + \frac{1}{2}\textrm{diam}(A)$`
But now we have an upper semicontinuous function which is $\leq$ a lower semicontinuous function. Thus by the katetov tong insertion theorem there is a continuous function $f$ with
`$g(x) - \frac{1}{2}\textrm{diam}(A) \leq f \leq h + \frac{1}{2}\textrm{diam}(A)$`
But this means that $A \subseteq B(f, \frac{1}{2}\textrm{diam}(A))$. Therefore $r(A) \leq \frac{1}{2}\textrm{diam}(A)$.
But we already know that $r(A) \geq \frac{1}{2}\textrm{diam}(A)$, so the two are equal and the result is proved.
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The supremum is lower semi-continuous, not upper. On $K=[-1,1]$, the supremum can be like this: $g(x)=0$ if $x\le 0$ and $g(x)=1$ if $x>0$. And the infimum can be like this: $h(x)=0$ if $x\ge 0$, and $h(x)=-1$ if $x<0$. While $0\le g-h\le 1$, there is no continuous funtion within the distance $1/2+\epsilon$ from them. – Sergei Ivanov Apr 12 2010 at 12:01
Hm. You're quite right. Got the definitions backwards. Sigh. – David R. MacIver Apr 12 2010 at 12:17
By the way, what are the assumptions of the insertion theorem? If it does apply to $\omega_1$, you can use "essential supremum" from my answer. – Sergei Ivanov Apr 12 2010 at 12:28
It does apply to $\omega_1$ yes. It works for any normal space. – David R. MacIver Apr 12 2010 at 12:30
Ok. I see why I got so confused. In the original case I applied this proof (which is to do with approximating discontinuous functions) the definitions of g and h were actually essentially those of your essential supremum and infimum! In my case it was for a single function which wasn't necessarily continuous, but the details work out much the same. So, using those definitions and the insertion theorem we can drop the conditions of compactness and locally countable bases. – David R. MacIver Apr 12 2010 at 12:40
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http://mathhelpforum.com/pre-calculus/102802-how-find-zeros-function.html
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# Thread:
1. ## How to find the zeros of this function
Find all the zeros of the function and write the polynomial as a product of linear factors.
f(x) = x^3 + 11x^2 + 39x + 29.
So I think that I might need to use synthetic division, but I'm not sure. The teacher didn't explian this to me. Any help?
2. Originally Posted by tsmith
Find all the zeros of the function and write the polynomial as a product of linear factors.
f(x) = x^3 + 11x^2 + 39x + 29.
So I think that I might need to use synthetic division, but I'm not sure. The teacher didn't explian this to me. Any help?
HI
Factorise f(x) , do you know how ?
$f(x)=(x+1)(x^2+10x+29)$
$x^2+10x+29$ is complex , you can always check its discriminant .
$x+1$ is a factor because $f(-1)=0$
Then you can do the long division to get its quadratic factor .
So the zero of this polynomial would be -1 because -1 makes $f(x)=0$
3. [I]factor: x^3 + 11x^2 + 39x + 29 = (x + 1)(x^2 + 10x + 29) = (x + 1)(x + (5 - 2i))(x + (5 + 2i))
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http://mathoverflow.net/questions/2360/any-work-on-the-adams-watters-triangle
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## Any work on the Adams-Watters triangle?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Does anyone know whether any arithmetical or asymptotic results have been obtained about the Adams-Watters triangle?
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I upvoted your question to balance out whoever voted it down. Thanks for adding the comment in which you added context and detail to the question. (I had no idea what you were talking about in your initial post.) – userN Oct 25 2009 at 0:14
## 1 Answer
There seems to be no response to this, but perhaps somebody knows something about it in another terminology.
Franklin T. Adams-Watters defined a triangle similar to Pascal's, but where the latter has c = a + b between and under a and b, Adams-Watters takes c = (a+b)/gcd(a,b). The first few rows look like this:
```` 1
1 1
1 2 1
1 3 3 1
1 4 2 4 1
1 5 3 3 5 1
1 6 8 2 8 6 1
1 7 7 5 5 7 7 1
1 8 2 12 2 12 2 8 1
````
Adams-Watters summed the rows and obtained a sequence a(n) which is numbered A125606 in the OEIS database. He conjectures that log(a(n))/n tends to log(2*zeta(3)/zeta(2)). That was what I meant when I asked about asymptotic information. (The analogous limit is log(2) for Pascal's triangle, of course)
I am actually more interested in whether the Adams-Watters triangle has any nontrivial arithmetic structure. When you add a and b, the sum is trivially divisible by gcd(a,b), so the operation (a+b)/gcd(a,b) removes this trivial information. Pascal's triangle has a lot of arithmetic structure, but I did not see any when I factored the entries in the A-W triangle out to the fiftieth row. Perhaps dividing by the gcd leaves nothing, or maybe I didn't think about it the right way.
Two caveats: The little piece above is not enough to see what happens. For example, the twelfth row is the first that has both odd and even entries (apart from the 1s). Also there seems no reason to think that the A-W triangle has any combinatorial significance.
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This is really obscure. There are no Google scholar hits and this post is the first Google hit. – Qiaochu Yuan Oct 25 2009 at 0:12
I'm assuming you mean log(a(n))/n tends to log(2*zeta(3)/zeta(2)), or a(n)^(1/n) tends to 2*zeta(3)/zeta(2). In any case, this doesn't seem to be all that reasonable; looking at the sums of the first 1000 rows, a(n)^(1/n) seems to fluctuate around 1.55 or so, while 2*zeta(3)/zeta(2) = 1.46. Barring some theoretical reason to expect an expression having this general form, I have trouble believing this conjecture. – Michael Lugo Oct 25 2009 at 0:37
Thanks for spotting the missing log. I never tried to check the asymptotics myself; I just factorized entries, looking for patterns. I agree that this may seem a bit obscure, but I suspect that the operation (a+b)/gcd(a,b) has arithmetic significance, because of the dividing away the trivial information. In fact I began from this quotient and found what Adams-Watters had done by searching OEIS with likely sequences. – engelbrekt Oct 25 2009 at 1:05
If we call the table entries $A(n,k)$, with the same numbering conventions as for Pascal's triangle, then the sequence $A(n,2)$ (which goes $1,3,2,3,8,7,2,\dots$) is A131134 in the Online Encyclopedia, research.att.com/~njas/sequences/A131134, but there's no significant information there other than the first 69 terms. Still, that list of terms raises some questions. $A(28,2)=88$, but $A(30,2)=2$; $A(36,2)=108$, but $A(37,2)=4$. $A(n,2)$ is unbounded above since if $p$ is prime then $A(p,2)\ge p$ or $A(p+1,2)>p$. But does $A(n,2)$ go to infinity? – Gerry Myerson Mar 19 2010 at 3:44
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http://mathoverflow.net/questions/60059?sort=newest
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## Why is the exterior differentiation operator sometimes visualized as the “boundary”?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I have been trying to get an intuitive grasp on exterior calculus and found this article particularly interesting :
http://homepage.mac.com/sigfpe/Mathematics/forms.pdf
I can understand why the exterior derivative of a function is visualized using the "level curves", since the derivative is supposed to show the growth of the function and the level curves are more tightly packed where the function grows faster. I can also understand why the wedge product is the intersection of the two sets of level curves, since it's pretty easy to see for the basis $dx \land dy$ and you can "expand" every 1-form in terms of dx and dy. (I find this one interesting since the wedge does look a little like an intersection symbol...)
Now my question comes with the "d" operator. In the link I gave the author says that it should be visualized as the "boundary" of the set of level curves, which makes sense when you try to interpret stoke's theorem, and also gives an intuitive sense to $d^2=0$, but I can't find an explanation as to why this would be a good visualization?
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4
$d$ is really a coboundary operator--the formal adjoint of $\partial$. This fact in the context of the de Rham complex is the content of Stokes' theorem. – Steve Huntsman Mar 30 2011 at 12:09
4
You might learn from this earlier MO question: "Is the boundary $\partial S$ analogous to a derivative?" mathoverflow.net/questions/46252 . See especially Terry Tao's enlightening answer. – Joseph O'Rourke Mar 30 2011 at 13:14
## 3 Answers
Whether a visualization is good really is a subjective thing. This visualization of $d$ works for the author, but it doesn't work for me. If you still find it perplexing after mulling it over, you're probably better off investing your energies into trying to understand $d$ directly rather than into trying to wrap your head around the visualization.
If you still seek visualizations, you could look to other properties of differential geometry to base them on instead of the idea of contour lines for a function.
One such property is that you can integrate an $n$-form over an $n$-dimensional region, so you can try and imagine $n$-forms as a way to measure things. If you can wrap your head around that, then, as in Steve's comment, you can try using Stokes' theorem visualize how $d$ must look.
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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'm not directly answering your question, but I would like to suggest two great books for you to learn the meaning of exterior calculus, integration and de Rham cohomology.
1. "Vector Analysis on Manifolds" by Janich. - I learned that stuff from this book and for me is still the best.
2. "Mathematical Methods in Classical Mechanics" by Arnold. - Well known for its great insight on the meaning of exterior derivative, Lie derivative, Stokes theorem, paralell transport, riemannian curvature, among other concepts.
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Steve has more or less already answered to your question.
Just to give you some more intuition, let $M$ be an oriented smooth real manifold of real dimension $m$ and consider the space $^s\mathcal E^p(M)$, `$s\in\mathbb N\cup\{+\infty\}$`, of differential forms $C^s(M,\Lambda^p T^*_M)$ endowed with the topology of uniform convergence of all the derivatives of order $\le s$ of the coefficients on every compact subset (contained in a coordinate chart) of $M$. If $K\subset M$ is a compact subset, let $^s\mathcal D^p(K)$ denote the subspace of elements $u\in ^s\mathcal E^p(M)$ with support contained in $K$, together with the induced topology. Finally, put ```$$
^s\mathcal D^p(M)=\bigcup_{K\subset\subset M} ^s\mathcal D^p(K).
$$``` The space $^s\mathcal D'_p(M)=^s\mathcal D'^{m-p}(M)$ is by definition the topological dual $(^s\mathcal D_p(M))'$ and is called the space of currents of dimension $p$ (or degree $m-p$) and order $s$ on $M$.
You should think of currents as a generalization of differential forms (locally they are differential forms with distributional coefficients), precisely how distributions generalize the concept of functions.
Example 1. If $f$ is a differential form of degree $q$ with $L^1_{\text{loc}}$ coefficients, we can associate to $f$ the current $T_f$ of dimension $m-q$ $$\langle T_f,u\rangle=\int_Mf\wedge u,\quad u\in ^0\mathcal D^{m-q}(M).$$ $T_f$ is of degree $q$ and of order $0$. The correspondence $f\mapsto T_f$ is injective.
Example 2. Let $Z\subset M$ be a closed oriented submanifold of $M$ of dimension $p$ and class $C^1$ (here $Z$ may have a boundary $\partial Z$). The current of integration over $Z$, denoted $[Z]$, is defined by $$\langle[Z],u\rangle=\int_Z u,\quad u\in ^0\mathcal D^p(M).$$ It is clear that $[Z]$ is a current of order $0$ on $M$ and that $\operatorname{Supp}(Z)=Z$ (the support of a current is the smallest closed subset $A\subset M$ such that the restriction of the current to forms with compact support contained in the complementary of $A$ is zero). Its dimension is $p=\dim Z$.
The exterior derivative `$dT\in ^{s+1}\mathcal D'^{q+1}(M)=^{s+1}\mathcal D'_{m-q-1}$` of a current $T\in ^s\mathcal D'^q(M)=^s\mathcal D'_{m-q}$ is defined by ```$$
\langle dT,u\rangle=(-1)^{q+1}\langle T,du\rangle,\quad u\in ^{s+1}\mathcal D^{m-q-1}(M).
$$``` For all forms $f\in ^1\mathcal E^q(M)$ and $u\in ^1\mathcal D^{m-q-1}(M)$, Stokes' formula implies $$0=\int_M d(f\wedge u)=\int_M df\wedge u+(-1)^qf\wedge du,$$ thus in our Example 1 we actually have $dT_f=T_{df}$ as it should be.
Finally, and this should clarify your question, in our Example $2$ another application of Stokes' formula gives $$\int_Z du=\int_{\partial Z}u,$$ therefore $\langle[Z],du\rangle=\langle[\partial Z],u\rangle$ and $d[Z]=(-1)^{m-p+1}[\partial Z]$.
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Hi diverietti, this is very nice stuff, can you recommend any book that develops these ideas? I want to learn some analysis on manifolds. – Olivier Bégassat Mar 30 2011 at 14:19
1
@ Olivier Bégassat Mike Spivak's "Calculus on Manifolds" is a classic source. – Dick Palais Mar 30 2011 at 14:25
Hi Olivier, it depends a lot on your kind of background. If you prefer a more geometrical/complex analytic flavor, then books as "Principle of algebraic geometry" by Griffiths and Harris, or "Complex Analytic and Differential Geometry" by Demailly are great. From a point of view of (real) differential geometry, Spivak's book is very well, but "Foundation of differentiable manifolds and Lie groups" by Warner is very good, too. – diverietti Mar 30 2011 at 15:03
2
To see how this stuff is used in differential topology, look at the book by Bott and Tu. – Deane Yang Mar 30 2011 at 15:47
Of course! Your are completely right, Deane! Bott and Tu is wonderful! – diverietti Mar 30 2011 at 15:48
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http://math.stackexchange.com/users/29566/anurag-kalia?tab=activity&sort=comments
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# Anurag Kalia
reputation
7
bio
website inchunksandbits.wordpress.com
location New Delhi, India
age 20
member for 1 year
seen May 14 at 2:24
profile views 29
Since this is an introduction that spans the entirety of the Stack Exchange network, I would just like to say... I love them all!
| | | bio | visits | | |
|----------|----------------|------------------|-------------------------------|----------------|--------|
| | 170 reputation | website | inchunksandbits.wordpress.com | member for | 1 year |
| 7 badges | location | New Delhi, India | seen | May 14 at 2:24 | |
# 7 Comments
| | | |
|-------|---------|---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
| Jan17 | comment | Volume between two paraboloidsTry curve tracing to visualize the problem in 3D. |
| Dec5 | comment | How can I introduce complex numbers to precalculus students?I always imagine complex numbers as a plane, only with one axis as the imaginary numbers. I thought it was the only way to do it, really! |
| Aug28 | comment | Number of points at which a tangent touches a curve@RobertIsrael - He said it exactly like I said. Maybe he didn't want me to get into detail - I'll give him benefit of doubt since he was at the end correct. Plus, he is otherwise an excelent teacher! |
| Aug24 | comment | Number of points at which a tangent touches a curveThis seems intriguing, and though I don't understand bits of it right now, I would have just have to study more. Thanks for your answer. I was expecting something of this kind from this site. :-) |
| Aug8 | comment | What are the points of discontinuity of $\tan x$?So you are saying any discreet function is continuous? That is logical if one comes to think about it. Plus, I am sorry I couldn't get the part of essential discontinuity. What does the notation D-dash mean? |
| Aug8 | comment | What are the points of discontinuity of $\tan x$?Actually, the definition in any textbook I referred to give the definition of points of continuity only. They are dead silent when talking about discontinuity, whether f(c) needs to be defined or not adding to the confusion. |
| Aug8 | comment | What are the points of discontinuity of $\tan x$?I read about the extended real line in calculus as $\mathbb R \bigcup \infty$. That is, they treat $\infty$ and $-\infty$ as the same. How can they be same? There is at least a minus sign as a point of difference between the two. Wikipedia on compactification asks to treat real line as a circle where its open ends meet at $\infty$. Yet real line is infinite in length! Do we treat the radius of this circle to be infinite too? Isn't it just more practical to add two points, $+\infty$ and $-\infty$ to real line, and more intuitive too? |
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http://www.mathplanet.com/education/pre-algebra/introducing-algebra/identify-properties
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# Identify properties
In this section, you´ll learn how to identify the properties of multiplication and addition and how you can use identification to help solve mathematical problems.
To solve rather hard problems without using a calculator you have to identify all expressions that have the same operation.
Example
$\begin{array}{lcl} 58+69+91=? \end{array}$
In this example, you can add in any order you prefer. The sum of the expression will not change if you prefer to add the numbers in a different order.
$\begin{array}{lcl} 91+58+69=218\\ 58+69+91=218 \end{array}$
The same is true for multiplication:
$\begin{array}{lcl} 5\cdot 4\cdot 30=600\\ 4\cdot 30\cdot 5=600 \end{array}$
Video lesson: Show that the following equation holds true
$\\3\cdot a\cdot 4=a\cdot 4\cdot 3$
Next Class: Introducing Algebra, Equations with variables
• Pre-Algebra
• Algebra 1
• Algebra 2
• Geometry
• Sat
• Act
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http://math.stackexchange.com/questions/170557/how-to-find-all-naturals-n-such-that-sqrt1-underbrace4-cdots4-n-text
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# How to find all naturals $n$ such that $\sqrt{1 {\underbrace{4\cdots4}_{n\text{ times}}}}$ is an integer?
How to find all naturals $n$ such that $\sqrt{1\smash{\underbrace{4\cdots4}_{n\text{ times}}}}$ is an integer?
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Do you mean $\sqrt{1\underbrace{4\cdots4}_{n-times}}$ or perhaps$\sqrt{\underbrace{1414\cdots14}_{n-times}}$? – Quixotic Jul 14 '12 at 1:24
I mean The first one . – Frank Jul 14 '12 at 1:27
1
There will be only two integer cases for $n = 2$ and $n = 3$ and for everything else $$\sqrt{1\underbrace{4\cdots4}_{n-times}}=2\sqrt{36\underbrace{11\cdots1}_{(n-2)-times}}$$ – Quixotic Jul 14 '12 at 1:37
I changed the TeX code from {}_{n-times} to {}_{n\text{ times}, so that instead of $1\underbrace{4\cdots4}_{n-times}$ we see $1\underbrace{4\cdots4}_{n\text{ times}}$. The hyphen looked like a minus sign (longer than a hyphen) since it was in math mode, and the "times" got italicized and needed something to artificially separate it from the $n$, since it was in math mode. – Michael Hardy Jul 14 '12 at 1:47
## 2 Answers
For $n \geq 4$ your number is equal to $4444$ modulo $10000$, and in particular modulo $16$. If it were a square, then $4444$ would be a square modulo $16$, implying $1111$ is a square modulo $4$. But $1111=3$ mod $4$, contradiction.
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Hint: the square root must end in $2$ or $8$ (except for $1$-do you permit no $4$'s?). You can see this by what digit squares end in $4$. You can extend this argument digit by digit, which will terminate by saying that some length of string of $4$'s can't be the end of a square. Then there won't be many possibilities to check.
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http://physics.stackexchange.com/questions/tagged/convection+homework
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# Tagged Questions
1answer
40 views
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### What is the characteristic length of a cylinder
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So I have faced a problem dealing with transient conduction and I need a little help with the problem solving concepts. I need to determine how long it would take to reach the final temperature but I ...
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### How would I go about solving this transient convection problem if the mean fluid temperature is constantly changing?
Let's say I have a ceramic slab on a conveyor belt that is initially at $450\,^{\circ}\mathrm{C}$ and there is air being blown over it at a speed of $35 \frac{m}{s}$ with an ambient temperature of ...
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### In what situations do I use the characteristic length of a fin to find the surface area?
So I'm learning about fins in heat transfer and it seems that there are two separate formulas for the surface area of a rectangular fin of length L, width w and thickness t. The fin is attached to a ...
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http://stats.stackexchange.com/questions/21207/handling-outliers-when-comparing-two-means-in-a-repeated-measures-design
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# Handling outliers when comparing two means in a repeated measures design
I am doing a simple study that involved taking a measure at time point 1 and time point 2 (12 weeks later). While the sample was a class, not all members were present at both time points, so I have 20 date points at time 1 and 21 date points at time 2. The measure has a score, and I am taking the means and doing a simple t-test to determine if the intervention caused any increase in the measure at time point 2.
Questions: Do I need to throw out outliers if they are more than 2 standard deviations higher than the mean?
When I do the t-test, do I need to look at one- or two-tailed distributions? My hypothesis is that the intervention will increase the mean at time 2, so I think I should consider a one-tailed distribution.
Lastly, I am assuming that I have to do a paired t-test since it is a repeated measures design.
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## 1 Answer
I will take these out of order. If it is possible to establish a correspondence between the measurements in the first set and the measurements in the second set (for example, Bob's score at time 1 and Bob's score at time 2 correspond because they both came from Bob), then you should do a paired t-test. That is, you should not calculate means for each time, but take differences, and calculate the mean and standard deviation of the differences. The standard error of the differences (i.e., the denominator of the t-statistic) is that standard deviation divided by $\sqrt{n}$. If some students did not participate at one of the occasions, then their scores should be set aside. Furthermore, you do not care if a score is more than 2 s.d.'s higher than the mean, although you may care if one of your differences is more than 2 s.d.'s above the mean of the differences.
The definition of an outlier is a data point that came from a different population than the one you want to be studying. The definition is not a data point that is far away from the rest of your data. However, we almost never know whether or not a data point came from a different distribution than the rest of our data, except that it looks really different. If you should ever spend much time conducting simulations, you will come to notice that every so often a data point that you know comes from the same distribution (because you wrote the simulation code) looks quite a bit different from the rest. This is an uncomfortable fact, but it is nonetheless true. Ultimately, you need to decide whether you believe that data point belongs there or not. There are some (potentially) helpful guidelines:
1. With ~20 data points, a z-score with an absolute value greater than 2 is pretty unlikely (although it wouldn't be if you had, say, 100 data points);
2. You can look at a plot of your data (e.g., a histogram) to see if the larger value is contiguous with the rest of your data, or if there is a large break between it and the rest;
3. It can help to run your analysis both with the potential outlier and without it (often, you will get the same answer both ways, and that's reassuring);
4. A final possibility is to use 'trimmed samples', that is, exclude the top and bottom 2 data points (given that you have ~20, this would be a 10% trimmed sample), note that this lessens your power, but many people think it's more even handed.
In the end, I'm afraid, you will still have to make a decision, however.
Lastly, you should know that the question of 2 vs. 1 - tailed t-tests has long been a contentious topic. It is probably not as important as people have made it out to be, but that is the nature of these things. Personally, I'm against 1-tailed tests, but my opinion is really unimportant. A question you could ask yourself is:
What if I find that the mean decreased by a very large amount? Would I say, 'Nope, there was no change', or would I say there was a change?
If it would be possible for you to believe a negative change if the data supported it, then you really should be using a 2-tailed test, but if there is no way you would ever believe the mean went down, then a 1-tailed test is probably fine, and you just let the old grumps (like me) harrumph about it. What you should not do is run the test both ways and pick the one that gives you the result you like best (or run a 1-tailed test, notice that the mean went down by a lot, then run a 2-tailed test and call it 'significant').
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http://math.stackexchange.com/questions/233793/basic-schubert-calculus-intuition
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Basic Schubert calculus intuition
I'm thinking about the famous problem from classical algebraic geometry of how many lines in $\mathbb{P}^3$ meet four given general lines. According to some lecture notes on intersection theory that I was reading, Schubert had the intuition that it's general enough to consider the case where the four lines are in fact two pairs of intersecting lines - a highly nongeneric setup. (And then of course, in this case the answer is easily seen to be two.)
My question is, can someone give a basic explanation of this intuition, and perhaps some explanation of which other scenarios admit this sort of logic (i.e. looking at non-generic-but-still-kind-of-generic cases)? I have a vague picture in my head of continuously varying one of the four lines and how a line meeting all four should vary continuously, but I'm not entirely convinced by it yet.
Thanks! (Also thanks to Michael Joyce for pointing me to this sort of problem in a previous answer.)
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What are you familiar with? My intuition behind this is that this corresponds to intersection theory on the Grassmannian, which is projective, and the dimension of the Grassmannian is 4, so as long as the setup ends with a finite number of lines satisfying the given conditions, that will be "the correct number". More concretely, this corresponds to some kind of "moving lemma": If you move a line to intersect another line in a way so that the setup never degenerates, there is a kind of "continuity" that keeps the number the same. But this is unclear to me without Grassmannians. – only Nov 9 '12 at 22:03
So you seem to be saying that in (projective) intersection theory problems, shifting around the setup may change the dimension of the space we're looking at (in this case, lines meeting the four given lines), but should never change the number of irreducible components? This sounds somewhat plausible. (And I hope someone weighs in on this philosophy.) But in this example, it's not clear (to me) that the number of lines that meet a generic set of 4 lines is nonempty. (I assume when you say "a finite number of lines" above, you mean nonzero.) Though I suppose I could perhaps just check that? – Rob Silversmith Nov 9 '12 at 23:53
You can create an intersection theory on Grassmannians. What this means here is that if you pass to rational equivalence, you can define the "intersection" of two rational equivalence classes in a way so that the intersection number is always of the proper dimension. (For example, you can intersect a curve on a surface with itself and get a "finite number of points", or rather, a dimension 0 rational equivalence class). Under rational equivalence on Grassmannians, which are projective, the number of points always stays the same. (continued in next comment) – only Nov 10 '12 at 2:52
So what's really happening is that whenever your intersection is actually a finite set of points, the number will coincide with the formal intersection number. When the intersection is not a finite set of points, the data from the intersection number is harder to use. If you accept the existence of such an intersection theory, then the specific example given proves the intersection number is 2, and proves that whenever you have an intersection consisting of a finite number of points, it will be 2 points. (to be continued) – only Nov 10 '12 at 2:56
The objects you are intersecting are in this context sections of line bundles. Any two sections are rationally equivalent, so you can pass to rational equivalence. To prove that the intersection is generically finite, I'm fairly sure there is some sheaf which will give you that after invoking the fact that ranks of sheaves are semicontinuous functions. On a sidenote, I realized that I never mentioned that "the number of points in the intersection" is with multiplicity. – only Nov 10 '12 at 2:59
1 Answer
(expanding on the above -- can't figure out how to edit, please delete earlier)
There's some nice discussion of this in Harris's book "3264 and all that", a (legitimately available) draft of which may be found on Google. Section 4.2 works out the Chow ring of $G(1,3)$. In particular, "A Specialization Argument" in Section 4.2.3 seems to be just the example you're after about degenerating to two intersecting lines. I gather that there is a generalization of this argument by Coskun and Vakil that might be of interest; the reference is in 3264.
To elaborate, let $\sigma_1$ be the cycle consisting of lines which meet a given line. What you're trying to do is compute $\sigma_1^4$, which will be the number of lines that meet four given lines. A sensible first step is to compute $\sigma_1^2$. This turns out to be $\sigma_{1,1} + \sigma_2$, where $\sigma_{1,1}$ is the set of lines in a given plane and $\sigma_2$ is the lines through a particular point. Here's the idea of how to get that answer via degeneration. Fix a line $L$ and a family of lines $M_t$ such that $L$ and $M_t$ do not meet, except when $t=0$ they intersection at a point. You're interested in the class of $\sigma_1(L) \cap \sigma_1(M_t)$, and it's sensible to look for the flat limit of this intersection as $t \to 0$. If $\ell$ meets both $L$ and $M_0$, either it is contained in the plane with both of them in it (there's the $\sigma_{1,1}$), or it goes through the point where they intersect (there's the $\sigma_2$).
Degenerating to two pairs of intersecting lines as you suggest lets you write $\sigma_1^4 = (\sigma_{1,1}+\sigma_2) \cdot (\sigma_{1,1} + \sigma_2) = 2$. To see why the above degeneration argument (I use the term loosely!) is actually legitimate, check out the referenced book.
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Look ^^ there. You should see `share` `edit` `flag` buttons. It's possible that you may not have enough reputation, yet, for all of those buttons, but you should have the first two. The `edit` button is (of course) the one you want. – Cameron Buie Nov 10 '12 at 5:44
I merged your accounts. That was probably the reason you couldn't delete the other post. – robjohn♦ Nov 10 '12 at 9:54
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http://math.stackexchange.com/questions/103052/prove-the-composition-of-these-map-objects-are-consistent/103094
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# Prove the composition of these map objects are consistent.
I have been working my way through Lawvere and Schanuel (1997) without too much trouble, but now that I am up to Article V, I am stumped. So, without further ado:
Exercise 6: In a category with products in which map objects exist for any two objects, there is for any three objects a standard map: $B^A \times C^B \stackrel{\gamma}{\longrightarrow} C^A$ which represents composition in the sense that $\gamma\langle\lceil f \rceil, \lceil g \rceil\rangle = \lceil gf \rceil$ for any $A \stackrel{f}{\longrightarrow} B \stackrel{g}{\longrightarrow} C.$
I understand how an exponential map object is like a product map, and I can diagram product maps just fine. They're pretty straightfoward. But for this problem, I think the key to solving the problem lies in isolating each of the $\lceil f \rceil, \lceil g \rceil$ terms in order to calculate the product $\lceil gf \rceil$. In the book, they only give them as parameters of the functions f and g taking A to B and B to C, respectively.
When I was first stuck, I looked up the definition of a exponential map object in wikipedia, but that didn't help me in the calculation. I can write the diagram from A to B and from B to C separately, but I am stuck at the point of putting them together into a composite function.
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## 1 Answer
By universal property we get that there are morphisms $\cdot_{A,B} \colon B^A \times A \to B$ and $\cdot_{B,C} \colon C^B \times B \to C$ which are universals.
Then you get the morphism $$\tilde \gamma \colon C^B \times B^A \times A \stackrel{1_{C^B} \times \cdot_{A,B}}{\longrightarrow} C^B \times B \stackrel{\cdot_{B,C}}{\longrightarrow} C$$ which is the map that let an element of $B^A$ act on $A$ and then let an element of $C^B$ act on the result.
Now consider the corresponding map $$\gamma \colon C^B \times B^A \to C^A$$ this is your composition (up the isomorphism $C^B \times B^A \cong B^A \times C^B$).
I'd rather left the details, but if you need them let me know and I'll try to add them (hoping I'll find the time).
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The level of detail in your answer was perfect! You used the UMP to set up the two maps, brought them together by means of substitution, made a projection from the product to the exponent, and then switched the order, which is correct up to isomorphism (why is that, btw?). – bwkaplan Jan 28 '12 at 20:26
@bwkaplan $C^B \times B^A$ and $B^A \times C^B$ are not necessarily the same object, some times are simply isomorphic, the arrow that I found is of type $C^B \times B^A \to C^A$ while you were looking for a morphism of type $B^A \times C^B \to C^A$, so in order to to obtain it you have to compose the morphism I've found with the isomorphism $B^A \times C^B \cong C^B \times B^A$. – ineff Jan 29 '12 at 16:39
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http://chorasimilarity.wordpress.com/2013/02/27/emergent-algebras-as-combinatory-logic-part-ii/
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# chorasimilarity
computing with space
Home > Uncategorized > Emergent algebras as combinatory logic (Part II)
## Emergent algebras as combinatory logic (Part II)
February 27, 2013
This post continues Emergent algebras as combinatory logic (Part I). My purpose is to introduce the calculus standing behind Theorem 1 from the mentioned post.
We have seen (Definition 2) that there are approximate sum and difference operations associated to an emergent algebra. Let me add to them a third operation, namely the approximate inverse. For clarity I repeat here the Definition 2, supplementing it with the definition of the approximate inverse. This gives:
Definition 2′. For any $\varepsilon \in \Gamma$ we give the following names to several combinations of operations of emergent algebras:
• the approximate sum operation is $\Sigma^{x}_{\varepsilon} (u,v) =$ $x \bullet_{\varepsilon} ((x \circ_{\varepsilon} u) \circ_{\varepsilon} v)$,
• the approximate difference operation is $\Delta^{x}_{\varepsilon} (u,v) = (x \circ_{\varepsilon} u) \bullet_{\varepsilon} (x \circ_{\varepsilon} v)$
• the approximate inverse operation is $inv^{x}_{\varepsilon} u = (x \circ_{\varepsilon} u) \bullet_{\varepsilon} x$.
The justification for these names comes from the explanations given in the post “The origin of emergent algebras (part II)“, where I discussed the sketch of a solution to the question “What makes the (metric) tangent space (to a sub-riemannian regular manifold) a group?”, given by Bellaiche in the last two sections of his article The tangent space in sub-riemannian geometry, in the book Sub-riemannian geometry, eds. A. Bellaiche, J.-J. Risler, Progress in Mathematics 144, Birkhauser 1996. We have seen there that the group operation (the noncommutative, in principle, addition of vectors) can be seen as the limit of compositions of intrinsic dilations, as $\varepsilon$ goes to $0$. It is important that this limit exists and that it is uniform, according to Gromov’s hint.
Well, with the notation $\delta^{x}_{\varepsilon} y = x \circ_{\varepsilon} y$, $\delta^{x}_{\varepsilon^{-1}} y = x \bullet_{\varepsilon} y$, it becomes clear, for example, that the composition of intrinsic dilations described in the figure from the post “The origin of emergent algebras (part II)” is nothing but the approximate sum from Definition 2′. (This is to say that formally, if we replace the emergent algebra operations with the respective intrinsic dilations, then the approximate sum operation $\Sigma^{x}_{\varepsilon}(y,z)$ appears as the red point E from the mentioned figure. It is still left to prove that intrinsic dilations from regular sub-riemannian spaces give rise to emergent algebras, this was done in arXiv:0708.4298.)
We recognize therefore the two ingredients of Bellaiche’s solution into the definition of an emergent algebra:
• approximate operations, which are just clever compositions of intrinsic dilations in the realm of sub-riemannian spaces, which
• converge in a uniform way to the exact operations which give the algebraic structure of the tangent space.
Therefore, a rigorous formulation of Bellaiche’s solution is Theorem 1 from the previous post, provided that we extract, from the long differential geometric work done by Bellaiche, only the part which is necessary for proving that intrinsic dilations produce an emergent algebra structure.
Nevertheless, Theorem 1 shows that the “emergence of operations” phenomenon is not at all specific to sub-riemannian geometry. In fact, once we get the idea of the right definition of approximate operations (from sub-riemannian geometry), we can simply try to prove the theorem by “abstract nonsense”, i.e. algebraically, with a dash of uniform convergence at the end.
For this we have to identify the algebraic relations which are satisfied by these approximate operations. For example, is the approximate sum associative? is the approximate difference the inverse of the approximate sum? is the approximate inverse of an element the inverse with respect to the approximate sum? and so on. The answer to these questions is “approximately yes”.
It is clear that in order to find the right relations (approximate associativity and so on) between these approximate operations we need to reason in a more clear way. Just by looking at the expressions of the operations from Definition 2′, it is obvious that if we start with a brute force “shut up and compute” approach then we will end rather quickly with a mess of parantheses and coefficients. There has to be a more easy way to deal with those approximate operations than brute force.
The way I have found has to do with a graphical representation of these operations, a way which eventually led me to graphic lambda calculus. This is for next time.
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For questions on writing functions (pure or using Set/SetDelayed) for any purpose, including the features that may be incorporated in those functions, such as options, patterns and conditions.
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### How to write a differential operator in Mathematica
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http://www.math.uni-bielefeld.de/sfb701/projects/view/11
|
Faculty of Mathematics
Collaborative Research Centre 701
Spectral Structures and Topological Methods in Mathematics
# Project B4
## Kolmogorov operators and SPDE
Principal Investigator(s) Other Investigators
## Summary:
The aim of the project is (a) to develop a theory providing analytic techniques to solve Kolmogorov and Fokker-Planck equations in infinite dimensions and reconstruct from their solutions a solution to the associated stochastic partial differential equations (SPDE), and (b) to solve and analyse the SPDE directly in case of more regular coefficients. Both will be done further developing several approaches which are in case (a) an approach via $L^p$-spaces with respect to an excessive measure of the Kolmogorov operator L and an approach based on a suitably newly formulated maximum principle for L on weighted spaces of weakly continuous functions, and in case (b) both the variational and semigroup (mild solution) approach. In particular, the spectral analysis and geometry of the Kolmogorov operators will be central points of the research. Among the main further issues are: existence and uniqueness of (infinitesimally) invariant measures, spectral properties and functional inequalities for L, large time asymptotics, jump type and other noises, small noise large deviations, finite speed of propagation, stochastic boundary dissipation, applications to SPDE from hydrodynamics and to Kolmogorov operators of particle systems.
## Recent Preprints:
| | | |
|-------|------------------------------------------------------------------------------------------------------------------------------------|-------------|
| 13039 | Stochastic nonlinear Schrödinger equations with linear multiplicative noise: the rescaling approach | PDF | PS.GZ |
| 13035 | On continuity equations in infinite dimensions with non-gaussian reference measure | PDF | PS.GZ |
| 13034 | On the absolute continuity of the distributions of smooth functions on infinite-dimensional spaces with measures | PDF | PS.GZ |
| 13033 | Stochastic Generalized Porous Media Equations with Reflection | PDF | PS.GZ |
| 13031 | Random attractor associated with the quasi-geostrophic equation | PDF | PS.GZ |
| 13019 | Finite speed of propagation for stochastic porous media equations | PDF | PS.GZ |
| 13011 | Exact null internal controllability for the heat equation on unbounded convex domains | PDF | PS.GZ |
| 13010 | On extensions of Sobolev functions on infinite-dimensional spaces | PDF | PS.GZ |
| 12144 | On estimates of solutions of Fokker–Planck–Kolmogorov equations with potential terms and non uniformly elliptic diffusion matrices | PDF | PS.GZ |
| 12143 | Nonlinear parabolic equations for probability measures | PDF | PS.GZ |
|
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http://en.wikipedia.org/wiki/K-d_tree
|
# k-d tree
KD-tree
Type Multidimensional BST
Invented 1975
Invented by Jon Louis Bentley
Time complexity
in big O notation
Average Worst case
Space O(n) O(n)
Search O(log n) O(n)
Insert O(log n) O(n)
Delete O(log n) O(n)
A 3-dimensional k-d tree. The first split (red) cuts the root cell (white) into two subcells, each of which is then split (green) into two subcells. Finally, each of those four is split (blue) into two subcells. Since there is no more splitting, the final eight are called leaf cells.
In computer science, a k-d tree (short for k-dimensional tree) is a space-partitioning data structure for organizing points in a k-dimensional space. k-d trees are a useful data structure for several applications, such as searches involving a multidimensional search key (e.g. range searches and nearest neighbor searches). k-d trees are a special case of binary space partitioning trees.
## Informal description
The k-d tree is a binary tree in which every node is a k-dimensional point. Every non-leaf node can be thought of as implicitly generating a splitting hyperplane that divides the space into two parts, known as half-spaces. Points to the left of this hyperplane represent the left subtree of that node and points right of the hyperplane are represented by the right subtree. The hyperplane direction is chosen in the following way: every node in the tree is associated with one of the k-dimensions, with the hyperplane perpendicular to that dimension's axis. So, for example, if for a particular split the "x" axis is chosen, all points in the subtree with a smaller "x" value than the node will appear in the left subtree and all points with larger "x" value will be in the right subtree. In such a case, the hyperplane would be set by the x-value of the point, and its normal would be the unit x-axis.[1]
## Operations on k-d trees
### Construction
Since there are many possible ways to choose axis-aligned splitting planes, there are many different ways to construct k-d trees. The canonical method of k-d tree construction has the following constraints:
• As one moves down the tree, one cycles through the axes used to select the splitting planes. (For example, in a 3-dimensional tree, the root would have an x-aligned plane, the root's children would both have y-aligned planes, the root's grandchildren would all have z-aligned planes, the root's great-grandchildren would all have x-aligned planes, the root's great-great-grandchildren would all have y-aligned planes, and so on.)
• Points are inserted by selecting the median of the points being put into the subtree, with respect to their coordinates in the axis being used to create the splitting plane. (Note the assumption that we feed the entire set of n points into the algorithm up-front.)
This method leads to a balanced k-d tree, in which each leaf node is about the same distance from the root. However, balanced trees are not necessarily optimal for all applications.
Note also that it is not required to select the median point. In that case, the result is simply that there is no guarantee that the tree will be balanced. A simple heuristic to avoid coding a complex linear-time median-finding algorithm, or using an O(n log n) sort of all n points, is to use sort to find the median of a fixed number of randomly selected points to serve as the splitting plane. In practice, this technique often results in nicely balanced trees.
Given a list of n points, the following algorithm uses a median-finding sort to construct a balanced k-d tree containing those points.
```function kdtree (list of points pointList, int depth)
{
// Select axis based on depth so that axis cycles through all valid values
var int axis := depth mod k;
// Sort point list and choose median as pivot element
select median by axis from pointList;
// Create node and construct subtrees
var tree_node node;
node.location := median;
node.leftChild := kdtree(points in pointList before median, depth+1);
node.rightChild := kdtree(points in pointList after median, depth+1);
return node;
}
```
It is common that points "after" the median include only the ones that are strictly greater than the median. For points that lie on the median, it is possible to define a "superkey" function that compares the points in all dimensions. In some cases, it is acceptable to let points equal to the median lie on one side of the median, for example, by splitting the points into a "less than" subset and a "greater than or equal to" subset.
The above algorithm implemented in the Python programming language is as follows:
```class Node: pass
def kdtree(point_list, depth=0):
if not point_list:
return None
# Select axis based on depth so that axis cycles through all valid values
k = len(point_list[0]) # assumes all points have the same dimension
axis = depth % k
# Sort point list and choose median as pivot element
point_list.sort(key=lambda point: point[axis])
median = len(point_list) // 2 # choose median
# Create node and construct subtrees
node = Node()
node.location = point_list[median]
node.left_child = kdtree(point_list[:median], depth + 1)
node.right_child = kdtree(point_list[median + 1:], depth + 1)
return node
```
Example usage would be:
```point_list = [(2,3), (5,4), (9,6), (4,7), (8,1), (7,2)]
tree = kdtree(point_list)
```
The generated tree is shown on the right, below.
k-d tree decomposition for the point set `(2,3), (5,4), (9,6), (4,7), (8,1), (7,2)`.
The resulting k-d tree.
This algorithm creates the invariant that for any node, all the nodes in the left subtree are on one side of a splitting plane, and all the nodes in the right subtree are on the other side. Points that lie on the splitting plane may appear on either side. The splitting plane of a node goes through the point associated with that node (referred to in the code as node.location).
A novel tree-building algorithm builds a balanced k-d tree in O(kn log n) time by sorting n points in k dimensions independently and prior to building the k-d tree.[2][3] A suitable sorting algorithm is Heapsort that creates a sorted array in O(n log n) time. Application of Heapsort to n points in each of k dimensions requires O(kn log n) time, and produces k sorted arrays of length n that contain references (or pointers) to the n points. These arrays are numbered from 0 to k-1. Each array represents the result of sorting the points in one of the k dimensions. For example, the elements of array 0, from first to last, reference the n points in order of increasing x-coordinate. Similarly, the elements of arrays 1, 2, and 3, from first to last, reference the n points in order of increasing y-, z- and w-coordinates, respectively.
In order to insert the first node into the k-d tree, the median element of array 0 is chosen and stored in the tree node. This median element splits array 0 into two subarrays. One subarray lies above the median element, and the other subarray lies below it. Also, the x-coordinate of the point that this median element references defines an x-aligned splitting plane that may be used to split each of the other k-1 arrays into two subarrays. The following procedure splits an array into two subarrays:
• Consider each element of the array in order from first to last.
• Test against the splitting plane the x-coordinate of the point that is referenced by the array element, and assign that element to one of two subarrays, depending on which side of the splitting plane the point lies.
• Ignore the array element that references the same point that the median element of array 0 references, because this point defines the splitting plane.
This procedure splits the arrays into two sets of subarrays while preserving the original sorted order within each subarray. These subarrays may then be used to insert nodes into the two subtrees at the next level of the tree in a recursive manner. However, if the subarrays comprise only one or two array elements, no further recursion is required because these cases may be solved trivially.
These guidelines will simplify creation of k-d trees:
• Arrays should be split into subarrays that represent "less than" and "greater than or equal to" partitioning. This convention requires that, after choosing the median element of array 0, the element of array 0 that lies immediately below the median element be examined to ensure that this adjacent element references a point whose x-coordinate is less than and not equal to the x-coordinate of the splitting plane. If this adjacent element references a point whose x-coordinate is equal to the x-coordinate of the splitting plane, continue searching towards the beginning of array 0 until the first instance of an array element is found that references a point whose x-coordinate is less than and not equal to the x-coordinate of the splitting plane. When this array element is found, the element that lies immediately above this element is the correct choice for the median element. Apply this method of choosing the median element at each level of recursion.
• This procedure for producing subarrays guarantees that the two subarrays comprise one less array element than the array from which these subarrays were produced. This characteristic permits re-use of the k arrays at each level of recursion as follows: (1) copy array 0 into a temporary array, (2) build the subarrays that are produced from array 1 in array 0, (3) build the subarrays that are produced from array 2 in array 1, (4) continue this pattern, and build the subarrays that are produced from array k-1 in array k-2, and finally (4) copy the temporary array into array k-1. This method permutes the subarrays so that at successive levels of the k-d tree, the median element is chosen from x-, y-, z- w-,... sorted arrays.
• The addresses of the first and last elements of the 2k subarrays can be passed to the next level of recursion in order to designate where these subarrays lie within the k arrays. Each of the two sets of k subarrays have identical addresses for their first and last elements.
This tree-building algorithm requires at most O([k-1]n) tests of coordinates against splitting planes to build each of the log n levels of a balanced k-d tree. Hence, building the entire k-d tree requires less than O([k-1]n log n) time, which is less than the O(kn log n) time that is required to sort the n points in k dimensions prior to building the k-d tree.
### Adding elements
This section requires expansion. (November 2008)
One adds a new point to a k-d tree in the same way as one adds an element to any other search tree. First, traverse the tree, starting from the root and moving to either the left or the right child depending on whether the point to be inserted is on the "left" or "right" side of the splitting plane. Once you get to the node under which the child should be located, add the new point as either the left or right child of the leaf node, again depending on which side of the node's splitting plane contains the new node.
Adding points in this manner can cause the tree to become unbalanced, leading to decreased tree performance. The rate of tree performance degradation is dependent upon the spatial distribution of tree points being added, and the number of points added in relation to the tree size. If a tree becomes too unbalanced, it may need to be re-balanced to restore the performance of queries that rely on the tree balancing, such as nearest neighbour searching.
### Removing elements
This section requires expansion. (February 2011)
To remove a point from an existing k-d tree, without breaking the invariant, the easiest way is to form the set of all nodes and leaves from the children of the target node, and recreate that part of the tree.
Another approach is to find a replacement for the point removed.[4] First, find the node R that contains the point to be removed. For the base case where R is a leaf node, no replacement is required. For the general case, find a replacement point, say p, from the subtree rooted at R. Replace the point stored at R with p. Then, recursively remove p.
For finding a replacement point, if R discriminates on x (say) and R has a right child, find the point with the minimum x value from the subtree rooted at the right child. Otherwise, find the point with the maximum x value from the subtree rooted at the left child.
### Balancing
Balancing a k-d tree requires care because k-d trees are sorted in multiple dimensions so the tree rotation technique cannot be used to balance them as this may break the invariant.
Several variants of balanced k-d trees exist. They include divided k-d tree, pseudo k-d tree, k-d B-tree, hB-tree and Bkd-tree. Many of these variants are adaptive k-d trees.
### Nearest neighbour search
Animation of NN searching with a k-d tree in two dimensions
The nearest neighbour search (NN) algorithm aims to find the point in the tree that is nearest to a given input point. This search can be done efficiently by using the tree properties to quickly eliminate large portions of the search space.
Searching for a nearest neighbour in a k-d tree proceeds as follows:
1. Starting with the root node, the algorithm moves down the tree recursively, in the same way that it would if the search point were being inserted (i.e. it goes left or right depending on whether the point is less than or greater than the current node in the split dimension).
2. Once the algorithm reaches a leaf node, it saves that node point as the "current best"
3. The algorithm unwinds the recursion of the tree, performing the following steps at each node:
1. If the current node is closer than the current best, then it becomes the current best.
2. The algorithm checks whether there could be any points on the other side of the splitting plane that are closer to the search point than the current best. In concept, this is done by intersecting the splitting hyperplane with a hypersphere around the search point that has a radius equal to the current nearest distance. Since the hyperplanes are all axis-aligned this is implemented as a simple comparison to see whether the difference between the splitting coordinate of the search point and current node is less than the distance (overall coordinates) from the search point to the current best.
1. If the hypersphere crosses the plane, there could be nearer points on the other side of the plane, so the algorithm must move down the other branch of the tree from the current node looking for closer points, following the same recursive process as the entire search.
2. If the hypersphere doesn't intersect the splitting plane, then the algorithm continues walking up the tree, and the entire branch on the other side of that node is eliminated.
4. When the algorithm finishes this process for the root node, then the search is complete.
Generally the algorithm uses squared distances for comparison to avoid computing square roots. Additionally, it can save computation by holding the squared current best distance in a variable for comparison.
Finding the nearest point is an O(log N) operation in the case of randomly distributed points, although analysis in general is tricky. However an algorithm has been given that claims guaranteed O(log N) complexity.[5]
In very high dimensional spaces, the curse of dimensionality causes the algorithm to need to visit many more branches than in lower dimensional spaces. In particular, when the number of points is only slightly higher than the number of dimensions, the algorithm is only slightly better than a linear search of all of the points.
The algorithm can be extended in several ways by simple modifications. It can provide the k-Nearest Neighbours to a point by maintaining k current bests instead of just one. Branches are only eliminated when they can't have points closer than any of the k current bests.
It can also be converted to an approximation algorithm to run faster. For example, approximate nearest neighbour searching can be achieved by simply setting an upper bound on the number points to examine in the tree, or by interrupting the search process based upon a real time clock (which may be more appropriate in hardware implementations). Nearest neighbour for points that are in the tree already can be achieved by not updating the refinement for nodes that give zero distance as the result, this has the downside of discarding points that are not unique, but are co-located with the original search point.
Approximate nearest neighbour is useful in real-time applications such as robotics due to the significant speed increase gained by not searching for the best point exhaustively. One of its implementations is best-bin-first search.
### Range search
This section requires expansion. (March 2013)
Analyses of binary search trees has found that the worst case time for range search in a k-dimensional KD tree containing N nodes is given by the following equation.[6]
$t_{worst} = O(k \cdot N^{1-\frac{1}{k}})$
## High-dimensional data
k-d trees are not suitable for efficiently finding the nearest neighbour in high dimensional spaces. As a general rule, if the dimensionality is k, the number of points in the data, N, should be N >> 2k. Otherwise, when k-d trees are used with high-dimensional data, most of the points in the tree will be evaluated and the efficiency is no better than exhaustive search,[7] and approximate nearest-neighbour methods should be used instead.
## Complexity
• Building a static k-d tree from n points takes:
• O(n log2 n) time if an O(n log n) sort such as Heapsort is used to compute the median at each level;
• O(n log n) time if a complex linear-time median-finding algorithm such as the one described in Cormen et al.[8] is used;
• O(kn log n) plus O([k-1]n log n) time if n points are sorted in each of k dimensions using an O(n log n) sort prior to building the k-d tree.
• Inserting a new point into a balanced k-d tree takes O(log n) time.
• Removing a point from a balanced k-d tree takes O(log n) time.
• Querying an axis-parallel range in a balanced k-d tree takes O(n1-1/k +m) time, where m is the number of the reported points, and k the dimension of the k-d tree.
• Finding 1 nearest neighbour in a balanced k-d tree with randomly distributed points takes O(log n) time on average.
## Variations
### Volumetric objects
Instead of points, a k-d tree can also contain rectangles or hyperrectangles.[9][10] Thus range search becomes the problem of returning all rectangles intersecting the search rectangle. The tree is constructed the usual way with all the rectangles at the leaves. In an orthogonal range search, the opposite coordinate is used when comparing against the median. For example, if the current level is split along xhigh, we check the xlow coordinate of the search rectangle. If the median is less than the xlow coordinate of the search rectangle, then no rectangle in the left branch can ever intersect with the search rectangle and so can be pruned. Otherwise both branches should be traversed. See also interval tree, which is a 1-dimensional special case.
### Points only in leaves
It is also possible to define a k-d tree with points stored solely in leaves.[11] This form of k-d tree allows a variety of split mechanics other than the standard median split. The midpoint splitting rule[12] selects on the middle of the longest axis of the space being searched, regardless of the distribution of points. This guarantees that the aspect ratio will be at most 2:1, but the depth is dependent on the distribution of points. A variation, called sliding-midpoint, only splits on the middle if there are points on both sides of the split. Otherwise, it splits on point nearest to the middle. Maneewongvatana and Mount show that this offers "good enough" performance on common data sets. Using sliding-midpoint, an approximate nearest neighbour query can be answered in $O \left ( \frac{ 1 }{ { \epsilon\ }^d } \log n \right )$. Approximate range counting can be answered in $O \left ( \log n + { \left ( \frac{1}{ \epsilon\ } \right ) }^d \right )$ with this method.
## References
1. I. Wald and V. Havran. On building fast kd-trees for ray tracing, and on doing that in O(NlogN) On building fast kd-trees for ray tracing, and on doing that in O(NlogN). IEEE Symposium on Interactive Ray Tracing, pp. 61-69, 2006.
2. Friedman, Jerome H., Bentley, Jon Louis, Finkel, Raphael Ari (sep 1977). "An Algorithm for Finding Best Matches in Logarithmic Expected Time". ACM Trans. Math. Softw. (ACM) 3 (3): 209–226. doi:10.1145/355744.355745. ISSN 0098-3500. Retrieved 29 March, 2013.
3. Lee, D. T.; Wong, C. K. (1977). "Worst-case analysis for region and partial region searches in multidimensional binary search trees and balanced quad trees". Acta Informatica 9 (1): 23–29. doi:10.1007/BF00263763.
4. Jacob E. Goodman, Joseph O'Rourke and Piotr Indyk (Ed.) (2004). "Chapter 39 : Nearest neighbours in high-dimensional spaces". Handbook of Discrete and Computational Geometry (2nd ed.). CRC Press.
5. Cormen, Thomas H.; Leiserson, Charles E., Rivest, Ronald L.. . MIT Press and McGraw-Hill. Chapter 10.
6. Rosenberg J. Geographical Data Structures Compared: A Study of Data Structures Supporting Region Queries. IEEE Transaction on CAD Integrated Circuits Systems 4(1):53-67
7. Houthuys P. Box Sort, a multidimensional binary sorting method for rectangular boxes, used for quick range searching. The Visual Computer, 1987, 3:236-249
8. de Berg, Mark et al. Computational Geometry: Algorithms and Applications, 3rd Edition, pages 99-101. Springer, 2008.
9. S. Maneewongvatana and D. M. Mount. It's okay to be skinny, if your friends are fat. 4th Annual CGC Workshop on Computational Geometry, 1999.
|
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|
http://mathoverflow.net/questions/44045
|
## `Naturally occuring' $K(\pi, n)$ spaces, for $n \geq 2$.
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
[edited!] Given a group $\pi$ and an integer $n>1$, what are examples of Eilenberg-Maclane spaces $K(\pi, n)$ that can be constructed as "known" manifolds? (or if not a manifold, say some space people had a pre-existing desire to study before $K(\pi,n)$ spaces were identified as being of interest)
Constructing $K({\bf Z}, 2)$ as ${\bf CP}^{\infty}$ is the only example I know - but there must be more out there.
I'm interested in concrete examples (like the one above) that could, e.g., be given in a Topics grad course for topology students. They seem to be scarse, so it would be nice to know what was known.
Note: I've excluded $n=1$ because most people know examples (or can figure them out) in this case.
-
9
In what sense is $\mathbb{C} \mathbb{P}^{\infty}$ a manifold? – Pete L. Clark Oct 29 2010 at 2:17
1
I believe there's a theorem to the effect that they will not be finite-dimensional manifolds, so one necessarily needs to consider Frechet, Banach etc. manfolds. – David Roberts Oct 29 2010 at 2:38
5
Given a nice inclusive definition of "manifold" that allows some examples, what would be an example of a weak homotopy type that is not represented by a manifold? – Tom Goodwillie Oct 29 2010 at 2:46
10
Any countable simplicial complex is homotopy-equivalent to a Hilbert manifold. The idea is to inductively embed the skeleton into the Hilbert cube in such a way that you have a regular neighbourhood, making your simplicial complex homotopy-equivalent to the open regular neighbourhood -- and since it's open in Hilbert space it's a manifold. – Ryan Budney Oct 29 2010 at 2:49
4
There's a lovely paper of Kodama and Michor (2006) where they show that the component of $Imm(S^1,\mathbb R^2)/Diff^+(S^1)$ corresponding to the the figure-8 immersion has that homotopy-type of a $K(\mathbb Z,2)$. Here $Imm(S^1,\mathbb R^2)$ denotes immersions of $S^1$ in the plane, and we're modding out by orientation-preserving reparametrizations. – Ryan Budney Oct 29 2010 at 4:37
show 10 more comments
## 5 Answers
Following up on Dai's answer, one can go a step further since $P U(H)$ is obviously a group. So if we can find a contractible space on which it acts freely, the quotient will be the next level up (namely, a $K(\mathbb{Z},3)$.
Such a space can be constructed as follows: take our favourite (separable, though that's not necessary) Hilbert space, $H$, and consider $HS(H)$, the space of Hilbert-Schmidt operators on $H$. This is isomorphic to the Hilbert tensor product $H^* \widehat{\otimes} H$ so is a Hilbert space. Its unitary group is thus contractible. The group $U(H)$ acts on $HS(H)$ by conjugation, and once we divide out by the centre this becomes free. Thus $P U(H)$ acts on $U(HS(H))$ freely and so the quotient is a $K(\mathbb{Z},3)$.
However, as $P U(H)$ does not act centrally on $U(HS(H))$, the iteration stops here.
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That is very nice! – Andreas Thom Oct 29 2010 at 7:37
Very cool, thanks for the details. – Romeo Oct 30 2010 at 15:27
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Let BTOP and BPL be the classifying spaces of topological/PL-sphere bundles and $PL/TOP$ the homotopy fiber of the map $BPL \to BTOP$. The $PL/TOP$ is a model for a $K(\mathbb{Z}/2\mathbb{Z},3)$ by Kirby and Siebenmann. This identifies a third cohomology class as obstruction to get a PL-structure on a topological sphere bundle.
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Ah, I had thought there was some result like this, but had only very vague recollections about it. Are there any proofs known other than Kirby and Siebenmann's? Are there any people working on this kind of thing in this decade? All the results I know in this direction are quite "old". – Romeo Oct 30 2010 at 15:26
I think, in the book by Madsen and Milgram are some results of this sort. And I don't really know, what you mean, but while there's certainly less activity in the PL-world today than a few decades ago, at least topological manifolds are a topic, quite a few people are still working on. And the quoted result is surely essential to compare the topological, PL and smooth world. – Lennart Meier Oct 31 2010 at 21:12
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I was just wondering who was carrying on the Kirby-Siebenmann, Ranicki, et all torch in the 21st century. A lot of topology grad students I know these days have never really heard the word "PL"... – Romeo Nov 4 2010 at 0:50
The following example appears in the definition of twisted $K$-theory.
Let $H$ be an infinite dimensional separable Hilbert space over $\mathbb{C}$. Since the unitary group $U(H)$ is contractible, the projective unitary group $PU(H)= U(H)/S^1$ has the homotopy type of $K(\mathbb{Z},2)$. The fact that $BPU(H)\simeq K(\mathbb{Z},3)$ and the fact that $PU(H)$ acts on the space of Fredholm operators $\mathrm{Fred}(H)$ are essential in the definition of twisted $K$-theory.
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Argh, was too slow in my comment to the question. – David Roberts Oct 29 2010 at 5:38
Very nice. Is there a place you recommend for an exposition on this? – Romeo Oct 29 2010 at 6:28
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Atiyah and Segal's paper on Twisted K-theory is where I would start reading about this. – Andrew Stacey Oct 29 2010 at 6:58
The first one... – David Roberts Oct 29 2010 at 10:11
If $M$ is a hyperfinite type III factor, then (at least conjecturally), its group of outer automorphisms is a $K(\mathbb Z,3)$.
This is based on the following three properties of that von Neumann algebra:
• The group of unitary central elements of $M$ is a circle, and thus a $K(\mathbb Z,1)$.
• The group of unitaries in $M$ is contractible.
• The automorphism group of $M$ is contractible (conjectural).
To see that $Out(M)\cong K(\mathbb Z,3)$, apply the long exact sequence of homotopy groups to the following two fiber sequences: $$U(Z(M)) \to U(M) \to Inn(M)$$ $$Inn(M) \to Aut(M) \to Out(M)$$
As a consequence, we also get that $BOut(M)\cong K(\mathbb Z,4)$.
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This sounds interesting, but is new to me - any beginning references to recommend? – Romeo Nov 19 2010 at 19:07
Popa, Sorin; Takesaki, Masamichi; The topological structure of the unitary and automorphism groups of a factor. ams.org/mathscinet/search/… – André Henriques Nov 19 2010 at 21:33
There is a very nice model of $K(\mathbb Z,n)$ which is given by the free abelian topological group on the pointed space $(S^n,\star)$, let us call that $F(S^n,\star)$. An element in $F(S^n,\star)$ is given by a finite set of points in $S^n \setminus \lbrace\star\rbrace$ such that each point in this finite carries a non-zero integer as a label with the obvious addition. The topology is more subtle to describe and made in such a way that $F(S^n,\star)$ is an abelian topological group, the inclusion $S^n \subset F(S^n,\star)$ is continuous and $\star=0$ in $F(S^n,\star)$.
Though, I am not sure whether $F(S^n,\star)$ is an infinite-dimensional manifold (I think not), it is still pretty regular being a topological group and a CW-complex at the same time.
This is all very classical and was studied in detail in
Dold, Albrecht; Thom, René, Quasifaserungen und unendliche symmetrische Produkte., Ann. of Math. (2) 67 1958 239–281.
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1
More generally, if M is any Moore space (many of these are finite dimensional manifolds!) then taking the geometric realization of the free abelian group on the singular simplices of M will give you the corresponding Eilenberg-MacLane space. – Saul Glasman Oct 29 2010 at 13:13
Or you could take symmetric prodcuts of $S^n$ labelled by elements of any abelian group $A$: that produces $K(A,n)$. But is that "geometric"? – Johannes Ebert Oct 29 2010 at 14:00
@Saul: Wow, nice, I've never seen that construction before. Where does it come from? – Romeo Oct 30 2010 at 15:28
@Andreas, cool, thanks, this kind of thing is perfect (and definitely wouldn't have found that on my own). Never read a paper of Dold before... – Romeo Oct 30 2010 at 15:30
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http://math.stackexchange.com/questions/253758/is-x44-an-irreducible-polynomial
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# Is $x^4+4$ an irreducible polynomial?
We know that $p(x)=x^4-4=(x^2-2)(x^2+2)$ is reducible over $\mathbb{Q}$ even not having roots there.
What about $q(x)=x^4+4\in \mathbb{Q}[x]$? Again, no roots.
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6
Notice that $x^4 + 4 = (x^2+2)^2 - (2x)^2$. – Erick Wong Dec 8 '12 at 19:13
## 5 Answers
$$\begin{eqnarray}x^4+4&=&(x^2+2i)\cdot (x^2-2i)\\ &=& (x-(1-i))\cdot (x+(1-i))\cdot (x-(1+i))\cdot(x+(1+i)) \\ &=& ((x-1)+i)\cdot ((x-1)-i)\cdot((x+1)-i)\cdot((x+1)+i) \\ &=& ((x-1)^2+1)\cdot((x+1)^2+1).\end{eqnarray}$$
Reducible.
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Great and thanks. Just one more question: is there some irreducibility criterion to answer this without trying to factor it? – Sigur Dec 8 '12 at 13:57
@Sigur: sometimes you can use Eisenstein's criterion, but it is generally a hard problem. – akkkk Dec 8 '12 at 14:03
@akkkk, it does not apply here. Any else? – Sigur Dec 8 '12 at 14:04
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@Sigur: no general one, because that would probably give you an efficient algorithm for prime factorization. – akkkk Dec 8 '12 at 14:05
2
I edited your answer so that it wasn't all on one line. – Fredrik Meyer Dec 8 '12 at 14:16
show 1 more comment
As Berci showed, this polynomial is indeed reducible over the rationals. One way to see it is to calculate its roots explicitly: $$x^4+4=0 \leftrightarrow x^2 = \pm 2i \leftrightarrow x = \pm \sqrt{2} (\frac{\sqrt{2}}{2}(1+i)) \vee x = \pm i\sqrt{2} (\frac{\sqrt{2}}{2}(1+i))$$ Or: $$x = \pm 1 \pm i$$ And since those roots are proper complex number in $\mathbb{Z}[i]$, you can pair $1+i$ with $\overline{1+i}=1-i$ and $-1+i$ with $\overline{-1+i} = -1-i$ and obtain the factorization $(x^2 - 2x + 2)(x^2 + 2x +2)$ (if $\alpha$ is a proper complex root of $p \in \mathbb{R}[x]$, then $\overline{\alpha}$ is another root, and $(x-\alpha)(x-\overline{\alpha}) = (x^2-2Re(\alpha) + |\alpha|^2)$ divides $p$.
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1
Amazing how a simple question can produce many interesting facts. – Sigur Dec 8 '12 at 14:20
$x^4+4 \cdot 1^4= x^4+ 2 \cdot 2 \cdot x^2+2^2 - (2x)^2$
Which is well known identity called Sophie Germain
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One may use the same version of completing the square that proves that $x+\dfrac1x \ge 2$ when $x>0$:
$$x+\frac1x = \left(x-2+\frac1x\right)+2 = \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2+2.$$
Similarly $$x^4+4 = \left( x^4 +4x^2 + 4 \right) - 4x^2 = \left(x^2+2\right)^2 - (2x)^2$$ the factor that as a difference of two squares.
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$$X^4+4=X^4+4X^2+4-4X^2 =(X+2)^2-(2X)^2 \,.$$
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http://mathhelpforum.com/algebra/38724-airplane-wind-speed-problem-print.html
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# Airplane and Wind speed problem
Printable View
• May 18th 2008, 07:22 AM
scrattorpedo
Airplane and Wind speed problem
An airplane averaged 160 miles per hour with the wind and 112 miles per hour against the wind. Determine the speed of the plane and the speed of the wind. Is this possible to calculate with the info presented?
• May 18th 2008, 07:48 AM
earboth
Quote:
Originally Posted by scrattorpedo
An airplane averaged 160 miles per hour with the wind and 112 miles per hour against the wind. Determine the speed of the plane and the speed of the wind. Is this possible to calculate with the info presented?
Let $v_p$ denote the speed of the plane and $v_w$ the speed of the wind. Then you have:
$\left|\begin{array}{l}v_p+v_w=160 \\ v_p-v_w=112\end{array}\right.$
Solve this system of simultaneous equations. I've got $v_p = 136\ \frac{mi}{h}$ and $v_w = 24\ \frac{mi}{h}$
All times are GMT -8. The time now is 10:15 AM.
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http://mathhelpforum.com/geometry/88924-find-area-polygon.html
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# Thread:
1. ## Find the area of the polygon
How do I find the area of the polygon to these 4?????????? I'm so lost
Attached Thumbnails
2. For the one with the sides = 11 and 21, remeber that you have a triangle and a rectangle, that when added together yield the total area.
so, the area of the rectangle is obviously 11*21 and to find the area of the triangle you gotta use a little trig to find the base before you can apply the formula $A=\frac{1}{2}bh$. So how do we find the base. Easy $\tan{5}=\frac{21}{b}$, therefore, $b=\frac{21}{\tan{5}}$. Got it?
3. You can pretty much use the same reasoning to solve the rest of the polygons. You know what I mean?
4. nope, not a clue.
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http://www.physicsforums.com/showthread.php?p=3272485
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Physics Forums
## a subspace has finite codimension n iff it has a complementary subspace of dim nu
1. The problem statement, all variables and given/known data
A subspace N of a vector space V has finite codimension n if the quotient space V/N is finite-dimensional with dimension n. Show that a subspace N has finite codimension n iff N has a complementary subspace M of dimension n. Do not assume V to be finite-dimensional.
2. The attempt at a solution
Let $$\left\{N+\alpha_i \right\}$$ ($$1\leq i \leq n$$) be the basis of V/N, I want to show the set spanned by $$\alpha_i$$ is the complementary subspace M.
First I show V=N+M:
since $$\left\{N+\alpha_i \right\}$$ are the basis, each v in V can be represented as $$\eta+\sum x_i \alpha_i, \eta \in N$$
Next I prove $$N\bigcap M$$ = {0}:
if this is not the case, there must be some element in N that can be represented as $$\sum x_i \alpha_i$$. Since N is a subspace, this means $$\alpha_i$$ must be in N. Therefore, $$\left\{N+\alpha_i \right\}$$ cannot be a basis for V/N
Am I correct?
Thanks
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Quote by yifli since $$\left\{N+\alpha_i \right\}$$ are the basis, each v in V can be represented as $$\eta+\sum x_i \alpha_i, \eta \in N$$
there must be some element in N that can be represented as $$\sum x_i \alpha_i$$.
Why should such a representation exists? You only know that $$\{N+\alpha_i\}$$ is a basis for V/N. This does not mean that $$\alpha_i$$ is a basis for V! (which seems like you're using!)
For example: Take $$V=\mathbb{R}^2$$ and $$N=\mathbb{R}\times\{0\}$$. Then $$\{N+(0,1)\}$$ is a basis for V/N. But (0,1) is not a basis for $$\mathbb{R}^2$$.
[QUOTE=micromass;3272539]Why should such a representation exists? You only know that $$\{N+\alpha_i\}$$ is a basis for V/N. This does not mean that $$\alpha_i$$ is a basis for V! (which seems like you're using!) I try to prove the complementary subspace M in question is the space spanned by $$\alpha_i$$. In orde to do this, I need to show $$M\cap N$$={0}. So suppose $$v \in M\cap N$$ and $$v \neq 0$$, that's why I said v can be represented as $$\sum x_i \alpha_i$$
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## a subspace has finite codimension n iff it has a complementary subspace of dim nu
I see, I misunderstood your proof. It seems to be correct though!!
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http://physics.stackexchange.com/questions/39256/closure-phase-measurement-in-astronomy-for-vlbi-technique
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closure phase measurement in astronomy for vlbi technique?
I have read this page on closure phase and could not understand the following: \begin{align} \psi_1 & = \phi_1 + e_B - e_C \\ \psi_2 & = \phi_2 - e_B \\ \psi_3 & = \phi_3 - e_C. \end{align} If the error in $\phi_3$ is $e_C$ and the error in $\phi_2$ is $e_B$, then why should the error in $\phi_1$ be connected with the errors of $\phi_2$ and $\phi_3$ in the way given?
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1 Answer
It's because the interference phase is the difference between two antennas, it is the difference of a quantity in two different places. To starting out, there are three different shifts $e_A$, $e_B$ and $e_C$, which are the unknown shifts in time at the three locations. This gives three phase errors.
$$\psi_1 = \phi_1 + e_B - e_C$$ $$\psi_2 = \phi_2 + e_A - e_B$$ $$\psi_3 = \phi_3 + e_A - e_C$$
But you just define $e_A=0$, by using the time of the reception at A as your time coordinate. Now you have only two errors and three measurements.
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Especially useful if you can't measure phase directly, as in optical interferometry – Martin Beckett Oct 8 '12 at 4:15
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http://physics.stackexchange.com/questions/51772/how-can-i-determine-the-type-of-magnet-used-in-a-dc-motor
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# How can I determine the type of magnet used in a DC motor?
I repair electrical motors in a workshop and sometimes I must repair permanent magnet DC motors and other motors that use permanent magnets. I need to specify type of permanent magnet (Ferrite or Alnico or rare earth, etc.) by a simple method. Is there a way to do this?
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– anna v Jan 21 at 11:37
Measure the density and look at the physical appearance of the surface. I think those two factors would probably narrow it down quite a bit. If that's not enough, maybe measure the dipole moment based on deflection of a compass, and determine the ratio of dipole moment to mass, $D/m$. – Ben Crowell May 3 at 2:39
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http://mathhelpforum.com/trigonometry/198926-trig-periods.html
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# Thread:
1. ## Trig periods
How do i figure out which period the function is in?
Im doing the problem 3 tan2 x-1=0
i got tan x= 1sqrt 3 i have to find a solution to x. How do i know tan is on the interval 0, pi and not 0, 2pi?
2. ## Re: Trig periods
The period of a periodic function $f(x)$ is the smallest positive real number $\alpha$ such that $f(x+\alpha)=f(x)$ for all $x$ in the domain of the function. In case of $\tan,$ $\tan(x+\pi)=\tan x$ for all real $x$ and no positive real number $\alpha$ smaller than $\pi$ satisfies $\tan(x+\alpha)=\tan x$ for all $x.$ (It's true that $\tan(x+2\pi)=\tan x$ for all $x$ but $2\pi$ is not the smallest such positive real number.)
3. ## Re: Trig periods
Originally Posted by noork85
How do i figure out which period the function is in?
Im doing the problem 3 tan2 x-1=0
i got tan x= 1sqrt 3 i have to find a solution to x. How do i know tan is on the interval 0, pi and not 0, 2pi?
$3\tan^2{x} - 1 = 0$
$\tan{x} = \pm \frac{1}{\sqrt{3}}$
since the solution interval is not given, then all possible solutions are ...
$x = \frac{\pi}{6} + k\pi ; k \in \mathbb{Z}$
$x = -\frac{\pi}{6} + k\pi ; k \in \mathbb{Z}$
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# Tagged Questions
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### Spin Glass Transitions in Random Bond Ising Model (RBIM)
In brief, is there a list of spin glass transition properties for the RBIM on different lattices? Is there any know results about the relationships between these probabilities for a graph and its ...
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### Is this a known entropy formula?
While playing around with a variant of the one-dimensional Ising model with periodic boundary conditions I came up with a formula, let's call it $F$, whose form looks suspiciously like an entropy ...
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http://www.physicsforums.com/showthread.php?s=0076ac1cd845fd0bf344561eb63ff067&p=4207228
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Physics Forums
## Not seeing the action of a free particle in the Path Integral Formulation
In the very first example of Feynman and Hibb's Path Integral book, they discuss a free particle with
$$\mathcal{L} = \frac{m}{2} \dot{x}(t)^2$$
In calculating it's classical action, they perform a simple integral over some interval of time $t_a \rightarrow t_b$.
$$S_{cl} = \frac{m}{2} \int_{t_b}^{t_a} \dot{x}(t)^2 \; dt = \frac{m}{2} \frac{(x_b - x_a)^2}{t_b - t_a}$$
I don't see how that result follows! Is there some nifty integration by parts that I'm missing?
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Quote by Shmi In the very first example of Feynman and Hibb's Path Integral book, they discuss a free particle with $$\mathcal{L} = \frac{m}{2} \dot{x}(t)^2$$ In calculating it's classical action, they perform a simple integral over some interval of time $t_a \rightarrow t_b$. $$S_{cl} = \frac{m}{2} \int_{t_b}^{t_a} \dot{x}(t)^2 \; dt = \frac{m}{2} \frac{(x_b - x_a)^2}{t_b - t_a}$$ I don't see how that result follows! Is there some nifty integration by parts that I'm missing?
It follows when the time interval is infinitesimal, where the velosity is considered constant (xb-xa)/(tb-ta) during the integration interval.
Recognitions: Science Advisor The "classical action" in the path integral means the actual value of the classical action functional for the trajectory of the particle, i.e., the solutions of the equations of motion with the boundary conditions at hand. To evaluate the propagator in position representation $U(t_a,x_a;t_b,x_b)$ you need to solve the equation of motion, which is for the free particle $$m\ddot{x}=0$$ with the boundary conditions $$x(t_a)=x_a, \quad x(t_b)=x_b.$$ The general solution is of course $$x(t)=a t+b$$ with the two integration constants chosen such that the boundary conditions are fulfilled, i.e., $$x_a=a t_a+b, \quad x_b=a t_b +b,$$ which leads to $$a=\frac{x_a-x_b}{t_a-t_b}, \quad b=\frac{x_b t_a-x_a t_b}{t_a-t_b}.$$ The action along the trajectory thus is $$S[x_{\text{xl}}]=\int_{t_b}^{t_a} \mathrm{d} t \frac{m}{2} \dot{x}^2 = \int_{t_b}^{t_a} \mathrm{d} t \frac{m}{2} a^2 = \frac{m}{2} a^2(t_a-t_b)=\frac{m}{2} \frac{(x_a-x_b)^2}{t_a-t_b}.$$ So there is a sign error in your result. For more on the evaluation of the path integral, see my QFT script, where in the first chapter I deal with non-relativistic quantum theory in the path-integral formalism: http://fias.uni-frankfurt.de/~hees/publ/lect.pdf
## Not seeing the action of a free particle in the Path Integral Formulation
perhaps the integration is from ta to tb.
Recognitions: Science Advisor No, you have to read the propagator from right to left (as the usual S-matrix element is $S_{fi}$), i.e., $U(t_a,x_a;t_b,x_b)$ is the transition-probability amplitude for the particle, starting at $x_b$ at time $t_b$ to a $x_a$ at time $t_a$. So you have to integrate from $t_b$ to $t_a$.
I checked the book and it is right as op has written. @vanhees71-How much sure are you that particle goes from b to a and not from a to b.
Recognitions: Science Advisor Let's do the calculation of the free-particle propagator in the operator formalism, which is much more convenient than the path-integral calculation. The propagator is defined as the solution of the time-dependent Schrödinger equation (initial-value problem): $$\psi(t,x)=\int_{\mathbb{R}} \mathrm{d} x' U(t,x;t',x') \psi(t',x').$$ Let's use the Schrödinger picture, where the time dependence is fully at the state vectors, i.e., $$|\psi(t) \rangle=\exp[-\mathrm{i} \hat{H}(t-t')] |\psi(t') \rangle.$$ The observable operators and thus also their eigenvectors are time-independent. Thus we have $$U(t,x;t',x')=\langle x|\exp[-\mathrm{i} \hat{H}(t-t')]|x' \rangle.$$ For the free particle $$\hat{H}=\frac{\hat{p}^2}{2m}$$ and thus it's convenient to write this in terms of the momentum-eigenstates, for which we know $$\langle{x}|{p}\rangle=u_{p}(x)=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} x p)$$ and the completeness relation $$\int_{\mathbb{R}} \mathrm{d} p |p\rangle \langle p|=1.$$ Inserting this completeness relation in the expression for the propagator, we find $$U(t,x;t',x')=\int_{\mathbb{R}} \mathrm{d} p \langle x|\exp[-\mathrm{i} \hat{H}(t-t')]|p \rangle u_{p}^*(x') = \int_{\mathbb{R}} \mathrm{d} p \frac{1}{2 \pi} \exp \left [-\frac{\mathrm{i}}{2m} p^2 (t-t') \right ] \exp[\mathrm{i} p(x-x').$$ To make sense out of this integral, we have to regularize this expression by substituting $$(t-t') \rightarrow t-t'-\mathrm{i} \epsilon, \quad \epsilon>0.$$ Doing the Gaussian integral and letting $\epsilon \rightarrow 0^+$ afterwards yields $$U(t,x;t',x')=\sqrt{\frac{m}{2 \pi \mathrm{i}(t-t')}} \exp \left (\frac{\mathrm{i} m(x-x')^2}{2 (t-t')} \right ).$$ The sign is thus clearly as specified in my previous posting. In the path-integral evaluation of the propagator the exponential is given by the classical action. This is the simple part of the path integral approach. The somewhat tedious part is to get the prefactor which is the path integral over all paths with the homogeneous boundary conditions $x(t)=x(t')=0$. For this calculation you have to go back to the descretized form of the path integral, evaluate the multi-dimensional Gauß integral and then take the continuum limit. You find the calculation (for the only slightly more difficult case of the harmonic oscillator) in my QFT manuscript, http://fias.uni-frankfurt.de/~hees/publ/lect.pdf
it's all green function derivation is o.k..But I was saying that x-x' appears with a square term so it does not matter for whether x-x' or x'-x.But with time suppose you write ψ(t',x') in place of ψ(t,x) and ψ(t,x) in place of ψ(t',x'). ψ(t',x')=∫dx U(t',x';t,x)ψ(t,x) so now t'-t would appear.All this means that t' is later than t.There is no subtle problem with this.
Recognitions: Science Advisor Sure, but the sign is determined uniquely by the choice of the sign in the time-evolution operator. The usual convention since the very beginning of quantum theory is the one given in my previous posting. If you interchange the time arguments, of course you get the opposite sign. It's also clear that the time dependence cannot be symmetric unter exchange of the time, because intrinsically you introduce a direction of time, given by the causal sequence of events: You prepare the system in a certain state at a certain time and then determine its evolution in the future. The same time, the Hamiltonian as the operator representing energy must be bounded from below, and thus time-reversal invariance must necessarily be represented by an antiunitary rather than a unitary operator. This can be seen on the example of this propagator too: If you interchange $t$ and $t'$ (i.e., initial and final time), you must take the conjugate complex to get the same propagator as before. This complex conjugation is due to the antiunitary nature of the time-reflection operation. This is different for parity or space reflection. Here, the Heisenberg algebra forces a unitary reprsentation of this symmetry, and that's why the propagator is symmetric under exchange of $x$ and $x'$.
I fully agree.I have a question what is time reversal matrix for dirac eqn.
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http://mathoverflow.net/questions/38671/how-to-recover-partition-from-its-multiset-of-hook-lengths/38808
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## How to recover partition from its multiset of hook lengths?
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One of the invariants associated to a partition is its multiset of hook lengths. For instance, as shown here, the partition (5,4,1) has hook lengths {1,1,1,2,3,3,4,5,5,7}. Is there a good way to go backwards (up to conjugation, of course)? An actual algorithm that does not involve brute forcing through a bunch of possibilities, and would give me one (or all) of the possible partitions?
What about the decision problem "Is this multiset the multiset of hook lengths of a partition?" For instance, for {1, 1, 1, 2, 3, 3, 4, 4, 5, 7, 8, 10} the answer is no, but I can only come up with very ad hoc ways to show that.
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The hook length formula, discussed further down the wikipedia page you linked, provides one systematic check that rules out your example. If we have a partition of $n$, the product of the hook lengths must divide $n!$, which is clearly false in your example. – Paul Johnson Sep 14 2010 at 13:21
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How so? In this case the product of purported hook lengths is 806400, which does divide 12!. The purported dimension of the associated character of $S_12$ would then be 594 (indeed an integer), which does not occur as the dimension of an irreducible representation. So this is indeed another way to rule it out, but it does require to know the dimensions of all those representations (and even then, other examples could be constructed where such a test fails). – Paul-Olivier Dehaye Sep 14 2010 at 15:06
Indeed, that's embarassing. Mentally divided by 10! instead of just 10. – Paul Johnson Sep 15 2010 at 10:02
For the record, another ad hoc way I have to exclude {1, 1, 1, 2, 3, 3, 4, 4, 5, 7, 8, 10} is as follows. Only 2 cells are not in the first hook and they thus have hook lengths 1 and 2. 8 has to be next to 10, either aligned with the 2,1 we just placed or not. Both choices quickly lead to contradiction, as we now know exactly how long the arm based at 8 is. – Paul-Olivier Dehaye Sep 15 2010 at 16:24
## 3 Answers
the same hook length multiset can be shared by arbitrarily many partitions, see http://plms.oxfordjournals.org/cgi/reprint/96/1/26.pdf and the references there.
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True. This is something I suspected, and thanks for the link. So it does make an algorithm appear unlikely, as its task would be much harder. It doesn't help for the decision problem question, though... – Paul-Olivier Dehaye Sep 14 2010 at 12:04
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See also arxiv.org/abs/0709.0897 for a free version. – Igor Pak Sep 14 2010 at 16:20
### 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 wanted to at least give a systematic way to show your given multiset is not a set of hook lengths after my flubbed comment. So: taking the $n$ quotients of a partition gives us a constraint on its possible hook lengths.
In particular, take the set of all hook lengths that are divisible by $n$, and then divide each of them by $n$. This new set of numbers will be the set of hook lengths of the $n$ different partitions that are its $n$-quotients.
If your multiset came from a partition, then together its two 2-quotients would have hook lengths 1,2,2,4,5, which can't be the hook lengths of two partitions.
Alternatively, since 5 of the hook lengths were divisible by 2, together the two 2-quotients will be a partition of 5, and when translated back to the original partition will account for 10 of the 12. Therefore, the 2-core must have had size 2. But the 2-cores are exactly the staircase positions, and so 2 isn't a 2-core.
I wish I had a good source for explaining cores and quotients. I think of them by translating through the Maya diagrams , as in Figure 5 on Page 49 of this paper. Maybe The description here helps, too.
As far as an algorithm to construct the possible partitions with a given hook length set thinking in terms of Maya diagrams and possibly cores and quotients could possibly provide a useful way to control the searching and branching.
I can imagine an algorithm that starts by immediately taking just the even hook lengths and dividing them all by 2, to find the hook-lengths of the 2-quotients. Then we can find the 2-core if it exists. There would now be a lot of branching: we have to distribute the hook-lengths of the 2-quotients over all possible splittings. But once we've chosen a splitting, we can recursively call our program again on the 2-quotients, which will be much smaller partitions.
When it reaches the end, you then have to glue the original partition back together from the core and quotient and check whether the parition has the right hook lengths -- this seems quite expensive.
I make no claims that this algorithm is any good -- I don't know whether the division helps compared to the branching. But as someone who hasn't programmed much the recursion seems relatively easy to code.
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Yes. this is the "very ad hoc" ways I was referring in my original post. Actually, the argument does not require branching if you take k=4, since you then have the multiset {1,1,2} which is supposed to be the union of 4 partitions. Since I know what those are (empty, empty, [1], ([2] or $[2]^t$)) and I know the core as well (empty), I "only" have 2 * 4! /2 possiblities to check, corresponding to the different orderings of the quotients. The hooks have to satisfy tremendous divisibility conditions, and it is precisely to clarify this that I am asking the question... – Paul-Olivier Dehaye Sep 15 2010 at 16:03
One condition on hooks that hasn't been mentioned yet is that $$h_{(1,1)} + h_{(i,j)} = h_{(i,1)} + h_{(1,j)}.$$ (and similarly, based at other points).
When $i<>1$ and $j<>1$, this means that the cells $(1,1)$ and $(i,j)$ in that property are corners of a (unique) square, whose other two corners are the other two points.
Since 10+4 is not expressible as the sum of any two of the other given hook lengths in my negative example, we know that 4 and 4 are on the same line as 10 (not that suprising given the large size of first hook length 10 compared to the size 12, but could be useful for bigger partitions).
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http://www.reference.com/browse/be+identical
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Definitions
# Identical particles
Identical particles, or indistinguishable particles, are particles that cannot be distinguished from one another, even in principle. Species of identical particles include elementary particles such as electrons, as well as composite microscopic particles such as atoms and molecules.
There are two main categories of identical particles: bosons, which can share quantum states, and fermions, which are forbidden from sharing quantum states (this property of fermions is known as the Pauli exclusion principle.) Examples of bosons are photons, gluons, phonons, and helium-4 atoms. Examples of fermions are electrons, neutrinos, quarks, protons and neutrons, and helium-3 atoms.
The fact that particles can be identical has important consequences in statistical mechanics. Calculations in statistical mechanics rely on probabilistic arguments, which are sensitive to whether or not the objects being studied are identical. As a result, identical particles exhibit markedly different statistical behavior from distinguishable particles. For example, the indistinguishability of particles has been proposed as a solution to Gibbs' mixing paradox.
## Distinguishing between particles
There are two ways in which one might distinguish between particles. The first method relies on differences in the particles' intrinsic physical properties, such as mass, electric charge, and spin. If differences exist, we can distinguish between the particles by measuring the relevant properties. However, it is an empirical fact that microscopic particles of the same species have completely equivalent physical properties. For instance, every electron in the universe has exactly the same electric charge; this is why we can speak of such a thing as "the charge of the electron".
Even if the particles have equivalent physical properties, there remains a second method for distinguishing between particles, which is to track the trajectory of each particle. As long as we can measure the position of each particle with infinite precision (even when the particles collide), there would be no ambiguity about which particle is which.
The problem with this approach is that it contradicts the principles of quantum mechanics. According to quantum theory, the particles do not possess definite positions during the periods between measurements. Instead, they are governed by wavefunctions that give the probability of finding a particle at each position. As time passes, the wavefunctions tend to spread out and overlap. Once this happens, it becomes impossible to determine, in a subsequent measurement, which of the particle positions correspond to those measured earlier. The particles are then said to be indistinguishable.
## Quantum mechanical description of identical particles
### Symmetrical and antisymmetrical states
We will now make the above discussion concrete, using the formalism developed in the article on the mathematical formulation of quantum mechanics.
For simplicity, consider a system composed of two identical particles. As the particles possess equivalent physical properties, their state vectors occupy mathematically identical Hilbert spaces. If we denote the Hilbert space of a single particle as H, then the Hilbert space of the combined system is formed by the tensor product $H otimes H$.
Let n denote a complete set of (discrete) quantum numbers for specifying single-particle states (for example, for the particle in a box problem we can take n to be the quantized wave vector of the wavefunction.) Suppose that one particle is in the state n1, and another is in the state n2. What is the quantum state of the system? We might guess that it is
$|n_1rang |n_2rang$
which is simply the canonical way of constructing a basis for a tensor product space from the individual spaces. However, this expression implies that we can identify the particle with n1 as "particle 1" and the particle with n2 as "particle 2", which conflicts with the ideas about indistinguishability discussed earlier.
Actually, it is an empirical fact that identical particles occupy special types of multi-particle states, called symmetric states and antisymmetric states. Symmetric states have the form
$|n_1, n_2; Srang equiv mbox\left\{constant\right\} times bigg\left(|n_1rang |n_2rang + |n_2rang |n_1rang bigg\right)$
Antisymmetric states have the form
$|n_1, n_2; Arang equiv mbox\left\{constant\right\} times bigg\left(|n_1rang |n_2rang - |n_2rang |n_1rang bigg\right)$
Note that if n1 and n2 are the same, our equation for the antisymmetric state gives zero, which cannot be a state vector as it cannot be normalized. In other words, in an antisymmetric state the particles cannot occupy the same single-particle states. This is known as the Pauli exclusion principle, and it is the fundamental reason behind the chemical properties of atoms and the stability of matter.
### Exchange symmetry
The importance of symmetric and antisymmetric states is ultimately based on empirical evidence. It appears to be a fact of Nature that identical particles do not occupy states of a mixed symmetry, such as
$|n_1, n_2; ?rang = mbox\left\{constant\right\} times bigg\left(|n_1rang |n_2rang + i |n_2rang |n_1rang bigg\right)$
There is actually an exception to this rule, which we will discuss later. On the other hand, we can show that the symmetric and antisymmetric states are in a sense special, by examining a particular symmetry of the multiple-particle states known as exchange symmetry.
Let us define a linear operator P, called the exchange operator. When it acts on a tensor product of two state vectors, it exchanges the values of the state vectors:
$P bigg\left(|psirang |phirang bigg\right) equiv |phirang |psirang$
P is both Hermitian and unitary. Because it is unitary, we can regard it as a symmetry operator. We can describe this symmetry as the symmetry under the exchange of labels attached to the particles (i.e., to the single-particle Hilbert spaces).
Clearly, P² = 1 (the identity operator), so the eigenvalues of P are +1 and −1. The corresponding eigenvectors are the symmetric and antisymmetric states:
$P|n_1, n_2; Srang = + |n_1, n_2; Srang$
$P|n_1, n_2; Arang = - |n_1, n_2; Arang$
In other words, symmetric and antisymmetric states are essentially unchanged under the exchange of particle labels: they are only multiplied by a factor of +1 or −1, rather than being "rotated" somewhere else in the Hilbert space. This indicates that the particle labels have no physical meaning, in agreement with our earlier discussion on indistinguishability.
We have mentioned that P is Hermitian. As a result, it can be regarded as an observable of the system, which means that we can, in principle, perform a measurement to find out if a state is symmetric or antisymmetric. Furthermore, the equivalence of the particles indicates that the Hamiltonian can be written in a symmetrical form, such as
$H = frac\left\{p_1^2\right\}\left\{2m\right\} + frac\left\{p_2^2\right\}\left\{2m\right\} + U\left(|x_1 - x_2|\right) + V\left(x_1\right) + V\left(x_2\right)$
It is possible to show that such Hamiltonians satisfy the commutation relation
$left\left[P, Hright\right] = 0$
According to the Heisenberg equation, this means that the value of P is a constant of motion. If the quantum state is initially symmetric (antisymmetric), it will remain symmetric (antisymmetric) as the system evolves. Mathematically, this says that the state vector is confined to one of the two eigenspaces of P, and is not allowed to range over the entire Hilbert space. Thus, we might as well treat that eigenspace as the actual Hilbert space of the system. This is the idea behind the definition of Fock space.
### Fermions and bosons
The choice of symmetry or antisymmetry is determined by the species of particle. For example, we must always use symmetric states when describing photons or helium-4 atoms, and antisymmetric states when describing electrons or protons.
Particles which exhibit symmetric states are called bosons. As we will see, the nature of symmetric states has important consequences for the statistical properties of systems composed of many identical bosons. These statistical properties are described as Bose-Einstein statistics.
Particles which exhibit antisymmetric states are called fermions. As we have seen, antisymmetry gives rise to the Pauli exclusion principle, which forbids identical fermions from sharing the same quantum state. Systems of many identical fermions are described by Fermi-Dirac statistics.
Parastatistics are also possible.
In certain two-dimensional systems, mixed symmetry can occur. These exotic particles are known as anyons, and they obey fractional statistics. Experimental evidence for the existence of anyons exists in the fractional quantum Hall effect, a phenomenon observed in the two-dimensional electron gases that form the inversion layer of MOSFETs. There is another type of statistic, known as braid statistics, which are associated with particles known as plektons.
The spin-statistics theorem relates the exchange symmetry of identical particles to their spin. It states that bosons have integer spin, and fermions have half-integer spin. Anyons possess fractional spin.
### N particles
The above discussion generalizes readily to the case of N particles. Suppose we have N particles with quantum numbers n1, n2, ..., nN. If the particles are bosons, they occupy a totally symmetric state, which is symmetric under the exchange of any two particle labels:
$|n_1 n_2 cdots n_N; Srang = sqrt\left\{frac\left\{prod_j N_j!\right\}\left\{N!\right\}\right\} sum_p |n_\left\{p\left(1\right)\right\}rang |n_\left\{p\left(2\right)\right\}rang cdots |n_\left\{p\left(N\right)\right\}rang$
Here, the sum is taken over all possible permutations p acting on N elements. The square root on the right hand side is a normalizing constant. The quantity Nj stands for the number of times each of the single-particle states appears in the N-particle state.
In the same vein, fermions occupy totally antisymmetric states:
$|n_1 n_2 cdots n_N; Arang = frac\left\{1\right\}\left\{sqrt\left\{N!\right\}\right\} sum_p mathrm\left\{sgn\right\}\left(p\right) |n_\left\{p\left(1\right)\right\}rang |n_\left\{p\left(2\right)\right\}rang cdots |n_\left\{p\left(N\right)\right\}rang$
Here, sgn(p) is the signature of each permutation (i.e. +1 if p is composed of an even number of transpositions, and −1 if odd.) Note that we have omitted the ΠjNj term, because each single-particle state can appear only once in a fermionic state.
These states have been normalized so that
$lang n_1 n_2 cdots n_N; S | n_1 n_2 cdots n_N; Srang = 1, qquad lang n_1 n_2 cdots n_N; A | n_1 n_2 cdots n_N; Arang = 1.$
### Measurements of identical particles
Suppose we have a system of N bosons (fermions) in the symmetric (antisymmetric) state
$|n_1 n_2 cdots n_N; S/A rang$
and we perform a measurement of some other set of discrete observables, m. In general, this would yield some result m1 for one particle, m2 for another particle, and so forth. If the particles are bosons (fermions), the state after the measurement must remain symmetric (antisymmetric), i.e.
$|m_1 m_2 cdots m_N; S/A rang$
The probability of obtaining a particular result for the m measurement is
$P_\left\{S/A\right\}\left(n_1, cdots n_N rightarrow m_1, cdots m_N\right) equiv bigg|lang m_1 cdots m_N; S/A ,|, n_1 cdots n_N; S/A rang bigg|^2$
We can show that
$sum_\left\{m_1 le m_2 le dots le m_N\right\} P_\left\{S/A\right\}\left(n_1, cdots n_N rightarrow m_1, cdots m_N\right) = 1$
which verifies that the total probability is 1. Note that we have to restrict the sum to ordered values of m1, ..., mN to ensure that we do not count each multi-particle state more than once.
### Wavefunction representation
So far, we have worked with discrete observables. We will now extend the discussion to continuous observables, such as the position x.
Recall that an eigenstate of a continuous observable represents an infinitesimal range of values of the observable, not a single value as with discrete observables. For instance, if a particle is in a state |ψ>, the probability of finding it in a region of volume d³x surrounding some position x is
$|lang x | psi rang|^2 ; d^3 x$
As a result, the continuous eigenstates |x> are normalized to the delta function instead of unity:
$lang x | x\text{'} rang = delta^3 \left(x - x\text{'}\right)$
We can construct symmetric and antisymmetric multi-particle states out of continuous eigenstates in the same way as before. However, it is customary to use a different normalizing constant:
$|x_1 x_2 cdots x_N; Srang = frac\left\{prod_j N_j!\right\}\left\{N!\right\} sum_p |x_\left\{p\left(1\right)\right\}rang |x_\left\{p\left(2\right)\right\}rang cdots |x_\left\{p\left(N\right)\right\}rang$
$|x_1 x_2 cdots x_N; Arang = frac\left\{1\right\}\left\{N!\right\} sum_p mathrm\left\{sgn\right\}\left(p\right) |x_\left\{p\left(1\right)\right\}rang |x_\left\{p\left(2\right)\right\}rang cdots |x_\left\{p\left(N\right)\right\}rang$
We can then write a many-body wavefunction,
$Psi^\left\{\left(S\right)\right\}_\left\{n_1 n_2 cdots n_N\right\} \left(x_1, x_2, cdots x_N\right)$ $equiv lang x_1 x_2 cdots x_N; S | n_1 n_2 cdots n_N; S rang$
$= sqrt\left\{frac\left\{prod_j N_j!\right\}\left\{N!\right\}\right\} sum_p psi_\left\{p\left(1\right)\right\}\left(x_1\right) psi_\left\{p\left(2\right)\right\}\left(x_2\right) cdots psi_\left\{p\left(N\right)\right\}\left(x_N\right)$
$Psi^\left\{\left(A\right)\right\}_\left\{n_1 n_2 cdots n_N\right\} \left(x_1, x_2, cdots x_N\right)$ $equiv lang x_1 x_2 cdots x_N; A | n_1 n_2 cdots n_N; A rang$
$= frac\left\{1\right\}\left\{sqrt\left\{N!\right\}\right\} sum_p mathrm\left\{sgn\right\}\left(p\right) psi_\left\{p\left(1\right)\right\}\left(x_1\right) psi_\left\{p\left(2\right)\right\}\left(x_2\right) cdots psi_\left\{p\left(N\right)\right\}\left(x_N\right)$
where the single-particle wavefunctions are defined, as usual, by
$psi_n\left(x\right) equiv lang x | n rang$
The most important property of these wavefunctions is that exchanging any two of the coordinate variables changes the wavefunction by only a plus or minus sign. This is the manifestation of symmetry and antisymmetry in the wavefunction representation:
$$
Psi^{(S)}_{n_1 cdots n_N} (cdots x_i cdots x_jcdots) = Psi^{(S)}_{n_1 cdots n_N} (cdots x_j cdots x_i cdots)
$$
Psi^{(A)}_{n_1 cdots n_N} (cdots x_i cdots x_jcdots) = - Psi^{(A)}_{n_1 cdots n_N} (cdots x_j cdots x_i cdots)
The many-body wavefunction has the following significance: if the system is initially in a state with quantum numbers n1, ..., nN, and we perform a position measurement, the probability of finding particles in infinitesimal volumes near x1, x2, ..., xN is
$N! ; left|Psi^\left\{\left(S/A\right)\right\}_\left\{n_1 n_2 cdots n_N\right\} \left(x_1, x_2, cdots x_N\right) right|^2 ; d^\left\{3N\right\}!x$
The factor of N! comes from our normalizing constant, which has been chosen so that, by analogy with single-particle wavefunctions,
$int!int!cdots!int; left|Psi^\left\{\left(S/A\right)\right\}_\left\{n_1 n_2 cdots n_N\right\} \left(x_1, x_2, cdots x_N\right)right|^2 d^3!x_1 d^3!x_2 cdots d^3!x_N = 1$
Because each integral runs over all possible values of x, each multi-particle state appears N! times in the integral. In other words, the probability associated with each event is evenly distributed across N! equivalent points in the integral space. Because it is usually more convenient to work with unrestricted integrals than restricted ones, we have chosen our normalizing constant to reflect this.
Finally, it is interesting to note that that antisymmetric wavefunction can be written as the determinant of a matrix, known as a Slater determinant:
$Psi^\left\{\left(A\right)\right\}_\left\{n_1 cdots n_N\right\} \left(x_1, cdots x_N\right)$
= frac{1}{sqrt{N!}} left| begin{matrix} psi_{n_1}(x_1) & psi_{n_1}(x_2) & cdots & psi_{n_1}(x_N) psi_{n_2}(x_1) & psi_{n_2}(x_2) & cdots & psi_{n_2}(x_N) cdots & cdots & cdots & cdots psi_{n_N}(x_1) & psi_{n_N}(x_2) & cdots & psi_{n_N}(x_N) end{matrix} right|
## Statistical properties
### Statistical effects of indistinguishability
The indistinguishability of particles has a profound effect on their statistical properties. To illustrate this, let us consider a system of N distinguishable, non-interacting particles. Once again, let nj denote the state (i.e. quantum numbers) of particle j. If the particles have the same physical properties, the nj's run over the same range of values. Let ε(n) denote the energy of a particle in state n. As the particles do not interact, the total energy of the system is the sum of the single-particle energies. The partition function of the system is
$Z = sum_\left\{n_1, n_2, cdots n_N\right\} expleft\left\{ -frac\left\{1\right\}\left\{kT\right\} left\left[epsilon\left(n_1\right) + epsilon\left(n_2\right) + cdots epsilon\left(n_N\right) right\right] right\right\}$
where k is Boltzmann's constant and T is the temperature. We can factorize this expression to obtain
$Z = xi^N$
where
$xi = sum_n expleft\left[- frac\left\{epsilon\left(n\right)\right\}\left\{kT\right\} right\right].$
If the particles are identical, this equation is incorrect. Consider a state of the system, described by the single particle states [n1, ..., nN]. In the equation for Z, every possible permutation of the n's occurs once in the sum, even though each of these permutations is describing the same multi-particle state. We have thus over-counted the actual number of states.
If we neglect the possibility of overlapping states, which is valid if the temperature is high, then the number of times we count each state is approximately N!. The correct partition function is
$Z = frac\left\{xi^N\right\}\left\{N!\right\}.$
Note that this "high temperature" approximation does not distinguish between fermions and bosons.
The discrepancy in the partition functions of distinguishable and indistinguishable particles was known as far back as the 19th century, before the advent of quantum mechanics. It leads to a difficulty known as the Gibbs paradox. Gibbs showed that if we use the equation Z = ξN, the entropy of a classical ideal gas is
$S = N k ln left\left(Vright\right) + N f\left(T\right)$
where V is the volume of the gas and f is some function of T alone. The problem with this result is that S is not extensive - if we double N and V, S does not double accordingly. Such a system does not obey the postulates of thermodynamics.
Gibbs also showed that using Z = ξN/N! alters the result to
$S = N k ln left\left(frac\left\{V\right\}\left\{N\right\}right\right) + N f\left(T\right)$
which is perfectly extensive. However, the reason for this correction to the partition function remained obscure until the discovery of quantum mechanics.
### Statistical properties of bosons and fermions
There are important differences between the statistical behavior of bosons and fermions, which are described by Bose-Einstein statistics and Fermi-Dirac statistics respectively. Roughly speaking, bosons have a tendency to clump into the same quantum state, which underlies phenomena such as the laser, Bose-Einstein condensation, and superfluidity. Fermions, on the other hand, are forbidden from sharing quantum states, giving rise to systems such as the Fermi gas. This is known as the Pauli Exclusion Principle, and is responsible for much of chemistry, since the electrons in an atom (fermions) successively fill the many states within shells rather than all lying in the same lowest energy state.
We can illustrate the differences between the statistical behavior of fermions, bosons, and distinguishable particles using a system of two particles. Let us call the particles A and B. Each particle can exist in two possible states, labelled $|0rangle$ and $|1rangle$, which have the same energy.
We let the composite system evolve in time, interacting with a noisy environment. Because the $|0rangle$ and $|1rangle$ states are energetically equivalent, neither state is favored, so this process has the effect of randomizing the states. (This is discussed in the article on quantum entanglement.) After some time, the composite system will have an equal probability of occupying each of the states available to it. We then measure the particle states.
If A and B are distinguishable particles, then the composite system has four distinct states: $|0rangle|0rangle$, $|1rangle|1rangle$, $|0rangle|1rangle$, and $|1rangle|0rangle$. The probability of obtaining two particles in the $|0rangle$ state is 0.25; the probability of obtaining two particles in the $|1rangle$ state is 0.25; and the probability of obtaining one particle in the $|0rangle$ state and the other in the $|1rangle$ state is 0.5.
If A and B are identical bosons, then the composite system has only three distinct states: $|0rangle|0rangle$, $|1rangle|1rangle$, and $frac\left\{1\right\}\left\{sqrt\left\{2\right\}\right\}\left(|0rangle|1rangle + |1rangle|0rangle\right)$. When we perform the experiment, the probability of obtaining two particles in the $|0rangle$ state is now 0.33; the probability of obtaining two particles in the $|1rangle$ state is 0.33; and the probability of obtaining one particle in the $|0rangle$ state and the other in the $|1rangle$ state is 0.33. Note that the probability of finding particles in the same state is relatively larger than in the distinguishable case. This demonstrates the tendency of bosons to "clump."
If A and B are identical fermions, there is only one state available to the composite system: the totally antisymmetric state $frac\left\{1\right\}\left\{sqrt\left\{2\right\}\right\}\left(|0rangle|1rangle - |1rangle|0rangle\right)$. When we perform the experiment, we inevitably find that one particle is in the $|0rangle$ state and the other is in the $|1rangle$ state.
The results are summarized in Table 1:
Table 1: Statistics of two particles
Particles Both 0 Both 1 One 0 and one 1
Distinguishable 0.25 0.25 0.5
Bosons 0.33 0.33 0.33
Fermions 0 0 1
As can be seen, even a system of two particles exhibits different statistical behaviors between distinguishable particles, bosons, and fermions. In the articles on Fermi-Dirac statistics and Bose-Einstein statistics, these principles are extended to large number of particles, with qualitatively similar results.
## The homotopy class
To understand why we have the statistics that we do for particles, we first have to note that particles are point localized excitations and that particles that are spacelike separated do not interact. In a flat d-dimensional space M, at any given time, the configuration of two identical particles can be specified as an element of M × M. If there is no overlap between the particles, so that they do not interact (at the same time, we are not referring to time delayed interactions here, which are mediated at the speed of light or slower), then we are dealing with the space [M × M]/{coincident points}, the subspace with coincident points removed. (x,y) describes the configuration with particle I at x and particle II at y. (y,x) describes the interchanged configuration. With identical particles, the state described by (x,y) ought to be indistinguishable (which ISN'T the same thing as identical!) from the state described by (y,x). Let's look at the homotopy class of continuous paths from (x,y) to (y,x). If M is Rd where $dgeq 3$, then this homotopy class only has one element. If M is R2, then this homotopy class has countably many elements (i.e. a counterclockwise interchange by half a turn, a counterclockwise interchange by one and a half turns, two and a half turns, etc, a clockwise interchange by half a turn, etc). In particular, a counterclockwise interchange by half a turn is NOT homotopic to a clockwise interchange by half a turn. Lastly, if M is R, then this homotopy class is empty. Obviously, if M is not isomorphic to Rd, we can have more complicated homotopy classes...
What does this all mean?
Let's first look at the case d $geq$ 3. The universal covering space of [M × M]/{coincident points}, which is none other than [M × M]/{coincident points} itself, only has two points which are physically indistinguishable from (x, y), namely (x, y) itself and (y, x). So, the only permissible interchange is to swap both particles. Performing this interchange twice gives us (x, y) back again. If this interchange results in a multiplication by +1, then we have Bose statistics and if this interchange results in a multiplication by −1, we have Fermi statistics.
Now how about R2? The universal covering space of [M × M]/{coincident points} has infinitely many points which are physically indistinguishable from (x,y). This is described by the infinite cyclic group generated by making a counterclockwise half-turn interchange. Unlike the previous case, performing this interchange twice in a row does not lead us back to the original state. So, such an interchange can generically result in a multiplication by exp(iθ) (its absolute value is 1 because of unitarity...). This is called anyonic statistics. In fact, even with two DISTINGUISHABLE particles, even though (x, y) is now physically distinguishable from (y, x), if we go over to the universal covering space, we still end up with infinitely many points which are physically indistinguishable from the original point and the interchanges are generated by a counterclockwise rotation by one full turn which results in a multiplication by exp(iφ). This phase factor here is called the mutual statistics.
As for R, even if particle I and particle II are identical, we can always distinguish between them by the labels "the particle on the left" and "the particle on the right". There is no interchange symmetry here and such particles are called plektons.
The generalization to n identical particles doesn't give us anything qualitatively new because they are generated from the exchanges of two identical particles.
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http://mathoverflow.net/questions/69035/the-category-of-l-adic-sheaves
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## The category of l-adic sheaves
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I'm currently trying to understand the construction of the category of l-adic constructible sheaves as in SGA5, and it seems that quite a lot of machinery (the MLAR condition, localization of the category of projective systems, etc.) has to be gone through before one can even construct this category and show that it's abelian, for instance. On the other hand it is not even true that the derived category of l-adic sheaves is defined in the obvious manner, since it is defined as a 2-limit of the derived categories of $\mathbb{Z}/l^n$ constructible sheaves.
I understand that the categorical machinery in existence today is a lot more powerful than it was in the 1970's, which makes me curious: is there a cleaner and more transparent way of doing this, and a more modern presentation than SGA or Frietag-Kiehl?
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9
Other than Ekedahl, "On the adic formalism", I recommend Laszlo-Olsson, "The six operations..., I, II", which made use of Gabber's recent result on finiteness in étale cohom. Their results are more general in that: 1. stacks, 2. over a base more general than a field with finite l-cd and 3. for unbounded derived categories. Note that SGA5 didn't do derived category: just l-adic sheaves. Same issue with SGA4.5 Rapport and Th. finitude; note that under the title "Th. de finitude en cohomologie l-adique" the results are not stated for adic case, but of course this happened after SGA5. – shenghao Jun 28 2011 at 20:36
Thanks for the references! – Akhil Mathew Jun 29 2011 at 0:52
Not sure how helpful this is, but I'm pretty sure the constructions in SGA only give the "correct" answer when the ground field is either finite or algebraically closed. So one might hope not only for a "cleaner and more transparent way" but also greater generality. – Justin Campbell Nov 12 2011 at 18:46
## 2 Answers
Section 1.4 in these notes of Brian Conrad is as nice as one could hope for, given the dryness of the adic formalism. I don't think the material differs substantially from Frietag-Kiehl, except in that the presentation is much cleaner. For the derived category stuff, the notes refer to Behrend's paper "Derived $\ell$-adic Categories for Algebraic Stacks" which I haven't looked at really, but a brief skim suggests it contains everything you might desire (constructions in extremely general situations), but nonetheless includes examples (!).
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If you ask Behrend for giving you an example, it will end up in sort of new prime numbers, which will be called Behrend's prime numbers! – Ehsan M. Kermani Nov 12 2011 at 10:06
Thanks. I had seen Conrad's notes earlier, but not Behrend's; this is very nice. – Akhil Mathew Nov 12 2011 at 13:43
I'm just going to point out, incidentally, that the (derived?) $\infty$-category of $\mathbb{Z}_l$-sheaves is apparently the homotopy limit (in the Joyal model structure) of the category of $\mathbb{Z}/l^n$-sheaves. This is one thing that I'd like to understand eventually, but I haven't seen it written up anywhere. – Akhil Mathew Nov 12 2011 at 13:44
1
@Akhil: Perhaps one would prove that (or, say, the $\mathbb{Q}_{\ell}$-analogue) by showing both sides are the derived categories of perverse sheaves, the classical definition having been done by Beilinson and the $\infty$-categorical version amenable to a similar treatment. But it would definitely be nice to have it written up. The limit Deligne takes appears (to me) to work only out of "good luck," (since it's not a real operation in the world of triangulated categories) whereas the homotopy limit is much more natural. – Moosbrugger Nov 12 2011 at 16:26
@Moosbrugger: I'm intrigued by this refinement of Beilinson's theorem, but have never heard it before. (It seems, though, to construct the abelian category perverse sheaves, one must first construct the derived category of $l$-adic sheaves the normal way -- so it would be some kind of bootstrapping.) – Akhil Mathew Nov 13 2011 at 1:52
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Zheng and Liu are using $\infty$-categories to study constructible sheaves on stacks, and they have a $\ell$-adic version too. (Though most of the details for the $\ell$-Adic version should appear in a second paper that is still in preparation, and I would not call their first paper easy to read. But it is certainly a modern presentation...) Reference : http://math.columbia.edu/~zheng/bc1.pdf
By the way, they use Gabber's finiteness results, and there is now a nice reference for these too ! (This is really cool.) http://www.math.polytechnique.fr/~orgogozo/travaux_de_Gabber/
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Very interesting. Thanks for these links. – Akhil Mathew Apr 10 2012 at 6:19
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http://terrytao.wordpress.com/2008/01/25/book-version-of-the-blog/
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What’s new
Updates on my research and expository papers, discussion of open problems, and other maths-related topics. By Terence Tao
# Book version of the blog
25 January, 2008 in admin, book | Tags: announcement, meta
It’s now been almost a year since I moved my “What’s new?” page from my home page to this blog. Since then, I’ve been quite happy with the directions this blog has been headed in (most of which I had not anticipated when I started), and also with the level of feedback, some of which has been extremely informative to me (and, I hope, to other readers as well).
Anyway, after discussing things with some of my friends and colleagues, I have decided to convert some of the posts here from 2007 into a book format, in order to place some of the mathematical content here in a more formal and traditional context (with accurate citations and references, etc.). After some thought, I decided not to transcribe all of my posts from last year (there are 93 of them!), but instead to restrict attention to those articles which (a) have significant mathematical content, (b) are not announcements of material that will be published elsewhere, and (c) are not primarily based on a talk given by someone else. As it turns out, this still leaves about 33 articles from 2007, leading to a decent-sized book of a couple hundred pages in length. For various reasons (including legal reasons), I have decided not to incorporate the comments to each post directly into the book format, although corrections, mention of relevant references, etc. will be added with acknowledgments in the endnotes to each article.
I’ve converted a couple articles into a book format, and also created a table of contents, to see what it would look like. (The format comes from the American Mathematical Society, with whom I am planning to publish the book.) It looks like it will be relatively straightforward to convert the rest (at least compared to writing books from scratch, which I know from experience to be quite time-consuming!). I’ll of course post updates here on this blog when the book is closer to completion.
At present, the structure and content of the book is still rather flexible; like all things related to this blog, it is an experiment. As such, I am open to suggestions on these matters. (For instance, I do not have any particularly imaginative title for the book, other than “What’s new – 2007″. )
## 25 comments
Everyone wants you to release the book under a creative commons license.
Perhaps a better title would be
What was new in 2007
because what was new in 2007 isn’t new in 2008!
Dear Prof. Tao,
that’s a very nice idea, and I would too suggest using a CC licence (or at least have an online copy available for private use, like Allen Hatcher’s topology book).
Also do you have a way of regularly archiving the whole blog content including comments (just in case the wordpress servers break down, it would be a huge loss!).
For the title I’d suggest to add an informative subtitle, and to think about it from a long-term perspective. For instance will there be such a book every year? If so will the $n+1$ book always systematically report on any significant advances made on the open problems presented in years $1$ to $n$ even if it is not the subject of a post of yours in year $n$? And so forth.
26 January, 2008 at 1:32 am
Attila Smith
Youpi!
which is French for
“What an excellent piece of news. I’m looking forward to buying this book
and devouring it, comfortably sitting in my best armchair, sipping cup after cup of moka coffee, Brahms gently played in the background ”
(French is a very concise language.)
26 January, 2008 at 4:44 am
Przemyslaw Chojecki
Superb idea. Would be also wonderful to see in the future an ongoing course in Dynamical Systems of yours in a book format, enriched maybe with solutions of some of exercises. Is it a chance for it?
Best Regards
26 January, 2008 at 11:42 am
DeadWolfe
I second the idea of making the dynamical systems notes into a book.
Hi,
Will the book be sold in bookstores?
KM
26 January, 2008 at 6:15 pm
Anonymous
Good job!I think the many will appreciate your work.
Dear all: thanks for the suggestions so far!
A subtitle does indeed resolve my concerns about the title being somewhat obscure (and, as one commenter pointed out, somewhat anachronistic).
I have not planned far ahead enough to see what to do about the (as yet mostly non-existent) 2008 posts, including the 254A lecture series notes, but presumably if the conversion of the 2007 posts proceeds painlessly and successfully then I will be encouraged to repeat the process in the future (though perhaps not exactly on an annual basis, depending on how the blog evolves; the idea of publishing one book a year, on top of all my other duties, does seem slightly excessive).
The reminder to backup this blog is also a good suggestion, and one which I will now do on a regular basis. [It turns out that it was just a matter of clicking a button from a menu; see
http://faq.wordpress.com/2005/09/28/backups/
. I now have a 981KB XML file which seems to contain all the essential content of the blog as of today.]
Judging by how my other books have been faring (with the possible exception of my problem solving book), I would doubt that you would see this book physically in a store, but with the magic of on-line retailing I would imagine that availability is unlikely to be a concern (as with other books published by the AMS). In any case, the original blog posts that the book is going to be almost entirely based on will be available at this site.
I am not an expert in licensing issues, but it seems that the usual academic model (in which the content of one’s publications can be used freely by later authors, with proper citation and attribution of course) would suffice here. (As said above, I am restricting the book to my own content, as opposed to that generated from the comments or from colleagues; the book will be based on the blog, but will not be an exact replica of it.)
I hope that this book will include the “career advice” posts, which I have found very useful in understanding how at least one “real mathematician” perceives mathematics. (I’m not sure if you consider them part of the blog proper, which is why I’m making this comment.)
Or, at the very least, I hope that more people read those posts. Such writings — by you and by other mathematicians — have made my transition from a student to a researcher (which is still in progress) easier to handle. The book venue may not be the right place for that, since it’s not obvious who exactly the market for the book is, and it may or may not include beginning graduate students, who are the ones who I’d think would benefit from such advice.
(I do not know whether this belongs here or on the original entry on the non-commutative Freiman theorem, which I did not read at the time; feel free to move it, if necessary.)
You say that the growth of subsets of G=SL_2(F_p) which do *not* generate A has not been studied. As it happens, any subset of G that does not generate A is either in a dihedral group or in a Borel subgroup. It should take about two seconds to deduce a Freiman theorem for a dihedral group from the Freiman theorem over Z/nZ.
The case of subsets of a Borel is more interesting. I think I told you (on this blog) some time ago that the “right” formulation of the sum-product theorem is not over fields – or, rather, the sum-product theorem is a “shadow” of a more general result about groups with commuting automorphisms. That same general result has an even more direct application to Borel subgroups. You see, the maximal torus T (the set of diagonal matrices, say) in a Borel subgroup B (the set of upper triangular matrices) acts by conjugation on the nilpotent subgroup U consisting of the unipotent elements of B (the set of upper triangular matrices). The actions commute because T is abelian.
(I am leaving out an important technical condition that is fulfilled trivially for SL_2 (as opposed to SL_n).)
The consequence is that, if a subset A of a Borel contains at least M diagonal elements and at least one unipotent element other than the identity, it will grow by a factor of at least M^{epsilon}, unless almost all the unipotents are already in the set. One can get plenty of diagonal elements by the same method I used in my SL_2 paper – unless A lies in a very small number of cosets of the unipotents.
There are plenty of little corners left over, as you can see. Would you think it worthwhile to write the complete “non-commutative Freiman theorem” for SL_2 down in its entirety? Do you know whether somebody is already doing that? I may be getting a graduate student soon, and this might be a nice warm-up problem; I do not know what you think.
In general, I’d expect something much like Freiman’s theorem for unipotent groups; I would guess growth is forced for subsets of non-solvable groups that do not lie in a solvable subgroup; and, finally, I’d expect growth in solvable groups to be
described by a mildly complicated version of Freiman’s theorem when the
set of eigenvalues of elements of the set A is small, and to be forced otherwise. It seems clear to me that the second task is the hardest (given that we already know Freiman over abelian groups). Even the third task might not be hard.
Tell me if there is something you know that does not fit in this division.
Dear Harald,
It would certainly be of interest for someone (such as your student) to write up a complete non-commutative Freiman theorem for $SL_2({\Bbb F}_p)$; I do not know of anyone who is already doing this. Besides the issue of subgroups, there is also the issue of what happens for very dense sets A (of size $p^{3-o(1)}$). Presumably what happens here is that iterated products of A soon fill up all of $SL_2$, because of the lack of low-dimensional representations (i.e. quasirandomness). In any case, sorting all this out would be an excellent data point with which to test various candidates for the conjectured “non-commutative Freiman theorem”, at least for nice classical groups such as $SL_n$.
Dear Isabel: Thanks for the suggestion. At this point it will depend to some extent on how much space everything else is going to take up. The case for making a print version of the career advice pages is a little less compelling to me than for the mathematically oriented pages, since the the latter is likely to continue to evolve slowly over time and is not really in need of a stable (and citeable) version. So I may save it for a future lean year in the blog if there is not enough other material to make a decent book.
Incidentally, I doubt that I will keep issuing updates to articles from year n in a book for year n+m; this is the kind of thing which is much better done in the online blog format than in the static book format.
Dear Terry,
Gowers worked out (in effect) what happens for very dense sets A: as pointed out by Nikolov and Pyber, the result in his paper on Quasirandom Groups has the very nice consequence that, when |A|>p^{2.5+epsilon} or so, A*A*A is the entire group SL_2(F_p).
From a talk by Babai, I know that Babai, Nikolov and Pyber now have a beautiful and completely general result. They work out that A*A*A is in fact equal to G for sufficiently large subsets of G, where G is any group that lacks low-dimensional representations. Here “sufficiently large” is tight. Moreover, they give bounds on the number of copies of A that are needed to cover G for softer senses of “sufficiently large”.
Dear Terry,
I would also like to hear your opinion on publishing the book under CC license and why do you have this opinion.
Thanks. :)
Oops sorry, I didn’t notice that you’ve already answered to this question.
29 January, 2008 at 3:29 pm
Roberta
Congratulations!!
“Why global regularity for Navier-Stokes is hard” and “The parity problem in sieve theory” are my favourite ones.
But why not “Einstein’s derivation of E=mc^2″? I really like this post. ( (a) restriction? =P)
Dear Roberta,
I drafted the contents list before the Einstein post (the last post of 2007!). But thanks for reminding me to put it in. I hope to put an updated version of the book here in a few weeks.
[...] Just recently, my friend Pablo left a comment on this blog with a link to mathematician’s Terence Tao’s blog, where he explains the process of making a book out of his blog posts. A professor of the [...]
If it is not secret what efficient method are you thinking of to convert the blog-formatted files to tex format?
Dear t8m8r,
I am mostly relying on plain old search-and-replace, which takes care of a lot of the conversion, but of course many finer details (particularly in converting links to references) have to be taken care of by hand (and I want to go through the material carefully in any case).
Thanks a lot. I am just starting a blog and thought it would be nice to have tex-version of some of the posts. At the same time I was a bit lazy to convert by hand. Your answer “in any case I want to go through the material carefully” kind of settles the issue for me too.
[...] In fact, let’s start with Terry Tao’s blog, which is on WordPress. In case you didn’t know, Terry Tao is an eminent and talented young mathematician who is extremely web-savvy. He decided to move from a static web-site to a blog for his news and discussion articles about a year ago. Original he moved to blogspot, but then soon moved on to WordPress, because it gave better LaTeX support. Last month he told us that he has decided to publish a book version of his blog. [...]
[...] April, 2008 in book, non-technical, update A few months ago, I announced that I was going to convert a significant fraction of my 2007 blog posts into book format. For [...]
[...] Tao is reworking some of his better blogposts into a book, to be published by the AMS (here’s a preliminary [...]
[...] Just recently, my friend Pablo left a comment on this blog with a link to mathematician’s Terence Tao’s blog, where he explains the process of making a book out of his blog posts. A professor of the [...]
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http://mathoverflow.net/users/10534?tab=recent
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# Derek Holt
6,838 Reputation
899 views
## Registered User
Name Derek Holt
Member for 2 years
Seen 3 hours ago
Website
Location University of Warwick
Age 64
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| 14h | comment | Why are Schur multipliers of finite simple groups so small?I remember asking Michael Aschbacher a similar question (I think it was actually why are the outer automorphism groups of all finite simple groups solvable) many years ago, and he opined that questions like that were pointless. Many common properties of the finite simple groups are just consequences of the classification. |
| 20h | comment | Are residually finite, perfect groups residually alternating?I don't think that example is perfect. |
| 20h | answered | Are residually finite, perfect groups residually alternating? |
| 2d | comment | How to find quotients of infinite triangle groups or von Dyck groups?Finite quotients of $G(2,3,7)$ are known as Hurwitz groups and have been much studied. It is known, for example, that the alternating group $A_n$ is a Hurwitz group for all sufficiently large $n$. |
| 2d | comment | A catalog of faithful representations of finite groups? |
| May15 | comment | A catalog of faithful representations of finite groups?It's also worth observing that the smallest dimensional faithful representation is not always irreducible (for a noncyclic abelian group for example) so, even if you know the character table, then the computations are not always completely routine. |
| May15 | comment | A catalog of faithful representations of finite groups?Perhaps you need to specify more precisely exactly what information you are looking for, and for which which groups. |
| May15 | comment | A catalog of faithful representations of finite groups?But the Atlas of Finite Groups only has information about groups that are "close" to being nonabelian simple. If all of the "noteworthy finite groups" have this property, then fine, but it would be no use at all for the many other types of interesting finite groups. |
| May13 | awarded | ● Nice Answer |
| May13 | accepted | about the non-solvable group of order $120$ |
| May13 | answered | about the non-solvable group of order $120$ |
| May11 | comment | The role of the Automatic Groups in the history of Geometric Group TheoryI am not aware of any specific examples at present. Many such examples in the past have turned out not to work, which may explain why people have given up! |
| May11 | answered | The role of the Automatic Groups in the history of Geometric Group Theory |
| May9 | comment | Generators of sections of free groupsThese numbers can certainly depend on the isomorphism type of $F/H$ (and not just on its order). |
| May9 | comment | Generators of sections of free groupsThe answer is yes, in the sense that there are algorithms to solve these problems, and it would not be particularly difficult to write programs in a language like GAP or Magma to do so. Is this what you are looking for, or is this more of a theoretical question? |
| May7 | awarded | ● Nice Answer |
| May2 | comment | Extensions with trivial induced outer actionThe assumptions are equivalent to $E = NC_E(N)$. |
| Apr28 | revised | element of order n such that $\pi(n)=\pi(G)$, where $\pi(n)$ denote the prime divisors of $n$Added group-theory tag |
| Apr28 | comment | element of order n such that $\pi(n)=\pi(G)$, where $\pi(n)$ denote the prime divisors of $n$You can easily constuct non-nilpotent examples by taking a direct product of an arbitrary group with a cyclic group of the required order. But I would be surprised if there were any nonabelian simple groups with this property. |
| Apr26 | comment | Do quasi convex hyperbolic subgroups remain quasi convex after adding redundant generators?@Alexey: Quantifiers! 1 says every geodesic, where 2 says there exists a geodesic. The subgroup $\langle xy \rangle$ of the free abelian group with free generators $x,y$ is qc using 2 but not using 1. |
| Apr25 | comment | Do quasi convex hyperbolic subgroups remain quasi convex after adding redundant generators?Definition 3 is interesting when the combing is regular and part of an automatic structure of $G$. In that case, the combing words that lie in $H$ form a regular set, and it is often not difficult, starting from the automatic structure of $G$, to compute a finite state automaton to recognise membership of combing words $H$. Handling intersections of such subgroups can then be done by standard operations on finite state automata. |
| Apr18 | comment | Normal subgroups of finite index in free groupsI am afraid that misread the definition of $H_{n,s}$ in your question. I took the definition to be $F_s/N$ where $N$ is the intersection of all normal subgroups of index exactly $n$ (rather than at most $n$, which is what you wrote). So Khalid Bou-Rabee's answer is more useful than mine for the question that you asked! |
| Apr18 | answered | Normal subgroups of finite index in free groups |
| Apr18 | comment | 2-sylow subgroupsYou take a set of permutations that generate a Sylow $2$-subgroup of the symmetric group $S_{2^{r-1}}$.Then make them into permutation matrices of degree $2^r$ that permute $2 \times 2$ block matrices. These permutation matrices, together with a Sylow $2$−subgroup of `${\rm GL}_2(q)$` acting on one of the $2 \times 2$ blocks, generate a Sylow $2$−subgroup of ${\rm GL}_{2^r}(q)$. |
| Apr17 | accepted | Classification of Special $p$-Groups |
| Apr17 | comment | Classification of Special $p$-GroupsI haven't heard of such a conjecture. Wuth current techniques it seems that there will always be a nonconstant error term in the exponent. So we cannot really justify conjecturing that "almost all $p$-groups are special", although that statement is somehow true logarithmically. |
| Apr17 | answered | Classification of Special $p$-Groups |
| Apr13 | comment | wreath product and matrix presentationI am not convinced that $H$ is uniquely defined. The wreath product is an associative operation on permutation groups, but not on abstract groups, and $Z_2$ looks like an abstract group rather than a permutatino group. |
| Apr12 | comment | The automorphisms of a 2-group of nilpotency class 2I'm impressed! I don't suppose many of us were knowledgeable enough to make that mistake! |
| Apr11 | comment | The automorphisms of a 2-group of nilpotency class 2In any nilpotent group of class 2, we have $[ab,c]=[a,c][b,c]$ and $[a,b]^{-1} =[b,a]$, so the commutator map is bilinear and alternating, and we get an induced map $G/Z(G) \times G/Z(G) \to Z(G)$. If $|Z(G)|=2$, then $[a^2,b]=1$ for all $a,b$, so $a^2 \in Z(G)$ and $G/Z(G)$ is elementary abelian. So $G$ is extraspecial. |
| Apr11 | accepted | The automorphisms of a 2-group of nilpotency class 2 |
| Apr11 | answered | The automorphisms of a 2-group of nilpotency class 2 |
| Apr10 | comment | When is Ad(pi) an irreducible representation ?No. It is a subgroup of ${\rm GL}_3({\mathbb R})$, which comes from the adjoint of the 2-dimensional representation of $2.A_5$. |
| Apr10 | comment | When is Ad(pi) an irreducible representation ?I think you want $G=2.A_5 = {\rm SL}_2(5)$ rather than $A_5$. |
| Apr10 | answered | When is Ad(pi) an irreducible representation ? |
| Apr10 | comment | Finite subgroups of $PGL(3,K)$ Yes you are right, I had miscalculated the character field! |
| Apr9 | answered | Finite subgroups of $PGL(3,K)$ |
| Apr6 | accepted | Nilpotency class of a certain finite 2-group |
| Apr1 | awarded | ● Enlightened |
| Apr1 | accepted | Are there any nontrivial ring homomorphisms $M_{n+1}(R)\rightarrow M_n(R)$? |
| Mar31 | awarded | ● Nice Answer |
| Mar31 | answered | Are there any nontrivial ring homomorphisms $M_{n+1}(R)\rightarrow M_n(R)$? |
| Mar30 | comment | Automorphism classes of the free groupYou can do this using the Whitehead Algorithm. More generally, it can be decided whether there is an automorphism mapping one $k$-tuple of elements onto another. |
| Mar29 | comment | P-group with abelian centralzerI understand what $cs(G)$ means but I do not understand "has at least three integer". Do you just mean $|cs(G)| \ge 3$. If so, why not write that? |
| Mar29 | comment | P-group with abelian centralzerIt is not clear what you mean by "$cs(G)$ has at least three integer". |
| Mar22 | awarded | ● Enlightened |
| Mar22 | awarded | ● Nice Answer |
| Mar20 | comment | Are all dinilpotent groups solvable, i.e., groups G=AB with nilpotent subgroups A, B ?As far as I know, this is still an open question. Why do you think that there is a counterexample? |
| Mar17 | comment | Subgroups of the general linear groupsIn general ${\rm GL}_{2n}(q)$ has an abelian subgroup of order $(q-1)q^{n^2}$. |
| Mar12 | comment | What are the known algorithms for computing the inverse of a group automorphism?It's an interesting question for free groups. When the number of generators i not too big (up to about 8), the Whitehead algorithm is quick, but it is not effective when there are more generators. I suspect that the modified coset enumeration procedure might be better. |
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http://mathoverflow.net/questions/1726/how-should-i-think-about-b-fields/1850
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## How should I think about B-fields?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
So, physicists like to attach a mysterious extra cohomology class in H^2(X;C^*) to a Kahler (or hyperkahler) manifold called a "B-field." The only concrete thing I've seen this B-field do is change the Fukaya category/A-branes: when you have a B-field, you shouldn't take flat vector bundles on a Lagrangian subvariety, but rather ones whose curvature is the B-field. How should I think about this gadget?
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## 5 Answers
Let me add a few words of explanation to Aaron's comment. Perturbative string theory is (at least at the level of caricature) concerned with describing small corrections to classical gravitational physics on the spacetime X. So, to do perturbative string theory on X, you need to choose a "background" metric on X. You might need to choose other fields as well, but we can assume for now that those are all set to zero.
Having chosen a metric, you can talk about strings moving in X. In the limit where the string length goes to zero, a single string will look like a particle. What sort of particle it looks like will depend on how it's vibrating inside X. In particular, a closed string has a set of vibrational states which a) appear massless in this limit, and b) fill out a representation R of the Lorentz group. Specifically, R is the representation induced from the tensor square V (x) V, where V is the standard representation of the little group that fixes some light-like vector. You can decompose V into a sum of traceless symmetric square, trace, and antisymmetric traceless square. The states in the first summand are states of the graviton, representing tiny quantum excitations of the metric in X. The states in the last summand, the antisymmetric representation, are tiny excitations of the B-field, which we set equal to zero. (The states in the trace representation are quanta of the "dilaton" field.)
So, we didn't give the B-field any respect when we started, but it turns out to part of the definition of a string background. And once you know about the B-field, it's easy to include it in the action for the sigma model to X: Add to your action the term i<[S],f*B>, where [S] is the fundamental class of the Riemann surface, and f: S -> X is the function embedding your string's worldsheet into X. Edit: Forgot a factor of i=root(-1), which is necessary to make the action real.
And I forgot to mention that Aaron's H is dB.
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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.
We like to do more than that, actually. The B-field is an element in the differential cohomology class $\check{H}^3(M)$, or, more geometrically, a connection on an abelian gerbe. Thus, there is a class $[H] \in H^3(M,Z)$ characterizing the gerbe. In the B-model, this twists the derived category. The connection is the part that changes the A-model, and when $[H] = 0$, you exactly get that the differential cohomology group is $H^2(X,U(1))$. In the geometric language, it's a flat connection on a trivial gerbe.
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I'm tempted to ask "Waht is a gerbe?", but the AMS Notices already answered: ams.org/notices/200302/what-is.pdf (and thanks for the nice answer!) – Konrad Voelkel Jun 18 2010 at 8:03
Let me try to add a different point of view on B-fields and mirror symmetry. Ideally in mirror symmetry, given a Calabi-Yau manifold X, you would like to "construct" its mirror X', where the symplectic form on X should give you the complex structure on X'. As already mentioned, classes of symplectic forms have moduli of real dimension $h^{1,1}(X)$ and complex structures on X' have moduli of complex dimension $h^{2,1}(X') = h^{1,1}(X)$. So the kahler class is not enough to determine all complex structures on X'. In the context of the Strominger-Yau-Zaslow conjecture there is a nice interpretation of the B-field. Suppose X = $T^*B / \Lambda$, where B is a smooth manifold and $\Lambda$ is locally the span over the integers of 1-forms $dy_1$, ..., $dy_n$ (here $y_1$, ..., $y_n$ are coordinates which change with affine transformations from one chart to the other). Then $X$ has a standard symplectic form. We can consider $X'= TB / \Lambda'$, where $\Lambda'$ is the dual lattice. Then X' has a natural complex structure defined as follows. In standard coordinates on TB, given by $(y,x)$ --> $x \partial_y$, the complex coordinates on X' are $z_k = e^{2\pi i(x_k + i y_k)}$, which are well defined due to the nature of the coordinates x and y. But the above complex coordinates can be twisted locally (on a coordinate patch) by $z_k (b) = e^{2\pi i(x_k + b_k + i y_k)}$, where $b = (b_1, \ldots, b_n)$ is some local data. But since on overlaps $U_i \cap U_j$ the coordinates have to match, we must have $b(i) - b(j) \in \Lambda$. It turns out that by putting $b_{ij} = b(i) - b(j)$ on overlaps, we get a cohomology class in $H^{1}(B, \Lambda)$, this is the B-field. The cohomology group $H^{1}(B, \Lambda)$ shoud coincide (in some cases at least) with $H^2(X, R/Z)$, which is what Kevin Lin mentioned. The elliptic curve case (mentioned by Kevin) can be seen from this point of view.
This point of view is also called "mirror symmetry without corrections" and it only approximates what happens in compact Calabi-Yaus. I have learned this in papers by Mark Gross (such as "Special lagrangian fibrations II: geometry") or the book "Calabi-Yau manifolds and related geometries" by Gross, Huybrechts and Joyce.
I would be interested to know how this interpretation connects to the other ones which have been described.
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Just to supplement Aaron's and A.J.'s comments: nLab:Kalb-Ramond field
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In general, you should think about B-field in quantum field theory as providing some noncommutativity.
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More details pls? – Kevin Lin Oct 23 2009 at 2:01
For physics point of view, e.g. arxiv.org/abs/hep-th/9912100. I'm not expert so others should know better about math books explaining this. – Ilya Nikokoshev Oct 23 2009 at 3:44
You might also want to look at the paper arXiv.org/pdf/hep-th/9903205 by Schomerus. He considers bosonic string theory in a D-brane sector (i.e., Dirichlet boundary conditions on some directions). If the ambient background has no B-field, the vertex operator algebra of the associated conformal field theory contains the commutative algebra of functions on the D-brane. Upon introduction of a background constant B-field, the algebra of functions deforms and in the limit where the B-field dominates, one recovers Kontsevich's expression for the Moyal product. – José Figueroa-O'Farrill Oct 26 2009 at 10:52
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http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.aoms/1177728845
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### A Single-Sample Multiple Decision Procedure for Ranking Means of Normal Populations with known Variances
Robert E. Bechhofer
Source: Ann. Math. Statist. Volume 25, Number 1 (1954), 16-39.
#### Abstract
This paper is concerned with a single-sample multiple decision procedure for ranking means of normal populations with known variances. Problems which conventionally are handled by the analysis of variance (Model I) which tests the hypothesis that $k$ means are equal are reformulated as multiple decision procedures involving rankings. It is shown how to design experiments so that useful statements can be made concerning these rankings on the basis of a predetermined number of independent observations taken from each population. The number of observations required is determined by the desired probability of a correct ranking when certain differences between population means are specified.
First Page:
Full-text: Open access
Permanent link to this document: http://projecteuclid.org/euclid.aoms/1177728845
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http://unapologetic.wordpress.com/2012/08/07/linear-lie-algebras/?like=1&source=post_flair&_wpnonce=c1c9c39881
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# The Unapologetic Mathematician
## Linear Lie Algebras
So now that we’ve remembered what a Lie algebra is, let’s mention the most important ones: linear Lie algebras. These are ones that arise from linear transformations on vector spaces, ’cause mathematicians love them some vector spaces.
Specifically, let $V$ be a finite-dimensional vector space over $\mathbb{F}$, and consider the associative algebra of endomorphisms $\mathrm{End}(V)$ — linear transformations from $V$ back to itself. We can use the usual method of defining a bracket as a commutator:
$\displaystyle [x,y]=xy-yx$
to turn this into a Lie algebra. When considered as a Lie algebra like this, we call it the “general linear Lie algebra”, and write $\mathfrak{gl}(V)$. Many Lie algebras are written in the Fraktur typeface like this.
Any subalgebra of $\mathfrak{gl}(V)$ is called a “linear Lie algebra”, since it’s made up of linear transformations. It turns out that every finite-dimensional Lie algebra is isomorphic to a linear Lie algebra, but we reserve the “linear” term for those algebras which we’re actually thinking of having linear transformations as elements.
Of course, since $V$ is a vector space over $\mathbb{F}$, we can pick a basis. If $V$ has dimension $n$, then there are $n$ elements in any basis, and so our endomorphisms correspond to the $n\times n$ matrices $\mathrm{Mat}_n(\mathbb{F})$. When we think of it in these terms, we often write $\mathfrak{gl}(n,\mathbb{F})$ for the general linear Lie algebra.
We can actually calculate the bracket structure explicitly in this case; bilinearity tells us that it suffices to write it down in terms of a basis. The standard basis of $\mathfrak{gl}(n,\mathbb{F})$ is $\{e_{ij}\}_{i,j=1}^n$ which has a $1$ in the $i$th row and $j$th column and $0$ elsewhere. So we can calculate:
$\displaystyle [e_{ij},e_{kl}]=\delta_{jk}e_{il}-\delta_{li}e_{kj}$
where, as usual, $\delta_{ij}$ is the Kronecker delta: $1$ if the indices are the same and $0$ if they’re different.
We can now identify some important subalgebras of $\mathfrak{gl}(n,\mathbb{F})$. First, the strictly upper-triangular matrices $\mathfrak{n}(n,\mathbb{F})$ involve only the basis elements $e_{ij}$ with $i<j$. If $i<j=k<l$ so the first term in the above expression for the bracket shows up, then the second term cannot show up, and vice versa. Either way, we conclude that the bracket of two basis elements of $\mathfrak{n}(n,\mathbb{F})$ — and thus any element of this subspace — involves only other basis elements of the subspace, which makes this a subalgebra.
Similarly, we conclude that the (non-strictly) upper-triangular matrices involving only $e_{ij}$ with $i\leq j$ also form a subalgebra $\mathfrak{t}(n,\mathbb{F})$. And, finally, the diagonal matrices involving only $e_{ii}$ also form a subalgebra $\mathfrak{d}(n,\mathbb{F})$. This last one is interesting, in that the bracket on $\mathfrak{d}(n,\mathbb{F})$ is actually trivial, since any two diagonal matrices commute.
As vector spaces, we see that $\mathfrak{t}(n,\mathbb{F})=\mathfrak{d}(n,\mathbb{F})+\mathfrak{n}(n,\mathbb{F})$. It’s easy to check that the bracket of a diagonal matrix and a strictly upper-triangular matrix is again strictly upper-triangular — we write $[\mathfrak{d}(n,\mathbb{F}),\mathfrak{n}(n,\mathbb{F})]=\mathfrak{n}(n,\mathbb{F})$ — and so we also have $[\mathfrak{t}(n,\mathbb{F}),\mathfrak{t}(n,\mathbb{F})]=\mathfrak{n}(n,\mathbb{F})$. This may seem a little like a toy example now, but it turns out to be surprisingly general; many subalgebras will relate to each other this way.
### Like this:
Posted by John Armstrong | Algebra, Lie Algebras
## 7 Comments »
1. [...] More examples of Lie algebras! Today, an important family of linear Lie algebras. [...]
Pingback by | August 8, 2012 | Reply
2. [...] the next three families of linear Lie algebras we equip our vector space with a bilinear form . We’re going to consider the endomorphisms [...]
Pingback by | August 9, 2012 | Reply
3. [...] we say — again in analogy with groups — that is abelian; this is the case for the diagonal algebra , for instance. Abelian Lie algebras are rather boring; they’re just vector spaces with [...]
Pingback by | August 13, 2012 | Reply
4. [...] obviously useful class of examples arises when we’re considering a linear Lie algebra . If is an invertible endomorphism of such that then the map is an automorphism of . Clearly [...]
Pingback by | August 18, 2012 | Reply
5. [...] is a very general phenomenon: if is any linear Lie algebra and is nilpotent, then conjugation by the exponential of is the same as applying the exponential [...]
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6. [...] some explicit examples, we look back at the algebras and . The second, as we might guess, is nilpotent, and thus solvable. The first, [...]
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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://physics.stackexchange.com/questions/tagged/quantum-mechanics
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# Tagged Questions
Quantum mechanics describes the microscopic properties of nature in a regime where classical mechanics no longer applies. It explains phenomena such as the wave-particle duality, quantization of energy and the uncertainty principle and is generally used in single body systems. Use the ...
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Bell's inequality defines a lower bound for agreement/disagreement between entangled particles. When the experiment is conducted it shows lower odds. What does this tell us? Is it possible that we ...
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### Change of basis in non-linear Schrodinger equation
At the mean-field level, the dynamics of a polariton condensate can be described by a type of nonlinear Schrodinger equation (Gross-Pitaevskii-type), for a classical (complex-number) wavefunction ...
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### Prove that the position operator is $\hat{x} = i\hbar \frac{d}{{dp}}$ in the momentum representation [closed]
Proof that: $x = i\hbar \frac{d}{{dp}}$ I did this, could you tell me if I am false or true \$\begin{array}{l} x{e^{\frac{{ipx}}{\hbar }}} = - i\hbar \frac{{d{e^{\frac{{ipx}}{\hbar }}}}}{{dp}} = ...
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### Is it possible to use quantum mechanics for an effective time based encryption?
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I'm slightly confused as to answer this question, someone please help: Consider a free particle in one dimension, described by the initial wave function \psi(x,0) = ...
3answers
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0answers
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### Question regarding operators and cylindrical coordinates
I have the following problem in my hand: I need to arrive from the Cartesian expression $$x_{j}{\partial_{k}}x_{j}{\partial_{k}}-x_{j}{\partial_{k}}x_{k}{\partial_{j}}$$ to this expression: ...
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### Does quantum mechanics depend solely on electromagnetic waves? [duplicate]
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### Matrix representation of state
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### How does a state vector change under an exchange of a boson and a fermion?
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1answer
68 views
### Can 3 photons be combined to give a spin-0 projection?
Motivation: The neutral pion decays to 2 photons ($\pi^0\to\gamma\gamma$) most of the time. For the decay of the neutral to 3 photons ($\pi^0\to 3\gamma$) we have an upper limit on the branching ...
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### How can superstring theories unify general relativity and quantum theory when no prediction can be made?
I am a newbie to superstring theories, but I came into this question: so superstring theories purport to unify general relativity and quantum theory. However, there is yet no definitive way to test ...
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### Grover algorithm $R_D$ Circuit
I need sketch two circuits to understand Grover algorithm. The first is the operator $R_f$ and another is the operator $R_D = H^{\otimes n}(2|0\rangle\langle0|-I)H^{\otimes n}$. I get the first ...
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### Why does quantum cryptography give us uncrackable codes?
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### What is three-photon interference?
Whilst reading this paper on a quantum processor that performs a type of matrix computation, I came across the concept of 'three-photon interference'. A quick Google search shows that this process is ...
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### Stark Effect on the 1st excited state of Hydrogen
I know the ground state of hydrogen is unaffected by the Stark effect to first order. And I also know that the 1st excited state is split from 4 degenerate states to 2 distinct, and 1 degenerate state ...
1answer
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### How do we know superposition exists?
How do we know superposition exists? Has it been observed, or has it been deduced, and how certain are we? The Copenhagen Interpretation seems to imply that superposition collapses into one state ...
1answer
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### Do electrons need specific energies to excite electrons
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2answers
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### Expanding two-variable function $f(x,y)$ over the complete sets $\{ g_{i}(x) \}$ and $\{ h_{j}(y) \}$
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### Why doublons and holons are not bounded in spin-1/2 Hubbard chain?
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1answer
84 views
### Some Dirac notation explanations
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### Quantum entanglement and speed of light $c$
On the topic of quantum entanglement, Wikipedia states: Repeated experiments have verified that this works even when the measurements are performed more quickly than light could travel between the ...
2answers
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### Your Mass is NOT from Higgs Boson
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2answers
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### Electron in an infinite potential well
Does this problem have any sense? Suppose an electron in an infinite well of length $0.5nm$. The state of the system is the superposition of the ground state and the first excited state. Find the ...
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### Motivation for Wigner Phase Space Distribution
Most sources say that Wigner distribution acts like a joint phase-space distribution in quantum mechanics and this is justified by the formula ...
3answers
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### Why is the Dirac equation not used for calculations?
From what I understand the Dirac equation is supposed to be an improvement on the Schrödinger equation in that it is consistent with relativity theory. Yet all methods I have encountered for doing ...
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### Degeneracy of orbitals in magenetic field
Why is that in an external magnetic field(uniform) the degeneracy of d,f orbitals is lost but the degeneracy of p orbitals remain intact assuming the main cause of losing degeneracy is the difference ...
3answers
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### Complex energy eigenstates of the harmonic oscillator
Given the Hamiltonian for the the harmonic oscillator (HO) as $$\hat H=\frac{\hat P^2}{2m}+\frac{m}{2}\omega^2\hat x^2\,,$$ the Schroedinger equation can be reduced to: \left[ ...
2answers
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### $\langle B|A \rangle$ expressed in terms of the Partition Function
Say you have an electron departing from point A and reaching poing B after a time t. According to some helping friend, the Partition Function for that electron going from point A to B can be written ...
1answer
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### Which is this formula Feynman talks about in the QED book?
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1answer
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### What experiments have been proposed to discriminate between interpretations of quantum mechanics?
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2answers
78 views
### Are there problems solvable with Newtonian physics, GR and QM?
First I must let you know that I don't have much understanding of neither GR nor quantum mechanics, and therefore this question. I've mentally pictured Newtonian physics, GR and quantum mechanics all ...
1answer
82 views
### Field operator eigenvalues
For an harmonic oscillator we can write the Hamiltonian eigenvalues in the basis of the amplitude eigenvalues : for example the ground state is a gaussian : $⟨x|0⟩=a.e^{-b.x^{2}}$. I was wondering ...
2answers
107 views
### How do we know that $\psi$ is the eigenfunction of an operator $\hat{H}$ with eigenvalue $W$?
I am kind of new to this eigenvalue, eigenfunction and operator things, but I have come across this quote many times: $\psi$ is the eigenfunction of an operator $\hat{H}$ with eigenvalue $W$. ...
1answer
45 views
### Is the de Broglie wavelength of a photon equal to the EM wavelength of the radiation?
Is the de Broglie (matter) wavelength $\lambda=\frac{h}{p}$ of a photon equal to the electromagnetic wavelength of the radiation? I guess yes, but how come that photons have both a matter wave and an ...
1answer
41 views
### Why doesn't intensity of light affect the emission of electrons?
So electrons of specific atoms have a minimum amount of energy needed to escape the atom, called the work function, W. Now let's say that you emit a certain frequency of light, and $hf<W$. However, ...
0answers
75 views
### Topological band theory [closed]
Why topological insulators were discovered so late? While the band theory was known long time ago! I mean why the topological properties of electronic bands were not noticed in the past?
1answer
46 views
### Possible states for two electrons in the helium atom
Consider the helium atom with two electrons, but ignore coupling of angular momenta, relativistic effects, etc. The spin state of the system is a combination of the triplet states and the singlet ...
1answer
37 views
### localized electrons in the crystals
Why electrons in low lying levels of individual atoms stay localized in their own atoms in a crystal? Doesn't this contradict Bloch's theorem?
2answers
143 views
### Where does quantum mechanics come from? [closed]
Where does quantum mechanics come from? If string theory is proved to be the correct quantum theory of gravity but it failed to explain where quantum mechanics came from can we still consider it a ...
3answers
587 views
### Is the universe fundamentally deterministic?
I'm not sure if this is the right place to ask this question. I realise that this maybe a borderline philosophical question at this point in time, therefore feel free to close this question if you ...
0answers
49 views
### Studying QM without math and physics background [duplicate]
I rode all posted answers about this topic but i need to ask you another information. I have done a semester course called "Principle of Physics" (i am studying Biotechnology) and one called ...
0answers
46 views
### What is the difference between various fields of physics? [closed]
what is the difference between the fields of physics? like high energy physics, particle physics, cosmology, quantum physics, quantum mechanics, experimental physics, theoretical physics, applied ...
1answer
92 views
### Positivity in the Pauli/Bloch/coherence vector representation
Suppose $\rho$ is an $n$-qubit state and $\vec{x}$ is a vector of coefficients in the Pauli representation (also called the Bloch or coherence vector). That is $$x_k = {\rm Tr}(\rho \sigma_k),$$ ...
0answers
51 views
### The gauge-invariance of the probability current
It is simple to show that under the gauge transformation \begin{cases}\vec A\to\vec A+\nabla\chi\\ \phi\to\phi-\frac{\partial \chi}{\partial t}\\ \psi\to \psi ...
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http://www.physicsforums.com/showthread.php?p=2339239
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Physics Forums
## differential equation problem :(
1. The problem statement, all variables and given/known data
I'm given a function acceleration a=-1.5*s, where s is a position. I need to find a in terms of t.
2. Relevant equations
3. The attempt at a solution
I know that a is also equal to d^2*s/dt^2. Therefore d^2*s/dt^2 is equal to -1.5*s. By dividing by s and multiplying by dt^2 I get d^2*s/s=-1.5*dt^2. At this point I'm not sure what to do. If I can figure out how to get rid of the dt^2 and the d^2*s, then I can probably solve the rest of the problem. I imagine I need to integrate but am not sure how that would work out.
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Quote by yoamocuy 1. The problem statement, all variables and given/known data I'm given a function acceleration a=-1.5*s, where s is a position. I need to find a in terms of t. 2. Relevant equations 3. The attempt at a solution I know that a is also equal to d^2*s/dt^2. Therefore d^2*s/dt^2 is equal to -1.5*s. By dividing by s and multiplying by dt^2 I get d^2*s/s=-1.5*dt^2. At this point I'm not sure what to do. If I can figure out how to get rid of the dt^2 and the d^2*s, then I can probably solve the rest of the problem. I imagine I need to integrate but am not sure how that would work out.
You cannot simply 'divide' and 'multiply' differentials like that. I know many non-mathematical sciences courses will tell you that it is fine to do so, but it is simply wrong.
I see that you are attempting to separate the variables, however, this method cannot generally be used for second order differential equations. What methods have you learnt for solving second order, homogeneous ODE's?
Well this is actually just a physics class so we haven't learned any methods for solving differential equations. I've taken calculus classes but won't be taking differential equations until next semester so I've just been looking through notes online trying to figure out how to solve this question.
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## differential equation problem :(
Quote by yoamocuy Well this is actually just a physics class so we haven't learned any methods for solving differential equations. I've taken calculus classes but won't be taking differential equations until next semester so I've just been looking through notes online trying to figure out how to solve this question.
The standard method of solving such problems is to take an Ansatz of the form
$$s = Ae^{\lambda t}$$
and substitute that into the ODE.
I'm sorry that I don't know of any good online references for DE's, but I'm sure someone here will be able to suggest a suitable reference.
Ok, thanks for the help. I'll keep searching and working with it.
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Quote by yoamocuy Ok, thanks for the help. I'll keep searching and working with it.
If you would like to post your working, I'd be more than happy to help you with it. All you need to do is substitute that Astatz I gave you into the ODE and you should find that you have a rather simple equation to solve.
Ok, I got an equation that looks like s^2*e^((lambda)*t)=-1.5, but when I try to solve for the roots it isn't possible. Am I still completely off on my attempt?
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Quote by yoamocuy Ok, I got an equation that looks like s^2*e^((lambda)*t)=-1.5, but when I try to solve for the roots it isn't possible. Am I still completely off on my attempt?
I'm not sure how you've managed to get that. Can you work out what
$$\frac{d^2 s}{dt^2} = \frac{d^2}{dt^2} Ae^{\lambda t}$$
is?
Ug I dont even know what I'm supposed to be solving for there. On the sites I've been looking at, it looks like they are just taking a Laplace transform and multiplying each differntial by e^-((lambda)*t)
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Quote by yoamocuy Ug I dont even know what I'm supposed to be solving for there. On the sites I've been looking at, it looks like they are just taking a Laplace transform and multiplying each differntial by e^-((lambda)*t)
I was simply asking what is the second derivative of $Ae^{\lambda t}$ with respect to t?
Of course, you can use Laplace transforms if you like, but it is much more straightforward (if a little inelegant) to us an Anstatz.
if you like, you can have a look at the following website, there are solved examples too, i hope it will help. http://www.math.sunysb.edu/~scott/mat127.spr06/DEnotes/
I took the second derivitive of that and got A*(lambda)^2*e^((lambda)*t). Would lambda be -2 for that? I also tried using Laplace transforms and ended up with s^2+1.5=0, which has imaginary roots.
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Quote by yoamocuy I took the second derivitive of that and got A*(lambda)^2*e^((lambda)*t). Would lambda be -2 for that?
You're on the right lines. So let's take a look at the over all equation,
$$A\lambda^2 e^{\lambda t} = -1.5Ae^{\lambda t}$$
Hence,
$$\lambda^2 = -1.5$$
Do you agree?
Yea, I agree with that, but if I solve for lambda I get j1.22. What can I do with that?
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Quote by yoamocuy Yea, I agree with that, but if I solve for lambda I get j1.22. What can I do with that?
Simple, you have determined lambda, so you now have the [general] solution,
$$s\left(t\right) = A\exp\left(i\sqrt{1.5}t\right)$$
That is it. You have now found the general solution to the ODE. You can use Euler's relation to write it in a 'nicer' form, but you have found a valid solution.
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http://www.euro-math-soc.eu/node/2786
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# The European Mathematical Society
Hosted by:
## The Ambient Metric
June 13, 2012 - 17:39 — Anonymous
Author(s):
Charles Fefferman, C. Robin Graham
Publisher:
Princeton University Press
Year:
2011
ISBN:
978-0-691-15313-1
Price (tentative):
\$55
Short description:
This monograph is devoted to the study of invariants of conformal classes of (semi-riemannian) metrics in differentiable manifolds. The title, "The Ambient Metric", refers to the object studied in the book. The ambient metric associated to a conformal structure on a manifold was originally introduced by the authors in a short article without complete proofs in 1985.
The book, of technical nature, and carefully written, includes full proofs. It is focused on a very particular problem in geometry. As such, it is addressed to researchers already interested in this area of mathematics.
URL for publisher, author, or book:
http://press.princeton.edu/catalogs/series/am.html
MSC main category:
53 Differential geometry
Review:
This monograph is devoted to the study of invariants of conformal classes of (semi-riemannian) metrics in differentiable manifolds. The title, "The Ambient Metric", refers to the object studied in the book. The ambient metric associated to a conformal structure on a manifold was originally introduced by the authors in a short article without complete proofs in 1985, see [C. Fefferman and C.R. Graham, Conformal invariants, in The Mathematical Heritage of Élie Cartan (Lyon, 1984), Astérisque, 1985, Numero Hors Serie, 95-116].
The book develops and applies the theory of the ambient metric in conformal geometry. The ambient metric is a (semi-riemannian) metric in $n+2$ dimensions that encodes the information of a conformal class of metrics in $n$ dimensions. The ambient metric has an alternative incarnation as the Poincaré metric, which is a metric in $n+1$ dimensions having the conformal manifold as its conformal infinity.
The existence and uniqueness of the ambient metric at the formal power series level is treated in detail. This includes the derivation of the ambient obstruction tensor and an explicit analysis of the special cases of conformally flat and conformally Einstein spaces. Then Poincaré metrics are introduced and shown to be equivalent to the ambient metric formulation. The special case of self-dual Poincaré metrics in four dimensions is considered, leading to a formal power series proof of LeBrun's collar neighborhood theorem, proved originally using twistor methods, for analytic metrics.
Conformal curvature tensors are introduced by using the ambient metric formulation, and their fundamental properties are established. The book concludes with the construction and characterization of scalar conformal invariants in terms of the curvature of the ambient metric.
The monograph is a text of technical nature, very carefully written, which includes full proofs. it is focused on a particular problem in geometry. As such, it is addressed to researchers already interested in this area of mathematics.
Reviewer:
Vicente Muñoz
Affiliation:
UCM
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http://mathoverflow.net/questions/64448/interesting-and-accessible-topics-in-graph-theory/85618
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## Interesting and Accessible Topics in Graph Theory
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
This summer, I will be teaching an introductory course in graph theory to talented high school seniors. The intent of the course is not to establish proficiency in graph theory, per se. Rather, I hope to use graph theory as a vehicle by which to convey a sense of developing "advanced" mathematics (remember, these students will have seen first-year calculus, at best).
What are you favorite interesting and accessible nuggets of graph theory?
"Interesting" could mean either the topic has a particularly useful application in the real-world or else is a surprising or elegant theoretical result. An added bonus would be if the topic can reveal gaps in our collective knowledge (for example, even small Ramsey numbers are still not known exactly). "Accessible" means that a bright, motivated student with no combinatorial background can follow the development of the topic from scratch, even if it takes several lectures.
-
More a suggestion than an answer: spend half a session highlighting the similarities and differences between theory of finite graphs and theory of infinite graphs. If you want an interesting tangent, the elementary first order theory of graphs is finitely axiomatizable and undecidable. This makes it handy to interpret into other theories to show those other theories are undecidable. Edited carefully, this tangent could be made accessible to your audience. Gerhard "Ask Me About System Design" Paseman, 2011.05.10 – Gerhard Paseman May 10 2011 at 20:48
1
Thanks for the many great suggestions. Reading all these has caused me to think that I could potentially structure the course in such a way as to introduce the widest number of adjectives that can precede "graph theory" or "combinatorics". For example, I see in the topics presented here: enumerative, extremal, geometric, computational, probabilistic, algebraic, and constructive (for lack of a better word - I'm referring to things like designs). As a sort of subquestion restricted to the comments, what other adjectives might I attempt to incorporate? – Austin Mohr May 14 2011 at 5:24
applied. Gerhard "Ask Me About System Design" Paseman, 2011.05.16 – Gerhard Paseman May 16 2011 at 20:06
@Austin: It would be nice if you let us know what you decide to teach, and how it goes this summer! – Joseph O'Rourke May 19 2011 at 23:34
@Joseph: I will certainly do so. Thanks for your interest. – Austin Mohr May 20 2011 at 5:17
show 1 more comment
## 18 Answers
I have found that the Art Gallery Problem engages middle- and high-school students, and quickly leads to the unknown, which itself can be eye-opening to students. (On the latter point, students tend to think of mathematics as settled, so it is nice for them to reach unsolved problems they can comprehend, which abound at the interface between geometry and graph theory.) Proving the traditional art gallery theorem (that $\lfloor n/3 \rfloor$ guards suffice and are sometimes necessary to cover an $n$-wall gallery) introduces triangulations and the chromatic number of a graph. There are many sources, including the recent book (if I may self-promote) Discrete and Computational Geometry.
Addendum. May I also recommend "How to Guard an Art Gallery and Other Discrete Mathematical Adventures", by T.S. Michael, whom I had the pleasure of teaching two decades before his book was published.
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2
May I chime in that Joseph's Art Gallery Theorems and Algorithms, though not as recent, remains nice to read. – J. M. May 10 2011 at 12:56
Steve Fisk has a beautiful solution to this problem, which I think might be accessible enough for Austin's purposes. But I'm biased because Fisk was a professor of mine in undergrad and died not too long ago. I've been thinking of giving a seminar talk on this theorem as a tribute. – David White May 10 2011 at 18:51
@David: Steve will live on through his beautiful proof! – Joseph O'Rourke May 19 2011 at 23:43
### 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.
Matching theory. This includes Hall's marriage theorem, Tutte's theorem and the Gale-Sharpley stable matching theorem.
One reason to teach this subject to undergrads is that it changes the way mathematicians think about algorithms. The standard algorithms one learns in high school and undergraduate studies (Dijkstra algorithm, quicksort and the likes, etc.) are all generally looked down on by mathematicians, as they (1) seem to be just particularly efficient methods to compute some objects which are already clear to exist, (2) are usually of low complexity (much lower than that of proofs in undergraduate mathematics), (3) add no theoretical value (in fact they do, but algorithmic content in mathematical proofs is often cleverly hidden by whoever writes up the proof, so it looks like they don't). In contrast, the perfect matching algorithm (by using augmenting paths) and the Gale-Sharpley algorithm are rather nontrivial and actually are major components in the proofs of the existence of perfect matchings rsp. stable matchings.
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If you're covering matching theory, I would add König's theorem (in a bipartite graph max matching + max independent set = #vertices), the theorem that a regular bipartite graph has a perfect matching, and Petersen's theorem that a bridgeless cubic graph has a perfect matching (e.g. a triangulated 2-manifold has a matching of its triangles). But I have to admit that last week in the matching part of my graph algorithms class I covered only König and Gale–Shapley because it's about graph algorithms not graph theory and I needed the remaining lecture time to cover the Hopcroft–Karp algorithm. – David Eppstein May 11 2011 at 4:29
One problem that I found quite interesting is the Hadwiger-Nelson problem for coloring the plane. The proofs that the answer is at most $7$ and at least $4$ are easy and elegant (but also quite different), and it has the added bonus that it is an open problem.
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The "$-1$ colors" theorem. This allows you to talk about coloring, about chromatic polynomials, and (for example) about hyperplane arrangements, their chambers and so on (Orlik+Terao's book on hyperplane arrangements has the details)
Doing everything in detail is a bit of work, but you probably don't need that.
Everyone with appreciation for beautiful math should fall for this!
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Could you provide an introductory link, please? – Hans Stricker Aug 15 at 20:28
The book by Orlik and Terao is quite accessible in fact. Richard Stanley has notes on his webpage on the subject, if I recall correctly. – Mariano Suárez-Alvarez Aug 15 at 23:26
1) As you say in the question, Ramsey numbers, both the (easy) upper bound - an elegant example of the power of combinatorics - and the lower bound, one of the rare proofs that is easy, short, and completely surpising.
2) Turan's theorem about the size of independent sets in graphs. Again, an easy application of the probabilistic method (though you don't even need the language of probability to state it). It's also interesting that this is an example, unlike Ramsey numbers, where the 'easy' proof actually gives the best possible bound (for general graphs).
Of course given the wide variety of problems graphs are applicable to, it's useful to be able to detect structured subgraphs, either complete graphs or independent sets. These are also good examples of non-constructive proofs.
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Planar graph duality. E.g. the facts that
• A set of edges forms a connected spanning subgraph of a planar graph G if and only if the complementary set of edges forms an acyclic subgraph of the dual, and vice versa.
• Since spanning trees are just connected acyclic subgraphs, it follows that a subgraph is a tree if and only if its complement is dual to a tree. Euler's formula follows immediately.
• The edges that are not in the minimum spanning tree of a planar graph G are the duals of the edges that are in the maximum spanning tree of its dual.
• A planar graph is bipartite if and only if its dual is Eulerian.
• A planar graph is 3-connected (polyhedral) if and only if its dual is.
• The graphic matroid of a planar graph is the dual of the graphic matroid of the dual graph. Planar graphs are the only graphs for which the dual of the graphic matroid is also graphic.
• A directed planar graph is acyclic if and only if its dual graph (with the dual edges oriented 90 degrees clockwise from the primal ones) is strongly connected.
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One could make an interesting module on Euler's formula alone, following David's 19(!) proofs of Euler's theorem: ics.uci.edu/~eppstein/junkyard/euler – Joseph O'Rourke May 10 2011 at 12:25
Thanks! Your comment reminds me that I need to update that site: there are a couple more proofs I've encountered (one in my own latest SoCG paper) that I haven't yet added. – David Eppstein May 10 2011 at 21:26
Here is a topic that he kept me wasting (?) many hours: there is a theorem that every embedding in 3-space of $K_7$ , the complete graph on 7 vertices, contains a knot. Draw such an embedding and find the knot(s)!
Every embedding of $K_6$ contains a link, but that is usually much easier to find.
If I can find the reference, I'll add that as a comment.
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The reference is Conway, J. H.; Gordon, C. McA. Knots and links in spatial graphs. J. Graph Theory 7 (1983), no. 4, 445–453. The proof of the first statement is interesting: they prove that the product of the Arf invariants of all embedded circles in the embedded $K_7$ is invariant under Reidemeister moves of the embedding. So only one example is needed! – Ronnie Brown Jan 13 2012 at 23:10
I think that presenting the connection between random walks and electrical networks (like in the classic text "Random walks and electric networks" by Doyle and Snell) is an interesting and feasible idea. Just a week ago I taught a 6-day course about this to talented high schoolers and it worked out very nicely. It's a good opportunity to show them interesting applications of probability and give a flavour of a vibrant field of mathematics. Plus, there is a quite a lot of room for digressions on Markov chains, spectral graph theory etc.
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I don't know how feasible it'll be to talk about spectral graph theory to students who may not have seen linear algebra before... – Qiaochu Yuan May 10 2011 at 9:51
Of course you can't treat this topic extensively, but mentioning expanders, the fact that their geometric properties are connected to the mixing rate of the random walk and can be analyzed algebraically is surely doable as a digression (as well as things like PageRank or random graphs). – Michal Kotowski May 10 2011 at 15:33
Coloring problems are beautiful and accessible. If you're mentioning Ramsey numbers (as you should!), it will be easy to segue into coloring problems (or vice versa).
Some topics worth including would be
• Bipartite iff 2-colorable
• Four/Five Color Theorems
• Chromatic number, basic bounds
• Edge chromatic number, Vizing's Theorem, König's Line Coloring Theorem
These were some of my favorite results in my first graph theory course.
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You beat me to it! I was going to say the four/five color theorem since this year, for a high school math circle ran at UBC, we had a lecture which stated the four color theorem, and proved the five color theorem. It went very well, and the students enjoyed it. – Eric Naslund May 10 2011 at 20:59
This is probably not what you're looking for, but if I were teaching such a course I'd focus on the interplay between algorithms for various graph problems and the more traditional mathematical approach of proving theorems in graph theory.
For example, you can prove the "max flow-min cut" theorem by developing an algorithm that simultaneously finds a maximum flow and a corresponding min-cut.
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I think the graph isomorphism problem is a good choice: it's a fundamental question and a good lead in to the notion of isomorphic and some complexity theory---it's not known if it's NP-complete. This should be doable in a couple of lectures, depending on how thorough you want to be.
There are also applications, but I'm not the one to tell you about them.
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1
The question is: what interesting theorems does this subject contain? As far as I understand, it's all open problems. – darij grinberg May 10 2011 at 8:59
Some very accessible and interesting content that I think would cause a positive impression can be found in the book "Graphs and their uses" by Oystein Ore.
If I was to give an introductory course on graph theory for such an audience, I would follow part of this book at the beginning and then I would complement with something else. I would start with a classic problem: is it possible to connect through paths in the plane each one of three wells to each one of three different houses without intersections? There is a very neat and elementary explanation on why this is not possible that only uses the Jordan's curve theorem.
After mentioning other classic examples such as the Königsberg bridge problem (and the criterion to find eulerian paths in a graph) and hamiltonian cycles (stressing the lack of an efficient general method to find these), I would move to a remarkable fact that is sometimes known as the "sports journalist paradox", according to which is quite common to find, in some sports tournaments where each team plays against each other, an oriented cycle that involves all teams; in other words, team A wins team B, which wins team C, and so on...till some team in the chain wins team A! It is not difficult to characterize when we can find such a behaviour, and the answer turns out to be quite often!
I would then mention map coloring, and give a complete proof of the five color theorem. But most important, I would discuss Euler's formula for poligonal nets and how to apply it to characterize all platonic solids. This achievement, while elementary, could be a nice way to keep the audience interested and pave the way to further deeper results.
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Take a look at my graph theory site, http://mathcove.net/petersen and let me know if the Petersen program would be useful. I would be willing to grant you a free license. Chris Mawata
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I'd suggest Turán's theorem, the bonus being the somehow surprising hardness of the corresponding problem for 3-uniform hypergraphs.
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On the application side, you can introduce graph theory as a way to model connected systems:
• web pages (familiar, has some nice visualizations, but hard to get your hands on),
• people a la Stanley Milgram and "six degrees of separation" (important but very "fuzzy"),
• actors connected by movies (precise via the Internet Movie Database and front-end Oracle of Bacon -- for ideas on using this, see my PRIMUS article "Kevin Bacon and Graph Theory," available through my web page)
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There are some nice pictures here, but any actual mathematical theorems? – darij grinberg May 10 2011 at 9:00
Nope, no theorems in these initial applications. Their purpose is to engage students early on and give contexts for distance, cliques, diameters, dynamic graphs, etc. I like theorems as well as the next mathematician, but they're not the only possible "nuggets." – Brian Hopkins May 11 2011 at 3:01
It seems to me that your goal should be to give them some elementary results with proofs (so they get used to that type of thinking) but also some high-level, cool, and surprising results to whet their appetite for more. In studying planarity you can find many cool and surprising results. I took a course from Laszlo Lovasz on such topics a few years ago, and he put his lecture notes online. You can link to them from this site: http://www.cs.elte.hu/~lovasz/geomgraph.html
Highlights I think high school students could understand and enjoy (perhaps skipping some of the proofs):
• Section 1.1. Kuratowski's Theorem: A graph $G$ is embeddable in the plane if and only if it does not contain a subgraph homeomorphic to the complete graph $K_5$ or the complete bipartite graph $K_{3,3}$.
• Section 1.3. Steinitz's Theorem: A simple graph is isomorphic to the skeleton of a 3-polytope if and only if it is 3-connected and planar.
• Section 2.1. Unit Distance Graphs
• Chapter 4. Turning the edges of $G$ into rubber-bands and talking about when they find equilibrium in the form of a planar representation of $G$
Finally, Chapter 6 has some really amazing results:
• The Cage Theorem: Every 3-connected planar graph is isomorphic to the 1-skeleton of a convex 3-polytope such that every edge of the polytope touches a given sphere.
• Koebe's Theorem: Let $G$ be a 3-connected planar graph. Then one can assign to each node $i$ a circle $C_i$ in the plane so that their interiors are disjoint, and two nodes are adjacent if and only if the corresponding circles are tangent.
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How about graphic sequences? This might be a nice introduction to the world of inverse problems - an important concept of whose existence most high school curricula wouldn't even hint.
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I believe the book
Hajnal Péter: Gráfelmélet. 1997, Polygon, Szeged.
is an extended answer to exactly this question. (There's a second edition from 2003, but apparently no translations to other languages.)
The writing style of this book makes it accessible to high school students (as opposed to the Lovász book whose concise style makes it ideal for research mathematicians). Thus, most of the material is accessible at high school level, while at the same time the book covers so many difficult topics that it'd be difficult to cover all the proofs even in a semester long high school course.
The book remains a useful reference for BSc combinatorics exams (not alone though, because some necessary topics are missing).
Here are some of the topics included (many of these were mentioned in other responses).
• Flow-cut theorem and algorithm, Menger's theorems, Kőnig-Hall, maximal matching algorithm for bipartite graphs, Tutte's theorem, and even Edmond's algorithm to find maximal matching in any graph.
• Euler circuits, then Dirac's sufficient condition for Hamiltonian circuit
• Graph coloring, Brook's theorem (when a graph's chromatic number reaches its maximum degree), high chromatic graphs without triangle, Hajós's characterisation of graphs with chromatic number (exercise 9.16 in Lovász), Vizing's theorem on edge coloring,
• Turán's theorem, Erdős-Stone theorem about the asymptotic on the number of edges of a graph not containing a particular non-bipartite graph, asymptotic for $C_4$-free graphs.
• Ramsey theorems.
• Complexity theory results about graph problems, including Karp reductions between Hamiltonian circuits and chromatic number and independence number. Does not include the Cook-Levin theorem so no problem is actually proven to be NP-complete.
• Planar graphs, duality, Whitney-duals, Euler's theorem, Kuratowski's theorem on the characterization of planar graphs (yes, with a proof), Robertson and Seymour's graph minor theorem without proof, the four color theorem without proof, the five color theorem and Kempe's proof, Hadwiger's conjecture.
• Perfect graphs, Lovász's theorem on the complement of perfect graphs (yes, with a proof), comparability graphs of posets are perfect.
Note finally that before any topic, you'd need to cover some of the first chapter which introduces basic terminology about graphs, which is important to plan if you are giving only a few lectures. (Have you told us about the total length of lectures you are planning to give?) Edit: ah, I see the lectures the question is referring to are now in the past, so there's no point to ask such a concrete question.
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http://www.aimath.org/ARCC/workshops/minimalmodcharp.html
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# The minimal model program in characteristic p
April 29 to May 3, 2013
at the
American Institute of Mathematics, Palo Alto, California
organized by
James McKernan and Chenyang Xu
This workshop, sponsored by AIM and the NSF, will be devoted to the minimal model program in characteristic $p$.
Despite recent progress in characteristic zero in all dimensions relatively little is known about the birational geometry of varieties in characteristic $p$, even for threefolds. Kawamata-Viehweg vanishing is one of the central results in characteristic zero but unfortunately it is known that Kodaira vanishing fails even for surfaces in characteristic $p$.
The singularities which appear in the minimal model program are adapted to the use of Kawamata-Viehweg vanishing. In characteristic $p$ there are some closely related singularities which arise naturally when considering the action of Frobenius. One aim of the workshop will be to understand how the two types of singularities compare.
Using ideas and techniques from characteristic zero coupled with some recent progress on alternatives to Kawamata-Viehweg vanishing in characteristic $p$, which use the action of Frobenius, one of the aims of the workshop will be to attack problems in the birational geometry of threefolds and possibly even higher dimensions in characteristic $p$.
The main topics of the workshop are
1. Vanishing theorems in finite characteristic.
2. The cone and base point free theorem in characteristic $p$.
3. Existence of three fold flips in characteristic $p$.
4. Semi-stable reduction for surfaces in characteristic $p$.
5. Boundedness of birational maps for threefolds.
6. The behavior of nef divisors modulo reduction to characteristic $p$.
The workshop will differ from typical conferences in some regards. Participants will be invited to suggest open problems and questions before the workshop begins, and these will be posted on the workshop website. These include specific problems on which there is hope of making some progress during the workshop, as well as more ambitious problems which may influence the future activity of the field. Lectures at the workshop will be focused on familiarizing the participants with the background material leading up to specific problems, and the schedule will include discussion and parallel working sessions.
The deadline to apply for support to participate in this workshop has passed.
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http://mathhelpforum.com/statistics/181313-monkeys-write-shakespear.html
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# Thread:
1. ## Monkeys to write shakespear.
I was curious about the actual amount of monkeys needed to produce something like hamlet, so I went and figured out how many letters there were: 194270
There are 26 letters in the alphabet so every possible combination of letter we can have in hamlet:
$26(194270)=5051020$
This problem produces a binomial distribution, so the mean of the normal curve would
be 1 because:
$\mu=np=(5051020)(5051020)^{-1}$
$\sigma = \sqrt{npq}=\sqrt{1(0.999999802020186)}$
$P(X\geqslant 1)=1-P(X\leqslant 1})$
$1-P(Z\leqslant \frac{1-1}{\sigma })$
$1-P(Z\leqslant 0})=1-.5=.5$
Therefore, if we have 5051020 monkeys, the probability of one of them producing hamlet (asssuming each only gets one try, the typing of one monkey does not effect the typing of another, and the probability of each monkey geting hamlet is the same) is on average 50%.
Is this correct?
2. im not sure i follow your first line. If you let a monkey type 194270 random characters the number of combinations is $26^{194270} \neq 26(194270)$
PS: what about punctuation and capitalisation?
3. Ha, never mind you would need 26^{194270} monkeys to have a 50% chance to write shakespeare...
4. Moving away from the hamlet problem, you are basically arguing that if:
$X \sim Bi(\frac{1}{p},p)$
then P(X>0) = 0.5
and hence
P(X=0) = 0.5
you can see this isn 't true by using $\frac{1}{p}$ that is small enough to fit in a calculator.
However it may be approximately true if $\frac{1}{p}$ gets sufficiently big (im not sure either way). i did check a pretty extreme case ( $\frac{1}{p} = 1,000,000,000$) and the probability was still less than 0.4
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http://nrich.maths.org/2401/index
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### Is There a Theorem?
Draw a square. A second square of the same size slides around the first always maintaining contact and keeping the same orientation. How far does the dot travel?
### Coins on a Plate
Points A, B and C are the centres of three circles, each one of which touches the other two. Prove that the perimeter of the triangle ABC is equal to the diameter of the largest circle.
### AP Rectangles
An AP rectangle is one whose area is numerically equal to its perimeter. If you are given the length of a side can you always find an AP rectangle with one side the given length?
# On the Edge
##### Stage: 3 Challenge Level:
Here are four tiles:
They can be arranged in a $2$ by $2$ square so that this large square has a green edge:
If the tiles are moved around, we can make a $2$ by $2$ square with a blue edge:
If I had nine tiles it would be quite easy to paint them so that, when they were arranged in a $3$ by $3$ square, the edge of this large square is green. I would need four tiles for the corners of the square, four tiles for the edges and one tile would go in the middle of the square so wouldn't need painting at all.
This is how the green-edged square would be made:
But I also want to be able to make a square with a blue edge and another square with a yellow edge.
How can the other sides of these tiles be painted so that all nine tiles can be rearranged to make two more $3$ by $3$ squares - one with a blue edge and one with a yellow edge?
Now try to colour sixteen tiles so that four $4$ by $4$ squares can be made - one with a green edge, one with a blue edge, one with a yellow edge and one with a red edge.
Find a way to colour $25$ tiles so that five $5$ by $5$ squares can be made, each with a differently coloured edge.
Do you think this is possible for $36$ tiles and six coloured edges?
Will it always be possible to add an extra colour as the squares get larger?
For a 3D version of this problem why not try " Inside Out "?
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http://mathoverflow.net/revisions/68905/list
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2 edited tags
1
# Are these vectors in the non-negative orthant of some $R^K$?
Given three sets of vectors $S_i=\left(V_{i1},V_{i2},\ldots,V_{in}\right),i=1,2,3$, s.t. the vectors within a set are pair-wise orthogonal and $V_{ij}\cdot V_{kl}\geq 0$ $\forall i,j,k,l$. Also, $\sum_jV_{ij}=w$ $\forall i$ where $w$ is some unit vector and $V_{ij}\cdot w=V_{ij}\cdot V_{ij}$ $\forall i,j$. It is given that the vectors in $S_i,S_j$ $\forall i\neq j$ lie in $R_{\geq 0}^k$ for some $k$. Can all the three sets of vectors be accommodated in $R_{\geq 0}^K$ for some $K$?
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http://mathhelpforum.com/pre-calculus/66893-functions-f-x.html
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# Thread:
1. ## Functions! f(x)=
ok so i have my core 1 exam on friday and am struggling with SOME functions.
i am at work and dont have the question for reference but its alogn the lines of
(x-2) is a factor of f(0)
find the values of k and c in f(x)=2x^2+kc+c
thanks in advance
2. I'm not sure but did you mistype your question, I think it should be something like:
$f(x)=2x^2+kx+c$
Can you remember what other information you are given about the roots of the equation?
With questions like these you usually use the discriminant, $b^2-4ac$.
For equal roots, $b^2-4ac=0$
For diffferent real roots, $b^2-4ac >0$
For no real roots, $b^2-4ac<0$
I know this doesn't answer your question but hope it helps you when you have the actual question in front of you
3. ## information
it very possibly said somethign along the lines of f(x)=6
if that doesn't mean anything i will have al ook when i get home and retype on this post with the whole question and how it is worded
4. Yeh post the actual question when you get home and I'll try and get my head round it then
5. ## the actual question!
You are given that f(x) = x^3 + kx +c. The value of f(0) is 6, and x-2 is a factor of f(x).
find the values of k and c
there we go, helps to have what your asking in front of you lol, thanks in advance
6. Originally Posted by coyoteflare
You are given that f(x) = x^3 + kx +c. The value of f(0) is 6, and x-2 is a factor of f(x).
find the values of k and c
there we go, helps to have what your asking in front of you lol, thanks in advance
f(0) = 6 => 6 = c.
(x - 2) is a factor => f(2) = 0 => 0 = 2^3 + 2k + c
And now you should be able to finish this.
7. thanks, can you explain to me why c is 6? is it to do with >0 and <0 ?
0 = 2^3 + 2k + 6
0 = 8 + 2k + 6
-14 = 2k
-7 = k ?
thanks again
8. Originally Posted by coyoteflare
thanks, can you explain to me why c is 6? is it to do with >0 and <0 ?
0 = 2^3 + 2k + 6
0 = 8 + 2k + 6
-14 = 2k
-7 = k ? Mr F says: yes.
thanks again
f(x) = x^3 + kx +c. The value of f(0) is 6
Therefore 6 = 0^3 + k(0) + c ......
9. Ah this isn't as hard as I first thought
You have the following equation
$x^3 + kx +c$
and you are told that $f(0) = 6$
You can therefore work out that then you sub 0 into the equation it equals 6, giving you:
$0^3 + 0k +c = 6$
therefore $c = 6$
Reading on from this you are told that x - 2 is a factor, therefore subbing 2 into the equation will give you zero.
$2^3 + 2k + 6 = 0$
The rest I'm sure is clear
Hope this explains it
Craig
10. Sorry I've got beaten to the answer
11. ## Thanks guys!
not as hard as i thought, just never had functions explained to me an got confused xD
thanks again!
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http://physics.stackexchange.com/questions/27099/some-questions-on-a-version-of-the-oraifeartaigh-model?answertab=votes
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Some questions on a version of the O'Raifeartaigh model
This form is taken from a talk by Seiberg to which I was listening to,
Take the Kahler potential ($K$) and the supersymmetric potential ($W$) as,
$K = \vert X\vert ^2 + \vert \phi _1 \vert ^2 + \vert \phi_2\vert ^2$
$W = fX + m\phi_1 \phi_2 + \frac{h}{2}X\phi_1 ^2$
• This notation looks a bit confusing to me. Are the fields $X$, $\phi_1$ and $\phi_2$ real or complex? The form of $K$ seems to suggest that they are complex - since I would be inclined to read $\vert \psi \vert ^2$ as $\psi ^* \psi$ - but then the form of $W$ looks misleading - it seems that $W$ could be complex. Is that okay?
Now he looks at the potential $V$ defined as $V = \frac{\partial ^2 K}{\partial \psi_m \partial \psi_n} \left ( \frac {\partial W}{\partial \psi_m} \right )^* \frac {\partial W}{\partial \psi_n}$
(..where $\psi_n$ and $\psi_m$ sums over all fields in the theory..)
For this case this will give, $V = \vert \frac{h}{2}\phi_1^2 + f\vert ^2 + \vert m\phi_1 \vert ^2 + \vert hX\phi_1 + m\phi_2 \vert ^2$
• Though for the last term Seiberg seemed to have a "-" sign as $\vert hX\phi_1 - m\phi_2 \vert ^2$ - which I could not understand.
I think the first point he was making is that it is clear by looking at the above expression for $V$ that it can't go to $0$ anywhere and hence supersymmetry is not broken at any value of the fields.
• I would like to hear of some discussion as to why this particular function $V$ is important for the analysis - after all this is one among several terms that will appear in the Lagrangian with this Kahler potential and the supersymmetry potential.
• He seemed to say that if *$\phi_1$ and $\phi_2$ are integrated out then in terms of the massless field $X$ the potential is just $f^2$"* - I would be glad if someone can elaborate the calculation that he is referring to - I would naively think that in the limit of $h$ and $m$ going to $0$ the potential is looking like just $f^2$.
• With reference to the above case when the potential is just $f^2$ he seemed to be referring to the case when $\phi_2 = -\frac{hX\phi_1}{m}$. I could not get the significance of this. The equations of motion from this $V$ are clearly much more complicated.
• He said that one can work out the spectrum of the field theory by "diagonalizing the small fluctuations" - what did he mean? Was he meaning to drop all terms cubic or higher in the fields $\phi_1, \phi_2, X$ ? In this what would the "mass matrix" be defined as?
The confusion arises because of the initial doubt about whether the fields are real or complex. It seems that $V$ will have terms like $\phi^*\phi^*$ and $\phi \phi$ and also a constant term $f^2$ - these features are confusing me as to what diagonalizing will mean.
Normally with complex fields say $\psi_i$ the "mass-matrix" would be defined the $M$ in the terms $\psi_i ^* M_{ij}\psi_j$ But here I can't see that structure!
• The point he wanted to make is that once the mass-matrix is diagonalized it will have the same number of bosonic and fermionic masses and also the super-trace of its square will be $0$ - I can't see from where will fermionic masses come here!
• If the mass-matrix is $M$ then he seemed to claim - almost magically out of the top of his hat! - that the 1-loop effective action is $\frac{1}{64\pi^2} STr \left ( M^4 log \frac{M^2}{M_{cut_off}^2} \right )$ - he seemed to be saying that it follows from something else and he didn't need to do any loop calculation for that!
I would be glad if someone can help with these.
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5
A brief comment on your first question: the superpotential is a holomorphic function, hence complex unless (locally) constant. It appears in the lagrangian density as the real part of its integral over "half the superspace". Chiral superfields are therefore complex: indeed they define holomorphic coordinates on a Kähler manifold. – José Figueroa-O'Farrill Jan 1 '12 at 6:05
Indeed, all the fields are complex. The rest is more or less standard in QFT, except for the minus sign error, which you got right. – Zohar Ko Jan 1 '12 at 19:08
Jose - Thanks for the explanation. @Zohar Ko I can understand that in principle this is standard QFT but I can't exactly do what he seems to say should be done. Like in what sense is the potential just $f^2$ and how is the mass-matrix (with fermionic masses!) going to emerge here. I have done this kind of calculation elsewhere but here things don't seem to fit in. And the last expression for the 1-loop effective action looks as mysterious. It would be great if you can sketch out the calculation. – user6818 Jan 2 '12 at 21:42
The mass matrix of the bosons and fermions coincides if $f=0$. Hence, in the super-trace over M^4\log M^2 there would be perfect cancelation if $f=0$. The answer is nonzero if $f\neq0$ and it scales like f^2. – Zohar Ko Jan 3 '12 at 12:55
@Zohar Can you give a reference for such an analysis about the supertrace of the mass-matrix of a theory? – user6818 Jan 7 '12 at 0:52
show 1 more comment
1 Answer
There are lots of questions here! I think I can answer at least some...
• First of all, you are aware that the fields in $W$ and $K$ are superfields? These contain the entire supermultiplet, so they must be complex in general. This is a short entry but it links to others: http://en.wikipedia.org/wiki/Superfield
• As mentioned by Jose in his comment, the superpotential is just a holomorphic function. Remember that the superpotential is in some sense not physical; it only shows up in the Lagrangian - the real physics of the theory - via things like $\frac{\partial W}{\partial \phi_m}^* \frac{\partial W}{\partial \phi_n}$ in your example. It is these that must be real, not $W$ itself, and they will be. I'm not sure why you say there would be terms like $\phi^*\phi^*$... Everything is mod-squared in that $V$, so it must be real.
• I agree with Zohar that the minus sign is probably a "typo" on Seiberg's part. My first thought when I skimmed your question was that there was $SU(2)$ contraction with an $\epsilon_{\alpha\beta}$, because one gets terms that look like that all the time, but after reading it, I don't think that's the case here.
• To understand the supertrace in the effective potential, see http://arxiv.org/pdf/hep-ph/0111209v2.pdf by S. Martin. Eq. 1.2 is exactly what Seiberg wrote, but with the supertrace written out explicitly with the $(-1)^{2s_n}(2s+1)$ prefactor. To understand why the one-loop effective potential looks like $Str(M^4 log\frac{M^2}{M_{cutoff}^2})$, consider a couple things.
• First, the thing inside the supertrace looks just like something you'd get doing a one-loop integral for some field, except that here $M$ is a matrix. Second, the prefactor takes care of what you get due to the multiplicity of the particles and the minus sign associated with fermion loops.
But let's say we don't believe this expression and calculate everything explicitly. We would diagonalize the mass matrix, thus finding the mass eigenvalues $m_i$ and then we'd put each of them one at a time into a loop diagram and calculate basically $\Sigma_i m_i^4 log\frac{m_i^2}{M_{cutoff}^2}$.
But notice that if we diagonalize $M$ to, say, $M'$, then $M'$ has the $m_i$ along it's diagonal. Then clearly
$\Sigma_i log\frac{m_i^2}{M_{cutoff}^2}=Tr(log\frac{M'^2}{M_{cutoff}^2})$.
Furthermore,
$Tr(log\frac{M'^2}{M_{cutoff}^2})= log(Det\frac{M'^2}{M_{cutoff}^2})$.
This follows because $M'$ is diagonal, so that obviously $Det(M'^2)=\Pi_i m_i^2$ and thus
$log(Det(M'^2))=log(\Pi_i m_i^2)=\Sigma_i log (m_i^2)=Tr(log(M'^2))$.
Now, if we diagonalized with some matrix $U$, then
$Tr(log\frac{M'^2}{M_{cutoff}^2})= log(Det\frac{M'^2}{M_{cutoff}^2}) = log(Det\frac{(UMU^\dagger)^2}{M_{cutoff}^2})$,
and since $U$ is unitary, $Det(UMU^\dagger)=Det(M)$. Thus,
$Tr(log\frac{M'^2}{M_{cutoff}^2})= log(Det\frac{M^2}{M_{cutoff}^2})=Tr(log\frac{M^2}{M_{cutoff}^2})$.
But from above
$Tr(log\frac{M'^2}{M_{cutoff}^2})=\Sigma_i log\frac{m_i^2}{M_{cutoff}^2}$.
Thus
$Tr(log\frac{M^2}{M_{cutoff}^2})=\Sigma_i log\frac{m_i^2}{M_{cutoff}^2}$.
The point of all this is to show that you can save yourself a lot of time in computing the effective potential by using the expression on the left rather than calculating each term individually.
• You ask where fermion masses will come from. They follow because the fields in $W$ and $K$ are superfields.
• Zohar already mentioned the mass matrices coinciding if $f=0$. I will just add that you know if SUSY is unbroken the superpartners have the same mass. Looking at that prefactor from Martin's eq. (1.2) that I quoted above should make clear how then their contributions to the supertrace cancel. The scalars are spin zero while the fermions are spin 1/2 and have a relative minus sign. There are twice as many scalars but the fermions get a factor of 2 in the prefactor of the supertrace. Thus they cancel.
• "Integrating out" is not the same as $m\rightarrow 0$, though in this case I think that still gave you the right idea. Generally when people talk about integrating out, it means they integrate out high-momentum modes, giving some effective theory valid up to $M_{cutoff}$, above which things are integrated out, and/or replacing particular Feynman diagrams with effective vertices, such as when people talk about a gluon-gluon-Higgs effective coupling, which really takes place through a loop involving other particles, particularly the (quite heavy) top quark. The fields that have masses above the cutoff will no longer appear as fields in the effective theory but will instead give effective operators in the Lagrangian or effective vertices in the Feynman rules. http://en.wikipedia.org/wiki/Effective_field_theory might be helpful.
• It's hard, bordering on impossible, to guess about the significance of the $\phi_2=-\frac{hX\phi_1}{m}$ without more context. Such a value obviously makes the last term in the $V$ you've written disappear, but why that is a big deal, I can't say. I have a feeling he was trying to say something about when nonzero vevs would break SUSY but I really have no context here.
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http://mathoverflow.net/questions/95274?sort=votes
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## homogenous bundles
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $G$ be a reductive algebraic group over $\mathbb{C}$ and $H$ be an algebraic subgroup of $G$. We suppose that $H$ acts on some scheme $S$, where $S$ is of finite type over $\mathbb{C}$. Then I would like to know what are the usual assumptions on $H$ and $S$ to have the existence of the scheme $$G \times^{H} S:=(G \times S)/rel$$ where $\forall g \in G, h \in H, s \in S$, we define the relation $rel$ by $$(g,s) \ rel \ (gh^{-1},h.s).$$ It is something well-known in the case where $S$ is an $H$-module, but for $S$ an arbitrary $H$-scheme I have been heard that it dosen't always exist. I would like to have references or precise response about the hypotheses I should do on $S$ and $H$. For instance, I think it's ok if $P$ is a parabolic subgroup of $G$ or even if it is a reductive subgroup but still in those cases I know no reference.
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$S$ doesn't need to have the specific form $\mathbf{V}(M)$ for an $H$-module $M$ - it could be any quasi-projective scheme with $H$-action. On the other hand, if $H =\mathbb{Z}/2\mathbb{Z}$, there is a counterexample in mathoverflow.net/questions/72152/… – S. Carnahan♦ Apr 27 2012 at 3:16
## 1 Answer
Section I.5 ("Quotients and associated sheaves") of Jantzen's Representations of Algebraic Groups is (at least in my mind) a standard resource for this question. (Here is a Google books link). He considers your question in full generality there. In particular, he proves (cf I.5.6.(8)) that if $G$ is an algebraic group over a field $k$ and $H$ is a closed subgroup scheme of $G$ then $G/H$ is a scheme. (Here the definition of $G/H$ agrees with what you think it should mean over a field, but in general the definition of $G/H$ is given categorically, cf the definition of the quotient faisceau $X/G$ for any $G$-space $X$ in I.5.5).
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http://mathoverflow.net/questions/71696/the-mackey-topology-on-a-von-neumann-algebra
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## The Mackey Topology on a Von Neumann Algebra
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Every von Neumann algebra $\mathcal M$ is the dual of a unique Banach space $\mathcal M_*$. The Mackey topology on $\mathcal M$ is the topology of uniform convergence on weakly compact subsets of $\mathcal M_*$. Is it known whether given a von Neumann subalgebra $\mathcal N \subseteq \mathcal M$, the Mackey topology on $\mathcal M$ restricts to the Mackey topology on $\mathcal N$?
The article below indicates that the answer was unknown at the time of its publication.
Aarnes, J. F., On the Mackey-Topology for a Von Neumann Algebra, Math. Scand. 22(1968), 87-107 http://www.mscand.dk/article.php?id=1864
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http://math.stackexchange.com/questions/127233/two-ring-questions/127282
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# Two ring questions
1. In the ring $R=\mathbb Z[X],$ is $(X)+(X^2)=(X)$?
2. It is known that if $R$ is a UFD, then $R[X]$ is a UFD. Is the converse true?
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Are these questions homework? Anyway, two hints: Q1) Note that for two ideals $I_1, I_2$, if $I_1 \supseteq I_2$, then $I_1 + I_2 = I_1$ (why?). Q2) The converse is true. Proof sketch: Take any non-unit element in $R$, consider prime factorization over $R[X]$. Show that this prime factorization uses elements from $R$ only. Furthermore, if $a \in R$ is prime in $R[X]$, it is prime in $R$ (why?), which proves the claim. – Johannes Kloos Apr 2 '12 at 14:11
2
Is this homework? If so, a homework tag is appropriate. Also, what have you tried so far? – Michael Joyce Apr 2 '12 at 14:11
1
Finally, aside from possibly appearing on the same homework assignment, these two questions are not obviously related and so should be posted as separate questions. (Since it would be completely reasonable to answer just one of the two) – Daniel Martin Apr 2 '12 at 14:13
Meta: I wonder if we should pre-emptively tag `(homework)` and let the OP overrule it and remove the tag if s/he wishes. – user2468 Apr 2 '12 at 14:19
1
@J.D. We don't generally do that. – Arturo Magidin Apr 2 '12 at 15:17
show 1 more comment
## 2 Answers
For $\rm(2)$ the key observation is that $\rm R$ is inertly embedded in $\rm R[X],$ i.e. factorizations in $\rm R[X]$ of $\rm r\in R^*$ already lie in $\rm R,$ i.e $\rm\: 0\ne r = fg,\ f,g\in R[X]\:$ $\rm\Rightarrow$ $\rm\:f,g\in R,\:$ by comparing degrees (and employing $\rm R$ is a domain). This implies the factorization theory of $\rm R[X]$ restricts faithfully to $\rm R$.
Thus, since $\rm R[X]$ is a UFD, any nonunit $\rm\:r\in R^*\:$ is a product of atoms in $\rm R[X],$ hence in $\rm R,$ by inertness. Further, such atoms $\rm\:p\:$ are prime in $\rm R[X]$ so also in $\rm R$ since $\rm p\ |\ a,b\:$ $\rm\Rightarrow$ $\rm\:p\ |\ a\:$ or $\rm\:p\ |\ b\:$ in $\rm R[X]$ so in $\rm R$, i.e. $\rm\:p\ |\ a\:$ in $\rm\:R[x]\:$ $\Rightarrow$ $\rm\:a = p\:\!f,\ f\in R[X]\:$ $\rm\Rightarrow$ $\rm\:f\in R\:$ $\Rightarrow$ $\rm\:p\ |\ a\:$ in $\rm R,\:$ by inertness. Hence $\rm R$ is a UFD, since prime factorizations of $\rm\:r\in R[X]\:$ pull back to prime factorizations in $\rm R.$
Similarly one can show that any inertly embedded subdomain of a GCD domain is also a GCD domain, and gcds remain the same in the subdomain. In a classic paper, Paul Cohn showed that any GCD domain can be inertly embedded in a Bezout domain, i.e. a domain $\rm D$ where gcds are linearly representable $\rm\gcd(a,b) = ac + bd,\ c,d\in D,\:$ see Cohn: Bezout rings and their subrings.
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## Did you find this question interesting? Try our newsletter
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Hints:
1. As was pointed out in the comments, if $I$ and $J$ are ideals of $R$ with $I\subseteq J$, then show that $I+J=J$.
2. Suppose that $R[x]$ is a UFD. If we have an irreducible element $\pi\in R$, then is $\pi$ irreducible in $R[x]$? If so, is $\pi$ then a prime element of $R$? Why?
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http://quant.stackexchange.com/questions/4512/is-it-possible-to-derive-the-risk-tolerance-from-the-portfolio-efficient-front/4527
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# Is it possible to derive the “risk tolerance” from the portfolio efficient frontier?
I am trying to solve the Portfolio Optimization Problem using a "Multi-objective Evolutionary Algorithm". After obtaining the efficient frontier, I would like to know if we can infer for each point of the efficient frontier the corresponding risk tolerance parameter (which is related to the investor's preference). Thanks.
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3
You mean the $\lambda$ in $\mu - \lambda \sigma$? – Bob Jansen Nov 12 '12 at 11:16
Exactly, known also as the "risk aversion coefficient". – Omar Nov 14 '12 at 4:22
## 1 Answer
Strictly speaking the risk aversion coefficient depends on the form of investor preferences. Your "multi-objective evolutionary algorithm" may or may not be easy to place in this format. However, it becomes easy if you think about the risk aversion coefficient in mean/variance space if you were a mean-variance variance investor.
In this case you would have utility $$U\left(w\right)\equiv w'\mu-\frac{1}{2}\lambda w'\Sigma w$$ with the first order condition $$\mu-\lambda\Sigma w=0$$ multiplying both sides by $w'$ and solving for $\lambda$ gives $$\lambda=\frac{w'\mu}{w'\Sigma w}=\frac{\mu_{p}}{\sigma_{p}^{2}}$$ or that the implied risk aversion coefficient given portfolio holdings if you were a mean-variance investor is the ratio between the portfolio mean to portfolio variance.
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http://mathhelpforum.com/geometry/6815-if-area-disc-print.html
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# If the area of a disc...
Printable View
• October 24th 2006, 12:16 PM
ceasar_19134
If the area of a disc...
If the area of a disc inscribed in a square is 36pi^2, what is the area of the square?
• October 24th 2006, 12:22 PM
topsquark
Quote:
Originally Posted by ceasar_19134
If the area of a disc inscribed in a square is 36pi^2, what is the area of the square?
If the disc is inscribed in the square, then we know that one side of the square is twice the radius of the circle. The area of the square is $36 \pi ^2$ (typo?) so the radius must be $6 \sqrt{ \pi }$, thus the side of the square is $12 \sqrt{ \pi }$ and thus the area of the square is $\left ( 12 \sqrt{ \pi } \right ) ^2 = 144 \pi$.
-Dan
• October 27th 2006, 07:24 AM
ceasar_19134
The area of the disc is 36pi^2, not the square.
• October 27th 2006, 07:52 AM
CaptainBlack
Quote:
Originally Posted by ceasar_19134
If the area of a disc inscribed in a square is 36pi^2, what is the area of the square?
If the pi^2 is a typo it is already answered here
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http://mathforum.org/mathimages/index.php?title=Volume_of_Revolution&oldid=20330
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# Volume of Revolution
### From Math Images
Revision as of 16:38, 14 June 2011 by Nordhr (Talk | contribs)
Solid of revolution
Field: Calculus
Image Created By: Nordhr
Website: [1]
Solid of revolution
This image is a solid of revolution
# Basic Description
When finding the volume of revolution of solids, in many cases the problem is not with the calculus, but with actually visualizing the solid. To find the volume of a solid like a cylinder, usually we use the formula ${\pi} {r^2} h$. Alternatively we can imagine chopping up the cylinder into thin cylindrical plates, much like slicing up bread, computing the volume of each thin slice, then summing up the volumes of all the slices.
The disc method is much like slicing up bread and computing the volume of each slice http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html
## =Disk Method
In general, given a function, we can graph it then revolve the area under the curve between two specific coordinates about a fixed axis to obtain a solid called the solid of revolution. The volume of the solid can then be computed using the disc method.
Note: There are other ways of computing the volumes of complicated solids other than the disc method.
In the disc method, we imagine chopping up the solid into thin cylindrical plates calculating the volume of each plate, then summing up the volumes of all plates.
For example, let's consider a region bounded by $y=x^2$, $y=0$,$x=0$ and $x=1$
<-------Plotting the graph of this area,
If we revolve this area about the x axis ($y=0$), then we get the main image on the right hand side of the page
This image shows a plane area being revolved to create a solid http://curvebank.calstatela.edu/volrev/volrev.htm
To find the volume of the solid using the disc method:
Volume of one disc = ${\pi} y^2{\Delta x}$ where $y$- which is the function- is the radius of the circular cross-section and $\Delta x$ is the thickness of each disc. Using the analogy of the bread, computing the volume of one disc would correspond to computing the volume of one slice of bread. With this in mind, the area of one disc would correspond to the area of a slice of bread, while the thickness of a disc would correspond to the thickness of a slice of bread. To find the total volume of the bread, we would have to sum up the volumes of each of the slices.
Volume of all discs:
Volume of all discs = ${\sum}{\pi}y^2{\Delta x}$, with $X$ ranging from 0 to 1
If we make the slices infinitesmally thick, the Riemann sum becomes the same as:
$\int_0^1 {\pi}y^2\,dx ={\pi}\int_0^1 (x^2)^2\, dx$
Evaluating this intergral,
${\pi}\int_0^1 x^4 dx$
=$[{{x^5\over 5} + C|}_0^1] {\pi}$
=$[{1\over 5} + {0\over 5}] {\pi}$
=${\pi}\over 5$
volume of solid= ${\pi\over 5} units^3$
In the example we discussed, the area is revolved about the $x$-axis. This does not always have to be the case. A function can be revolved about any fixed axis. Also, given a different function, to find the volume of revolution about the $x$-axis, we can substitute it in the place of $x^2$. Note: we would also need to change the bounds as per the given information. The method discussed in the example works for all functions that have bounds and are revolved about the $x$-axis.
## References
Bread image http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html
Revolving image http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html
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# References
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http://mathoverflow.net/questions/120612?sort=votes
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## Trichotomies in mathematics
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Added. Thanks to all who participated! Let me humbly apologize to those who were annoyed (quite understandably) by this thread, deeming it nothing more than an exercise in futility. If you thought the question, if legitimate at all, should have been restricted to interesting manifestations of a hyperbolic-parabolic-elliptic subdivision, then I can fully agree (although part of the idea was to interpret the question as you see fit); I left it open ended primarily because of the Weil trichotomy, which is of completely different kind and is so much more than a hierarchy, and in relation to which I was interested in hearing other people's opinions and elaborations. See, for instance, how Edward Frenkel, in a fascinating Bourbaki talk, builds upon the Weil trichotomy to introduce a parallel between Langlands and electro-magnetic dualities, which he uses as a springboard for the ideas from physics that have entered the arena of the geometric Langlands program. Or take the cherished three-sided parallel between the basic three-dimensional (from the point of view of etale cohomology) objects and their branched coverings: $\mathbb{P}_{\mathbb{F}_q}^1$, $\mathrm{Spec}(\mathbb{Z})$, and $S^3$, with primes in number fields corresponding to knots in threefolds, $\log{p}$ corresponding to hyperbolic length, etcetera.
To those who were not convinced that there was a neat trichotomy of algebraic surfaces (arguing they should instead form a tetrachotomy by Kodaira dimension), let alone in higher dimensional algebraic geometry, I refer to Sándor Kovács's answer here, which demonstrates rather eloquently the fundamental trichotomy of birational geometry:
http://mathoverflow.net/questions/81913/how-frequent-are-smooth-projective-varieties-with-anti-ample-canonical-bundle
Original post. For many purposes, notably in classification hierarchies or in Weil's "big picture" of the fundamental unity in mathematics, it seems as if mathematical reality is more accurately captured by trichotomies than by two-sided dictionaries or questions of "either/or." The most basic is of course the trichotomy negative - zero - positve embodied by the complete ordered field $(\mathbb{R},<)$ --- this is the Arrow of Time, if you will, or the conditioning of a dynamical system into states of past/present/future. As evidenced by some of the examples below, this trichotomy underlies varied, if crude, classification schemes in mathematics.
Other trichotomies arise from closer examinations of a mathematical parallel. Mathematicians have always been fond of discovery by analogy; they take very seriously the intuitions supplied by different yet loosely connected fields. In doing so, they are guided by a tacit, platonic belief in the fundamental Unity of mathematics. An example is the similarity between finite geometries and Riemann surfaces. To explain this parallel, indeed to make sense of it, it is necessary to provide a "middle column" in the dictionary: the arithmetic geometry of number fields and arithmetic surfaces. This leads to the trichotomy that Weil explained so lucidly in a letter (which he wrote in 1940 in prison for his refusal to serve in the army) to his sister, the philosopher Simone Weil. This point of view led, as we know, to an entire new field of mathematical inquiry.
Below I have listed some other cherished mathematical trichotomies. I am interested in seeing yet others, perhaps more specialized. This is my question: add more trichotomies to the list. Furthermore, I am interested in any reflections anyone might have, such as pertaining, for instance, to any of the following questions. Is 3 the most ubiquitous number in coarse classification schemes? Is it fair to say that a given trichotomy echoes the primeval trichotomy $(-,0,+)$? In a given trichotomy, is there a natural "middle column" of a corresponding three-sided dictionary? Is this "middle column" in any way the most fundamental, the most interesting, or the most elusive?
Trichotomies in mathematics: some examples.
• The fabric of topology, geometry, and analysis is the real line $\mathbb{R}$. Tarski's eight axioms characterize it in terms of a complete binary total order <, a binary operation +, and a constant 1. (Multiplication comes afterwards - it is implied by Tarski's axioms - and so does the Bourbaki definition of the reals as the complete ordered field). The sign trichotomies $(<,=,>)$ and $(-,0,+)$ ensuing from those axioms have repercussions throughout all of mathematics.
• For example, there are three constant curvature spaces, leading to the three maximally symmetric geometries: hyperbolic, flat (or Euclidean), and elliptic (e.g. spherical forms).
• Locally symmetric spaces fall into three types: non-compact type, flat, and compact type.
• In complex analysis, there are three simply connected cloths: the Riemann surfaces $\Delta$, $\mathbb{C}$, and $\hat{\mathbb{C}}$.
• The connected component of the group of conformal automorphisms of a compact Riemann surface is one of the following three: trivial, $S^1 \times S^1$, $\mathrm{PGL}_2(\mathbb{C})$.
• The complexity of fundamental groups, as showcased first of all by topological surfaces: genuinely non-abelian (perhaps we could say: anabelian) - abelian (or more generally, containing a finite index nilpotent subgroup) - and trivial (or more generally, finite). This is of course related to the subject of growth of finitely generated groups, brought forward by Lee Mosher's answer.
• In dynamics, a fixed point (or a periodic cycle) can be either repelling, indifferent, or attracting.
• In Thurston's work on surface homeomorphisms, elements of the mapping class group are classified according to dynamics into three types: pseudo-Anosov, reducible, and finite-order.
• In algebraic geometry, the positivity of the canonical bundle is central to the classification and minimal model problems. More generally, positivity is a salient feature of algebraic geometry. For a delightful discussion, see Kollar's review of Lazarsfeld's book "Positivity in algebraic geometry." (Bull. AMS, vol. 43, no. 2, pp. 279-284). The most basic example is the trichotomy of algebraic curves (rational, elliptic, general type).
• In birational algebraic geometry, at a very coarse level, there are three kinds of varieties out of which a general variety is made: rational curves, Calabi-Yau manifolds, and varieties of general type (or hyperbolic type, if you prefer). For example, an algebraic surface either: 1) admits a pencil of rational curves; or 2) admits a pencil of elliptic curves or is abelian or K3 (or a double quotient of a K3); or else 3) it is of general type. Abelian and K3 are examples of Calabi-Yau manifolds.
• More concretely, consider smooth hypersurfaces $X \subset \mathbb{P}^n$. They divide into three types, according to how their degree $d$ compares with the dimension. If $d \leq n$, they contain plenty of rational curves (certainly uncountably many). If $d = n+1$, they are an example of a Calabi-Yau manifold, and typically contain a countably infinite number of rational curves. (The generating function of the number of rational curves of a given degree is then a very interesting function, of significance in the physics of quantum gravity.) And if $d \geq n+2$, then $X$ is of general type, and it is conjectured to typically contain only finitely many rational curves. (More precisely, Bombieri and Lang have conjectured that a variety of general type contains only finitely many maximal subvarieties not of general type).
• In diophantine geometry, rational points are supposed to come from rational curves and abelian varieties. The sporadic examples are believed to be finitely many. This leads to the following trichotomy for the growth rate of the number of rational points of bounded (big, i.e. exponential) height: polynomial growth - logarithmic growth - $O(1)$. Furthermore, even in dimension 1, it is for abelian varieties that the situation is the deepest and the most mysterious.
• In topology, it seems as if the interesting dimensions fall into three qualitatively different ranges: $d = 3$, $d = 4$, and $d \geq 5$. (Although this might be stretching it a bit too much). Of these, four dimensions -- the "middle column" -- is the most mysterious, and also the most relevant for physics.
• The "Weil trichotomy," of course, goes at least as far back to Kronecker and Dedekind: curves over $\mathbb{F}_q$ - number fields - Riemann surfaces. Class field theory and Iwasawa theory are particularly eloquent examples of this trichotomy. Another example is of course the zeta function and the Riemann hypothesis.
• One would be tempted to extend the latter trichotomy to [non-Archimedean world ($p$-adic, profinite) - global arithmetic - Archimedean world (geometry, topology, complex variables)], if the middle column did not subsume (much of) the flanking columns. Likewise the triple [$l$-adic cohomology-motive-Hodge structure] would probably not be admissible. Here is a variation on the theme (you may find it to be rubbish, in which case throw it away). There are two ways of completing (or taking limits of) the regular polygons $C_n$. The first is to think of $C_n$ as $\frac{1}{n}\mathbb{Z}/\mathbb{Z}$ and take the direct limit (in this case, union, or synthesis: $\rightarrow$), which is $\mathbb{Q}/\mathbb{Z}$. Completing, we get the circle $S^1 = \mathbb{R}/\mathbb{Z}$, which is the simplest manifold. The second is to think of $C_n$ as $\mathbb{Z}/n$ and take the projective limit (or deconstruction: $\leftarrow$), which is $\hat{\mathbb{Z}} = \prod_p \mathbb{Z}_p$, the profinite version of the circle. In this way, Archimedean (continuous) objects and $p$-adic objects may be seen as the two possible different limits (synthesis and deconstruction) of the same finite objects. Taking $C_n$ to be more general finite groups, we get essentially all the Lie groups, on the one hand; and all the profinite groups, on the other hand.
• That we live in three perceptible spatial dimensions does not, of course, fit our bill. But in 1984, Manin published an article ("New dimensions in geometry") in which, guided by ideas from number theory (Arakelov geometry) and physics (supersymmetry), he proposed that there are three kinds of geometric dimensions, modeled on the affine superscheme $\mathrm{Spec} \mathbb{Z}[x_i;\xi_j]$, an "object of the category of topological spaces locally ringed by a sheaf of $\mathbb{Z}/2$-graded supercommutative rings." Here, $\xi_j$ are "odd," anticommuting variables, commuting with the "even" variables $x_i$. See the three coordinate axes $x, \xi$ and $\mathrm{Spec} \mathbb{Z}$ in his picture of "three-space-2000." The arithmetic axis $\mathrm{Spec} \mathbb{Z}$ is implicit in complex algebraic geometry, and is essential in problems such as the Ax-Grothendieck theorem and the construction of rational curves in Fano manifolds.
• In the theory of linear groups there is, loosely speaking, a trichotomy: $\mathbb{G}_m$ (linear tori) - semisimple - $\mathbb{G}_a$ (unipotent).
• Algebraic groups: reductive - abelian variety - unipotent. Especially, the classification of one-dimensional groups: $\mathbb{G}_m$ - $E$ - $\mathbb{G}_a$. (Thanks, Terry Tao!)
• Variant: among commutative algebraic groups, there are: multiplicative type - abelian varieties - additive type (unipotent).
• $\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$ are the only finite-dimensional associative division algebras over the continuum. (Thanks Paul Reynolds, Teo B, and Sam Lewallen!)
• The most basic PDEs of physics: the wave equation (hyperbolic) - the heat and Schrodinger equations (parabolic) - the Laplace equation (elliptic). (Thanks Alexandre Eremenko!)
• An infinite finitely-generated group has $1,2$ or $\infty$ ends. (Thanks shane.orourke and Artie Prendergast-Smith!)
• A random walk is either transient, null recurrent, or positive recurrent. (Thanks Vaughn Climenhaga!)
• Zeta functions can by dynamical (Artin-Mazur); arithmetical on schemes of finite type over $\mathbb{Z}$ (Riemann and Hasse-Weil); and geometric (Selberg's zeta function of a hyperbolic surface).
• In Model theory, there is an important trichotomy between super-stable theories, strict-stable (stable but not superstable) theories, and non stable theories.
• It seems fair to say that there are three kinds of three-dimensional simply connected spaces: $\mathbb{P}_{\mathbb{F}_q}^1$, $\mathbb{Spec}(\mathbb{Z})$ compactified at archimedean infinity, and $S^3$. This brings about the Mazur knotty dictionary and the fruitful analogy between primes and knots (especially hyperbolic knots).
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A finitely-generated infinite group has 1, 2, or infinitely many ends. (But if you leave out the word "infinite", it becomes a tetrachotomy; this seems to be a weakness of the question.) – Artie Prendergast-Smith Feb 2 at 21:14
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From the movie $\pi$: "Hold on. You have to slow down. You're losing it. You have to take a breath. Listen to yourself. You're connecting a computer bug I had with a computer bug you might have had and some religious hogwash. You want to find the number 216 in the world, you will be able to find it everywhere. 216 steps from a mere street corner to your front door. 216 seconds you spend riding on the elevator. When your mind becomes obsessed with anything, you will filter everything else out and find that thing everywhere." – Benjamin Dickman Feb 3 at 10:05
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This feels like an awfully arbitrary and contentless question to me. This seems like asking, what are some interesting sets with 3 elements? Voting to close. A more reasonable question would be to ask for examples where there is an interesting hyperbolic-parabolic-elliptic trichotomy and connections between these examples. – Eric Wofsey Feb 3 at 13:28
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Let's please not be quick to close this question. There are some intriguing clusters of answers, some along a (negative, 0, positive) axis, and others along a (closed, open, closed) axis [if I may put it crudely], and others along other axes still. There is a certain amount of erudition in both the question and answers, and there is some potential to glimpse "analogies between analogies" in the words of Ulam. It could be a useful exercise to pursue this. – Todd Trimble Feb 4 at 2:29
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Meta thread here: meta.mathoverflow.net/discussion/1525/…. Please upvote this comment so it appears "above the fold" – David White Feb 4 at 16:36
show 28 more comments
## 30 Answers
Almost everything in mathematics, indeed, can be "hyperbolic", "parabolic" or "elliptic". Like PDE's, Riemann surfaces, or manifolds of higher dimension, fractional-linear transformations, fixed points of a map, etc.
Not even mentioning the 3 kinds of the conic sections:-)
Of course this can be traced back to the fundamental trichotomy "positive", "zero" and "negative".
In differential geometry we have three great areas: "positive curvature", "negative curvature" and "zero curvature".
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The curvature trichotomy indeed comes up in many different ways, e.g. look at the striking differences between positive, $0$ (hyper-Kaehler) and negative quaternionic-Kaehler geometry. – Paul Reynolds Feb 3 at 12:50
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The trichotomy elliptic-parabolic-hyperbolic holds for Riemann surfaces, but in higher dimensions there are far more possibilities. In dimension 3, there are Thurston's 8 geometries ( and usually 3-manifolds have to be decomposed into pieces to be geometric). In dimension 4, Wallach had a list of 17 geometries, but one item in his list actually contains an infinite number of geometries. (And certainly not all 4-manifolds are locally homogeneous.) – unknown (google) Feb 4 at 12:31
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Of course, in higher dimensions we have more possibilities, but still they are frequently roughly classified into "hyperbolic", "parabolic" and "elliptic" types. – Alexandre Eremenko Feb 5 at 12:52
### 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.
1. After passing to a subsequence if necessary, a sequence of real numbers either (a) converges to a real number; (b) diverges to $+\infty$; or (c) diverges to $-\infty$. In a similar vein, a sequence of positive real numbers either (a) converges to a positive real; (b) diverges to $+\infty$; or (c) diverges to $0$. In nonstandard analysis, these trichotomies become those of being bounded, negative unbounded, or positive unbounded, or of being infinitesimal, unbounded, or neither (i.e. both bounded, and bounded away from zero). These are of course variants of the basic $(-,0,+)$ trichotomy.
2. Up to isomorphism, there are only three types of (connected) one-dimensional algebraic groups over an algebraically closed field: the additive group of the field, the multiplicative group of that field, and the elliptic curves over that field. (This last family would definitely be the "middle column".) This is of course connected to many of the other trichotomies previously mentioned. On the Riemann surface side, it comes from the fact that all one-dimensional connected complex groups are isomorphic to ${\bf C}/\Gamma$ for some discrete subgroup $\Gamma$ of ${\bf C}$, which can have rank 0 (additive case), 1 (multiplicative case), or 2 (elliptic curve case).
3. If one squints at it in just the right way, the classification of finite simple groups is a trichotomy: cyclic, Lie type (including Lie over F_1, i.e. alternating group), or sporadic. (Of course, it can be sliced in many other ways; counting the items in this Wikipedia list, for instance, would make it a tetratetracontachotomy.) Sometimes it is conceptually useful to split up the large Lie type groups into three regimes: large characteristic but bounded rank; large rank but bounded characteristic (including the alternating groups); and large characteristic and large rank. (Alternatively, one can partition into bounded rank, alternating, and unbounded rank.) One can debate as to which of these categories is the "middle column".
4. If $\xi_1,\xi_2,\xi_3$ are three frequencies with $\xi_3 = \xi_1+\xi_2$, then we have the Littlewood-Paley trichotomy: (a) "high-low" interactions with $|\xi_1| \gg |\xi_2|$ and $|\xi_1| \sim |\xi_3|$; (b) "low-high" interactions with $|\xi_1| \ll |\xi_2|$ and $|\xi_2| \sim |\xi_3|$; and (c) "high-high" interactions with $|\xi_1| \sim |\xi_2|$ and $|\xi_1| \gg |\xi_3|$. (One has to carefully demarcate the boundaries between these three possibilities to ensure it is a true trichotomy.) To an algebraic geometer, this would reflect the Y-shaped nature of the amoeba of the set $\{ (\xi_1,\xi_2,\xi_3): \xi_3 = \xi_1 + \xi_2 \}$. This trichotomy is important in harmonic analysis and PDE, and in particular in the paradifferential calculus of products and paraproducts (see e.g. this blog post of mine). Often, one of the three interactions will be the most dominant, reflecting either a high-to-low frequency cascade or a low-to-high frequency cascade, but it depends heavily on the situation. Note that this trichotomy is basically a variant of the $(<,=,>)$ trichotomy.
5. (ADDED LATER) Another variant of the $(<,=,>)$ trichotomy: most basic examples of semilinear PDE (or more precisely, a semilinear PDE problem, such as an initial value problem in a certain function space) can be classified as subcritical, critical, or supercritical, depending on whether the nonlinear component of the PDE is "weaker than", "comparable to", or "stronger than" the linear component in a suitable asymptotic limit (usually the fine scale/high frequency limit, although for scattering theory the coarse scale/low frequency limit is the relevant one instead). This distinction (which can usually be made precise through a scaling analysis or dimensional analysis) is often decisive in determining the difficulty level of the PDE problem. For instance, the regularity problem for 3D Navier-Stokes is supercritical and thus considered close to intractable, but 2D Navier-Stokes is critical and was solved decades ago. The global analysis of Ricci flow (with surgery) was considered supercritical until Perelman discovered new monotone quantities that made it critical, which was absolutely necessary for Perelman to be able to execute the rest of Hamilton's program and solve the Poincare and geometrisation conjectures. In this trichotomy, the critical (or scale-invariant) case is generally viewed as the most interesting and delicate, with some very nice mathematical tools coming into play to control the interaction between different scales. Perhaps it should also be pointed out that this trichotomy is orthogonal to the elliptic/parabolic/hyperbolic trichotomy, which only concerns the linear component of the PDE and not the nonlinear component, and all nine combinations (critical elliptic PDE, supercritical parabolic PDE, etc.) are studied in the literature.
6. (ADDED YET LATER) In analysis, there are basically three scenarios that prevent a weakly convergent sequence $f_n$ of functions in some function space from being strongly convergent in that space: (a) escape to "horizontal infinity" (basically, the support of the function runs off to spatial infinity, i.e. moving bump type examples); (b) escape to "vertical infinity" (the peaks of the function go to infinity, e.g. a sequence of approximations to the identity converging weakly but not strongly to a delta function); and (c) escape to "frequency infinity" (the functions become increasingly oscillatory). If one can shut down all three modes of escape then one can recover strong convergence, and thus also strong (pre)compactness, cf. the Arzela-Ascoli theorem which has three hypotheses (compact domain, pointwise boundedness, equicontinuity) to shut down (a), (b), and (c) respectively. In Section 2.9 of Lieb-Loss, these three scenarios are called "wanders off to infinity", "goes up the spout", and "oscillates to death" respectively.
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+1 for tetratetracontachotomy. I think I typed that correctly. – Lee Mosher Feb 3 at 14:41
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Perhaps one should mention that with the classification of finite simple groups, things are usually in **dichotomies**: for instance, the dichotomy theorem (odd or even-type characteristic), small vs. big (quasithin or large), etc. – Steve D Feb 4 at 4:23
Each set in a topological space partitions the space into three parts: interior, boundary, exterior.
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I don't understand why I am downvoted. I think my example is most appropriate, it is one of the fundamental trichotomies in mathematics. In fact it is directly connected to some other fundamental examples in the original post and in other responses such as "negative, zero, positive" or "hyperbolic, parabolic, elliptic" or "false, undecidable, true". – GH Feb 4 at 8:24
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+1. I think your example is good, and indeed does resonate with these other trichotomies, in fact builds a bridge between "negative, zero, positive" and "false, between-false-and-true, true" that I hadn't guessed. Indeed, it should be compared to the MO question on why we use the open-set formulation of 'topology', especially in answers such as sigfpe's (Dan Piponi's) and the summary given by Marcos here: mathoverflow.net/questions/19152/…. Compare also Tarski's interpretation of intuitionistic logic in a topology of a space. – Todd Trimble Feb 4 at 10:32
@Todd: Thank you for your positive feedback and support! – GH Feb 4 at 11:57
A simple yet useful one: An irreducible complex representation of a compact Lie group is either 'real', 'quaternionic', or 'complex'. That is, it is the complexification of a real irreducible, or it can be considered quaternionic through the existence of an equivariant conjugate-linear (real-)automorphism $j$ that squares to $-I$, or it is neither.
The statement combines Schur's Lemma and the fact that there are three associative real division algebras, here seen through complex eyes.
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This theorem of Soler is just great! Thank you for this link! – Vesselin Dimitrov Feb 3 at 7:53
There are exactly three finite-dimensional, associative division algebras over $\mathbb R$: $$\mathbb R, ~~ \mathbb C,~~\mathbb H$$
This was also mentioned in Paul Reynolds' answer.
Arnol'd thought this was a particularly fundamental trichotomy, and he has a fascinating table of "related" trichotomies in his book "Arnol'd's Problems" (sorry for the apostrophe catastrophe). Here's a picture I uploaded a few years ago:
Arnol'd's table
I wonder if these triples are in the same "family" as $(-1 ,0 , 1)$. At first they seem to be different because there is no distinguished element, analogous to $0$ - but perhaps $\mathbb C$ plays this role? I think the proof that there are three division algebras might actually use the $(-1,0,1)$ trichotomy, but I don't have a moment to think about it right now - maybe someone can leave a comment confirming this? (Edit: I could also imagine $\mathbb R$ playing the role of $0$, because it is the identity element for the tensor product, for example - any opinions either way?)
Another question I find interesting, which I'd hoped to think about at some point for fun, asks not just for (related) occurrences of 3-element sets in mathematical classifications, but more generally, for related occurrences of $n$-element sets for small $n$. I'd always dreamed of taking the number $5$ of regular platonic solids, for example, and trying to deduce from this as many other "small, finite number classifications" as possible.
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I'm glad Arnol'd puts Pontryagin classes in the quaternionic column. This point seems to often go unmentioned. – Paul Reynolds Feb 3 at 11:37
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@Sam, in the setting of my answer, if you integrate $\chi(g^2)$ over all $g$ in the group $G$ where $\chi$ is the character of the rep, you get $1,0,-1$ when the rep is 'real', 'complex' or 'quaternionic' respectively. Another manifestation of the $1,0,-1$ trichotomy is in the behaviour of tensor products over $\mathbb{C}$ of the different types. – Paul Reynolds Feb 3 at 14:12
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A longer version of this table with commentary is in Polymathematics : is mathematics a single science or a set of arts? V. I. Arnold. basepub.dauphine.fr/bitstream/handle/123456789/… – unknown (yahoo) Feb 4 at 2:21
An infinite finitely generated group $G$ has either 1, 2 or infinitely many ends; the last case is equivalent to saying that $G$ splits as an amalgamated free product or an HNN extension over a finite subgroup, by a theorem of Stallings. (Of course the assumption that $G$ is infinite brushes a fourth possibility under the carpet, namely that the number of ends is zero.)
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Snap!.......... – Artie Prendergast-Smith Feb 2 at 21:31
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I've just seen that you beat me to it in your comment to the original question! – shane.orourke Feb 2 at 21:53
Your answer is much better than my lazy comment, though, because of the extra details about Stalling's theorem. – Artie Prendergast-Smith Feb 2 at 22:09
Finite, countable, and uncountable.
Those are the three distinctions we have for cardinalities. For most people uncountable would usually mean $2^{\aleph_0}$. But even for set theorists, given a model of ZFC, the finite sets are finite, the countable sets are countable, and the rest is madness$^*$.
Replacing cardinality by topological-measure theoretic properties of subsets of an ordinal $\kappa$, there are non-stationary sets (small); stationary sets (big, but not too big); and clubs (which is practically everything).
$^*$ Good madness! Like a smile without a cat.
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'...and the rest is madness' +! :-) – David Roberts Feb 5 at 1:26
The representation type of a finite dimensional algebra is either finite, tame or wild. This is a theorem of Drozd.
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Every isometry of a proper $CAT(-1)$ space is either elliptic, parabolic, or hyperbolic: elliptic means fixes a finite point; hyperbolic means fixes two infinite points connected by a translation axis, equivalently translation distance bounded away from zero; parabolic means translation distance limiting to zero but no fixed point. There are versions for proper Gromov hyperbolic spaces, and even for the nonproper case, if you are willing to "quasify" the statements of the cases, and if you are willing to let the trichotomy degenerate to a dichotomy.
Every isometry of Teichmuller space is either elliptic, parabolic, or hyperbolic. This is Bers' form of Thurston's trichotomy for mapping classes: finite order, reducible, pseudo-Anosov. This trichotomy also has an interpretation in terms of the action of the mapping class group on the curve complex which is a nonproper Gromov hyperbolic space by a theorem of Masur and Minsky.
For elements of $Out(F_n)$, the outer automorphism group of a rank $n$ free group, there are related trichotomies and other -otomies coming from the work of Bestvina, Feighn, and Handel on relative train track theory. The simplest one is that every element of $Out(F_n)$ is either of finite order, or of polynomial growth, or of exponential growth.
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Previous MO questions:
• rational/trigonometric/elliptic trichotomy http://mathoverflow.net/questions/58040/groups-quantum-groups-and-fill-in-the-blank
• trichotomy of interrelated model structures: h-model, q-model, m-model http://mathoverflow.net/questions/86942/is-the-category-of-metric-spaces-and-continuous-maps-quillen-equivalent-to-top
• "such "log-exp functions" are either eventually positive, eventually zero, or eventually negative. ... It guarantees that the germs at infinity of such functions do indeed form a field K." http://mathoverflow.net/questions/45284/examples-of-sequences-whose-asymptotics-cant-be-described-by-elementary-function
• a function of a complex variable with an algebraic addition theorem must be: 1) A rational function, 2) A rational function of e^px, or 3) A rational function of the Weierstrass elliptic function and its derivative. http://mathoverflow.net/questions/96452/trig-functions-based-on-convex-curves
• "Every finitely generated infinite profinite group has a just infinite quotient. There is a trichotomy due to Wilson (and refined by Grigorchuk) describing what they can look like." http://mathoverflow.net/questions/49591/what-is-the-virtue-of-profinite-groups-as-mathematical-objects/68895
• "there is a trichotomy of curves given by g=0, g=1, and g≥2. If you look at topological, geometric, arithmetic properties of these curves, their properties align very strongly with these classes." http://mathoverflow.net/questions/56011/why-should-i-believe-the-mordell-conjecture
• Kodaira dimension. κ(Y)<0, κ(Y)=0, κ(Y)=dimY. http://mathoverflow.net/questions/81913/how-frequent-are-smooth-projective-varieties-with-anti-ample-canonical-bundle
• "Rank and period of primes in the Fibonacci sequence; a trichotomy," Fib. Quart., 45 (No. 1, 2007), 56-63). http://mathoverflow.net/questions/84797/can-the-difference-of-two-distinct-fibonacci-numbers-be-a-square-infinitely-often
M.SE:
• "The set-theoretic setup of Categories for the working mathematician is somewhat subtle. ... There is therefore a trichotomy of small sets, large sets, and proper classes. This is not the usual practice: we normally think of all sets as being small." http://math.stackexchange.com/questions/201062/confusion-over-the-use-of-universes-in-category-theory
• "There are three distinct aspects of schemes that each have their own purpose" http://math.stackexchange.com/questions/99605/why-study-schemes/99615
TCS.SE:
• "one of the most amazing facts about logic is that consistency strength boils down to the question "what is the fastest-growing function you can prove total in this logic?" As a result, the consistency of many classes of logics can be linearly ordered! If you have an ordinal notation capable of describing the fastest growing functions your two logics can show total, then you know by trichotomy that either one can prove the consistency of the other, or they are equiconsistent." http://cstheory.stackexchange.com/questions/4816/axioms-necessary-for-theoretical-computer-science/4821
A frequently cited paper: "A trichotomy theorem in natural models of AD+", in "Set Theory and Its Applications", Contemporary Mathematics, vol. 533, Amer. Math. Soc., Providence, RI, 2011, pp. 227-258.
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"A frequently cited paper" Oh, that's nice to know. – Andres Caicedo Feb 4 at 22:44
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The first and fourth bullet points are essentially the same trichotomy :) – Gjergji Zaimi Feb 5 at 2:54
The reduction (special fiber) $E_{\mathfrak p}$ of (the Neron model of) an elliptic curve $E$ modulo a prime ${\mathfrak p}$ is one of:
1. good reduction = stable reduction = $E_{\mathfrak p}$ is non-singular
2. multiplicative reduction = semi-stable reduction = $E_{\mathfrak p}$ is a product of the multiplicative group times a finite group
3. additive reduction = unstable reduction = $E_{\mathfrak p}$ is a product of the additive group times a finite group
Of course, this trichotomy is a reflection of the fact that there are only three sorts of connected one-dimensional Lie groups, namely the additive group, the multiplicative group, and the compact case (elliptic curves).
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Knots are either torus, satellite or hyperbolic.
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Do you mean prime knots? – Bruce Westbury Feb 5 at 6:07
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A connected sum is a satellite knot, albeit in a somewhat trivial way, cf. en.wikipedia.org/wiki/Satellite_knot. So there's no need to restrict to prime knots for this trichotomy to work. – Danny Ruberman Feb 6 at 1:41
In algebraic topology, the sequence $$\text{ordinary cohomolgy},\qquad K\text{-theory},\qquad \text{elliptic cohomology}$$ closely parallels the trichotomy in the classification of algebraic groups ($\mathbb G_a$, $\mathbb G_m$, elliptic curves).
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Every finitely generated group is either of polynomial growth, intermediate growth, or exponential growth.
As a statement, there is not much to this, the only mathematical content is that the growth function of every finitely generated group has an exponential upper bound.
But as a method of classifying finitely generated groups, it has been very fruitful: Gromov's theorem on groups of polynomial growth; the incredibly rich theory that arose from Grigorchuk's original construction of an intermediate growth group; and the emergence of rich classes of exponential growth groups such as word hyperbolic groups.
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Every logical statement is either true, undecidable or false.
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This is wrong because you are mixing apples and organges. You should say "provable", "undecidable", or "refutable". – Andrej Bauer Feb 2 at 21:41
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But also, there is the fourth possibility of "both provable and refutable", which arises when the underlying theory is inconsistent. – Joel David Hamkins Feb 2 at 21:57
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I'm not a logician. I knew I was going to get slated on this one but for some reason I wrote it anyway... – Simon Lyons Feb 2 at 23:10
Primes of the form 4n+1, of the form 4n+3, and 2 Primes over a prime p which are unramifed, tamely ramified, or wildly ramified – Gene Ward Smith Feb 3 at 0:13
@Andrej Bauer: like other commenters, I'm not a logician; but I think Simon Lyons' answer does not mix apples and oranges, as it probably really meant: «(provably) true, undecidable (in the sense of independent from the axioms), or (provably) false» . Otherwise, if not "provably true", what does it mean to be "true" for a bare logical statement? Nothing (except if you're assuming a metatheory of some kind in which you can talk of semantics). Anyway, JDHamkins's comment remarks there's actually a "quadrichotomy" so the answer wasn't fully in topic anyway. – Qfwfq Feb 5 at 14:27
show 2 more comments
Let me point out that Vladimir Arnold was quite interested in similar question. He called subj. "mathematical trinities", see e.g. his paper "Symplectization, Complexification and Mathematical Trinities". As far as I remember from his lectures, his ideas were that many of these "trinities" are actually related to each other; and he also considered subj. as a tool to invent to theories: see question marks at already cited "Arnold's table": jpg.
Let me also mention some "trinities" which occur in my own research related to Capelli identities (which are some non-commutative analogs of det(AB)=det(A)det(B) ).
Matrix trinity - a) generic b) symmetric c) antisymmetric
Here how it goes in Capelli (and related Cayley) identities:
a) generic matrices - original Capelli identity has been discovered by Capelli in 19-th century - it is for "generic matrices" $A=x_{ij}$ $B = \partial_{ji}$
b) symmetric matrices - analog of the Capelli identity has been discovered by Turnbull around 1940-ies - here $A=(x_{ij}+x_{ji})$ $B= \partial_{ji} + \partial_{ij}$.
c) antisymmetric matrices - analog has been found by Howe-Umeda and Kostant-Sahi around 1990, here $A=(x_{ij}-x_{ji})$ $B= \partial_{ji} - \partial_{ij}$.
Similar generalization were found for Cayley identity respectively: a) attributed to Cayley b) Garding 1948 c) Shimura 1984 - see arXiv:1105.6270 for quite a complete information.
My question: is it really trinity ? Or you can propose some analogs of Cayley-Capelli for some other matrices, say "symplectic" ?
It is might be strange, but other trinities like R,C,H also appears in the Capelli story - and they give different identities. Moreover trinities can be combined and we might get trinity^trinity^trinity...
Actually H-analog of the Capelli identity is not fully known for the momemnt - only analog for 1x1 matrices has been discovered quite recently by student of R. Borcherds, An Huang. Looking at this example I proposed some C-analogs of Capelli identities. Actually all generic/symmetric/antisymmetric can be complexified, hopefully there should exist quaternionic analogs and thus we might have trinity^trinity. Some partial results of trinity^trinity spirit for Cayley identity contained in loc. cit.
There are certain analogs of Capelli identities for classical Lie algebras: this can seen as gl/so/su trinity, well probably it is not the trinity in some strict sense. I have no idea can we have something like trinity^trinity^trinity ...
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The Gelfand-Kirillov dimension of an algebra is either zero, one, or at least $2$.
This sounds silly, but one has to remember that the GKdim is a real number (or $\infty$) and that there are algebras with GKdim equal to $r$ for all real $r\geq2$.
This is a combination of results of several people. See the books by McConnell and Robson, or by Krause and Lenagan.
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The endomorphism ring of an elliptic curve is either $\mathbb{Z}$, an order in a quadratic field, or an order in a quaternion algebra (ranks $1,2$, and $4$, respectively).
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In thermodynamic formalism for dynamical systems, a Hölder continuous potential function $\phi$ on a countable state topological Markov chain $(X,\sigma)$ is either positive recurrent, null recurrent, or transient. These correspond to the three possibilities for equilibrium states (shift-invariant measures maximising the quantity $h(\mu) + \int_X \phi\,d\mu$): existence of a finite equilibrium state is equivalent to positive recurrence; null recurrence is the boundary case where the equilibrium state becomes $\sigma$-finite but not finite, and transience is the case where there is no equilibrium state (all the weight has gone to infinity).
These can be characterised in terms of a particular sequence $a_n>0$: positive recurrence is equivalent to $\limsup a_n > 0$, null recurrence is equivalent to $a_n\to 0$ and $\sum a_n=\infty$, and transience is equivalent to $\sum a_n<\infty$. I imagine this trichotomy for sequences appears in other places as well.
Edit: It's worth mentioning that this trichotomy is also true for random walks on directed graphs (weighted or unweighted) -- historically I believe this is where it was first studied and where the terminology came from, but as a dynamicist I more immediately think of the interpretation above. In this setting the interpretations are as follows:
• Positive recurrent -- with probability 1, a random walk returns to where it started, and the expected return time is finite.
• Null recurrent -- the walk returns to the starting position with probability 1, but the expected return time is infinite.
• Transient -- with probability 1, the walk never returns to its starting position.
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Did anybody mention yet the different arithmetic-wise behaviour of algebraic curves of genus $0$, $1$, or $\geq2$?
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Over on Riemann surface land, this is the positive curvature / zero curvature / negative curvature trichotomy. (So this is the translate of one trichotomy by another (the Weil trichotomy)!) – Terry Tao Feb 4 at 16:29
In number theory, there is the trichotomy $\infty$, odd $p$, and $2$ (the oddest prime).
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A local domain can be pure characteristic $0$, pure positive characteristic, or mixed characteristic. (Meaning that the algebra has characteristic $0$ but its residue field has positive characteristic.)
Of course, this being a trichotomy relies on the fact that it's a domain; otherwise you could also have characteristic $p^n$ (with the residue field having characteristic $p$). I don't really know how that case is classified (I guess it's a form of mixed characteristic)...
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If the ring contains a field, then it can't be mixed characteristic. Better would be to say that any local domain has this trichotomy. – Dustin Cartwright Feb 5 at 14:42
Oops, you are absolutely right. Silly me. I will go fix this. – Harry Altman Feb 6 at 1:03
Here is a cluster of examples with a common theme, based partly on comments here and in an MO thread on whether an empty space should be considered connected, and partly on an article in the nLab, "too simple to be simple".
One should note that some of these trichotomies were once-upon-a-time considered dichotomies; for example, for many people in the past, 1 was a prime number. Also one should notice that there are a number of cross-connections (homomorphisms, if you will) between these examples, and that list is by no means complete (this is CW, so feel free to add more!).
• A module can be reducible, irreducible, or zero.
• A filter in a Boolean algebra can be "submaximal" (faute de mieux!), a maximal filter = ultrafilter, or an improper filter.
• An element in a p.i.d. is composite, prime, or a unit.
• A topological space (or a graph) can be disconnected, connected, or "unconnected" (empty).
• As for elements satisfying a predicate, one can have a multiplicity, unique existence, or nonexistence.
The last trichotomy is often considered a dichotomy: nonuniqueness vs. uniqueness. (I.e., nonexistence falls under the scope of uniqueness = "at most one".) But experience in mathematics, e.g. in category theory and its focus on universal properties, shows that unique-existence deserves to be considered in a category of its own.
• Theories can be incomplete but consistent, complete, or inconsistent.
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Unfortunately, the p.i.d. example has the blemish that one has to exclude the zero element (or add it to make it a tetrachotomy). – Terry Tao Feb 6 at 18:43
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Oh, thanks for pointing that out. I guess one could say instead that an ideal in a commutative ring is either submaximal, maximal, or unit, but this makes it look a lot like the previous example (in fact, the previous example becomes a special case in some sense). Perhaps I'll edit; I don't know. – Todd Trimble Feb 6 at 18:52
There are three types of subgroups of $PGL_2(\mathbb{C})$ that act on $\mathbb{P}^1$ non-transitively but with finitely many orbits:
(1) Type $T$: a one-dimensional torus
(2) Type $N$: the normalizer of a one-dimensional torus
(3) Type $U$: containing a non-trivial one-dimensional unipotent subgroup
This trichotomy plays a key role in the study of the geometry of spherical varieties, a class of algebraic varieties that includes grassmannians, flag varieties, toric varieties, algebraic monoids and symmetric spaces. It is particularly important in understanding the analogues of Schubert subvarieties (i.e. closures of orbits of a Borel subgroup) of a spherical variety.
In this example, there is no "middle" case as there is no intrinsic order to the three types.
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This is related to the trichotomy of elements of $PGL_2$ which are either elliptic, parabolic, or hyperbolic. – Agol Feb 3 at 3:36
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So I really should have ordered the list in the more fun manner of N-U-T. As far as I know, there is no use made of the ordering of this trichotomy in the study of spherical varieties, at least not explicitly. – Michael Joyce Feb 3 at 14:11
Every system of linear equations over the reals say has either no solution, or one solution, or infinitely many solutions.
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Suppose that a group $G$ has an action on a tree with no inversions and no global fixed point. Then either (a) $G$ is expressible as a (non-trivial) amalgamated free product; (b) $G$ is indicable (i.e. it maps onto an infinite cyclic group); or (c) $G$ is expressible as the union of a strictly ascending sequence of subgroups.
The negation of this property (every action of $G$ on a tree without inversions has a fixed point) is called FA by (J.-P.) Serre.
A more mundane, but related, example: a single automorphism of a tree either fixes a point, inverts an edge, or has an invariant (doubly-infinite) line contained in the tree on which the automorphism acts by translations.
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The classification theorem of closed surfaces states that any connected closed surface is homeomorphic to some member of one of these three families: the $sphere$; the connected sum of $g$ $tori$, ; the connected sum of $k$ $real$ $projective$ $planes$. this is a simple example of the trichotomy.sphere can be taken as $0$ tori. so $SPHERE$ serves the middle column.
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Every prime of a field either ramifies, splits or is inert in a field extension.
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NP-hard, Intermediate(conjectural), Polynomial time classes will separate $P$ from $NP$.
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Everything has an end, except for a sausage, which has two.
good-bye.
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You are right. +1 from me. – Vesselin Dimitrov Feb 9 at 10:50
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http://homotopical.wordpress.com/2009/12/10/ordinary-vs-generalized-cohomology-theories/
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# Motivic stuff
## Ordinary vs generalized cohomology theories
Posted by Andreas Holmstrom on December 10, 2009
(Post updated on 7 Jan 2010, based on some useful feedback from Tobias Barthel)
Introduction
When trying to classify or organize cohomology theories “found in nature” within (commutative) algebraic geometry, one realizes that there is a fundamental divide between generalized cohomology theories and the more restrictive notion of ordinary cohomology theories. This can result in confusion, for example when speaking of “universal” cohomology theories. Today I will try to give some explanation of what the difference is, although I am still very far from understanding this completely. These two notions of cohomology can be given precise definitions in topology, and there are several ways one could imagine making them precise also in algebraic geometry.
First a word about terminology: The phrase “generalized cohomology” has been used in several different ways in the literature. For example, the excellent article Motivic sheaves and filtrations on Chow groups by Jannsen, in the Motives volumes, uses a definition of generalized cohomology which corresponds to what I want to call ordinary cohomology. My choice of terminology seems more common nowadays, and it also corresponds to well-established practice in topology.
Cohomology theories in algebraic topology
In topology, a cohomology theory is by definition a sequence of functors on a suitable category of topological spaces, which satisfy the Eilenberg-Steenrod axioms. If we include the so called dimension axiom, we get the definition of ordinary cohomology, and if we exclude it, we get the definition of generalized cohomology. The classical Brown representability theorem says essentially that a generalized cohomology in the above sense is exactly the same thing as a functor represented by a spectrum. For precise statements, and details on the Eilenberg-Steenrod axioms and Brown representability, see Kono and Tamaki: Generalized cohomology (Google Books link), or May: A concise course in algebraic topology (pdf available here). Recall that for any abelian group G, we can define singular cohomology with coefficients in G; these cohomology theories are the only examples of ordinary cohomology theories in algebraic topology. There are many examples of generalized (non-ordinary) cohomology theories, for example complex cobordism, elliptic cohomology/topological modular forms, Brown-Peterson cohomology, and various forms of K-theory. Stable homotopy theory is the study of the category of spectra, i.e. the objects representing such generalized cohomology theories.
A very important class of generalized cohomology theories is the class of oriented theories. Roughly speaking, these are the cohomology theories which admit a reasonable theory of characteristic classes of line bundles. To any oriented cohomology theory one can associate a formal group, and there is a converse to this for formal groups which are “Landweber exact”. There is a “universal” oriented cohomology theory, namely complex cobordism. Some other examples: Singular cohomology with coefficients in any ring (or maybe Q-algebra) corresponds to the additive formal group. Complex K-theory corresponds to the multiplicative formal group. An elliptic cohomology theory corresponds to a formal group law coming from an elliptic curve. See Kono and Tamaki, or Lurie’s Survey of elliptic cohomology, for definitions and more details.
(I feel I should apologize for being so brief in this section, but there are already good references for algebraic topology, and today I want to get to some interesting algebraic geometry before I die. Hopefully I will find time to post in the future on things such as background on formal groups, and various approaches to defining the category of spectra.)
Algebraic geometry background
Although I did not give precise definitions in the algebraic topology discussion, such definitions can be found in the references, and with these definitions it is completely clear what one means by “ordinary” and “generalized” cohomology. In algebraic geometry, the same distinction clearly exists “in nature”, but to give precise definitions is more difficult. Roughly speaking, ordinary cohomology theories can be understood using only homological algebra, while generalized cohomology theories need the more flexible language of homotopical algebra.
Some examples of ordinary cohomology theories: Etale cohomology, Deligne cohomology, motivic cohomology, crystalline cohomology, de Rham cohomology, Betti cohomology. Most ordinary cohomology theories are defined as the sheaf cohomology of some sheaf of abelian groups (or more generally in terms of hypercohomology of some complex of sheaves of abelian groups). Here the notion of sheaf depends on some choice of Grothendieck topology. Many ordinary cohomology theories come in pairs of an “absolute” and a “geometric” theory (more about this in a future post!). Any Weil cohomology theory is ordinary, as well as any Bloch-Ogus theory. In algebraic topology, we have mentioned that ordinary theories correspond to abelian groups. The picture in algebraic geometry is more complicated, partly because we want to apply our cohomology theories to categories which are more complicated and more varied (see earlier posts on varieties and Weil cohomology for some examples).
Some examples of generalized (non-ordinary) cohomology theories: algebraic cobordism, algebraic K-theory, Witt groups. Theories of this kind are never defined in terms sheaf cohomology of abelian sheaves as above. However, there are more general notions of sheaf cohomology which do apply in most cases, using simplicial sheaves/presheaves in some form.
Remark 1: To speak of “universal” cohomology theories, it is necessary to specify what one means by “cohomology theory”. If we want to talk about all generalized cohomology theories, I guess the only thing that could be universal is some good notion of “stable homotopy type”. However, when it comes to more restrictive notions of cohomology, I am quite sure the following three statements can be made precise: (1) Algebraic cobordism is universal among oriented theories. (2) Pure motives is the universal ordinary (Weil) cohomology theory for smooth projective varieties over a field. (3) Motivic cohomology is the universal ordinary (Bloch-Ogus) cohomology for general varieties over a field.
Edit: Algebrac K-theory also has a universal property, at least when regarded as a functor on symmetric monoidal categories, but I am not sure about the details of this. There are some hints in Tyler Lawson’s answer to this MathOverflow question, and some more details in another answer of Clark Barwick.
Remark 2: It seems like the noun “motive” is used exclusively in connection with the world of ordinary cohomology theories (pure motives, mixed motives, triangulated categories of motives, etc). However, the adjective “motivic” is used in settings related to ordinary as well as generalized cohomology, e.g. motivic homotopy theory (generalized), motivic cohomology (ordinary).
Remark 3: There is a general heuristic principle which says that “any cohomology becomes ordinary after tensoring with Q” (i.e. killing all torsion). Some examples: (1) Algebraic K-theory cannot be defined in terms of homological algebra/abelian sheaf cohomology, but after tensoring with Q this becomes possible. (2) Homotopy groups in topology are very hard to compute in general, and homological algebra doesn’t help you at all, but after tensoring with Q, everything can be described in terms of (differential graded) homological algebra, thanks to rational homotopy theory. (3) The classical Grothendieck school in the 60s never really bothered about homotopical algebra – this seems related to the fact that they were always studying cohomology theories with coefficients in Q-algebras only. (4) In topology, and maybe also in algebraic geometry, any generalized cohomology theory $E$ has an associated Atiyah-Hirzebruch spectral sequence, which relates the ordinary cohomology with coefficients in $E^*(point)$ to $E$ itself, and I believe this spectral sequence tends to be complicated, but degenerate after tensoring with Q.
Edit: For oriented cohomology theories, this phenomenon is probably related to the fact that any formal group law is isomorphic to the additive one over the rationals.
Characterizing ordinary cohomology theories
I am not aware of any completely satisfactory definition of generalized cohomology theory in algebraic geometry (the best candidate would be “something represented by a spectrum in the sense of motivic homotopy theory”). However, we could pretend for a moment that such a definition exists, and then ask for a characterization/definition of ordinary cohomology theories. I can imagine four approaches to this question, but I have no idea if any of them can be made precise in any reasonable way.
Candidate 1: “A cohomology theory is ordinary if it is functorial not only with respect to maps in the classical sense, but also with respect to correspondences”. Recall that a correspondence from $X$ to $Y$ is a cycle on $X \times Y$, a special case being the graph of a map $X \to Y$. In his Motives volume article mentioned above, Jannsen remarks that any cohomology satisfying his axioms is functorial with respect to correspondences. Voevodsky also discusses this in the introduction to Homology of schemes I , claiming that in topology this functoriality property actually characterizes ordinary cohomology theories among all generalized theories. This surrounding discussion on what Voevodsky calls his underlying “simple topological intuition” was unfortunately not included in the preprint but only in the published version.
Candidate 2: “A cohomology theory is ordinary if it is oriented and its associated formal group is the additive formal group”. This definition makes sense if we have a notion of oriented theory in algebraic geometry, as well as a correspondence with formal group laws. There are definitions of oriented theories which might be suitable for our purpose here, but I am not sure about the formal groups bit. For closely related ideas, see for example this preprint of Naumann, Østvær, and Spitzweck.
Candidate 3: “A cohomology theory is ordinary if it factors through some suitable triangulated category of motives”. One reason for thinking that this might be a reasonable definition is that any theory I know of with this property (i.e. any theory which corresponds to a “realization functor”) should be thought of as being ordinary. Another reason is some very vague feeling that for a nice base scheme S, maybe the triangulated category DM(S) of motives relates to the motivic stable homotopy category SH(S) roughly as the derived category of abelian groups relates to the classical stable homotopy category (this might be complete nonsense though).
Edit: The right way of making this precise might be by using the motivic Eilenberg-MacLane functor. I hope to come back to this in a future post.
Candidate 4: Urs Schreiber advocates the viewpoint that every cohomology theory can be expressed in terms of a mapping space Map(X,K) in some higher topos in the sense of Lurie. Whenever this point of view is valid, one could probably define the cohomology to be ordinary if the target object K is in the essential image of some kind of Dold-Kan correspondence. I believe this idea is excellent, but it is not always clear which higher categories to work with in a concrete algebro-geometric problem. See the nLab pages on cohomology and generalized cohomology for more details.
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http://mathforum.org/mathimages/index.php?title=Euclidean_Algorithm&oldid=25220
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# Euclidean Algorithm
### From Math Images
Revision as of 20:00, 14 July 2011 by PhoebeJiang (Talk | contribs)
Euclid's Method to find the gcd
Fields: Number Theory and Algebra
Created By: Phoebe Jiang
Website: [ ]
Euclid's Method to find the gcd
This image shows Euclid's method to find the greatest common divisor (gcd) of two integers. The greatest common divisor of two numbers a and b is the largest integer that divides the numbers without a remainder.
Here I use 52 and 36 as an example to show you how Euclid found the gcd, so you have a sense of the Euclidean algorithm in advance. As you have probably noticed already, Euclid uses lines, defined as multiples of a common unit length, to represent numbers. First, use the smaller integer of the two, 36, to divide the bigger one, 52. Use the remainder of this division, 16, to divide 36 and you get the remainder 4. Now divide the last divisor, 16, by 4 and you find that they divide exactly. Therefore, 4 is the greatest common divisor. For every two integers, you will get the gcd by repeating the same process until there is no remainder.
You may have many questions so far: "What is going on here?" "Are you sure that 4 is the gcd of 52 and 36?" Don't worry. We will talk about them precisely later. This brief explanation is just to preheat your enthusiasm for Euclidean Algorithm! It is amazing to see that he explains and proves his algorithm relying on visual graphs, which is different from how we treat number theory now.
# Basic Description
We all know that 1 divides every number and that no positive integer is smaller than 1, so 1 is the smallest common divisor for any two integers a and b. Then what about the greatest common divisor? Finding the gcd is not as easy as finding the smallest common divisor.
When asked to find the gcd of two integers, a possible way we can think of is to prime factorize each integer and see which factors are common between the two integers, or we could simply try different numbers and see which number works. However, both approaches could be very sophisticated and time consuming as the two integers become relatively large.
About 2000 years ago, Euclid, one of the greatest mathematician of Greece, devised a fairly simple and efficient algorithm to determine the gcd of two integers, which is now considered as one of the most efficient and well-known early algorithms in the world. The Euclidean algorithm hasn't changed in 2000 years and has always been the the basis of Euclid's number theory.
Euclidean algorithm (also known as Euclid’s algorithm) describes a procedure for finding the greatest common divisor of two positive integers. This method is recorded in Euclid’s Elements Book VII. This book contains the foundation of number theory for which Euclid is famous.
The Euclidean algorithm comes in handy with computers because large numbers are hard to factor but relatively easy to divide. It is used in many other places and we’ll talk about its applications later.
# A More Mathematical Explanation
Note: understanding of this explanation requires: *Number Theory, Algebra
[Click to view A More Mathematical Explanation]
## The Description of Euclidean Algorithm
### Mathematical definitions and their abbreviations
• ' [...]
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## The Description of Euclidean Algorithm
### Mathematical definitions and their abbreviations
• a mod b is the remainder when a is divided by b (where mod = modulo).
Example: 7 mod 4 = 3; 4 mod 2 = 0; 5 mod 9 = 5
• a $\mid$ b means a divides b exactly or b is divided by a without any remainder.
Example: 3 $\mid$ 6 ; 4 $\mid$ 16
• gcd means the greatest common divisor, also called the greatest common factor (gcf), the highest common factor (hcf), and the greatest common measure (gcm).
• gcd(a, b) means the gcd of two positive integers a and b.
Keep those abbreviations in mind; you will see them a lot later.
### Precondition
The Euclidean Algorithm is based on the following theorem:
Theorem: $gcd(a, b) = gcd(b,~ a~mod~b)$ where $a > b$ and $a~ mod~ b ~\ne 0$
Proof: Since $a > b$, $a$ could be denoted as $a = kb + r$ with $0 \leqslant r < b$. Then $r = a~mod~b$. Assume $d$ is a common divisor of $a$ and $b$, thus $d \mid a , d\mid b$, or we could write them as $a= q_1 d, b = q_2 d.$ Because of $r = a - kb$, $r = q_1 d - k q_2 d = (q_1 - k q_2) d$ and we will get$d \mid r$. Therefore $d$ is also a common divisor of $(b, r) = (b, a~mod~b)$. Hence, the common divisors of $(a, b)$ and $(b, a~mod~b)$ are the same. In other words, $(a, b)$ and $(b, a~mod~b)$ have the same common divisors, and so they have the same greatest common divisor.
### Description
The description of the Euclidean algorithm is as follows:
• Input two positive integers, a,b (a > b)
• Output g, the gcd of a, b
• Internal Computation
1. Divide a by b and get the remainder r.
2. If r=0, report b as the gcd of a and b. If r $\ne$0, replace a by b and replace b by r. Go back to the previous step.
The algorithm process is like this:
$a = k_0b + r_1, \quad 0 < r_1 < b$
$b = k_1r_1 + r_2 , \quad 0 < r_2 < r_1$
$r_1 = k_2r_2 + r_3, \quad 0 < r_3 < r_2$
$r_2 = k_3r_3 + r_4, \quad 0 < r_4 < r_3$
... ...
$r_{n -2} = k_{n - 1}r_{n -1} + r_n,\quad 0 < r_{n -1} < r_n$
$r_{n -1} = k_nr_n, \quad\quad\quad\quad r_n = 0$
To sum up,
$(a, b) = (b, r_1) = (r_1, r_2) = (r_2, r_3) = (r_3, r_4) = ... ... =(r_{n- 2}, r_{n -1}) = (r_{n- 1}, r_n)$
$r_n$ is the gcd of a and b.
Note: The Euclidean algorithm is iterative, meaning that the next step is repeated using the result from the last step until it reaches the end.
### Example
An example will make the Euclidean algorithm clearer. Let's say we want to know the gcd of 168 and 64.
168 = 2 $\times$ 64 + 40
64 = 1 $\times$ 40 + 24
40 = 1 $\times$ 24 + 16
24 = 1 $\times$ 16 + 8
16 = 2 $\times$ 8
(168, 64) = (64, 24) = (24, 16) = (16, 8)
Therefore, 8 is the gcd of 168 and 64.
• Here's an applet for you to play around with finding the gcd by using the Euclidean algorithm.
## Proof of the Euclidean Algorithm
### Modern Proof
In order to prove that Euclidean algorithm works, the first thing is to show that the number we get from this algorithm is a common divisor of a and b. Then we will show that it is the greatest. Recall that
$a = k_0b + r_1, \quad a\,\bmod\,b = r_1, \quad 0 < r_1 < b$
$b = k_1r_1 + r_2 , \quad b\,\bmod\,r_1 = r_2, \quad 0 < r_2 < r_1$
$r_1 = k_2r_2 + r_3, \quad r_1\,\bmod\,r_2 = r_3, \quad 0 < r_3 < r_2$
$r_2 = k_3r_3 + r_4, \quad r_2\,\bmod\,r_3 = r_4, \quad 0 < r_4 < r_3$
... ...
$r_{n -2} = k_{n - 1}r_{n -1} + r_n,\quad r_{n-2}\,\bmod\,r_{n- 1} = r_n, \quad 0 < r_{n -1} < r_n$
$r_{n -1} = k_nr_n, \quad\quad\quad\quad r_{n- 1}\,\bmod\, \quad r_n = 0$
Based on the last two equations, we substitute $r_{n -1}$ with $k_nr_n$ in the last to second equation such that $r_{n -2} = k_{n -1}r_{n -1} + r_n = k_{n -1}k_nr_n + r_n = (k_{n -1}k_n + 1) r_n$.
Thus we have $r_n \mid r_{n -2}$.
From the equation before those two, we repeat the steps we did just now: $r_{n -3} = k_{n -2}r_{n -2} + r_{n -1} = k_{n -2} \Big( (k_{n -1} k_n + 1)r_n \Big) + k_nr_n = (k_{n -2} k_{n -1} k_n + k_{n - 2} + k_n) r_n$.
Now we know $r_n \mid r_{n -3}$.
Continue this process and we will find that $r_n \mid a, r_n \mid b$, so $r_n$, the number we get from Euclidean algorithm, is indeed a common divisor of a and b.
Second, we need to show that $r_n$ is the greatest among all the common divisors of a and b. To show that $r_n$ is the greatest, let's assume that there is another common divisor of a and b, d, where d is a positive integer. Then we could rewrite a and b as a = dm , b = dn, where m and n are also positive integers. This second part of the proof is going to be similar to the first part because they both repeat the same steps and eventually get the result, but this time we start from the first equation of the Euclidean algorithm:
Because $a = k_0b + r_1$ Therefore $r_1 = a - k_0b = dm - k_0 dn = (m - k_0 n) d$ (substitute dm for a and dn for b)
Therefore, $d \mid r_1$. Let $r_1 = d_1 d$.
Consider the second equation. Solve for $r_2$ in the same way. Because $b = k_1r_1 + r_2$ Therefore $r_2 = b - k_1r_1= dn - k_1d_1 d = (n - k_1d_1) d$ Thus, $d \mid r_2$
Continuing the process until we reach the last equation, we will get $d \mid r_n$. Since we pick d to represent any possible common divisor of a and b except $r_n, d \mid r_n$ means that $r_n$ divides any other common divisor of a and b, meaning that $r_n$ must be greater than all the other common divisors. Therefore, the number we get from the Euclidean Algorithm, $r_n$, is indeed the greatest common divisor of a and b.
### Euclid's Proof
Now let's look at Euclid's proof. Since Euclid's method of finding the gcd is based on several definitions, I quote the first 15 definitions in Book VII of his Elements for you.
Definitions
1. A unit is that by virtue of which each of the things that exist is called one.
2. A number is a multitude composed of units.
3. A number is a part of a number, the less of the greater, when it measures the greater.
4. but parts when it does not measure it.
5. The greater number is a multiple of the less when it is measured by the less.
6. An even number is that which is divisible into two equal parts.
7. An odd number is that which is not divisible into two equal parts, or that which differs by an unit from an even number.
8. An even-times even number is that which is measured by an even number according to an even number.
9. An even-times odd number is that which is measured by an even number according to an odd number.
10. An odd-times odd number is that which is measured by an odd number according to an odd number.
11. A prime number is that which is measured by an unit alone.
12. Numbers prime to one another are those which are measured by an unit alone as a common measure.
13. A composite number is that which is measured by some number.
14. Numbers composite to one another are those which are measured by some number as a common measure.
15. A number is said to multiply a number when that which is multiplied is added to itself as many times as there are units in the other, and thus some number is produced.[1]
NOTE:
In a nutshell, Euclid's one unit is the number 1 in algebra. He uses lines to represent numbers; the longer the line the greater the number.
In Def.3, "measure" means "divide."
Proposition 1. (See Image 1)
Two unequal numbers being set out, and the less being continually subtracted in turn from the greater, if the number which is left never measures the one before it until an unit is left, the original numbers will be prime to one another.
For, the less of two unequal numbers AB, CD being continually subtracted from the greater, let the number which is left never measure the one before it until an unit is left;
image 1
I say that AB, CD are prime to one another, that is, that an unit alone measures AB, CD.
For, if AB, CD are not prime to one another, some number will measure them.
Let a number measure them, and let it be E; let CD, measuring BF, leave FA less then itself,
let, AF measuring DG, leave GC less than itself, and let GC, measuring FH, leave an unit HA.
Since, then, E measures CD, and CD measure BF, therefore E also measures BF.
But it also measures the whole BA;
therefore it will also measure the remainder AF.
But AF measures DG;
therefore E also measures DG.
But it also measures the whole DC;
therefore it will also measure the remainder CG.
But CG measures FH;
therefore E also measures FH.
But it also measures the whole FA;
therefore it will also measure the remainder, the unit AH, though it is a number: which is impossible.
Therefore no number will measure the numbers AB, CD; therefore AB, CD are prime to one another. [1] [VII.Def.12] Q.E.D.
NOTE :
Here is a translation of Proposition 1.
Euclid wants to show that a and b must be prime to each other if we get 1 left instead of 0. Why?
Recall a > b. Write Euclid's proof in equations and we will get:
$a = kb + r, \quad 0 < r < b$
$b = qr + t, \quad 0 < t < r$
$r = lt + {\color{Red}1}, \quad 1 < r$
Assume a and b have a common measure e with e greater than one. Then e measures or divides r based on the first equation above, and e divides t based on the second equation above. Hence, e divides r and 1, but e cannot divide 1. In other words, 1 cannot be divided by e without any remainder because e is greater than 1. Therefore, a and b are prime to each other.
Proposition 2. (See Image 2)
Given two numbers not prime to one another, to find their greatest common measure.
Let AB, CD be the two given numbers not prime to one another.
Thus it is required to find the greatest common measure of AB, CD.
If now CD measures AB - and it also measures itself - CD is a common measure of CD, AB.
And it is manifest that it is also the greatest; for no greater number than CD will measure CD.
But, if CD does not measure AB, then, the less of the numbers AB, CD being continually subtracted from the greater, some number will be left which will measure the one before it.
For an unit will not be left; otherwise AB, CD will be prime to one another [VII, I], which is contrary to the hypothesis.
image 2
Therefore, some number will be left which will measure the one before it.
Now let CD, measuring BE, leave EA less than itself, let EA, measuring DF, leave FC less than itself, and let CF measure AE.
Since then, CF measures AE, and AE measures DF,
therefore CF will also measure DF.
But it also measures itself;
therefore it will also measure the whole CD.
But CD measures BE;
therefore CF also measures BE.
But it also measures EA;
therefore it will also measure the whole BA.
But it also measures CD;
therefore CF measures AB, CD.
Therefore CF is a common measure of AB, CD.
I sat next that it is also the greatest.
For, if CF is not the greatest common measure of AB, CD, some number which is greater than CF will measure the numbers AB, CD.
Let such a number measure them, and let it be G.
Now, since G measures CD, while CD measures BE, G also measures BE.
But it also measures the whole BA;
therefore it will also measure the remainder AE.
But AE measures DF;
therefore G will also measure DF.
But it also measures the whole DC;
therefore it will also measure the remainder CF, that is, the greater will measure the less: which is impossible.
Therefore no number which is greater than CF will measure the number AB, CD;
therefore CF is the greatest common measure of AB, CD.
PORISM. From this it is manifest that, if a number measure two numbers, it will also measure their greatest common measure. [1] Q.E.D
NOTE :
Prop.2 is pretty self-explanatory, proved in a similar way as Prop.1.
Comparing the modern proof with Euclid's proof, it is not hard to notice that the modern proof is more about algebra, while Euclid did his proof of his algorithm using geometry because, at that time, algebra had not been invented yet. However, the main idea is pretty much the same. They both prove that the result is a common divisor first and then show that it is the greatest among all the common divisors.
# Extended Euclidean Algorithm
Expand the Euclidean algorithm and you will be able to solve '''Bézout's identity''' for x and y where d = gcd(a, b):
$ax +by = gcd(a, b).$
Note: Usually either x or y will be negative since a, b and gcd(a, b) are positive and both a and b are usually greater than gcd(a, b).
## Lead In
Recall that
$a = k_0b + r_1, \quad 0 < r_1 < b$
$b = k_1r_1 + r_2 , \quad 0 < r_2 < r_1$
$r_1 = k_2r_2 + r_3, \quad 0 < r_3 < r_2$
$r_2 = k_3r_3 + r_4, \quad 0 < r_4 < r_3$
... ...
$r_{n -2} = k_{n - 1}r_{n -1} + r_n,\quad r_{n-2}\,\bmod\,r_{n- 1} = r_n, \quad 0 < r_{n -1} < r_n$
$r_{n -1} = k_nr_n, \quad\quad\quad\quad r_{n- 1}\,\bmod\, \quad r_n = 0$
Solve for $r_n$ using the second to last equation and we get:
$r_n = r_{n-2} - k_{n-1}r_{n-1}$
Because $r_n = gcd(a, b)$ by Euclidean algorithm,
$gcd(a, b) = r_{n -2} - k_{n -1}r_{n-1}$
Now let's solve for $r_{n-1}$ in the same way:
$r_{n -1} = r_{n -3} - k_{n-2}r_{n-2}$
Substitute Eq. 2 into Eq. 1:
$gcd(a, b) = r_{n -2} - k_{n -1}r_{n-1}$
$gcd(a, b) = r_{n -2} - k_{n -1}(r_{n -3} - k_{n-2}r_{n-2})$
$gcd(a, b) = r_{n -2} - k_{n -1}r_{n -3} + k_{n-1}k_{n-2}r_{n-2}$
${\color{Blue}gcd(a, b)} = (1 + k_{n-1}k_{n-2}){\color{Blue}r_{n-2}} - k_{n-1} {\color{Blue}r_{n - 3}}$
Now you can see gcd(a, b) is expressed by a linear combination of $r_{n-2}$ and $r_{n-3}$. If we continue this process by using the previous equations from the list above, we could get a linear combination of $r_{n-3}$ and $r_{n-4}$ with $r_{n-3}$ representing $r_{n-2}$ and $r_{n-4}$ representing $r_{n-3}$. If we keep going like this till we hit the first equation, we can express gcd(a, b) as a linear combination of a and b, which is what we intend to do.
## Description
The description of the extended Euclidean algorithm is:
Input: Two non-negative integers a and b ( $a \leqslant b$).
Output: d = gcd(a, b) and integers x and y satifying ax + by = d.
Computation:
• If $b = 0,$ set $d = a, x = 1, y = 0,</ math> and return <math>(d, x, y).$
• If not, set$x_2 = 1, x_1 = 0, y_2 = 0, y_1 = 1$
• While $b > 0$, do
$q = floor(\frac{a}{b}), r = a - qb, x = x_2 - qx_1, y = y_2 - q y_1.$
$a = b, b = r, x_2 = x_1, x_1 = x, y_2 = y_1, y_1 = y.$
• Set $d = a, x = x_2, y = y_2,$ and return $(d, x, y).$
## Example
This linear equation is going to be very complicated with all these notations, so it is much easier to understand with an example:
Solve for integers x and y such that 168x + 64y = 8.
• Apply Euclidean algorithm to compute gcd(168, 64), and we have a list of the following equations:
$168 = 2 \times 64 + 40$
$64 = 1 \times 40 + 24$
$40 = 1 \times 24 + 16$
$24 = 1 \times 16 + 8$
$16 = 2 \times 8 + 0$
Thus gcd(168, 64) = 8. So we know that we can solve for x and y by extended Euclidean algorithm.
• Use the extended Euclidean algorithm to get x and y:
From the fourth equation we get
$8 = 24 - 1 \times 16.$
From the third equation we get
$16 = 40 - 1 \times 24$.
• Substitute Eq. 4 into Eq. 3:
$8 = 24 - 1 \times (40 - 1 \times 24)$
$8 = 24 - 1 \times 40 + 1 \times 24$
$8 = 2 \times 24 - 1 \times 40$
• Do the same steps for the second equation in the list: $24 = 64 - 1 \times 40$
$8 = 2\times (64 - 1 \times 40) - 1\times 40$
$8 = 2\times 64 - 3\times 40$
For the first equation in the list, we get $40 = 168 - 2 \times 64$
$8 = 2\times 64 - 3 \times (168 - 2 \times 64)$
$8 = -3 \times 168 + 8 \times 64$
$\therefore x = -3, y = 8$
Euclidean algorithm and extended Euclidean algorithm makes it elegantly easy to compute the two Bézout's coefficients.
# Efficiency
How efficient could Euclidean algorithm be? Is it always perfect? Does Euclidean algorithm have shortcomings?
## Number of Steps - Lamé's Theorem
Gabriel Lamé is the first person who shows the number of steps required by the Euclidean algorithm. Lamé's theorem states that the number of steps in Euclidean algorithm for gcd(a,b) is at most five times the number of digits of the smaller number b. Thus, the Euclidean algorithm is linear-time in the number of digits in b.
Proof
Recall the division equations from the Euclidean algorithm,
$a = k_0b + r_1, \quad 0 < r_1 < b$
$b = k_1r_1 + r_2 , \quad 0 < r_2 < r_1$
$r_1 = k_2r_2 + r_3, \quad 0 < r_3 < r_2$
$r_2 = k_3r_3 + r_4, \quad 0 < r_4 < r_3$
... ...
$r_{n -2} = k_{n - 1}r_{n -1} + r_n,\quad 0 < r_n < r_{n -1}$
$r_{n -1} = k_nr_n, \quad r_{n+1} = 0$
We can tell from the equations that the number of steps is $n+1$ with n being the same n as in the division equations. So we want to prove that $n + 1 \leqslant 5k$ where k is the number of digits of b.
Notations:
1. a and b are integers and we assume a is bigger than b, so $a > b \geqslant 1$.
2. The Fibonacci Numbers are 1, 1, 2, 3, 5, 8, 13, ... , where every later number is the sum of the two previous numbers.
3. Denote $Fn$ as the nth Fibonacci number (i.e. $F_2 = 1, F_3 = 2, F_4 = 3, F_5 = 8$).
4. All the numbers in the division equations, $a, b, r_n, k_n$, are positive integers.
Analyze the division equations and we will have three conclusions:
• $r_n$ couldn't be 0, as otherwise all the remainders would be 0. Hence, $r_n \geqslant 1 = F_2$.
• We know that $r_{n- 1} > r_n$. Thus, according to the last equation $r_{n -1} = k_nr_n$, $k_n$ should be greater than 1: $k_n \geqslant 2$. Therefore, $r_{n -1} = k_nr_n \geqslant 2 r_n \geqslant 2F_2 = 2 = F_3$.
• $k_{n -1}$ is an integer, so $k_{n -1} \geqslant 1$. Thus, $r_{n -2} \geqslant r_n + r_{n -1}$. Since $r_{n -1} \geqslant F_3$ and $r_{n} \geqslant F_2,$ we have $r_{n-2} \geqslant F_2 + F_3 = F_4$.
Simplify the three conclusions:
$r_n \geqslant 1 = F_2$
$r_{n -1} = k_nr_n \geqslant 2 r_n \geqslant 2F_2 = F_3$
$r_{n-2} \geqslant r_{n} + r_{n -1} \geqslant F_2 + F_3 = F_4$
According to induction,
$r_{n -3} \geqslant r_{n -1} + r_{n -2} \geqslant F_3 + F_4 = F_5$
...
$r_1 \geqslant r_3 + r_2 \geqslant F_{n- 1} + F_n = F_{n+1}$
$b \geqslant r_2 + r_1 \geqslant F_n + F_{n+1} = F_{n+2}$
Therefore, $b \geqslant F_{n + 2}$
A theorem about the lower bound of Fibonacci numbers states that for all integers $n \geqslant 3$, it is true that $F_n > \alpha ^{n -2}$ where $\alpha = \frac{1+\sqrt{5}}{2} \approx 1.61803$ ( the sum of the Golden Ratio and 1).
Therefore, $b \geqslant F_{n+2} > \alpha ^n$
Thus, $\log_{10} b > \log_{10} \alpha ^n = n \log_{10} \alpha \approx n~0.208 \approx \frac{n}{5}$
Because b has k digits, $k > \log_{10} b$. Then
$k > \log_{10} b > \frac{n}{5}$
$k > \frac{n}{5}$
$5k > n$
$5k +1 > n + 1$
$5k \geqslant n + 1$
$n + 1 \leqslant 5k$.
Therefore, the number of steps $n + 1$ required by Euclidean algorithm for gcd(a,b) is no more than five times the number of digits of b $5k$.
## Shortcomings of the Euclidean Algorithm
The Euclidean algorithm is an ancient but good and simple algorithm to find the gcd of two nonnegative integers; it is well designed both theoretically and practically. Due to its simplicity, it is widely applied in many industries today. However, when dealing with really big integers (prime numbers over 64 digits in particular), finding the right quotients using the Euclidean algorithm adds to the time of computation for modern computers.
Stein's algorithm (also known as the binary GCD algorithm) is also an algorithm to compute the gcd of two nonnegative integers brought forward by J. Stein in 1967. This alternative is made to enhance the efficiency of the Euclidean algorithm, because it replaces complicated division and multiplication with addition, subtraction and shifts, which make it easier for the CPU to compute large integers.
The algorithm has the following conclusions:
• gcd(m, 0) = m, gcd(0, m) = m. It is because every number except 0 divides 0 and m is the biggest number that can divide itself.
• If e and f are both even integers, then gcd(e, f) = 2 gcd($\frac{e}{2}, \frac{f}{2}$), because 2 is definitely a common divisor of two even integers.
• If e is even and v is odd, then gcd(e, f) = gcd($\frac{e}{2}$, f), because 2 is definitely not a common divisor of an even integer and an odd integer.
• Otherwise both are odd and gcd(e, f) = gcd($\frac{|e-f|}{2}$, the smaller one of e and f). According to Euclidean algorithm, the difference of e and f could also divide the gcd of e and f. And Euclidean algorithm with a division by 2 results in an integer because the difference of two odd integers is even.
The description of Stein's algorithm:
Input:$u, v ( 0 < u \leqslant v )$;
Output: g = gcd(u, v)
Inner Computation:
1. g = 1.
2. While both u and v are even integers, do $u = \frac{u}{2}, v = \frac{v}{2}, g = 2g$.
3. While $u > 0$, do:
1. While u is even, do: $u =\frac{u}{2}$.
2. While v is even, do: $v = \frac{v}{2}$.
3. $t = \frac{\left\vert u - v \right\vert }{2}$.
4. If $u \leqslant v$, u = t; else, v = t.
4. Return ($g \cdot v$)
Example: $u=168, v=64$
1. $u = \frac{168}{2} = 84, v = \frac{64}{2} = 32, g = 2$; $u = \frac{84}{2} = 42, v = \frac{32}{2} = 16, g = 4$; $u = \frac{42}{2} = 21, v = \frac{16}{2} = 8, g =8$;
2. $u = 21 > 0;$
1. $v = \frac{8}{2} = 4; v = \frac{4}{2} = 2; v = \frac{2}{2} = 1;$
2. $t =\frac{\left\vert 21- 1 \right\vert}{2} = \frac{20}{2} = 10; t = 10$,
3. $\because u = 21 > 1 = v, \therefore u = t =10;$
3. $u = 10 > 0, v =1;$
1. $u = \frac{10}{2} = 5;$
2. $t = \frac{\left\vert 5- 1 \right\vert}{2} = \frac{4}{2} = 2; t = 2,$
3. $\because u = 5 > 1 = v, \therefore u = t = 2;$
4. $u = 2 > 0, v =1;$
1. $u = \frac{2}{2} = 1;$
2. $t = \frac{\left\vert 1 - 1 \right\vert}{2} = \frac{0}{2} = 0; t = 0,$
3. $\because u = 2 > 1 = v, \therefore u = t = 0;$
5. $g = g \cdot v = 8 \times 1 = 8.$
Now you may have a better understanding of the efficiency of Stein's algorithm, which substitutes divisions with faster operations by exploiting the binary representation that real computers use nowadays.
# Why It's Interesting
The Euclidean algorithm is a fundamental algorithm for other mathematical theories and various subjects in different areas. Please see The Application of Euclidean Algorithm to learn more about the Euclidean algorithm.
# Teaching Materials
There are currently no teaching materials for this page. Add teaching materials.
# References
[3] Artmann, Benno. (1999) ‘‘ Euclid-the creation of mathematics.’’ New York: Springer-Verlag.
[4] Weisstein, Eric W. Euclidean Algorithm. From MathWorld--A Wolfram Web Resource. Retrieved from http://mathworld.wolfram.com/EuclideanAlgorithm.html.
[6] Health, T.L. (1926) Euclid The Thirteen Books of the Elements. Volume 2, Second Edition. London: Cambridge University Press.
[8] Ranjan, Desh. Euclid’s Algorithm for the Greatest Common Divisor. Retrieved from http://www.cs.nmsu.edu/historical-projects/Projects/EuclidGCD.pdf.
[11] Gallian, Joseph A. (2010) Contemporary Abstract Algebra Seventh Edition. Belmont: Brooks/Cole, Cengage Learning.
[14] Wikipedia (Binary GCD Algorithm). (n.d.). Binary GCD Algorithm. Retrieved from http://en.wikipedia.org/wiki/Binary_GCD_algorithm.
# Future Directions for this Page
1. More applets or animations of Euclidean algorithm.
2. More pictures if possible.
3. Worst case of Euclidean algorithm.
Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.
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http://math.stackexchange.com/questions/56112/general-term-of-sequence
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General term of sequence
In my work on number of partitions of natural numbers in parts non greater than 5 arise the sequence $$775,1015,1285,1585,1915,2275,2665,3085,3535,4015,4525,5065,5635,6235,6865,7525,8215,8935,9685$$ I now first 60 terms of this sequence and all can be divided by 5. Any help on general term.
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Could you explain how, precisely, you obtained the number $775$? It obviously isn't the number of ways of partitioning $1$ into parts of size at most $5$... – Zhen Lin Aug 7 '11 at 12:16
775 is first term of an additional sequence that arise in context. – Adi Dani Aug 7 '11 at 12:24
1 Answer
Hint: Consider the corresponding sequence of differences: $240,270,300,330,\ldots$.
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Using this hint you can find that $a_n = 5(3n^2 + 39n + 113)$, $n \geq 1$. – Shai Covo Aug 7 '11 at 13:01
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http://en.wikipedia.org/wiki/Central_Limit_Theorem
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Central limit theorem
(Redirected from Central Limit Theorem)
In probability theory, the central limit theorem (CLT) states that, given certain conditions, the mean of a sufficiently large number of independent random variables, each with a well-defined mean and well-defined variance, will be approximately normally distributed.[1] The central limit theorem has a number of variants. In its common form, the random variables must be identically distributed. In variants, convergence of the mean to the normal distribution also occurs for non-identical distributions, given that they comply with certain conditions.
In more general probability theory, a central limit theorem is any of a set of weak-convergence theories. They all express the fact that a sum of many independent and identically distributed (i.i.d.) random variables, or alternatively, random variables with specific types of dependence, will tend to be distributed according to one of a small set of attractor distributions. When the variance of the i.i.d. variables is finite, the attractor distribution is the normal distribution. In contrast, the sum of a number of i.i.d. random variables with power law tail distributions decreasing as |x|−α−1 where 0 < α < 2 (and therefore having infinite variance) will tend to an alpha-stable distribution with stability parameter (or index of stability) of α as the number of variables grows.[2]
Central limit theorems for independent sequences
A distribution being "smoothed out" by summation, showing original density of distribution and three subsequent summations; see Illustration of the central limit theorem for further details.
Classical CLT
Let {X1, ..., Xn} be a random sample of size n—that is, a sequence of independent and identically distributed random variables drawn from distributions of expected values given by µ and finite variances given by σ2. Suppose we are interested in the sample average
$S_n := \frac{X_1+\cdots+X_n}{n}$
of these random variables. By the law of large numbers, the sample averages converge in probability and almost surely to the expected value µ as n → ∞. The classical central limit theorem describes the size and the distributional form of the stochastic fluctuations around the deterministic number µ during this convergence. More precisely, it states that as n gets larger, the distribution of the difference between the sample average Sn and its limit µ, when multiplied by the factor √n (that is √n(Sn − µ)), approximates the normal distribution with mean 0 and variance σ2. For large enough n, the distribution of Sn is close to the normal distribution with mean µ and variance σ2/n. The usefulness of the theorem is that the distribution of √n(Sn − µ) approaches normality regardless of the shape of the distribution of the individual Xi’s. Formally, the theorem can be stated as follows:
Lindeberg–Lévy CLT. Suppose {X1, X2, ...} is a sequence of i.i.d. random variables with E[Xi] = µ and Var[Xi] = σ2 < ∞. Then as n approaches infinity, the random variables √n(Sn − µ) converge in distribution to a normal N(0, σ2):[3]
$\sqrt{n}\bigg(\bigg(\frac{1}{n}\sum_{i=1}^n X_i\bigg) - \mu\bigg)\ \xrightarrow{d}\ N(0,\;\sigma^2).$
In the case σ > 0, convergence in distribution means that the cumulative distribution functions of √n(Sn − µ) converge pointwise to the cdf of the N(0, σ2) distribution: for every real number z,
$\lim_{n\to\infty} \Pr[\sqrt{n}(S_n-\mu) \le z] = \Phi(z/\sigma),$
where Φ(x) is the standard normal cdf evaluated at x. Note that the convergence is uniform in z in the sense that
$\lim_{n\to\infty}\sup_{z\in{\mathbf R}}\bigl|\Pr[\sqrt{n}(S_n-\mu) \le z] - \Phi(z/\sigma)\bigr| = 0,$
where sup denotes the least upper bound (or supremum) of the set.[4]
Lyapunov CLT
The theorem is named after Russian mathematician Aleksandr Lyapunov. In this variant of the central limit theorem the random variables Xi have to be independent, but not necessarily identically distributed. The theorem also requires that random variables |Xi| have moments of some order (2 + δ), and that the rate of growth of these moments is limited by the Lyapunov condition given below.
Lyapunov CLT.[5] Suppose {X1, X2, ...} is a sequence of independent random variables, each with finite expected value μi and variance σ 2
i
. Define
$s_n^2 = \sum_{i=1}^n \sigma_i^2$
If for some δ > 0, the Lyapunov’s condition
$\lim_{n\to\infty} \frac{1}{s_{n}^{2+\delta}} \sum_{i=1}^{n} \operatorname{E}\big[\,|X_{i} - \mu_{i}|^{2+\delta}\,\big] = 0$
is satisfied, then a sum of (Xi − μi)/sn converges in distribution to a standard normal random variable, as n goes to infinity:
$\frac{1}{s_n} \sum_{i=1}^{n} (X_i - \mu_i) \ \xrightarrow{d}\ \mathcal{N}(0,\;1).$
In practice it is usually easiest to check the Lyapunov’s condition for δ = 1. If a sequence of random variables satisfies Lyapunov’s condition, then it also satisfies Lindeberg’s condition. The converse implication, however, does not hold.
Lindeberg CLT
Main article: Lindeberg's condition
In the same setting and with the same notation as above, the Lyapunov condition can be replaced with the following weaker one (from Lindeberg in 1920). For every ε > 0
$\lim_{n \to \infty} \frac{1}{s_n^2}\sum_{i = 1}^{n} \operatorname{E}\big[(X_i - \mu_i)^2 \cdot \mathbf{1}_{\{ | X_i - \mu_i | > \varepsilon s_n \}} \big] = 0$
where 1{…} is the indicator function. Then the distribution of the standardized sums $\frac{1}{s_n}\sum_{i = 1}^n \left( X_i - \mu_i \right)$ converges towards the standard normal distribution N(0,1).
Multidimensional CLT
Proofs that use characteristic functions can be extended to cases where each individual X1, ..., Xn is an independent and identically distributed random vector in Rk, with mean vector μ = E(Xi) and covariance matrix Σ (amongst the individual components of the vector). Now, if we take the summations of these vectors as being done componentwise, then the multidimensional central limit theorem states that when scaled, these converge to a multivariate normal distribution.[6]
Let
$\mathbf{X_i}=\begin{bmatrix} X_{i(1)} \\ \vdots \\ X_{i(k)} \end{bmatrix}$
be the i-vector. The bold in Xi means that it is a random vector, not a random (univariate) variable. Then the sum of the random vectors will be
$\begin{bmatrix} X_{1(1)} \\ \vdots \\ X_{1(k)} \end{bmatrix}+\begin{bmatrix} X_{2(1)} \\ \vdots \\ X_{2(k)} \end{bmatrix}+\cdots+\begin{bmatrix} X_{n(1)} \\ \vdots \\ X_{n(k)} \end{bmatrix} = \begin{bmatrix} \sum_{i=1}^{n} \left [ X_{i(1)} \right ] \\ \vdots \\ \sum_{i=1}^{n} \left [ X_{i(k)} \right ] \end{bmatrix} = \sum_{i=1}^{n} \left [ \mathbf{X_i} \right ]$
and the average will be
$\left (\frac{1}{n}\right)\sum_{i=1}^{n} \left [ \mathbf{X_i} \right ]= \frac{1}{n}\begin{bmatrix} \sum_{i=1}^{n} \left [ X_{i(1)} \right ] \\ \vdots \\ \sum_{i=1}^{n} \left [ X_{i(k)} \right ] \end{bmatrix} = \begin{bmatrix} \bar X_{i(1)} \\ \vdots \\ \bar X_{i(k)} \end{bmatrix}=\mathbf{\bar X_n}$
and therefore
$\frac{1}{\sqrt{n}} \sum_{i=1}^{n} \left [\mathbf{X_i} - E\left ( X_i\right ) \right ]=\frac{1}{\sqrt{n}}\sum_{i=1}^{n} \left [ \mathbf{X_i} - \mu \right ]=\sqrt{n}\left(\mathbf{\overline{X}}_n - \mu\right)$.
The multivariate central limit theorem states that
$\sqrt{n}\left(\mathbf{\overline{X}}_n - \mu\right)\ \stackrel{D}{\rightarrow}\ \mathcal{N}_k(0,\Sigma)$
where the covariance matrix Σ is equal to
$\Sigma=\begin{bmatrix} {\color{Red}Var \left (X_{1(1)} \right)} & {\color{OliveGreen}Cov \left (X_{1(1)},X_{1(2)} \right)} & Cov \left (X_{1(1)},X_{1(3)} \right) & \cdots & Cov \left (X_{1(1)},X_{1(k)} \right) \\ {\color{OliveGreen}Cov \left (X_{1(2)},X_{1(1)} \right)} & {\color{Turquoise}Var \left (X_{1(2)} \right)} & {\color{RubineRed}Cov \left(X_{1(2)},X_{1(3)} \right)} & \cdots & Cov \left(X_{1(2)},X_{1(k)} \right) \\ Cov \left (X_{1(3)},X_{1(1)} \right) & {\color{RubineRed}Cov \left (X_{1(3)},X_{1(2)} \right)} & Var \left (X_{1(3)} \right) & \cdots & Cov \left (X_{1(3)},X_{1(k)} \right) \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ Cov \left (X_{1(k)},X_{1(1)} \right) & Cov \left (X_{1(k)},X_{1(2)} \right) & Cov \left (X_{1(k)},X_{1(3)} \right) & \cdots & Var \left (X_{1(k)} \right) \\ \end{bmatrix}.$
Central limit theorems for dependent processes
CLT under weak dependence
A useful generalization of a sequence of independent, identically distributed random variables is a mixing random process in discrete time; "mixing" means, roughly, that random variables temporally far apart from one another are nearly independent. Several kinds of mixing are used in ergodic theory and probability theory. See especially strong mixing (also called α-mixing) defined by α(n) → 0 where α(n) is so-called strong mixing coefficient.
A simplified formulation of the central limit theorem under strong mixing is:[7]
Theorem. Suppose that X1, X2, … is stationary and α-mixing with αn = O(n−5) and that E(Xn) = 0 and E(Xn12) < ∞. Denote Sn = X1 + … + Xn, then the limit
$\sigma^2 = \lim_n \frac{E(S_n^2)}{n}$
exists, and if σ ≠ 0 then $S_n / (\sigma \sqrt n)$ converges in distribution to N(0, 1).
In fact,
$\sigma^2 = E(X_1^2) + 2 \sum_{k=1}^{\infty} E(X_1 X_{1+k}),$
where the series converges absolutely.
The assumption σ ≠ 0 cannot be omitted, since the asymptotic normality fails for Xn = Yn − Yn−1 where Yn are another stationary sequence.
There is a stronger version of the theorem:[8] the assumption E(Xn12) < ∞ is replaced with E(|Xn|2 + δ) < ∞, and the assumption αn = O(n−5) is replaced with $\sum_n \alpha_n^{\frac\delta{2(2+\delta)}} < \infty.$ Existence of such δ > 0 ensures the conclusion. For encyclopedic treatment of limit theorems under mixing conditions see (Bradley 2005).
Martingale difference CLT
Main article: Martingale central limit theorem
Theorem. Let a martingale Mn satisfy
• $\frac1n \sum_{k=1}^n \mathrm{E} ((M_k-M_{k-1})^2 | M_1,\dots,M_{k-1}) \to 1$ in probability as n tends to infinity,
• for every ε > 0, $\frac1n \sum_{k=1}^n \mathrm{E} \Big( (M_k-M_{k-1})^2; |M_k-M_{k-1}| > \varepsilon \sqrt n \Big) \to 0$ as n tends to infinity,
then $M_n / \sqrt n$ converges in distribution to N(0,1) as n → ∞.[9][10]
Caution: The restricted expectation[clarification needed] E(X; A) should not be confused with the conditional expectation E(X|A) = E(X; A)/P(A).
Remarks
Proof of classical CLT
For a theorem of such fundamental importance to statistics and applied probability, the central limit theorem has a remarkably simple proof using characteristic functions. It is similar to the proof of a (weak) law of large numbers. For any random variable, Y, with zero mean and a unit variance (var(Y) = 1), the characteristic function of Y is, by Taylor's theorem,
$\varphi_Y(t) = 1 - {t^2 \over 2} + o(t^2), \quad t \rightarrow 0$
where o (t2) is "little o notation" for some function of t that goes to zero more rapidly than t2. Letting Yi be (Xi − μ)/σ, the standardized value of Xi, it is easy to see that the standardized mean of the observations X1, X2, ..., Xn is
$Z_n = \frac{n\overline{X}_n-n\mu}{\sigma \sqrt{n}} = \sum_{i=1}^n {Y_i \over \sqrt{n}}.$
By simple properties of characteristic functions, the characteristic function of Zn is
$\left[\varphi_Y\left({t \over \sqrt{n}}\right)\right]^n = \left[ 1 - {t^2 \over 2n} + o\left({t^2 \over n}\right) \right]^n \, \rightarrow \, e^{-t^2/2}, \quad n \rightarrow \infty.$
But this limit is just the characteristic function of a standard normal distribution N(0, 1), and the central limit theorem follows from the Lévy continuity theorem, which confirms that the convergence of characteristic functions implies convergence in distribution.
Convergence to the limit
The central limit theorem gives only an asymptotic distribution. As an approximation for a finite number of observations, it provides a reasonable approximation only when close to the peak of the normal distribution; it requires a very large number of observations to stretch into the tails.
If the third central moment E((X1 − μ)3) exists and is finite, then the above convergence is uniform and the speed of convergence is at least on the order of 1/n1/2 (see Berry-Esseen theorem). Stein's method[11] can be used not only to prove the central limit theorem, but also to provide bounds on the rates of convergence for selected metrics.[12]
The convergence to the normal distribution is monotonic, in the sense that the entropy of Zn increases monotonically to that of the normal distribution.[13]
The central limit theorem applies in particular to sums of independent and identically distributed discrete random variables. A sum of discrete random variables is still a discrete random variable, so that we are confronted with a sequence of discrete random variables whose cumulative probability distribution function converges towards a cumulative probability distribution function corresponding to a continuous variable (namely that of the normal distribution). This means that if we build a histogram of the realisations of the sum of n independent identical discrete variables, the curve that joins the centers of the upper faces of the rectangles forming the histogram converges toward a Gaussian curve as n approaches infinity, this relation is known as de Moivre–Laplace theorem. The binomial distribution article details such an application of the central limit theorem in the simple case of a discrete variable taking only two possible values.
Relation to the law of large numbers
The law of large numbers as well as the central limit theorem are partial solutions to a general problem: "What is the limiting behavior of Sn as n approaches infinity?"[citation needed] In mathematical analysis, asymptotic series are one of the most popular tools employed to approach such questions.
Suppose we have an asymptotic expansion of f(n):
$f(n)= a_1 \varphi_{1}(n)+a_2 \varphi_{2}(n)+O(\varphi_{3}(n)) \qquad (n \rightarrow \infty).$
Dividing both parts by φ1(n) and taking the limit will produce a1, the coefficient of the highest-order term in the expansion, which represents the rate at which f(n) changes in its leading term.
$\lim_{n\to\infty}\frac{f(n)}{\varphi_{1}(n)}=a_1.$
Informally, one can say: "f(n) grows approximately as a1 φ1(n)". Taking the difference between f(n) and its approximation and then dividing by the next term in the expansion, we arrive at a more refined statement about f(n):
$\lim_{n\to\infty}\frac{f(n)-a_1 \varphi_{1}(n)}{\varphi_{2}(n)}=a_2 .$
Here one can say that the difference between the function and its approximation grows approximately as a2 φ2(n). The idea is that dividing the function by appropriate normalizing functions, and looking at the limiting behavior of the result, can tell us much about the limiting behavior of the original function itself.
Informally, something along these lines is happening when the sum, Sn, of independent identically distributed random variables, X1, ..., Xn, is studied in classical probability theory.[citation needed] If each Xi has finite mean μ, then by the law of large numbers, Sn/n → μ.[14] If in addition each Xi has finite variance σ2, then by the central limit theorem,
$\frac{S_n-n\mu}{\sqrt{n}} \rightarrow \xi ,$
where ξ is distributed as N(0, σ2). This provides values of the first two constants in the informal expansion
$S_n \approx \mu n+\xi \sqrt{n}. \,$
In the case where the Xi's do not have finite mean or variance, convergence of the shifted and rescaled sum can also occur with different centering and scaling factors:
$\frac{S_n-a_n}{b_n} \rightarrow \Xi,$
or informally
$S_n \approx a_n+\Xi b_n. \,$
Distributions Ξ which can arise in this way are called stable.[15] Clearly, the normal distribution is stable, but there are also other stable distributions, such as the Cauchy distribution, for which the mean or variance are not defined. The scaling factor bn may be proportional to nc, for any c ≥ 1/2; it may also be multiplied by a slowly varying function of n.[16][17]
The law of the iterated logarithm specifies what is happening "in between" the law of large numbers and the central limit theorem. Specifically it says that the normalizing function $\sqrt{n\log\log n}$ intermediate in size between n of the law of large numbers and √n of the central limit theorem provides a non-trivial limiting behavior.
Alternative statements of the theorem
Density functions
The density of the sum of two or more independent variables is the convolution of their densities (if these densities exist). Thus the central limit theorem can be interpreted as a statement about the properties of density functions under convolution: the convolution of a number of density functions tends to the normal density as the number of density functions increases without bound. These theorems require stronger hypotheses than the forms of the central limit theorem given above. Theorems of this type are often called local limit theorems. See,[18] Chapter 7 for a particular local limit theorem for sums of i.i.d. random variables.
Characteristic functions
Since the characteristic function of a convolution is the product of the characteristic functions of the densities involved, the central limit theorem has yet another restatement: the product of the characteristic functions of a number of density functions becomes close to the characteristic function of the normal density as the number of density functions increases without bound, under the conditions stated above. However, to state this more precisely, an appropriate scaling factor needs to be applied to the argument of the characteristic function.
An equivalent statement can be made about Fourier transforms, since the characteristic function is essentially a Fourier transform.
Extensions to the theorem
Products of positive random variables
The logarithm of a product is simply the sum of the logarithms of the factors. Therefore when the logarithm of a product of random variables that take only positive values approaches a normal distribution, the product itself approaches a log-normal distribution. Many physical quantities (especially mass or length, which are a matter of scale and cannot be negative) are the products of different random factors, so they follow a log-normal distribution.
Whereas the central limit theorem for sums of random variables requires the condition of finite variance, the corresponding theorem for products requires the corresponding condition that the density function be square-integrable.[19]
Beyond the classical framework
Asymptotic normality, that is, convergence to the normal distribution after appropriate shift and rescaling, is a phenomenon much more general than the classical framework treated above, namely, sums of independent random variables (or vectors). New frameworks are revealed from time to time; no single unifying framework is available for now.
Convex body
Theorem. There exists a sequence εn ↓ 0 for which the following holds. Let n ≥ 1, and let random variables X1, …, Xn have a log-concave joint density f such that f(x1, …, xn) = f(|x1|, …, |xn|) for all x1, …, xn, and E(Xk2) = 1 for all k = 1, …, n. Then the distribution of
$\frac{X_1+\dots+X_n}{\sqrt n}$
is εn-close to N(0, 1) in the total variation distance.[20]
These two εn-close distributions have densities (in fact, log-concave densities), thus, the total variance distance between them is the integral of the absolute value of the difference between the densities. Convergence in total variation is stronger than weak convergence.
An important example of a log-concave density is a function constant inside a given convex body and vanishing outside; it corresponds to the uniform distribution on the convex body, which explains the term "central limit theorem for convex bodies".
Another example: f(x1, …, xn) = const · exp( − (|x1|α + … + |xn|α)β) where α > 1 and αβ > 1. If β = 1 then f(x1, …, xn) factorizes into const · exp ( − |x1|α)…exp( − |xn|α), which means independence of X1, …, Xn. In general, however, they are dependent.
The condition f(x1, …, xn) = f(|x1|, …, |xn|) ensures that X1, …, Xn are of zero mean and uncorrelated;[citation needed] still, they need not be independent, nor even pairwise independent.[citation needed] By the way, pairwise independence cannot replace independence in the classical central limit theorem.[21]
Here is a Berry-Esseen type result.
Theorem. Let X1, …, Xn satisfy the assumptions of the previous theorem, then [22]
$\bigg| \mathbb{P} \Big( a \le \frac{ X_1+\dots+X_n }{ \sqrt n } \le b \Big) - \frac1{\sqrt{2\pi}} \int_a^b \mathrm{e}^{-t^2/2} \, \mathrm{d} t \bigg| \le \frac C n$
for all a < b; here C is a universal (absolute) constant. Moreover, for every c1, …, cn ∈ R such that c12 + … + cn2 = 1,
$\bigg| \mathbb{P} ( a \le c_1 X_1+\dots+c_n X_n \le b ) - \frac1{\sqrt{2\pi}} \int_a^b \mathrm{e}^{-t^2/2} \, \mathrm{d} t \bigg| \le C ( c_1^4+\dots+c_n^4 ).$
The distribution of $(X_1+\dots+X_n)/\sqrt n$ need not be approximately normal (in fact, it can be uniform).[23] However, the distribution of c1X1 + … + cnXn is close to N(0, 1) (in the total variation distance) for most of vectors (c1, …, cn) according to the uniform distribution on the sphere c12 + … + cn2 = 1.
Lacunary trigonometric series
Theorem (Salem - Zygmund). Let U be a random variable distributed uniformly on (0, 2π), and Xk = rk cos(nkU + ak), where
• nk satisfy the lacunarity condition: there exists q > 1 such that nk+1 ≥ qnk for all k,
• rk are such that
$r_1^2 + r_2^2 + \cdots = \infty \text{ and } \frac{ r_k^2 }{ r_1^2+\cdots+r_k^2 } \to 0,$
• 0 ≤ ak < 2π.
Then[24][25]
$\frac{ X_1+\cdots+X_k }{ \sqrt{r_1^2+\cdots+r_k^2} }$
converges in distribution to N(0, 1/2).
Gaussian polytopes
Theorem Let A1, ..., An be independent random points on the plane R2 each having the two-dimensional standard normal distribution. Let Kn be the convex hull of these points, and Xn the area of Kn Then[26]
$\frac{ X_n - \mathrm{E} X_n }{ \sqrt{\operatorname{Var} X_n} }$
converges in distribution to N(0, 1) as n tends to infinity.
The same holds in all dimensions (2, 3, ...).
The polytope Kn is called Gaussian random polytope.
A similar result holds for the number of vertices (of the Gaussian polytope), the number of edges, and in fact, faces of all dimensions.[27]
Linear functions of orthogonal matrices
A linear function of a matrix M is a linear combination of its elements (with given coefficients), M ↦ tr(AM) where A is the matrix of the coefficients; see Trace_(linear_algebra)#Inner product.
A random orthogonal matrix is said to be distributed uniformly, if its distribution is the normalized Haar measure on the orthogonal group O(n, R); see Rotation matrix#Uniform random rotation matrices.
Theorem. Let M be a random orthogonal n × n matrix distributed uniformly, and A a fixed n × n matrix such that tr(AA*) = n, and let X = tr(AM). Then[28] the distribution of X is close to N(0, 1) in the total variation metric up to[clarification needed] 2√3/(n−1).
Subsequences
Theorem. Let random variables X1, X2, … ∈ L2(Ω) be such that Xn → 0 weakly in L2(Ω) and Xn2 → 1 weakly in L1(Ω). Then there exist integers n1 < n2 < … such that $( X_{n_1}+\cdots+X_{n_k} ) / \sqrt k$ converges in distribution to N(0, 1) as k tends to infinity.[29]
Q-analogues
A generalized q-analog of the classical central limit theorem has been described by Umarov, Tsallis and Steinberg[30] in which the independence constraint for the i.i.d. variables is relaxed to an extent defined by the q parameter, with independence being recovered as q->1. In analogy to the classical central limit theorem, such random variables with fixed mean and variance tend towards the q-Gaussian distribution, which maximizes the Tsallis entropy under these constraints. Umarov, Tsallis, Gell-Mann and Steinberg have defined q-analogs of all symmetric alpha-stable distributions, and have formulated a number of conjectures regarding their relevance to an even more general Central limit theorem.[31]
Applications and examples
Simple example
Comparison of probability density functions, p(k) for the sum of n fair 6-sided dice to show their convergence to a normal distribution with increasing n, in accordance to the central limit theorem. In the bottom-right graph, smoothed profiles of the previous graphs are rescaled, superimposed and compared with a normal distribution (black curve).
A simple example of the central limit theorem is rolling a large number of identical, unbiased dice. The distribution of the sum (or average) of the rolled numbers will be well approximated by a normal distribution. Since real-world quantities are often the balanced sum of many unobserved random events, the central limit theorem also provides a partial explanation for the prevalence of the normal probability distribution. It also justifies the approximation of large-sample statistics to the normal distribution in controlled experiments.
This figure demonstrates the central limit theorem. The sample means are generated using a random number generator, which draws numbers between 1 and 100 from a uniform probability distribution. It illustrates that increasing sample sizes result in the 500 measured sample means being more closely distributed about the population mean (50 in this case). It also compares the observed distributions with the distributions that would be expected for a normalized Gaussian distribution, and shows the chi-squared values that quantify the goodness of the fit (the fit is good if the reduced chi-squared value is less than or approximately equal to one). The input into the normalized Gaussian function is the mean of sample means (~50) and the mean sample standard deviation divided by the square root of the sample size (~28.87/√n), which is called the standard deviation of the mean (since it refers to the spread of sample means).
Real applications
A histogram plot of monthly accidental deaths in the US, between 1973 and 1978 exhibits normality, due to the central limit theorem
Published literature contains a number of useful and interesting examples and applications relating to the central limit theorem.[32] One source[33] states the following examples:
• The probability distribution for total distance covered in a random walk (biased or unbiased) will tend toward a normal distribution.
• Flipping a large number of coins will result in a normal distribution for the total number of heads (or equivalently total number of tails).
From another viewpoint, the central limit theorem explains the common appearance of the "Bell Curve" in density estimates applied to real world data. In cases like electronic noise, examination grades, and so on, we can often regard a single measured value as the weighted average of a large number of small effects. Using generalisations of the central limit theorem, we can then see that this would often (though not always) produce a final distribution that is approximately normal.
In general, the more a measurement is like the sum of independent variables with equal influence on the result, the more normality it exhibits. This justifies the common use of this distribution to stand in for the effects of unobserved variables in models like the linear model.
Other illustrations
Main article: Illustration of the central limit theorem
Given its importance to statistics, a number of papers and computer packages are available that demonstrate the convergence involved in the central limit theorem.[34]
History
Tijms writes:[35]
The central limit theorem has an interesting history. The first version of this theorem was postulated by the French-born mathematician Abraham de Moivre who, in a remarkable article published in 1733, used the normal distribution to approximate the distribution of the number of heads resulting from many tosses of a fair coin. This finding was far ahead of its time, and was nearly forgotten until the famous French mathematician Pierre-Simon Laplace rescued it from obscurity in his monumental work Théorie Analytique des Probabilités, which was published in 1812. Laplace expanded De Moivre's finding by approximating the binomial distribution with the normal distribution. But as with De Moivre, Laplace's finding received little attention in his own time. It was not until the nineteenth century was at an end that the importance of the central limit theorem was discerned, when, in 1901, Russian mathematician Aleksandr Lyapunov defined it in general terms and proved precisely how it worked mathematically. Nowadays, the central limit theorem is considered to be the unofficial sovereign of probability theory.
Sir Francis Galton described the Central Limit Theorem as:[36]
I know of scarcely anything so apt to impress the imagination as the wonderful form of cosmic order expressed by the "Law of Frequency of Error". The law would have been personified by the Greeks and deified, if they had known of it. It reigns with serenity and in complete self-effacement, amidst the wildest confusion. The huger the mob, and the greater the apparent anarchy, the more perfect is its sway. It is the supreme law of Unreason. Whenever a large sample of chaotic elements are taken in hand and marshaled in the order of their magnitude, an unsuspected and most beautiful form of regularity proves to have been latent all along.
The actual term "central limit theorem" (in German: "zentraler Grenzwertsatz") was first used by George Pólya in 1920 in the title of a paper.[37][38] Pólya referred to the theorem as "central" due to its importance in probability theory. According to Le Cam, the French school of probability interprets the word central in the sense that "it describes the behaviour of the centre of the distribution as opposed to its tails".[38] The abstract of the paper On the central limit theorem of calculus of probability and the problem of moments by Pólya[37] in 1920 translates as follows.
The occurrence of the Gaussian probability density 1 = e−x2 in repeated experiments, in errors of measurements, which result in the combination of very many and very small elementary errors, in diffusion processes etc., can be explained, as is well-known, by the very same limit theorem, which plays a central role in the calculus of probability. The actual discoverer of this limit theorem is to be named Laplace; it is likely that its rigorous proof was first given by Tschebyscheff and its sharpest formulation can be found, as far as I am aware of, in an article by Liapounoff. [...]
A thorough account of the theorem's history, detailing Laplace's foundational work, as well as Cauchy's, Bessel's and Poisson's contributions, is provided by Hald.[39] Two historical accounts, one covering the development from Laplace to Cauchy, the second the contributions by von Mises, Pólya, Lindeberg, Lévy, and Cramér during the 1920s, are given by Hans Fischer.[40] Le Cam describes a period around 1935.[38] Bernstein[41] presents a historical discussion focusing on the work of Pafnuty Chebyshev and his students Andrey Markov and Aleksandr Lyapunov that led to the first proofs of the CLT in a general setting.
A curious footnote to the history of the Central Limit Theorem is that a proof of a result similar to the 1922 Lindeberg CLT was the subject of Alan Turing's 1934 Fellowship Dissertation for King's College at the University of Cambridge. Only after submitting the work did Turing learn it had already been proved. Consequently, Turing's dissertation was never published.[42][43][44]
See also
• Asymptotic equipartition property
• Delta method – to compute the limit distribution of a function of a random variable.
• Illustration of the central limit theorem
• Theorem of de Moivre–Laplace
• Fisher–Tippett–Gnedenko theorem – limit theorem for extremum values (such as max{Xn})
• Central limit theorem for directional statistics - Central limit theorem applied to the case of directional statistics
• Erdős-Kac theorem - connects the number of prime factors of an integer with the normal probability distribution
• Stable distribution - distributions such that linear combinations of i.i.d. samples lead to samples with the same distribution
• Tweedie convergence theorem - A theorem that can be considered to bridge between the central limit theorem and the Poisson convergence theorem[45]
Notes
1. Rice, John (1995), Mathematical Statistics and Data Analysis (Second ed.), Duxbury Press, ISBN 0-534-20934-3 )[]
2. Voit, Johannes (2003), The Statistical Mechanics of Financial Markets, Springer-Verlag, p. 124, ISBN 3-540-00978-7
3. Billingsley (1995, p.357)
4. Bauer (2001, Theorem 30.13, p.199)
5. Billingsley (1995, p.362)
6. Van der Vaart, A. W. (1998), Asymptotic statistics, New York: Cambridge University Press, ISBN 978-0-521-49603-2, LCCN .V22 1998 QA276 .V22 1998
7. Billingsley (1995, Theorem 27.4)
8. Durrett (2004, Sect. 7.7(c), Theorem 7.8)
9. Durrett (2004, Sect. 7.7, Theorem 7.4)
10. Billingsley (1995, Theorem 35.12)
11. Stein, C. (1972), "A bound for the error in the normal approximation to the distribution of a sum of dependent random variables", Proceedings of the Sixth Berkeley Symposium on Mathematical Statistics and Probability: 583–602, MR 402873, Zbl 0278.60026
12. Chen, L.H.Y., Goldstein, L., and Shao, Q.M (2011), Normal approximation by Stein's method, Springer, ISBN 978-3-642-15006-7
13. Artstein, S.; Ball, K.; Barthe, F.; Naor, A. (2004), "Solution of Shannon's Problem on the Monotonicity of Entropy", Journal of the American Mathematical Society 17 (4): 975–982, doi:10.1090/S0894-0347-04-00459-X
14. Rosenthal, Jeffrey Seth (2000) A first look at rigorous probability theory, World Scientific, ISBN 981-02-4322-7.(Theorem 5.3.4, p. 47)
15. Johnson, Oliver Thomas (2004) Information theory and the central limit theorem, Imperial College Press, 2004, ISBN 1-86094-473-6. (p. 88)
16. Vladimir V. Uchaikin and V. M. Zolotarev (1999) Chance and stability: stable distributions and their applications, VSP. ISBN 90-6764-301-7.(pp. 61–62)
17. Borodin, A. N. ; Ibragimov, Il'dar Abdulovich; Sudakov, V. N. (1995) Limit theorems for functionals of random walks, AMS Bookstore, ISBN 0-8218-0438-3. (Theorem 1.1, p. 8 )
18. Petrov, V.V. (1976), Sums of Independent Random Variables, New York-Heidelberg: Springer-Verlag
19. Rempala, G.; Wesolowski, J.(2002) "Asymptotics of products of sums and U-statistics", Electronic Communications in Probability, 7, 47–54.
20. Klartag (2007, Theorem 1.2)
21. Durrett (2004, Section 2.4, Example 4.5)
22. Klartag (2008, Theorem 1)
23. Klartag (2007, Theorem 1.1)
24. Zygmund, Antoni (1959), Trigonometric series, Volume II, Cambridge . (2003 combined volume I,II: ISBN 0-521-89053-5) (Sect. XVI.5, Theorem 5-5)
25. Gaposhkin (1966, Theorem 2.1.13)
26. Barany & Vu (2007, Theorem 1.1)
27. Barany & Vu (2007, Theorem 1.2)
28. Meckes, Elizabeth (2008), "Linear functions on the classical matrix groups", Transactions of the American Mathematical Society 360 (10): 5355–5366, arXiv:math/0509441, doi:10.1090/S0002-9947-08-04444-9
29. Gaposhkin (1966, Sect. 1.5)
30. Umarov, Sabir; Tsallis, Constantino and Steinberg, Stanly (2008), "On a q-Central Limit Theorem Consistent with Nonextensive Statistical Mechanics", Milan j. Math. (Birkhauser Verlag) 76: 307–328, doi:10.1007/s00032-008-0087-y, retrieved 2011-07-27.
31. Umarov, Sabir; Tsallis, Constantino, Gell-Mann, Murray and Steinberg, Stanly (2010), "Generalization of symmetric α-stable Lévy distributions for q>1", J Math Phys. (American Institute of Physics) 51 (3): 033502, doi:10.1063/1.3305292, PMC 2869267, PMID 20596232.
32. Dinov, Christou & Sanchez (2008)
33. Marasinghe, M., Meeker, W., Cook, D. & Shin, T.S.(1994 August), "Using graphics and simulation to teach statistical concepts", Paper presented at the Annual meeting of the American Statistician Association, Toronto, Canada.
34. Henk, Tijms (2004), Understanding Probability: Chance Rules in Everyday Life, Cambridge: Cambridge University Press, p. 169, ISBN 0-521-54036-4
35. Galton F. (1889) Natural Inheritance[]
36. ^ a b Pólya, George (1920), "Über den zentralen Grenzwertsatz der Wahrscheinlichkeitsrechnung und das Momentenproblem", (in German) 8 (3–4): 171–181, doi:10.1007/BF01206525
37. ^ a b c Le Cam, Lucien (1986), "The central limit theorem around 1935", Statistical Science 1 (1): 78–91, doi:10.2307/2245503
38. Hald, Andreas History of Mathematical Statistics from 1750 to 1930, Ch.17.[]
39. Fischer, Hans (2011), A History of the Central Limit Theorem: From Classical to Modern Probability Theory, Sources and Studies in the History of Mathematics and Physical Sciences, New York: Springer, doi:10.1007/978-0-387-87857-7, ISBN 978-0-387-87856-0, MR 2743162, Zbl 1226.60004 (Chapter 2: The Central Limit Theorem from Laplace to Cauchy: Changes in Stochastic Objectives and in Analytical Methods, Chapter 5.2: The Central Limit Theorem in the Twenties)
40. Bernstein, S.N. (1945) On the work of P.L.Chebyshev in Probability Theory, Nauchnoe Nasledie P.L.Chebysheva. Vypusk Pervyi: Matematika. (Russian) [The Scientific Legacy of P. L. Chebyshev. First Part: Mathematics, Edited by S. N. Bernstein.] Academiya Nauk SSSR, Moscow-Leningrad, 174 pp.
41. Hodges, Andrew (1983) Alan Turing: the enigma. London: Burnett Books., pp. 87-88.[]
42. Zabell, S.L. (2005) Symmetry and its discontents: essays on the history of inductive probability, Cambridge University Press. ISBN 0-521-44470-5. (pp. 199 ff.)
43. Aldrich, John (2009) "England and Continental Probability in the Inter-War Years", Electronic Journ@l for History of Probability and Statistics, vol. 5/2, Decembre 2009. (Section 3)
44. Jørgensen, Bent (1997). The theory of dispersion models. Chapman & Hall. ISBN 978-0412997112.
References
• Barany, Imre; Vu, Van (2007), "Central limit theorems for Gaussian polytopes", The Annals of Probability (Institute of Mathematical Statistics) 35 (4): 1593–1621, arXiv:0610192, doi:10.1214/009117906000000791
• Bauer, Heinz (2001), Measure and Integration Theory, Berlin: de Gruyter, ISBN 3110167190
• Billingsley, Patrick (1995), Probability and Measure (Third ed.), John Wiley & sons, ISBN 0-471-00710-2
• Bradley, Richard (2007), Introduction to Strong Mixing Conditions (First ed.), Heber City, UT: Kendrick Press, ISBN 0-9740427-9-X
• Bradley, Richard (2005), "Basic Properties of Strong Mixing Conditions. A Survey and Some Open Questions", Probability Surveys 2: 107–144, arXiv:math/0511078v1, doi:10.1214/154957805100000104
• Dinov, Ivo; Christou, Nicolas; Sanchez, Juana (2008), "Central Limit Theorem: New SOCR Applet and Demonstration Activity", Journal of Statistics Education (ASA) 16 (2)
• Durrett, Richard (2004), Probability: theory and examples (4th ed.), Cambridge University Press, ISBN 0521765390
• Gaposhkin, V.F. (1966), "Lacunary series and independent functions", Russian Math. Surveys 21 (6): 1–82, doi:10.1070/RM1966v021n06ABEH001196 .
• Klartag, Bo'az (2007), "A central limit theorem for convex sets", Inventiones Mathematicae 168, 91–131.doi:10.1007/s00222-006-0028-8 Also arXiv.
• Klartag, Bo'az (2008), "A Berry-Esseen type inequality for convex bodies with an unconditional basis", Probability Theory and Related Fields. doi:10.1007/s00440-008-0158-6 Also arXiv.
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http://math.stackexchange.com/questions/234586/use-the-given-graph-f-over-the-interval-0-7-to-find-the-following
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# Use the given graph $f$ over the interval (0, 7) to find the following:
(a) The open intervals on which $f$ is increasing.
I answered $(0, 1), (3, 5), (5,7)$
(b) The open intervals on which $f$ is concave upward.
I answered $(1, 4)$
(c) The open intervals on which $f$ is concave downward.
I answered $(0, 1)$
(d) The coordinates of the points of inflection.
I answered:
$(1,3)$ smallest $x$-value
$(3,1)$
$(5,4)$ largest $x$-value
WebAssign said that I got all of these wrong. I am trying to understand why. Please help!
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Have you studied derivatives? – amWhy Nov 11 '12 at 2:19
Yes I have and I know them very well. I'm just not good at analyzing graphs. – dsta Nov 11 '12 at 2:21
## 2 Answers
For a, it is increasing on $(3,7)$. True, it has zero derivative at $5$, but $f(5) \gt f(x)$ for all $x$ close to and less than $5$. For b, it is clearly concave downward at $1$. It looks concave upward starting at $2$ for an interval $(2,4)$ and again on $(5,7)$ to me. Then for c, concave downward is $(0,2)$ and $(4,5)$. For d, the points of inflection are the changes in concavity, so $(2,2), (4,3), (5,4)$
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For question a, think of what it means to be increasing. When you take the derivative, and you find critical values, you have to test points around those critical values, to determine if it is a minimum or maximum. Take the function $f(x)=x^3$. The derivative is$f(x)=3x^2$, and the critical value is $x=0$. The test intervals to find the relative extrema, are $(-\infty,0)$ and $(0, \infty)$. If you take test values, you'll see that the derivative is positive on the two intervals, meaning that that tangent lines drawn at any point on the graph will have a positive slope; furthermore, implying that the function is increasing on $(-\infty, \infty)$ So there is relationship between the original function, and it's derivative: the function is increasing at a particular x-value when it's derivative is positive at that x-value; and decreasing at a particular x-value when it's derivative is negative at that x-value.
The function you give is indeed increasing on $(0,1)$ But look back at your two other intervals, $(3,5)$ and $(5,7)$ and draw tangent lines to the graph. What are the slopes of the tangent lines? Positive, negative?
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## The higher Van Kampen Theorems and computation of the unstable homotopy groups of spheres
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Since Ronnie Brown and his collaborators have come up with a general proof of the higher Van Kampen theorems, what impediments are there to using these to compute the unstable homotopy groups of spheres?
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Which result of Brown are you referring to? The higher Van Kampen theorem is usually called "Stover's spectral sequence" no? I haven't seen a computation of anything using any of these techniques, but I'm not sure which result of Brown you're referring to. – Ryan Budney Sep 24 2010 at 4:14
The "Higher Homotopy van Kampen Theorem" described in his book Nonabelian Algebraic Topology. – Harry Gindi Sep 24 2010 at 4:23
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@Ryan.Have a look at Ronnie Brown's web pages for a history of the van Kampen theorem in its higher dimensional forms and an extensive bibiography including calculations.Yes,there are spectral sequences that give some (sometimes poor) information in van Kampen type situations, for instance a paper in Topology by Artin and Mazur as well as the one that you mention, but the original vKT allows the complete calculation of a homotopy type of a union from the parts. The first Brown-Higgins generalisation allowed calculations that were difficult to do otherwise, but this is low dim. homotopy. – Tim Porter Sep 24 2010 at 6:56
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I think that it's worth pointing out that algorithms existed long before Brown's work for computing homotopy groups of spheres. The problem is that those algorithms are too slow to be of much practical use. I expect that the same is true for any answer coming from some kind of "higher van Kampen" theorem. There are patterns in the homotopy groups of spheres (eg the chromatic picture of the stable stems), but they are too complicated to arise from any simple description. – Nikita Sep 24 2010 at 13:53
@Tim Porter, can you provide a direct link? There's a lot of information on Brown's webpage and I'm not seeing "a history of the van Kampen Theorem" anywhere. – Ryan Budney Sep 24 2010 at 17:55
show 3 more comments
## 3 Answers
Here are some answers on the HHSvKT - I have been persuaded by a referee that we ought also to honour Seifert.
These theorems are about homotopy invariants of structured spaces, more particularly filtered spaces or n-cubes of spaces. For example the first theorem of this type
Brown, R. and Higgins, P.~J. On the connection between the second relative homotopy groups of some related spaces. Proc. London Math. Soc. (3) \textbf{36}~(2) (1978) 193--212.
said that the fundamental crossed module functor from pairs of pointed spaces to crossed modules preserves certain colimits. This allows some calculations of homotopy 2-types and then you need further work to compute the 1st and 2nd homotopy group; of course these two homotopy groups are pale shadows of the 2-type. As example calculations I mention
R. Brown Coproducts of crossed $P$-modules: applications to second homotopy groups and to the homology of groups'', {\em Topology} 23 (1984) 337-345.
(with C.D.WENSLEY), `Computation and homotopical applications of induced crossed modules', J. Symbolic Computation 35 (2003) 59-72.
In the second paper some computational group theory is used to compute the 2-type and so 2nd homotopy groups of some mapping cones of maps $Bf: BG \to BH$ where $f:G \to H$ is a morphism of groups.
For applications of the work with Loday I refer you for example to the bibliography on the nonabelian tensor product
www.bangor.ac.uk/r.brown/nonabtens.html
which has 100 items, and also
Ellis, G.~J. and Mikhailov, R. A colimit of classifying spaces. {Advances in Math.} (2010) arXiv: [math.GR] 0804.3581v1 1--16.
So in the tensor product work, we determine $\pi_3 S(K(G,1))$ as the kernel of a morphism $\kappa: G \otimes G \to G$ (the commutator morphism!).
These theorems have connectivity conditions which means they are restricted in their applications, and don not solve all problems! There is still some interest in computing homotopy types of some complexes which cannot otherwise be computed. It is also of interest that the calculations are generally nonabelian ones. So the aim is to make some aspects of higher homotopy theory more like the theory of the fundamental group(oid), which is why I coined the term `higher dimensional group theory' as indicating new structures underlying homotopy theory.
Even the 2-dimensional theorem on crossed modules seems little known or referred to! The proof is not so hard, but requires the notion of the homotopy double groupoid of a pair of pointed spaces. See also some recent presentations available on my preprint page.
Further comment: 11:12 24 Sept.
The HHSvKT's have two roles. One is to allow some calculations and understanding not previously possible. People concentrate on the homotopy groups of spheres but what about the homotopy types of more general complexes? One aim is to give another weapon in the armoury of algebraic topology.
The crossed complex work applies nicely to filtered spaces. The new book (pdf downloadable) gives an account of algebraic topology on the border between homotopy and homology without using singular or simplicial homology, and allows for some calculations for example of homotopy classes of maps in the non simply connected case. It gets some homotopy groups as modules over the fundamental group.
I like the fact that the Relative Hurewicz Theorem is a consequence of a HHSvKT, and this suggested a triadic Hurewicz Theorem which is one consequence of the work with Loday. Another is determination of the critical group in the Barratt-Whitehead n-ad connectivity theorem - to get the result needs the apparatus of cat^n-groups and crossed n-cubes of groups (Ellis/Steiner).
The hope (expectation?) is also that these techniques will allow new developments in related fields - see for example work of Faria Martins and Picken in diff geom.
Developments in algebraic topology have had over the decades wide implications, eventually in algebraic number theory. People could start by trying to understand and apply the 2-dim HHSvKT!
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Can't get any more definitive than this! – Harry Gindi Sep 24 2010 at 16:36
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@ Harry. I thought that the best answer to your question was two fold. Firstly to e-mail Ronnie as he knows the stuff inside out and has lots of links to his pages available, so I did that. (You see the result. (We're quick off the mark in North Wales, :-))) The second stage should be to up-date the n-Lab entry and bring in some of the other stuff that RB did, e.g. the material with Loday. That will take time and hopefully others will contribute to it as well. – Tim Porter Sep 24 2010 at 16:49
I figured you had a hand in this! =), indeed! – Harry Gindi Sep 24 2010 at 16:55
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I think it is important to remember that the Brown-Loday theorem concerns colimits of cat-n groups obtained from an n stage filtration of the underlying space. Moreover, cat-n groups can only provide information on the n-type.
So, if you wanted to compute pi_200 of the two sphere, you would need a cat-200 group (at least). And if you wanted to apply the Brown_Loday VK theorem you would need a 200 stage topological filtration of the two sphere. Any thoughts ? or do I have this wrong?
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@Carl. Hi. No you are exactly right, but pessimistically so. :-) (See my reply to Tom above) The theory has not been developed that far as to be able to see the ways of extracting information from the gadget. There is probably too much data in the model and one has to see how to use algebraic methods to prise out of it information that is needed. That is not that unusual in algebra. With some of the gigantic simple groups there is a lot of energy that goes into, for instance, understanding what subgroups they have and that sort of effort leads to new tools for use elsewhere. – Tim Porter Sep 24 2010 at 18:12
I feel this is a question of horses for courses. If you wanted to compute pi_200 of the 2-sphere this way you would need a 200-cube structure on the 2-sphere, e.g. from 200 subspaces, and it is not so clear how to get this. However SX does have a convenient triad structure, with two cones, and this is used in classical homotopy theory. The traditional SvKT for the fundamental group does not compute the fundamental group of the circle. The HHSvKT computes some things not otherwise computable, e.g. some nonabelian n-ad groups. Better to look at what these do rather than don't do! – Ronnie Brown Sep 27 2010 at 9:42
Carl writes cat-n groups can only provide information on the n-type'. I wonder about the word only'! The HHSvKT computes (when it works) the n-type of a colimit from the n-types of the pieces and the gluing information. Sometimes (this has been done for cat$^2$-groups) this allows a lot of information on homotopy groups of the colimit, Whitehead products, composition operators, ... Is not that amazing? Also not many have worked on these gadgets. – Ronnie Brown Sep 27 2010 at 20:04
Ronnie and collaborators' HHvK theorems are essentially all for crossed complexes and similar. These are, as I'm sure you're aware, partially linearised homotopy types. In particular, for a simply connected space, they are just chain complexes. So for $S^k, k>1$ it's not more useful than the homology.
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@David You are forgetting the version that RB proved with Loday. The end result there is a cat^n group and hence a model of the n+1 type of the 'union'/colimit. This leads to the non-Abelian tensor product and that has been a great idea, but they do not relevant to the unstable homotopy groups of spheres at least as far as I know. – Tim Porter Sep 24 2010 at 6:43
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Hmm, good question. The problem I think moves from calculating homotopy groups of spheres to calculating colimits of cat^n groups. I don't think there is a very big literature on cat^n groups (but I note there is a lot on the nonabelian tensor product - RB has collected them on his website) and tools to deal with them. – David Roberts Sep 24 2010 at 8:28
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@ Harry I will do but it will be a bit later on as I have to think about how best to answer e.g. to look up one of RB and J-LLs examples! – Tim Porter Sep 24 2010 at 8:33
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I suppose that the theorem in question turns the computation of, say, $\pi_{200}(S^2)$ into mere algebra: it tells you how to write down generators and relations for some complicated algebraic object called a cat_n group and then you just have to learn how to extract your answer from that. – Tom Goodwillie Sep 24 2010 at 13:41
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@Tom My own take on it might be a bit different. There is a curiosity in knowing the homotopy groups of spheres, but that is partially a curiosity in studying the way in which the algebra reflects homotopy structure, so modelling the homotopy 200-type of $S^2$ is perhaps in the long run more central to the understanding of the total homotopy type of $S^2$, than 'merely' knowing the value of $\pi_{200}(S^2)$. The $cat^n$ group ($cat^{200}$-group of $S^2$ would contain an awesome amount of information on certain homotopy operations. I cannot start to think what they might be! – Tim Porter Sep 24 2010 at 18:02
show 4 more comments
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http://mathoverflow.net/questions/95576/contraction-by-the-fundamental-form-of-a-hermitian-metric
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## Contraction by the Fundamental Form of A Hermitian Metric
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $M$ be a complex manifold, with a Hermitian metric $g$ which we will consider as a $C^\infty(M)$-bi-module map $$g:\Omega^1(M) \otimes_{C^{\infty}(M)} \Omega^1(M) \to C^\infty(M),$$ where $\Omega^1(M)$ is the module of one forms of $M$. Moreover, let $\omega$ be the fundamental form of $g$. Now contraction by $\omega$ is a map $$\omega \llcorner :\Omega^1(M) \wedge \Omega^1(M) \to C^\infty(M).$$ Am I naive in thinking that there might be some simple relationship between these two maps? For example, $$(\omega \llcorner) \circ \wedge = g,$$ where $$\wedge:\Omega^1(M) \otimes_{C^{\infty}(M)} \Omega^1(M) \to \Omega^1(M) \wedge \Omega^1(M)$$ is the obvious map. Does it help if I assume that $g$ is Kaehler, ie d$\omega = 0$?
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Sorry but this isn't really appropriate for this site. You should simply write out the definition of the two maps in, say, local co-ordinates and compare them. And, if you still have any questions, I would recommend posting them on math.stackexchange.com. – Deane Yang Apr 30 2012 at 14:24
Ok. But can I take that as a yes? – Ago Szekeres Apr 30 2012 at 14:26
No, you can't take my answer as a yes. – Deane Yang Apr 30 2012 at 15:02
By the way, what do you get if you apply your map to $(v,v)$, where $v$ is a non-vanishing vector field? – Deane Yang Apr 30 2012 at 15:11
Yes I see now, a zero on the left side, and a non-zero value on the right side. Thanks. – Ago Szekeres Apr 30 2012 at 15:58
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http://mathhelpforum.com/number-theory/15549-prove-mathematical-induction.html
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# Thread:
1. ## Prove by Mathematical Induction
Hi everyone I need a little help, Im stuck.
I have to prove by induction that:
$<br /> 2^{n+2} + 3^{2n+1}<br />$
is exactly divisible by 7 for all positive integers of n.
Now what I have done is:
1)
If $2^{n+2} + 3^{2n+1}$ is divisible by 7
=>( $2^{n+2} + 3^{2n+1}=7a$ )-----> Pn statement
where a is a positive integer
2) $P_1$ statement where n=1
=>
LHS: $2^{1+2} + 3^{2(1)+1} = 35$
RHS: $7(5) = 35$
Therefore $P_1$ is true.
3) Assuming $P_k$ to be true where n=k
=> $P_k = 2^{k+2} + 3^{2k+1} = 7a$
4) Therefore
$P_{k+1} = 2^{(K+1)+2} + 3^{2(k+1)+1}$
Now here is where Im stuck, Im not getting to factor out a 7 i.e.
getting 7(some integer) to prove that $P_{k+1}$ is divisible by 7.
Any help would be greately appreciated.
2. Originally Posted by princemac
Hi everyone I need a little help, Im stuck.
I have to prove by induction that:
$<br /> 2^{n+2} + 3^{2n+1}<br />$
is exactly divisible by 7 for all positive integers of n.
Now what I have done is:
1)
If $2^{n+2} + 3^{2n+1}$ is divisible by 7
=>( $2^{n+2} + 3^{2n+1}=7a$ )-----> Pn statement
where a is a positive integer
2) $P_1$ statement where n=1
=>
LHS: $2^{1+2} + 3^{2(1)+1} = 35$
RHS: $7(5) = 35$
Therefore $P_1$ is true.
3) Assuming $P_k$ to be true where n=k
=> $P_k = 2^{k+2} + 3^{2k+1} = 7a$
4) Therefore
$P_{k+1} = 2^{(K+1)+2} + 3^{2(k+1)+1}$
Now here is where Im stuck, Im not getting to factor out a 7 i.e.
getting 7(some integer) to prove that $P_{k+1}$ is divisible by 7.
Any help would be greately appreciated.
for (4), we will do a trick often used in math--add zero in a weird way, that way we don't change anything, but are able to factor things in a convenient way.
$P(k+1): 2^{k+3} + 3^{2k + 3} = 2^{k + 3} \underbrace { + 2 \cdot 3^{2k + 1} - 2 \cdot 3^{2k + 1} } + 3^{2k + 3}$
Note that what is underbraced is zero
$= 2 \left( 2^{k + 2} + 3^{2k + 1} \right) - 3^{2k + 1} \left( 2 - 3^2\right)$
$= 2 \cdot 7a + 7 \cdot 3^{2k + 1}$
$= 7 \left( 2a + 3^{2k + 1} \right)$
which is divisible by 7
3. Ahhh many thanks, now that Ive seen what you did I can see another way.
$P_{k+1}= 2^{k+3} + 3^{2k + 3} = (2)(2^{k+2}) + (9)3^{2k + 1}<br />$
$2^{k+2}= 7a - 3^{2k + 1}$
Therefore
$P_{k+1} = 14a + 3^{2k + 1}(9 -2)= 7(2a +$ $3^{2k+1})$
Thats pretty cool with the zero, I'll have to remeber that.
Thanks again.
4. Originally Posted by princemac
Hi everyone I need a little help, Im stuck.
I have to prove by induction that:
$<br /> 2^{n+2} + 3^{2n+1}<br />$
is exactly divisible by 7 for all positive integers of n.
Now what I have done is:
1)
If $2^{n+2} + 3^{2n+1}$ is divisible by 7
=>( $2^{n+2} + 3^{2n+1}=7a$ )-----> Pn statement
where a is a positive integer
2) $P_1$ statement where n=1
=>
LHS: $2^{1+2} + 3^{2(1)+1} = 35$
RHS: $7(5) = 35$
Therefore $P_1$ is true.
3) Assuming $P_k$ to be true where n=k
=> $P_k = 2^{k+2} + 3^{2k+1} = 7a$
4) Therefore
$P_{k+1} = 2^{(K+1)+2} + 3^{2(k+1)+1}$
Now here is where Im stuck, Im not getting to factor out a 7 i.e.
getting 7(some integer) to prove that $P_{k+1}$ is divisible by 7.
Any help would be greately appreciated.
Hello,
I take over your result and transform it a little bit:
$P_{k+1} = 2^{(K+1)+2} + 3^{2(k+1)+1} =$
$2 \cdot 2^{k+2} + 9 \cdot 3^{2k+1} =$
$2^{k+2} + 2^{k+2} + 3^{2k+1} + 3^{2k+1} + 7 \cdot 3^{2k+1}$ rearrange the summands:
$\underbrace{2^{k+2} + 3^{2k+1}}_{\text{divisible by 7}} + \underbrace{2^{k+2} + 3^{2k+1}}_{\text{divisible by 7}} + \underbrace{7 \cdot 3^{2k+1}}_{\text{divisible by 7}}$
5. once again, it seems i was thinking too hard. both your methods seem nicer than mine for some reason
6. Originally Posted by Jhevon
once again, it seems i was thinking too hard. both your methods seem nicer than mine for some reason
Disagree, yours seems more elegant to me.
RonL
7. Originally Posted by Jhevon
once again, it seems i was thinking too hard. both your methods seem nicer than mine for some reason
Originally Posted by CaptainBlack
Disagree, yours seems more elegant to me.
RonL
I am with the Captain all the way on this one!
Jhevon, yours is the preferred way because it is easier to remember after one has done several of these.
8. Hello, princemac!
I have to prove by induction that:
$2^{n+2} + 3^{2n+1}$ is divisible by 7 for all positive integers $n$.
Your work is correct . . .
Verify $P(1):\;\;2^{k+2} + 3^{2\cdot1+1} \:= \:35$ . . . True!
Assume $P(k)$ is true: . $2^{k+2} + 3^{2k+1} \:= \:7a$ .for some integer $a$.
We want to prove $P(k+1):\;2^{k+3} + 3^{2k + 3} \:=\:7b$ .for some integer $b$.
[I like to write this statement now, so I know what I'm aiming for.]
Begin with $P(k):\;2^{k+2} + 3^{2k+1} \:=\:7a$
Add $2^{k+2} + 8\!\cdot\!3^{2k+1}$ to both sides:
. . $\underbrace{2^{k+2} + 2^{k+2}} + \underbrace{3^{2k+1} + 8\!\cdot\!3^{2k+1}} \;=\;7a + 2^{k+2} + 8\!\cdot\!3^{2k+1}$
- . - $2\!\cdot\!2^{k+2} \quad+ \quad(1 + 8)\!\cdot\!3^{2k+1} \;= \;7a + \underbrace{2^{k+2} + 3^{2k+1}}_{\text{this is }7a} + 7\!\cdot\!3^{2k+1}$
. . . . . . . . . . $2^{k+3} \;+ \;9\!\cdot\!3^{2k+1} \;=\;7a + 7a + 7\!\cdot\!3^{2k+1}$
. - . . - . . . . . $2^{k+3} + 3^2\!\cdot\!3^{2k+1} \;=\;14a + 7\!\cdot\!3^{2k+1}$
. . . . . . . . . . . $2^{k+3} + 3^{2k+3} \;=\;7\underbrace{(2a + 3^{2k+1})}_{\text{This is an integer}}$
Therefore: . . . $2^{k+3} + 3^{2k+3} \;=\;7b$
We have proved $P(k+1)$ . . . The inductive proof is complete.
9. Is it a general rule that one should read the entire thread before responding?
If not, there ought to be such a rule!
Is the pervious post in this thread a case in my point?
10. Absolutely right, Plato!
Guilty as charged . . .
I thought I read through the posts (evidently not)
. . and thought I had something original to offer (wrong again).
11. Thank you for that.
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http://mathoverflow.net/questions/11964/strong-induction-without-a-base-case/12054
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## Strong induction without a base case
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Strong induction proves a sequence of statements $P(0)$, $P(1)$, $\ldots$ by proving the implication
"If $P(m)$ is true for all nonnegative integers $m$ less than $n$, then $P(n)$ is true."
for every nonnegative integer $n$. There is no need for a separate base case, because the $n=0$ instance of the implication is the base case, vacuously. But most strong induction proofs nevertheless seem to involve a separate argument to handle the base case (i.e., to prove the implication for $n=0$).
Can you think of a natural example of a strong induction proof that does not treat the base case separately? Ideally it should be a statement at the undergraduate level or below, and it should be a statement for which strong induction works better than ordinary induction or any direct proof.
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1
Interesting question. (I confess that, in the introduction to proofs class I taught twice in the last year, I didn't want to address the logical superfluity of the base case in strong induction for fear it would confuse my students. But I guess MIT students are not so easily confused.) Why is it tagged combinatorics? – Pete L. Clark Jan 16 2010 at 6:47
3
This comment is too facetious to be an answer, but you could let P(n) be the statement "P(t) is true for all 0<=t<n" ;-) – Kevin Buzzard Jan 16 2010 at 8:35
3
Bjorn, this phenomenon is not limited to induction on the natural numbers, but arises naturally in transfinite recursion also. Indeed, it is usually considered a feature of a properly performed transfinite recursion if it has the property you state. Perhaps you want to generalize your question? – Joel David Hamkins Jan 16 2010 at 23:03
1
Yes, I'm aware of transfinite induction, but since the thrust of my question is the same in that setting as in the natural number setting, I phrased my question in the most elementary terms. – Bjorn Poonen Jan 16 2010 at 23:33
1
@Pete: As for the combinatorics tag: I couldn't find a better one. I was thinking that "discrete math" might be appropriate, and combinatorics seemed to be closest to this. Feel free to re-tag the question. – Bjorn Poonen Jan 16 2010 at 23:49
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## 15 Answers
My example is the classical proof that sqrt(2) is irrational.
More generally, many proofs that proceed by showing that there are no minimal counterexamples exemplify your phenomenon. The method of no-minimal-counterexamples is exactly the same as strong induction, but where one proves the required implication by contradiction. In many applications of this method, it is often clear that the smallest numbers are not counterexamples, and this would not ordinarily regarded as a separate base "case".
In the classical proof that sqrt(2) is irrational, for example, we suppose sqrt(2) = p/q, where p is minimal. Now, square both sides and proceed with the usual argument, to arrive at a smaller counterexample. Contradiction! This amounts to a proof by strong induction that no rational number squares to 2, and there seems to be no separate base case here.
People often carry out the classical argument by assuming p/q is in lowest terms, but the argument I just described does not need this extra complication. Also, in any case, the proof that every rational number can be put into lowest terms is itself another instance of the phenomenon. Namely, if p/q is a counterexample with p minimal, then divide by any common factor and apply induction. There seems to be no separate base case here where it is already in lowest terms, since we were considering a minimal counterexample. Perhaps someone objects that there is no induction here at all, since one can just divide by the gcd(p,q). But the usual proof that any two numbers have a gcd is, of course, also inductive: considering the least linear combination xq+yp amounts to strong induction, again with no separate base case.
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I like this lead. Fermat's proof that $x^4 - y^4 = z^2$ has no nontrivial integer solutions might be a more convincing example. See http%3A%2F%2Fen.wikipedia.org%2Fwiki%2FProof_of_Fermat%27s_Last_Theorem_for_specific_exponents%23n.C2.A0.3D.C2.A04 – François G. Dorais♦ Jan 17 2010 at 2:03
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Why, any proof that uses Fermat's method of infinite descent invariably uses the strong induction. Or am I committing a blunder here? – Abhishek Parab Jan 17 2010 at 4:01
1
Infinite descent was Fermat's signature tool. He at least deserves credit for popularizing the method. Earlier uses of induction that I'm aware of were either "Greek style" (case 1, case 2, case 3, end with etc.) or the more sophisticated: start with n and keep going down by 1 until you reach the base case. The first occurrence of real induction that I know is from Pascal, so shortly after Fermat. – François G. Dorais♦ Jan 17 2010 at 5:01
1
It doesn't work since verifying descent requires proving p/2 < p so requires proof that p isn't 0, i.e. disproving the base case P(0). Even if you start at 1 vs. 0 there are still analogous objections. – Bill Dubuque Jul 7 2010 at 21:10
1
@Joel Yes, without a rigorous definition of what it means to "not treat the base case separately" the question cannot be answered. I've seen many dozens of similar questions in various forums and a consensus is never reached due to the lack of such precision. – Bill Dubuque Jul 22 2010 at 22:13
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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 hope the following will satisfy Bjorn. It is a proof by induction which naturally skips over the base case and is at the undergraduate level. I saw the argument for the first time today in the paper containing 10 proofs in Russian of the Fundamental Theorem of Algebra which Ilya Nikokoshev made a link to in his answer to a question asking for lots of different proofs of that theorem. Here we go:
Claim: A nonzero polynomial (over a field) has no more roots than its degree.
Proof: We prove this by induction on the degree $n$ of the polynomial. Assume that the polynomial $$p(X) = a_nX^n + \cdots + a_1X + a_0$$ of degree $n$ has at least $n+1$ different roots $r_1,\dots, r_{n+1}$. Consider the polynomial $$q(X) = a_n(X-r_1)\cdots (X-r_n).$$ We have $p(X) \not= q(X)$ since $p(r_{n+1}) = 0$ and $q(r_{n+1}) \not= 0$. The difference $d(X) = p(X) - q(X)$ is a nonzero polynomial of degree less than $n$ having at least $n$ roots $r_1,\dots,r_n$. This contradicts the inductive hypothesis. QED
[EDIT: IGNORE what follows in the next paragraph, which was in the original post, since I confused myself about strong vs. ordinary induction. The above proof is by strong induction since the degree of d(X) is merely less than n and not necessarily n-1 itself.]
One aspect of this which does not fit Bjorn's request is that this argument uses ordinary induction, not strong induction. But really, is that such a big deal? I suspect his main interest is seeing an inductive argument at all where the base case is naturally not mentioned, rather than specifically one using strong induction.
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You're ok, this does use strong induction. – François G. Dorais♦ Jan 17 2010 at 2:10
Really? Although one could run through the argument as a strong induction type statement, it only involves two adjacent cases. I tend to think of true strong inductive arguments as those (like factorization into primes) where you prove one case by appealing to some earlier case, but it in practice is not going to be the immediately preceding case. – KConrad Jan 17 2010 at 2:29
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It's not the immediately preceding case: $d(X)$ has degree less than $n$, not necessarily $n-1$. (The $n+1$ is misleading, it's just the smallest integer bigger than $n$.) – François G. Dorais♦ Jan 17 2010 at 2:32
Oh, duh. I was reading "degree than less than n" at the end of the proof as "degree n-1". – KConrad Jan 17 2010 at 4:10
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Good example! One comment: The claim is stronger, and the inductive step clearer, if in the claim you change "polynomial of nonzero degree" to "nonzero polynomial". – Bjorn Poonen Jan 18 2010 at 6:07
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I think the best example of this will be the fundamental theorem of aithmetic: the assertion that every natural number greater than 1 is either prime or the product of primes (and uniquely so).
Proof of existence is by strong induction. Assume true below n. If n is prime, we're done. Otherwise, n = ab for some a,b < n, and by induction these are products of primes, so n is also. QED (and no need for special anchor case).
Proof of uniqueness does not use induction.
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Neither of those cases hold if n = 1, so I guess you want to start the induction at n = 2. In that case the existence proof with "prime" replaced by "irreducible" works in any Noetherian commutative ring and this Noetherian induction also doesn't have a base case, I guess. – Qiaochu Yuan Jan 16 2010 at 23:15
Nitpick: Perhaps replace positive by $\geq 2$ in the statement, otherwise the base case is different, though clearly not an "anchor case." – François G. Dorais♦ Jan 16 2010 at 23:18
Thanks, I corrected. @Qiaochu: Yes, but the question wants the most elementary example, so I complied. – Joel David Hamkins Jan 16 2010 at 23:25
@Joel: That is a nice example of strong induction, but I am really hoping for an argument that does not need a separate case to get started (here the separate case is the case where n is prime). Sorry if my question is vague; I'm not sure I can formalize what I want. – Bjorn Poonen Jan 16 2010 at 23:26
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OK, I'm beginning to understand what you want. So I posted another example, on the cumulative hierarchy (see below....or above?) – Joel David Hamkins Jan 17 2010 at 0:24
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The cumulative hierarchy in set theory is the hierarchy Vα that is often defined by the transfinite recursion:
• V0 = emptyset
• Vα+1 = P(Vα), taking the power set at successor stages
• Vλ = U{ Vα | α<λ }, taking unions at limits.
And if one wants to consider only the class HF of hereditarily finite sets, one restricts to the natural number induction α=n, leading to HF = Vω = U Vn. These definitions, however, split into separate cases for zero, successor and limit.
An equivalent definition, however, completely avoids this split. Namely:
• For any ordinal α, let Vα = U { P(Vβ) | β < α }.
This is easily seen to be equivalent to the previous definition.
Similarly, one can define the hereditary finite sets HF as the union of Vn, where Vn = U { P(Vm | m < n }. This definition needs no base case, and does not split into cases.
One can now prove all the basic facts about the Vα hierarchy, also without splitting into zero, successor and limit cases. For example, every Vα is transitive, since by induction it is the union of transitive sets. Similarly, the hierarchy is increasing, etc.
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It seems to me that, when you use the improved (non-case-splitting) definition of the hierarchy, the particular facts you mention at the end --- transitivity and monotonicity --- can be proved without any induction at all (provided you know that the < relation on ordinals is transitive). But I agree that other facts about the hierarchy become provable by non-case-splitting inductions. – Andreas Blass Dec 25 2010 at 0:11
One context in which one does inductive proofs without a base case is that of Conway's games/numbers, which are defined inductively without a base case.
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Any constructively valid proof by well-founded induction, specialized to the natural numbers as a particular well-founded set, will be an example. Recall that a set $A$ with a relation < is well-founded if for any $S\subseteq A$, if $(\forall y\lt x)(y\in S) \Rightarrow x\in S$, then $S=A$. A proof by well-founded induction proceeds by proving that the set $S$ of "all $x\in A$ such that blah" satisfies that condition, i.e. that if blah is true for all $y\lt x$, then it is also true for $x$. In classical logic, one can separate out such a proof into "case 1: there are no $y\lt x$" and "case 2: there are some $y\lt x$", but in constructive logic this is not generally possible. (It is still possible in the particular case of strong induction on the naturals, since equality of naturals is decidable.) However, interesting proofs by well-founded induction do still exist constructively.
For example, if $A$ is well-founded, and $B$ is equipped with a function $t\colon P B \to B$, then one can prove by well-founded induction on $A$ that there is a unique function $f\colon A\to B$ defined by well-founded recursion, i.e. such that for all $x\in A$ we have $f(x) = t(\lbrace f(y) \mid y \in A \wedge y \lt x\rbrace)$. Define an attempt to be a partial function $A\rightharpoonup B$ whose domain is down-closed for < and which satisfies the desired condition insofar as it is defined. We can then prove by well-founded induction that any two attempts are equal on the intersection of their domains, and that for all $x$ there exists an attempt whose domain contains $x$, hence the union of all attempts is the desired function.
In classical logic, where emptiness of a set is decidable, we could separate out the empty set as a special case in this argument (or any other), but it wouldn't gain us anything; it would be just as "non-genuine" as Tran Chieu Minh's construction in the other direction.
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There is another of this same fact called noetherian induction by applying the descending chain condition on the spectrum of a noetherian ring. – Harry Gindi Jan 17 2010 at 1:06
This is perhaps a rather strange example. There is a paper of Andreas Blass An induction principle and pigeonhole principle for K-finite sets (J. Symbolic Logic 59, 1995, 1186-1193) where the goal is to give an intuitionistic proof of that there is no surjection from $X$ onto $X+1$ when $X$ is finite (Theorem 2). Before proving this result, Blass proves a strong induction principle for finite sets (Theorem 1). The proof of Theorem 1 uses ordinary induction with a base case, but the proof of Theorem 2 uses the strong induction principle of Theorem 1 instead. Blass' proof of Theorem 2 very straightforward, but I think that a direct proof of Theorem 2 (along the same lines) would be unbearably long.
I guess it's hard to understand this one without seeing the proofs. So I'll give a similar proof (without the intuitionistic fuss, for clarity). The strong induction on finite sets I will use is the following.
(SI) If $\mathcal{K}$ is a family of sets such that if all proper subsets of $A$ are in $\mathcal{K}$ then $A$ is in $\mathcal{K}$ too, then $\mathcal{K}$ contains all finite sets.
Let $\mathcal{K}$ be the family of all sets $A$ such that if $A$ is a proper subset of $B$ and $f:A \to B$ then the image $f[A]$ is not all of $B$. I claim that $\mathcal{K}$ satisfies the hypothesis of (SI).
Suppose all proper subsets of $A$ are in $\mathcal{K}$. Let $B$ be a proper superset of $A$. Given $f:A \to B$, we consider two cases.
If $f^{-1}[A] = A$, then $f[A] \subseteq A$ and so $f[A]$ is certainly not all of $B$.
If $A' = f^{-1}[A]$ is a proper subset of $A$. Then $A' \in \mathcal{K}$ and hence $f[A']$ is not all of $A$. But then $A \setminus f[A'] = A \setminus f[A] \subseteq B \setminus f[A]$, which shows that $f[A]$ is not all of $B$.
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Here's another possible example. Let S(0), S(1), S(2), ..., be the sequence of numbers defined by the formula $S(n) = 1 + \sum_{k=0}^{n-1} S(k)$ for every nonnegative integer n. Then we can show that $S(n) = 2^n$ by strong induction on $n$.
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1
This doesn't seem to be the kind of example that was requested. First, the argument here seems more naturally carried out with (weak) induction, rather than strong induction, since the induction appeals to the immediately preceding case. Secondly, the nature of the argument at n=0, even if you do it with strong induction, is not the same as for other n, so it treats what amounts to the base case differently. – Joel David Hamkins Jan 21 2010 at 13:19
Here is the proof by strong induction I had in mind. Suppose by strong induction that $S(k) = 2^k$ for all $k < n$. Then $S(n) = 1 + \sum_{k=0}^{n-1} S(k) = 1 + \sum_{k=0}^{n-1} 2^k = 1 + (2^n - 1) = 2^n$. (The third equation is using the usual formula for geometric series.) The proof seems to be using strong induction, not weak induction. I don't think I needed to handle n = 0 separately. If you'd rather avoid the appeal to the formula for geometric series, then the examples in my other answer might be better. – Ravi Boppana Jan 21 2010 at 13:59
1
@Ravi: Yes, I thought that is what you had had in mind. My point was that your defining recurrence formula simplifies in one step to S(n+1)=S(n)+S(n), which avoids the series and seems naturally treated with weak induction. – Joel David Hamkins Jan 21 2010 at 19:10
It's difficult to answer such questions without a rigorous definition of what it means for a proof to 'treat the base case separately'. What if the base case is proved as the first step in a lemma that is invoked? Or explicitly somewhere way down the line in some long chains of lemmas. Does that count or not? It's really a subtle proof-theoretical question. @Born: did any reply satisfy you? If so, I'd be very interested to know which one(s).
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@steve: I'm not sure I understand your objection, but how about this modification. Let A be defined by $A(n) = \sum_{k=0}^{n-1} A(k)$ for every nonnegative integer n. Then by strong induction we can show that A(n) = 0. That's a trivial example, but if we wanted to, we could make it less trivial: $B(n) = 1 - n + \sum_{k=0}^{n-1} B(k)$, which has solution $B(n) = 1$.
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This may qualify, though there is a special case hidden inside the argument.
Consider a simplified game of nim in which there are $n>0$ matchsticks, a player may remove $1$, $2$, or $3$ matchsticks each turn, and the player who takes the last matchstick wins.
Theorem. The first player has a winning strategy if $n\not\equiv 0\pmod{4}$; the second player has a winning strategy if $n\equiv 0\pmod{4}$.
Proof. Strong induction on $n$. Assume the result holds for all $k\lt n$. If $n\equiv 0\pmod{4}$, then after player 1's turn there will be $k\lt n$ matchsticks left, with $k\not\equiv 0 \pmod{4}$. By the induction hypothesis, the first person to play at this point has a winning strategy, this being player 2; thus, player 2 has a winning strategy.
If $n\not\equiv 0\pmod{4}$, then write $n=4\ell + t$ with $1\leq t\leq 3$. Have player 1 take $t$ matchsticks, leaving $4\ell$ matchsticks. If $\ell=0$, player 1 just won. If $\ell>0$, then there are $k\lt n$ matchsticks left, with $k\equiv 0\pmod{4}$. By the induction hypothesis, the player who moves second has a winning strategy, this being the original player 1. So player 1 has a winning strategy in this case.
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One may transform any strong induction proof into one which (legalistically speaking!) doesn't explicitly treat the base case.
Proving $P(n)$ by induction, one assumes `$ \forall k<n \ P(k)$` which weakens to `$\forall k<n\ P(0)\implies P(k)$.` Now one may adapt whatever proof one had for the induction step to establish from this that $P(0)\implies P(n)$. Now one proves $P(0)$ and concludes $P(n)$ by modus ponens.
Yes, I know, morally the base case still got special treatment, but formally now that happens in the induction step. Thus finding a formal distinction between theorems that require special treatment for the base case and those that don't seems impossible.
In any case, if you can prove $\forall n\geq 0\ P(n)$ by strong induction, you can prove $\forall n>0 \ P(0)\implies P(n)$ by strong induction with no special treatment for the base case, morally or formally. So examples abound.
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I posted an answer on sci.math.research recently where I used induction with a base case, and then thought I could use strong induction instead. Later another poster came up with a nice argument that hid or did away with the induction. Perhaps you can frame it to fit your needs.
The problem: Given an ascending sequence b_i and descending sequence a_i, both containing n positive numbers, show that for every permutation p on n letters that max (a_i*b_i, 1 <=i <= n) <= max (a_i*b_p(i), 1 <= i <= n) .
Proof sketch (by strong induction?): Assume the result is true for all m < n. If p fixes n, we're done. Otherwise, swap b_n and b_p(n) and note that the maximum will not increase. Done.
The above may not be the most natural example nor the best proof of the result.
It may suggest something that will help. Although it could be argued that this is a "strong inductive" example of a different proof, I think it could also be argued that the different proof is one that hides the inductive principle at work. Now let the community judge.
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@RaviB: You still got the "base case" issue, namely, 2^0 =1; @poonen: structurally speaking, we will always be in need of a base case regardless of whether strong or ordinary reduction is used.
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@RaviB: how on earth did your comment get minused ? I understand where trying to get. The problem that I am referring to is one of systemic level. The way we build up definitions for various systems necessitates base cases. – steve Jan 22 2010 at 6:22
I'm confused a little about not proving base case in strong induction. Can someone point out the error in following proof( I'm using strong induction)?
Theorem: Prove that for all $n \in N$, $1.3^0 + 3.3^1 + 5.3^2 + .. + (2n+1)3^n = n3^{n+1}$
Proof:
Let n be arbitrary natural number. Suppose, for all natural numbers k smaller than n, $1.3^0 + 3.3^1 + 5.3^2 + .. + (2k+1)3^k = k3^{k+1}$
Since (n-1) < n, Then it follows from our assumption that $1.3^0 + 3.3^1 + 5.3^2 + .. (2n-1)3^{n-1} = (n-1)3^n$
So, $1.3^0 + 3.3^1 + 5.3^2 + .. + (2n+1)3^n$ = $1.3^0 + 3.3^1 + 5.3^2 + .. (2n-1)3^{n-1} + (2n+1)3^n$ = $(1.3^0 + 3.3^1 + 5.3^2 + .. (2n-1)3^{n-1}) + (2n+1)3^n$ = $(n-1)3^n + (2n+1)3^n$ = $3n.3^n$ = $n3^{n+1}$
Hence, by the assumptions of strong induction, $1.3^0 + 3.3^1 + 5.3^2 + .. + (2n+1)3^n = n3^{n+1}$
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2
To apply the induction hypothesis "for all natural numbers k smaller than n ..." to n-1, you need not only that (as you said) n-1 < n but also that n-1 is a natural number. That fails when n=0, so your argument doesn't cover the case n=0. – Andreas Blass Jul 3 2010 at 4:45
Right, so basically we need to check the base case, n = 0. I'm feeling that, the assumption that $\forall k < 0 P(k)$ is true leads to these kind of problems and even in the strong induction I should check the base case. That said, I don't understand why strong induction says that you don't need to prove the base case. – Himanshu Jul 3 2010 at 5:04
1
Strong induction doesn't require you to prove the base case provided you really prove all the things that strong induction does require you to prove, namely the "strong induction step" for all n, not just (as you did) for n not equal to 0. In some cases, like your example, a complete proof of the strong induction step may involve treating the n=0 case separately, and then it looks like a base case. Nevertheless, it's a part of the strong induction step. – Andreas Blass Jul 3 2010 at 17:18
Somehow Andreas's explanation reminds me of the "proof" that all horses are of the same colour. – Willie Wong Jul 7 2010 at 21:25
1
In broad terms, this is the same story as the horses: An alleged proof by induction on n fails to handle one value of n, and thus gives nonsense for that value and for larger n. The difference is that Himanshu's argument failed to cover the case n=0, while the horse argument fails to cover the case n=2. – Andreas Blass Jul 8 2010 at 1:09
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http://mathoverflow.net/questions/79322?sort=oldest
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## Finding the greatest (smallest) factor of a number smaller (greater) than another number
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Instead of iterating through all the possible numbers, is there a better way to find the greatest factor of a number $n$, such that it is less than $m$ ($m$ < $n$). Similarly how does one find the smallest factor of $n$ that is greater than $m$?
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4
This can be hard even if $n$ is squarefree with known but numerous prime factors, because it reduces to a knapsack problem (given items of size $c_i$, find $s$ large enough to approximate $\exp sc_i$ sufficiently well by primes $p_i$, and take $n = \prod_i \phantom. p_i$, etc.). – Noam D. Elkies Oct 28 2011 at 4:22
## 2 Answers
Here's a way to find small factors. I think that finding large factors (up to the square root of $n$) is as hard as factoring in general.
Edit: Maybe the method in the paper I linked is not so great because it looks for small factors in a set of number. But it may cite more relevant papers.
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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.
You can certainly use Pollard's Rho algorithm to probabalistically compute the greatest factor of $n$ smaller than $m$ in $O(min(n^\frac{1}{4}, m^\frac{1}{2}) polylog(n))$ time. Your other question is actually the same as the first. That is $k>m$ is a factor of $n$ iff $\frac{n}{k}$ is a factor of $n$ and smaller than $\frac{n}{m}$.
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http://nrich.maths.org/2684/index?nomenu=1
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## 'Golden Construction' printed from http://nrich.maths.org/
### Show menu
In this problem you start with a square, construct a golden rectangle and calculate the value of the golden ratio.
When you cut a square off a golden rectangle you are left with another rectangle whose sides are in the same ratio. In the diagram below, if you remove the orange square from rectangle $AEFD$, you are left with the yellow rectangle whose sides are in the same ratio as $AEFD$.
(1) Follow the instructions for drawing the rectangle $AEFD$. You can make the most accurate drawing by using a ruler and compasses. Draw a square $ABCD$ of side length 10 cm. Bisect $AB$ at $M$ and draw an arc of radius $MC$ to meet $AB$ produced at $E$. If you prefer you can just measure $MC$ and mark $E$ on $AB$ so that $ME=MC$. Draw $EF$ perpendicular to $AB$ to meet $DC$ produced at $F$.
Measure $AE$ and $BE$. From your measurements calculate the ratios $AE$/$AD$ and $BC/BE$. What do you notice?
(2) Calculate the exact lengths of $MC$, $AE$ and $BE$. Calculate the exact value of the ratios $AE$/$AD$ and $BC$/$BE$ and prove that they are equal. [Note: To get exact values you must work with surds and you will not be able to use a calculator.]
(3) Suppose this ratio is denoted by $\phi$ and take 10 cm as 1 unit then $AE$ is $\phi$ units. Show that $BE$ is $1/\phi$ units and hence $$\phi = 1 + {1\over \phi}\quad (1)$$
Explain just from equation (1), and without solving the equation, why the equation must have a solution between 1 and 2.
(4) Draw the graphs of $y=x$, $y=1/x$ and $y=1+1/x$ on the same axes and use your graph to find an approximate value for $\phi$.
(5) Solve equation (1) to find the value of $\phi$.
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http://math.stackexchange.com/questions/43157/does-this-equality-always-hold
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# Does this equality always hold?
Is it true in general that $\displaystyle\frac{\mathrm{d}}{\mathrm{d}x} \int_0^{x} f(u,x) \mathrm{d}u = \int_0^{x} \left( \frac{\mathrm{d}}{\mathrm{d}x} f(u,x) \right)\mathrm{d}u +f(x,x )$ ?
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1
– Fabian Jun 4 '11 at 10:02
@Fabian: thank you – ghshtalt Jun 4 '11 at 11:08
## 2 Answers
Yes, it does, under the conditions indicated below.
Let $$I(x)=\displaystyle\int_{0}^{x}f(u,x)\; \mathrm{d}u.\qquad(\ast)$$
If $f(u,x)$ is a continuous function and $\partial f/\partial x$ exists and is continuous, then
$$I^{\prime }(x)=\displaystyle\int_{0}^{x}\dfrac{\partial f(u,x)}{\partial x}\; \mathrm{d}u+f(x,x)\qquad(\ast\ast)$$
follows from the Leibniz rule and chain rule.
Note: the integrand of $(\ast\ast)$ is a partial derivative.
It generalizes to the integral
$$I(x)=\displaystyle\int_{u(x)}^{v(x)}f(t,x)\; \mathrm{d}t.$$
Under suitable conditions ($u(x),v(x)$ are differentiable functions, $f(t,x)$ is a continuous function and $\partial f/\partial x$ exists and is continuous), we have
$$I^{\prime }(x)=\displaystyle\int_{u(x)}^{v(x)}\dfrac{\partial f(t,x)}{\partial x}\; \mathrm{d}t+f(v(x),x)v^{\prime }(x)-f(u(x),x)u^{\prime }(x).$$
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excellent, thank you – ghshtalt Jun 4 '11 at 11:09
You are welcome! – Américo Tavares Jun 4 '11 at 11:40
In short, it is true if $f$ and the partial derivative with respect to $f$ are continuous in the region of differentiation.
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thank you mixedmath – ghshtalt Jun 4 '11 at 11:10
@ghshtalt: oh, I read your comment just to see Americo's far superior answer! Well, it happens. – mixedmath♦ Jun 4 '11 at 11:19
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http://crypto.stackexchange.com/questions/3286/relation-between-threshold-cryptosystem-and-secure-multiparty-computation/3287
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# Relation between Threshold Cryptosystem and Secure Multiparty Computation ?
Is there any relation between secure multiparty computation and threshold cryptosystem ?
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## 2 Answers
Yes and no.
A threshold cryptosystem means the decryption key can be split into $n$ shares such that only $t\leq n$ are required to recover it. That property in isolation is not useful for multiparty computation.
However when you combine a threshold cryptosystem with one that is at least partially homomorphic (meaning you can do some operation, like addition or multiplication, under encryption), then the two properties combined can make a useful basis for multiparty computation.
The model is that each party encrypts their inputs, you use the homomorphic property to do some computation on the ciphertexts while they are still encrypted, and the threshold property enforces that only the final value is decrypted assuming less than $t$ dishonest key share holders. Without the threshold (or at least a distributed $n$-out-of-$n$ key), the keyholder could simply decrypt the inputs and learn everyone's value.
The most famous paper on this is "Multiparty Computation from Threshold Homomorphic Encryption." This approach is used in lots of papers. A different approach in the same model is "Mix and Match: Secure Function Evaluation via Ciphertexts." It is less used (I think maybe a bit overlooked) but you can do interesting things with it.
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Yes. Threshold cryptosystems come under the secure multiparty computation branch of cryptography.
Essentially, threshold cryptosystems require multiple parties to cooperate in order to decrypt a ciphertext.
Secure multiparty computation is a field of study involving cryptosystems that require more than one party to compute an operation. As such, threshold cryptosystems fall into this field.
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but any formal study of the relation is made ? any references? – sashank Jul 20 '12 at 14:56
There's no formal study of the relation, because that's like asking for a study on the relationship between basketball and sports. Threshold cryptosystems are a type of secure multiparty computation. – Polynomial Jul 20 '12 at 15:30
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http://mathhelpforum.com/calculus/61993-not-so-simple-equation-solve-print.html
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# not-so-simple equation to solve
Printable View
• November 28th 2008, 05:39 AM
Simo
not-so-simple equation to solve
Dear all,
I have a tricky problem; I need to solve this equation for $N$
$\binom {L} {L-R}\sum_{r=1}^{R}\binom {R} {r} (-1)^{R-r}\left(\frac{r}{L}\right)^N\ln\frac{r}{L} =0<br />$
Is it possible in some way?
The only constraints are $R \leqslant L$, $R,L>0$
Thanks a lot
Simo
• November 28th 2008, 06:54 AM
shawsend
1 Attachment(s)
Going gets tough, tough resort to numerical methods. The following is Mathematica code to calculate the value of n for the supplied values of R and L. First out is the expression for the sum, second out is the numerically calculated value of n which makes the sum equal to zero, and the third out is a back-substitution of the the root for a check.
Code:
```In[234]:=
R = 3;
L = 5;
formula1 = Sum[Binomial[R, r]*(-1)^(R - r)*
(r/L)^n*Log[r/L], {r, 1, R}]
root = n /. FindRoot[formula1 == 0,
{n, 1.2}]
formula1 /. n -> root
Out[236]= -(3/5)^n Log[5/3] + 3 (2/5)^n Log[5/2] - 3 5^-n Log[5]
Out[237]= 1.38015
Out[238]= -5.55112*10^-17```
• November 28th 2008, 07:07 AM
Simo
Ok,
I already have a numerical solution like that, but it is not interesting for me.
Is not possible to find an analytical solution, without resort to numerical methods?
All times are GMT -8. The time now is 10:34 AM.
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http://physics.stackexchange.com/questions/tagged/cosmology+dark-energy
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http://mathhelpforum.com/advanced-statistics/51507-sampling-distribution-print.html
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# Sampling distribution
Printable View
• October 1st 2008, 06:40 AM
mirrormirror
Sampling distribution
Suppose that X1, X2,....Xm and Y1,Y2,....,Yn are independent random samples, with the variables Xi normally distributed with mean $\mu _1$ and variance $\sigma{_1}^{2}$ and the variables Yi normally distributed with mean $\mu_2$ and variance $\sigma{_2}^{2}$. The difference between the sample means, $\bar{X} - \bar{Y}$ is then a linear combination of m + n normally distributed random variables and is itself normally distributed.
Suppose that $\sigma{_1}^{2}$ = 2, $\sigma{_2}^{2}$ = 2.5, and m=n. Find the sample sizes so that ( $\bar{X} - \bar{Y}$) will be within 1 unit of ( $\mu _1 - \mu _2$) with probability .95.
I don't know how to solve this one. I've tried to set up P(( $\mu _1 - \mu _2) - 1$ $\preceq$ $(\bar{X} - \bar{Y} )$ $\preceq$ $(\mu _1 - \mu _2) + 1)$ = .95
but I don't if that's right. And, if so, how do I go on?
• October 1st 2008, 09:23 AM
CaptainBlack
Quote:
Originally Posted by mirrormirror
Suppose that X1, X2,....Xm and Y1,Y2,....,Yn are independent random samples, with the variables Xi normally distributed with mean $\mu _1$ and variance $\sigma{_1}^{2}$ and the variables Yi normally distributed with mean $\mu_2$ and variance $\sigma{_2}^{2}$. The difference between the sample means, $\bar{X} - \bar{Y}$ is then a linear combination of m + n normally distributed random variables and is itself normally distributed.
Suppose that $\sigma{_1}^{2}$ = 2, $\sigma{_2}^{2}$ = 2.5, and m=n. Find the sample sizes so that ( $\bar{X} - \bar{Y}$) will be within 1 unit of ( $\mu _1 - \mu _2$) with probability .95.
I don't know how to solve this one. I've tried to set up P(( $\mu _1 - \mu _2) - 1$ $\preceq$ $(\bar{X} - \bar{Y} )$ $\preceq$ $(\mu _1 - \mu _2) + 1)$ = .95
but I don't if that's right. And, if so, how do I go on?
The variance of $\bar X-\bar Y$ is $\sigma_1^2/m+\sigma_2^2/n$ and its mean is $\mu_1-\mu_2$
RonL
All times are GMT -8. The time now is 10:27 AM.
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http://math.stackexchange.com/questions/93293/algorithm-for-a-deck-manipulation?answertab=votes
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# Algorithm for a deck manipulation
Let's say you have a randomised deck of $N$ different cards. An $M$-action ($M\le N$) is defined as follows: you look at the top $M$ cards of the deck, put as many of them as you choose on top of the deck, and the rest on the bottom, each time in any order you choose. How many $M$-actions does it take to arrange an $N$-card deck into a given order?
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– Andrew Dec 21 '11 at 20:36
2
I'm sorry you asked this question, because I have paying work to do. I must be strong... – TonyK Dec 21 '11 at 21:08
Say M=4 and N is large. If I find 1 and 2 in the first look, put them on top and put the others on the bottom, which cards do I look at for the next try? The 3rd, 4th, 5th and 6th? If I look at the top 4 again I can never get more than the first 4 sorted. – Ross Millikan Dec 21 '11 at 21:40
@Ross: I think you still look at the 1st, 2nd, 3rd, and 4th. So you may want to send the first few to the bottom of the deck for a while (keeping them in order whenever they rise to the top again) to get to some fresh cards. – mjqxxxx Dec 21 '11 at 22:05
1
@TonyK: 1234 --> 3421 --> 3214 --> 1432. – mjqxxxx Dec 21 '11 at 22:56
show 4 more comments
## 1 Answer
Let's call the operation of looking at the top $M$ cards and placing them on the top and/or bottom of the deck in arbitrary order scrying. There are $N!$ different states that the deck can be in initially, and each iteration of scry M presents $(M+1)!$ choices of where to place the cards. So a simple lower bound is that, in the worst case, it must take $\lceil{\log(N!)/\log((M+1)!)}\rceil$ iterations to sort the deck. This is conservative, as seen by considering the $M=1$ case, where the task may not even be achievable.
A better lower bound is obtained by picturing the cards in a circle, with a pointer between the bottom and top cards. Then the operation scry M becomes: rearrange the $M$ cards clockwise from the pointer, in place, then move the pointer up to $M$ places clockwise. To place the cards in the correct order, ignoring the position of the pointer, must take $\lceil{\log((N-1)!)/\log(M!)}\rceil$ operations in the worst case.
On the other hand, it can't take more than $N^2/(M-1)$ iterations, as demonstrated by the following algorithm. Assume without loss of generality that the target order is $\{1,2,...N\}$. For $i=1,2,...N$: move the pointer clockwise by $M$ positions per iteration until card $i$ is in the window, then move card $i$ and the pointer clockwise by $M-1$ positions per iteration until the $i$-th position is in the window, then deposit card $i$ in the $i$-th position and increment $i$. This takes at most $N/(M-1)$ iterations per correctly placed card. This can be improved by carrying along the next $k$ cards instead of just one (reducing the distance the pointer can be moved per iteration to $M-k$). With this variation, it takes at most $N/(M-k)$ iterations per $k$ correctly placed cards, for a total count of $N^2/(k(M-k))$ iterations, which is minimized at $4N^2/M^2$ by taking $k=M/2$.
Because of the finite "range" of the operation, I would conjecture that the actual complexity is $\Theta(N^2)$, i.e., that this upper bound is asymptotically tight, and that it is achieved when the deck is shuffled into the reverse of its target order. I'm less sure about the exact dependence on $M$.
Update: For $M=2$ and $2\le N\le10$, the maximum number of operations required is $1,1,3,5,7,11,14,19,23.$ This is reasonably well fit by $\alpha N^2$, where $\alpha$ is around $0.2-0.25$. The entropy lower bound $\lceil{\log((N-1)!)/\log(M!)}\rceil$ with the same parameters is $0,1,3,5,7,10,13,16,19.$
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3
Goodness, what a curiously specific choice of name... :-) – Steven Stadnicki Dec 21 '11 at 22:59
1
Also, I believe the proof that the complexity is $\Theta(N^2)$ is pretty straightforward, along the lines you suggested (although I'm less sure about the maximum): any operation on $M$ cards can only change the number of inversions of the deck's permutation by $O(f(M))$ for some $f$ - specifically, some function of $M$ but not of $N$ - and since a permutation can have $\Theta(N^2)$ inversions we need to do $\Theta(N^2)$ of these operations to get them all. – Steven Stadnicki Dec 21 '11 at 23:03
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http://mathoverflow.net/questions/7746/periodic-mapping-classes-of-the-genus-two-orientable-surface
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## Periodic mapping classes of the genus two orientable surface
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Please, any information on the periodic mapping classes of the genus two orientable surface, $O_2$, will be greatly thanked. We had been studying the topological structure of 3d surface bundles and reintrepreting them as a circle bundles over orbifolds.
In the http://web.archive.org/web/20070316045651/http://www.smm.org.mx/SMMP/html/modules/Publicaciones/AM/Cm/35/artExp08.pdf -work you would like see the cases $O_1$, among $N_1$ and $N_2$, solved. Any feedback on the results and conjectures, some of them obviously false, will bring a lot of happiness :)
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## 2 Answers
If you want to enumerate the finite-order automorphisms (up to conjugacy) I suggest the following exercise. The associated 3-manifold is Seifert fibred. So determine how the genus 2 surface is sitting in the Seifert manifold (horizontal incompressible surface).
This will give you a formula relating the various branch points of the monodromy to the Seifert data. Moreover, you should be able to go back-and-forth between the description of the Seifert-fibred space (unnormalized Seifert data, fibred over a genus 0, 1 or 2 surface) and the monodromy of the surface. So the classification of Seifert-fibred spaces basically gives you a dictionary for walking-through the finite-order automorphisms of a mapping class group.
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your answer induce me a headache... nah! it is a joke :), what I´m going to do is to trace your programme in my solved cases, after that I'll tell you, thanks! In the other hand as you might see professor Hatcher has enlighted us very sharp... – ivane Dec 4 2009 at 6:03
2
I think you'll have to "get your hands dirty" a little bit but once you do I hope you'll like my explanation. – Ryan Budney Dec 4 2009 at 6:22
do you mean $\chi(F)=n[\chi(B)-m+\sum_i{q_i}^{-1}]$, right? – ivane Dec 5 2009 at 17:25
1
Yes, so $n$ is the order of the monodromy, $F$ is the horizontal incompressible surface, $B$ is the quotient of $F$ by the automorphism you're interested in, $m$ is the number of non-free orbits of the automorphism's action and the $q_i$'s are the orders of the stabilizers of points in the non-free orbits. In hatcher's notion if you have a seifert-fibred space $M[g,0;p_i/q_i]$ you have a horizontal incompressible surface if and only if the sum $\sum_i p_i/q_i = 0$. – Ryan Budney Dec 5 2009 at 18:49
1
In this case your $n = LCM\{q_i : \forall i\}$ -- think about how the horizontal incompressible surface is sitting in the "model Seifert manifolds" (chapter 2 Hatcher's notes) to see this. – Ryan Budney Dec 5 2009 at 18:49
### 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 paper listed below there is a calculation of all the finite group actions on a genus 2 surface. There are 20 of them, with the groups ranging from order 2 to order 48. Nine of the actions are of cyclic groups, of orders 2,2,3,4,5,6,6,8,10 respectively. The paper also does the genus 3 case. The techniques are mostly algebraic. It is an interesting exercise to try to find nice geometric pictures of all the actions.
S.A.Broughton, Classifying finite group actions on surfaces of low genus, J.Pure Appl.Alg. 69 (1991), 233-270.
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2
Alternatively, Proposition 1.11 in Allen Hatcher's 3-manifolds notes available on his web-page, together with the formula on the next page relating the Euler characteristic of the horizontal surface to the euler characteristic of the base. This is what you need to get to the algebra Allen refers to. – Ryan Budney Dec 4 2009 at 6:05
professor Hatcher: thank you very much... perhaps this scarse letters are to far from reflecting how happy I am, but believe me: I realy am!. My next work is to tell which SFS arise... – ivane Dec 4 2009 at 6:10
Prof.Ryan: it greatly pleased me the arousal interest of my quizz, thanks again. – ivane Dec 4 2009 at 6:14
Ah, the orbifold Euler-characteristic gonna rule the possibilities inter Faser-fläche und Zerlegungfläche... thank-x – ivane Dec 5 2009 at 17:18
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