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http://math.stackexchange.com/questions/249262/find-all-functions-for-which-x-cdot-fxyf-y-x-cdot-fx?answertab=votes
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# Find all functions for which $\ x \cdot f(xy)+f(-y)=x \cdot f(x)$
Does anyone have idea for solution (for all non-zero numbers)?
$$\ x ≠ 0,y ≠ 0$$
$$\ f: R \setminus \{0\} → R$$
$$\ x \cdot f(xy)+f(-y)=x \cdot f(x)$$
Thanks!
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3
Is that equation supposed to be true for all $x$ and $y$? For some pair $x$ and $y$? For some function $y$ of $x$? For some function $x$ of $y$? – Chris Eagle Dec 2 '12 at 15:56
for all non-zero numbers – Miloš Havlíček Dec 2 '12 at 15:58
1
Right, now put that in the question. Important facts should not lurk in the comments. – Chris Eagle Dec 2 '12 at 15:59
1
The nil function works! Do I get a candy for finding a solution? – gniourf_gniourf Dec 2 '12 at 16:41
If you will say me how to solve it, I will be incredibly thankful :-) – Miloš Havlíček Dec 2 '12 at 16:48
show 1 more comment
## 3 Answers
I can solve this completely given the additional assumption that $f$ is continuously differentiable on its domain. I suspect that it's possible to relax this to merely "continuous", but you'll probably need some sort of niceness assumption to get anywhere.
If we set $x=1$, the functional equation becomes $$f(y)+f(-y)=f(1) \, .$$ So the equation is equivalent to $$x \, f(xy)-f(y)+f(1)=x \, f(x) \, ,$$ which I will suggestively rewrite as $$x[f(xy)-f(x)]=f(y)-f(1) \, .$$
Now, restrict to the case where $y$ is positive and set $y=1+\epsilon$. Since $f$ is differentiable, we can apply the mean value theorem to find $c_\epsilon \in [x, xy]$, $d_\epsilon \in [1, y]$ such that $$f(xy)-f(x)=x \epsilon \, f'(c_\epsilon)\\ f(y)-f(1)= \epsilon f'(d_\epsilon)$$ (where we take the convention that interval endpoints are commutative: that is, if $a>b$, $[a,b]=[b,a]$).
So our equation becomes $$x^2\epsilon f'(c_\epsilon)=\epsilon f'(d_\epsilon) \, .$$ But $f'$ is continuous by assumption, and we have $\lim_{\epsilon \to 0} c_\epsilon=x$, $\lim_{\epsilon \to 0} d_\epsilon=1$. So, passing to the limit, $$x^2 f'(x)=f'(1) \, ;$$ that is, $f'(x)=\frac{C}{x^2}$ for all $x$ and some constant $C$, and so $f(x)=\frac{C}{x}+D$ for all $x$ and some constants $C,D$.
Now, we can plug this $f$ back into the original functional equation to find which values of $C,D$ actually work. After a little bit of algebra, we see that the left-hand side of the equation reduces to $D+Dx$, while the right-hand side reduces to $C+Dx$. So we must have $C=D$.
That is, the continuously differentiable $f$ which satisfy the equation are precisely those of the form $f(x)=C(1+\frac{1}{x})$ for some constant $C$.
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Partial Solution
$$x \cdot f(xy)+f(-y)=x \cdot f(x)$$
Multiplying by $y$ we get
$$xy \cdot f(xy)+yf(-y)=xy \cdot f(x) \,.$$
We define a new function $g(t):= tf(t)$.
Then, the functional equation becomes
$$g(xy)-g(-y)=yg(x) \,.$$
Thus
$$g(xy)=yg(x)+g(-y)$$
In particular, we get
$$\frac{g(xy)-g(-y)}{xy+y}=\frac{g(x)}{x+1} \,.$$
It is easy to see that $g(-1)=0$. Thus, with the extra assumption that $g(t)$ is differentiable at $-1$, by setting $x \to -1$, we get that $g$ is differentiable at $-y$ and $g'(-y)=g'(-1)$. Let $g'(-1)=C$, then $g(t)= Ct+D$ for some constants $C,D$, and the functional equation Yields $c=d$.
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I guess there doesn't lies any solution to this one . Replacing all y's by x. We get x∗f(x^2)+f(−x)=x∗f(x)....(1)in this equation if we put x=1,we get f(-1)=0 .So let us consider there lies a function Q(x) such that , f(x)=(x+1)Q(x). Replacing the f(x) in equation (1) we get one more equation, if we consider the lading coefficient of Q(x) be a1, and it be of degree n.....we see that n=0. so there doesn't lie a function as Q(x). we get f(x)=(x+1);which actually does't satisfy equation 1. so we conclude that f(x) is a constant function and is identically equal to zero.
ALTERNATIVELY considering the given functional equation if we differentiate both sides with respect to x we get
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The first solution works only for polynomials... The second one only for differentiable functions.... – N. S. Dec 2 '12 at 17:17
I am sure that there is at least one possible solution. – Miloš Havlíček Dec 2 '12 at 17:20
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http://mathoverflow.net/questions/40773/cohomological-dimension-of-the-push-forward-functor
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## cohomological dimension of the push-forward functor
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $\ell$ be a prime number and let $f:X \to Y$ be a morphism of schemes of finite type over the complex numbers (or a regular scheme of dimension at most 1, in which $\ell$ is invertible). How to prove that the $\ell$-cohomological dimension of $Rf_*$ in the etale topology is finite? I don't understand the proof in Kiehl and Weissauer's book "Weil Conjectures, Perverse Sheaves and l’adic Fourier Transform", Appendix D, since there is no base change for $f_*$. The crucial case seems to be when $f$ is an open immersion.
And what about the analytic version? $f:X \to Y$ is a morphism of complex analytic spaces, and $F$ is a sheaf of abelian groups on $X$ (in particular, no $\ell$).
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3
I suppose $f$ is separated. Read Deligne's marvelous "Th. finitude" in SGA 4.5 (source of all ideas in the K-W argument). Deligne's inductive methods prove that $f_{\ast}$ has cohom. dimension (on torsion-sheaves with torsion-orders invertible on the regular base of dim. at most 1) at most $2 {\rm{dim}}(X) + {\rm{dim}}(Y)$. For the analytic version $2 {\rm{dim}}(X)$ works assuming $X$ paracompact & Hausdorff: use topological dim. theory & link of higher direct images with sheafified cohom. & link of derived functor cohom. with Cech theory on paracompact Hausdorff spaces. – BCnrd Oct 1 2010 at 17:16
Thanks, Brian. Now I seem to remember you told me this before...haha. Do you think the proof of KW is correct? It only refers to SGA4, Exp X. Maybe I missed something, but it doesn't make any sense to me. BTW, I think we can remove the separated assumption. For a general morphism f, let U be an open affine in X with complement Z. Then U is separated over Y. Consider the exact triangles f_*j_!j^*F \to f_*F \to (fi)_*i^*F and f_*j_!G \to (fj)_*G \to (fi)_*i^*j_*G, and use the coh. dim. of j_*, where the open immersion j is separated.... – shenghao Oct 1 2010 at 22:07
Dear Shenghao: well, when I presented this result in the Weil II seminar a couple of years ago I didn't see any problems with K-W, perhaps in part because anything which was unclear I knew how to handle by looking back at Deligne's article. What you say about removing separatedness sounds superficially OK, but I haven't sat down to think it through carefully. – BCnrd Oct 1 2010 at 22:31
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http://mathhelpforum.com/algebra/94010-another-inequality-question.html
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# Thread:
1. ## another inequality question
Solve this inequality .
$<br /> |x-2|<\frac{1}{x}<br />$
case 1 :
$-\frac{1}{x}<x-2$
i got $0<x<1$and $x>1$
case 2 : $x-2<\frac{1}{x}$
i got $x<-0.414$or $0<x<2.414$
How can i combine the results to get the solution set ?? THanks for your explaination .
2. The only valid solution set is $0<x<2.414<br />$
Graph the inequality and you'll see why.
$\frac{1}{x}$ takes on only negative values for all negative x
3. Hello, thereddevils!
I-Think is absolutely correct . . .
Solve this inequality: . $|x-2| \:<\:\tfrac{1}{x}$
The graph of $f(x) \:=\:|x-2|$ is a V-shaped graph with its vertex at (2,0).
The graph of $g(x) \:=\:\tfrac{1}{x}$ is a hyperbola in the 1st and 3rd quadrants.
Code:
``` |
\ |* /
\| /
\ * /
|\ * /
| \ oP
| \ / *
- - - - - - - + - * - - - - -
* |
* |
* |
* |
|
*|
|```
The graphs intersect at: $P\left(1\!+\!\sqrt{2},\:\sqrt{2}\!-\!1\right)$
We see that $f(x) < g(x)$ on the interval: . $\left(0,\:1\!+\!\sqrt{2}\right)$
4. Originally Posted by I-Think
The only valid solution set is $0<x<2.414<br />$
Graph the inequality and you'll see why.
$\frac{1}{x}$ takes on only negative values for all negative x
But $|1-2|<\frac{1}{1}$
1<1 ???
Or maybe there's a simpler approach to this question .. anyone ?? Thanks .
5. Originally Posted by Soroban
Hello, thereddevils!
I-Think is absolutely correct . . .
The graph of $f(x) \:=\:|x-2|$ is a V-shaped graph with its vertex at (2,0).
The graph of $g(x) \:=\:\tfrac{1}{x}$ is a hyperbola in the 1st and 3rd quadrants.
Code:
``` |
\ |* /
\| /
\ * /
|\ * /
| \ oP
| \ / *
- - - - - - - + - * - - - - -
* |
* |
* |
* |
|
*|
|```
The graphs intersect at: $P\left(1\!+\!\sqrt{2},\:\sqrt{2}\!-\!1\right)$
We see that $f(x) < g(x)$ on the interval: . $\left(0,\:1\!+\!\sqrt{2}\right)$
Thanks a lot Soroban , but is it possible for me to solve it without using the graphical method ?? Just out of curiousity .
6. Of course it's possible and quite straightforward:
$\left| x-2 \right|<\frac{1}{x}\Rightarrow -\frac{1}{x}<x-2<\frac{1}{x}.$
Eeach one of this must be solved separately.
First inequality: $x-2+\frac{1}{x}=\frac{x^{2}-2x+1}{x}=\frac{(x-1)^{2}}{x}>0.$ In order so that this works, we require that $x>0$ and, of course $x\ne1.$
Second inequality: $x-2-\frac{1}{x}=\frac{x^{2}-2x-1}{x}=\frac{(x-1)^{2}-2}{x}<0\implies\left| x-1 \right|<\sqrt{2}.$ But why we're allowed to do this, because in the original inequality we need to have $x>0,$ otherwise there's no solution to this. How can a negative number be greater than a positive one? That's why we have $x>0;$ thus the solution to this remaining inequality is $1-\sqrt{2}<x<1+\sqrt{2}.$
Almost full solution set will be given by $(0,\infty )\cap \left( 1-\sqrt{2},1+\sqrt{2} \right)=\left( 0,1+\sqrt{2} \right),$ and since we stated that it's $x\ne1$ our full solution set is actually $x\in(0,1)\cup \left( 1,1+\sqrt{2} \right).$
7. Originally Posted by Krizalid
Of course it's possible and quite straightforward:
$\left| x-2 \right|<\frac{1}{x}\Rightarrow -\frac{1}{x}<x-2<\frac{1}{x}.$
Eeach one of this must be solved separately.
First inequality: $x-2+\frac{1}{x}=\frac{x^{2}-2x+1}{x}=\frac{(x-1)^{2}}{x}>0.$ In order so that this works, we require that $x>0$ and, of course $x\ne1.$
Second inequality: $x-2-\frac{1}{x}=\frac{x^{2}-2x-1}{x}=\frac{(x-1)^{2}-2}{x}<0\implies\left| x-1 \right|<\sqrt{2}.$ But why we're allowed to do this, because in the original inequality we need to have $x>0,$ otherwise there's no solution to this. How can a negative number be greater than a positive one? That's why we have $x>0;$ thus the solution to this remaining inequality is $1-\sqrt{2}<x<1+\sqrt{2}.$
Almost full solution set will be given by $(0,\infty )\cap \left( 1-\sqrt{2},1+\sqrt{2} \right)=\left( 0,1+\sqrt{2} \right),$ and since we stated that it's $x\ne1$ our full solution set is actually $x\in(0,1)\cup \left( 1,1+\sqrt{2} \right).$
THanks a lot Krizalid .. yeah that makes sense .
But how come the answer got from graphical method and analytical method is different ?? Shouldnt they be the same ?
The answer got from graphical method includes 1 which makes the inequality invalid .
8. Of course graphical method should yield the same answer, then I think Soroban messed up somewhere.
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http://nrich.maths.org/7302/note
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### Chocolate
There are three tables in a room with blocks of chocolate on each. Where would be the best place for each child in the class to sit if they came in one at a time?
### Plants
Three children are going to buy some plants for their birthdays. They will plant them within circular paths. How could they do this?
### Gran, How Old Are You?
When Charlie asked his grandmother how old she is, he didn't get a straightforward reply! Can you work out how old she is?
# Count the Digits
## Count the Digits
We can do all sorts of things with numbers - add, subtract, multiply, divide ...
Most of us start with counting when we are very little. We usually count things, objects, people etc. In this activity we are going to count the number of digits that are the same.
Rule 1 - The starting number has to have just three different digits chosen from $1, 2, 3, 4$.
Rule 2 - The starting number must have four digits - so thousands, hundreds, tens and ones.
For example, we could choose $2124$ or $1124$.
So when we've got our starting number we'll do some counting. Here is a worked example.
Starting Number
We then count in order the number of $1$s, then the number of $2$s, then $3$s and lastly $4$s, and write it down as shown here.
So the first count gave one $1$, one $3$ and two $4$s.
You may see that this has continued so the third line shows that the line above had three $1$s, one $2$, one $3$ and one $4$.
The fourth line counts the line above giving four $1$s, one $2$, two $3$s and one $4$.
And so it goes on until ... until when?
Your challenge is to start with other four digit numbers which satisfy the two rules and work on it in the way I did.
Tell us what you notice.
What happens if you have five digits in the starting number?
Got a solution? Then click on Submit a Solution above.
### Why do this problem?
This activity, in line with the theme for this month, offers an 'action' to perform on a group of numbers which pupils can continue and explore. Or, you could think of the writing down of the 'description' of a sequence as an action performed on that sequence. It might particularly appeal to those pupils who enjoy number work but who are perhaps not used to succeeding in this area.
### Possible approach
Introduce the task by taking an example and work it through with the group/class of pupils, emphasising how careful we have to be with the simple act of counting.
Give them plenty of time to explore their own choice of numbers before bringing them together to share findings. Depending on the age and experience of the learners, you may like to give them a separate sheet of paper simply to note down anything they notice as they work. Encourage them not to rub out as they go along so they have a record of their thoughts, to some extent.
A whole-group discussion could focus on what they notice and what other questions they might have as a result of working on this task. Some children might be keen to try to explain their findings. Do encourage them in this, even if you are not sure of the reasons yourself. Admitting your possible uncertainty will spur them on!
### Key questions
Tell me about what you see happening.
What will you do now?
Can you make any predictions before you start the next one?
### Possible extension
Change the rules so that only odd numbers are available, for example:
Rule 1 - The starting number must have just three different digits chosen from $1, 3, 5, 7$ Rule 2 - The starting number should have four digits, so thousands, hundreds, tens and ones.
For example, $3155$ or $1135$.
Some children might like to find out about 'Golomb sequences' which are related to this task.
### Possible support
Some pupils may need help in carefully counting the number of occurrences of each digit. It might, therefore, be useful for children to work in pairs so that someone else is always checking the counting.
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.
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http://math.stackexchange.com/questions/142217/so3-lie-group
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# $SO(3)$ Lie group
I'm a little stuck at the moment where to go next with this. I know that there is a fact that there is a curve in $SO(3)$, beginning and ending at the identity which cannot be deformed to the constant curve, but when traversed twice can be deformed.
I want to be able to explain this clearly. Can someone explain this to me or help me out a little please?
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1
Well, this is just a consequence of the fact that the fundamental group of SO(3) is the cyclic group of order 2. Equivalently, the universal cover of SO(3) is the group SU(2), and the kernel is $\{\pm 1\}$. – M Turgeon May 7 '12 at 14:25
2
– M Turgeon May 7 '12 at 14:25
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This has nothing to do with Lie algebra... – Olivier Bégassat May 7 '12 at 14:32
3
– t.b. May 7 '12 at 15:17
1
@MTurgeon Touche. – Neal May 7 '12 at 16:25
show 4 more comments
## 2 Answers
I find it easiest to see that $SO(3)$ has a nontrivial order-two element of its homotopy group by thinking concretely $SO(3)$. A rigid motion of $\mathbb{R}^3$ that fixes the origin may be described by an axis of rotation and a (positively oriented) angle of rotation between $-\pi$ and $\pi$. Note that if we rotate about an axis by an angle greater than $\pi$, this is the same as rotating about the same axis by a negative angle greater than $-\pi$.
We may thus parametrize $SO(3)$ by an axis through the origin and a distance along that axis in $[-\pi,\pi]$. This is a three-ball --- but wait, there's more! A rotation of $\pi$ about an axis is the same as a rotation by $-\pi$ about the axis, so to complete the picture of $SO(3)$, we need to quotient the ball by antipodal identification.
Modulo concerns about topological structure (take it for granted that two rotations about similar axis, of similar angles, are "nearby" in $SO(3)$), we have constructed a homeomorphism between $SO(3)$ and the three-ball of radius $\pi$ modulo antipodal boundary identification, $\mathbb{R}P^3$.
Now "it's clear"* why we have a nontrivial order-two element of $\pi_1SO(3)$. Take the circle defined by traversing $\mathbb{R}P^3$ from the south pole to the north pole through the origin. It is already "stretched taut" --- no deformation can pull it down to a point. But if the path traversed twice can be contracted to a point; it's a quick exercise to draw a path homotopic to the loop traveled twice and contract it down to the north/south pole.
This may also give some motivation for why Dirac's belt trick works. The belt represents a loop in $SO(3)$ and the "twist" at any point along the length of the belt indicates the angle of rotation of the element of $SO(3)$ through which the loop passes. We start with a flat belt (a path mapping onto the origin) and then begin twisting one end while keeping the other fixed, until the second end has passed through one full rotation. Now the second end is as flat as the first, indicating that the path is closed, but the the belt is twisted in the middle.
Try as you might, you cannot get the loops out of the belt without rotating either end. But if you do the same thing again, rotate the second end through a second full rotation, now you can untwist the belt without rotating either end.
• Perhaps intuitively clear, but described precisely in BenjaLim's answer.
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Neal has already explained to you why $SO(3) \cong \Bbb{R}P^3$. Therefore to understand why there is a loop $a$ in the fundamental group of $SO(3)$, it suffices to prove that $\pi_1(\Bbb{R}P^3)$ has an element of order 2. To do this view $\Bbb{R}P^3$ with the usual $CW$ - structure. Now by the Van - Kampen theorem we know that attaching a 3-cell and higher to a path connected space $X$ does not change the fundamental group, so that $\pi_1(\Bbb{R}P^3) \cong \pi_1(\Bbb{R}P^2)$. Applying the Van Kampen theorem again shows that $\pi_1(\Bbb{R}P^2)$ is the free group on one generator $a$ subject to the relation $a^2 =1$, which is nothing more than $\Bbb{Z}/2\Bbb{Z}$. Choosing your basepoint to be the identity shows that indeed there is a non-contractible loop at the identity in $SO(3)$.
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http://math.stackexchange.com/questions/3627/diagonal-update-of-the-inverse-of-xdxt
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# Diagonal update of the inverse of $XDX^T$
I have a matrix $F=XDX^T$, where $D$ is $m\times m$ and diagonal and $X$ is $n\times m$.
Now, I compute $F^{-1}$.
Is there an efficient method to update $F^{-1}$ if $D$ is updated by $D'=D+G$, where $G$ is a sparse diagonal matrix?
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Note: "sparse diagonal matrix" is a very redundant expression. ;) – J. M. Aug 30 '10 at 11:23
How else would I imply a diagonal matrix with only very few of the diagonal nonzero though? – Projectile Fish Aug 30 '10 at 23:53
I would probably have used "low-rank diagonal matrix". :) – J. M. Aug 31 '10 at 20:38
Ah thanks, that makes more sense. – Projectile Fish Aug 31 '10 at 22:09
## 1 Answer
Yes, there is a very efficient way to compute the updated $F^{-1}$. Begin by writing $X$ in terms of its columns: $X = [x_1 x_2 \dots x_m]$. It follows that: $$XGX^T = \sum_{i=1}^{m} G_{ii} x_ix_i^T = \sum_{i\in S} G_{ii} x_ix_i^T$$ where $S \subset \{1,\dots,m\}$ is the set of indices for which $G_{ii}$ is nonzero. Suppose that $S$ contains $r$ indices. Since $G$ is sparse by assumption, $r$ is much smaller than $m$. If we assemble the columns of $X$ corresponding to indices in $S$, we obtain a new matrix $Y$, such that: $$XGX^T = YHY^T$$ where $Y$ is $m \times r$, and $H$ is a diagonal matrix whose diagonal consists of all the nonzero entries along the diagonal of $G$. Now apply the Woodbury matrix identity: $$(F+YHY^T)^{-1} = F^{-1} - F^{-1}Y(H^{-1}+Y^TF^{-1}Y)^{-1}Y^TF^{-1}$$ Computing the left-hand side directly is costly, as it requires inverting an $m\times m$ matrix. However, since $F^{-1}$ has already been computed, we can compute the right-hand side efficiently; the only inverse required is that of an $r\times r$ matrix, and $r$ is much smaller than $m$.
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Note, however, that though $D$, $XDX^T$, and their inverses have the same inertia (via Sylvester), the "corrected" inverse may or may not have the same inertia (this depends on the nature of G). Though, it should hopefully be cheap to permute the entries of G (and the corresponding columns of X) such that all zero entries of G are in the trailing portion of the matrix, making SMW applicable. – J. M. Aug 30 '10 at 11:35
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http://unapologetic.wordpress.com/2007/08/30/extraordinary-naturality/?like=1&_wpnonce=81a70c627a
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# The Unapologetic Mathematician
## Extraordinary Naturality
Now that we’ve gone back and rewritten the definition of naturality, let’s push it a bit.
First, notice that if we’re enriching over $\mathbf{Set}$ (in “ordinary” categories) then $\hom_\mathcal{D}(1_{S(A)},\eta_{B})$ means “take a morphism from $S(A)$ to $S(B)$ and follow it with $\eta_{B}$“. On the other hand, $\hom_\mathcal{D}(\eta_A,1_{T(B)})$ means “first do $\eta_A$, then follow it with a morphism from $T(A)$ to $T(B)$“. This recipe gives us back exactly the old naturality square, so $\mathbf{Set}$-natural transformations are exactly the ordinary natural transformations we’re familiar with!
So let’s take this reformulation of the naturality condition and tweak it. Instead of considering a family of arrows (in $\mathcal{D}_0$) $\eta_C:S(C)\rightarrow T(C)$, let’s move the variable over from the left to the right and consider a family $\beta_C:K\rightarrow T(C,C)$. Here, $K$ is an object of $\mathcal{D}$, and $T:\mathcal{C}^\mathrm{op}\otimes\mathcal{C}\rightarrow\mathcal{D}$ is a bifunctor. Now we say that the $\beta_C$ are the components of an “extraordinary $\mathcal{V}$-natural transformation” if the following diagram commutes:
This looks bizarre at first, though clearly it’s related to our revision of the enriched naturality diagram. It turns out that we’ve seen this sort of naturality before, though. If we read the diagram in $\mathbf{Set}$, consider a monoidal category $\mathcal{M}$ with duals, and use the functor $T(A,B)=A^*\otimes B$, then this is exactly the sort of naturality we find in the duality arrows $\eta_M:\mathbf{1}\rightarrow M^*\otimes M$!
Dually, we can define extraordinary $\mathcal{V}$-naturality for a family of morphisms $\gamma_C:T(C,C)\rightarrow K$. Write out this diagram, and show that the duality arrows $\eta_M:M\otimes M^*\rightarrow\mathbf{1}$ provide an example.
As another exercise, take these extraordinary naturality diagrams and work out the interpretation in $\mathbf{Set}$ explicitly. That is, actually start with some morphism $f:A\rightarrow B$ in the upper left-hand corner, and evaluate it all around. When we did this for our new $\mathcal{V}$-naturality diagram above we got our old naturality squares back. What do we get for extraordinary $\mathbf{Set}$-naturality?
### Like this:
Posted by John Armstrong | Category theory
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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://nrich.maths.org/2742
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### Sport Collection
This is our collection of favourite mathematics and sport materials.
### Speed-time Problems at the Olympics
Have you ever wondered what it would be like to race against Usain Bolt?
### The Olympic Torch Tour
Imagine you had to plan the tour for the Olympic Torch. Is there an efficient way of choosing the shortest possible route?
# High Jumping
##### Stage: 4 and 5
Article by John Barrow
If you are training to be good at any sport then you are in the business of optimisation - doing all you can to enhance anything that will make you do better and minimise any faults that hinder your performance. This is one of the areas of sports science that relies on the insights that are possible by applying a little bit of mathematics. Here we are going to think about two athletics events where you try to launch the body over the greatest possible height above the ground: high jumping and pole vaulting.
This type of event is not as simple as it sounds. Athletes must first use their strength and energy to launch their body weight into the air. If we think of a high jumper as a projectile of mass M launched vertically upwards at speed $U$ then the height $H$ that can be reached is given by the formula $$U^{2}=2g H$$ where $g$ is the acceleration due to gravity. Alternatively we can think in terms of energy conservation. The kinetic energy of the jumper at take-off is $\frac{1}{2}M U^{2}$ and this will be transformed into the potential energy $M g H$ gained by the jumper at the maximum height $H$ when he is instantaneously at rest at the highest point. Equating the two gives $U^{2}=2g H$ again.
All this sounds straightforward but the tricky point is the quantity $H$ - what exactly is it? It is not the height that is cleared by the jumper. Rather, it is the height that the jumper's centre of gravity is raised, and that is rather a subtle thing because it makes it possible for a high jumper's body to pass over the bar even though his centre of gravity passes under the bar. When an object has a bendy shape it is possible for its centre of gravity to lie outside of the body. One way to locate the centre of gravity of an object is to hang it up from any point on the object and drop a weighted string from the same point, marking where the string drops. Then repeat this by hanging the object up from another point. Draw a second line where the hanging string now falls. The centre of gravity is where the lines of the two strings cross. If the object is a square then the centre of gravity will lie at the geometrical centre but if it is L-shaped or U-shaped the centre of gravity will not lie inside the boundary of the body at all.
It is this possibility that allows a high jumper to control where his centre of gravity lies and what trajectory it follows when he jumps. The aim of our high-jumper is to get his body to pass over the bar whilst making his centre of gravity pass as far underneath the bar as possible. In this way he will make optimal use of his explosive take-off energy.
The simple high-jumping technique that you first learn at school, called the 'scissors' technique is far from optimal. In order to clear the bar your centre of gravity, as well as your whole body, must pass over the bar. In fact your centre of gravity probably goes close to 30 centimetres higher than the height of the bar. This is a very inefficient way to clear a high-jump bar. The high-jumping techniques used by top athletes are much more elaborate. The old 'straddle' technique involved the jumper rolling around the bar with their chest always facing the bar. This was the favoured technique of world-class jumpers up until 1968 when the American Dick Fosbury amazed everyone by introducing a completely new technique which involved a backwards flop over the bar and won him the Gold Medal at the 1968 Olympics in Mexico City. This method was only safe when inflatable landing areas became available. Fosbury's technique was much easier for high jumpers to learn than the straddle and it is now used by every serious high jumper. It enables a high jumper to send their centre of gravity well below the bar even though their body curls over and around it. The more flexible you are the more you can curve your body around the bar and the loweryour centre of gravity will be. The 2004 Olympic men's high-jump champion Stefan Holm, from Sweden, is rather small by the standards of high jumpers but is able to curl his body to a remarkable extent. His body is very U-shaped at his highest point. He sails over 2m 37 cm but his centre of gravity goes well below the bar.
##### The photographer Peter Kjelleras captures the Olympic high-jump champion Stefan Holm jumping at the World Athletics Championships in Paris, in 2003. Holm dramatically demonstrates his ability to send his centre of gravity far below the bar he is clearing 2.32 metres above the ground.
The high-jumper's centre of mass is about two-thirds of the way up his body when he is standing or running in towards the take off point. He needs to increase his launch speed to the highest possible by building up his strength and speed, and then use his energy and gymnastic skill to raise his centre of gravity by $H$, which is the maximum that the formula $U^{2}=2g H$ will allow. Of course there is a bit more to it in practice! When a high jumper runs in to launch himself upwards he will only be able to transfer a small fraction of his best possible horizontal sprinting speed into his upward launch speed. He has only a small space for his approach run and must turn around in order to take off with his back facing the bar. The pole vaulter is able to do much better. He has a long straight run down the runway and, despite carrying a long pole, the world's best vaulters can achieve speeds of close to $10$ metres per second at launch. The elastic fibre glass pole enables them to turn the energy of their horizontal motion $\frac{1}{2}M U^{2}$ into vertical motion much more efficiently than the high jumper. Vaulters launch themselves vertically upwards and perform all the impressive gymnastics necessary to curl themselves in an inverted U-shape over the bar,sending their centre of gravity as far below it as possible.
Let's see if we can get a rough estimate of how well we might expect them to do. Suppose they manage to transfer all their horizontal running kinetic energy of $\frac{1}{2}MU^{2}$ into vertical potential energy of $MgH$ then they will raise their centre of mass a height of:
$$H=\frac{U^{2}}{2g}$$
If the Olympic champion can reach $9\mathrm{\ ms^{-1}}$ launch speed then since the acceleration due to gravity is $g=10\mathrm{\ ms^{-2}}$ we expect him to be able to raise his centre of gravity height of $H=4$ metres. If he started with his centre of gravity about $1.5$ metres above the ground and made it pass $0.5$ metres below the bar then he would be expected to clear a bar height of $1.5+4+0.5=6$ metres. In fact, the American champion Tim Mack won the Athens Olympic Gold medal with a vault of $5.95$ metres (or $19^{\prime }6\frac{1}{4}"$ in feet and inches) and had three very close failures at $6$ metres, knowing he had already won the Gold Medal, so our very simple estimates turn out to be surprisingly accurate.
John D. Barrow is Professor of Mathematical Sciences and Director of the Millennium Mathematics Project at Cambridge University.
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.
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http://mathhelpforum.com/math-topics/117589-use-subscript-print.html
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# Use of a subscript?
Printable View
• November 30th 2009, 08:12 AM
Nyx-
Use of a subscript?
I don't get the point of it? So, I'm not sure so please correct me if I'm wrong (This is my question)
Say I have, 3 crates of apples but I want to show how many per crate would I do?
Like A (Apples) and a 3 below it slightly to show that I have 3 of those? and a five as the exponent to show that there are 3 crates with 5 apples in each?
• November 30th 2009, 08:59 AM
Bill Zimmerly
Quote:
Originally Posted by Nyx-
I don't get the point of it? So, I'm not sure so please correct me if I'm wrong (This is my question)
Say I have, 3 crates of apples but I want to show how many per crate would I do?
Like A (Apples) and a 3 below it slightly to show that I have 3 of those? and a five as the exponent to show that there are 3 crates with 5 apples in each?
Subscripts are used for many purposes. (An example being the sequential nature of the elements of a set.)
Sequence - Wikipedia, the free encyclopedia
You could represent your apple crate set like this...
{ $5_{1}, 7_{2}, 4_{3}$}
Which shows that crate #1 has 5 apples, crate #2 has 7, etc.
• November 30th 2009, 11:21 AM
Jameson
You technically could use subscripts like the above poster said. He was using them to fit the situation you proposed. However I find using subscripts in this way confusing and not a common use. Subscripts are very useful in numbering the order of things, but not so much with the quantity of things. If A represents an apple, then you can just use multiplication to show more than one, i.e. 2A, 10A, etc.
My comments are about convention and how easy notation is to read by others.
All times are GMT -8. The time now is 10:06 AM.
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http://math.stackexchange.com/questions/147101/is-there-a-theorem-or-conjecture-that-specifies-a-balanced-prime-must-be-multipl/147289
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# Is there a theorem or conjecture that specifies a balanced prime must be multiple of 3# distance from neighboring primes?
I hope I have written this question in an answerable form. Basically, I am assuming this theorem or conjecture exists, but I can't find it from my reading / searching (I'm not a mathematician). What I know:
• A "balanced prime" is the middle prime number in a sequence of three consecutive primes in arithmetic progression. This would make the number "CPAP-3", and thus the minimum separation is 6 = 3# = 3 * 2.
• What I've noticed from just analyzing balanced primes (see my post on GooglePlus: I am too new of a user to post an image) is that not only is their minimum separation 6 = 3# = 3 * 2 (see the lowest magenta "line" in picture). But greater separations must be a multiple of 6 (by observation)
So, my question is: is there a named theorem or conjecture that describes the isolation of balanced primes must be a multiple of 3#? I'm not a mathematician, but I'm guessing this exists and I just can't find it.
I'm also very interested to know if there's a theorem or conjecture about how the maximum isolation of a balanced prime grows. It appears to grow faster than log(Pn).
Thanks!
PS: I labeled the graph with terms like "prime loneliness" before I was pointed to better terminology. "swiss primes" are actually "balanced primes."
EDIT (again): From comments: this is a proven (and easily explained) theorem (much more elementary than the Green-Tao theorem), which covers the more general case of primes in arithmetic progression (AP-K; not necessarily consecutive primes). Thank you to all for pointing me to the obvious! :)
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The article you gave a link to mentions early on, with reference, a more general result about $k$ primes in AP. – André Nicolas May 19 '12 at 17:50
3,5,7 appears to be a sequence which has difference which is not a multiple of 3. – Mark Bennet May 19 '12 at 17:51
@MarkBennet yes, 3,5,7 is an exception, but I think that would be covered in the theorem for minimum separation of a balanced prime (besides 5, all others have a minimum separation of 3#) – Steve Koch May 19 '12 at 19:02
@AndréNicolas Thank you! I had read that, but didn't properly realize that CPAP-3 is a special case of AP-3. And apparently the Green-Tao theorem says: "If an AP-k does not begin with the prime k, then the common difference is a multiple of the primorial k# = 2·3·5·...·j, where j is the largest prime ≤ k." – Steve Koch May 19 '12 at 19:06
@SteveKoch: That's not the Green-Tao Theorem, it is far more elementary. Green-Tao is the assertion that there are arbitrarily long (finite!) arithmetic sequences of primes. – André Nicolas May 19 '12 at 19:09
show 2 more comments
## 2 Answers
I don't think anyone has addressed here the question OP raised of the maximum isolation of a "balanced prime". It is conjectured (but I'm not sure whether it has been proved) that for any odd prime $p$ there are infinitely many three-term arithmetic progressions of primes with $p$ the smallest member. So, for example, there should be infinitely many primes $q$ such that $2q-3$ is prime, making $q$ a balanced prime with isolation $q-3$. In short, the isolation of a balanced prime can, conjecturally, be practically as big as the prime itself.
EDIT: On re-reading the question, I'm no longer sure what's being asked. Maybe OP means primes $p$ and positive integers $a$ such that $p\pm a$ are both prime and there are no other primes between $p-a$ and $p+a$. In that case, we're asking about gaps between successive primes, and there is a lot of literature on that; there are results, and widely-believed conjectures, and lots of room between the two. It's generally believed the gap between $p$ and the next prime maxes out at something like $(\log p)^2$. I see no reason why the isolation of "balanced" primes should be any different.
MORE EDIT: I notice that these primes are tabulated at the Online Encyclopedia of Integer Sequences. However I don't see anything there likely to aid in the estimation of maximum isolation.
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Thanks, @Gerry Myerson! I don't have reputation to upvote these answers, but thank you everyone for the links and help. – Steve Koch Jul 31 '12 at 13:28
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http://mathoverflow.net/questions/70651/computing-simultaneous-hamming-neighborhood-for-a-set-of-strings/70687
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## Computing Simultaneous Hamming Neighborhood for a Set of Strings
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $S = \lbrace s_1, s_2 \ldots s_n \rbrace$ be a set of strings each of length $k$ from an alphabet $\Sigma$, $h(s_i, s_j)$ denote the hamming distance between two strings. The simultaneous hamming neighborhood is defined as $N_{\alpha} = \lbrace s' | h(s',s_j) \leq \alpha, \forall s_j \in S , s' \in \Sigma^k \rbrace$, $1\leq \alpha \leq k$.
I would like to know if this problem (i.e. computing $N_{\alpha}$ efficiently) has been considered earlier ? -- By efficiently I mean the running time of the algorithm should be something like $O(|N_{\alpha}|)$, when $|N_{\alpha}|$ is much larger than $n$.
Thank you very much for your help.
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## 1 Answer
I assume that you require the running time of an algorithm to be polynomial also in k. Then this is impossible even with the binary alphabet unless P=NP by the result by Frances and Litman [FL97].
Consider the easier task of deciding, given k-bit strings s1,…,sn∈{0,1}k and an integer α, whether Nα contains any element or not. This problem is equivalent to what is called the Minimum Radius problem in [FL97], where it is proved to be NP-complete.
[FL97] M. Frances and A. Litman. On covering problems of codes. Theory of Computing Systems, 30(2):113–119, March 1997. http://dx.doi.org/10.1007/BF02679443
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Hello Tsuyoshi, Thank you for that reference its useful. However I'm interested in the complexity when $k$ is a fixed constant. Thank you, Vamsi. – Vamsik Jul 18 2011 at 23:53
@Vamsik: If k is fixed, then it is straightforward to check all k-letter strings in poly(n) time because, as long as every letter in Σ appears in some string s_i, there are only |Σ|^k≤(kn)^k=poly(n) strings of length k. The case where Σ contains unused letters can be also handled easily. – Tsuyoshi Ito Jul 19 2011 at 0:41
@Tsuyoshi: $\Sigma$ is part of input, so any bivariate polynomial in $n$ and $\Sigma$ would be efficient. On the other hand if $\Sigma$ were to be constant I'm wondering if exists is an $O(|N_{\alpha}|)$ algorithm. – Vamsik Jul 19 2011 at 0:44
@Vamsik: If you fix k and you are fine with a polynomial in n and |Σ|, then check all the k-letter strings. There are |Σ|^k = poly(|Σ|) strings, and each string can be tested in poly(n) time for membership in N_α, which means that this algorithm runs in poly(n,|Σ|) time. – Tsuyoshi Ito Jul 19 2011 at 0:49
1
@Vamsik: (1) In that case, it is trivially impossible because you have to distinguish between the case where N_α is empty and the case where |N_α|≤100 (say) in a constant time, in which you cannot even read the whole input. (2) Even if you change the question again, I will no longer try to answer. I cannot keep chasing the moving target. – Tsuyoshi Ito Jul 19 2011 at 2:43
show 2 more comments
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http://mathoverflow.net/questions/111833/simplicity-of-eigenvalues-of-an-elliptic-operator-under-a-symmetry-assumption
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## Simplicity of eigenvalues of an elliptic operator under a symmetry assumption
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
A striking difference in the spectral analysis of 2nd order elliptic boundary-value problems between one and several space dimensions is the following. In one space dimension, the eigenvalues are simple (Sturm-Liouville theory) though they can be multiple in several space dimensions. For instance the eigenfunctions of the Laplacian in the disk $D_2\subset{\mathbb R}^2$ have the form $f(r)\cos n(\theta-\phi)$ in polar coordinates; if $n\ge1$, then the corresponding eigenvalue has double multiplicity.
Let me assume that the data (domain $\Omega\subset{\mathbb R}^2$, operator, boundary conditions) be invariant under a symmetry, say $x_1\leftrightarrow-x_1$. Then one may consider the restriction of the boundary-value problem to either the space $E_+$ of even functions ($u(x)=u(-x_1,x_2)$) or the space $E_-$ of odd functions ($u(x)=-u(-x_1,x_2)$). The full spectrum is then the union of the even and odd spectra, with addition of multiplicities. Here is an example of such a problem $$\Omega=D_2,\qquad \Delta u=\lambda u\quad\hbox{ in }D_2,\qquad\frac{\partial u}{\partial r}=\rho(\theta)u\quad\hbox{ on }r=1,$$ where $\rho$ is an even function. For an other example, one may replace the boundary condition by $$\frac{\partial u}{\partial r}=\rho(\theta)\frac{\partial u}{\partial \theta}.$$
Is it true or not that the eigenvalues of each restricted problem (even or odd) are all simple ?
At least, the Krein-Rutman theory tells us that the first eigenvalue is simple, hence the first even eigenvalue is simple.
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## 1 Answer
For example, consider the Laplacian on $[-1,1] \times [-1,1]$ with Dirichlet boundary conditions. If $m$ and $n$ are distinct positive integers, $\sin(m \pi x) \sin(n \pi y)$ and $\sin(n \pi x) \sin(m \pi y)$ are both eigenfunctions for the same eigenvalue, and are both odd. Similarly $\cos((m+1/2) \pi x) \cos((n+1/2) \pi y)$ and $\cos((n+1/2) \pi x) \cos((m+1/2) \pi y)$ which are both even.
EDIT: Cases where a positive integer can be represented in many ways as a sum of squares present even more multiplicity.
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http://physics.stackexchange.com/questions/tagged/electromagnetic-radiation?page=3&sort=newest&pagesize=15
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# Tagged Questions
Propagating solutions to Maxwell’s equations in classical electromagnetism and real photons in quantum electrodynamics. A superset of thermal-radiation.
1answer
99 views
### Commercial Infrared lights
I purchased an infrared light. It's a 100 W Philips infrared lightbulb. Says it's infrared, but I haven't done any spectrum analysis so I don't know for sure if it's just red or really infrared. As I ...
2answers
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### Radiated power and energy density for a black-body
I am having an hard time trying to understand why the radiated power per unit area $P$ of a black body is given by $$P=\frac{c}{4} u$$ in terms of the energy density $u$ and the velocity of light. I ...
1answer
110 views
### Spectral radiance unit conversion [closed]
I have spectral radiance data in SRUs (spectral radiance units), as a function of wavelength: $$a = \mu W cm^{-2} sr^{-1} nm^{-1}$$ However, I am working with software which requires my data in the ...
3answers
370 views
### Intensity of light
If we have 2 beams of light with equal intensities, but with different frequencies, wouldn't the one with the higher frequency generate more power? If so, how come the intensity, which is in $W/m^2$, ...
2answers
197 views
### Jones vector and matrices
With Jones vectors and matrices one can describe the change in polarization of a EM wave. What is the convention of the reference coordinate system; Is it fixed or does it change whenever the ...
1answer
72 views
### Can I find the frequency generated by tapping a battery on a coin?
According To How Stuff Works , if you tap a 9 volt battery on a coin it will transmit Radio waves. I was wondering, because I don't have a radio with me at the moment, Is there a way I could discover ...
3answers
760 views
### Is there something special in the visible part of electromagnetic spectrum?
I always wondered how much information we get from color. Things we see have different colors; edible products change color when began to spoil so we have a notion what color a fresh product should ...
1answer
123 views
### RF WiFi Radiation
I would like to extrapolate my current router wifi radiation from my phone. If I know that my router is transmitting at 300mW and my phone displays the strength in -dbm (from 0 to -100 scale) if I ...
1answer
66 views
### Signal emission and detection with certain parameters; Electrical Engineering [closed]
Could a micro-transmittting device, smaller than, say, a golf ball be employed to emit an ultra-narrow detectable beam, (e.g. 1mm) with range of one or 1.5 decimeter to 2 or less centimeter? The ...
0answers
52 views
### Maximum acceleration [closed]
Assume that a tiny and shiny (perfect conductor) Au metal in free space has a shape of a thin cylinder of thickness 1 m (like a tiny coin of thickness 1m) and end cap area of 500 m2. A ...
3answers
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### Magnetic fields and gravitational waves. How far do they reach?
I read that magnetic fields perpendicular to a current shoot out and expand all the way to infinity. Additionally a gravitational wave, no matter how small will also expand to infinity at the velocity ...
3answers
712 views
### Amplitude of an electromagnetic wave containing a single photon
Given a light pulse in vacuum containing a single photon with an energy $E=h\nu$, what is the peak value of the electric / magnetic field?
1answer
226 views
### Optical constants of noble metals: the Drude model for microwave modelling
I have a question regarding the optical constants of noble metals. According to Johnson and Christy's paper Optical Constants of Noble Metals (Phys. Rev. B 6, 4370–4379 (1972), ...
1answer
232 views
### How EM waves are produced by accelerating charged particles?
How the electro-magnetic waves are produced by the accelerating charged particle? Graphical explanations are most welcomed. Is the explanation given by the below mentioned article correct regarding ...
1answer
318 views
### How do EM waves get detached from an antenna?
How does an electro-magnetic waves get detached from an antenna and spread to the space? While an antenna receives an EM wave, which quantity of the EM wave (electric or magnetic) is used for ...
0answers
76 views
### How to calculate radiative transition rate of exciton in a quantum dot with specific dimension?
I am writing rate equations for a nanophotonic system including three quantum dots. I need to calculate that radiative transition rates of exciton in ground state in those quantum dots. In the paper ...
2answers
240 views
### What is the physical meaning of retarded time?
Consider this figure Now, when I measure a field produced by the charge $e$ at the point $\mathbf r$, at the time $t=t_1$, it means that the charge sent the signal field at the time $t=t_r$, where ...
1answer
65 views
### Idea of precursors of the electro-magnetic waves
The idea of the material Maxwell equation is almost clear. But I'm curious about the idea that except for material equation the pure Maxwell equation should work, but in harder sense: more currents ...
0answers
57 views
### Formula for transverse decrease in signal strength?
Alice is transmitting an electromagnetic beam to Bob. Assume it is a well collimated beam, like a laser beam, or a maser beam. As Bob's distance from Alice increases, his signal strength decreases ...
2answers
216 views
### Why does the electric field dominate in light?
I read a book on the wave property of light where the author mentioned that the electric field, instead of magnetic field, dominates the light property. I don't understand why. In Maxwell's theory, a ...
1answer
531 views
### Phase and Group Velocity of Electromagnetic Waves
Moving charges produce oscillating electric and magnetic fields -we have an electromagnetic wave. In terms of moving charges or at the level of charges, what is phase velocity and group velocity of ...
1answer
75 views
### Units of azimuthally integrated radiant intensity
If the radiant intensity $I(\phi,\theta)$ of electromagnetic radiation is given in spherical coordinates ($\phi$ the azimuth, $\theta$ the polar angle, indenpendent of $R$) with units $Js^{-1}sr^{-1}$ ...
1answer
119 views
### Finding the energy levels of an electron in a plane perpendicular to a uniform magnetic field
Suppose we have an electron, mass $m$, charge $-e$, moving in a plane perpendicular to a uniform magnetic field $\vec{B}=(0,0,B)$. Let $\vec{x}=(x_1,x_2,0)$ be its position and $P_i,X_i$ be the ...
3answers
156 views
### Difference in velocity of light in change in medium
It is often seen that according to physics the light changes it's velocity according to the medium through which it is traveling. So can it be explained that why so happen?
1answer
64 views
### Getting a given wavelength radio signal given an antenna with real-world constraints
Supposing you are given a transmitting antenna of whatever type of metal is most commonly used these days, and supposing that you are applying an AC current with the intent of transmitting a 1 m ...
2answers
2k views
### Why is Near Field Communication (NFC) range limited to about 20cm?
Near Field Communication (NFC) operates at 13.56 MHz. Near Field is the region situated at a distance r << λ λ = c/f ...
2answers
268 views
### Can electromagnetic fields be used to deconstruct and reconstruct atoms?
I was thinking one day and came up with a theory after reading about how scientists were studying anti-matter by using electro magnetic fields to separate matter from the anti-matter they made. It ...
1answer
394 views
### Force on Earth due to Sun's radiation pressure
I have been asked by my Classical Electrodynamics professor to calculate the force that the Sun exerts in the Earth's surface due to its radiation pressure supposing that all radiation is absorbed and ...
2answers
277 views
### why dosen't a charged particle radiate energy in circular motion in a uniform magnetic field?
I have studied in my Physics course that one of the drawbacks of Rutherford's atomic model was that when an electron will revolve around the nucleus, it is undergoing acceleration and so it should ...
1answer
49 views
### Range of electromagnetic waves
I was reading this article. There is a statement "It is a well-known fact that the telecom towers mounted with antennas in the lower frequency bands can cover far greater areas than those using the ...
1answer
149 views
### Does a magnetic field have gravity?
Re-reading http://physics.stackexchange.com/a/33156/5265; I find myself thinking if light, being EM in the humanly visible spectrum, may possess gravity - does a magnetic field also possess gravity?
1answer
120 views
### What is the ion drag mechanism in dielectric heating?
While reading about dielectric heating on Wikipedia, I read about the ion drag mechanism but there wasn't enough information about. I know there is another Phys.SE question talking about the ion drag ...
1answer
64 views
### How does the specific frequency of EM Radiation relate to displacing electrons from their orbits?
I've only a general grasp on how all this works, so it could be I'm asking this poorly or misunderstanding what happens. With that said: The energy of EM radiation is a function of its frequency. ...
2answers
191 views
### Radio communication and antennas
I'm trying to understand how an antenna is able to understand, but have a few quesions: I understand that for an antenna to distinguish between 2 (or more) signals, they have to be in different ...
1answer
158 views
### Does $\lambda\nu = c$ hold for all the waves in the universe?
Are all waves in the universe the same as electromagnetic waves? Basically, my question arises from an equation I found in my chemistry textbook: $$\lambda \nu ~=~ c.$$ This states that the ...
10answers
3k views
### Why is light called an 'electromagnetic wave' if it's neither electric nor magnetic?
How can light be called electromagnetic if it doesn't appear to be electric nor magnetic? If I go out to the sunlight, magnets aren't affected (or don't seem to be). And there is no transfer of ...
0answers
84 views
### Wireless signal strength
My question is possibly somewhat misplaced, but I'll try to explain as best as I can. Suppose I have a transmitter with a frequency of 2500MHz and a power of 1W. It radiates uniformly in all ...
1answer
61 views
### Does every wavenumber of IR result in a different kind of vibration?
Does every wavenumber of IR result in a different kind of vibration? If that is true, what if a molecule absorb 2 different wavenumbers (which cause different rocking and symmetrical stretching for ...
2answers
106 views
### Power due to dipole radiation and time reversal symmetry in classical E&M
The dipole formula for the power loss emitted by a time varying electric dipole is (in natural units) $P = \frac{\dot d_i^2}{6 \pi}$. This is clearly even under time reversal symmetry $T$, but a ...
0answers
110 views
### Antenna Power and gain calculation [closed]
I have a wireless security related question, the second part confused me: Your wireless network usually has a range of 100 feet. However you are having a (confidential) meeting in a 10’ x 10’ x ...
1answer
58 views
### Could we really charge metal plates using microwaves?
While skimming through Dielectric heating, I read that they use microwaves to charge the plates. How do they do that?
3answers
64 views
### What kind of “camera” and “light” source should I use to detect the path along which the light moves?
I would like to have an image ( in any kind of space ), where I see the path of a "light" source. In my understanding the most common, directed source would be a laser pointer. ...
1answer
158 views
### Velocity of electron in electrostatic field. Does radiation matter?
There's a voltage difference of 1000 Volts between two points 2 meters apart. An electron starts at the point of lower potential and is left to travel alone in a straight line until it reaches the ...
0answers
20 views
### Why can radio wave travel through walls while imfrared can't? [duplicate]
Possible Duplicate: If both radio waves and gamma rays can travel through walls Does it have anything to do with the materials used to build the wall? What determines if a material can ...
0answers
51 views
### Electromagnetic charge [duplicate]
Possible Duplicate: If electromagnetic field give charge to particles, does photon carry charge? Is it possible to charge photons I was wondering if light when it's in wave form (seeing ...
2answers
105 views
### Xray compression of secondary in hydrogen bomb
Accounts of the "secret" of the hydrogen bomb describe Xrays from a primary fission explosion reflecting off of the bomb case (occasionally passing through polystyrene foam) and compressing and ...
3answers
159 views
### What if $\gamma$-rays in Electron microscope?
I was referring Electron microscopes and read that the electrons have wavelength way less than that of visible light. But, the question I can't find an answer was that, If gamma radiation has the ...
0answers
128 views
### Transmission formula for two near field polarizing filters?
Consider two parallel wire "comb" filters oriented at 45 degrees relative "tooth" angle. The comb tooth spacing spacing is much less than one wavelength, and the comb tooth length is much more than ...
2answers
328 views
### How objects have different color in different lights?
As everybody knows that Black objects are black because almost all the light that falls on them is absorbed into the material. Little or no light is reflected back toward our eyes, therefore we see ...
1answer
64 views
### 1MeV ray hits a band gap of 2eV : trying to understand X-rays
I am trying to understand why radio waves pass though the human body, light does not, but X-rays still pass. In my mind I imagine that radio waves don't supply enough energy and don't excite any ...
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http://mathoverflow.net/questions/97872/quaternion-ring
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## Quaternion ring
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $\mathbb Z_n$ denote the integers modulo $n$. Let $\mathbb Z_n[i, j, k]$ be the quaternionic ring over $\mathbb Z_n$, that is, the free module over $\mathbb Z_n$ with basis ${1, i, j, k}$ and multiplication defined by $$i^2=j^2=k^2=ijk=-1.$$
It is well-known that if $n=p$ where $p$ is a odd prime then $\mathbb Z_p[i, j, k]$ is isomorphic to the full matrix ring of order $2$ over $\mathbb Z_p$. What can be said about the structure of $\mathbb Z_n[i, j, k]$ for a composite $n$. Is also a full matrix ring?
Note:If we can prove that $\mathbb Z_n[i, j, k]$ is semi-simple then it is necessarily a direct sum of matrix rings over a field by a theorem of Wedderburn.
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## 1 Answer
Yes, for odd $n$, the ring $(\mathbb Z/n)[i,j,k]$ is isomorphic to the ring of two-by-two matrices over $\mathbb Z/n$.
To explain this, it is better to write $\mathbb Z/n$ and $\mathbb Z/p$ rather than $\mathbb Z_n$ and $\mathbb Z_p$, because, in fact, the decisive statement concerns $p$-adic integers $\mathbb Z_p$, and we should reserve the notation for that.
What is true is that, using the notation for $p$-adic integers, for odd prime $p$, $\mathbb Z_p[i,j,k]$ is the full matrix ring over the $p$-adic integers $\mathbb Z_p$. Granting this for a moment, $(\mathbb Z/p^\ell)[i,j,k]$ is the image of $\mathbb Z_p[i,j,k]$ by mapping $p^\ell \mathbb Z_p$ to $0$, since $\mathbb Z_p/p^\ell \mathbb Z_p\approx \mathbb Z/p^\ell \mathbb Z$.
By Sun-Ze's theorem, the general $\mathbb Z/n$ is the sum of the $\mathbb Z/p^\ell$ where $p^\ell$ are the prime powers dividing $n$. Thus, for odd $n$, $(\mathbb Z/n)[i,j,k]$ is the sum of the corresponding rings for the $p^\ell$.
To prove that $\mathbb Z_p[i,j,k]$ is the full matrix ring, start with the point that $\mathbb Q_p[i,j,k]$ is the full matrix ring, for odd $p$. There are various proofs of this... For example, the surjectivity of norms on finite fields, together with Hensel's Lemma, proves that the quadratic form $a^2+b^2+c^2+d^2=\hbox{norm}(a+bi+cj+dk)$ has a non-trivial zero, so $\mathbb Q_p[i,j,k]$ is not a division ring.
The additional ingredient is that the ring of "Hurwitz integers" consisting of $\mathbb Z[i,j,k]$ with $(1+i+j+k)/2$ adjoined is Euclidean, in the appropriate sense for a non-commutative ring. "Locally" at odd primes $p$, the $1/2$ in the definition of the Hurwitz integers is a unit, so, locally, at odd primes, the $p$-adic version of the Hurwitz integers is the naive notion of local quaternion integers, $\mathbb Z_p[i,j,k]$. The rest of the discussion is just clean-up.
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3
In addition, note that if $\mathbb H$ denotes the Hurwitz integers then $\mathbb H\otimes \mathbb Q_2$ is a field of quaternions (the underlying quadratic form being anisotropic). Thus $\mathbb H\otimes \mathbb Z/2^k$ will be anisotropic for $k$ big enough ($k>3$ should do the job) and won't be a subring of the ring of two by two matrices over $\mathbb Z/2^k$. – G.C. May 24 at 23:58
thanks for your answer, is very useful for me. – miguel May 25 at 13:32
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http://physics.stackexchange.com/questions/13127/simulating-eye-diagrams?answertab=oldest
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Simulating eye diagrams
I'm trying to figure out how to simulate eye diagrams for communications systems using Python. I'm not sure I have the theory down completely, though. From what I could figure out using some old LabWindows C code, the process is something like this:
1. Take an input pattern and convert it into a discrete signal with the same time step between points as the system's step data
2. Perform an FFT on the input pattern and convert the system's step data to an S-parameter
3. Multiply the transformed pattern and the S-parameter pointwise (equivalent to convolution in the time domain)
4. Take the inverse FFT of product.
This is essentially all the old code seems to be doing, but unfortunately at the key steps there are calls to LabWindows functions that have poor online documentation, and I don't have access to the development libraries myself.
Any help or knowledge of eye diagram theory would be a tremendous help.
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It seems to me that this should be migrated to Stack Overflow site. – AdamRedwine Oct 3 '11 at 21:22
2 Answers
I think the sequence of steps should be:
1. convert input-data into analogue signal (modulation)
2. apply distortion and noise to simulate the effects of the channel (e.g. using convolution with impulse response)
3. demodulate the analogue signal into the out-data
Then you plot input-data against out-data.
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My problem is that while there are plenty of software packages out there that can do the simulation, I need to integrate it into an automated test system. I can't seem to find any literature online detailing how to actually do this simulation. I was hoping that there would be someone who had worked with this sort of metric before and could aid me in my development. – bheklilr Aug 3 '11 at 13:31
– whoplisp Aug 3 '11 at 13:47
I found the solution. I was not making my pattern long enough to get the frequency content necessary from the FFT. – bheklilr Aug 3 '11 at 14:01
@bheklilr: You should post that as an answer and accept it. – Kevin Reid Oct 2 '11 at 13:16
I needed to ensure that the pattern itself was long enough to get the correct $\Delta f$ in the frequency domain, since $\Delta f = \frac{1}{N}$ where $N$ is the number of points in the pattern signal. If this $\Delta f$ does not match that of the transformed data, then there is no real way to combine the two.
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http://nrich.maths.org/2357/solution
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Efficient Packing
How efficiently can you pack together disks?
Perfect Eclipse
Use trigonometry to determine whether solar eclipses on earth can be perfect.
Construct the Solar System
Make an accurate diagram of the solar system and explore the concept of a grand conjunction.
Far Horizon
Stage: 4 Challenge Level:
The solution below is based on the one sent in by Barinder of Langley Grammar School. We had quite a number of correct, well laid out solutions to this problem this month including those from Roy of Allerton High School, Dan (no address given) and Calum of Wayland High School.
Although definitely not in proportion, this makes the problem seem a lot easier. The question is asking for the length of the arc I have coloured red. To get this, I decided to find the angle $\theta$ on the diagram, and use the equation
Arc Length $= \frac {\theta}{360} \times 2 \pi r$ where $\theta$ is measured in degrees and r is the radius.
$\angle OAB = 90^\circ$, since it is where a tangent and a radius of a circle meet - it is a circle theorem.
Thus, the triangle AOB can be drawn as follows:
We can now use trigonometry to find $\theta$:
$$\begin{align*} \cos \theta &= \frac {6367000}{6367025} \\ \cos \theta &= 0.99999607 \\ \theta &= \cos^{-1} (0.99999607) = 0.1606^\circ \mbox{(4 d.p.)}\end{align*}$$
Substitute this into the equation for the arc length of a circle earlier to obtain the length required:
Arc Length $= \frac {0.1606}{360} \times 2\pi r.$
Arc Length $= 0.000446 \times 2 \times \pi \times 6367000 = 17,842.3m = 17.8 km$
For this next part, we are given the arc length, since this corresponds to the distance between England and France. The diagram is therefore:
This is essentially the reverse of the previous question. We need to find the angle $\alpha$ first, and to do this, we consider the arc length of the sector OAD of the circle:
Arc Length $= \frac {\alpha}{360} \times 2 \pi r$.
So $32000 = \frac {\alpha}{360} \times 2 \times \pi \times 6367000$.
Then $32000 \times 360 = \alpha \times 2 \times \pi \times 6367000$.
So $\alpha = 0.288^\circ \mbox{(3 d.p.)}$
Since we now have the angle $\alpha$, we can consider the triangle AOB:
$\cos \alpha = \frac {6367000}{6367000 + h}$.
So $6367000 + h = \frac {6367000}{\cos (0.288)}$.
So $6367000 + h = 6367080.415$
$h = 6367080.415 - 6367000 = 80.415$ m high.
Thus, the cliffs of Dover are $80.4$ metres high.
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.
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http://math.stackexchange.com/questions/tagged/integral-transforms+harmonic-analysis
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# Tagged Questions
1answer
74 views
### Decaying Fourier transform and smoothness
Suppose that $f\in L^1 (\mathbb{R})$ and that for any $n\in \mathbb{N}$ there is $C_n > 0$ such that its Fourier transform satisfies $$|\hat{f}(\xi )| \le C_n(1+|\xi |^2)^{-n}.$$ I want to show ...
1answer
118 views
### Convolution square root of $\delta$
I want to somehow classify the distributional solutions of the equation $$f \ast f = \delta$$ where $\delta = \delta _0$ is the Dirac delta distribution. Clearly, by Fourier transformation, we ...
0answers
65 views
### Different proofs of support theorem for Radon Transform
Let $\mathcal{H}$ be a set of all hyperplanes in $\mathbb{R}^n$. Radon transform of function $f(x) \colon \mathbb{R}^n \to \mathbb{R}$ is defined as function $R[f] \colon \mathcal{H} \to \mathbb{R}$ ...
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http://mathoverflow.net/questions/28241/lattice-stick-number-vs-stick-number-of-knot/33616
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## Lattice Stick Number vs. Stick Number of Knot
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Can the lattice stick number of a knot be bounded by the stick number of the knot?
The stick number $S(K)$ of a knot $K$ is the fewest number of segments needed to realize it by a simple 3D polygon. The lattice stick number $S_L(K)$ is the fewest segments in a realization in the cubic lattice, with all segments parallel to a coordinate axis. For example, the stick number of the trefoil knot $K=3_1$ is 6, and its lattice stick number is 12 (the latter a result of Huh and Oh from 2005).
My question is whether it is possible to bound $S_L(K)$ by $m S(K)$, where $m$ is some multiplier factor. Ideally $m$ would be a constant, but perhaps it is more realistic to expect it to depend on the complexity of the knot (e.g., on its crossing number $cr(K)$). What I have in mind is replacing each stick in a stick realization by a bounded lattice path.
Addendum. Tracy Hall's clever example below indicates that it is unlikely that $m$ could be a constant.
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Clarification: for lattice stick number, can a "segment" by arbitrarily long, or are you counting the total number of unit segment lengths? (Both questions look equally interesting a priori.) – Cap Khoury Jun 15 2010 at 12:50
@Cap Khoury: The former. In general, the length of the sticks is not part of the definition of $S_L(K)$. One stick can consist of several collinear unit-length segments. – Joseph O'Rourke Jun 15 2010 at 12:57
## 2 Answers
I wouldn't be surprised by something like a quadratic bound, or possibly something reasonable in terms of another complexity measure for the knot, but I see no hope for making $m$ constant. Consider the following construction: given $m$, choose some large number like $N=(10m)^6$ of points uniformly at random in the unit sphere, and connect them sequentially in a cycle with straight line segments to define a knot $K$. By construction $K$ has stick number no more than $N$, but each stick has a long narrow tunnel that it must traverse in a very precise direction, which is difficult to do with only $m$ lattice sticks. Of course any one tunnel can be made shorter and wider with an affine transformation (or any small collection of tunnels, with a piecewise affine transformation) but I am convinced (without attempting a rigorous proof) that with probability approaching $1$ a knot so constructed has a lattice stick number much higher than $mS_L(K)$.
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This is a convincing example, Tracy! Thanks! – Joseph O'Rourke Aug 4 2010 at 13:12
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The lattice stick number is obviously bounded by some function of the stick number: there are finitely many graphs with stick number at most k (because there are finitely many line segment arrangements in the plane and choices of over-under relationships on each crossing) so the lattice stick number of stick-number-k graphs is just the max of the lattice stick number of this finite set of graphs. This argument doesn't give an explicit or good bound on the number, though.
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Thanks, David, for that clear argument! – Joseph O'Rourke Aug 5 2010 at 16:22
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http://mathoverflow.net/revisions/114233/list
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2 edited title
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# For which semisimple element $s$ in finite group of lie type $C_{G}(s)$ is a Levi subgroup ?
Let $G$ be a finite simple group of lie type. Let $s$ be a semisimple element lying in maximal torus $T_{w}$ for $w\in W$ where $W$ is the Weyl group of $G$. Can we say that $C_{G}(s)$ is Levi subgroup of a Parobolic containing $s$ by looking just conjugacy class of $w$?
I assume that $G'$ is simply connected.
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http://www.scratchapixel.com/lessons/3d-basic-lessons/lesson-4-geometry/conventions-again-row-major-vs-column-major-vector/
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• Row Major vs Column Major Vector
# Row Major vs Column Major Vector
Earlier in this lesson, we have explained that vectors (or points) can be written down as [1x3] matrices (one row, three columns). Note however that we could have also written them down as [3x1] matrices (three rows, one column). Technically, these two ways of expressing points and vectors as matrices are perfectly valid and choosing one mode or the other is just a matter of convention.
Vector written as [1x3] matrix: $$\scriptsize V=\begin{bmatrix}x & y & z\end{bmatrix}$$
Vector written as [3x1] matrix: $$\scriptsize V=\begin{bmatrix}x\\y\\z\end{bmatrix}$$
In the first example ([1x3] matrix) we have expressed our vector or point in what we call the row-major order. Simply the vector (or point) is written as a row of three numbers. In the second example, we say that that points or vectors are written in the column-major order (that is we write the three coordinates of the vector or point vertically, as a column).
Remember that we express points and vectors as matrices to multiply them by [3x3] transformation matrices (for the sake of simplicity we will work with [3x3] rather than [4x4] matrices). We have also learned that we can only multiply matrices when the number of columns from the left matrix and the number of rows from the right matrix are the same. In other words the matrices [m x p] and [p x n] can be multiplied with each other but the matrices [p x m] and [p x n] can't. Note that if we write a vector as a [1x3] matrix we can multiply it by a [3x3] matrix (assuming this [3x3] matrix is on the right inside of the multiplication), but if we write this vector as a [3x1] matrix then we can't multiply it by a [3x3] matrix. This is illustrated in the following examples. The inner dimensions (3 and 3) of the matrices involved in the multiplication are the same (in green) so this multiplication is valid (and the result is a transformed point written in the form of a [1x3] matrix):
$$\scriptsize [1 \times \color{\green}{3}]*[\color{\green}{3} \times 3] \rightarrow \\ \scriptsize \begin{bmatrix}x & y & z\end{bmatrix} * \begin{bmatrix} c_{00}&c_{01}&{c_{02}}\\ c_{10}&c_{11}&{c_{12}}\\ c_{20}&c_{21}&{c_{22}}\\ \end{bmatrix} =\begin{bmatrix}x'&y'&z'\end{bmatrix}$$
The inner dimensions (1 and 3) of the matrices involved in the multiplication are not the same (in red) so this multiplication is not possible:
$$\scriptsize [3 \times \color{\red}{1}]*[\color{\red}{3} \times 3] \rightarrow \begin{bmatrix}x\\ y\\z\end{bmatrix} * \begin{bmatrix} c_{00}&c_{01}&{c_{02}}\\ c_{10}&c_{11}&{c_{12}}\\ c_{20}&c_{21}&{c_{22}}\\ \end{bmatrix}$$
So what do we do? The solution to this problem is not to multiply the vector or the point by the matrix, but the matrix to the vector. In other words we move the point or vector to the right inside of the multiplication:
$$\scriptsize [{3} \times \color{\green}{3}]*[\color{\green}{3} \times {1}] \rightarrow \begin{bmatrix} c_{00}&c_{01}&{c_{02}}\\ c_{10}&c_{11}&{c_{12}}\\ c_{20}&c_{21}&{c_{22}}\\ \end{bmatrix} * \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}x'\\y'\\z'\end{bmatrix}$$
Note that the result of this operation is a transformed point written in the form of a [3x1] matrix. So we get a point to start with and we finish with a transformed point which is what we want. Problem solved. To summarize, when by convention we decide to express vectors or points in row-major order ([1x3]), we need to put the point on the left inside of the multiplication and the [3x3] on the right inside of the multiplication sign. This is called in mathematics, a left or pre-multiplication. If you decide to write the vectors in column-major order instead ([3x1]), the [3x3] matrix needs to be of the left inside of the multiplication and the vector or point on the right inside. This is called a right or post-multiplication.
We need to be careful about how these terms are actually used. For instance the Maya documentation says "the matrices are post-multiplied in Maya. For example, to transform a point P from object-space to world-space (P') you would need to post-multiply by the worldMatrix. (P' = P x WM)", which is confusing because it's actually a pre-multiplication but they are speaking about the position of the matrix in regards to the point here. That's actually an imprecise (incorrect) use of the terminology. It should have been written that in Maya, points and vectors are expressed as row-major vectors and that they are therefore pre-multiplied (meaning the transformation matrix is on the right of the point or vector).
The following table summarizes the differences between the two conventions (where P, V and M respectively stands for Point, Vector and Matrix).
| | | | |
|--------------------|-------------------------------------------------------------|------------------------------|---------|
| Row-major order | $$\scriptsize P/V=\begin{bmatrix}x & y & z\end{bmatrix}$$ | Left or pre-multiplication | P/V * M |
| Column-major order | $$\scriptsize P/V=\begin{bmatrix}x \\ y \\ z\end{bmatrix}$$ | Right or post-multiplication | M * P/V |
Now that we have learned about these two conventions you might ask "isn't that just about writing things on paper?". We know how to compute the product of two matrices A and B: multiply each coefficient within A's current row by the associated elements within B's current column and sum up the result. Lets apply this formula using the two conventions and lets compare the results:
Row-major order
$$\scriptsize { \begin{bmatrix}x & y & z\end{bmatrix} * \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} }$$ $$\scriptsize { \begin{array}{l}x' = x * a + y * d * y + z * g\\y' = x * b + y * e * y + z * h\\z' = x * c + y * f * y + z * i\end{array} }$$
Column-major order
$$\scriptsize { \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} * \begin{bmatrix}x\\y\\z\end{bmatrix} }$$ $$\scriptsize { \begin{array}{l}x' = a * x + b * y + c * z\\y' = d * x + e * y + f * z\\z' = g * x + h * y + i * z\end{array} }$$
Multiplying a point or a vector by a matrix should give us the same result whether we use row- or colum-major order. If you use a 3D application to rotate a point by a certain angle around the z-axis, you expect the point the be in certain position after the rotation no matter what internal convention the developer used to represent points and vectors. However as you can see from looking at the table above, multiplying a row-column and column-major point (or vector) by the same matrix clearly wouldn't give us the same result? To get back on our feet, we would actually need to transpose the [3x3] matrix used in the column-major multiplication to be sure that x', y' and z' are the same (if you need to remember what the transpose of a matrix is, check the chapter on Matrix Operations). Here is what we get:
Row-major order
$$\scriptsize { \begin{bmatrix}x & y & z\end{bmatrix} * \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} }$$ $$\scriptsize { \begin{array}{l}x' = x * a + y * d * y + z * g\\y' = x * b + y * e * y + z * h\\z' = x * c + y * f * y + z * i\end{array} }$$
Column-major order
$$\scriptsize { \begin{bmatrix} a & d & g \\ b & e & h \\ c & f & i \end{bmatrix} * \begin{bmatrix}x\\y\\z\end{bmatrix} }$$ $$\scriptsize { \begin{array}{l}x' = a * x + d * y + g * z\\y' = b * x + e * y + h * z\\z' = c * x + f * y + i * z\end{array} }$$
In conclusion, going from row-major order to column-major order not only involve to swap the point or vector and the matrix in the multiplication but also to transpose the [3x3] matrix, to guarantee that both conventions give the same result (and vice versa).
From these observations, we can see that any series of transformations applied to a point or a vector when a row-major convention is used can be written in sequential order (or reading order). Imagine for instance that you want to translate point P with matrix T then rotate it around the z-axis with Rz then around the y-axis with Ry. You can write:
$$\scriptsize P'=P * T * R_z * R_y$$
If you were to use a column-major notation you would need to call the transform in reverse order (which one might find counter-intuitive):
$$\scriptsize P'=R_y * R_z * T * P$$
So you may think, "there must be some a reason to prefer one system to another". In fact, both conventions are correct and give us the same result, but for some technical reasons, Maths and Physics texts generally treat vectors as column vectors.
Order of transformation when we use colum-major matrices is more similar in mathematics to the way we write function evaluation and composition.
The row-major matrix convention however makes matrices easier to teach which is the reason we use it for Scratchapixel (as well as Maya, DirectX. They are also defined as the standard in the RenderMan specifications). However some 3D APIs such as OpenGL, use a column-major convention.
## Implication in Coding: Does it Impact Performance?
There is another potentially very important aspect to take into consideration if you need to choose between row-major and column-major, but this has nothing to do really with the conventions themselves and how practical one is over the other. It has more to do with the computer and the way it works. Remember that we will be dealing with [4x4] matrices. Typically the implementation of a matrix in C++ looks like this:
class Matrix { ... float m[4][4]; };
As you can see the 16 coefficients of the [4x4] matrix are stored in a two-dimensional array of floats (or doubles depending on the precision you need. Our C++ Matrix class is a template). Which means that in memory the 16 coefficients will be laid out in the following manner: c00, c01, c02, c03, c10, c11, c12, c13, c20, c21, c22, c23, c30, c31, c32, c33. In other words, they are laid out contiguously in memory. Now lets see how these coefficients are accessed in a vector-matrix multiplication where vectors are written in row-major order:
// row-major order x' = x * c00 + y * c10 + z * c20 y' = x * c01 + y * c11 + z * c21 z' = x * c02 + y * c12 + z * c22
As you can see the elements of the matrix for x' are not access sequentially. In other words to compute x' we need the 1st, 5th and 9th float of the matrix 16 floats array. To compute y' we need to access the 2nd, 6th and 10th float of this array. And finally for z' we need the 3rd, 7th and 11th float from the array. In the world of computing, accessing elements from an array in a non-sequential order, is not necessarily a good thing. It actually potentially degrades the cache performance of the CPU. We won't go into too much details here, but lets just say that the closest memory that the CPU can access to is called a cache. This cache is very fast to access to but can only store a very limited number of data. When the CPU needs to access some data, it first check if it exists in the cache. If it does the CPU access this data right away (cache hit), but it doesn't (cache miss), it first needs to create an entry in the cache for it, then copy to this location the data from the main memory. This process is obviously more time consuming than when the data already exists in the cache so ideally we want to avoid cache misses as much as possible. Additionally to copying the particular data from main memory, the CPU also copies a chunk of the data that lives right next to it (for instance the next 24 bytes), because hardware engineers figured that if your code needed to access an element of an array for instance, it was likely to access the elements following it soon after. Indeed, in programs, we often loop over elements of an array in sequential order and this assumption is therefore likely to be true. Applied to our matrix problem, accessing the coefficients of the matrix in non sequential order can therefore be a problem. Assuming the CPU loads the requested float in the cache plus the 3 floats next to it, our current implementation might lead to many cache misses, since the coefficients used to compute x' y' and z' are 5 floats apart in the array. On the other hand, if you use a column-major order notation, computing x' for instance require to access the 1st, 2nd and 3rd elements of the matrix.
// column-major order x' = c00 * x + c01 * y + c02 * z y' = c10 * x + c11 * y + c12 * z z' = c20 * x + c21 * y + c22 * z
The coefficients are accessed in sequential order which also means that we making a good use of the CPU cache mechanism (only 3 cache misses instead of 9 in our example). In conclusion we can say that from a programming point of view, implementing our point- or vector-matrix multiplication using a colum-major order convention might be better, performance wise, than the version using the row-major order convention. Practically though, we haven't been able to demonstrate that this was actually the case (when you compile your program using the optimisation flags -O, -O2 or -O3, the compiler can do the work for you by optimising loops over multi-dimensionals arrays) and we have been successfully using the row-major order version without any lose of performance compared to a version of the same code using a column-major order implementation.
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template<typename T> class Vec3 { public: Vec3(T xx, T yy, T zz) : x(xx), y(yy), z(zz) {} T x, y, z, w; }; template<typename T> class Matrix44 { public: T m[4][4]; Vec3<T> multVecMatrix(const Vec3<T> &v) { #ifdef ROWMAJOR return Vec3<T>( v.x * m[0][0] + v.y * m[1][0] + v.z * m[2][0], v.x * m[0][1] + v.y * m[1][1] + v.z * m[2][1], v.x * m[0][2] + v.y * m[1][2] + v.z * m[2][2]); #else return Vec3<T>( v.x * m[0][0] + v.y * m[0][1] + v.z * m[0][2], v.x * m[1][0] + v.y * m[1][1] + v.z * m[1][2], v.x * m[2][0] + v.y * m[2][1] + v.z * m[2][2]); #endif } }; #include <cmath> #include <cstdlib> #include <cstdio> #include <ctime> #define MAX_ITER 10e8 int main(int argc, char **argv) { clock_t start = clock(); Vec3<float> v(1, 2, 3); Matrix44<float> M; float *tmp = &M.m[0][0]; for (int i = 0; i < 16; i++) *(tmp + i) = drand48(); for (int i = 0; i < MAX_ITER; ++i) { Vec3<float> vt = M.multVecMatrix(v); } fprintf(stderr, "Clock time %f\n", (clock() - start) / float(CLOCKS_PER_SEC)); return 0; }
## Row-major and Column-Major Order in Computing
For the sake of completeness, lets just mention as well, that terms row-major and column-major order can also be used in computing to describe the way elements of multidimensional arrays are laid out in memory. In row-major order, the elements of a multi-dimensional array are laid out one after the other, from the left to right, top to bottom. This is the method used by C/C++. For example the matrix:
$$\scriptsize M = \begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}$$
could be written in C/C++ as:
float m[2][3]={{1, 2, 3}, {4, 5, 6}};
and the elements of this array would be laid out contiguously in linear memory as:
1 2 3 4 5 6
In column-major order, which is used by languages such as FORTRAN and MATLAB, elements of the matrix are stored in memory from top to bottom, left to right. Using the same matrix example, the elements of the matrix would be stored (and accessed) in memory in the following way:
1 4 2 5 3 6
Knowing how the elements of a matrix are laid out in memory is important especially when you try to access them using pointer offset and for loop optimisation (we have explained previously in this chapter that it could affect the CPU cache performance). However since we will only be considering C/C++ as our programming language, column-major ordering (applied to computing) is of no great interest to us. We are only mentionning what the terms mean in computing, so that you are aware that they might describe two different things depending of the context in which they are used. You should be careful to not mix them up. In the context of mathematics, they describe whether you treat vectors (or points) as rows of coordinates or as columns and the second, and in the context of computing, they describe the way a certain programming language stores and accesses elements of multi-dimensional array (which matrices are) in memory.
OpenGL is an interesting case in that regard. When GL was initially created, the developers chose the row-major vector convention. Developers who extended OpenGL though thought they should go back to to column-major vector which they did. However for compatibility reasons, they didn't want to change the code for the point-matrix multiplication and decided instead to change the order in which the coefficients of the matrix were stored in memory. In other words OpenGL stores the coefficients in column-major order which means that the translation coefficients m03, m13 and m23 from a matrix using column-major vector have indices 13, 14, 15 in the float array as would the translation coefficients m30, m31 and m32 from a matrix using row-major vector. XX explain here that the translation coeff in column major matrix are at 03 13 23 XX Add picture? XX.
## Summary
The differences between the two conventions are summarised in the following table:
Row-major vector (Mathematics) Column-major vector (Mathematics)
$$\scriptsize P/V=\begin{bmatrix}x & y & z\end{bmatrix}$$ $$\scriptsize P/V=\begin{bmatrix}x \\ y \\ z\end{bmatrix}$$
Pre-multiplication $$\scriptsize vM$$ Post-multiplication $$\scriptsize Mv$$
Call order and the order the transforms are applied is the same: "take P, transform by T, transform by Rz, transform by Ry" is written as $$\scriptsize P'=P*T*R_z*R_y$$ Call order is the reverse of the order the transforms are applied: "take P, transform by T, transform by Rz, transform by Ry" is written as $$\scriptsize P'=R_y*R_z*T*P$$
API: Direct X, Maya API: OpenGL, PBRT, Blender
The rows of the matrix represent the bases (or axes) of a coordinate system (red: x-axis, green: y-axis, blue:z-axis) $$\scriptsize { \begin{bmatrix} \color{red}{c_{00}}& \color{red}{c_{01}}&\color{red}{c_{02}}&0\\ \color{green}{c_{10}}& \color{green}{c_{11}}&\color{green}{c_{12}}&0\\ \color{blue}{c_{20}}& \color{blue}{c_{21}}&\color{blue}{c_{22}}&0\\0&0&0&1 \end{bmatrix} }$$ The columns of the matrix represent the bases (or axes) of a coordinate system (red: x-axis, green: y-axis, blue:z-axis) $$\scriptsize { \begin{bmatrix} \color{red}{c_{00}}& \color{green}{c_{01}}&\color{blue}{c_{02}}&0\\ \color{red}{c_{10}}& \color{green}{c_{11}}&\color{blue}{c_{12}}&0\\ \color{red}{c_{20}}& \color{green}{c_{21}}&\color{blue}{c_{22}}&0\\0&0&0&1\end{bmatrix} }$$
The translation values are stored in the c30, c31 and c32 elements. $$\scriptsize { \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\Tx&Ty&Tz&1\end{bmatrix} }$$ The translation values are stored in the c03, c13 and c23 elements. $$\scriptsize { \begin{bmatrix}1&0&0&Tx\\0&1&0&Ty\\0&0&1&Tz\\0&0&0&1\end{bmatrix} }$$
Transpose the matrix to use it as a column-major ordered matrix Transpose the matrix to use it as a row-major ordered matrix
Row-major matrix (Computing) Column-major matrix (Computing)
API: Direct X, Maya, PBRT API: OpenGL
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http://mathhelpforum.com/calculus/82335-help-integrating-e-x-over-interval-0-1-a.html
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# Thread:
1. ## help with integrating e^-x over the interval [0,1]
integrate e^-xdx over the interval [0,1]
let u=?
du= ?
when x= 0, u=?
when x = 1, u = ?
thanks!
2. $-x=u$
$dx=-du$
$x=0\Rightarrow u=0$
$x=1\Rightarrow u=-1$
3. First, you may note that $\frac{d}{dx}e^x=e^x$, and therefore that $\frac{d}{dx}e^{kx}=ke^{kx}$ by the Chain Rule.
When $k=-1$ and $f'(x)=e^{-x}$, we want the derivative $f'(x)$ to end up with the positive coefficient $1$ instead of $-1$. To do this, we add a $-1$ as a coefficient to the antiderivative so that we will end up with $1$:
$\begin{aligned}<br /> \frac{d}{dx}(-1)e^{-x}&=\frac{d}{dx}(-1)e^{(-1)x}\\<br /> &=(-1)\frac{d}{dx}e^{(-1)x}\\<br /> &=(-1)(-1)e^{(-1)x}\\<br /> &=1\cdot e^{(-1)x}\\<br /> &=e^{(-1)x}\\<br /> &=e^{-x}.<br /> \end{aligned}$
The general antiderivative is therefore $f(x)=(-1)e^{-x}+C=-e^{-x}+C$.
Alternatively, we could let
$<br /> \begin{aligned}<br /> u&=-x\\<br /> du&=-dx<br /> \end{aligned}$
to obtain
$-\int_0^{-1} e^u\,du =\int_{-1}^0 e^u\,du.$
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http://planetmath.org/SimplicialCategory
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# simplicial category
The simplicial category $\Delta$ is defined as the small category whose objects are the totally ordered finite sets
$[n]=\{0<1<2<\ldots<n\},\quad n\geq 0,$ (1)
and whose morphisms are monotonic non-decreasing (order-preserving) maps. It is generated by two families of morphisms:
| | | |
|----------------------------------------------------------------------------|----------------------------------------------------------------------------|----------------------------------------------------------------------------|
| $\displaystyle\delta^{n}_{i}$ | $\displaystyle\colon$ | $\displaystyle[n-1]\to[n]\quad\mbox{is the injection missing\ }i\in[n],$ |
| $\displaystyle\sigma^{n}_{i}$ | $\displaystyle\colon$ | $\displaystyle[n+1]\to[n]\quad\mbox{is the surjection such that\ }\sigma^{n}_{i% }(i)=\sigma^{n}_{i}(i+1)=i\in[n].$ |
The $\delta^{n}_{i}$ morphisms are called face maps, and the $\sigma^{n}_{i}$ morphisms are called degeneracy maps. They satisfy the following relations,
| | | | |
|-----------------------------------------------------------------------------|-----------------------------------------------------------------------------|-----------------------------------------------------------------------------|-----|
| $\displaystyle\delta^{{n+1}}_{j}\,\delta^{n}_{i}$ | $\displaystyle=$ | $\displaystyle\delta^{{n+1}}_{i}\,\delta^{n}_{{j-1}}\quad\mbox{for\ }i<j,$ | (2) |
| $\displaystyle\sigma^{{n-1}}_{j}\,\sigma^{n}_{i}$ | $\displaystyle=$ | $\displaystyle\sigma^{{n-1}}_{i}\,\sigma^{n}_{{j+1}}\quad\mbox{for\ }i\leq j,$ | (3) |
| $\displaystyle\sigma^{n}_{j}\,\delta^{{n+1}}_{i}$ | $\displaystyle=$ | $\displaystyle\left\{\begin{array}[]{ll}\delta^{n}_{i}\,\sigma^{{n-1}}_{{j-1}}&% \mbox{if\ }i<j,\\ \mathrm{id}_{n}&\mbox{if\ }i=j\mbox{\ or\ }i=j+1,\\ \delta^{n}_{{i-1}}\,\sigma^{{n-1}}_{j}&\mbox{if\ }i>j+1.\end{array}\right.$ | (4) |
All morphisms $[n]\to[0]$ factor through $\sigma^{0}_{0}$, so [0] is terminal.
There is a bifunctor $+\colon\Delta\times\Delta\to\Delta$ defined by
| | | | |
|-----------------------------------------------------------------------------|-----------------------------------------------------------------------------|-----------------------------------------------------------------------------|-----|
| $\displaystyle[m]+[n]$ | $\displaystyle=$ | $\displaystyle[m+n+1],$ | (5) |
| $\displaystyle(f+g)(i)$ | $\displaystyle=$ | $\displaystyle\left\{\begin{array}[]{ll}f(i)&\mbox{if\ }0\leq i\leq m,\\ g(i-m-1)+m^{{\prime}}+1&\mbox{if\ }m<i\leq(m+n+1),\end{array}\right.$ | (6) |
where $f\colon[m]\to[m^{{\prime}}]$ and $g\colon[n]\to[n^{{\prime}}]$. Sometimes, the simplicial category is defined to include the empty set $[-1]=\emptyset$, which provides an initial object for the category. This makes $\Delta$ a strict monoidal category as $\emptyset$ is a unit for the bifunctor: $\emptyset+[n]=[n]=[n]+\emptyset$ and $\mathrm{id}_{\emptyset}+f=f=f+\mathrm{id}_{\emptyset}$. Further, $\Delta$ is then the free monoidal category on a monoid object (the monoid object being [0], with product $\sigma^{0}_{0}\colon[0]+[0]\to[0]$).
There is a fully faithful functor from $\Delta$ to $\mathord{\mathbf{Top}}$, which sends each object $[n]$ to an oriented $n$-simplex. The face maps then embed an $(n-1)$-simplex in an $n$-simplex, and the degeneracy maps collapse an $(n+1)$-simplex to an $n$-simplex. The bifunctor forms a simplex from the disjoint union of two simplicies by joining their vertices together in a way compatible with their orientations.
There is also a fully faithful functor from $\Delta$ to $\mathord{\mathbf{Cat}}$, which sends each object $[n]$ to a pre-order $\mathord{\mathbf{n+1}}$. The pre-order $\mathord{\mathbf{n}}$ is the category consisting of $n$ partially-ordered objects, with one morphism $a\to b$ if and only if $a\leq b$.
Type of Math Object:
Definition
Major Section:
Reference
## Mathematics Subject Classification
18G30 Simplicial sets, simplicial objects (in a category)
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## Info
Owner: mhale
Added: 2002-08-27 - 23:50
Author(s): mhale
## Versions
(v8) by mhale 2013-03-22
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http://mathoverflow.net/questions/28595/does-there-exist-a-riemann-surface-corresponding-to-every-field-extension-any-ot/28600
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## Does there exist a Riemann surface corresponding to every field extension? Any other hypothesis needed?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
My question is a doubt I had in the last point to the first answer to this MO question - http://mathoverflow.net/questions/14314/algebraic-topologies-like-the-zariski-topology
Can one associate a Riemann surface to any arbitrary field extension? The statement there says its true, it is the (equivalence class of) valuations which fix the base field. The result seems too fascinating to be true. Are there any extra hypotheses?
I am aware of the result that the valuations of C(x)/C form the Riemann sphere (counting the exponent of any (x-a) in each rational function gives a correspondence between points on the complex plane (minus infinity) and valuations.)
Any references?
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1
There are good answers below, but briefly: the "Zariski Riemann surface" is usually not a Riemann surface (sorry!): it is a top. space. When $K/k$ is f.g. of trdeg $1$, the Z.R.S. is essentially the Zariski topology on the unique complete, normal curve over $k$ with function field $K$. (When $k = \mathbb{C}$, there really is a compact Riemann surface here. Otherwise not.) If the trdeg is greater than one, the Z.R.S. is more complicated than any one variety with the given function field: it is keeping track of all models at once. If tr.deg. is zero, the Z.R.S. is a single point. – Pete L. Clark Jun 18 2010 at 11:22
## 3 Answers
Zariski introduced an abstract notion of Riemann surface associated to, for example, a finitely generated field extension $K/k$. It's a topological space whose points are equivalence classes of valuations of $K$ that are trivial on $k$, or equivalently valuation rings satisfying $k\subset R_v\subset K$. If $A$ is a finitely generated $k$-algebra inside $K$ then those $R_v$ which contain $A$ form an open set.
In the case of a (finitely generated and) transcendence degree 1 extension all of these valuation rings are the familiar DVRs -- local Dedekind domains -- and they serve to identify the points in the unique complete nonsingular curve with this function field. (There is also the trivial valuation with $R_v=K$, which corresponds to the generic point of that curve.)
In higher dimensions there are lots of complete varieties to contend with -- you can keep blowing up. Also there are more possibilities for valuations. Most of the valuation rings are not Noetherian. A curve in a surface gives you a discrete valuation ring, consisting of those rational functions which can meaningfully be restricted to rational functions on the curve: those which do not have a pole there. A point on a curve on a surface gives you a valuation whose ring consists of those functions which do not have a pole all along the curve, and which when restricted to the curve do not have a pole at the given point. The value group is $\mathbb Z\times \mathbb Z$ lexicographically ordered. A point on a transcendental curve in a complex surface, or more generally a formal (power series) curve in a surface gives you a valuation by looking at the order of vanishing; the value group is a subgroup of $\mathbb R$.
This space of valuations has something of the flavor of Zariski's space of prime ideals in a ring: it is compact but not Hausdorff, for example. It can be thought of as the inverse limit, over all complete surfaces $S$ with this function field, of the space (Zariski topology) of points $S$.
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On the point of the general relationship to Spec of a ring. I once had a conversation with an expert on Zariski's construction with valuations rings, where it seemed that a locale could immediately be written down as a presentation. There is also a well-known presentation of Spec as a locale, by divisibility basically. But the former point I hadn't seen mentioned anywhere (whether or not it is useful). – Charles Matthews Jun 18 2010 at 9:28
### 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 need the field extension to have transcendence degree 1 over $\mathbb{C}$ to get a Riemann surface. More generally, you can get an algebraic curve at least for every transcendence degree 1 extension over an algebraically closed field. Chapter I of Hartshorne does this in detail, I don't know an analytic approach.
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Chapter I, Section 1.6. – H. Hasson Jun 18 2010 at 4:16
1
The extension needs to be finitely generated, so you can avoid things like $\mathbb{C}(t,t^{1/2},t^{1/3},\ldots)$. – S. Carnahan♦ Jun 18 2010 at 4:35
2
If $K/k$ is any extension of fields one can define the Riemann-Zariski space ${\rm{RZ}}(K/k)$ to consist of all valuations on $K$ trivial on $k$. It has a natural quasi-compact "Zariski topology". These were used by Zariski mainly for $K/k$ finitely generated, but make sense in general. Variants on this have been extremely useful in recent work in non-archimedean geometry as well as resolution of singularities (cf. work of M. Temkin). – BCnrd Jun 18 2010 at 4:42
@Scott, Ahh, I always forget finitely generated hypotheses, they're just completely internalized to me. I should be more careful. – Charles Siegel Jun 18 2010 at 6:09
For compact Riemann surface, the algebraic approach and the analytic approach is the same(Chow's lemma), so the algebraic answer is sufficient for you. i.e. For every extension finitely generated over \mathbb{C} which has transcendental degree 1(up to isomorphism), there is a unique Riemann surface you want.
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http://mathhelpforum.com/number-theory/150373-every-prime-3-congruent-pm-1-mod-6-a.html
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Thread:
1. Every prime > 3 is congruent to \pm 1 mod 6
Every prime > 3 is congruent to $\pm 1 \ \mbox{mod 6}$
Induction maybe?
$5\equiv\pm 1\ \mbox{(mod 6)}$ This isn't true though
2. any number can be
0,1,2,3,4,5 mod 6
because the number is prime - 0,2,3,4 can be ruled out
3. $5\equiv -1\ \mbox{(mod 6)},$ since $5-(-1)=6\times 1$.
You could try arguing along these lines: all primes greater than 3 are odd. Therefore, they must be congruent to either -1, 1, or 3 mod 6. (5 and -1 are equivalent). However, any number greater than 3 that is congruent to 3 mod 6 would be divisible by 3, and hence not prime. Therefore, all primes greater than 3 are congruent to 1 or -1 mod 6.
[EDIT]: This is essentially the same as what aman_cc said.
4. Originally Posted by dwsmith
Every prime > 3 is congruent to $\pm 1 \ \mbox{mod 6}$
Induction maybe?
$5\equiv\pm 1\ \mbox{(mod 6)}$ This isn't true though
You can generalize and consider integers $a$ that are relatively prime to $6$ and the result will follow, in particular, to any prime $p>3$.
Let $a=6q+r$, with $0\leq r<6$. Then $gcd(a,6)=gcd(6,r)$, so we need only to look at values $r$ such that $gcd(6,r)=1$. These are $r=1,5$.
Now, $a\equiv 1(mod\ 6)$,or $a\equiv 5\equiv -1(mod\ 6)$.
5. Keep it simple ... I assume that $p > 3$ in the whole post by the way ... since the fact $3 | 3$ does not contradict the primality of $p$ ... just adjust what is below as you wish ...
Let $p \equiv x \pmod{n} \Rightarrow p = kn + x$, $k \in \mathbb{Z}$, $x \in \mathbb{Z}/n\mathbb{Z}$
Then it follows that $\gcd{(n, x)} | p$
But if $p$ is prime we must have $\gcd{(n, x)} = 1$ (otherwise there is a contradiction with respect to the definition of a prime number)
Applying this to the current problem with $n = 6$, the only $x$ that are coprime to $6$ (in the congruence ring $\mathbb{Z}/6\mathbb{Z}$ of course) are, surprisingly enough, $1$ and $5$ (which turns out to be congruent to $-1 \pmod{6}$).
It follows that if $p$ is indeed prime then it must be congruent to $\pm 1 \pmod{6}$
...
Interestingly, it is also possible to prove that this implies that $\gcd{(k, n)} = 1$ for any prime $p$ ... This gives me an idea, I'll be right back !
6. To Bacterius
gcd(n,x) = 1 or p
i.e. p = p mod p^2, so gcd(p^2,p) = p, and p|p
But since n is 6, then we can say that gcd(n,x) should be 1
7. I don't understand what you mean, can you develop ?
8. I was just pointing out that gcd(n,x) isn't necessarily 1, it could be p also, at least in general ... for this particular case, since n is 6, then gcd(n,x) would be 1 ...
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http://mathhelpforum.com/algebra/212279-rational-function-sum-partial-fractions.html
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Thread:
1. rational function as a sum of partial fractions
Express the rational function as a sum of partial fractions: f(x) = (3x^2+5x+4)\(x^3+x^2+x+1)
2. Re: rational function as a sum of partial fractions
Originally Posted by brianbrian
Express the rational function as a sum of partial fractions: f(x) = (3x^2+5x+4)\(x^3+x^2+x+1)
Have a look at this.
3. Re: rational function as a sum of partial fractions
huh, thanks but i guess that teacher want me to do this with this A,B,C thing, so how can I compromise that result with this A,B,C (i hope you know what i am talking about, its hard to explain for me)
as ABC thing I mean situation when partial consists of letter in nominator and some (numbers with x) in denominator
4. Re: rational function as a sum of partial fractions
Hello, brianbrian!
$\text{Express }\,\frac{3x^2+5x+4}{x^3+x^2+x+1}\,\text{ as a sum of partial fractions.}$
We have: . $\frac{3x^2+5x+4}{(x+1)(x^2+1)} \;=\;\frac{A}{x+1} + \frac{Bx}{x^2+1} + \frac{C}{x^2+1}$
. . . . . . . . . . $3x^2+5x+4 \;=\;A(x^2+1) + Bx(x+1) + C(x+1)$
Let $x = \text{-}1: \;\;2 \;=\;2(A) + B(0) + C(0) \quad\Rightarrow\quad A \,=\,1$
Let $x = 0:\;\;4 \:=\:A + B(0) + C \quad\Rightarrow\quad 4 \:=\:1 + C \quad\Rightarrow\quad C \,=\,3$
Let $x = 1\!:\;\;12 \:=\:A(2) + B(2) + C(2) \quad\Rightarrow\quad 12 \:=\:2 + 2B + 6 \quad\Rightarrow\quad B \,=\,2$
Therefore: . $\frac{3x^2 + 5x + 4}{x^3+x^2+x+1} \;=\;\frac{1}{x+1} + \frac{2x}{x^2+1} + \frac{3}{x^2+1}$
5. Re: rational function as a sum of partial fractions
Great! Thanks so much. Unfortunatly small question appeared. Why there is a x next to B? I think that i understand the whole idea now, but i can't get why this x stands next to B (and why B, not A or C).
6. Re: rational function as a sum of partial fractions
Any polynomial can be factored (at least theoretically- it might be extremely difficult to find the coefficients) into linear or quadratic factors (with real coefficients). Since we want "common fractions", we need the degree of the numerator to be less than that of the denominator. So the numerator of a fraction having a linear denominator is a number and the numerator of a fraction having a quadratic denominator is linear. Soroban separated the fraction having linear numerator and quadratic denominator into two fractions: $\frac{Bx+ C}{x^2+ 1}= \frac{Bx}{x^2+ 1}+ \frac{C}{x^2+1}$.
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http://math.stackexchange.com/questions/113286/limit-of-int-infty-infty-fx-sintxdx-as-t-to-infty
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# Limit of $\int_{-\infty}^{\infty} f(x)\sin(tx)dx$ as $|t|\to\infty$
Let $f$, a Lebesgue integrable function in $\mathbb{R}$ ($\int_{\mathbb{R}}|f| < \infty$). Let: $$g(t) := \int_{-\infty}^{\infty} f(x)\sin(tx)dx$$ Show that $g$ is continuous (which I did), and that: $$\lim_{|t| \rightarrow \infty} g(t) = 0$$
Why is the second part correct?
Thanks!
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1
– David Mitra Feb 25 '12 at 17:55
1
This is so called Riemann-Lebesgue Lemma! – беркай Mar 20 '12 at 20:58
## 1 Answer
The result is trivial if $f$ is the characteristic function of an interval, and therefore also if it is a step function. Can you take it the rest of the way from there?
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http://quant.stackexchange.com/questions/1157/why-is-the-ratio-of-hi-low-range-to-open-close-range-close-to-2
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# Why is the ratio of Hi-Low range to Open-Close range close to 2?
I tried it in several symbols and timeframes with the same result:
$$\frac {mean(HIGH-LOW)}{mean(|CLOSE-OPEN|)}$$
````Symbol Result
------ ------
EURUSD:W1 1.9725
EURUSD:D1 2.0023
EURUSD:H1 2.1766
USDJPY:W1 1.9949
USDJPY:D1 2.0622
USDJPY:H1 2.2327
SAN.MC:D1 2.0075
BBVA.MC:D1 2.0075
REP.MC:D1 2.1320
````
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Are W1, D1, H1 time frames? You might want to make that clearer. – Louis Marascio Jun 16 '12 at 13:21
## 2 Answers
There is a very good reason why the ratio $$\frac {mean(HIGH-LOW)}{mean(|CLOSE-OPEN|)} \approx 2$$ on various financial series. If the price of a security evolves according to a Wiener process beginning at the opening bell and throughout the day, and the drift is negligible for that period of time, i.e.$\mu=0$, then the denominator of the above ratio closely approximates the average absolute deviation, $$AAD=\frac{2\sigma}{\sqrt{2\pi}}\int_0^\infty xe^{-x^2/2}dx=\sqrt {2/\pi}\cdot\sigma$$ for a normal distribution, where $\sigma$ is the standard deviation. On the other hand $$\mathbb E(HIGH-OPEN) = \sqrt{2/\pi}\cdot\sigma$$ $$\mathbb E(LOW-OPEN) = -\sqrt{2/\pi}\cdot\sigma$$(See the running maximum of a Wiener process on Wikipedia.) So we have for such an idealized Wiener process: $$\frac {\mathbb E(HIGH-LOW)}{\mathbb E(|CLOSE-OPEN|)} = \frac{\sqrt{2/\pi}\cdot\sigma-\left(-\sqrt{2/\pi}\cdot\sigma\right)}{\sqrt {2/\pi}\cdot\sigma} = 2.$$ It should not be too surprising to see this more or less borne out by observation.
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Unfortunately, I cannot answer fully your question. Though I'll give you my partial answer.
First of all, using entire price history of an index (from Yahoo), this is what I got:
```` Daily Weekly Monthly
DJIA 2.91 2.37 2.33
NASDAQ 100 1.74 1.94 1.91
NYSE Composite 1.61 1.59 1.85
S&P 100 1.75 1.90 1.97
S&P 600 Small Cap 1.59 1.82 1.99
````
Based on this, I don't think we can claim that the ratio is always 2.
So you agree, that the ratio is not always 2. But still, you want to know why it is equal to 2.91 or 1.59 or whatever.
This is how I would proceed in answering the question. First, the ratio can be expressed as
$$\frac{E[max(P_1, P_2,...,P_n)-min(P_1, P_2,...,P_n)]} {E[|P_1 - P_n|]}$$
Second, I would start expanding the fraction in order to get a better picture of what exactly influences the ratio. I hope to expand the fraction, and then have some terms cancel each other out and then obtain as an answer 2, $\sigma / \mu$, or something else concise and beautiful. The problem is, it is extremely hard (at least for me) to obtain analytical expression for the expected value of a maximum (or minimum) of a sequence of correlated non-normal variables--the prices. I do not think anyone can give you the analytical expression for that. So this is where it ends as for analytical answer.
You can also use numerical methods, something like Monte Carlo simulation. Assume some model for prices, simulate them, and do some sensitivity analysis in parameters of the model to see how they affect the ratio.
Finally, one thing I do not get is why you are interested in that ratio. Shoundn't you instead be interested in $$E(\frac{max(P_1, P_2,...,P_n)-min(P_1, P_2,...,P_n)]} {|P_1 - P_n|})$$
For example if the true expected value of the above ratio is equal to 10, and during a trading day the price is such that the ratio is 20, you would start buying the asset because you expect the close price to be higher than the current price. You would then sell the asset for a profit. Using your definition of the ratio, I see no usefulness in it. Care to explain?
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I didn't try indexes, but DJIA components are closer than the index itself (2.91) : ```MMM 2.0118
AA 1.8223
AXP 1.9819
T 1.9381
BAC 1.8917
BA 1.9329
CAT 1.9299
CVX 1.9701
CSCO 1.9582
KO 1.8967
DD 1.8921
XOM 1.9578
GE 1.9642
HPQ 1.8642
HD 1.9523
INTC 1.9294
IBM 1.8367
JNJ 1.9227
JPM 1.9468
KFT 2.0584
MCD 1.9899
MRK 1.9604
MSFT 1.9045
PFE 1.9475
PG 1.9921
TRV 1.8728
UTX 1.9770
VZ 1.9525
WMT 2.0081
DIS 1.8454``` – jla May 16 '11 at 11:57
Did you use the entire price history? You might have used a period in which the ratio was close to 2. Why are you interested in that ratio? – Dmitrii I. May 16 '11 at 20:19
I took the entire daily price history available from Yahoo. I'm not interested in that ratio in itself, only as property value that many symbols have in common. – jla May 17 '11 at 11:13
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http://mathoverflow.net/questions/12020/proving-two-partizan-games-are-equivalent/12562
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## proving two partizan games are equivalent
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Is there any equivalent version of the Sprague-Grundy theorem (that states that every impartial game under the normal play convention is equivalent to nim) for partizan games?
More specifically, are there any "non-trivial" examples of partizan games that are known to be equivalent?
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## 4 Answers
In partizan theory, the theorem is: G=H if and only if G-H is a second player win.
A Nim position with two heaps of size 1 is equal to the Hackenbush position with one red and one blue edge each connected to the ground. Both are second player win positions and so are both equal to 0. They are not trivially equivalent since either player moving in Nim leaves a Next-player win game whereas Left playing in Hackenbush leaves a Right-win game and similarly Right plays to a Left-win game. In symbols, Nim: {$$|$$}=0; Hackenbush: {-1|1}=0.
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I think the closest thing to the Sprague-Grundy theorem in the partizan normal-play case is the corollary of Theorem 69 in On Numbers and Games. Every short game* has a unique canonical/normal form where all "dominated options" are deleted and all "reversible options" have been replaced and all options are written in their canonical forms. For the games which happen to be impartial, their canonical forms are the standard forms for Nim heaps: `$\left\{\left\{|\right\}|\left\{|\right\}\right\}$` for the Nim heap of size 1 (often denoted $*$), etc.
If you don't have ONAG at hand, you can read about dominated and reversible options in An Introduction to Conway's Games and Numbers. Their theorem 2.25 is essentially Theorem 69 from ONAG, but has a subtle error: it claims that there is a unique example in each equivalence class with no dominated or reversible options (as opposed to a good unique canonical choice).
Here is a counter-example: `$\left\{ \left\{|\right\} | \right\}$` is `$\left\{0|\right\}=1$` and has no dominated or reversible options. But `$\left\{ \left\{*|*\right\} | \right\}=\left\{ \left\{\left\{\left\{|\right\}|\left\{|\right\}\right\}|\left\{\left\{|\right\}|\left\{|\right\}\right\}\right\} | \right\}$` is equivalent to `$\left\{0|\right\}=1$`, and it also has no dominated/reversible options.
*A game is short if it has only finitely many positions. The Sprague-Grundy theorem usually applies to the short impartial games, stating that they're each equivalent to a single finite Nim heap.
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Partizan generalizations of Sprague-Grundy are the subject of Conway's book On Numbers and Games. His later work, Winning Ways for your Mathematical Plays with Berkelamp and Guy, gives examples of how to apply the theory to games that human beings might actually want to play. I'm afraid that it's been too long since I've read it to remember any nontrivial examples.
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Partizan misere games have also been studied. See Misère canonical forms of partizan games. A misere game is where the last person to move loses rather than wins.
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http://math.stackexchange.com/questions/179644/generalization-of-metric-spaces
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# Generalization of metric spaces
The Wikipedia article for metrics mentions several generalizations of metric spaces, but all of them seem to have the property that the metric must be non-negative for all x and y. To me it seems like a space where distances don't have to be non-negative would be an obvious generalization (removing the symmetry property would also be necessary). For example, consider the real line with the "distance function"
$$y-x$$
under which "distances" to numbers on one side of y would be negative while "distances" to numbers on the other side would be positive.
Does this generalization have a name, or is it useless enough not to have been studied?
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What properties would you want the distance functions to have, to distinguish them among arbitrary functions $X\times X\to\Bbb R$? – anon Aug 6 '12 at 20:42
$d(x,y) = 0$ iff $x = y$ would still work. I am unsure if an analog to the triangle inequality would work. – Aqwis Aug 6 '12 at 20:44
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We call that distance function subtraction! That is, we don't call it a distance function at all. Are subtractions the only kind of examples you care about? – Qiaochu Yuan Aug 6 '12 at 20:53
Not sure if I asked the question properly. Basically I'm wondering whether spaces where non-negativity and the "identity of indiscernibles" don't hold (of which subtraction of real numbers is an example) have been studied or are worth studying at all. – Aqwis Aug 6 '12 at 20:59
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Look up Minkowski space and Lorentz manifolds in general. – Will Jagy Aug 6 '12 at 21:02
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## 1 Answer
Will Jagy's comment deserves to be an answer:
Look up Minkowski space and Lorentz manifolds in general.
Minkowski space in $n+1$ dimensions is $\mathbb R^{n+1}$ with the "distance function" $$d(\langle x_1,\ldots,x_n, t\rangle, \langle y_1,\ldots,y_n,u\rangle) = \sqrt{(t-u)^2-(x_1-y_1)^2-(x_2-y_2)^2\cdots-(x_n-y_n)^2}$$
Here $d(x,y)=d(y,x)$ and distances cannot be negative, but they can be null or purely imaginary! (For mathematical sanity, one usually considers the square of this distance function, such as not to be troubled with the multi-valuedness of the square root, though).
Minkowski space is the basic fabric of relativity. Indeed the fundamental postulate of the Special Theory of Relativity could be phrased as:
The stage on which physics plays out can be given the structure of $3+1$-dimensional Minkowski space, such that all fundamental laws of nature are preserved by every Minkowski isometry. (Or at least by every Minkowski isometry that can be "smoothly" turned into the identity).
(And, by the way, light rays connect points whose mutual Minkowski distance is 0).
Beware, however, that the Minkowski distance is not used to define the topology of Minkowski space. One uses the ordinary Euclidean topology on $\mathbb R^{n+1}$.
Minkowski isometries are better known as Lorentz transformations, though often that name is only used for isometries that fix the point $\langle 0,\ldots,0,0\rangle$.
Lorentzian manifolds generalize Minkowski space to curved spacetimes for General Relativity.
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http://physics.stackexchange.com/questions/53794/identity-as-a-trivial-reducible-representation
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Identity as a trivial reducible representation
In particle physics, I was taught that a representation of a group is a function $r: group \rightarrow matrices\,(n\times n)$ such that $r(g_1)r(g_2)=r(g_1g_2)$ and $r(e)=I_{n\times n}$. Then, that a representation is reducible when you can find a matrix $A$ such that $Ar(g)A^{-1}$ is in diagonal-block form for every element of the group.
Then the professor tried to find in complicated ways reducible representations of $SO(N)$, $SU(N)$ and so on. But the trivial function that assigns $I_{n\times n}$ to every value of $g$ is not already a reducible transformation? I know it must be somehow useless, but what did I lose?
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If $n>1$ then yes, the identity representation is trivially reducible and irreducibly trivial. :) You are saying that if you have a bunch of objects which don't transform at all under the group then they don't transform into each other. This is a true statement but we tend to leave such trivial cases out of the discussion. – Michael Brown Feb 13 at 1:31
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Yes that would be a reducible representation decomposed into direct sums of the trivial (singlet) representation $1\oplus\dots\oplus 1$. What do you mean by what you lost? You can construct reducible representations by taking direct sums of irreducible ones in all sort of ways, so this is just the most trivial example of them all (using only the singlet representation). We are typically much more interested in irreducible representations. – Heidar Feb 13 at 1:31
2 Answers
What you have constructed is a representation, but not a faithful one. Since your homomorphism $r$ is not injective, you lose some of the structure of the group. In fact, since $r$ is trivial, you lose all the structure of the group. While most useful statements about $G$ apply to $r(G)$ equally well, you cannot pull back anything useful from $r(G)$ to $G$, so your representation doesn't tell you anything about $G$, defeating the whole purpose of using representations in the first place.
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For example, addition of angular momentum in quantum mechanics involves writing reducible representations of $\mathrm{SU}(2)$ as directs sums of irreducible ones.
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http://mathoverflow.net/questions/48771/proofs-that-require-fundamentally-new-ways-of-thinking/73973
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## Proofs that require fundamentally new ways of thinking
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I do not know exactly how to characterize the class of proofs that interests me, so let me give some examples and say why I would be interested in more. Perhaps what the examples have in common is that a powerful and unexpected technique is introduced that comes to seem very natural once you are used to it.
Example 1. Euler's proof that there are infinitely many primes.
If you haven't seen anything like it before, the idea that you could use analysis to prove that there are infinitely many primes is completely unexpected. Once you've seen how it works, that's a different matter, and you are ready to contemplate trying to do all sorts of other things by developing the method.
Example 2. The use of complex analysis to establish the prime number theorem.
Even when you've seen Euler's argument, it still takes a leap to look at the complex numbers. (I'm not saying it can't be made to seem natural: with the help of Fourier analysis it can. Nevertheless, it is a good example of the introduction of a whole new way of thinking about certain questions.)
Example 3. Variational methods.
You can pick your favourite problem here: one good one is determining the shape of a heavy chain in equilibrium.
Example 4. Erdős's lower bound for Ramsey numbers.
One of the very first results (Shannon's bound for the size of a separated subset of the discrete cube being another very early one) in probabilistic combinatorics.
Example 5. Roth's proof that a dense set of integers contains an arithmetic progression of length 3.
Historically this was by no means the first use of Fourier analysis in number theory. But it was the first application of Fourier analysis to number theory that I personally properly understood, and that completely changed my outlook on mathematics. So I count it as an example (because there exists a plausible fictional history of mathematics where it was the first use of Fourier analysis in number theory).
Example 6. Use of homotopy/homology to prove fixed-point theorems.
Once again, if you mount a direct attack on, say, the Brouwer fixed point theorem, you probably won't invent homology or homotopy (though you might do if you then spent a long time reflecting on your proof).
The reason these proofs interest me is that they are the kinds of arguments where it is tempting to say that human intelligence was necessary for them to have been discovered. It would probably be possible in principle, if technically difficult, to teach a computer how to apply standard techniques, the familiar argument goes, but it takes a human to invent those techniques in the first place.
Now I don't buy that argument. I think that it is possible in principle, though technically difficult, for a computer to come up with radically new techniques. Indeed, I think I can give reasonably good Just So Stories for some of the examples above. So I'm looking for more examples. The best examples would be ones where a technique just seems to spring from nowhere -- ones where you're tempted to say, "A computer could never have come up with that."
Edit: I agree with the first two comments below, and was slightly worried about that when I posted the question. Let me have a go at it though. The difficulty with, say, proving Fermat's last theorem was of course partly that a new insight was needed. But that wasn't the only difficulty at all. Indeed, in that case a succession of new insights was needed, and not just that but a knowledge of all the different already existing ingredients that had to be put together. So I suppose what I'm after is problems where essentially the only difficulty is the need for the clever and unexpected idea. I.e., I'm looking for problems that are very good challenge problems for working out how a computer might do mathematics. In particular, I want the main difficulty to be fundamental (coming up with a new idea) and not technical (having to know a lot, having to do difficult but not radically new calculations, etc.). Also, it's not quite fair to say that the solution of an arbitrary hard problem fits the bill. For example, my impression (which could be wrong, but that doesn't affect the general point I'm making) is that the recent breakthrough by Nets Katz and Larry Guth in which they solved the Erdős distinct distances problem was a very clever realization that techniques that were already out there could be combined to solve the problem. One could imagine a computer finding the proof by being patient enough to look at lots of different combinations of techniques until it found one that worked. Now their realization itself was amazing and probably opens up new possibilities, but there is a sense in which their breakthrough was not a good example of what I am asking for.
While I'm at it, here's another attempt to make the question more precise. Many many new proofs are variants of old proofs. These variants are often hard to come by, but at least one starts out with the feeling that there is something out there that's worth searching for. So that doesn't really constitute an entirely new way of thinking. (An example close to my heart: the Polymath proof of the density Hales-Jewett theorem was a bit like that. It was a new and surprising argument, but one could see exactly how it was found since it was modelled on a proof of a related theorem. So that is a counterexample to Kevin's assertion that any solution of a hard problem fits the bill.) I am looking for proofs that seem to come out of nowhere and seem not to be modelled on anything.
Further edit. I'm not so keen on random massive breakthroughs. So perhaps I should narrow it down further -- to proofs that are easy to understand and remember once seen, but seemingly hard to come up with in the first place.
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Perhaps you could make the requirements a bit more precise. The most obvious examples that come to mind from number theory are proofs that are ingenious but also very involved, arising from a rather elaborate tradition, like Wiles' proof of Fermat's last theorem, Faltings' proof of the Mordell conjecture, or Ngo's proof of the fundamental lemma. But somehow, I'm guessing that such complicated replies are not what you have in mind. – Minhyong Kim Dec 9 2010 at 15:18
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Of course, there was apparently a surprising and simple insight involved in the proof of FLT, namely Frey's idea that a solution triple would give rise to a rather exotic elliptic curve. It seems to have been this insight that brought a previously eccentric seeming problem at least potentially within the reach of the powerful and elaborate tradition referred to. So perhaps that was a new way of thinking at least about what ideas were involved in FLT. – roy smith Dec 9 2010 at 16:21
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Never mind the application of Fourier analysis to number theory -- how about the invention of Fourier analysis itself, to study the heat equation! More recently, if you count the application of complex analysis to prove the prime number theorem, then you might also count the application of model theory to prove results in arithmetic geometry (e.g. Hrushovski's proof of Mordell-Lang for function fields). – D. Savitt Dec 9 2010 at 16:42
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In response to edit: On the other hand, I think those big theorems are still reasonable instances of proofs that are difficult to imagine for a computer! Incidentally, regarding your example 2, it seems to me Dirichlet's theorem on primes in arithmetic progressions might be a better example in the same vein. – Minhyong Kim Dec 9 2010 at 17:34
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I agree that they are difficult, but in a sense what I am looking for is problems that isolate as well as possible whatever it is that humans are supposedly better at than computers. Those big problems are too large and multifaceted to serve that purpose. You could say that I am looking for "first non-trivial examples" rather than just massively hard examples. – gowers Dec 9 2010 at 18:04
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## 59 Answers
Grothendieck's insight how to deal with the problem that whatever topology you define on varieties over finite fields, you never seem to get enough open sets. You simply have to re-define what is meant by a topology, allowing open sets not to be subsets of your space but to be covers.
I think this fits the bill of "seem very natural once you are used to it", but it was an amazing insight, and totally fundamental in the proof of the Weil conjectures.
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Obviously, I agree that this was fundamental. But since we're speaking only about Grothendieck topologies and not the eventual proof of the Weil conjectures, there could be a curious sense in which this idea might be particularly natural to computers. Imagine encoding a category as objects and morphisms, which I'm told is quite a reasonable procedure in computer science. You'll recall then that it's somewhat hard to define a subobject. – Minhyong Kim Dec 10 2010 at 1:20
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That is, it's easier to refer directly to arrows $A\rightarrow B$ between two of the symbols rather than equivalence classes of them. In this framework, a computer might easily ask itself why any reasonable collection of arrows might not do for a topology. Grothendieck topologies seem to embody exactly the kind of combinatorial and symbolic thinking about open sets that's natural to computers, but hard for humans. We are quite attached to the internal 'physical' characteristics of the open sets, good for some insights, bad for others. – Minhyong Kim Dec 10 2010 at 1:26
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Although this has already been said elsewhere on MathOverflow, I think it's worth repeating that Gromov is someone who has arguably introduced more radical thoughts into mathematics than anyone else. Examples involving groups with polynomial growth and holomorphic curves have already been cited in other answers to this question. I have two other obvious ones but there are many more.
I don't remember where I first learned about convergence of Riemannian manifolds, but I had to laugh because there's no way I would have ever conceived of a notion. To be fair, all of the groundwork for this was laid out in Cheeger's thesis, but it was Gromov who reformulated everything as a convergence theorem and recognized its power.
Another time Gromov made me laugh was when I was reading what little I could understand of his book Partial Differential Relations. This book is probably full of radical ideas that I don't understand. The one I did was his approach to solving the linearized isometric embedding equation. His radical, absurd, but elementary idea was that if the system is sufficiently underdetermined, then the linear partial differential operator could be inverted by another linear partial differential operator. Both the statement and proof are for me the funniest in mathematics. Most of us view solving PDE's as something that requires hard work, involving analysis and estimates, and Gromov manages to do it using only elementary linear algebra. This then allows him to establish the existence of isometric embedding of Riemannian manifolds in a wide variety of settings.
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Generating functions seem old hat to those who have worked with them, but I think their early use could be another example. If you did not have that tool handy, could you create it?
Similarly, any technique that has been developed and is now widely used is made to look natural after years of refining and changing the collected perspective, but might it not have seemed quite revolutionary when first introduced? Perhaps the question should also be about such techniques.
Gerhard "Old Wheels Made New Again" Paseman, 2010.12.09
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I'd like to add to generating functions the idea that you can use singularity analysis to determine the coefficient growth. But I don't know how unexpected this was when first used... – Martin Rubey Dec 9 2010 at 17:24
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"any technique that has been developed and is now widely used is made to look natural after years of refining and changing the collected perspective, but might it not have seemed quite revolutionary when first introduced?" It surely was, and it is exactly why it is widely used now: it allowed a lot of things that were impossible previously and we are still trying to figure out how much is "a lot". Also note, that shaping an idea and recognizing its power is a long process, so "unexpected" means that 20 years ago nobody would have thought of that, not that it shocked everyone on one day. – fedja Dec 17 2010 at 12:53
The method of forcing certainly fits here. Before, set theorists expected that independent results would be obtained by building non-standard, ill-founded models, and model theoretic methods would be key to achieve this. Cohen's method begins with a transitive model and builds another transitive one, and the construction is very different from all the techniques being tried before.
This was completely unexpected. Of course, in hindsight, we see that there are similar approaches in recursion theory and elsewhere happening before or at the same time.
But it was the fact that nobody could imagine you would be able to obtain transitive models that mostly had us stuck.
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I took the last set theory course that Cohen taught, and this isn't how he presented his insight at all (though his book takes this approach). The central problem is "how do I prove that non-constructible [sub]sets [of N] are possible without access to one?", and his solution is "don't use a set; use an adaptive oracle". Once that idea is present, the general method falls right into place. The oracle's set of states can be any partial order, generic filters fall right out, names are clearly necessary, everything else is technical. The hardest part is believing it will actually work. – Chad Groft Dec 14 2010 at 2:07
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@Chad : Very interesting! Curious that his description is so "recursion-theoretic." Do you remember when was this course? – Andres Caicedo Dec 14 2010 at 2:48
Do Cantor's diagonal arguments fit here? (Never mind whether someone did some of them before Cantor; that's a separate question.)
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I vote yes. Look how hard Liouville had to work to find the first examples of transcendental numbers, and how easy Cantor made it to show that there are scads of them. – Gerry Myerson Dec 9 2010 at 22:07
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While Cantor's argument is amazing, and certainly produces scads, Liouville didn't have to work that hard; his approach is also very natural, and doesn't rely on much more than the pigeon-hole principle. – Emerton Dec 10 2010 at 3:15
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Cantor's whole idea of casting mathematics in the language of set theory is now so pervasive we don't even think about it. It dominated our subject until the category point of view. So to me these are the two most basic insights, viewing mathematics in terms of sets, and then in terms of maps. Etale topologies are just one example of viewing maps as the basic concept. – roy smith Dec 13 2010 at 17:55
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My favorite example from algebraic topology is Rene Thom's work on cobordism theory. The problem of classifying manifolds up to cobordism looks totally intractable at first glance. In low dimensions ($0,1,2$), it is easy, because manifolds of these dimensions are completely known. With hard manual labor, one can maybe treat dimensions 3 and 4. But in higher dimensions, there is no chance to proceed by geometric methods.
Thom came up with a geometric construction (generalizing earlier work by Pontrjagin), which is at the same time easy to understand and ingenious. Embed the manifold into a sphere, collapse everything outside a tubular neighborhood to a point and use the Gauss map of the normal bundle... What this construction does is to translate the geometric problem into a homotopy problem, which looks totally unrelated at first sight.
The homotopy problem is still difficult, but thanks to work by Serre, Cartan, Steenrod, Borel, Eilenberg and others, Thom had enough heavy guns at hand to get fairly complete results.
Thom's work led to an explosion of differential topology, leading to Hirzebruch's signature theorem, the Hirzebruch-Riemann-Roch theorem, Atiyah-Singer, Milnor-Kervaire classification of exotic spheres.....until Madsen-Weiss' work on mapping class groups.
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What about Euler's solution to the Konigsberg bridge problem? It's certainly not difficult, but I think (not that I really know anything about the history) it was quite novel at the time.
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@Kimball: It was so novel that Euler didn't even think the problem or its solution were mathematical. See the extract from a letter of Euler on the page en.wikipedia.org/wiki/Carl_Gottlieb_Ehler. – KConrad Jan 2 2012 at 15:46
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I don't know who deserves credit for this, but I was stunned by the concept of view complicated objects like functions simply as points in a vector space. With that view one solves and analyzes PDEs or integral equations in Lebesgue or Sobolev spaces.
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I think one can credit this point of view to Fréchet, who introduced metric spaces for applications to functional analysis. – Qiaochu Yuan Jan 18 2011 at 16:26
Technically, the following are not proofs, or even theorems, but I think they count as insights that have the quality that it's hard to imagine computers coming up with them. First, there's:
Mathematics can be formalized.
Along the same lines, there's:
Computability can be formalized.
If you insist on examples of proofs then maybe I'd be forced to cite the proof of Goedel's incompleteness theorem or of the undecidability of the halting problem, but to me the most difficult step in these achievements was the initial daring idea that one could even formulate a mathematically satisfactory definition of something as amorphous as "mathematics" or "computability." For example, one might argue that the key step in Turing's proof was diagonalization, but in fact diagonalization was a major reason that Goedel thought one couldn't come up with an "absolute" definition of computability.
Nowadays we are so used to thinking of mathematics as something that can be put on a uniform axiomatic foundation, and of computers as a part of the landscape, that we can forget how radical these insights were. In fact, I might argue that your entire question presupposes them. Would computers have come up with these insights if humans had not imagined that computers were possible and built them in the first place? Less facetiously, the idea that mathematics is a formally defined space in which a machine can search systematically clearly presupposes that mathematics can be formalized.
More generally, I'm wondering if you should expand your question to include concepts (or definitions) and not just proofs?
Edit. Just in case it wasn't clear, I believe that the above insights have fundamentally changed mathematicians' conception of what mathematics is, and as such I would argue that they are stronger examples of what you asked for than any specific proof of a specific theorem can be.
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There's something amusing about the idea of a computer coming up with the idea that computability can be formalized. – ndkrempel Dec 10 2010 at 13:58
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This is an example that has bothered me in the past, and I have to admit that I don't have a good answer to it. The ability to introspect seems to be very important to mathematicians, and it's far from clear how a computer would do it. One could perhaps imagine a separate part of the program that looks at what the main part does, but it too would need to introspect. Perhaps this infinite regress is necessary for Godelian reasons but perhaps in practice mathematicians just use a bounded number of levels of navel contemplation. – gowers Dec 10 2010 at 14:29
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Conversely, this type of introspection and formalisation is much less effective outside of mathematics (Weinberg has called this the "unreasonable ineffectiveness of philosophy".) Attempts to axiomatise science, the humanities, etc., for instance, usually end up collapsing under the weight of their own artificiality (with some key exceptions in physics, notably relativity and quantum mechanics). The fact that mathematics is almost the sole discipline that actually benefits from formalisation is indeed an interesting insight in my opinion. – Terry Tao Dec 11 2010 at 16:59
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But if you axiomatize some portion of some other science, doesn't that axiomatization constitute mathematics? So it seems almost tautologous to say that only mathematics "benefits" from formalization. – Michael Hardy Dec 11 2010 at 17:11
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The use of spectral sequences to prove theorems about homotopy groups. For instance, until Serre's mod C theory, nobody knew that the homotopy groups of spheres were even finitely generated.
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Not sure whether to credit Abel or Galois with the "fundamental new way of thinking" here, but the proof that certain polynomial equations are not solvable in radicals required quite the reformulation of thinking. (I'm leaning towards crediting Galois with the brain rewiring reward.)
P.S. Is it really the case that no one else posted this, or is my "find" bar not working properly?
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"Use of group theory to prove insolvability of 5th degree equation" is part of an earlier answer. – Gerry Myerson Dec 12 2010 at 11:14
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I think that Eichler and Shimura's proof of the Ramanujan--Petersson conjecture for weight two modular forms provides an example. Recall that this conjecture is a purely analytic statement: namely that if $f$ is a weight two cuspform on some congruence subgroup of $SL_2(\mathbb Z)$, which is an eigenform for the Hecke operator $T_p$ ($p$ a prime not dividing the level of the congruence subgroup in question) with eigenvalue $\lambda_p$, then $| \lambda_p | \leq 2 p^{1/2}.$ Unfortunately, no purely analytic proof of this result is known. (Indeed, if one shifts one's focus from holomorphic modular forms to Maass forms, then the corresponding conjecture remains open.)
What Eichler and Shimura realized is that, somewhat miraculously, $\lambda_p$ admits an alternative characterization in terms of counting solutions to certain congruences modulo $p$, and that estimates there due to Hasse and Weil (generalizing earlier estimates of Gauss and others) can be applied to show the desired inequality.
This argument was pushed much further by Deligne, who handled the general case of weight $k$ modular forms (for which the analogous inequality is $| \lambda_p | \leq 2 p^{(k-1)/2}$), using etale cohomology of varieties in characteristic $p$ (which is something of a subtle and more technically refined analogue of the notion of a congruence mod $p$). (Ramanujan's original conjecture was for the unique cuspform of weight 12 and level 1.)
The idea that there are relationships (some known, others conjectural) between automorphic forms and algebraic geometry over finite fields and number fields has now become part of the received wisdom of algebraic number theorists, and lies at the heart of the Langlands program. (And, of course, at the heart of the proof of FLT.) Thus the striking idea of Eichler and Shimura has now become a basic tenet of a whole field of mathematics.
Note: Tim in his question, and in some comments, has said that he wants "first non-trivial instances" rather than difficult arguments that involve a whole range of ideas and techniques. In his comment to Terry Tao's answer regarding Perelman, he notes that long, difficult proofs might well include within them instances of such examples. Thus I am offering this example as perhaps a "first non-trivial instance" of the kind of insights that are involved in proving results like Sato--Tate, FLT, and so on.
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Another example from logic is Gentzen's consistency proof for Peano arithmetic by transfinite induction up to $\varepsilon_0$, which I think was completely unexpected, and unprecedented.
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It seems that certain problems seem to induce this sort of new thinking (cf. my article "What is good mathematics?"). You mentioned the Fourier-analytic proof of Roth's theorem; but in fact many of the proofs of Roths' theorem (or Szemeredi's theorem) seem to qualify, starting with Furstenberg's amazing realisation that this problem in combinatorial number theory was equivalent to one in ergodic theory, and that the structural theory of the latter could then be used to attack the former. Or the Ruzsa-Szemeredi observation (made somewhat implicitly at the time) that Roth's theorem follows from a result in graph theory (the triangle removal lemma) which, in some ways, was "easier" to prove than the result that it implied despite (or perhaps, because of) the fact that it "forgot" most of the structure of the problem. And in this regard, I can't resist mentioning Ben Green's brilliant observation (inspired, I believe, by some earlier work of Ramare and Ruzsa) that for the purposes of finding arithmetic progressions, that the primes should not be studied directly, but instead should be viewed primarily [pun not intended] as a generic dense subset of a larger set of almost primes, for which much more is known, thanks to sieve theory...
Another problem that seems to generate radically new thinking every few years is the Kakeya problem. Originally a problem in geometric measure theory, the work of Bourgain and Wolff in the early 90s showed that the combinatorial incidence geometry viewpoint could lead to substantial progress. When this stalled, Bourgain (inspired by your own work) introduced the additive combinatorics viewpoint, re-interpreting line segments as arithmetic progressions. Meanwhile, Wolff created the finite field model of the Kakeya problem, which among other things lead to the sum-product theorem and many further developments that would not have been possible without this viewpoint. In particular, this finite field version enabled Dvir to introduce the polynomial method which had been applied to some other combinatorial problems, but whose application to the finite field Kakeya problem was hugely shocking. (Actually, Dvir's argument is a great example of "new thinking" being the key stumbling block. Five years earlier, Gerd Mockenhaupt and I managed to stumble upon half of Dvir's argument, showing that a Kakeya set in finite fields could not be contained in a low-degree algebraic variety. If we had known enough about the polynomial method to make the realisation that the exact same argument also showed that a Kakeya set could not have been contained in a high-degree algebraic variety either, we would have come extremely close to recovering Dvir's result; but our thinking was not primed in this direction.) Meanwhile, Carbery, Bennet, and I discovered that heat flow methods, of all things, could be applied to solve a variant of the Euclidean Kakeya problem (though this method did appear in literature on other analytic problems, and we viewed it as the continuous version of the discrete induction-on-scales strategy of Bourgain and Wolff.) Most recently is the work of Guth, who broke through the conventional wisdom that Dvir's polynomial argument was not generalisable to the Euclidean case by making the crucial observation that algebraic topology (such as the ham sandwich theorem) served as the continuous generalisation of the discrete polynomial method, leading among other things to the recent result of Guth and Katz you mentioned earlier.
EDIT: Another example is the recent establishment of universality for eigenvalue spacings for Wigner matrices. Prior to this work, most of the rigorous literature on eigenvalue spacings relied crucially on explicit formulae for the joint eigenvalue distribution, which were only tractable in the case of highly invariant ensembles such as GUE, although there was a key paper of Johansson extending this analysis to a significantly wider class of ensembles, namely the sum of GUE with an arbitrary independent random (or deterministic) matrix. To make progress, one had to go beyond the explicit formula paradigm and find some way to compare the distribution of a general ensemble with that of a special ensemble such as GUE. We now have two basic ways to do this, the local relaxation flow method of Erdos, Schlein, Yau, and the four moment theorem method of Van Vu and myself, both based on deforming a general ensemble into a special ensemble and controlling the effect on the spectral statistics via this deformation (though the two deformations we use are very different, and in fact complement each other nicely). Again, both arguments have precedents in earlier literature (for instance, our argument was heavily inspired by Lindeberg's classic proof of the central limit theorem) but as far as I know it had not been thought to apply them to the universality problem before.
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But, Terry, are the adjectives "radical" or "fundamentally new" really justified in the description of any of these examples? and of our business as a whole? – Gil Kalai Dec 10 2010 at 13:30
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In other words, his contribution was that D:A=C:B (algebraic topology is to continuous incidence geometry as algebraic geometry is to discrete incidence geometry), which was definitely a very different way of thinking about these four concepts that was totally absent in previous work. (After Guth's work, it is now "obvious" in retrospect, of course.) – Terry Tao Dec 11 2010 at 16:48
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Perhaps what this example shows is that a computer trying to generate mathematical progress has to look at more than just the 1-skeleton of mathematics (B is solved by C; A is close to B; hence A might be solved by C) but also at the 2-skeleton (B is solved by C; D is to A as C is to B; hence A might be solved by D) or possibly even higher order skeletons. It seems unlikely though that these possibilities can be searched through systematically in polynomial time, without the speedups afforded by human insight... – Terry Tao Dec 11 2010 at 17:12
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Gromov's use of J-holomorphic curves in symplectic topology (he reinterpreted holomorphic functions in the sense of Vekua) as well as the invention of Floer homology (in order to deal with the Arnol'd conjecture).
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I'm a little surprised no one has cited Thurston's impact on low-dimensional topology and geometry. I'm far from an expert, so I'm reluctant to say much about this. But I have the impression that Thurston revolutionized the whole enterprise by taking known results and expressing them from a completely new perspective that led naturally both new theorems and a lot of new conjectures. Perhaps Thurston himself or someone else could say something, preferably in a separate answer so I can delete mine.
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Quillen's construction of the cotangent complex used homotopical algebra to find the correct higher-categorical object without explicitly building a higher category. This may sound newfangled and modern, but if you read Grothendieck's book on the cotangent complex, his explicit higher-categorical construction was only able to build a cotangent complex that had its (co)homology truncated to degree 2. Strangely enough, by the time Grothendieck's book was published, it was already obsolete, as he notes in the preface (he says something about how new work of Quillen (and independently André) had made his construction (which is substantially more complicated) essentially obsolete).
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Emil Artin's solution of Hilbert's 17th problem which asked whether every positive polynomial in any number of variables is a sum of squares of rational functions.
Artin's proof goes roughly as follows. If $p \in \mathbb R[x_1,\dots,x_n]$ it not a sum of squares of rational functions, then there is some real-algebraically closed extension $L$ of the field of rational functions in which $p$ is negative with respect to some total ordering (compatible with the field operations), i.e. there exists a $L$-point of $R[x_1,\dots,x_n]$ at which $p$ is negative. However, using a model theoretic argument, since $\mathbb R$ is also a real-closed field with a total ordering, there also has to be a real point such that $p<0$, i.e. there exists $x \in \mathbb R^n$ such that $p(x)< 0$. Hence, if $p$ is everywhere positive, then it is a sum of squares of rational functions.
The ingenius part is the use of a model theoretic argument and the bravery to consider a totally ordered real-algebraic closed extension of the field of rational functions.
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Gromov's proof that finitely generated groups with polynomial growth are virtually nilpotent. The ingenious step is to consider a scaling limit of the usual metric on the Cayley graph of the finitely generated group.
Of course the details are messy and to get the final conclusion one has to rely on a lot of deep results on the structure of topological groups. However, already the initial idea is breathtaking.
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Topological methods in combinatorics (started by Lovasz' proof of the Kneser conjecture, I guess).
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Heegner's solution to the Gauss class number 1 problem for imaginary quadratic fields, by noting that when the class number is 1 then a certain elliptic curve is defined over Q and certain modular functions take integer values at certain quadratic irrationalities, and then finding all the solutions to Diophantine equations that result, seems to me equally beautiful and unexpected. Maybe its unexpectedness kept people from believing it for a long time.
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I don't know how good an example this is. The Lefschetz fixed point theorem tells you that you can count (appropriately weighted) fixed points of a continuous function $f : X \to X$ from a compact triangulable space to itself by looking at the traces of the induced action of $f$ on cohomology. This is a powerful tool (for example it more-or-less has the Poincare-Hopf theorem as a special case).
Weil noticed that the number of points of a variety $V$ over $\mathbb{F}_{q^n}$ is the number of fixed points of the $n^{th}$ power of the Frobenius map $f$ acting on the points of $V$ over $\overline{\mathbb{F}_q}$ and, consequently, that it might be possible to describe the local zeta function of $V$ if one could write down the induced action of $f$ on some cohomology theory for varieties over finite fields. This led to the Weil conjectures, the discovery of $\ell$-adic cohomology, etc. I think this is a pretty good candidate for a powerful but unexpected technique.
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Sometimes mathematics is not only about the methods of the proof, it is about the statement of the proof. E.g., it is hard to imagine an theorem-searching algorithm ever finding a proof of the results in Shannon's 1948 Mathematical Theory of Communication, without that algorithm first "imagining" (by some unspecified process) that there could BE a theory of communication.
Even so celebrated a mathematician as J. L. Doob at first had trouble grasping that Shannon's reasoning was mathematical in nature, writing in his AMS review (MR0026286):
[Shannon's] discussion is suggestive throughout, rather than mathematical, and it is not always clear that the author's mathematical intentions are honorable.
The decision of which mathematical intentions are to be accepted as "honorable" (in Doob's phrase) is perhaps very difficult to formalize.
[added reference]
One finds this same idea expressed in von Neumann's 1948 essay The Mathematician:
Some of the best inspirations of modern mathematics (I believe, the best ones) clearly originated in the natural sciences. ... As any mathematical discipline travels far from its empirical source, or still more, if it is a second or third generation only indirectly inspired by ideas coming from "reality", it is beset by very grave dangers. It becomes more and more purely aestheticizing, more and more l`art pour le art. ... Whenever this stage is reached, the only remedy seems to me to be the rejuvenating return to the source: the reinjection of more or less directly empirical ideas.
One encounters this theme of inspiration from reality over-and-over in von Neumann's own work. How could a computer conceive theorems in game theory ... without having empirically played games? How could a computer conceive the theory of shock waves ... without having empirically encountered the intimate union of dynamics and thermodynamics that makes shock wave theory possible? How could a computer conceive theorems relating to computational complexity ... without having empirically grappled with complex computations?
The point is straight from Wittgenstein and E. O. Wilson: in order to conceive mathematical theorems that are interesting to humans, a computer would have to live a life similar to an ordinary human life, as a source of inspiration.
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Donaldson's idea of using global analysis to get more insight about the topology of manifolds. Nowadays it is clear to us that (non-linear) moduli spaces give something new, and more than linear (abelian) Hodge theory, for example, but I think at that time this was really new.
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I fully agree with this. I was a graduate student at Harvard, when Atiyah came and described Donaldson's thesis (Donaldson got his Ph.D. the same year as me). Before that, we all thought we were trying to understand Yang-Mills, because it connected geometric analysis to physics and not because we thought it would prove topological theorems. As I recall it, Atiyah said that when Donaldson first proposed what he wanted to do, Atiyah was skeptical and tried to convince Donaldson to work on something less risky. – Deane Yang Dec 11 2010 at 5:10
And how about Perelman's proof of Poincare's conjecture?
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It would be a better example if the proof were easier to understand ... – gowers Dec 9 2010 at 16:38
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I think there are at least two aspects of the Perelman-Hamilton theory that fit the bill. One is Hamilton's original realisation that Ricci flow could be used to at least partially resolve the Poincare conjecture (in the case of 3-manifolds that admit a metric with positive Ricci curvature). There was some precedent for using PDE flow methods to attack geometric problems, but I think this was the first serious attempt to attack the manifestly topological Poincare conjecture in that fashion, and was somewhat contrary to the conventional wisdom towards Poincare at the time. [cont.] – Terry Tao Dec 9 2010 at 17:30
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The other example is when Perelman needed a monotone quantity in order to analyse singularities of the Ricci flow. Here he had this amazing idea to interpret the parabolic Ricci flow as an infinite-dimensional limit of the elliptic Einstein equation, so that monotone quantities from the elliptic theory (specifically, the Bishop-Gromov inequality) could be transported to the parabolic setting. This is a profoundly different perspective on Ricci flow (though there was some precedent in the earlier work of Chow) and it seems unlikely that this quantity would have been discovered otherwise. – Terry Tao Dec 9 2010 at 17:32
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Terry's answer illustrates a principle relevant to this question: even if a proof as a whole is too complex to count as a good example, there are quite likely to be steps of the proof that are excellent examples. – gowers Dec 9 2010 at 20:54
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Shigefumi Mori's proof of Hartshorne's conjecture (the projective spaces are the only smooth projective varieties with ample tangent bundles). In his proof, Mori developed many new techniques (e.g. the bend-and-break lemma), which later became fundamental in birational geometry.
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The use of ideals in rings, rather than elements (in terms of factorization, etc...).
This was followed by another revolutionary idea: using radical (Jacobson radical, etc...) instead of simple properties on elements.
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Use of Lagrange theorem (group theory) to prove Fermat's small theorem?
Use of fixed point methods (and completeness) to prove existence of solutions to differential equations?
Use of Fields theory to prove the impossibility of the trisection of the angle?
Use of group theory to prove insolvability of 5th degree equation?
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Fermat´s small theorem is in its nature group-theoretic. I don´t see anything surprising in proving it via Lagrange. Application of fixed point methods to differential equations is applying analysis to analysis. It only asks one to realize that differential and integral operators are maps... :-) But I think the 3rd and 4th point of your answer definitely suit to the original question. – ex falso quodlibet Dec 9 2010 at 19:00
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Lobachevsky and Bolyai certainly introduced a fundamentally new way of thinking, though I'm not sure it fits the criterion of being a proof of something - perhaps a proof that a lot of effort had been wasted in trying to prove the parallel postulate.
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Morse theory is another good example. Indeed it is the inspiration for Floer theory, which has already been mentioned.
Atiyah-Bott's paper "Yang-Mills equations on a Riemann surface" and Hitchin's paper "Self-duality equations on a Riemann surface" both contain rather striking applications of Morse theory. The former paper contains for example many computations about cohomology rings of moduli spaces of holomorphic vector bundles over Riemann surfaces; the latter paper proves for instance that moduli spaces of Higgs bundles over Riemann surfaces are hyperkähler.
Note that these moduli spaces are algebraic varieties and can be (and are) studied purely from the viewpoint of algebraic geometry. But if we look at things from an analytic point of view, and we realize these moduli spaces as quotients of infinite dimensional spaces by infinite dimensional groups, and we use the tools of analysis and Morse theory, as well as ideas from physics(!!!), then we can discover perhaps more about these spaces than if we viewed them just algebraically, as simply being algebraic varieties.
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http://physics.stackexchange.com/questions/4108/examples-where-momentum-is-not-equal-to-mv/4114
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# Examples where momentum is not equal to $mv$?
I am aware that momentum is the thing which is conserved due to symmetries in space (rotational symmetry, translaitonal symmetry, etc). I am aware that in some systems, the generalized momentum,
$$p_j = {\partial \mathcal{L} \over \partial \dot{q}_j}$$
is not simply $mv$. One example of this might be a charged particle in an electric field,
$$p_j = mv_i + {q \over c} A_i$$
Can anyone provide other examples where this is the case (simple, preferabily) and an intuition perhaps of why the "old" definition of momentum breaks down?
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Wikifying since this is a list question. (Honestly, I'm not entirely sure what the value is in having just a list of examples where momentum is not equal to $mv$...) – David Zaslavsky♦ Jan 28 '11 at 22:33
The conservation law of center of mass (coming from Lorentz boosts in the Noether framework) guarantees that the momentum over the energy is always the center of mass velocity. So you can't have momentum and energy without a velocity equal to their ratio. – Ron Maimon May 2 '12 at 15:06
## 4 Answers
the momentum is the quantity that is conserved as a result of the symmetry with respect to translations. This is the so-called Emmy Noether's theorem.
The only reason why we had $p=mv$ in classical mechanics was that the Lagrangian only depended on $\dot q$ by the kinetic energy $m\dot q^2/2$ term. By the differentiation, you may see that $p=m\dot q$. However, whenever the Lagrangian is different, the relationship between momentum and velocity gets modified.
You quoted one example. Another example is any theory in relativity where $$p = \frac{m_0 v}{\sqrt{1-v^2/c^2}}$$ which may also be derived from the Lagrangian. Note that in the relativistic case, the Lagrangian is (minus) the proper time of the world line.
In some cases, the momentum gets modified but people have disagreed what it is. A recently discussed example here was the Abraham-Minkowski controversy about the momentum of a photon in a dielectric material:
Is the Abraham-Minkowski controversy resolved?
For photons, the momentum is $p=E/c$ in the vacuum - this itself is different from massive slow particles - but this relationship can get multiplied or divided by the index of refraction $n$, and arguments arose from this point.
In field theory, one has totally different formulae for the momentum. There are no particles i.e. no velocities of a few objects. Instead, the momentum is distributed in the field - e.g. electromagnetic field - and every region of space carries a momentum density. For example, the momentum density of the electromagnetic field is $\vec E\times \vec H$ or $\vec D\times \vec B$ - which of them is "true" was a famous controversy, equivalent to the controversy about the momentum of a single photon.
Other fields obviously have other formulae for the momentum density.
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This is well and good, but I'm not sure the OP was asking about SR at all... – Noldorin Jan 29 '11 at 16:42
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Well, it doesn't matter whether OP was explicitly asking about SR; SR clearly has to be a part of a correct answer because it's the most important framework in which $p$ isn't just $m_0 v$. – Luboš Motl Feb 3 '11 at 14:25
I think if you use non Cartesian coordinates in the expression p = m v, you will obtain something more generalized.
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I don't see how it can be any more general using another coordinate system... – Noldorin Jan 28 '11 at 22:38
OK, in another coordinate system the unities are different. – Vladimir Kalitvianski Jan 28 '11 at 22:44
Dear downvoters, tell me where I am wrong, please. I would like to learn. – Vladimir Kalitvianski Jan 31 '11 at 16:21
Momentum defined as $mv$ isn't conserved in collisions approaching $c$, whereas relativistic momentum, $\gamma mv$, is. Hence the redefinition of momentum.
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Generalized momentum and momentum are two separate things.
In an electromagnetic field, the mechanical (also known as the kinetic) momentum of a non-relativistic charged particle is still $\vec{p}=m\vec{v}$.
$\vec{P}=m\vec{v}+q\vec{A}$ is called the canonical momentum.
And in analytical mechanics, the generalized momentum can be almost everything, including position coordinates like $x,y,z$ or $r,\theta,\phi$.
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– Mark Eichenlaub Jun 22 '11 at 5:45
@Mark Eichenlaub: The momentum of electromagnetic field is not part of the momentum of any charged particles inside it. $qA$ is not the momentum of electromagnetic field; that would be $\vec{g}=\varepsilon_0\vec{E}\times\vec{B}$. – C.R. Jun 24 '11 at 11:34
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http://physics.stackexchange.com/questions/27665/the-role-of-rigor
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The Role of Rigor
The purpose of this question is to ask about the role of mathematical rigor in physics. In order to formulate a question that can be answered, and not just discussed, I divided this large issue into five specific questions.
1) What are the most important and the oldest insights (notions, results) from physics that are still lacking rigorous mathematical formulation/proofs.
2) The endeavor of rigorous mathematical explanations, formulations, and proofs for notions and results from physics is mainly taken by mathematicians. What are examples that this endeavor was beneficial to physics itself.
3) What are examples that insisting on rigour delayed progress in physics.
4) What are examples that solid mathematical understanding of certain issues from physics came from further developements in physics itself. (In particular, I am interested in cases where mathematical rigorous understanding of issues from classical mechanics required quantum mechenics, and also in cases where progress in physics was crucial to rigorous mathematical solutions of questions in mathematics not originated in physics.)
5) The role of rogor is intensly discussed in popular books and blogs. Please supply references (or better annotated references) to academic studies of the role of mathematical rigour in modern physics.
(Of course, I will be also thankful to answers which elaborate on a single item related to a single question out of these five questions.)
Related Math Overflow questions: Examples-of-non-rigorous-but-efficient-mathematical-methods-in-physics (related to question 1); Examples-of-using-physical-intuition-to-solve-math-problems; Demonstrating-that-rigour-is-important
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This is interesting, but I am not sure it is within the purview of this forum. As I understand it (or at least this is my opinion), this is a place to ask specific and answerable questions, usually of a technical nature. Note also that you are unlikely to find real experts in the history and philosophy of science here, so the quality of the discussion will not likely exceed that of those popular books and blogs you refer to. – user566 Sep 24 '11 at 18:01
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– Michael Sep 24 '11 at 18:31
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– Gil Kalai Sep 24 '11 at 18:44
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I am not sure either. There certainly could be an interesting discussion on this subject somewhere, but my feeling it is unlikely here because of lack of the appropriate expertise, and because this topic can be divisive among mathematical and theoretical physicists. I may well be wrong, this is just my gut feeling. – user566 Sep 24 '11 at 18:52
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@UGPhysics: I really don't see how it would get meaningful answers there. He is asking questions that are probably best answerable by either physicists or historians of science. – Joe Fitzsimons Sep 25 '11 at 3:36
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6 Answers
I still think that it's not the right place for this type of questions. Nevertheless, the topic itself is interesting, and I'll also have a shot at it. Since I'm neither a philosopher of science, nor an historian (and there are probably very few such people on this site, one of the reasons this question might not be suitable), I'll focus on my own restricted field, statistical physics.
1) There are many. For example, a satisfactory rigorous derivation of Boltzmann equation, the best result to this day remaining the celebrated theorem of Lanford proved in the late 1970s. In equilibrium statistical mechanics, one of the major open problems is the proof that the two-dimensional $O(N)$ models have exponentially decaying correlations at all temperatures when $N>2$ (there is supposedly a close relationship between such models and four-dimensional gauge models, and this problem might shed light on the issue of asymptotic freedom in QCD, see this paper for a critical discussion of these issues). Of course, there are many others, such as trying to understand why naive real-space renormalization (say, decimation) of lattice spin systems provides reasonably accurate results (even though such transformations are known to be generally ill-defined mathematically); but it seems to me that it's unlikely to happen, which does not mean that the philosophy of the renormalization group cannot find uses in mathematical physics (it already has led to several profound results).
2) Well, one major example was Onsager's rigorous computation of the free energy of the 2d Ising model, which showed that all approximation schemes used by physicists at that time were giving completely wrong predictions. Rigorous results can also lead to (i) new approaches to old problems (this is the case recently with SLE), (ii) new results that were not known to physicists (this is the case with, e.g., the results of Johansson and others on growth models), (iii) a much better understanding of some complicated phenomena (e.g., the equilibrium properties of fixed magnetization Ising models), (iv) settling controversies in the physics literature (a famous example was the problem of determining the lower critical dimension of the random-field Ising model, which was hotly debated in the 1980s, and was rigorously settled by Bricmont and Kupiainen).
3) None that I know of. Although, one might say that the "paradoxes" raised against Boltzmann's theory by Zermelo and Loschmidt were both of mathematical nature (and thus criticized the apparent lack of of rigour of Boltzmann's approach), and did delay the acceptance of his ideas.
4) Not sure about this point. Certainly the numerous conjectures originating from physics, in particular striking predictions, provide both motivation, and sometimes some degree of insight to the mathematicians... But I am not sure that's what you're asking for.
5) There are many papers discussing such issues, e.g.:
• "Theoretical mathematics'': Toward a cultural synthesis of mathematics and theoretical physics, by Arthur Jaffe and Frank Quinn, Bull.Am.Math.Soc. 29 (1993) 1-13.
• Is Mathematical Rigor Necessary in Physics?, by Kevin Davey, The British Journal for the Philosophy of Science (2003), Volume: 54, Issue: 3, Pages: 439-463.
and references therein.
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2) is a great example. This is why this question should stay... – András Bátkai Sep 25 '11 at 14:54
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@András Bátkai : yes, I find it one of the most compelling example. But it should be pointed out that it took until the 1960s for (most) physicists to take it really seriously (remember that to most people, this was a non-realistic, two-dimensional toy model, and they were more willing to associate the disagreements with predictions from their approximation schemes to pathologies of the model rather than to failure of these approximations. – Yvan Velenik Sep 27 '11 at 14:58
Rigor is clarity of concepts and precision of arguments. Therefore in the end there is no question that we want rigor.
To get there we need freedom for speculation, first, but for good speculation we need...
...solid ground, which is the only ground that serves as a good jumping-off point for further speculation.
Sometimes physicists behave is if rigor is all about replacing an obvious but non-precise argument with a tedious and boring proof. But more often than not rigor is about identifying the precise and clear definitions such that the obvious argument becomes also undoubtly correct.
There are many historical examples.
For instance the simple notion of differential forms and exterior derivatives. It's not a big deal in the end, but when they were introduced into physics they not only provided rigor for a multitude of vague arguments about infinitesimal variation and extended quantity. Maybe more importantly, they clarified structure. Maxwell still filled two pages with the equations of electromagnetism at a time when even the concepts of linear algebra were an arcane mystery. Today we say just $d \star d A = j_{el}$ and see much further, for instance derive the charge quantization law rigorously with child's ease. The clear and precise concept is what does this for us.
And while probaby engineers could (and maybe do?) work using Maxwell's original concepts, the theoreticians would have been stuck. One can't see the subtleties of self-dual higher gauge theory, for instance, without the rigorous concept of de Rham theory.
There are many more examples like this. Here is another one: rational CFT was "fully understood" and declared solved at a non-rigorous level for a long time. When the rigorous FRS-classification of full rational CFT was established, it not onyl turned out that some of the supposed rational CFT construction in the literature did not actually exist, while other existed that had been missed, more importantly was: suddenly it was very clear why and which of these examples exist. Based on the solid ground of this new rigor, it is now much easier to base new non-rigorous arguments that go much further than one could do before. For instance about the behaviour of rational CFT in holography.
Rigor is about clarity and precision, which is needed for seeing further. As Ellis Cooper just said elsewhere:
Rigor cleans the window through which intuition shines.
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Great asnwer! ${}$ – Mariano Oct 5 '11 at 2:04
I can by no means claim to give a full answer on this question, but perhaps a partial answer is better than no answer at all.
As regards (1) perhaps the most famous example is the Navier-Stokes equation. We know it produces extremely good results for modeling fluid flow, but we can't even show that there always exists a solution. Indeed, there is a Clay prize going for proving the existence of smooth solutions on $\mathbb{R}^3$ (problem statement here).
An example of (2) is that the study of topological quantum field theory has been motivated at least in part by mathematics.
As regards (3) I don't really think this has ever happened. However, by this, I do not mean that demanding rigor would not prevent or slow the progression of physics, but rather that it seems extremely hard to find an example of a case where a relatively large community has not simply ignored any such demand. Certainly it is true that mathematically rigorous formulations often follow far behind the current state of the art in physics, but there is nothing unexpected about this.
I do not currently have any good answers as regards the remainder of your question.
There is a relatively interesting essay on this (C. Vafa - On the future of mathematics/physics interaction) in Mathematics: Frontiers and Perspectives, which also mentions the TQFT example.
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Rigorous arguments are very similar to computer programming--- you need to write a proof which can (in principle) ultimately be carried out in a formal system. This is not easy, and requires defining many data-structures (definitions), and writing many subroutines (lemmas), which you use again and again. Then you prove many results along the way, only some of which are of general usefulness.
This activity is extremely illuminating, but it is time consuming, and tedious, and requires a great deal of time and care. Rigorous arguments also introduce a lot of pedantic distinctions which are extremely important for the mathematics, but not so important in the cases one deals with in physics.
In physics, you never have enough time, and we must always have a only just precise enough understanding of the mathematics that can be transmitted maximally quickly to the next generation. Often this means that you forsake full rigor, and introduce notational short-cuts and imprecise terminology that makes turning the argument rigorous difficult.
Some of the arguments in physics though are pure magic. For me, the replica trick is the best example. If this ever gets a rigorous version, I will be flabbergasted.
1) What are the most important and the oldest insights (notions, results) from physics that are still lacking rigorous mathematical formulation/proofs.
Here are old problems which could benefit from rigorous analysis:
• Mandelstam's double-dispersion relations: The scattering amplitude for 2 particle to 2 particle scattering can be analytically expanded as an integral over the imaginary discontinuity $\rho(s)$ in the s parameter, and then this discontinuity $\rho(s)$ can be written as an integral over the t parameter, giving a double-discontinuity $\rho(s,t)$ If you go the other way, expand the discontinuity in t first then in s, you get the same function. Why is that? It was argued from perturbation theory by Mandelstam, and there was some work in the 1960s and early 1970s, but it was never solved as far as I know.
• The oldest, dating back centuries: Is the (Newtonian, comet and asteroid free) solar system stable for all time? This is a famous one. Rigorous bounds on where integrability fails will help. The KAM theorem might be the best answer possible, but it doesn't answer the question really, since you don't know whether the planetary perturbations are big enough to lead to instability for 8 planets some big moons, plus sun.
• continuum statistical mechanics: What is a thermodynamic ensemble for a continum field? What is the continuum limit of a statistical distribution? What are the continuous statistical field theories here?
• What are the generic topological solitonic solutions to classical nonlinear field equations? Given a classical equation, how do you find the possible topological solitons? Can they all be generated continuously from given initial data? For a specific example, consider the solar-plasma--- are there localized magneto-hydrodynamic solitons?
There are a bazillion problems here, but my imagination fails.
2) The endeavor of rigorous mathematical explanations, formulations, and proofs for notions and results from physics is mainly taken by mathematicians. What are examples that this endeavor was beneficial to physics itself.
There are a few examples, but I think they are rare:
• Penrose's rigorous proof of the existence of singularities in a closed trapped surface is the canonical example: it was a rigorous argument, derived from Riemannian geometry ideas, and it was extremely important for clarifying what's going on in black holes.
• Quasi-periodic tilings, also associated with Penrose, first arose in Hao and Wang's work in pure logic, where they were able to demonstrate that an appropriate tiling with complicated matching edges could do full computation. The number of tiles were reduced until Penrose gave only 2, and finally physicists discovered quasicrystals. This is spectacular, because here you start in the most esoteric non-physics part of pure mathematics, and you end up at the most hands-on of experimental systems.
• Kac-Moody algebras: These came up in half-mathematics, half early string theory. The results became physical in the 1980s when people started getting interested in group manifold models.
• The ADE classificiation from Lie group theory (and all of Lie group theory) in mathematics is essential in modern physics. Looking back further, Gell-Mann got SU(3) quark symmetry by generalizing isospin in pure mathematics.
• Obstruction theory was essential in understanding how to formulate 3d topological field theories (this was the subject of a recent very interesting question), which have application in the fractional quantum hall effect. This is very abstract mathematics connected to laboratory physics, but only certain simpler parts of the general mathematical machinery are used.
3) What are examples that insisting on rigour delayed progress in physics.
This has happened several times, unfortunately.
• Statistical mechanics: The lack of rigorous proof of Boltzmann ergodicity delayed the acceptance of the idea of statistical equilibrium. The rigorous arguments were faulty--- for example, it is easy to prove that there are no phase transitions in finite volume (since the Boltzmann distribution is analytic), so this was considered a strike against Boltzmann theory, since we see phase transitions. You could also prove all sorts of nonsense about mixing entropy (which was fixed by correctly dealing with classical indistinguishability). Since there was no proof that fields would come to thermal equilibrium, some people believed that blackbody light was not thermal. This delayed acceptance of Planck's theory, and Einstein's. Statistical mechanics was not fully accepted until Onsager's Ising model solution in 1941.
• Path integrals: This is the most notorious example. These were accepted by some physicists immediately in the 1950s, although =the formalism wasn't at all close to complete until Candlin formulated Grassman variables in 1956. Past this point, they could have become standard, but they didn't. The formalism had a bad reputation for giving wrong results, mostly because people were uncomfortable with the lack of rigor, so that they couldn't trust the method. I heard a notable physicist complain in the 1990s that the phase-space path integral (with p and q) couldn't possibly be correct because p and q don't commute, and in the path integral they do because they are classical numbers (no, actually, they don't--- their value in an insertion depends discontinuously on their time order in the proper way). It wasn't until the early 1970s that physicists became completely comfortable with the method, and it took a lot of selling to overcome the resistance.
• Quantum field theory construction: The rigorous methods of the 1960s built up a toolbox of complicated distributional methods and perturbation series resummation which turns out to be the least useful way of looking at the thing. It's now C* algebras and operator valued distributions. The correct path is through the path integral the Wilsonian way, and this is closer to the original point of view of Feynman and Schwinger. But a school of rigorous physicists in the 1960s erected large barriers to entry in field theory work, and progress in field theory was halted for a decade, until rigor was thrown out again in the 1970s. But a proper rigorous formulation of quantum fields is still missing.
In addition to this, there are countless no-go theorems that delayed the discovery of interesting things:
• Time cannot be an operator (Pauli): this delayed the emergence of the path integral particle formulation due to Feynman and Schwinger. Here, the time variable on the particle-path is path-integrated just like anything else.
• Von-Neumann's proof of no-hidden variables: This has a modern descendent in the Kochen Sprecher theorem about entangled sets of qubits. This delayed the Bohm theory, which faced massive resistance at first.
• No charges which transform nontrivially under the Lorentz group(Coleman-Mandula): This theorem had both positive and negative implications. It killed SU(6) theories (good), but it made people miss supersymmetry (bad).
• Quasicrystal order is impossible: This "no go" theorem is the standard proof that periodic order (the general definition of crystals) is restricted to the standard space-groups. This made quasicrystals bunk. The assumption that is violated is the assumption of strict periodicity.
• No supergravity compactifications with chiral fermions (Witten): this theorem assumed manifold compactification, and missed orbifolds of 11d SUGRA, which give rise to the heterotic strings (also Witten, with Horava, so Witten solved the problem).
4) What are examples that solid mathematical understanding of certain issues from physics came from further developements in physics itself. (In particular, I am interested in cases where mathematical rigorous understanding of issues from classical mechanics required quantum mechenics, and also in cases where progress in physics was crucial to rigorous mathematical solutions of questions in mathematics not originated in physics.)
There are several examples here:
• Understanding the adiabatic theorem in classical mechanics (that the action is an adiabatic invariant) came from quantum mechanics, since it was clear that it was the action that needed to be quantized, and this wouldn't make sense without it being adiabatic invariant. I am not sure who proved the adiabatic theorem, but this is exactly what you were asking for--- an insightful classical theorem that came from quantum mechanics (although some decades before modern quantum mechanics)
• The understanding of quantum anomalies came directly from a physical observation (the high rate of neutral pion decay to two photons). Clarifying how this happens through Feynman diagrams, even though a naive argument says it is forbidden led to complete understanding of all anomalous terms in terms of topology. This in turn led to the development of Chern-Simons theory, and the connection with Knot polynomials, discovered by Witten, and earning him a Fields medal.
• Distribution theory originated in Dirac's work to try to give a good foundation for quantum mechanics. The distributional nature of quantum fields was understood by Bohr and Rosenfeld in the 1930s, and the mathematics theory was essentially taken from physics into mathematics. Dirac already defined distributions using test functions, although I don't think he was pedantic about the test-function space properties.
5) The role of rigor is intensly discussed in popular books and blogs. Please supply references (or better annotated references) to academic studies of the role of mathematical rigour in modern physics.
I can't do this, because I don't know any. But for what it's worth, I think it's a bad idea to try to do too much rigor in physics (or even in some parts of mathematics). The basic reason is that rigorous formulations have to be completely standardized in order for the proofs of different authors to fit-together without seams, and this is only possible in very long hindsight, when the best definitions become apparent. In the present, we're always muddling through fog. So there is always a period where different people have slightly different definitions of what they mean, and the proofs don't quite work, and mistakes can happen. This isn't so terrible, so long as the methods are insightful.
The real problem is the massive barrier to entry presented by rigorous definitions. The actual arguments are always much less daunting than the superficial impression you get from reading the proof, because most of the proof is setting up machinery to make the main idea go through. Emphasizing the rigor can put undue emphasis on the machinery rather than the idea.
In physics, you are trying to describe what a natural system is doing, and there is no time to waste in studying sociology. So you can't learn all the machinery the mathematicians standardize on at any one time, you just learn the ideas. The ideas are sufficient for getting on, but they aren't sufficient to convince mathematicians you know what you're talking about (since you have a hard time following the conventions). This is improved by the internet, since the barriers to entry have fallen down dramatically, and there might be a way to merge rigorous and nonrigorous thinking today in ways that were not possible in earlier times.
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+ You could be teaching - college level IMHO. Anyway, I've proved a few things, but the insight precedes the proof sometimes by years, and the proof goes through versions. (Differential execution, performance sampling, correctness of sorting algorithms.) – Mike Dunlavey May 29 '12 at 0:56
Well written answer! – A New Guy Nov 15 '12 at 14:42
1) Phase transition of the quantum Heisenberg ferromagnet (long range order at a positive temperature in more than 2 space dimensions). To my knowledge, this is still an open problem (in contrast to the quantum Heisenberg antiferromagnet).
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The AdS/CFT correspondence (conjecture) is not proved mathematically, yet.
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It is difficult to prove such a thing--- better to say it has not been fully constructed yet, and it hasn't been shown to pass all the insane consistency checks it requires. There is really no better definition than AdS/CFT for string theory on AdS. – Ron Maimon May 29 '12 at 4:29
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http://mathforum.org/mathimages/index.php?title=Volume_of_Revolution&diff=21266&oldid=8347
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# Volume of Revolution
### From Math Images
(Difference between revisions)
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| | {{Image Description Ready | | {{Image Description Ready |
| | |ImageName=Solid of revolution | | |ImageName=Solid of revolution |
| - | |Image=VolumeOfRev.jpg | + | |Image=Revolutionvolume1.gif |
| | |ImageIntro=This image is a solid of revolution | | |ImageIntro=This image is a solid of revolution |
| - | |ImageDescElem=This image shows the solid formed after revolving the region bounded by <math>y=x^2</math>, <math>y=0</math>,<math>x=0</math> and <math>x=1</math>, about the <math>x</math>-axis | + | |ImageDescElem=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 <math>{\pi} {r^2} h</math>. 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.[[Image:bread.gif|left|thumb| The disc method is much like slicing up bread and computing the volume of each slice http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html]] |
| - | |ImageDesc=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 <math>{\pi} {r^2} h</math>. 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.[[Image:bread.gif|left|thumb| The disc method is much like slicing up bread and computing the volume of each slice http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html]] | + | |ImageDesc===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'''. <br> | | 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'''. <br> |
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| - | If we revolve this area about the x axis (<math>y=0</math>), then we get the main image on the right hand side of the page[[Image:Revolution.gif|Right|thumb|This image shows a plane area being revolved to create a solid http://curvebank.calstatela.edu/volrev/volrev.htm]] | + | If we revolve this area about the x axis (<math>y=0</math>), then we get the image below to the left. [[Image:Revolution.gif|Right|thumb|This image shows a plane area being revolved to create a solid http://curvebank.calstatela.edu/volrev/volrev.htm]] |
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| | To find the volume of the solid using the disc method: | | To find the volume of the solid using the disc method: |
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| | Volume of one disc = <math>{\pi} y^2{\Delta x}</math> where <math>y</math>- which is the function- is the radius of the circular cross-section and <math>\Delta x</math> 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 one disc = <math>{\pi} y^2{\Delta x}</math> where <math>y</math>- which is the function- is the radius of the circular cross-section and <math>\Delta x</math> 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. |
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| | | + | [[Image:VolumeOfRev.jpg|left|thumb| [http://wikis.swarthmore.edu/miwiki/index.php/User:Lmasis1 Lmasis1] ]] |
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| | Volume of all discs: | | Volume of all discs: |
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| | Volume of all discs = <math>{\sum}{\pi}y^2{\Delta x}</math>, with <math>X</math> ranging from 0 to 1 | | Volume of all discs = <math>{\sum}{\pi}y^2{\Delta x}</math>, with <math>X</math> ranging from 0 to 1 |
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| - | If we make the slices infinitesmally thick, the Riemann sum becomes the same as: | + | If we make the slices infinitesimally thick, the Riemann sum becomes the same as: |
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| | <math>\int_0^1 {\pi}y^2\,dx ={\pi}\int_0^1 (x^2)^2\, dx</math> | | <math>\int_0^1 {\pi}y^2\,dx ={\pi}\int_0^1 (x^2)^2\, dx</math> |
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| - | Evaluating this intergral, | + | <br><br> |
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| | | + | Evaluating this integral, |
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| | In the example we discussed, the area is revolved about the <math>x</math>-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 <math>x</math>-axis, we can substitute it in the place of <math>x^2</math>. 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 <math>x</math>-axis. | | In the example we discussed, the area is revolved about the <math>x</math>-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 <math>x</math>-axis, we can substitute it in the place of <math>x^2</math>. 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 <math>x</math>-axis. |
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| - | ==References== | + | ==Washer Method== |
| - | Bread image http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html<br> | + | |
| - | Revolving image http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html | + | |
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| | | + | [[Image:cross_section.jpg|Cross Section|right]] |
| | | + | The washer method can be used when the rotated plane does not touch the axis around which it is being rotated. One instance in which the plane isn't touching the rotational axis is when the plane is not just bounded by one function, but instead two. For now we'll assume that one function is consistently smaller than the other, so there is a 'smaller function' and a 'larger function.' The main image on this page is an example of when the washer method is used. The top curve (which we will call f(x) ) is proportional to the square root of x, and the bottom curve (which we will call g(x) )is linear. The boundaries for the functions are x =2 and x = 10. A cross section is shown to the right. |
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| - | |AuthorName=Lizah Masis | + | The basic philosophy behind the washer method is the same as behind the disk method. We still must integrate around the rotational axis. The difference is that we cannot just use one radius (ie Radius = R - r). This wouldn't work because then two sections with the same area would necessarily have to have the same volume, but this is not the case. If two circles of the same radius are rotated around the same axis, if one is farther away, it will create more volume, as demonstrated in the animation below. Thus, instead of subtracting the radius of the smaller function from the bigger function, we subtract the volume the rotated smaller function would create from the volume the bigger function creates. The formula for the washer method is: <br /> |
| - | |Field=Algebra | + | <math> V = \pi \times \int(f(x)^2 - g(x)^2)dx </math> |
| - | |InProgress=No | + | |
| - | volume of solid= <math>{\pi\over 5} units^3</math> | + | |
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| | | + | <pausegif id="1" wiki="no">Rotating_circles.gif</pausegif> |
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| | | + | If one function is not consistently smaller than the other, we can break up the problem into two smaller problems. If the functions f(x) and g(x) cross at some arbitrary value c, we use f(x) as the larger function from our start value to c, but as the smaller function from c to the end value. If our start and end values are a and b respectively, the formula is: |
| | | + | <math> V = \pi \times \Big( \int_a^c (f(x)^2 - g(x)^2)dx + \int_c^b (g(x)^2 - f(x)^2)dx \Big) </math> |
| | | + | |
| | | + | ==References== |
| | | + | Bread image http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html<br> |
| | | + | Revolving image http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html |
| | | + | |AuthorName=Nordhr |
| | | + | |AuthorDesc=made in OpenGL |
| | | + | |SiteName=Nordhr |
| | | + | |SiteURL=http://wikis.swarthmore.edu/miwiki/index.php/User:Nordhr |
| | | + | |Field=Calculus |
| | |Pre-K=No | | |Pre-K=No |
| | |Elementary=No | | |Elementary=No |
| | |MiddleSchool=No | | |MiddleSchool=No |
| | |HighSchool=Yes | | |HighSchool=Yes |
| - | |other=Pre-calculus and elementary calculus | + | |HigherEd=Yes |
| - | |AuthorName=Lizah Masis | + | |
| - | |SiteURL=http://wikis.swarthmore.edu/miwiki/index.php/User:Lmasis1 | + | |
| - | |Field=Calculus | + | |
| | }} | | }} |
## Current revision
Solid of revolution
Field: Calculus
Image Created By: Nordhr
Website: Nordhr
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
# A More Mathematical Explanation
[Click to view A More Mathematical Explanation]
## Disk Method
In general, given a function, we can graph it then revolve the area under the curve b [...]
[Click to hide A More Mathematical Explanation]
## 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 image below to the left.
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 infinitesimally 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 integral,
${\pi}\int_0^1 x^4 dx$
=$[{{x^5\over 5} + C|}_0^1] {\pi}$
=$[{1\over 5} + {0\over 5}] {\pi}$
=${\pi}\over 5$
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.
## Washer Method
The washer method can be used when the rotated plane does not touch the axis around which it is being rotated. One instance in which the plane isn't touching the rotational axis is when the plane is not just bounded by one function, but instead two. For now we'll assume that one function is consistently smaller than the other, so there is a 'smaller function' and a 'larger function.' The main image on this page is an example of when the washer method is used. The top curve (which we will call f(x) ) is proportional to the square root of x, and the bottom curve (which we will call g(x) )is linear. The boundaries for the functions are x =2 and x = 10. A cross section is shown to the right.
The basic philosophy behind the washer method is the same as behind the disk method. We still must integrate around the rotational axis. The difference is that we cannot just use one radius (ie Radius = R - r). This wouldn't work because then two sections with the same area would necessarily have to have the same volume, but this is not the case. If two circles of the same radius are rotated around the same axis, if one is farther away, it will create more volume, as demonstrated in the animation below. Thus, instead of subtracting the radius of the smaller function from the bigger function, we subtract the volume the rotated smaller function would create from the volume the bigger function creates. The formula for the washer method is:
$V = \pi \times \int(f(x)^2 - g(x)^2)dx$
Click to stop animation.
If one function is not consistently smaller than the other, we can break up the problem into two smaller problems. If the functions f(x) and g(x) cross at some arbitrary value c, we use f(x) as the larger function from our start value to c, but as the smaller function from c to the end value. If our start and end values are a and b respectively, the formula is: $V = \pi \times \Big( \int_a^c (f(x)^2 - g(x)^2)dx + \int_c^b (g(x)^2 - f(x)^2)dx \Big)$
## References
Bread image http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html
Revolving image http://mathdemos.gcsu.edu/mathdemos/sectionmethod/sectionmethod.html
# Teaching Materials
There are currently no teaching materials for this page. Add teaching materials.
# About the Creator of this Image
made in OpenGL
Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.
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http://mathhelpforum.com/math-topics/200929-i-am-having-problems-related-series-sequences-here-those-questions.html
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1Thanks
• 1 Post By richard1234
# Thread:
1. ## I am having problems related to series and sequences and here are those questions.
Q1. The sum of first n terms(n>1) of an A.P. is 153 and c.d. is 2. If the first term is an integer, the number of possible values of n is?
Q2. The minimum number of terms of 1+3+5+7+ adds up to a number exceeding 1357 is?
Q3.The number of terms common to two A.P.s 3,7,11,...,407 and 2,9,16,...,709 is?
Q4. An A.P whose first term is unity and in which the sum of the first half of any number of terms to that of the second half of the same number of terms is in constant ratio, the c.d d is?
Q.5 If x,y,z are real numbers satisfying the expression 25(9x² + y²) + 9z² -15(5xy + yz + 3zx) = 0 then x,y,z are in?
a)AP
b)GP
c)HP
d)none
Here are some questions to which I really need an explained answer. I hope I will get answer soon
2. ## Re: I am having problems related to series and sequences and here are those questions
Originally Posted by lord287
Q1. The sum of first n terms(n>1) of an A.P. is 153 and c.d. is 2. If the first term is an integer, the number of possible values of n is?
Q2. The minimum number of terms of 1+3+5+7+ adds up to a number exceeding 1357 is?
Q3.The number of terms common to two A.P.s 3,7,11,...,407 and 2,9,16,...,709 is?
Q4. An A.P whose first term is unity and in which the sum of the first half of any number of terms to that of the second half of the same number of terms is in constant ratio, the c.d d is?
Q.5 If x,y,z are real numbers satisfying the expression 25(9x² + y²) + 9z² -15(5xy + yz + 3zx) = 0 then x,y,z are in?
a)AP
b)GP
c)HP
d)none
Here are some questions to which I really need an explained answer. I hope I will get answer soon
What have you tried?
3. ## Re: I am having problems related to series and sequences and here are those questions
Q1: You have $a + (a+2) + \dots + (a + 2n-2) = 153$. There are n terms, so $a$ appears n times. Also, $2 + 4 + \dots + (2n - 2) = 2(1 + 2 + \dots + (n-1) = n(n-1)$. So the LHS is equivalent to
$an + n(n-1) = 153 \Rightarrow n(a + n-1) = 153$. Positive factors of 153 are 1,3,9,17,51,153. However n cannot equal 1, but n can equal any other factor of 153, and there would be an integer value for $a$. So there are five values.
Q2: $1 + 3 + 5 + \dots + 2k-1 = k^2$. What is the smallest k such that $k^2 > 1357$?
Q3: Solve the congruence $n \equiv 3 (\mod 4)$ and $n \equiv 2 (\mod 7)$. By the Chinese remainder theorem, all solutions are congruent mod 77.
Q4: I don't really understand that question, but what if $d = 0$?
Q5: Expand everything out and multiply by 2:
$450x^2 + 50y^2 + 18z^2 - 150xy - 30yz - 90zx = 0$. Note that this factors to
$(15x - 5y)^2 + (15x - 3z)^2 + (5y - 3z)^2 = 0$
It is obvious that $15x = 5y$, $15x = 3z$, $5y = 3z$, because each of the squares must be zero. Solving for the variables yields $x = x$, $y = 3x$, $z = 5x$, so they must be in an arithmetic progression.
4. ## Re: I am having problems related to series and sequences and here are those questions
Originally Posted by richard1234
Q3: Solve the congruence $n \equiv 3 (\mod 4)$ and $n \equiv 2 (\mod 7)$. By the Chinese remainder theorem, all solutions are congruent mod 77.
Q4: I don't really understand that question, but what if $d = 0$?
THANX SO MUCH
And if u want I can give u the answer to the 4th question the answer to 4th is d=2, If u can do it now plz help me.
And can u explain Q3 with a different method I have not done the Chinese remainder theorem. I am really not understanding what u did in Q3. And given answer of Q3 is 14. Plz give a little more details for these two answers if u can.
Also plz tell why u took out the positive factors of 153 in question 5 and will the integer value of 'a' be fixed or the 5 values of 'n' will be satisfied for any value of 'a'?
5. ## Re: I am having problems related to series and sequences and here are those questions
Q1: $n(a + n - 1) = 153$. Both numbers are (positive) integers so you want to find factors of 153 that satisfy n (greater than 1). Once you fix a factor for n, there is an integer value for $a$.
Q3: You know mods, right? If not, I suggest you look them up.
You can solve the congruence, or you can simply find the first integer that is both 3 mod 4 and 2 mod 7. This happens to be 23. By the Chinese remainder theorem (see link below), all other solutions must be congruent mod 28 (Idk what I was thinking when I said 77...sorry about that). Therefore the solutions are 23, 51, 79, 107, ..., 387 (415 is too large). How many numbers are in the set {23, 51, 79, ..., 387}?
Chinese remainder theorem - Wikipedia, the free encyclopedia
$Q4$: So if you have 2n terms with common difference d, you're saying that $k(1 + (1+d) + \dots + (1+(n-1)d)) = 1 + nd + 1 + (n+1)d + \dots + (1 + (2n-1)d)$?
If so, I can see how d = 2 (since the sum 1 + 3 + ... + 2n - 1 is a perfect square; the first half of the terms sum up to ~1/4 of the total sum). But what about d = 0? 1,1,1,1... is technically considered an arithmetic sequence.
6. ## Re: I am having problems related to series and sequences and here are those questions
More questions:-
Q. If a + b + c = 3, a>0, b>0, c>0, then the greatest value of a²b³c² is
Q. The sum of 0.2 + 0.004 + 0.00006 +...... ∞: is
Ans. 2000/9801
Q. The sum pf all the numbers of the form n3 which lie between and 10000 is?
Ans. 53261
Q. If harmonic progression Hn = 1 + 1/2 +.......+ 1/n then value of 1 + 3/2 + 5/3 +.....+ 2n-1/n is
a) Hn + n
b) 2n- Hn
c) (n-1) + Hn
d) Hn + 2n
Sorry if u people are finding these questions by me easy but they r not easy for me since I am only in grade 10+1
7. ## Re: I am having problems related to series and sequences and here are those questions
Originally Posted by lord287
More questions:-
Q. If a + b + c = 3, a>0, b>0, c>0, then the greatest value of a²b³c² is
Q. The sum of 0.2 + 0.004 + 0.00006 +...... ∞: is
Ans. 2000/9801
Q. The sum pf all the numbers of the form n3 which lie between and 10000 is?
Ans. 53261
Q. If harmonic progression Hn = 1 + 1/2 +.......+ 1/n then value of 1 + 3/2 + 5/3 +.....+ 2n-1/n is
a) Hn + n
b) 2n- Hn
c) (n-1) + Hn
d) Hn + 2n
Sorry if u people are finding these questions by me easy but they r not easy for me since I am only in grade 10+1
Q2.
$\displaystyle \begin{align*} 0.2 + 0.004 + 0.00006 + \dots &= \frac{2}{10} + \frac{4}{1000} + \frac{6}{100000} + \dots \\ &= \frac{2}{10}\left(\frac{1}{1} + \frac{2}{100} + \frac{3}{10000} + \dots \right) \\ &= \frac{1}{5}\sum_{n = 0}^{\infty}\frac{n + 1}{100^n} \\ &= \frac{1}{5}\left[ \sum_{n = 0}^{\infty}\frac{n}{100^n} + \sum_{n = 0}^{\infty}\left(\frac{1}{100}\right)^n \right] \end{align*}$
The second sum is geometric. I'm not sure how to evaluate the first...
8. ## Re: I am having problems related to series and sequences and here are those questions
Q3 should be pretty simple. What are all the cubes from 1 to 10000, and how do you sum the first n cubes?
I'm still figuring out Q1 but it probably involves AM-GM inequality.
9. ## Re: I am having problems related to series and sequences and here are those questions
Q2: Series obviously converges. Also, note that
$\frac{2}{10} + \frac{4}{1000} + \frac{6}{100000} + \dots = (\frac{2}{10} + \frac{2}{1000} + \dots) + (\frac{2}{1000} + \frac{2}{100000} + \dots) + \dots$
Factor stuff out.
$= (\frac{2}{10} + \frac{2}{1000} + \dots)(1 + \frac{1}{100} + \frac{1}{10000} + \dots)$
Both series are geometric, it shouldn't be hard to compute the sum.
10. ## Re: I am having problems related to series and sequences and here are those questions
Originally Posted by richard1234
Q2: Series obviously converges. Also, note that
$\frac{2}{10} + \frac{4}{1000} + \frac{6}{100000} + \dots = (\frac{2}{10} + \frac{2}{1000} + \dots) + (\frac{2}{1000} + \frac{2}{100000} + \dots) + \dots$
Factor stuff out.
$= (\frac{2}{10} + \frac{2}{1000} + \dots)(1 + \frac{1}{100} + \frac{1}{10000} + \dots)$
Both series are geometric, it shouldn't be hard to compute the sum.
Didn't understand how did u factor that stuff out. I had been trying to factor it but could not and I am not understanding how come the way u have written it is the factor of that equation??
11. ## Re: I am having problems related to series and sequences and here are those questions
Originally Posted by richard1234
Q3 should be pretty simple. What are all the cubes from 1 to 10000, and how do you sum the first n cubes?
I'm still figuring out Q1 but it probably involves AM-GM inequality.
Oh I am sorry I wrote wrongly the real question was sum of cubes between 100 and 10000, not 1 to 10000
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http://math.stackexchange.com/questions/63969/is-it-possible-to-have-a-field-without-an-additive-identity
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# Is it possible to have a field without an additive identity?
If I drop the axiom that Zero is the identity of an addition what consequences does this entail? What do I need to change to my axiomatization?
By definition it is not possible, but are there mathematical structures like a field but without an additive identity?
I don't have a concrete example/issue but I am interested in this question.
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Needs are relative to purposes. Need to do what? – Qiaochu Yuan Sep 12 '11 at 19:58
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Without an identity element, you cannot define inverses. – Rasmus Sep 12 '11 at 20:00
Do you want an additive identity, or a multiplicative identity? You original post had $0$ in it, now it says "identity element". – Arturo Magidin Sep 12 '11 at 20:13
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@Stephan: in your last sentence, what is "it"? – Qiaochu Yuan Sep 12 '11 at 21:10
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Is it so hard to imagine a "field" without a zero? Why is this such a silly question to you. I wanted to know if, like @Zhen wrote, there is a mathematical structure like a field but without a zero. So if I add x and -x it would disappear instead of becoming a 0. Is it possible to have such a structure and how does it look like and what cant you do with it. In my humble opinion some of you take this Q&A site too serious and should stick to questions of their level instead of making some novices feel dumb and get lost in personal issues... – Stephan Schielke Sep 13 '11 at 6:03
show 9 more comments
## 6 Answers
Yes. It is part of the definition of a field (and, more generally, a ring).
People do study various alterations to the concept of field, such as
but, I think it is worth mentioning, all of these have an identity element for the $+$ operation. It is a pretty fundamental thing to want one's mathematical structures to have. I'm not saying that structures without an additive identity aren't worth studying, just that they would behave so differently from fields I don't see any reason to even say "It's like a field, but..."
Edit: Looking at the page on semifields, it appears that an alternative definition of the term does in fact allow for not having an additive identity. So much for objects getting names that make sense...
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Do you get advertising revenue from wikipedia or something ??! ;) – The Chaz 2.0 Sep 14 '11 at 1:17
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Haha I wish :) I just usually err on the side of putting more links in my answers, than fewer (as long as they're all relevant). – Zev Chonoles♦ Sep 14 '11 at 1:26
And I appreciate it! I recently saw a diagram that related (if memory serves...) monoids, fields, groups, etc. Maybe I can find the link... – The Chaz 2.0 Sep 14 '11 at 1:29
The usual definition of a field is:
Definition. A set $F$ together with two binary operations $+$ and $\times$ is a field if and only if:
1. $+$ is associative;
2. $+$ is commutative;
3. $+$ has a neutral element $0$;
4. For every $a\in F$ there exists $b\in F$ such that $a+b=0$.
5. $\times$ is associative;
6. $\times$ is commutative;
7. $\times$ distributes over $+$;
8. There is an element $1\neq 0$ such that $a\times 1=a$ for all $a\in F$;
9. For each $a\in F$, if $a\neq 0$ then there exists $x\in F$ such that $a\times x = 1$.
Now, you can certainly ask whether the axioms are independent. For example, it is easy to see that if you drop the assumption of commutativity for $+$, then you can deduce it from the other eight axioms; on the other hand, the real quaternions show that you cannot drop the assumption that $\times$ is commutative and deduce it from the other axioms.
So you could be asking whether Axiom 3 is independent of the other axioms. The main difficulty with dropping Axiom 3 is that without it Axiom 4 becomes unintelligible, and Axiom 9 is also problematic. So before we drop Axiom 3, we need to replace Axioms 4 and 9 with something else that, together with Axiom 3 give a field, but which make sense in the absence of Axiom 3.
There are plenty of ways of defining "abelian group" without specifying the existence of a neutral element (e.g., "for every $a,b\in F$ there exists $x\in F$ such that $a+x=b$") replacing 4 with something like this will automatically imply the existence of a $0$. Do you have something specific in mind?
(For a similar question, see for example this sci.math post by Dave Rusin where he discusses the independence of the axioms of a vector space, where he faces a similar problem with dropping "existence of $0$")
Added. While I was writing this, the title of the post was changed to "without an identity" instead of "without a $0$". You run into similar problems: if you drop Axiom 8, you need to replace Axiom 9 with something that still makes sense; depending on what you replace it with, it may or may not imply the existence of a multiplicative identity in the presence of the other axioms. Again, the question is whether you have something in mind or not.
If the question is just a poorly phrased way of asking if the one element ring is a field, the answer is that it is not considered to be a field. There are good reasons for this, even though the one element ring satisfies all the axioms except for the $1\neq 0$ clause of Axiom 8.
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Hmmm. If Axiom 3 is dropped and Axiom 4 is replaced with 4a: 'For every $a, b\in F$ there exists $c\in F$ s.t. $a+c=b$' then you still recover 3 by letting $b=a$ in 4a. But if we restrict 4a to $b\neq a$ then there aren't any immediately obvious 'hooks' to get back a zero... – Steven Stadnicki Jul 1 '12 at 16:10
This depends on the axioms, of course. There are two common ways to axiomatize fields. I was taught using the second approach. In both a field is composed of $\langle F,0,1,+,\cdot\rangle$
The First Way: Groups.
1. $\langle F,0,+\rangle$ is an abelian group;
2. $\langle F\setminus\{0\},1,\cdot\rangle$ is an abelian group;
3. Distributivity, that is $a\cdot(b+c)=a\cdot b+a\cdot c$
In this axiomatization, since a group cannot be empty, we have to have $0\neq 1$, otherwise the second condition will not hold.
The Second Way: Specifying the axioms.
1. $\forall a\forall b(a+b=b+a)$ (Commutativity of addition)
2. $\forall a(a+0=a)$ (Zero is the identity of addition)
3. $\forall a\exists b(a+b=0)$ (Addition is invertible)
4. $\forall a\forall b\forall c(a+(b+c)=(a+b)+c)$ (Addition is associative)
5. $\forall a\forall b(a\cdot b=b\cdot a)$ (Multiplication is commutative)
6. $\forall a(a\cdot 1=a)$ (One is the identity of multiplication)
7. $\forall a\exists b(a\neq 0\rightarrow a\cdot b=1)$ (Non-zero elements are multiplication-invertible)
8. $\forall a\forall b\forall c(a\cdot(b\cdot c)=(a\cdot b)\cdot c))$ (Multiplication is associative)
9. $\forall a\forall b\forall c(a\cdot(b+c)=(a\cdot b)+(a\cdot c))$ (Distributivity of multiplication over addition)
Note that these axioms are satisfied by $\{e\}$, by interpreting $e=0=1$. This sort of axiomatization allows the field with one element to exist.
If however, to the second approach we add $0\neq 1$ then we have equivalence between the two ways.
In both these cases we have $1$ defined in the language, and so we have to interpret it somehow, even if we do allow $0=1$, we still have $1\in F$. Simply because it is embedded into the definition of a field and cannot be avoided.
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I am confused, isn't this answer essentially the same as those in math.stackexchange.com/questions/164811 which garnered over 30 upvotes combined, and not a single downvote? Why is this answer being downvoted? – Asaf Karagila Jun 30 '12 at 23:11
@Ben: there will be a edit log! But okay... – Asaf Karagila Jul 1 '12 at 15:20
Yes, by definition. The need for an identity element is deeply entrenched in group theory (otherwise there could be no distinction between groups and monoids), and the fact that the additive group of a field is, indeed, a group, is similarly entrenched in field theory.
Of course one could investigate the consequences of the field axioms without an additive identity -- but one would have to be careful about selecting which axioms that means; the usual textbook presentations rarely distinguish between variants of the axioms that can be shown to be equivalent given the existence of an additive identity. In any case, whatever results you reach then would certainly not be results about fields as everyone understand them. Unless, that is, you happened to pick a selection of axioms that imply that zero must exist anyway. For example, it seems to be hard to require any useful properties of subtraction without allowing a proof that $x-x=y-y$ for all $x$ and $y$, and that this common value is an additive identity.
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Suppose that we talk about a unary function "-" defined on F with -(0)=0, and -(-(x))=x. Note that instead of talking about a "inverses" for a field or group, we could just talk about those structures with such a unary function. Now let ⟨F,+, -⟩ satisfy the following properties:
1. For all x, y, (x+y)=(y+x)
2. For all x, y, z ((x+y)+z)=(x+(y+z))
3. For all x, (x+-(x))=0
I simply don't see how this implies that "for all x, (x+0)=x", so I don't see why we can't consider a field-like structure which drops only the axiom of an additive identity. How does this work? We basically have a commutative, associative structure which has an inverse also, since 3. comes as basically equivalent to the notion of having an additive inverse. So, now look at Asaf's way of defining a field. Instead of having ⟨F,0,+⟩ as an Abelian group, we have ⟨F,0,+⟩ as a structure where + satisfies association, commutation, and property 3. above. We leave the rest of Asaf's way of defining a field intact, so it comes as possible to only drop the axiom of an additive inverse for a field.
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Of course your axioms do not imply $x+ 0 = x$ for all $x$. Take for instance any set $X$ with more than one element: call one of the elements $0$. Define $-x = x$ for all $x \in X$ and $x+y = 0$ for all $x,y \in X$. Then $0$ is an absorbing element, not an identity....But this structure is nothing like a field -- indeed, you only have one binary operation! So what is your point here? – Pete L. Clark Sep 14 '11 at 1:18
@Pete As I try to clear up in the edit, we now have a structure which satisfies the property of having an inverse (3. comes as equivalent to such a property, as I understand it), and satisfies addition and commutation. In other words, it satisfies all axioms of an additive Abelian group without the existence of an additive identity, and it comes as consistent. So, if you join in the other axioms for a field under a formulation like that of Asaf's, the set of axioms for a field without that of an additive inverse is NOT inconsistent. So, such structures DO come as possible. – Doug Spoonwood Sep 14 '11 at 1:48
$@$Doug: if you are saying that your modified set of axioms is consistent, then this is obvious because any field satisfies these axioms. If you are saying this axiom system together with the additional axiom that $0$ is not an additive identity is consistent: yes, this is true. Take any field $F$ and redefine the addition operation to be identically $0$ as above, define $-x = x$ for all $x$, and keep the same multiplication operation. – Pete L. Clark Sep 14 '11 at 5:11
Two questions: 1) Who said that such structures were not possible? Certainly not me: I was the one who gave you a model! (You said "I simply don't see how this implies...", which is not a proof by anyone's standard.) 2) In what sense is a structure in the above example "like a field" as the OP requests? What results can you prove about such structures? Being given by a similar list of axioms is a very superficial similarity between structures. – Pete L. Clark Sep 14 '11 at 5:11
@Pete No you didn't say such weren't possible. Sorry if it seemed like I implied that you did. The OP as it stands currently says this... and yes I certainly did not give a proof. Yes, this answer doesn't give as much information as the OP requests. But, I'd think you'll probably have to use a formulation like this, because of what Arturo says (if he's correct) "So before we drop Axiom 3, we need to replace Axioms 4 and 9 with something else that, together with Axiom 3 give a field, but which make sense in the absence of Axiom 3." – Doug Spoonwood Sep 14 '11 at 12:32
If I understand the definition of a field via abelian groups and distributivity as in Asaf's answer correctly, the perhaps surprising answer is no, you don't need to have an identity element in your axiom set for a field. The trick lies in that you don't need an identity element in the axioms for a group. Group theory, in fact, can get founded upon a single axiom, or several axioms which make no reference to a neutral element, if you really want that. I've only glanced at things here, so see 1 2 and 3 for particular axiom sets. Once you pick an appropriate axiom set for groups, you use the first part of Asaf's procedure for defining a field, which implies no mention of an identity element in the axiom set for fields. The existence of inverses also doesn't mean one needs to talk about the identity element in the axiom set. One could rewrite such existence statements via unary functions as one might do in the context of universal algebra, and this does capture the concept here, since inverses can get proved unique.
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The idea that axioms in a formal language "make reference" to semantic concepts like the existence of an additive identity is a somewhat naive philosophical stance that seems entirely antithetical to the relentless syntacticism you have been espousing on this site. If I wanted to play the usual Doug Spoonwood role I would say: "The existence of an additive identity comes as a certain WFF. You either have that WFF as an axiom or you don't. Other axioms might imply the existence of an additive identity, but that doesn't come as the same thing as having the additive identity axiom." – Pete L. Clark Sep 13 '11 at 14:59
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I confess that I am not too surprised that I inadvertently clarified your position. But...you now believe that informal mathematical statements may be formalized as WFFs in multiple ways and that this is not problematic? That's good news. – Pete L. Clark Sep 14 '11 at 0:17
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By the way, the standard symbols for universal and existential quantifiers are $\forall$ and $\exists$, not ! and @. The advantage of using $\forall$ and $\exists$ is that a worldwide mathematical/philosophical audience will understand you without your having to define your symbols in each comment...and incidentally, you will look less silly. – Pete L. Clark Sep 14 '11 at 0:20
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I intentionally did not put all the parentheses in to see if you now understand that people refer to WFFs by standard shorthands rather than writing them out formally. (Why did you not complain about the $\ldots$? That's definitely forbidden in WFFs.) If you are still hung up on these syntactic points, then I suspect that what I am actually trying to say is falling on deaf ears. (By the way, your language is not very polite. I have a PhD in mathematics and have published a paper in mathematical logic. Are you sure you want to accuse me of not understanding WFFs?) – Pete L. Clark Sep 14 '11 at 1:30
1
$@$Doug: do you really not understand that I can unambiguously specify a WFF without writing it out in full, and that reasoning with and proving things about WFFs is not the same as writing them down? And why do you refuse to use $\forall$ and $\exists$? I am really having a hard time believing that you are making these comments in good faith. – Pete L. Clark Sep 14 '11 at 5:18
show 7 more comments
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http://stats.stackexchange.com/questions/5747/if-a-and-b-are-correlated-with-c-why-are-a-and-b-not-necessarily-correlated
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# If A and B are correlated with C, why are A and B not necessarily correlated?
I know empirically that is the case. I have just developed models that run into this conundrum. I also suspect it is not necessarily a yes/no answer. I mean by that if both A and B are correlated with C, this may have some implication regarding the correlation between A and B. But, this implication may be weak. It may be just a sign direction and nothing else.
Here is what I mean... Let's say A and B both have a 0.5 correlation with C. Given that, the correlation between A and B could well be 1.0. I think it also could be 0.5 or even lower. But, I think it is unlikely that it would be negative. Do you agree with that?
Also, is there an implication if you are considering the standard Pearson Correlation Coefficient or instead the Spearman (rank) Correlation Coefficient? My recent empirical observations were associated with the Spearman Correlation Coefficient.
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19
An example is to take $A=X$, $B=Y$, and $C=X+Y$. We can take $X$ and $Y$ to be independent, yet both $A$ and $B$ are correlated (positively, Pearson) with $C$. – G. Jay Kerns Dec 25 '10 at 23:51
Thanks, that's actually a great comment. Short, but it captures the essence of the reason why it is so. – Gaetan Lion Dec 26 '10 at 0:07
## 6 Answers
Because correlation is a mathematical property of multivariate distributions, some insight can be had purely through calculations, regardless of the statistical genesis of those distributions.
For the Pearson correlations, consider multinormal variables $X$, $Y$, $Z$. These are useful to work with because any non-negative definite matrix actually is the covariance matrix of some multinormal distributions, thereby resolving the existence question. If we stick to matrices with $1$ on the diagonal, the off-diagonal entries of the covariance matrix will be their correlations. Writing the correlation of $X$ and $Y$ as $\rho$, the correlation of $Y$ and $Z$ as $\tau$, and the correlation of $X$ and $Z$ as $\sigma$, we compute that
• $1 + 2 \rho \sigma \tau - \left(\rho^2 + \sigma^2 + \tau^2\right) \ge 0$ (because this is the determinant of the correlation matrix and it cannot be negative).
• When $\sigma = 0$ this implies that $\rho^2 + \tau^2 \le 1$. To put it another way: when both $\rho$ and $\tau$ are large in magnitude, $X$ and $Z$ must have nonzero correlation.
• If $\rho^2 = \tau^2 = 1/2$, then any non-negative value of $\sigma$ (between $0$ and $1$ of course) is possible.
• When $\rho^2 + \tau^2 \lt 1$, negative values of $\sigma$ are allowable. For example, when $\rho = \tau = 1/2$, $\sigma$ can be anywhere between $-1/2$ and $1$.
These considerations imply there are indeed some constraints on the mutual correlations. The constraints (which depend only on the non-negative definiteness of the correlation matrix, not on the actual distributions of the variables) can be tightened depending on assumptions about the univariate distributions. For instance, it's easy to see (and to prove) that when the distributions of $X$ and $Y$ are not in the same location-scale family, their correlations must be strictly less than $1$ in size. (Proof: a correlation of $\pm 1$ implies $X$ and $Y$ are linearly related a.s.)
As far as Spearman rank correlations go, consider three trivariate observations $(1,1,2)$, $(2,3,1)$, and $(3,2,3)$ of $(X, Y, Z)$. Their mutual rank correlations are $1/2$, $1/2$, and $-1/2$. Thus even the sign of the rank correlation of $Y$ and $Z$ can be the reverse of the signs of the correlations of $X$ and $Y$ and $X$ and $Z$.
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whuber, what are "multinormal variables"? – Gaetan Lion Dec 26 '10 at 0:19
– whuber♦ Dec 26 '10 at 0:20
whuber, thanks the link was really helpful. – Gaetan Lion Dec 28 '10 at 22:36
As usual, a most thorough explanation you get a well deserved "Best Answer" check mark. – Gaetan Lion Dec 30 '10 at 21:17
@Gaetan Lion You are very kind. I have enjoyed reading all the answers to this question (and marked them all up). – whuber♦ Dec 30 '10 at 22:13
I'm on an annual fishing trip right now. There is a correlation between the time of day I fish and the amount of fish I catch. There is also a correlation between the size of the bait I use and the amount of fish I catch. There is no correlation between the size of the bait and the time of day.
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Basil, I love it! +1 for a plain English explanation. – Gaetan Lion Dec 28 '10 at 22:35
Best. Answer. On stats.stackexchange. Ever – Chris Beeley Feb 9 '12 at 20:22
I will leave the statistical demonstration to those who are better suited than me for it... but intuitively say that event A generates a process X that contributes to the generation of event C. Then A is correlated to C (through X). B, on the other hand generates Y, that also shapes C. Therefore A is correlated to C, B is correlated to C but A and B are not correlated.
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1
@Nice. I think you mean "A and B are not correlated" in the very last part of your last sentence. – suncoolsu Dec 25 '10 at 21:05
Yes, Nico with suncoolsu correction... this is a reasonably good explanation. You are partially describing Path Analysis. – Gaetan Lion Dec 25 '10 at 23:10
Yes, sorry, I got mixed up with the letters ;) – nico Dec 26 '10 at 8:17
Correlation is the cosine of the angle between two vectors. In the situation described, (A,B,C) is a triple of observations, made n times, each observation being a real number. The correlation between A and B is the cosine of the angle between $V_A=A-E(A)$ and $V_B=B-E(B)$ as measured in n-dimensional euclidean space. So our situation reduces to considering 3 vectors $V_A$, $V_B$ and $V_C$ in n dimensional space. We have 3 pairs of vectors and therefore 3 angles. If two of the angles are small (high correlation) then the third one will also be small. But to say "correlated" is not much of a restriction: it means that the angle is between 0 and $\pi/2$. In general this gives no restriction at all on the third angle. Putting it another way, start with any angle less than $\pi$ between $V_A$ and $V_B$ (any correlation except -1). Let $V_C$ bisect the angle between $V_A$ and $V_B$. Then C will be correlated with both A and B.
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+1 correlation in terms of an angle between multi-dimensional vectors is intuitive for me. – pate Dec 26 '10 at 23:55
+1: Nice geometric explanation. – whuber♦ Dec 28 '10 at 22:52
I think it's better to ask "why SHOULD they be correlated?" or, perhaps "Why should have any particular correlation?"
The following R code shows a case where x1 and x2 are both correlated with Y, but have 0 correlation with each other
````x1 <- rnorm(100)
x2 <- rnorm(100)
y <- 3*x1 + 2*x2 + rnorm(100, 0, .3)
cor(x1,y)
cor(x2,y)
cor(x1,x2)
````
The correlation with Y can be made stronger by reducing the .3 to .1 or whatever
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Unfortunately, I am not an R user. So, the codes above mean less to me than they mean to you. – Gaetan Lion Dec 25 '10 at 23:07
1
@Gaetan Lion: in this code, $x_1$ and $x_2$ are independent root normals, and $y = 3x_1 + 2x_2$ plus a normal noise term with standard deviation of 0.3. Clearly $y$ is positively correlated to $x_1$ and $x_2$, which are independent. – shabbychef Dec 26 '10 at 6:08
As an add-on to whuber's answer: The presented formula
$1 + 2 \rho \sigma \tau - \left(\rho^2 + \sigma^2 + \tau^2\right) \ge 0$.
can be transformed into following inequality (Olkin, 1981):
$\sigma\tau - \sqrt{(1-\sigma^2)(1-\tau^2)} \le \rho \le \sigma\tau + \sqrt{(1-\sigma^2)(1-\tau^2)}$
A graphical representation of the upper and lower limits for $\rho$ looks like:
Olkin, I. (1981). Range restrictions for product-moment correlation matrices. Psychometrika, 46, 469-472. doi:10.1007/BF02293804
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http://math.stackexchange.com/questions/91350/multiple-solutions-for-both-primal-and-dual
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# Multiple solutions for both primal and dual
If matrix $A$ in an LP (or $A^T$ in its dual) has full row (column- in dual) rank, is it possible that both primal and dual have multiple solutions?
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## 1 Answer
According to the table you have mentioned, there is no record for multiple implies multiple, every multiple solution will be expressed as degenerate in dual. Therefore, this is impossible that both have multiple solution.
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Thank you for your reply. The question now is if it is impossible to have multiple solutions for both primal and dual, why the first line of the table is not: --Multiple implies "Unique" and degenerate'-- (like the third line) – Quick Dec 15 '11 at 1:41
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http://mathhelpforum.com/statistics/178389-how-set-up-t-test-print.html
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# How to set up this T-test
Printable View
• April 23rd 2011, 10:10 AM
bodhisattva87
How to set up this T-test
Need some help setting up a statistical analysis for a project. I'm going to try to just give an example similar to my project rather than take the time to explain it (not that it's that complicated. it would just require a lengthy explanation). Here goes.
Let's say I have an instrument that's measuring the distance from itself to an object that is in motion. The instrument takes measurements every second and records it. I run the exact same test four times with the object following exactly the same path every time. Unfortunately, I wind up with four slightly different sets of data. I want to evaluate the performance of the instrument by comparing these four sets.
My professor told me I should use a T-Test. I'm fairly certain that I have to pick two sets of data at a time and acquire the difference between the two at each point in time in order to eliminate one of the variables (instead of time and the actual location of the object, the only variable will be time). I want to see if the difference between two (or more) of these sets is significant and use either a 90% or 95% confidence interval. Can anybody help me set this up? I haven't done this kind of stuff in a long time and I can't quite figure it out.
My data is all in excel and ready to go. I just need some guidance. Btw, there are about 1500 points in each set so I wouldn't be able to post my info.
Thanks!
• April 23rd 2011, 04:19 PM
Effendi
You can only check two sets of data against each other at a time. You need to have a parameter in mind, are the data sets matched (would a point on one data set correspond to a point on another set)? You might want to check the data, to see if it's normally distributed, it doesn't need to be perfect or even remotely symmetric, just so long as it's single peaked and doesn't have any outliers. For a matched pairs T-Test you find T using the formula: $\bar{x}$/( $s$/ $\sqrt{n}$) where $\bar{x}$ is the mean difference between the sets. You would have 1499 degrees of freedom, I don't think there are any tables that can precisely match your t-value to a precise p value when you have that many degrees of freedom. If unmatched you want a two sample T-test, to calculate T: ( $\bar{x}$1- $\bar{x}$2)/squarerootof(s1/n1 + s2/n2) where s1 and s2 are the standard deviations of the data sets. I can explain the rest when I know if the data sets are matched.
All times are GMT -8. The time now is 05:46 PM.
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http://physics.stackexchange.com/questions/62769/volume-of-gas-at-which-relative-fluctuation-of-gas-density-occurs/62787
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# Volume of gas at which relative fluctuation of gas density occurs
I have the following question:
In what volume of gas occurs 10 % relative fluctuation of gas density under pressure of $10^5\text{ Pa}$ and temperature of $293.15\text{ K}$?
I don't understand the topic but I assume this is about ideal gas. Can you please explain this to someone who has just high school physics knowledge?
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## 1 Answer
You can compute relative fluctuation of gas volume (which is the same as fluctuation of gas density here) by computing probabilities using entropy (or equivalently Gibbs energy) difference. The following page has explained all the steps.
The final formula would be: $\delta = \langle {(V - V_0)^2 \over V^2} \rangle = {1 \over N}$
In which $N$ is number of atoms which we can compute from ideal gas state equation: $N = {PV \over kT}$
So $\delta = \langle {(V - V_0)^2 \over V^2} \rangle = {k T \over P V}$
$\delta = \langle {(V - V_0)^2 \over V^2} \rangle = 0.01$
$V = {k T \over P \delta} = 4\times 10^{-24} ~\text m^3 = 4000 ~\text{nm}^3$
This volume is small; you need very few atoms to have such a huge fluctuations.
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Can you please explain in detail where did the original values go? (Or how did we get to 0.01 and 4*10^-24.) The page you are linking to is above my level of understanding :- / – Andrew123321 Apr 30 at 14:04
${(V - V_0) \over V} = 0.1$ So approximately $\langle {(V - V_0)^2 \over V^2} \rangle = 0.01$ – Azad Apr 30 at 14:39
@Andrew123321 May I ask what grade are you in and where you see the question? – Azad Apr 30 at 14:42
I see. Can you please also expand on 4*10^-24? I am an undergraduate student of computer science and this is an excercise from physics class I've taken. – Andrew123321 Apr 30 at 14:52
Well, $k$ is the Boltzmann constant $k = 1.38 \times 10^{-23} J/K$ , $T = 293.15 K$ , $P=10^5 Pa$ , $\delta = 0.01$. Just put them in the last formula. – Azad Apr 30 at 16:22
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http://physics.stackexchange.com/questions/31211/rotation-of-parabola?answertab=active
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Rotation of parabola
I'm designing a parabolic solar concentrator and am doing ray-tracing on Matlab from scratch. I'm beginning to look into compound concentrators and would need to have an equation for a parabola rotated by x degrees centered at a specific (non origin) point. The equation I am using for my parabolas is $a*(x-X)^2 - Y$.
I am not sure how to rotate this equation besides using a rotation matrix, solving for y using quadratic equation and choosing one term. Even this method gives me another parabola in $y(x)$.
Any suggestions on a simple way to get half of a parabola rotated by $\theta$? Should I be looking into parameterization?
Thanks
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Do you want it rotated around the central axis of the parabola, or an arbitrary axis? I would think that rotation around the central axis would be relatively easy, and that rotation around an arbitrary axis would be much more difficult. – Colin McFaul Jul 3 '12 at 19:05
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A rotated parabola is not a function and therefore you only choice is to use a rotation matrix. – ja72 Jul 3 '12 at 19:06
I think it is easier to rotate the rays, and keep the paraboloid fixed. – Bernhard Jul 3 '12 at 19:21
Why are you ray tracing in the first place? – Colin K Jul 3 '12 at 19:25
– Shrim Jul 3 '12 at 21:11
show 2 more comments
1 Answer
I don't know if this is useful, but I would proceed with the parametrization and the rotation matrix, anyway.
Let us rename $x-X\rightarrow x$. Then, notice that the equation of the parabola $y = a x^2$ can be parametrized by $x = t$, $y = a t^2$, as $t$ goes from $-\infty$ to $\infty$; or, as a vector,
$$(x(t), y(t))=(t,a t^2)$$
To rotate the graph of the parabola about the origin, you must rotate each point individually. Rotation clockwise by an angle $\theta$ is a linear transformation with matrix
$$\left( \begin{array}{ccc} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \\ \end{array} \right)$$
Thus, if we apply this linear transformation to a point $(t, t^2)$ on the graph of the parabola, we get
$$\left( \begin{array}{ccc} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \\ \end{array} \right) \left( \begin{array}{ccc} t \\ a t^2\\ \end{array} \right) = \left( \begin{array}{ccc} t\cos\theta +a t^2\sin\theta\\ -t\sin\theta+a t^2\cos\theta\\ \end{array} \right)$$
So, as $t$ goes from $-\infty$ to $\infty$, this is a parametrization of the graph of the rotated parabola.Then you have to convert back to $x$ and $y$, put them in the equation $y=x^2$ and that's it.
To get a cartesian equation for the new parabola, you can just solve for $t$ in the first line $a t^2 + t cot\theta = x/\sin\theta$ and put the expression for $t$ in the second one. Doing this, you have an equation for $x$ and $t$ that corresponds to the "constraint" $x$ and $y$ must satisfy to be on the new parabola!
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so from the rotated parameterized matrix at the end, converting this back to x and y simply gives [xcos() + ysin(), -xsin() + ycos()], which is just [x,y] rotated by theta, correct? Is there a specific way to convert this back into cartesian to be useful? – Shrim Jul 3 '12 at 20:44
Thanks for the response, btw – Shrim Jul 3 '12 at 20:44
I edited, maybe this is more clear! Isn't it? – usumdelphini Jul 3 '12 at 21:09
yea, that makes sense to me. Thanks for the help – Shrim Jul 3 '12 at 23:07
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Good idea! I'd add that leaving things in parametric form may be useful, as it makes computing a surface normal trivial. – Colin K Jul 4 '12 at 1:54
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http://crypto.stackexchange.com/questions/5475/looking-for-examples-for-proof-by-reduction
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# Looking for examples for “proof by reduction”
Im looking for examples for a proof by reduction. For example:
Let $A=(Gen, H)$ be a hash function. We define a new Hash function $A'=(Gen',H')$ with
Gen=Gen'
$H'_s(x)=H_s(H_s(x))$
It should be shown that if A is collision resistant, then A' is collision resistant too.
• I DONT LOOK FOR A SOLUTION FOR THIS QUESTION! *
Im looking for an idea how to construct a reduction so that i can solve this problem on my own.
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(Gen, H) $\:$ seems like it would be a hash family, rather than a hash function. $\hspace{1.5 in}$ – Ricky Demer Nov 25 '12 at 12:07
@Ricky: You are (formally) right. I have no idea how to add an upper s to H;-) But it is correct: in the line with H'(x)=H(H(x)) there are upper s missing on H. In "Katz/Lindell: Introduction to modern cryptography (Page 129)" (Gen, H) is called a Hash function in the definition; but its mentioned in the text that it is a familiy of Hash functions and s picks one function. – twallutis Nov 25 '12 at 12:19
I may run the danger to be a bit out of topic but this is an interesting approach on rigorous proofs by reductions usually employed by cryptographers. It criticizes the too much effort on proofs by reductions which sometimes doesn't take into account all the attacker's window and consequently "tends to kick dust to eyes" – curious Nov 26 '12 at 22:58
Also known as a keyed hash function. – Paŭlo Ebermann♦ Nov 29 '12 at 19:01
Did you find your solution? If so, please add it as an answer, so this question is more complete. Also, if one of the answers here has helped you, consider accepting it. – Paŭlo Ebermann♦ Nov 29 '12 at 19:03
## 2 Answers
In order to show that A' is collision resistant, you start with an attacker on the collision resistance of A' and show that you can build an attacker which will break the collision resistance of A. Specifically, you're given an oracle B' which will give values x' and y' such that H(H(x')) = H(H(y')) and you then need to construct an attacker B with the use of B' such that B returns two values x and y where H(x) = H(y).
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When you try to reduce the security of one primitive (H') to the security of another primitive (H), you assume that there exists an adversary that could break H'. Then you show that existence of said adversary implies an adversary against H by describing an algorithm (the reduction) that uses the adversary to break H.
as a very simple example, consider the hash function $H$ and the hash function $H'(x) = H (x \oplus 1^{|x|})$ (with $\oplus$ being XOR and $1^{n}$ denoting the string of $n$ ones).
Now assume there exists an adversary $\mathcal{A}$ against the collision resistance of $H'$, i.e. there exists a ppt algorithm $\mathcal{A}$ that on input $s$ (and potentially a security parameter) will output $x$, $x'$ such that $H'(x)=H'(x')$ (with non-negligible probability).
Now your reduction needs to transform such a collision into a collision for $H$. If you look at how $H'$ is constructed, you can easily see that this can be done by computing $y = x\oplus 1^{|x|}$ and $y' = x'\oplus 1^{|x|}$ and outputting $y,y'$.
So your reduction is the algorithm that runs $\mathcal{A}$, flips all the bits in $\mathcal{A}$'s output and returns it. You can easily verify, that this reduction is successful whenever $\mathcal{A}$ is successful. As $H$ is collision resistant, the probability that $\mathcal{A}$ is successful must therefore be negligible and thus $H'$ is collision resistant.
So the problem you need to solve for your case is basically the question "How can I turn a collision of $H'$ into a collision of $H$?" When you think a moment about the way $H'$ is constructed, I am sure you will see an obvious way to do that.
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Sorry for the delay; end-of-year stress;-) Here is my answer (so far i have no response; so i dont know if it is correct): Lets assume that A' is not collision-resistent. We write H'=H(y) instead of H'=H(H(x)). Then "collision-resistant" means: there exists y,y' with H(y)=H(y'), y<>y' When we now look at A, then H: {0,1}* -> {0,1}^l (arbitrary length -> fixed-length l) H': {0,1}^l -> {0,1}^l (fixed-length -> fixed-length) If H' is not collision-resistant, then we have y,y' with H(y)=H(y'). That is a special case of H and that would mean that H is not collision-resistant too. – twallutis Dec 23 '12 at 14:58
The only problem: this solution looks too easy... – twallutis Dec 23 '12 at 14:58
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http://mathhelpforum.com/advanced-applied-math/173068-haar-wavelets.html
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# Thread:
1. ## Haar Wavelets
Show that the rescaled Haar wavelets $\psi_{jk}(x)=2^{j/2}\psi(2^jx-k)$ form an orthonormal basis basis for $L^2(\mathbb{R})$.
So I know that what I have to show is that:
$\int_{\mathbb{R}}\psi_{jk}(x)\psi_{lm}(x)dx=\delta _{jl}\delta_{km}$
I'm just stuck as to how to simplify that integral.
2. Hmm. Maybe this might help:
$\psi_{jk}(x)=2^{j/2}\begin{cases}<br /> 1,&\quad 0\le 2^{j}x-k<1/2\\<br /> -1,&\quad 1/2\le 2^{j}x-k<1\\<br /> 0,&\quad\text{otherwise}<br /> \end{cases}=2^{j/2}\begin{cases}<br /> 1,&\quad k\le 2^{j}x<1/2+k\\<br /> -1,&\quad 1/2+k\le 2^{j}x<1+k\\<br /> 0,&\quad\text{otherwise}<br /> \end{cases}$
$=2^{j/2}\begin{cases}<br /> 1,&\quad 2^{-j}k\le x<2^{-j}(1/2+k)\\<br /> -1,&\quad 2^{-j}(1/2+k)\le x<2^{-j}(1+k)\\<br /> 0,&\quad\text{otherwise}<br /> \end{cases}.$
Here I'm using the mother wavelet function $\psi(x)$ as defined in the wiki. So the only points at which this function is nonzero are in the half-open interval $[2^{-j}k,2^{-j}(k+1)).$
Now, what if you could show that $[2^{-j}k,2^{-j}(k+1))\cap[2^{-l}m,2^{-l}(m+1))=\varnothing$ if either $j\not= l$ or $k\not=m?$ That would certainly be sufficient to show the zero part of the delta functions, wouldn't it? Because then, under the integral sign, each function would drag the other one down to zero.
What happens when both $j=l$ and $k=m?$ What does the integrand do?
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http://ams.org/bookstore-getitem/item=GSM-76
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New Titles | FAQ | Keep Informed | Review Cart | Contact Us Quick Search (Advanced Search ) Browse by Subject General Interest Logic & Foundations Number Theory Algebra & Algebraic Geometry Discrete Math & Combinatorics Analysis Differential Equations Geometry & Topology Probability & Statistics Applications Mathematical Physics Math Education
Measure Theory and Integration
Michael E. Taylor, University of North Carolina, Chapel Hill, NC
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Graduate Studies in Mathematics
2006; 319 pp; hardcover
Volume: 76
ISBN-10: 0-8218-4180-7
ISBN-13: 978-0-8218-4180-8
List Price: US\$61
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See also:
An Introduction to Measure Theory - Terence Tao
A First Course in Sobolev Spaces - Giovanni Leoni
A (Terse) Introduction to Lebesgue Integration - John Franks
Introduction to Differential Equations - Michael E Taylor
This self-contained treatment of measure and integration begins with a brief review of the Riemann integral and proceeds to a construction of Lebesgue measure on the real line. From there the reader is led to the general notion of measure, to the construction of the Lebesgue integral on a measure space, and to the major limit theorems, such as the Monotone and Dominated Convergence Theorems. The treatment proceeds to $$L^p$$ spaces, normed linear spaces that are shown to be complete (i.e., Banach spaces) due to the limit theorems. Particular attention is paid to $$L^2$$ spaces as Hilbert spaces, with a useful geometrical structure.
Having gotten quickly to the heart of the matter, the text proceeds to broaden its scope. There are further constructions of measures, including Lebesgue measure on $$n$$-dimensional Euclidean space. There are also discussions of surface measure, and more generally of Riemannian manifolds and the measures they inherit, and an appendix on the integration of differential forms. Further geometric aspects are explored in a chapter on Hausdorff measure. The text also treats probabilistic concepts, in chapters on ergodic theory, probability spaces and random variables, Wiener measure and Brownian motion, and martingales.
This text will prepare graduate students for more advanced studies in functional analysis, harmonic analysis, stochastic analysis, and geometric measure theory.
Readership
Graduate students interested in analysis.
Reviews
"Taylor's treatment throughout is elegant and very efficient ... I found reading the text very enjoyable."
-- MAA Reviews
"The book is very understandable, requiring only a basic knowledge of analysis. It can be warmly recommended to a broad spectrum of readers, to graduate students as well as young researchers."
"This monograph provides a quite comprehensive presentation of measure and integration theory and of some of their applications."
-- Mathematical Reviews
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http://nrich.maths.org/431/clue
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nrich enriching mathematicsSkip over navigation
### Real(ly) Numbers
If x, y and z are real numbers such that: x + y + z = 5 and xy + yz + zx = 3. What is the largest value that any of the numbers can have?
### Cubics
Knowing two of the equations find the equations of the 12 graphs of cubic functions making this pattern.
### Patterns of Inflection
Find the relationship between the locations of points of inflection, maxima and minima of functions.
# Cubic Spin
##### Stage: 5 Challenge Level:
If the graph of the cubic polynomial has rotational symmetry then a maximum point must be rotated to become a minimum and vice versa so the center of rotation will be the midpoint of the line joining the maximum and minimum points. If there are no maximum and minimum points then consider the point of inflexion.
The graph of a function has rotational symmetry about the origin if and only if $f(-x) = -f(x)$. You can do this question without calculus if you can find the transformation of coordinates that removes the quadratic term from the polynomial equation.
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.
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http://physics.stackexchange.com/questions/tagged/history+newtonian-gravity
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# Tagged Questions
1answer
490 views
### Combining Proportions to get Newton's Law of Universal Gravitation
I've read a little on the history of Newton's Law of Gravitation and noticed that the formula can be separated into 3 distinct parts that lead to the end result of $F_g = G \frac{m_1 m_2}{r^2}$; the ...
2answers
139 views
### Were there any efforts made by early physicists to discover and explain how composite bodies fall?
At the dawn of the modern era, Galileo discovered and described how composite bodies fall through the air (or at least the discovery is usually attributed to him). I'm interested in whether this had ...
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http://mathhelpforum.com/math-challenge-problems/175180-divergence-series.html
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# Thread:
1. ## Divergence of a series
Here's a challenge problem I just made up:
Let $\{a_n\}$ be a sequence of positive real numbers such that $\sum a_n = \infty$. Show that
$\frac{a_1}{a_0}+\frac{a_2}{a_0+a_1}+\frac{a_3}{a_0 +a_1+a_2}+ \dots = \infty$.
2. Let $S_n = \sum_{k=0}^{n}a_k$.
We want to show that $\sum_{k=0}^n\frac{\Delta S_k}{S_k}}$ diverges as $n\to \infty$.
( I chose this notation to suggest the similarity with $\int_{ [ a,\infty)} \tfrac{f'(x)}{f(x)}dx = +\infty$ , if $\displaystyle\lim_{x\to\infty} f(x) = +\infty$ )
We assume that $\frac{\Delta S_k}{S_k}} \to 0$ otherwise the series would diverge trivially.
Let $L_k = \log S_k$, then $\Delta L_k = \log S_{k+1} - \log S_k = \log \left( 1 +\frac{\Delta S_k}{S_k} \right) =\frac{\Delta S_k}{S_k} + O\left( \left(\frac{\Delta S_k}{S_k}\right)^2 \right)$
We observe that $\sum_{k=0}^n \Delta L_k = \log S_{n+1} - \log S_0$ diverges since $\log S_n \to \infty$ and now by the limit comparison test - both differences are always positive-, we get that $\sum_{k=0}^n\frac{\Delta S_k}{S_k}}$ diverges too.
Notation: $\Delta a_n = a_{n+1}-a_n$
3. Very nice Paul!
Here's my solution:
It's well known that if $x_n>0$ and $\sum {x_n}$ converges , then $\prod (1+x_n)$ converges also. Let $A_n = \sum_{k=0}^n a_k$. We have
$\prod_{k=1}^n (1+\frac{a_k}{A_{k-1}}) = \prod_{k=1}^n\frac{A_k}{A_{k-1}} = A_n/a_0$. Now since $A_n \to \infty$, we see that $\prod (1+\frac{a_k}{A_{k-1}})$ diverges, which implies that $\sum \frac{a_k}{A_{k-1}}$ diverges.
4. Originally Posted by Bruno J.
Very nice Paul!
Here's my solution:
It's well known that if $x_n>0$ and $\sum {x_n}$ converges , then $\prod (1+x_n)$ converges also. Let $A_n = \sum_{k=0}^n a_k$. We have
$\prod_{k=1}^n (1+\frac{a_k}{A_{k-1}}) = \prod_{k=1}^n\frac{A_k}{A_{k-1}} = A_n/a_0$. Now since $A_n \to \infty$, we see that $\prod (1+\frac{a_k}{A_{k-1}})$ diverges, which implies that $\sum \frac{a_k}{A_{k-1}}$ diverges.
Your solution is same as what came to my mind when I saw the question.
For those who might want to see a reference for the relation between the sum and the product used by Bruno J, please see [Theorem 7.4.6, 1].
I quote the result for convenience.
Lemma. Let $\{x_{n}\}_{n\in\mathbb{N}}$ be a sequence of nonnegative numbers, then
$\sum_{n\in\mathbb{N}}x_{n}$ and $\prod_{n\in\mathbb{N}}(1+x_{n})$ converge or diverge together.
References
[1] L.S. Hahn and B. Epstein, Classical Complex Analysis, Jones and Bartlett Publishers, Inc., London, 1996.
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http://mathoverflow.net/users/1409?tab=recent
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Mariano Suárez-Alvarez
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| 57m | comment | What is an interpretation of the relation in the cohomology of the pure braid groups?One you notice that those forms are closed (and considering those forms is very natural once you see that you are working with an arrangement of hyperplanes), the relations of Arnold are the very first that come to mind. Why would one conjecture that these are all relations, I don't know, but that these forms generate the whole thing is a sensible optimistic guess —although Arnold surely had reasons too. |
| 1d | comment | What is the 31th homotopy group of the 2 - sphere ?«For separable states, the original Hilbert space $S^7$ simplifies to $S^2\times S^2$» Reading physics papers requires a lot of restraint :-) |
| 1d | comment | Reference request: Riemannian manifold of linear isometries from $\mathbb{C}^n$ into $\mathbb{C}^m$@Tysin, it is clear that your question is not sufficiently transparent as to what exactly it is asking for —maybe elaborating more on what you are looking for will make it easier for you to find it. Your comment here is also rather opaque :-) |
| 1d | comment | What is the 31th homotopy group of the 2 - sphere ?But there is no fourth Hopf fibration! |
| 1d | comment | What is the 31th homotopy group of the 2 - sphere ?In what way the two things you list as motivation are motivation for asking what the 31st homotopy group is? |
| 2d | comment | A catalog of faithful representations of finite groups?There is no need of the «with repetitions» part: the smallest faithful reps are always multiplicity free. |
| May15 | comment | A catalog of faithful representations of finite groups?At the very least, explain what you mean by noteworthy... |
| May15 | comment | Special automorphisms of extraspecial groups @John, as soon as you get 50 points of reputation you will be able to add comments. Using the answer box to respond to someone's answer does not work at all. |
| May14 | comment | Awfully sophisticated proof for simple facts(You can avoid the strange substraction of two equal numbers in the end by using the reduced Euler characteristic; this is a standard trick) This of course subjective, as it depends on one's background, but this does not seem awfully sophisticated at all to me and, instead, appears quite natural! :-) |
| May13 | comment | What is the “fundamental theorem of invariant theory” ?This question reminds me of the Jabberwocky poem :-) |
| May13 | comment | Proper actions on unitary spheres of a Hilbert spaceEvery torsion free discrete group $G$ acts properly and discontinuousy on the unit sphere of $L^2(G)$, no? |
| May13 | comment | Isomorphic maximal commutative semi-simple sub algebras of M_n(C). Don't ask questions in answer boxes (!) If you want another question, ask a new question. |
| May13 | comment | If $M_n(R)$ and $M_m(R)$ satisfy the same polynomial identities is it true that $m=n$?@Thiago, I was just providing an example to Pasha's second paragraph, quite independently of what the question asked for. |
| May12 | comment | Is there an analogue of curvature in algebraic geometry?Well, in order to get an isomorphism, you changed both the domain and the codomain of the map. The original domain and codomains had certain interest of their own, so the fact that after redefining things you do get an isomorphism may be not as useful as it may appear initially when seem from the point of view of the contexts where the question originally arose :-) |
| May12 | comment | Normal regular sequence in noncommutative algebrasIf you explained what is that «some result» in the homogeneous case, it might be easier to see what you are after in the general case. |
| May12 | comment | Is there an easy way to write down the singular cohomology of a hypersurface in a toric variety?Your question specializes to «is there a simple way of writing down the singular cohomology of a hypersurface in $P^n$?» One can compute the dimensions of the rational cohomology groups if the surface is smooth, I think, but I don't know if the ring structure comes out as easily. |
| May12 | comment | H^d[U(1)^n,U(1)] of the Borel cohomology and Chern-Simons theoryI for one still don't know what cohomology you are talking aboout —Konrad asked you to be specific about this a few weeks ago but you haven't answered that afaict. Without knowing that, it is simply impossible to even make sense of what you are asking! For example, the answer by Henr.L below seems to have decided that you are taking about singular cohomology of the space $U(1)^n$, which I think is quite surely no the case... |
| May11 | comment | A characterization of Hilbert spaces?Dear Wlodzimierz, you can find formatting tips on this page htpp://mathoverflow.net/editing-help; in particular, a common way to get a link you have to use the syntax [some text](the actual URL). I hope your time on MO is not that bad, because I for one enjoy very much reading your contributions! |
| May11 | comment | Resolutions of Lie algebrasCan't you just copy what Tate did? Pick a presentation; start with the free Lie algebra on the generators, add a generators $y$ of degree $1$ per relation $r$, and define $d(y)=r$. Find generators for the homology of this dg, lift them to cycles, add generators to kill them and so on. |
| May11 | comment | non-convex Polytope definition.There are various definitions of what regularity should mean in Coxeter's book Regular polytopes. |
| May11 | comment | If $M_n(R)$ and $M_m(R)$ satisfy the same polynomial identities is it true that $m=n$?If $R$ is a ring of row-finite column finite infinite matrices, then $R\cong M_2(R)$. |
| May11 | comment | Awfully sophisticated proof for simple factsThis was popular among whom? The book by Cartan and Eilenberg, the very first textbook on the subject, already has the computation done in terms of the usual very small periodic projective resolution: after that, using anything else to compute this seems pretty weird! |
| May10 | comment | A characterization of Hilbert spaces?Adding a link to the earlier question (specially if you use the notation introduced there) would be useful. |
| May10 | comment | Tenacious structure Notice that if one projectivizes $\mathbb A^d$ (with origin set at $p$) then one knows that the resulting projective space $\mathbb P^{d-1}(\mathbb F_3)$ does have a «cellular decomposition» and the question is asking if one can find one such decomposition which lifts to a decomposition of $X$ —the converse is obvious. |
| May10 | asked | Tenacious structure |
| May10 | comment | Functional equationsUmar did not say pretty much anything, really! |
| May9 | comment | Binary Operation on a Cubic Surface(You should get the second edition) |
| May9 | comment | Binary Operation on a Cubic SurfaceThere is a book by Manin on, more or less, this subject: Cubic forms. |
| May5 | comment | What arithmetic information is contained in the algebraic K-theory of the integersBeautiful answer! |
| May4 | comment | Can you compute the quotient set below? Your comment makes me think that you misunderstand what I wrote. I did noy say that the two polynomials $X^2$ and $X(X+1)$ are in the same orbit ---I said exactly the opposite. The rest of the comment, I don't understand it. |
| May4 | revised | Can you compute the quotient set below? added 270 characters in body; added 1 characters in body |
| May4 | revised | Can you compute the quotient set below? added 184 characters in body |
| May4 | comment | Can you compute the quotient set below? I did not say that you would arrive at a different problem: but the version with polynomials is expressed in terms of familiar things —polynomials and linear changes of variables— instead of weird ordered pairs under an unmotivated equivalence relation! |
| May4 | answered | Can you compute the quotient set below? |
| Apr27 | revised | Quotients of classifying spacesadded 414 characters in body |
| Apr27 | revised | Quotients of classifying spacesadded 48 characters in body |
| Apr27 | answered | Quotients of classifying spaces |
| Apr23 | asked | What is flexible about flexible algebras? |
| Apr20 | accepted | Associative algebras with Jacobson radical of codimension 1 |
| Apr20 | answered | Associative algebras with Jacobson radical of codimension 1 |
| Apr15 | revised | The intersection complex and the Cohen-Macaulay propertyedited title |
| Apr14 | revised | complete or open Kähler manifold and simply connected edited body; edited title |
| Apr10 | awarded | ● Popular Question |
| Apr6 | awarded | ● Nice Answer |
| Apr2 | awarded | ● Popular Question |
| Mar31 | awarded | ● Nice Question |
| Mar31 | revised | The octonions on a bad dayadded 121 characters in body; edited body |
| Mar31 | asked | The octonions on a bad day |
| Mar20 | awarded | ● Good Answer |
| Mar19 | awarded | ● Popular Question |
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http://mathoverflow.net/revisions/40953/list
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## Return to Answer
4 In my example, the isotropy group is $A_n$, not $S_n$.
One situation in which it is essential to use groupoids is the study of orbifolds.
Slogan: The set of points of an orbifold is a groupoid.
Here's a concrete problem that is illuminated by the language of groupoids. Suppose I have an orbifold $M$ with a singuar stratum $X$. The stratum $X$ is isomorphic to $S^1$, and its isotropy group is some finite group $G$. Let's also assume that $X$ is oriented.
Question: What is the "monodromy" of going around that stratum?
At first glance, one might guess that it's an element of $Aut(G)$.
That's wrong! The monodromy is an element of $Out(G)$.
So we have a somewhat paradoxical situation in front of us: there is a group associated to every point of $X$. Yet, the monodromy is not acting by automorphisms of that group.
Here's an example of orbifold that nicely illustrates the kind if situation that can occur: $$M = (S^1\times V )/S_n,$$ where $S_n$ is the symmetric group and $V$ is a faithful representation. The group $S_n$ acts on the circle $S^1$ via the projection $S_n\twoheadrightarrow\mathbb Z_2$, and then the antipodal map. The representation $V$ of $S_n$ is just put there so that the orbifold isn't too degenerate (it can be omitted if you don't mind working with non-effective orbifolds).
In that example, the manifold $X$ is $S^1/\mathbb Z_2$. The isotropy group is the alternating group $A_n$. The monodromy is computed in the following way. Go half way around $S^1$, and then identifying "$S_n$ $A_n$ at point -1" with "$S_n$ $A_n$ at point +1" via any group element of $S_n$ that sends -1 to +1. A choice of such an element yields an automorphism of $S_n$. A_n$. But since there is no best way making such a choice, the only canonical thing is its class in$Out(S_n)$.Out(A_n)$.
Ok. Maybe now is good moment to try to remove some of the confusion.
It all becomes more clear once you realize that the thing that is associated to a point of $X$ is not a group. It's a groupoid:
If $[M/G]$ is an orbifold and $x$ is a point in $M/G$, then the groupoid that lives above $x$ has objects given by points $m\in M$ mapping to $x$. An arrow from $m$ to $m'$ is given by a element of $G$ that sends $m$ to $m'$.
The monodromy is then simply an automorphism of that groupoid (so now there's nothing weird any more). But this automorphism might fail to fix any of the objects of the groupoid. And so it can't be viewed as an automorphism of the corresponding group, unless you make some unnatural choices.
3 deleted 2 characters in body; deleted 3 characters in body
One situation in which it is essential to use groupoids is the study of orbifolds.
Slogan: The set of points of an orbifold is a groupoid.
Here's a concrete problem that is illuminated by the language of groupoids. Suppose I have an orbifold $M$ with a singuar stratum $X$. The stratum $X$ is isomorphic to $S^1$, and its isotropy group is some finite group $G$. Let's also assume that $X$ is oriented.
Question: What is the "monodromy" of going around that stratum?
At first glance, one might guess that it's an element of $Aut(G)$.
That's wrong! The monodromy is an element of $Out(G)$.
So we have a somewhat paradoxical situation in front of us: there is a group associated to every point of $X$. Yet, the monodromy is not acting by automorphisms of that group.
Here's an example of orbifold that nicely illustrates the kind if situation that can occur: $$M = (S^1\times V )/S_n,$$ where $S_n$ is the symmetric group and $V$ is a faithful representation. The group $S_n$ acts on the circle $S^1$ via the projection $S_n\twoheadrightarrow\mathbb Z_2$, and then the antipodal map. The representation $V$ of $S_n$ is just put there so that the orbifold isn't too degenerate (it can be omitted if you don't mind working with non-effective orbifolds).
In that example, the manifold $X$ is $S^1/\mathbb Z_2$. The monodromy is computed in the following way. Go half way around $S^1$, and then identifying "$S_n$ at point -1" with "$S_n$ at point +1" via any group element that sends -1 to +1. A choice of such an element yields an automorphism of $S_n$. But since there is no best way making such a choice, the only canonical thing is the element of its class in $Out(S_n)$.
Ok. Maybe now is good moment to try to remove some of the confusion.
It all becomes more clear once you realize that the thing that is associated to a point of $X$ is not a group. It's a groupoid:
If $[M/G]$ is an orbifold and $x$ is a point in $M/G$, then the groupoid that lives above $x$ has objects given by points $m$ of $m\in M$ mapping to $x$. A morphis An arrow from $m$ to $m'$ is given by a element of $G$ that sends $m$ to $m'$.
The monodromy is then simply an automorphism of that groupoid (so now there's nothing weird any more). But this automorphism might fail to fix any of the objects of the groupoid. And so it can't be viewed as an automorphism of the corresponding group, unless you make some unnatural choices.
2 added 2000 characters in body; deleted 15 characters in body; deleted 4 characters in body; added 1 characters in body; edited body; added 12 characters in body
One topic situation in which it is essential to use groupoids is the study of orbifolds.
Slogan: the The set of points of an orbifold is a groupoid.Here's a concrete classification problem that is illuminated by the language of groupoids.Suppose I have an orbifold $M$ and with a singuar stratum $X$. The stratum $X$ that is isomorphic to $S^1$, and whose its isotropy group is some finite group $G$. Let's also assume that $X$ is oriented.
Question: What is the "monodromy" of going around that stratum?
At first glance, one might guess that it's an element of $Aut(G)$.
That's wrong! The monodromy is an element of $Out(G)$.
So we have a somewhat paradoxical situation in front of us: there is a group associated to every point of $X$. Yet, the monodromy is not acting by automorphisms of that group.
Here's an example of orbifold that nicely illustrates the kind if situation that can occur: $$M = (S^1\times V )/S_n,$$ where $S_n$ is the symmetric group and $S_n$.
[Sorry, I pressed to quickly V$is a faithful representation. The group$S_n$acts on the circle$S^1$via the projection$S_n\twoheadrightarrow\mathbb Z_2$, and then the antipodal map. The representation$V$of$S_n$is just put there so that the orbifold isn't too degenerate (it can be omitted if you don't mind working with non-effective orbifolds). In that example, the manifold$X$is$S^1/\mathbb Z_2$. The monodromy is computed in the following way. Go half way around$S^1$, and then identifying "post" button..$S_n$at point -1" with "$S_n$at point +1" via any group element that sends -1 to +1. A choice of such an element yields an automorphism of$S_n$. But since there is no best way making such a choice, the only canonical thing is the element of$Out(S_n)\$.
Ok. I'm still writing up my answer]Maybe now is good moment to try to remove some of the confusion.
It all becomes more clear once you realize that the thing that is associated to a point of $X$ is not a group. It's a groupoid:
If $[M/G]$ is an orbifold and $x$ is a point in $M/G$, then the groupoid that lives above $x$ has objects given by points $m$ of $M$ mapping to $x$. A morphis from $m$ to $m'$ is given by a element of $G$ that sends $m$ to $m'$.
The monodromy is then simply an automorphism of that groupoid (so now there's nothing weird any more).But this automorphism might fail to fix any of the objects of the groupoid. And so it can't be viewed as an automorphism of the corresponding group, unless you make some unnatural choices.
1
One topic in which it is essential to use groupoids is the study of orbifolds.
Slogan: the set of points of an orbifold is a groupoid.
Here's a concrete classification problem that is illuminated by the language of groupoids.
Suppose I have an orbifold $M$ and a singuar stratum $X$ that is isomorphic to $S^1$, and whose isotropy group is the symmetric group $S_n$.
[Sorry, I pressed to quickly on the "post" button... I'm still writing up my answer]
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http://physics.stackexchange.com/questions/41406/is-it-true-that-an-isolated-fundamental-particle-does-not-decay/41407
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# Is it true that an isolated fundamental particle does not decay?
Is it true that an isolated fundamental/elementary particle does not decay?
It seems logical to me.
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What exactly is a "fundamental particle" to you? – ungerade Oct 22 '12 at 13:13
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Fundamental = elementary – mick Oct 22 '12 at 13:22
Concerning the edit made , i assumed it ( fundamental explained now by the then missing link ) was well known on this site to all. – mick Oct 22 '12 at 13:25
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– ungerade Oct 22 '12 at 13:34
re: 2 comments up: (some) people here will get in all sorts of arguments about what it means for a particle to be fundamental. But the definition on Wikipedia that you linked to is the commonly accepted one. – David Zaslavsky♦ Oct 22 '12 at 14:10
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## 3 Answers
No, because not all the fundamental particles (in the sense of the standard model) are stable. In particular, electrons and photons are stable; muons and tau leptons are not, and will decay into lighter leptons, e.g. $\mu^-\to e^- \bar\nu_e \nu_\mu$.
Neutrinos are kind of funny because, while they are prevented from decaying in the traditional sense by conservation of energy, momentum, and lepton family number, they do oscillate - so if you start with an electron neutrino, for example, it will turn into something that can be observed as any flavor of neutrino, and in this way you can find yourself measuring something like $\nu_e\to\nu_\tau$. But then you can just as well measure $\nu_\tau\to\nu_e$, and if you have two (actually three) particles which can all decay into each other, does it even make sense to call it a decay?
Quarks try to be funny but actually wind up just being annoying, because they are never found in isolation so nobody is exactly sure how an isolated quark would behave if you could put one in a universe by itself. That being said, heavy quarks (charm, bottom, theoretically top) are routinely observed to decay in high-energy collisions, where asymptotic freedom presumably applies, so it's not much of a stretch to imagine that an isolated heavy quark would decay into lighter quarks plus a pion or leptons.
Likewise, the weak gauge bosons decay all the time in collisions, so if you managed to create a universe that contained only a $W^\pm$ or $Z^0$ and nothing else, it would presumably decay quickly into a lepton and antineutrino or into some combination of hadrons. Same goes for the Higgs, except with different possible decay products.
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For now the best and accepted answer , although a bit above me for now. maybe some wiki-ing. – mick Oct 22 '12 at 13:59
However what about diffusion of light ? – mick Oct 22 '12 at 14:00
@mick what about it? Diffusion isn't really decay. Also I could edit in some Wikipedia links or more information if you let me know what parts are above you. – David Zaslavsky♦ Oct 22 '12 at 14:01
Im confused about diffusion then. If a wave diffuses , how about the particle interpretation then ? Isnt it particle decay in the particle interpretation ?? – mick Oct 22 '12 at 14:20
No, that's just the probability amplitude for measuring the particle in a particular position spreading out (assuming we're talking about diffusion in position space). – David Zaslavsky♦ Oct 22 '12 at 14:23
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It's only true if you take "isolated" to mean "isolated for an infinitely long time", in which case, an unstable particle would never form either. An isolated particle is exactly stable, this is the S-matrix state, and the exactly stable particles in our world are the electron/positron, the neutrino, the photon, the graviton and some dark matter.
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What about decays in waves or virtual particles ? or diffusion of light ? – mick Oct 22 '12 at 13:23
The proton is, to experimental evidence, exactly stable, but not fundamental. (I know you know this) – Jerry Schirmer Oct 22 '12 at 13:57
Im no expert Jerry. – mick Oct 22 '12 at 14:19
@JerrySchirmer: The proton can't be exactly stable because of proton decay in the standard model, and gravitational proton decay (really neutron decay) in the formation of black holes from neutron stars and their subsequent evaporation. The photon is exactly stable and not fundamental (meaning it is a mixture of SU(2) and U(1) gauge bosons), and I thought you said "photon" not "proton" at first. The S-matrix definition of "fundamental" is "exactly stable", though, so the photon is S-matrix fundamental, although field theoretically not fundamental. The proton is not fundamental in either view. – Ron Maimon Oct 22 '12 at 18:11
@RonMaimon: there is no proton decay in the standard model. The formation of neutron stars involves electron capture and are not proton decay, but an interaction with electrons. Most GUTs predict proton decay, but this has never been observed. – Jerry Schirmer Oct 22 '12 at 21:04
Presumably if it was fundemental there would be nothing for it to decay into ?
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there is no conservation law of amount of particles or waves as far as I know. – mick Oct 22 '12 at 13:22
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http://physics.stackexchange.com/questions/tagged/observables+uncertainty-principle
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Tagged Questions
2answers
344 views
Hamiltonian of oscillators quantized proof
https://docs.google.com/open?id=0BxrBcN1-BZWUOXNxR1l4S0l2MjQ http://www.2shared.com/complete/Qjy1_uzp/Quantum_Mechanics_in_Simple_Ma.html (I uploaded a pdf file that contains the parts of the ...
3answers
1k views
Proof of Canonical Commutation Relation (CCR)
I am not sure how $QP-PQ =i\hbar$ where $P$ represent momentum and $Q$ represent position. $Q$ and $P$ are matrices. The question would be, how can $Q$ and $P$ be formulated as a matrix? Also, what is ...
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160 views
Why does $i ( LK-KL )$ represent a real quantity?
According to my textbook, it says that $i( LK-KL )$ represents a real quantity when $K$ and $L$ represent a real quantity. $K$ and $L$ are matrices. It says that this is because of basic rules. ...
3answers
988 views
How does non-commutativity lead to uncertainty?
I read that the non-commutativity of the quantum operators leads to the uncertainty principle. What I don't understand is how both things hang together. Is it that when you measure one thing first ...
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http://crypto.stackexchange.com/questions/2412/finding-the-lfsr-and-connection-polynomial-for-binary-sequence?answertab=votes
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# Finding the LFSR and connection polynomial for binary sequence.
I have written a C implementation of the Berlekamp-Massey algorithm to work on finite fields of size any prime. It works on most input, except for the following binary GF(2) sequence:
$0110010101101$ producing LFSR $\langle{}7, 1 + x^3 + x^4 + x^6\rangle{}$
i.e. coefficients $c_1 = 0, c_2 = 0, c_3 = 1, c_4 = 1, c_5 = 0, c_6 = 1, c_7 = 0$
However, when using the recurrence relation
\begin{equation} s_j = (c_1s_{j-1} + c_2s_{j-2} + \cdots + c_Ls_{j-L}) \mbox{ for } j \geq L. \end{equation} to check the result, I get back:
0110010001111, which is obviously not right.
Using the Berlekamp-Massey Algorithm calculator they say the (I believe) characteristic polynomial should be $x^7 + x^4 + x^3 + x^1$. Which, according to my paper working, the reciprocal should indeed be $1 + x^3 + x^4 + x^6$.
What am I doing wrong? Where is my understanding lacking?
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## 2 Answers
This is probably just a difference in notation than any failure in understanding or implementation. Some people define recurrence relations with subscripts in reverse order than others; the original description given by Berlekamp in his 1968 book Algebraic Coding Theory began counting from $1$ instead of $0$ etc. Observe that $$x^7 + x^4 + x^3 + x^1 = x^7(1 + x^{-3} + x^{-4} + x^{-6})$$ in comparison to your $1 + x^3 + x^4 + x^6$ which you say is the correct reciprocal of what the web site's answer should be. So I would say that the web site seems to be following Berlekamp's original description and giving you an answer that is "off-by-one".
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Unfortunately, there was an implementation problem.
see http://math.stackexchange.com/questions/134545/finding-the-lfsr-and-connection-polynomial-for-binary-sequence for the full explanation
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Hi jamesj629 and welcome to crypto. Please provide a short summary of the document you linked. – Hendrik Brummermann♦ Aug 29 '12 at 22:12
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http://mathoverflow.net/questions/49722/a-question-about-hodge-structures-associated-to-spinor-groups/80554
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## a question about Hodge structures associated to spinor groups
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In Deligne's 1972 article on the Weil conjecture for K3 surfaces, he essentially constructed an inclusion of Shimura data $(GSpin(V),X)\subset(GSp(W),H(W))$, where $V$ is a vector space over $\mathbb{Q}$ of dimension $n+2$ endowed with a quadratic form $q$ of signature $(n,2)$, $C(V)$ the Clifford algebra associated to $(V,q)$, $C^+(V)$ the even part of $C(V)$, $GSpin(V)$ the reductive $\mathbb{Q}$-group of invertible elements in $C^+(V)$ that preserve $V$ under conjugation in $C(V)$. The Hermitian symmetric domain $X$ can be identified as the space of $q$-isotropic negative planes in $V_\mathbb{R}$, as is described in Kudla's article "algebraic cycles in Shimura varieties of orthogonal type". The inclusion of Shimura data mentioned above is given by the representation $(W,\rho_W)$ by left translation of $GSpin(V)\subset C^+(V)$ on $W=C^+(V)$, which respects a canonically defined symplectic pairing up to scalar. Finally $H(W)$ is the Siegel double space associated to the above symplectic structure.
Note that for any $x\in X$ (or in $H(W)$), $(W,\rho\circ x)$ is a Hodge structure of type $(-1,0),(0,-1)$, hence $W\otimes W$ underlies a Hodge structure of type $(-2,0),(-1,-1),(0,-2)$. On the other hand, the canonical representation $(V,\rho_V)$ of $GSpin(V)$ gives Hodge structures $(V,\rho_V\circ x)$ of type $(-2,0),(-1,-1),(0,-2)$.
My question : is there a natural embedding of $V$ into $W\otimes W$ as a subrepresentation of $GSpin(V)$? if there is, then I can naturally understand $V$ as a Hodge substructure of $W\otimes W$. The Hodge types already coincile, but I don't know if it follows from some simple universal construction.
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note that in Deligne's 1972 article he hasn't modified yet the sigh rules as his later articles on Shimura varieties and Hodge structures. He showed that for the canonical representation of $SO(V)$ on $V$, one can find Hodge structure of type $(-1,1),(0,0),(1,-1)$ on $V$ through the Shimura datum $(SO(V),X')$, the later being deduced from $(GSpin(V),X)$ by taking adjoint group. If one considers the action of $GSpin(V)$ on $V$, one finds a non-trivial action of the center $\mathbb{G}_m$ on $V$, and simple calculations shows that in this case the Hodge types are shifted by $(-1,-1)$. – genshin Dec 17 2010 at 11:31
sign rules instead of sigh rules... – genshin Dec 17 2010 at 11:34
## 2 Answers
Hello,
Indeed, there does exist an embedding of Hodge structures as you mention. The thing works as follows.
Fix an element $v_0\in V$. Then $V$ acts on the vector space $C^+(V)$ by $v\mapsto(x\mapsto vxv_0)$. This induces an embedding $V \hookrightarrow End_k(C^+(V)$. This is equivariant with respect to the actions of $CSpin(V)$, where $CSpin(V)$ acts on $V$ by conjugation, hence by definition of the Hodge structures, it gives a morphism of weight zero Hodge structures $V(1) \hookrightarrow End_k(C^+(V)$.
Now a polarization of $C^+(V)$ identifies its dual with $C^+(V)(-1)$, hence a embedding of weight $2$ Hodge structures $$V\hookrightarrow C^+(V)\otimes C^+(V).$$
All the best
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Thanks a great deal, but i'm a little confused. It is clear that $V$ acts on the total spinor algebra $C(V)$ by multiplication on the left. But how should this preserve the grading? $V$ is contained in the odd part $C^_(V)$ while $C^+(V)$ is even. Do I need to find a natural identification of $End(C^+(V))$ with something else? Maybe I misunderstood your explanations. – genshin Dec 18 2010 at 0:15
Sorry for the mistake, and thanks for pointing it out ! Hopefully the answer is closer to correct now. – F C Dec 18 2010 at 8:43
Dear F C, I am probably just confused, but don't we need some property of $v_0$ for this to work as written? (Because if $v_0$ is random, then after conjugating on the RHS by an element $u$ in $CSpin(V)$, we would obtain not the morphism $x \mapsto uvu^{-1} x v_0,$ but rather $x \mapsto u v u^{-1} x u v_0 u^{-1}$.) – Emerton Dec 18 2010 at 9:07
Dear Emerton, the thing is that the action of $CSpin(V)$ on $C^+(V)$ that you want to consider to get the weight one HS is by multiplication on the left, and not by conjugation. The action on $End_k(C^+(V))$ is thus given by $(u.f)(x)=uf(u^{-1}x)$, which gives the correct formula. – F C Dec 18 2010 at 9:15
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I have a related question. Is it possible to characterise the image of V inside W\otimes W ?
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http://mathoverflow.net/questions/77655?sort=newest
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Number of points on a complex sphere with pairwise inner product restriction
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Considered the following inner products:
$(1)$ $\langle x,y \rangle = \sum_{t=1}^{n}x_{t}y_{t}$
$(2)$ $\langle x,y \rangle_{c} = \sum_{t=1}^{n}x_{t}\bar{y}_{t}$
consider the following surfaces:
$\underline{Surface (a)}$: $\langle x, x \rangle = 1$
$\underline{Surface (b)}$: $\langle x, x \rangle = \mathbb{i} = \sqrt{-1}$
$\underline{Surface (c)}$: $\langle x, x \rangle_{c} = 1$
In each of the above surfaces, how many points can one place so that the inner product (defined in both $(1)$ and $(2)$) between any pair of the points is purely imaginary of form $0 + \mathbb{i}r$ where $\mathbb{i}=\sqrt{-1}$ and $r \in \mathbb{R}$ and how many points are there so that the pairwise product is purely real of form $r \in \mathbb{R}$?
The case when we seek the pairwise inner product(both $(1)$ and $(2)$ to be purely real is infinite for surfaces $(a)$ and $(c)$ (Just restrict your sphere to have purely real coordinates and search among those points).
Likewise, the case when we seek the pairwise inner product(both $(1)$ and $(2)$ to be purely imaginary is infinite for surface $(b)$ (Just restrict your sphere to have purely imaginary coordinates and search among those points).
What happens in the following combinations?
$\underline{A}$:$(b)$ when we seek pure imaginary inner products (both $(1)$ and $(2)$).
$\underline{B}$:$(a)$ and $(c)$ when we seek pure real inner products (both $(1)$ and $(2)$).
$\underline{A}$ has been shown to have finitely many points ($O(n)$ atmost) by unknown(google) below.
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1 Answer
It's easy to see you can't have infinitely many points: there would be two that are within $\epsilon>0$ of each other, and thus would have inner product very close to $\langle z,z\rangle=1$ (which would then not be purely imaginary).
Since $\Re\langle u,v\rangle$ forms a genuine inner product on $\mathbb C^n$, two vectors whose inner product is purely imaginary would be orthogonal. Thus you can have at most $2n$ such vectors. In the other direction, it's easy to see that the following is a collection of $2n$ vectors where every pairwise inner product is purely imaginary: $$(1,0,\ldots,0),(i,0,\ldots,0),(0,1,0,\ldots,0),(0,i,0,\ldots,0),\ldots$$
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I looked at example $c)$ here on page $1$. It gives example of $x=1$ and $y=i$ and provides inner product which is purely imaginary and states that $x$ and $y$ are not orthogonal. faculty.iu-bremen.de/stoll/teaching/… – unknown (google) Oct 10 2011 at 3:43
Your example isn't quite right; note that the problem wants a set of vectors such that the pairwise inner product is strictly non-zero. – ARupinski Oct 10 2011 at 3:44
It seems just a bit odd that we have infinite points with pairwise real inner products and only finitely many imaginary inner products. – unknown (google) Oct 10 2011 at 3:56
I think you are correct. – unknown (google) Oct 10 2011 at 3:57
I am including a variant of the inner product on which as well I am interested. – unknown (google) Oct 10 2011 at 4:02
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http://unapologetic.wordpress.com/2011/02/08/semistandard-generalized-tableaux/?like=1&source=post_flair&_wpnonce=4ca8fc8c32
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# The Unapologetic Mathematician
## Semistandard Generalized Tableaux
We want to take our intertwinors and restrict them to the Specht modules. If the generalized tableau $T$ has shape $\lambda$ and content $\mu$, we get an intertwinor $\bar{\theta}_T\in\hom(S^\lambda,M^\mu)$. This will eventually be useful, since the dimension of this hom-space is the multiplicity of $S^\lambda$ in $M^\mu$.
Anyway, if $t$ is our standard “reference” tableau, then we can calculate
$\displaystyle\begin{aligned}\bar{\theta}_T(e_t)&=\bar{\theta}_T(\kappa_t\{t\})\\&=\kappa_t\theta_T(\{t\})\\&=\kappa_t\left(\sum\limits_{S\in\{T\}}S\right)\\&=\sum\limits_{S\in\{T\}}\kappa_t(S)\end{aligned}$
We can see that it will be useful to know when $\kappa_t(S)=0$. It turns out this happens if and only if $S$ has two equal elements in some column.
Indeed, if $\kappa_t(S)=0$, then
$\displaystyle S+\sum\limits_{\substack{\pi\in C_t\\\pi\neq e}}\mathrm{sgn}(\pi)\pi S=0$
Thus for some $\sigma\in C_t$ with $\mathrm{sgn}(\sigma)=-1$ we must have $S=\sigma S$. But then we must have all the elements in each cycle of $\sigma$ the same, and these cycles are restricted to the columns. Since $\sigma$ is not the identity, we have at least one nontrivial cycle and at least two elements the same.
On the other hand, assume $S(i)=S(j)$ in the same column of $S$. Then $[e-(i\,j)](S)=0$. But then the sign lemma tells us that $(e-(i\,j))$ is a factor of $\kappa_t$, and thus $\kappa_t(S)=0$.
This means that we can eliminate some intertwinors $\bar{\theta}_T$ from consideration by only working with things like standard tableaux. We say that a generalized tableau is semistandard if its columns strictly increase (as for standard tableaux) and its rows weakly increase. That is, we allow repetitions along the rows, but only so long as we never have any row descents. The tableau
$\displaystyle\begin{array}{ccc}1&1&2\\2&3&\end{array}$
is semistandard, but
$\displaystyle\begin{array}{ccc}2&1&1\\3&2&\end{array}$
is not.
## 1 Comment »
1. [...] of Intertwinors from Semistandard Tableaux Let’s start with the semistandard generalized tableaux and use them to construct intertwinors . I say that this collection is [...]
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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://quant.stackexchange.com/questions/841/software-for-decomposing-structured-products-into-plain-vanilla-products/1206
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Software for decomposing structured products into plain vanilla products
Nowadays structured products (or packages) with complex payoff diagrams are omnipresent.
Do you know of any software, add-ons, apps, code whatever, that enables you to enter a payoff diagram or a cashflow profile which gives you the basic building blocks like the underlying, zero coupon bonds and esp. all the option components with their different strikes to replicate this payoff?
EDIT: Because some people asked what the input of such a tool could be, have a look at this example - I am asking for a software that is able to do this kind of decomposition automatically: http://www.risklatte.com/Articles_new/Exotics/exotic_28.php
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What would the input to this software be? – barrycarter Mar 29 '11 at 14:12
@barrycarter: e.g. the kinks of the payoff diagram by strike and payout level – vonjd Mar 29 '11 at 14:20
I cannot get, how they can plot $\frac{169}{S}$ like a linear function. Edited: they even cover it with a put. Wow. It's a kind of magic. Edited: "This payoff exactly replicates the payoff of the structure." Indian mystery... – Ilya Mar 30 '11 at 12:40
I can't think of a piece of software per se, but if you look at the payoff diagram, every discontinuity is a binary option and every linear payoff is a vanilla option. Do you have a more complex example? Once you have a uniform input format, I don't think it'd be hard to write software that does this. – barrycarter Apr 6 '11 at 15:51
4 Answers
I do not know such a software - but we can think about the code. There are tow points which you have to define properly:
1. which assets (correspondently, payoffs) are you allowed to replicate the complicated option?
2. as barrycarter has already asked - what should be the form of the input?
Further procedure should be quite easy. You are trying to find a linear combination $\lambda$ of basic assets $s_1,s2_,...$ (because in practice this is the only possibility for you to "combine" it) which fits the complex payoff $\gamma$. It's just a peace-wise affine optimization problem. Once you minimize the difference $|\lambda - \gamma|$ you have either zero (so you have found the replication formula) or smth greater than zero (which means that there is no replication formula which perfectly covers this complicated payoff).
Once you will determine the points I've mentioned - I believe I will be able to help you to solve this problem.
Edited: Let us call your payoff $P(S)$ and simple payoff functions are $P_1(S,\theta_1),P_2(S,\theta_2),...$, where $\theta$ are parameters, e.g. strike for Call or Put.
Then you would like to check if there exist $a_1,a_2,...$ such that $$P(S) = \sum\limits_i a_i P_i(S,\theta_i).$$
You can solve this problem by defining $$J(a,\theta) = ||P(\cdot) - \sum\limits_i a_i P_i(\cdot,\theta_i)||$$ where you can use any norm - and in fact due to the structure of payoffs, this norm should be defined only on some finite interval $[0,S']$. Then you solve $$J(a,\theta)\to \min$$ and if the extremum value is $0$ - you can cover your exotic payoff with simple ones, if non-zero - you cannot cover it perfectly, but the obtained values of $a,\theta$ will be optimal.
If you need more details about the solution of optimization problem -just tell me.
P.S. I think the paper you have refereed to is not correct - the payoff is not peace-wise affine while they plot it (and considered it) as peace-wise affine function.
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@Gortaur: Thank you - I edited the question to qualify what I mean exactly. – vonjd Mar 30 '11 at 12:16
@vonjd - it was clear. Ok, so we should talk about two questions which I left in my answer. If I can leave here my e-mail, or there are private messages? – Ilya Mar 30 '11 at 13:19
@Gortauer: 1. underlying, zero coupon bond, call, put: any number long/short, 2. see my editing the question for an example. I would appreciate it if you could edit your answer to incoporate the solution - so everybody can learn from your expertise - thank you! – vonjd Mar 30 '11 at 17:53
@vonjd - I don't want to hide anything - just I thought it can be more convenient to discuss it by e-mails. – Ilya Mar 31 '11 at 14:41
@vonjd - I've edited my answer. – Ilya Mar 31 '11 at 14:49
show 3 more comments
There are already quite a lot of softwares that do that. Quite expensive however for most of them. Then it depends whether you're interested into a trading software (trade capture and stuff) or a pricing engine.
Trading softwares : murex, misys summit, calypso ... provide tools to structure deals and value them. Then they are processed front to back. Pricing engines : NumeriX, Pricing partners ... are able to define payoff scripts and value them.
disclaimer : I used to work for one of these vendors, but I don't think my answer is biased.
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the example uses an european payoff. every european payoff can be decomposed by call and put combinations. just hold for each strike a qtty = to the 2nd derivative of the payoff....
in the general case, that is non european payoff, there is not always a perfect hedge.
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This sounds like a perfect application for genetic algorithms to me.
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Thank you - I don't think that you need a random component for that, a deterministic algorithm should do the job just fine! – vonjd Mar 31 '11 at 14:28
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Could you elaborate? This answer isn't very helpful on its own. For example, just how would someone apply GA to the problem described in the question? – chrisaycock♦ Mar 31 '11 at 15:51
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http://mathhelpforum.com/statistics/184322-combination-question.html
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# Thread:
1. ## Combination question?
A jar contains ten blue balls, ten red ones, ten white ones, and ten green ones. Each set of ten balls of the same colour is numbered from 1 to 10. How many different selections of four balls are possible if exactly three of the balls bear the same numbers?
The answer's supposed to be 1440.
Can someone show me the steps of solving this problem? Thanks!
2. ## Re: Combination question?
Originally Posted by wabbt
A jar contains ten blue balls, ten red ones, ten white ones, and ten green ones. Each set of ten balls of the same colour is numbered from 1 to 10. How many different selections of four balls are possible if exactly three of the balls bear the same numbers?
The answer's supposed to be 1440.
Can someone show me the steps of solving this problem? Thanks!
CASE (i)
suppose the three balls bearing the same number bear the number 1.
there are four subcases:
1) those three balls are B,R,G
2) those three balls are B,R,W
3) those three balls are R,G,W
4) those three balls are B,G,W
for subcase 1) there are 36 possible choices. (why?)
So for CASE(i) there are 4*36 choices.
Hence the total number of choices are 4*10*36.
3. ## Re: Combination question?
Originally Posted by wabbt
A jar contains ten blue balls, ten red ones, ten white ones, and ten green ones. Each set of ten balls of the same colour is numbered from 1 to 10. How many different selections of four balls are possible if exactly three of the balls bear the same numbers? The answer's supposed to be 1440.
There ten ways to pick the number that occurs three times.
The there are thirty-six ways we can pick the odd-man out.
There are four ways each of those selections can be made.
4. ## Re: Combination question?
How come it's 36 instead of 37? Since we only need 3 balls of the same color and there would be 37 left to choose from?
5. ## Re: Combination question?
Originally Posted by wabbt
How come it's 36 instead of 37? Since we only need 3 balls of the same color and there would be 37 left to choose from?
Once we have the number to be used three times in effect we have ruled out four balls because we cannot the forth ball of that number. That leaves thirty-six to choose.
6. ## Re: Combination question?
Hello, wabbt!
A jar contains ten blue, ten red, ten white, and ten green balls.
Each set of ten balls of the same colour is numbered from 1 to 10.
How many different selections of four balls are possible
if exactly three of the balls bear the same numbers?
The answer's supposed to be 1440.
Consider this analogous problem . . .
Remove the face cards from a standard deck of cards.
You 40 cards, from Ace to Ten in four suits.
You draw four cards.
In how many ways can you get Three-of-a-Kind?
The are 10 choices for the value of the triple.
There are ${4\choose3} = 4$ ways to get the triple.
There are 36 choices for the fourth card.
. . (We do not want Four-of-a-Kind.)
Therefore, there are: . $10\cdot4\cdot 36 \:=\:1440$ ways.
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http://mathoverflow.net/questions/21950/a-infinity-structure-on-the-ribbon-graph-complex-and-more-general-graph-complexes/22082
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## A-infinity structure on the ribbon graph complex and more general graph complexes
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Moduli spaces of curves (with nonempty boundary or at least one marked point) admit cell decompositions in which the cells are labelled by ribbon graphs. In fact, the moduli space of normalised metric ribbon graphs with $n$ boundary components is homeomorphic to the product of the moduli space of curves and an $(n-1)$-simplex.
From this geometric statement one can show that the ribbon graph complex (a chain complex spanned by ribbon graphs, with the differential given by summing over all edge contractions) computes the cohomology of moduli spaces of curves.
My question is: does anyone know how to describe the cochain level cup product structure on the ribbon graph complex? By general machinery, if a complex computes the cohomology of a space then it carries an $A_\infty$ structure for which it is equivalent to the cochains on the space with their usual $A_\infty$ structure.
I would like to know if there is a way to write down a combinatorial formula for this $A_\infty$ structure on the ribbon graph complex.
Similarly, the Lie graph complex (in which vertices are labelled by words in a generic Lie algebra) computes the cohomology of the spaces $BOut(F_n)$ (the classifying spaces of outer automorphism groups of free groups). Is there a way to describe the resulting $A_\infty$ structure on the Lie graph complex?
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Are moduli spaces of curves (rationally) formal? DM space should be formal, right? – Kevin Lin Apr 20 2010 at 10:23
2
Rational formality for moduli of smooth curves is not known, although Voronov showed roughly that they are formal in the Harer stable range. The DM compactifications, being compact Kahler orbifolds are certainly formal, and in fact, there is a nice paper where the modular operad they form is shown to be formal. – Jeffrey Giansiracusa Apr 21 2010 at 7:17
## 1 Answer
There is a way to obtain these kinds of spaces as geometric realizations of categories of graphs. See Igusa's book on Reidemeister torison. This gives a simplicial model. Basically its the barycentric subdivision. There is a standard formula for the cup product in this context which will probably work.
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yes, one can write down cup products by choosing a simplicial approximation of the diagonal, but then for the full A-infinity structure one needs some choices of explicit homotopies and higher homotopies to make the diagonal coassociative. A brute force approach becomes pretty much impossible and I was wondering if someone might have managed to do something more clever. – Jeffrey Giansiracusa Apr 21 2010 at 19:11
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http://math.stackexchange.com/questions/73949/why-is-0-0-not-a-minimum-of-fx-y-y-3x2y-x2
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# Why is $(0, 0)$ not a minimum of $f(x, y) = (y-3x^2)(y-x^2)$?
There is an exercise in my lists about those functions: $$f(x, y) = (y-3x^2)(y-x^2) = 3 x^4-4 x^2 y+y^2$$
$$g(t) = f(vt) = f(at, bt); a, b \in \mathbf{R}$$
It asks me to prove that $t = 0$ is a local minimum of $g$ for all $a, b \in \mathbf{R}$
I did it easily: $$g(t) = 3 a^4 t^4-4 a^2 t^2 b t+b^2 t^2$$ $$g'(t) = 2 b^2 t-12 a^2 b t^2+12 a^4 t^3$$ $$g''(t) = 2 b^2-24 a^2 b t+36 a^4 t^2$$
It is a critical point: $$g'(0) = 0; \forall a, b$$
Its increasing for all a, b: $$g''(0) = 2b^2 > 0; \forall b \ne 0$$ and $$b = 0 \implies g(t) = 3 a^4 t^4$$ which has only one minimum, at $0$, and no maximum
However, it also asks me to prove that $(0, 0)$ is not a local minimum of $f$. How can this be possible? I mean, if $(0, 0)$ is a minimum over every straight line that passes through it, then, in this point, $f$ should be increasing in all directions, no?
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4
You've tried plotting your function? – J. M. Oct 19 '11 at 7:46
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Approach $(0,0)$ along the parametric path $x=t$, $y=2t^2$. – André Nicolas Oct 19 '11 at 7:51
Yes, and this seems inconsistent with what I found about $g$ – rogi Oct 19 '11 at 7:52
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Even a simple-looking function of two variables can do weird things. – André Nicolas Oct 19 '11 at 7:55
2
It is zero on the curves $y=x^2$ and $y=3x^2$. It is negative when $x^2 < y < 3x^2$. – AD. Oct 19 '11 at 7:58
show 2 more comments
## 3 Answers
Using the factorisation you have we note that $f$ is:
• zero on the curves $y=x^2$ and $y=3x^2$
• negative when $x^2<y<3x^2$
• positive otherwise.
Since there are points arbitrary close to $o=(0,0)$ that satisfy $x^2<y<3x^2$ (say $y=2x^2$ where $x$ is small) and points arbitrary close to $o$ that do not satisfy $x^2<y<3x^2$ (say $y=0$ and small $x$) we see that $o$ is not a local minimum.
EDIT Note that in any neighbourhood of $o$ there is no line segment through $o$ which is contained in the region $x^2<y<3x^2$.
2nd Edit Draw a line through $o$, then there is a line segment surrounding $o$ such that the line segment has no point $(x,y)$ such that $x^2<y<3x^2$. I hope this is sufficiently clear - otherwise just ask!
Descriptive picture
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Ok, but if there is an "arbitrarily close" to $(0, 0)$ that is negative, can't we trace a straight line through it and teh origin? – rogi Oct 19 '11 at 8:20
Pick a definite point $(a,b)$, very very close to (0,0)$, at which the function is negative. Join$(a,b)$to the origin by a line. By what you proved, as you approach the origin along that line, after a while the function will be positive. However, that does not change the fact that, to be fancy, for any$\epsilon>0$, there is a point$(a_\epsilon,b_\epsilon)$, at distance$<\epsilon\$ from the origin, at which the function is negative. – André Nicolas Oct 19 '11 at 8:36
What I don't get is this "after a while teh function will be positive". It should be positive immediately, otherwise, $(0, 0)$ would not be minimum of $g$ – rogi Oct 19 '11 at 17:41
@AYGHOR What do you mean? – AD. Oct 19 '11 at 18:10
What I mean is: $P$ is a point as close as it can be to $(0, 0)$. Then there is a line from $(0, 0)$ to $P$. If $f(P)$ is negative, then $(0, 0)$ is not a minimum of $f$ restricted to this line. EDIT: I mean straight lines. – rogi Oct 19 '11 at 19:17
show 6 more comments
Saying that "$(0,0)$ is a local minimum of $f$ restricted to any line through the origin" means that, for every such line, there might be some bad points on the line at which $f$ takes negative values, but none of them are "close to the origin". Note that "close to the origin" can depend on the line we're talking about.
However, there are lots of lines through the origin, and each one can have some bad points. It's quite possible for the bad points on all the lines together to approach the origin, which can make $(0,0)$ not a local minimum of $f$.
That's the idea how the seeming paradox can be resolved. As for proving that it really happens: can you characterize the points $(x,y)$ at which $f(x,y) < f(0,0) = 0$? Are those points staying far away from $(0,0)$, or are they all around the origin?
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Sorry, that "bad point" idea did not make any sense to me – rogi Oct 19 '11 at 8:04
If "$f$ restricted to every straight line passing through teh origin" has a minimum at $(0, 0)$, then there should be at least circle, even if ridiculously small, around $(0, 0)$ over which teh values of $f$ are $> 0$, no? – rogi Oct 19 '11 at 8:14
@AYGHOR: No, that's not true. (This particular function $f$ is a counterexample!) – Hans Lundmark Oct 19 '11 at 9:27
This is the problem. Such a circle do not exists for your function. – xen Oct 19 '11 at 9:28
OMG I'M BLIND I CAN'T SEE WHY O_O – rogi Oct 19 '11 at 17:43
show 1 more comment
Draw the set of points in the $xy$-plane where $f(x,y) = 0$. Then look at the regions of the plane that are created and figure out in which of them $f(x,y)$ is positive and in which $f(x,y)$ is negative. From there, you should be able to prove that $(0,0)$ is neither a local maximum nor a local minimum point. (Hint: Using your sketch, is there any disk centered at $(0,0)$ in which $f(x,y)$ takes its minimum value at $(0,0)$?)
As important as understanding why the phenomenon you are observing is happening (no local minimum at the origin even though the function restricted to any straight line through the origin has a local minimum there) is to figure out how you would construct such a function and why the example given is, in some sense, the simplest kind of example you could construct. The idea used here is very similar to creating an example which shows that $\lim_{(x,y) \rightarrow (0,0)} f(x,y)$ may not exist even if the limit exists along all straight lines approaching the origin.
What you're really learning in this example is that the seemingly reasonable intuition that we would naturally have that we can understand a function of two variables near a point by understanding all of the functions of one variable obtained by restricting the function to lines through that point is misguided -- in particular, it fails if we are trying to find local maxima and minima.
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http://mathoverflow.net/revisions/85016/list
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## Return to Answer
1 [made Community Wiki]
This comment serves to record a partial attempt, which didn't get very far but might be useful to others. Following a suggestion of Mark Wildon and Arthur B, define `$$f_n(\alpha) := \sum (-1)^r \binom{n}{r}^{\alpha}.$$` This is zero for $n$ odd, so we will assume $n$ is even from now on.
Mark Wildon shows that it would be enough to show that $f_n(1/2) \geq 0$ for all $n$. It is easy to see that $f_n(0) = 1$ and $f_n(1)=0$. Arthur B notes that, experimentally, $f_n(\alpha)$ appears to be decreasing on the interval $[0,1]$. If we could prove that $f_n$ was decreasing, that would of course show that $f_n(1/2) > f_n(1) =0$.
I had the idea to break this problem into two parts, each of which appears supported by numerical data:
1. Show that $f_n$ is convex on $[0,1]$.
2. Show that $f'_n(1) < 0$.
If we establish both of these, then clearly $f_n$ is decreasing.
I have made no progress on part 1, but here is most of a proof for part 2. We have `$$f'_n(1) = \sum (-1)^r \binom{n}{r} \log \binom{n}{r} = \sum (-1)^r \binom{n}{r} \left( \log(n!) - \log r!- \log (n-r)! \right)$$` `$$=-2 \sum (-1)^r \binom{n}{r} \left( \log(1) + \log(2) + \cdots + \log (r) \right)$$` `$$=-2 \sum (-1)^r \binom{n-1}{r} \log r.$$` At the first line break, we combined the $r!$ and the $(n-r)!$ terms (using that $n$ is even); at the second, we took partial differences once.
This last sum is evaluated asymptotically in this math.SE thread. The leading term is $\log \log n$, so the sum is positive for $n$ large, and $f'_n$ is negative, as desired. The sole gap in this argument is that the math.SE thread doesn't give explicit bounds, so this proof might only be right for large enough $n$.
This answer becomes much more interesting if someone can crack that convexity claim.
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http://www.sacredduty.net/2012/04/13/mop-block-calculations-part-2/
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A Protection Paladin Blog
# MoP Block Calculations – Part 2
Posted on April 13, 2012 by
Suggested subtitles:
How to make String Theory look easy
Why I quit theorycrafting
101 reasons I should have majored in History
In the last post, we worked out the basic math modeling of our new block mechanics. That was actually the hard part, believe it or not – getting the model right is the biggest part of the job, and the most likely place to make mistakes. And we did have to make a few approximations to get there, but they should all reasonably good approximations.
Now, we need to take that modeling and calculate relationships between the different inputs. That takes a bunch of calculus, but it’s mostly straightforward plug-and-chug operations. So while it’s a lot of equations and math, it’s not terribly complicated or interesting math. In fact, after going through it, I’ll tell you that it’s incredibly tedious and ugly as hell. But we’re going to have to do it anyway, so let’s get down to it.
To recap from last time, we have the following equations:
$D = D_0 F_{a} F_{\rm av} F_{b}$
$F_{\rm av} = (1 – A)$
$F_b = 1 – GB_v” – (1-G) B_c S B_v’ – (1-G)B_c(1-S)B_v$
$G = R_{\rm SotR}/R_{\rm att}$
$S = R_{\rm SotR}(R_{\rm att}T_{\rm buff} – 1)/(R_{\rm att}-R_{\rm SotR})$
$R_{\rm SotR} = R_{\rm HPG}/3(1-\alpha_{\rm DP})$
$F_a$, $F_{\rm av}$, and $F_b$ are the armor, avoidance, and block mitigation factors, respectively. $G$ is the probability of any given attack being a guaranteed block due to Shield of the Righteous (SotR). $S$ is the probability that an attack which isn’t a guaranteed block occurs during the uptime of the other SotR buff, which gives increased block value. $B_c$ is our character sheet block chance, $B_v$ is our character sheet block value (30%), $B_v’$ is our block value during the SotR buff (50%), and $B_v”$ is our block value for the guaranteed block (75%). $R_{\rm SotR}$ is the rate at which we cast SotR, $R_{\rm att}$ is the incoming attack rate (which we’ll see shortly is actually a little different than the swing timer because of avoidance), $T_{\rm buff}$ is the duration of the SotR block value buff, $R_{\rm HPG}$ is our Holy Power generation rate, and $\alpha_{\rm DP}$ is the Divine Purpose proc chance (15%).
Got all that? Good. There will be a quiz on it later (not really).
A Word on Assumptions
In the comments of the last post, Weebey pointed out a few aspects of the derivation that looked fishy. And rightfully so, I did have to make a few assumptions, and I wasn’t as clear as I could have been about what they were. Any good derivation should come part and parcel with a good degree of error analysis.
Let’s quickly go over the assumptions we had to make in the previous post too. First, we’re assuming that we never waste a SotR cast – that every cast gives us a guaranteed block. However, if we cast SotR twice in a row without blocking something in-between, we don’t get 2 blocks later – the buff just refreshes, it doesn’t stack. So there’s a chance we’ll get some clipping, which will cause some attrition in $G$. In other words, we’d need to scale $G$ down by an attrition factor. That’s going to be tedious though, and it shoudn’t happen very often since SotR is off-GCD and can be timed based on Holy Power pooling, so we’re ignoring that for now. When we get around to the numerical simulations (Part 4, at this rate), we’ll see how big an effect this is, and whether that’s a reasonable approximation.
Technically our expression for $S$ is a small approximation too, because we’ve based it on a model of a single application of the buff. Since subsequent applications add duration rather than refreshing it, it should be a very good approximation, but the same variability that applies to $G$ will have a subsequent effect on $S$. Again, we’ll have to wait for the numerical simulations (or do some ugly combinatorics) to see how much error this introduces.
In general, we’ve treated everything in this derivation as nice and regular – boss attacks and SotR casts come in nicely-spaced intervals. That in itself is an approximation, because in combat scenarios those things vary quite a bit. That leads to bunching, which leads to the same sort of attrition seen in $G$ and causes a greater variance in the values of $G$ and $S$ from fight to fight. You might get two or three SotR casts in a row, or you might get none for 10 seconds if you miss all of your Crusader Strikes. We should have a pretty good estimate of the stochastic behavior, but if the variances are large we might find that individual encounter-length data sets deviate significantly. The numerical sims will give us some information on that detail as well.
Finally, in the last post (which I’ve since updated) I originally said that $R_{\rm att}$ was the boss attack rate, or the inverse of the boss’s swing timer $T_{\rm att}$. That’s not strictly true, because we’ll avoid some of those attacks, and since the SotR buff ignores avoidance, the effective attack rate as it pertains to blocking will actually be lower (or equivalently, the effective swing timer will be higher). The reason I ignored this in the last post is that $R_{\rm att}$ only shows up in the expressions for $G$ and $S$, so those two parameters encapsulate the effect. That made it a little easier while working through the derivation by hand, but looking at it now I realize that it makes the whole thing a lot more confusing to understand.
Thus, I’ve gone back and edited Part 1 to make it clear that $R_{\rm att}$ is the effective attack rate rather than the “true” incoming attack rate $R^{(0)}_{\rm att}$, and similarly for the “effective” and “true” boss swing timers $T_{\rm att}$ and $T^{(0)}_{\rm att}$. The rigorous definition of $R_{\rm att}$ is then
$R_{\rm att} = (1-A) R_{\rm att}^{(0)} = F_{\rm av}/T^{(0)}_{\rm att}$.
This is the definition we’ll use when we plug in numbers.
The Incoming Damage Formula
OK then, it’s time to start the calculus portion of the derivation. What we’re interested in is the change in incoming damage, which is the derivative of $D$. We’ve never specified whether $D$ represented raw damage or damage per second, and it doesn’t actually matter which – the equation is the same either way. Taking our very first equation and performing implicit differentiation, we have:
$dD/D_0 = dF_a F_{\rm av} F_b + F_a dF_{\rm av} F_b + F_a F_{\rm av} dF_b$
Each of these terms represents a relative change in damage taken due to a small change in that factor, either armor ($dF_a$), avoidance ($dF_{\rm av}$), or block ($dF_b$). We’ll have to evaluate each of those differentials to relate the change in the factor to the underlying change in armor, avoidance rating, and mastery, respectively.
Armor Factor
The armor factor is easy – it’s unchanged from how it currently works. The armor mitigation factor is just one minus the amount of mitigation $M_a$ we get from armor, and $M_a$ has the form
$M_a = (Ar)/(Ar + K)$
where $Ar$ is your armor and $K$ is the armor constant, which is 32573 for a level 88 boss. Therefore,
$F_a = 1 – M_a = 1 – Ar/(Ar+K) = K/(Ar + K)$
and
$dF_a = -dAr K / (Ar+K)^2 = – dAr F_a / (Ar+K)$.
Simple enough! We can use this equation to directly relate a change in armor $dAr$ to a change in the armor mitigation factor $dF_a$.
Avoidance Factor
Next up is avoidance, which is a little trickier for two reasons: diminishing returns (DR) and rating conversions. First of all, we’ll make the assumption that we only care about one type of avoidance – say dodge. The equations are symmetric in both, so finding the results for one is equivalent to finding the results for the other. We need to express the avoidance mitigation factor a little more specifically though, to isolate the part of avoidance affected by DR. So we’ll define it like this:
$F_a = (1 – A_0 – A_d)$.
Here, $A_0$ is all of your avoidance from miss and parry, as well as any dodge sources that aren’t affected by DR (ex: base dodge). $A_d$ is the post-DR dodge value, so we need to relate that to the pre-DR dodge value, which we’ll call $a_d$. The relationship between those two is,
$1/A_d = 1/C_d + k_d/a_d$,
where $C_d$ and $k_d$ are the diminishing returns coefficients for dodge (or parry). I’ve given them subscripts because block uses the same DR equation, but with different coefficients – this way we can keep them straight. Differentiating this equation gives us,
$-dA_d/A_d^2 = -k da_d/a_d^2$
which can be re-arranged to give
$dA_d = da_d k_d (A_d/a_d)^2$.
and if we solve the DR equation for $A/a$ and substitute, we get the most convenient form:
$dA_d = (da_d/k_d)(1-A_d/C_d)^2$
This is convenient because it expresses the change in post-DR avoidance $dA_d$ in terms of the change in pre-DR avoidance $da_d$, the post-DR dodge value $A_d$, and the constants $k_d$ and $C_d$. We know the constants, and we can read $A_d$ off of the character sheet. The pre-DR dodge value is very simply linked to rating:
$da_d = dr_d/f_d$
where $dr_d$ is the amount of dodge rating we’re adding and $f_d$ is the conversion factor relating dodge rating to dodge percentage (176.72 at level 85 in Cataclysm).
Differentiating $F_{\rm av}$ and plugging all of this in, we get:
$dF_{\rm av} = – dA_d = – (da_d/k_d)(1-A_d/C_d)^2 =- dr_d (1-A_d/C_d)^2/(k_d f_d)$
Which is our final expression for $dF_{\rm av}$
Block Factor
Finally, we need to attack the block differential. This is by far the most complicated because there are so many factors to consider. We see why when we differentiate $F_b$:
$dF_b = -dG B_v” + dG B_c S B_v’ – (1-G)dB_c S B_v’ \\ \hspace{0.5in} – (1-G)B_c dS B_v’ +dG B_c (1-S) B_v \\ \hspace{0.5in} -(1-G)dB_c (1-S) B_v + (1-G) B_c dS B_v$
Grouping terms, we can simplify this to
$dF_b = -dG [B_v'' -B_c S B_v' - B_c (1-S) B_v] \\ – dB_c (1-G)[S B_v' + (1-S) B_v] \\ – dS (1-G) B_c (B_v’ – B_v)$
To continue, we need the derivatives $dB_c$, $dG$, and $dS$. The first is easy, the last two are a bit ugly. Let’s do the easy one first. Block uses the same diminishing returns formula as avoidance, but with different coefficients. Since $dB_c$ is our post-DR change in block chance, we’ll let our pre-DR change in block chance be $db_c$, and the DR coefficients will be $C_B$ and $k_B$. We then have:
$dB_c = \frac{db_c}{k_B}(1-B^{\rm (DR)}_c/C_B)^2$
where $B^{\rm (DR)}_c = B_c-B^{(0)}_c$ is the portion of our block which is subject to diminishing returns (at this point, everything but our base block chance).
$db_c$ is related to mastery rating very simply:
$db_c = dr_m/f_m$
Thus, our final form is
$dB_c = \frac{dr_m}{f_m}\frac{(1-B^{\rm (DR)}_c/C_B)^2}{k_B}&s=1$.
$dB_c$ depends on the amount of mastery rating added $dr_m$, the rating-to-percentage conversion factor for mastery $f_m$ (which is 179.28/2.25 = 79.68 at level 85), our post-DR block chance, and the DR coefficients for block. Yes, that’s the simple one.
Now for the tricky ones. What does $dG$ look like?
$dG = dR_{\rm SotR}/R_{\rm att} – dR_{\rm att}R_{\rm SotR}/R_{\rm att}^2$
So for $dG$, we need to go further and find the change in SotR cast rate due to all potential factors – that means haste, expertise, and hit will show up in there. We also need the change in effective incoming attack rate, which will depend on avoidance. We’ll come back to those two later. Continuing on, the differential $dS$ ends up looking like this:
$dS = dR_{\rm SotR} \frac{R_{\rm att}}{R_{\rm SotR}} \frac{S}{(R_{\rm att}-R_{\rm SotR})} + dR_{\rm att} \frac{1-R_{\rm SotR}^2 T_{\rm buff}}{(R_{\rm att}-R_{\rm SotR})^2}&s=1$
where I’ve saved you a bit of tedious algebra and simplified the form some. Or as a textbook would say, “the derivation is left as an exercise for the reader” (man I hated when textbooks did that). So $dS$ also depends on the differentials of $R_{\rm SotR}$ and $R_{\rm att}$, as expected.
$R_{\rm att}$ will depend on avoidance, and it’s actually fairly straightforward, so we’ll do that first. $T^{(0)}_{\rm att}$ is the “true” boss swing timer, so that’s invariant, giving us $dR_{\rm att} = dF_{\rm av} / T^{(0)}_{\rm att}$. Using our earlier definition of $dF_{\rm av}$, we have:
$dR_{\rm att} = – dr_d (1-A_d/C_d)^2/(k_d f_d T^{(0)}_{\rm att})$
Which relates $dR_{\rm att}$ to our fundamental avoidance quantity, $dr_d$ dodge rating.
$R_{\rm SotR}$ depends on $R_{\rm HPG}$. So we need to find out how $R_{\rm HPG}$ depends on haste, expertise, and hit. That expression looks something like this:
$R_{\rm HPG} = [1 - (m-h) - (d+p-e)](1+\alpha_{\rm GC})(1+s)R_{\rm CS} \\ + [1 - (2m-h-e)](1+s)R_{\rm J} + R_{\rm BL}$
Where I’ve used $m$ to represent the hit cap, $h$ for hit, $d$ for dodge cap, $p$ for parry cap, $e$ for expertise, and $s$ for haste. $R_{\rm CS}$ and $R_{\rm J}$ are the cast rates for Crusader Strike and Judgement, respectively, before the effects of haste rating. The first term is the HPG rate due to Crusader Strike (which can miss, dodge, or be parried) and Grand Crusader. Since Grand Crusader grants HP on-cast rather than on-hit, it’s simply a linear scaling of our CS cast rate, so I’ve used $\alpha_{\rm GC}=0.2$ to represent that. Note that this makes another assumption – that we don’t waste any GC procs. That’s a fairly safe assumption in the 4.5-second CS system, though.
The second term is HPG due to Judgment, which is on spell miss, but gets contributions from both hit and expertise anyway. The final term is any income we might get from the Glyph of Blessed Life, but it could also incorporate any other constant income that’s unrelated to hit, expertise, or haste.
The reason I say it will look “something like this” is because I haven’t been careful to enforce hit and expertise caps in these expressions, partly because that would make the expressions a lot uglier to read. We don’t really need to worry about that though, as we can just assume we’re not silly enough to be stacking hit or expertise past their relevant caps. There are edge cases where this could happen (e.g. you have 7.5% hit and more than 7.5% exp, in which case exp suddenly becomes half as effective as it normally would be), but we’ll assume we’re not in those conditions (and handling that special case is easy to do in post-processing anyway). Differentiating, we have:
$dR_{\rm HPG} = (dh + de)(1+s)\left [ (1+\alpha_{\rm GC})R_{\rm CS} + R_{\rm J}\right ] \\ + ds [1-(m-h)-(d+p-e)](1+\alpha_{\rm GC})R_{\rm CS} \\ + ds [1-(2m-h-e)] R_{\rm J}$
Or, using our definition of $R_{\rm HPG}$ to simplify a bit,
$dR_{\rm HPG} = (dh + de)(1+s)[(1+\alpha_{\rm GC})R_{\rm CS} + R_{\rm J}] + ds(R_{\rm HPG}-R_{\rm BL})/(1+s)$.
$dh$, $de$, and $ds$ still need to be expressed in terms of their rating equivalents, but that’s fairly straightforward since they’re all linear:
$dh = dr_h/f_h$
$de = dr_e/f_e$
$ds = dr_s/f_s$
And there we have it. From here we can substitute backwards to express $dG$ and $dS$ in terms of the rating quantities $dr_h$, $dr_e$, and $dr_s$. That’s another task I’ll leave as an exercise for the reader, but I’ll quote you the results:
$dR_{\rm SotR} = \left ( \frac{dr_h}{f_h} + \frac{dr_e}{f_e} \right ) \frac{(1+s)[(1+\alpha_{\rm GC})R_{\rm CS}+R_{\rm J}]}{3(1-\alpha_{\rm DP})} + \frac{dr_s}{f_s}\frac{R_{\rm HPG} – R_{\rm BL}}{3(1-\alpha_{\rm DP})(1+s)}&s=2$
$dG =\left ( \frac{dr_h}{f_h}+\frac{dr_e}{f_e}\right ) \frac{(1+s)[(1+\alpha_{\rm GC})R_{\rm CS}+R_{\rm J}]}{3 R_{\rm att} (1-\alpha_{\rm DP})} + \frac{dr_s}{f_s}\frac{R_{\rm HPG}-R_{\rm BL}}{3 R_{\rm att}(1-\alpha_{\rm DP})(1+s)} \\ – \frac{dr_d}{f_d}\frac{R_{\rm SotR}(1-A_d/C_d)^2}{k_d T^{(0)}_{\rm att}R_{\rm att}^2}&s=2$
$dS =\left ( \frac{dr_h}{f_h}+\frac{dr_e}{f_e}\right ) \frac{(1+s)[(1+\alpha_{\rm GC})R_{\rm CS}+R_{\rm J}]S}{3(1-\alpha_{\rm DP})R_{\rm SotR}(R_{\rm att}-R_{\rm SotR})} \\ + \frac{dr_s}{f_s}\frac{(R_{\rm HPG}-R_{\rm BL})R_{\rm att} S}{3(1-\alpha_{\rm DP})(1+s)R_{\rm SotR}(R_{\rm att}-R_{\rm SotR})} \\ – \frac{dr_d}{f_d}\frac{(1-R_{\rm SotR}^2T_{\rm buff})}{(R_{\rm att}-R_{\rm SotR})^2}\frac{(1-A_d/C_d)^2}{k_d T^{(0)}_{\rm att}}&s=2$
We have one final task: substitute these expressions into the one for $dF_b$ so that we have one expression that links $dF_b$ to $dr_h$, $dr_e$, and $dr_s$. This is not a pretty job, so forgive me if I skip the messy algebra and quote you the result:
$dF_b =-\left ( \frac{dr_h}{f_h} + \frac{dr_e}{f_e}\right ) \left ( \frac{(1+s)[(1+\alpha_{\rm GC})R_{\rm CS} + R_{\rm J}]}{3(1-\alpha_{\rm DP})} \right ) \left [\frac{B_v'' - B_c S B_v' - B_c (1-S)B_v}{R_{\rm att}} + \frac{S(1-G)B_c(B_v'-B_v)}{R_{\rm SotR}(R_{\rm att}-R_{\rm SotR})} \right ] \\ – \frac{dr_s}{f_s}\left ( \frac{(R_{\rm HPG}-R_{\rm BL})}{3 (1-\alpha_{\rm DP})(1+s)} \right ) \left [ \frac{B_v'' - B_c S B_v' - B_c (1-S)B_v}{R_{\rm att}} +\frac{S(1-G)B_c(B_v'-B_v)}{R_{\rm SotR}(R_{\rm att}-R_{\rm SotR})}\right] \\ – \frac{dr_d}{f_d} \left ( \frac{(1-A_d/C_d)^2}{k_d T^{(0)}_{\rm att}}\right ) \left [ \frac{R_{\rm SotR}[B_v'' - B_c S B_v' - B_c (1-S)B_v ]}{R_{\rm att}^2} + \frac{(1-R_{\rm SotR}^2 T_{\rm buff}) (1-G) B_c (B_v’-B_v) }{(R_{\rm att}-R_{\rm SotR})^2} \right ] \\ – \frac{dr_m}{f_m}\frac{(1-B^{\rm (DR)}_c/C_B)^2}{k_B}(1-G)[S B_v' + (1-S) B_v]&s=1$
…. W T F. So much for the nicer, simpler, easier-to-understand block mechanics, eh?
This is, frankly, way too much algebra to conveniently lug around. We’re going to need a more compact way to write this if we’re going to get anywhere with it. So we’re going to define a few constants to help us with that. They are:
$F_{bG} = B_v” – B_c S B_v’ – B_c(1-S)B_v&s=1$
$F_{bB} = (1-G)[S B_v' + (1-S)B_v]&s=1$
$F_{bS} = (1-G)B_c(B_v’-B_v)&s=1$
$\beta_h = \frac{(1+s)[(1+\alpha_{\rm GC})R_{\rm CS}+R_{\rm J}]}{3(1-\alpha_{\rm DP})} \left [ \frac{F_{bG}}{R_{\rm att}}+ \frac{S F_{bS}}{(R_{\rm att}-R_{\rm SotR})R_{\rm SotR}} \right ]&s=2$
$\beta_s = \frac{(R_{\rm HPG}-R_{\rm BL})}{3(1-\alpha_{\rm DP})(1+s)} \left [ \frac{F_{bG}}{R_{\rm att}} + \frac{S F_{bS}}{(R_{\rm att}-R_{\rm SotR})R_{\rm SotR}} \right ]&s=2$
$\beta_d = \frac{(1-A_d/C_d)^2}{k_d T^{(0)}_{\rm att}} \left [ \frac{R_{\rm SotR} F_{bG}}{R_{\rm att}^2} + \frac{(1-R_{\rm SotR}^2T_{\rm buff}) F_{bS}}{(R_{\rm att}-R_{\rm SotR})^2} \right ]&s=2$
$\beta_m = \frac{(1-B^{\rm (DR)}_c/C_B)^2}{k_B}F_{bB}&s=2$
The factors $F_{bG}$, $F_{bB}$, $F_{bS}$ allow us to put our original equation for $dF_b$ in a simpler form:
$dF_b = -dG F_{bG} – dB_c F_{bB} – dS F_{bS}&s=1$.
And the factors $\beta_h$, $\beta_s$, $\beta_d$, and $\beta_m$, which represent the block scaling factors for hit/expertise, haste, dodge/avoidance, and mastery, respectively, allow us to put our final expression for $dF_b$ in a very simple form:
$dF_b = -\left (\frac{dr_h}{f_h}+\frac{dr_e}{f_e}\right ) \beta_h – \frac{dr_s}{f_s}\beta_s – \frac{dr_d}{f_d}\beta_d -\frac{dr_m}{f_m}\beta_m&s=2$
This is going to be much easier to work with. With this expression for $dF_b$, along with the previously-derived expressions for $dF_{\rm av}$, $dF_a$, $F_a$, $F_{\rm av}$, $F_b$, and all of the sub-factors we’ve defined in this post, we can finally start making comparisons between the different stats.
Unfortunately, that’s going to have to wait until next time, because this blog post is already much too long. Next time (Part 3), we’ll summarize the equations we have, and start plugging in reasonable numbers to see how the different stats stack up against one another. Then, in Part 4, we’ll use a simple Monte-Carlo simulation to see how well this analytical model holds up to a simple numerical model, which will tell us how well the assumptions we’ve made in this analytical model hold up under more realistic conditions.
Editor’s note: Since publication, I have found one or two minor errors in the calculation – mostly transcription errors that occurred, either in my hand-calculations or when converting into LaTeX format for the blog post. I have updated the post with these corrections without annotation.
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### 10 Responses to MoP Block Calculations – Part 2
1. I don’t undestand a single piece of any of that… but man I love that you do that
gronc
2. Weebey says:
I really hope you didn’t have to do all that algebra by hand; I know I would have gone crazy, and the odds of an error would have been in excess of 90%.
This is largely semantic, but I don’t think it’s the blocking change per se that makes it so nasty–moving to a second roll, in my opinion, actually makes more intuitive sense than the current system. In fact, when I started playing, I assumed it was a one roll system, and was mildly surprised when I found out how it really worked. Most of the nastiness seems to come the new version of SotR.
I have one small request for the next post: I would like to see the sign of the cross term d^2 F_{b}/(dr_{h}dr_{m}). Intuitively, this should split into two components with opposite signs: a positive component due to higher hit increasing G, which depresses the value of mastery; and a negative part that comes from higher hit leading to larger S. My guess is that with the new implementation the first term is going to be larger than the second, so that hit and mastery do not “help” each other (recall that, since we are looking to minimize damage taken, we actually want negative derivatives), but I’m not sure.
• Weebey says:
EDIT: I meant to say I assumed it was a two roll system!
• Theck says:
You’re right, most of the nastiness is due to SotR, and not the two-roll system itself.
And yes, I did all of the algebra by hand, and double-checked most of it by doing it again. I did actually miss something, as my proofreader pointed out to me (although a fat lot of good that does NOW) – Grand Crusader. Luckily that’s easily fixed by treating it as an extension of $R_{rm CS}$. I’m going to have to go back and edit that in, because it’s a legitimate oversight, and not one I can easily hand-wave away as “well, we’re just encapsulating that in $R_{rm CS}$.” Or rather, I could do that, but it would be disingenuous and not terribly intuitive.
• Theck says:
There, it should be properly updated for Grand Crusader now.
3. Zaephod says:
Theck, did you take into account that the new expertise is going to take away from dodge first, then from parry? If not, will it change anything about your model for holy power generation due to expertise? And it can’t be said enough, Thank you for all that you do!
src: http://us.battle.net/wow/en/blog/4544194/Dev_Watercooler_–_Mists_of_Pandaria_Stat_Changes-3_1_2012
• Theck says:
Yes, that’s taken into account. that’s the (d+p-e) term. If it were the old system, it would have to be something like (d-e)+(p-e). With all of those quantities being zero-bounded, of course.
4. queldan says:
Theck, you are one VERY insane man – and I don’t think the community would love any more any other way
5. Rizamp says:
/em’s brain explodes
6. Pingback: Warrior Block Calculations – Part 1 | Sacred Duty
• ### Theck’s Twitter
• @Slootbag I could maybe see an efficiency argument though. Large DPS gain for small relative HPG impact due to proc 1 day ago
• @Slootbag it is a sizable damage increase though, but so would just prioritizing AS with or w/o proc 1 day ago
• @Slootbag it's actually an HPG loss though. The probability of wasting a proc is very small, so the loss incurred by CS pushback is greater 1 day ago
• @Slootbag @ConverttoRaid TLDR rarely worth separating AS from ASGC, and none of the edge cases boost one over CS/J but not the other 1 day ago
• @Slootbag @ConverttoRaid or you want max DPS, in which case AS> all regardless of proc status. 1 day ago
• ### Ana’s Twitter
• Third time I've heard Clarity in the last hour. Good thing I still love this track! 6 hours ago
• Hey @riftmaker I made you assist again- I might be late and miss the start! Getting in the car soon, 2.5 hrs away 8 hours ago
• My grandfathers old friend, taking a photo of me, my two sisters and my mom, calls out: "OK ladies, tits up!" ......... What. 23 hours ago
• Turned my radio back on at the awesome \$25/6mo deal. Live EDC NYC playing!! Oh my xm, I missed you! 1 day ago
• Only 30 min into this car trip and already I desperately miss my satellite radio. I wonder how fast it would be active if I called to resub. 1 day ago
• ### Mel’s Twitter
• @Esoth @KihraOfTemerity @Kerriodos @methodtreckie @Anafielle Say what now? That sounds almost exactly like my philosophy. 4 days ago
• @SerrinneWoW Co-incidence, or the start of a pattern? :P 1 month ago
• @Rhidach @_vidyala @SerrinneWoW Clearly Sara isn't a foodie. Anyone who knows me knows how I go out of my way to avoid a good meal. 1 month ago
• @SerrinneWoW Are you stalking me? 1 month ago
• @_vidyala @SerrinneWoW @Jasyla_ @Kaleri_ @Chronis_ Liekwise! Sorry I had to miss dinner yesterday. 1 month ago
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http://math.stackexchange.com/questions/tagged/correlation?page=3&sort=newest&pagesize=15
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# Tagged Questions
For questions about correlation of two random variables. Use it with [tag: random-variables] and [tag: probability].
3answers
276 views
### Correlation between variables
I asked this question on stats SE but did not find a suitable answer so far. Maybe someone can help. Given n random variables x1,...,xn (one-dimensional). The following is known (corr() = Pearson ...
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### Relationship between x and y
Studying a financial model from a third party we see: 36month - 1.2 multipler 48month - 1.3333 multipler 60month - 1.5 multipler I am trying to determine if ...
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### Use Pearson's correlation coefficient on a matrix
I have a problem to interpret the following formula which is said to be the Pearson's correlation coefficient: r = \frac{N \left(\sum XY\right) - \left(\sum X\right) \left(\sum ...
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### If two Gaussian random variables are uncorrelated, they are statistically independent
I read in a textbook that when two gaussian variables are uncorrelated, then they are statistically independent? How can I prove that?
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### Sampling a distribution with restrictions: eliminating the correlation between two variables
I have a collection of 400.000+ word-pairs. Each word-pair has an association strength, which is a measure of how related the two words are to each other (as in cow-milk). Each word-pair also has a ...
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### Correlations between neighboring Voronoi cells
For a sequence $X_1,X_2,X_3,\ldots$ of random variables, what it means to say $X_1$ is correlated with $X_2$ is unambiguous. It may be that the bigger $X_1$ is, the bigger $X_2$ is likely to be. If, ...
1answer
222 views
### Expectation product of pairwise uncorrelated variables
Suppose I have three uncorrelated random variables $X, Y$ and $Z$ (discrete or continuous) such that $$\newcommand{\Cov}{\mathrm{Cov}}\Cov(X,Y)=0;\quad \Cov(Y,Z)=0;\quad \Cov(X,Z)=0 \tag{$\ast$}$$ I ...
3answers
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### correlated or independent
Let $(X_1, X_2)$ be a randomly chosen pair out of $\{1,2, \ldots, 20\}$ (draw without repetition). Are both events $$E_1:=\{X_1 \geq 8\}$$ and $$E_2:=\{X_2 \geq 12\}$$ positive or negative correlated. ...
1answer
186 views
### indicator variable are uncorrelated, if they are independent?
Is it true that if two indicator variables are independent then they are uncorrelated? If covariance =0 $\Rightarrow$ uncorrelated Does the covariance between indicator variables exist? thx
0answers
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### Correlation and Regression Question
Two separate tests are designed to measure a students ability to solve problems. Several students are randomly selected to take both tests and the results are: \begin{matrix} \text{Test A}(x) ...
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### Correlated Random Numbers Between A Range
I am trying to generate random numbers within a range say 57 to 107 which are correlated by a factor of 0.8. The numbers are for illustration simplicity only.
1answer
207 views
### 3D correlation visualization?
Incomer per person (x axis) correlates with life expectancy (y axis). These two indicators change over time (z axis). x correlates with y. Moreover, x and y both correlate with z. The question is: ...
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### Long time scale correlation
I have some EXTREMELY noisy data (standard deviation a is greater than the mean), but plotting it with a 15 data point running average does well to get a visual indication of the trending. I want to ...
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558 views
### Generalized variance
Generalized variance is the determinant of correlation matrix. Does increasing the off-diagonal entries (correlation coefficients) decreases the determinant? Is a proof available? All elements are ...
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### Pearson's Correlation, for comparing a PRNG?
Other than uniformity tests on random numbers of which can be done with other methods, I had wondered if the result of the Pearson Product Moment Correlation function would be an effective means to ...
3answers
404 views
### Bounds on off-diagonal entries of a correlation matrix
Assume that all the entries of an $n \times n$ correlation matrix which are not on the main diagonal are equal to $q$. Find upper and lower bounds on the possible values of $q$. I know that the ...
1answer
1k views
### What is the standard deviation of multiple correlated random variables subtracted from another random variable?
Wiki states that standard deviation of $X-Y$ is: $$\sigma_{x-y} = \sqrt { \sigma_x^2 + \sigma_y^2 - 2\rho\sigma_x\sigma_y }$$ I have a number (say 3) correlated random variables to be subtracted ...
2answers
136 views
### Calculating the correlation between subsequent values in a stream of numbers
I have a stream of integer values being generated $V_1,\cdots, V_n$ and want to calculate how the value of $V_{n+1}$ is correlated to $V_n.$ I would also like to calculate this at run time as ...
3answers
165 views
### Using Correlation for mouse gesture recognition
I am in need to build a mouse gesture recognition system which will compare given recognition to the the gestures in training data and will say where a given gesture best fits. I am planning to use ...
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### Calculating the variance of the ratio of random variables
I want to calculate $\newcommand{\var}{\mathrm{var}}\var(X/Y)$. I know that the solution is $$\var(X) + \var(Y) - 2 \var(X) \var(Y) \mathrm{corr}(X,Y) \>,$$ but, how do I derive it from "common" ...
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### Intuitive meaning of Pearson Product-moment correlation coefficient Formula
I can't understand the intuition behind Pearson Product-moment correlation coefficient Formula for bivariate data. The formula is : $\rho$ = cov(X,Y)/($S_x$ * $S_y$) where cov is covariance. $S_x$ and ...
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### How to 'minimize' correlation between series
Hi fellow mathemagicians, let's say that I have 3 series of numerical results (they represent 'drawdowns') : ...
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### Proving correlation coefficient = 1 or -1 given X and Y=a + bX
Given $X$ and $Y = a+bX$, I have to prove that: If $b \lt 0$, then $\rho = -1$. If $b \gt 0$, then $\rho = 1$. I've gotten to the point where I have: \rho = \frac{b \cdot \sigma_x }{ ...
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### Cross correlation in MATLAB HELP
I have values from two sensors stored in two vectors A and B. They both represent values of the sensors at times TA and TB which is stored in two other vectors(since it is not uniform sampling) Both A ...
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### example on variance of stochastic processes
I saw this expression in a book and I cannot understand how did he get this expression. Suppose $Z_t$ and $D_t$ are some stochastic processes and we have these expressions, \$Z_{t_k} - Z_{t_{k-1}} = ...
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### Correlation between Beta distributions
I have a Computer Science background and not very knowledgeable in Probability and Statistics. So excuse me if my question,notation, or language is flawed. Anyways, the problems is that we have two ...
1answer
135 views
### What is the characteristic formula for the addition of two lognormal distributions?
If I have two lognormal processes (X and Y) with mean and volatiilty for each and also correlation between the two, what is the characteristic formula of X + Y (i.e. what is the new mean and new ...
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231 views
### Calculate Correlation between 2 values
I have a random vector $X = ( X_1 , X_2 )$ that has a bidimensional normal repartition with mean $0$ and covariation matrix : \Sigma = \left( \begin{array}{ccc} 1 & q \\ q & 1 ...
4answers
666 views
### Going back from a correlation matrix to the original matrix
I have N sensors which are been sampled M times, so I have an N by M readout matrix. If I want to know the relation and dependencies of these sensors simplest thing is to do a Pearson's correlation ...
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### Statistics - Correlation between 2 sets of data
If i know the means and standard deviations of 2 sets of data, and i know the slope of the regression line, how can I find the correlation? edit Sample 1 SD: 12.37 Sample 2 SD: 7.00 Slope of ...
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http://math.stackexchange.com/questions/69828/asymptotics-of-lcm?answertab=active
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# Asymptotics of LCM
Let $\operatorname{LCM}(x_1,x_2,\ldots,x_n)$ be the least common multiple of the integers $x_i$.
How can one find the asymptotics of $\operatorname{LCM}(f(1),f(2),\dots,f(n))$ as $n$ approaches infinity if one knows the asymptotics of the strictly increasing function on the integers $f(n)$?
Edit: Is there some result if one assumes $f(n)$ has natural density in the primes ie is prime with probability $1/\ln(f(n))$, has an average of $ln(f(n))$ factors, $ln(ln(f(n))$ prime factors, and a $6/\pi^2$ probability to be squarefree.
And how to prove these average properties ?
Edit2: Instead of above estimates use: For every strictly increasing $f(n)$, consider instead the question for $g(f(n))$, which is a uniformly random integer in the range $(0.99f(n),1.01f(n))$ say, alternatively $(f(n)-1000,f(n)+1000)$.
Then im looking for asymptotics of $LCM[g(f(1)),g(f(2))...g(f(x))]$ as $x\rightarrow\infty$, given $f(n)$.
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7
Asymptotics for $\mathrm{LCM}(1,2,3...,n)$ is already interesting! – GEdgar Oct 4 '11 at 18:43
2
– lhf Oct 4 '11 at 20:33
Experiment: $(1,2,3\ldots,n)$ $(1,2,3...,n)$ These do look different from each other. That's why \ldots exists. (I know of a philosophy professor's web site where the advice to students about grammar, spelling, usage, punctuation, etc., says to put spaces between the dots in an ellipsis. Same thing here, but it's built in.) – Michael Hardy Oct 5 '11 at 0:56
So in that reference, LCM($1,2,\cdots,n)$ is asymptotically greater than $e^{ax}$ if $a<1$ and asymptotically smaller than $e^{ax}$ if $a>1$. But that is not yet an asymptotic estimate for this! – GEdgar Oct 6 '11 at 18:28
1
– anon Oct 6 '11 at 19:00
## 2 Answers
The paper The least common multiple of a quadratic sequence by Javier Cilleruelo, Compositio Mathematica (2011), 147: 1129-1150 gives the answer when $f$ is a quadratic polynomial. From the abstract:
For any irreducible quadratic polynomial $f(x)$ in $\mathbb{Z}[x]$ we obtain the estimate $\log\mathrm{LCM} (f(1),...,f(n)) = n\log n + B\,n + o(n)$ where $B$ is a constant depending on $f$.
This is a link to the paper in Arxix.
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Konowing asymptotics of $f$ is not enough. Consider functions $f_1(n)=2^n$ and $f(n)=p_1p_2\ldots p_n\;$, where $p_n$ is the $n$-th prime number.
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I don't understand your example. Those functions have different asymptotics and LCM(f(1),f(2),...,f(n)) has a different asymptotic for each of them. – Oscar Cunningham Oct 4 '11 at 19:05
10
Consider $f_1(n) = 2^n$ and $f_2(n) = 2^n-1$, then. Then the LCM of the second sequence grows much faster than that of the first. – Michael Lugo Oct 4 '11 at 19:10
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http://physics.stackexchange.com/questions/12983/the-philosophy-behind-the-mathematics-of-quantum-mechanics
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# The philosophy behind the mathematics of quantum mechanics
My field of study is computer science, and I recently had some readings on quantum physics and computation.
This is surely a basic question for the physics researcher, but the answer helps me a lot to get a better understanding of the formulas, rather than regarding them "as is."
Whenever I read an introductory text on quantum mechanics, it says that the states are demonstrated by vectors, and the operators are Hermitian matrices. It then describes the algebra of vector and matrix spaces, and proceeds.
I don't have any problem with the mathematics of quantum mechanics, but I don't understand the philosophy behind this math. To be more clear, I have the following questions (and the like) in my mind (all related to quantum mechanics):
• Why vector/Hilbert spaces?
• Why Hermitian matrices?
• Why tensor products?
• Why complex numbers?
(and a different question):
• When we talk of an n-dimensional space, what is "n" in the nature? For instance, when measuring the spin of an electron, n is 2. Why 2 and not 3? What does it mean?
Is the answer just "because the nature behaves this way," or there's a more profound explanation?
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2
These are the axioms of quantum mechanics, and axioms are by definition not deducible from "more profound explanation". Any interpretation of these things is just a change of words. For example, the vector space notion is roughly equivalent to superposition principle. – C.R. Jul 30 '11 at 13:09
The "philosphy" behind this Math is called Physics, usually. You try to ride the horse backwards, which is understandable for someone coming from math. – Georg Jul 30 '11 at 15:09
– genneth Jul 30 '11 at 22:05
1
– Kostya Jul 31 '11 at 12:40
## 7 Answers
Vector spaces because we need superposition. Tensor product because this is how one combines smaller systems to obtain a bigger system when the systems are represented by vector space. Hermitation operator because this allows for the possibility of having discrete-valued observables. Hilbert space because we need scalar products to get probability amplitudes. Complex numbers because we need interference (look up double slit experiment).
The dimension of the vector space corresponds to the size of the phase space, so to speak. Spin of an electron can be either up or down and these are all the possibilities there are, therefore the dimension is 2. If you have $k$ electrons then each of them can be up or down and consequently the phase space is $2^k$-dimensional (this relates to the fact that the space of the total system is obtained as a tensor product of the subsystems). If one is instead dealing with particle with position that can be any $x \in \mathbb R^3$ then the vector space must be infinite-dimensional to encode all the independent possibilities.
Edit concerning Hermitation operators and eigenvalues.
This is actually where the term quantum comes from: classically all observables are commutative functions on the phase space, so there is no way to get purely discrete energy levels (i.e. with gaps in-between the neighboring values) that are required to produce e.g. atomic absorption/emission lines. To get this kind of behavior, some kind of generalization of observable is required and it turns out that representing the energy levels of a system with a spectrum of an operator is the right way to do it. This also falls in neatly with rest of the story, e.g. the Heisenberg's uncertainty principle more or less forces one to have non-commutative observables and for this again operator algebra is required. This procedure of replacing commutative algebra of classical continuous functions with the non-commutative algebra of quantum operators is called quantization. [Note that even on quantum level operators can still have continuous spectrum, which is e.g. required for an operator representing position. So the word "quantum" doesn't really imply that everything is discrete. It just refers the fact that the quantum theory is able to incorportate this possibility.]
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You can get interference without complex numbers. Sound waves interfere, but they're real-valued. In a bound system of spin-1/2 particles such as a nucleus, you can actually use real-valued wavefunctions; it works fine, and wave interference effects do occur. What you can't have without complex numbers is a traveling wave representing a spin-1/2 particle. – Ben Crowell Jul 30 '11 at 17:48
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@Ben: I don't agree. Waves live naturally in the complex domain: they carry both amplitude and phase. You can pretend that these are two real numbers but this is actually not the case: the phase is $2\pi$-periodic. I.e. this is nothing else than polar decomposition of a complex number. – Marek Jul 30 '11 at 17:56
Thanks for the good answer. Just one point: "eigenvalues of operators represent physically possible values that can be measured." That's another fact that I don't have an intuition for. Eigenvalues is a mathematical concept, while observables and measurement are physical notions. How do they relate? – Sadeq Dousti Jul 30 '11 at 19:13
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Your first paragraph is a great summary of something I've been annoyed with ever since I opened my first QFT book. I think I have 5 of those books and not in a single one of them do they actually write why they need the math, they just STATE the equations needed and go along. But for us who are not mathematicians but for example programmers or engineers we want to know the why's :) Someone should write a QFT book that concentrates on the physical intuition, sort of like Feynmans "strange theory of light and matter" but more ambitious and math is allowed if it is explained :) – Bjorn Wesen Jul 30 '11 at 21:21
Scott Aaronson, himself a (quantum) computer scientist, thinks and writes about a number of these subjects in his paper Is Quantum Mechanics An Island In Theoryspace? - at least the "why complex numbers and not the reals or the quaternions?", and I'm pretty sure he mentions it in his 'Democritus' lectures as well.
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The lectures were fantastic, thanks for the link. – recipriversexclusion Aug 1 '11 at 19:13
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@recipriversexclusion: I have a CS background, and those lectures really made it click what quantum computing was about, in terms of computation (as opposed to the physics aspect). I hope it's the same for people coming from physics. Glad you liked them, in any case! – yatima2975 Aug 1 '11 at 21:25
First of all, the philosophy of quantum mechanics is hardly straightforward to most physicists. This has spawned an entire industry of "quantum interpretations" to the "quantum measurement problem".
Quantum mechanics is probably one of the best solutions to the problem of making mysticism mathematically precise. Quantum mechanics puts mysticism on a firm mathematically footing. Quantum mechanics is about Cosmic Consciousness and Reality.
Really, you can read about it straight from the horse's mouth in the book Quantum Questions, which is a compilation of mystical writings by the founders of quantum mechanics themselves. People like Werner Heisenberg, Erwin Schroedinger, Albert Einstein, Louis de Broglie, Jeans, Max Planck, Wolfgang Pauli and Arthur Eddington. They're not mere "fringe" physicists, even though "fringe" physicists like Jack Sarfatti, David Bohm, Amit Goswami, John Hagelin and Frank Tipler often do have a point...
Another book you can read is Quantum Enigma.
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Is the answer just "because the nature behaves this way," or there's a more profound explanation
I would say , yes to "because nature behaves this way". It is the most economical description of experimental data using mathematics, to date.
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It is strange that you are not very comfortable with Hilbert spaces because actually continuous functions form Hilbert space and physics is all about functions. The general difference with classical mechanics is in the fact that classical mechanics was formulated long before people who used and learned (and even developed) it could understand manifolds and Lie algebras while for the time of quantum mechanics the idea of Hilbert space and all that stuff became more or less natural.
The same with the rest. You could formulate QM without complex numbers (in principle), but this would be the same as formulation of Maxwell's equations without vectors. People say that Maxwell's equations actually were formulated before physicist were familiar enough with the vectors and it was a nightmare.
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Hilbert spaces are not necessarily composed of continuous functions. For example, all Euclean spaces are Hilbert spaces, but the vectors inside are not continuous functions. – C.R. Jul 30 '11 at 15:41
@Karsus Ren, If you read my answer carefully you'd mention that I did not say that Hilbert spaces are necessarily composed of continuous functions. Probably I should had used 'e.g.' instead of 'actually'. – Misha Jul 30 '11 at 15:49
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@Misha: "e.g." wouldn't be much better. Continuous functions are just bad example. Prime examples of Hilbert spaces come from measure theory ($L^2(X, \Omega, \mu)$ spaces) and measurable functions (built on the Borel sigma algebra) are in general much worse behaved than continuous ones; but one sacrifices this in order to obtain better topological properties of the space -- one can take limits and stay in the same space. This is in fact the key difference between Riemann and Lebesgue integral. – Marek Jul 30 '11 at 16:30
@Marec, when one says "actually sheep are mammal" it is obvious that not every mammal is sheep. Surprisingly, it does not work when you speak of Hilbert spaces. I was trying to point out that Hilbert space is not something complex and/or artificial. It is something which is studied at school but at the different level of abstraction. – Misha Jul 30 '11 at 16:41
@Misha: but you are not saying continuous functions are measurable functions. You are explicitly saying that they form a Hilbert space (implying there's nothing else in it). This is true about as much as saying that rational numbers form the real line $\mathbb R$, while in fact in both cases these are only dense subspaces. I agree with your sentiment that Hilbert space is nothing artificial. I just wanted you to be bit more precise. – Marek Jul 30 '11 at 18:00
show 1 more comment
I asked the same questions some time back.
Read :-)
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Quantum mechanics is a real time solution of Newton's equation. If you solve Newton's equation in real numbers you get Newton's solution. If you solve Newton's equation in complex numbers you get quantum solution and the difference between the two solutions is relativistic. Google into Roger Anderton's time dependent Newton's equation and see that the difference between Newton's mechanics in that Newton used real numbers and when Newton's equation is solved in complex numbers it gives quantum mechanics and the difference between quantum mechanics and newton's mechanics is relativistic mechanics or quantum = newton + relativiity. It is quite ineteresting article written by an english engineer
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– Heidar Jul 30 '11 at 16:44
– Sadeq Dousti Jul 30 '11 at 16:51
Joe, Welcome to Physics.SE. Our participants here range from professional physicists to beginning students, but we all have an interest in the topic. I've deleted your other answers up to now because they seem be philosophical with little physical content. – dmckee♦ Jul 30 '11 at 16:51
Thanks for the Link Sadeq, kheyli bahale! ;) – Heidar Jul 30 '11 at 17:12
– Heidar Jul 30 '11 at 17:13
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http://physics.stackexchange.com/questions/22575/how-many-photons-does-it-take-to-measure-a-linear-polarization/22629
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# How many photons does it take to measure a linear polarization?
A star emits perfectly (100%) linearly polarized light at an arbitrary angle.
How many photons must you detect to measure this angle to a precision of n binary digits? (with greater than 50% probability of being correct.)
What is the best strategy for performing this measurement?
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1
Is this by any chance motivated by some sort of homework-like exercise? The first part in particular reads like a homework question. – David Zaslavsky♦ Mar 19 '12 at 17:37
What do you know about the measuring device? If it is perfect then you can get an exact result (as many bits as you like) with two measurements. – yohBS Mar 19 '12 at 17:42
@David Zaslavsky: Not homework, but a very basic question. If this is a known result, please give references. But really, I don’t know if the answer is n, 2n, $n^2$, or $2^n$. – Jim Graber Mar 19 '12 at 21:24
@ yohBS: I am assuming the measurement device is an ideal Malus law calcite crystal type splitter followed by a pair of detectors. If a better measuring device is available, I’m not aware of it. – Jim Graber Mar 19 '12 at 21:26
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@yohBS : an ideal de vice can give you an exact measurement with two measurement provided you have infinitely many photons. If the number of photons is limited, quantum mechanics comes into play and forbids a perfect measurement. For example with a single photon, a measurement output is binary, and there is no way to get more than one bit of information out of the polarization of a photon. Jim Graber's question is about this transition from this single photon regime to the bright light classical regime you're speaking about. – Frédéric Grosshans Mar 22 '12 at 20:45
show 1 more comment
## 3 Answers
Edited to add part III
$\newcommand\ket[1]{\left| #1 \right>} \newcommand\ketbra[1]{\ket{#1}\left< #1 \right|}$ If you have $N=2^n$ possible polarization at angle $k\frac\pi N$ with $k\in[ 0 .. N-1 ]$, the possible states of $\nu$ photons can be written $\ket{k,\nu}=\left(\frac{\ket0+e^{2ik\pi/N}\ket1}{\sqrt2}\right)^{\otimes\nu}$.
A key parameter for the probability of confusing two different angle $k$ and $k'$ is the scalar product $$\left<k',\nu\middle|k,\nu\right>=\left(\cos\left([k-k']\frac\pi N\right)\right)^\nu$$
## I. Discrimination between two states
If I restrict the problem to the discrimination between two angles $k$ and $k+1$, the optimal theoretical measurement is well known and is given by the Helstrom discrimination. The success probability is given by $$P_c=\frac12\left[1+\sqrt{1-\left|\left<k',\nu\middle|k,\nu\right>\right|^2}\right]$$ If you want $P_c\ge 1-\varepsilon$ with $\varepsilon\ll1$, this becomes $$\begin{align} \left(\cos\frac\pi N\right)^\nu&\le2\sqrt{\varepsilon(1-\varepsilon)}\\ \nu&\ge\frac{\frac12\log\varepsilon+\log2+\frac12\log(1-\varepsilon)}{\log\left(\cos\frac\pi N\right)} \end{align}$$ If $N=2^n\gg1$, we have $\cos\frac{\pi}{N}\simeq 1-\frac12\left(\frac\pi N\right)^2$ and therefore $\log\left(\cos\frac\pi N\right)\simeq-\frac{\log e}{2}\left(\frac\pi N\right)^2$ We have then $$\nu\ge N^2\frac2{\pi^2}(-\ln\varepsilon-\ln2-\ln(1-\varepsilon))$$ So in this case, the number of needed photons in of the order of $N^2=2^{2n}$, with a prefactor depending on $\varepsilon$.
The Helstrom measurement itself seems difficult to perform, but an adaptative set-up (the Dolinar receiver) compining interferometric displacement and single photon detectors allows to achieve this performance when the states are coherent states. I suppose a similar set-up can be built for polarizations.
## II. The more general problem, estimated through entropy
Your problem is not a discrimination problem, but the distinction among many angles. The amount of information one can encode in $\nu$ photons identically polarized along an arbitrary angle $\theta$ is given by by the von Neuman entropy of the mixed state $\rho_\nu=\int d\theta\ketbra{\theta,\nu}$. Rewriting $\ket{k,\nu}$ and some algebra can show that this entropy is the entropy of the binomial distribution of parameters $p=\frac12$ and $n=\nu$.
This entropy is $\frac12\log\frac{\pi e\nu}{2} + O\left(\frac1\nu\right)$. An order or magnitude of the $\nu$ needed to distinguish between $2^{2n}$ polarization is then given by $$\begin{align} n\simeq\frac12\log_2\frac{\pi e\nu}{2}\\ \nu\simeq\frac{2^{2n+1}}{e\pi}, \end{align}$$ Which is of the same order of magnitude than the discrimination.
## III. Unambiguous discrimination
Another way to look at the problem is the number of photon needed to perform a na unambiguous discrimination (UD) measurement. A UD measurement is a measurement which never makes an error, but sometimes fails. Of course, given the non-orthogonality of the states, this failure probability can never be zero, and is 100% when the number of photon is too small. The problem is a special case of the one Chefles and Barnett studied here here. The first non-zero success probability is for $\nu=N-1$, but this probability is small ($\nu2^{-\nu}$). It increases every time $\nu$ increases by 2, but I've found no simple expression (and I've tried some time for a paper). I guess (from the previous parts) the probability becomes non-negligible only when $\nu\sim N^2$.
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After thinking about it overnight, I think I can prove an upper limit of order 2^($2n+1$). Early in the search, I think it is advantageous to search at multiple angles, say 0, 45, 90 and 135 degrees equally until you now approximately where the angle is. Thereafter, you can search primarily at 45 degrees away from the polarization angle. That way your probability of success is approximately 50% for each photon, and you gain the maximum amount of information per measurement. Since there are $2^n$ equally spaced possibilities and the standard deviation decreases with the square root, you need to make 2^($2n$) measurements. The extra plus one allows you to measure on both sides and also allows for the greater inefficiency of the early part of the search.
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I'm going to have to agree with Jim Graber's ballpark calculation. I ran a couple of special cases assuming you just leave the polarizer fixed and take the ratio of the detection events as the tangent of the polarization angle. Ignoring the ambiguity relating to the polarity of the tangent (we can't readily distinguish +20 degrees from -20 degrees e.g.) I found the least favorable case to be the 45 degree case. In that example, one million detection events would give us a Poisson expectation of 500,000 with a standard deviation of 700 in each detector. (I think that's right: variance = expectation?) Examining this graphically gives me a range of .001 radians, which is 10 binary digits of accuracy. Of course to double the accuracy you need four times the hit count. This is pretty close to Jim's formula.
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http://mathhelpforum.com/pre-calculus/113611-rational-zero-print.html
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# Rational Zero
Printable View
• November 9th 2009, 10:16 PM
flexus
Rational Zero
list all possible rational zeros of f.
f(x)=3x^4 + 4x^3 -5x^2 -8
• November 10th 2009, 12:22 AM
Bacterius
Do you know what a rational zero is ? It actually is a non-complex root of the equation. Let us solve $f(x) = 0$ :
$3x^4 + 4x^3 -5x^2 -8 = 0$
Can you solve it now ?
• November 10th 2009, 01:02 AM
CaptainBlack
Quote:
Originally Posted by Bacterius
Do you know what a rational zero is ? It actually is a non-complex root of the equation. Let us solve $f(x) = 0$ :
$3x^4 + 4x^3 -5x^2 -8 = 0$
Can you solve it now ?
WRONG!, that would make $\sqrt{2}$ a rational zero of $x^2-2$
A rational number is a number that can be written as the ratio of two integers.
This question is meant to be addressed using the rational root theorem, which tells you that any rational root of a polynomial equation with integer coefficients is the ratio of a factor of the constant term to a factor of the coefficient of the highest power occuring.
So for:
$3x^4 + 4x^3 -5x^2 -8 = 0$
the rational roots are amoung $\pm1,\ \pm 2,\ \pm 4,\ \pm 8,\ \pm1/3,\ \pm 2/3,\ \pm 4/3,\ \pm 8/3$
You try each of these to find which are actual roots.
CB
• November 10th 2009, 01:07 AM
Bacterius
Aah
Sorry sorry ! Here, a thanks to forgive me (Nod)
All times are GMT -8. The time now is 12:48 AM.
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http://physics.stackexchange.com/questions/478/is-it-possible-to-project-a-magnetic-field-at-a-location-in-space/979
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# Is it possible to project a magnetic field at a location in space?
A magnetic field strength drops-off quickly as the distance from a magnet increases.
Is there any way to use electromagnetic fields to create a magnetic field at a location. For example, if there are strong electromagnetic fields intersecting at a location from many strong transmitters, could a constructive interference be created which creates a local magnetic field?
The engineering idea is that if you create an array of radio transmitters at a strong strength working together, could there be a interference pattern at a point far from the transmitters (maybe 100 meters away) where the fields produce a magnetic field at that location.
I think this could be done as a EM field is composed of a electric field and a magnetic field, shouldn't there be a way to make the magnetic field strong at a location in space.
A positive consequence of this would be that a magnetic placed at that location 100 meters away could be moved by the field. Also, if the field was alternating, rather than steady, then a electromagnet at that location could be alternating, and the force on the electromagnet could provide motive force.
I guess since I have never heard of this, it cannot be done. But then again, it was once said that levitating a normal magnet could not be done, then someone did it with a spinning magnet which levitates above another magnet, now it is a popular toy. The reason it works is that the magnet is spinning so that it cannot flip over. See spin stabilized magnetic levitation.
--- Update
Hi, I would like to clarify, I wish to move a physical object, a small magnet at a distance, not to transmit power to the object. Thanks!
--- Update 2
My engineering goal for this is a small magnetic, spin stabilized, being pushed from a distant location. A small spin stabilized magnet pushed from below by some "projected" magnetic field could potentially overcome the force of gravity. Imagine a thin lightweight spinning disk with a magnet in the center. If a projected magnetic field is pushing it upwards then it would stay stable and could be projected from ground into space.
This might be done with an array of thousands of transmitters on the ground, somewhere in the desert far away from people. All transmitters could focus a small magnetic field at this magnet.
Alternatively, if the magnetic field alternates with a frequency, then the magnet in the levitated device could also be alternating via electromagnetism, it could be powered by an battery or microwave power transmitted to the object. This is more or less an engineering problem if there is some basic science which can allow this to be a potential reality.
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## 6 Answers
This is possible, it's just very very difficult to do.
People regularly do this with the electric field of light to move dielectric particles (insulators) in the lab, and the technique is known as "optical tweezers":
http://en.wikipedia.org/wiki/Optical_tweezers
The reason you don't want to try this with magnetic particles is that the magnetic field of light is much weaker than the electric field, or rather that it doesn't interact very much with most magnetic materials.
Greg's answer above is half correct - The fact that these fields oscillate in time means that the applied force would oscillate as well, however, the field gradient is exploited instead to make optical tweezers work.
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Could powerful lasers pointed up at some object scale up the optical tweezers effect by creating millions of optical tweezers, lift an object upwards? – Phil Feb 3 at 14:10
billions, trillions, of optical tweezers all over the bottom of a spin stabilized object – Phil Feb 3 at 14:43
Nope, this method is not feasible.
You are right that the EM field is "Electro"+"Magnetic", but these contributions are oscillating in time. Every interference they produce will oscillate in time as well, even if the peak magnitude of these oscillation can (theoretically) be high (that is the constructive interference you mentioned).
Let's say you just take the magnetic field vector of your $i$th beam at the position of your (pointlike) object:
$B_i(t) = B_{0i} \sin(2 \pi \nu_i t + \phi_i)$
If you have a $n$ waves that constructively interfere, which have to have the same frequency ($\nu_i = \nu$), and the the total magnetic field will be
$B(t) = (\sum_{i=1}^n B_i) \sin(2 \pi \nu t) = n B \sin(2 \pi \nu t)$
with the final, very simplified step being true if all the field magnitudes are the same. Thus your field will still have the oscillation at the original frequency (eg. radio frequency (~ tens-hundreds of MHz) or laser beam (hundreds of THz). No macroscopic object would react to this high frequency perturbation due to inertia. And even if they would, the resulting motion wouldn't be a push or a pull, but oscillation on the time scale of $1/\nu$. That's why in general, motion is induced with stationary or slowly moving fields.
... and this haven't even taken into account a lot of practical limitations to this scheme.
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What about an opposing oscillating magnetic field at the same frequency generated by the object which is to be moved? – Phil Nov 18 '10 at 4:09
I mean - north/north repels, south/south repels. Therefore a push. – Phil Nov 18 '10 at 4:26
Electromagnetic fields (under these conditions we discussed here) follow the rules of superposition: the field strength is just the sum of the different waves. If your object generates waves that is just one more $B_i$ in the above sum. Also, these waves don't "move" each other, they affect the material: the field changes the motion of a charged particle/piece of object. – Greg Nov 18 '10 at 6:24
Hi Greg, what I'm trying to do is find a condition under which I can get the object to move under this strong electromagnetic field where most likely there is a strong sum of the different waves, a strong constructive interference. – Phil Nov 18 '10 at 14:34
So what I was thinking is that this is a magnetic field at a location in space (at the constructive interference point), and the field is alternating + then - very quickly. So, if I had a electromagnet which is alternating + and - quickly in a opposite phase to the field at this point, then couldn't this electromagnet feel a force acting upon it in a linear direction, if the + and - were opposite to the + and - of the field at that point. – Phil Nov 18 '10 at 14:36
show 2 more comments
Sure, it's possible. That's how sets of cell phone radio towers work. In a typical setup, several of these towers are placed in a line, some distance outside of a town, such that the line is roughly perpendicular to the direction to the town. The placement of the towers is designed so that the signals reinforce each other as much as possible only in the town. That way, most of the cell towers' transmitted power gets directed toward the area where most people are actually using their cell phones, and you don't waste a whole lot of power by transmitting into open wilderness.
With the cell towers, though, the only things that really get pushed around by the EM waves are electrons in people's cell phones. Theoretically, there's no reason the same concept couldn't be used to move something larger, like a magnet, but you'd need vastly more powerful transmitters - perhaps more powerful than anyone knows how to build with current technology (depending on how large of a magnet you're trying to move, and how far away it is). Still, it's just an engineering problem.
Also, a side note: it's generally the electric field that's responsible for moving things around, not the magnetic field. Magnetic fields themselves don't transfer energy, they only change the direction of existing motion. But the two kinds of fields mix when you change your reference frame, so it's kind of a semi-arbitrary distinction anyway.
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I wish to move a small magnet at a large distance from the transmitters. – Phil Nov 10 '10 at 13:59
An optical lens concentrates radiating electromagnetic fields and these fields (including the magnetic B-field) may be much higher than at the source, so yes it is possible. For more constant and strong B-fields, it also depends on the type of field you want to generate (cannot disobey Earnshaw theorem, Maxwell equations, etc.). In general today, static and dynamic magnetic fields are usually generated in a more simple fashion with fields at the source higher than at the target. That doesn't mean it is impossible the other way around, and if doable and practical it would be a great breakthrough, since generating magnetic fields in a region higher than in the source, would allow for us to generate very high B-fields in general, fields of strength that would be impossible to generate today because they would destroy the source by electrical breakdown. Such high B-fields could have major applications for nuclear fusion, ion rockets, nuclear magnetic resonance, and particle accelerators. It is not a trivial problem. Possibly a solution could be found considering dynamic non-radiating fields (Schott discovered these) or long wavelength evanescent wave emissions (wireless energy transfer) with some kind of wave interference/cancelation in the source. Go for it.
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This can certainly be done, as induction chargers work pretty much that way. That said, this technique is not very efficient and it is hard to scale at large distances.
For large distances, one of the most popular ideas is using microwave transmission.
There's a very good article over at wikipedia with a round-up of all these technologies with pros and cons.
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Hi, I would like to clarify, I wish to move a physical object, a small magnet at a distance, not to transmit power to the object. Thanks! – Phil Nov 10 '10 at 13:58
1
Ok, microwave transmission is not very relevant then, but induction chargers do work by "shooting" a magnetic field and would move a small magnet. As I said, though, there are practical problems of efficiency and design which make this infeasible (but not impossible). – Sklivvz♦ Nov 10 '10 at 14:26
Voted up for the links. :) Thanks. – Robin Maben Nov 27 '10 at 5:42
projecting a magnetic field under a magnet to force movement and levatison can give forward inertia to produce perpetual Motion
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the magnetic feild is projected from with in the ship and each time its projected it makes the ship move and a smaller magnetic feild makes it stay stationary – user14045 Oct 14 '12 at 22:41
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http://oeis.org/wiki/User%3APeter_Luschny/SchinzelSierpinskiConjectureAndCalkinWilfTree
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This site is supported by donations to The OEIS Foundation.
# The Schinzel-Sierpinski conjecture and the Calkin-Wilf tree.
KEYWORDS: Schinzel-Sierpinski conjecture, Calkin-Wilf tree, semiprimes, supersingular primes.
Concerned with sequences: A000790, A001358, A002267, A060324, A062251, A108574, A108764, A182914, A182915.
## Introduction
A conjecture of Schinzel and Sierpinski asserts that every positive rational number r can be represented as a quotient of shifted primes, that
$r = \frac{p+1}{q+1} \quad (p, q$ primes).
In fact we would expect there to be not just one such pair of primes, but infinitely many pairs. Schinzel and Sierpinski wrote:
Tout nombre rationnel positif peut être représenté d'une infinité de manières sous la forme (p+1)/(q+1) ainsi que sous la forme (p-1)/(q-1), où p et q sonst des nombres premiers.
We will determine (computationally) the smallest prime p so that r = (p+1) / (q+1) for some q prime. This pair of primes (p,q) we will call the Schinzel-Sierpinski primes of r (if such primes exist). Thus we assume a map
$\mathbb{Q}^{*} \rightarrow \mathbb{P}^2, \quad r \mapsto (p,q) \, .$
Additionally we define an encoding with sgn(p,q) = 1 if p < q and -1 otherwise
$\mathbb{P}^2 \rightarrow \mathbb{Z}; \ (p,q) \mapsto \pm pq \, .$
This function will be called the Schinzel-Sierpinski encoding of the (positive) rational number r = (p, q), p, q primes. In the unlikely case that no pair of prime numbers corresponding to a given r can be found we set by convention p = q = 1. Additionally we define 0 → 0.
Conversely, let us for instance decode 39. The prime factors of 39 are 3 and 13, thus 39 represents 7/14 = 2/7 and -39 represents 7/2.
## The case of integers
The Schinzel-Sierpinski primes can be easily computed by a simple search.
```SchinzelSierpinskiPrimes := proc(a, type)
local p,q,P; P := select(isprime,[$1..4000]):
if a = 1 then RETURN(2) fi;
for p in P do for q in P do
if (p+1)/(q+1) = a then
`if`(type="numer", p, q); RETURN(%) fi
od od; print("More primes needed!") end:
```
Alois Heinz pointed out that there is a much more efficient and elegant way to compute the denominators (q) returned by this function if the input 'a' is an integer 'n'.
```a := proc(n) local q;
q := 2;
while not isprime (n*(q+1)-1) do
q := nextprime(q);
od; q
end:
```
However, as long as the Schinzel-Sierpinski conjecture is not proved this is a classical implementation of a potential infinite loop. Wikipedia: "An infinite loop is a sequence of instructions in a computer program which loops endlessly, either due to the loop having no terminating condition, or having one that can never be met."
The full and efficient implementation can be found in the appendix. Note that we use a single function for both the numerator and the denominator of r.
```A062251 := n -> SchinzelSierpinskiPrimes(n,"numer"):
A062251 = 2, 5, 11, 11, 19, 17, 41, 23, 53, 29, 43, 47, 103, 41,
59, 47, 67, 53, 113, 59, 83, 131, 137, 71, 149, 103, 107, 83,
A060324 := n -> SchinzelSierpinskiPrimes(n,"denom"):
A060324 = 2, 2, 3, 2, 3, 2, 5, 2, 5, 2, 3, 3, 7, 2, 3, 2,
3, 2, 5, 2, 3, 5, 5, 2, 5, 3, 3, 2, 5, 2, 13, 3
```
## The Calkin-Wilf tree
We give the two functions needed to compute the Calkin-Wilf tree. For details and generalizations we refer the reader to the blog "Stern's Diatomic Array" (see the link below ).
```DijkstraFusc := proc(m) option remember;
local a, b, n; a := 1; b := 0; n := m;
while n > 0 do
if type(n, odd) then b := a+b else a := a+b fi;
n := iquo(n,2) od; b end:
// returns the n-th level of the Calkin-Wilf tree
CalkinWilfTree := proc(n) local k;
seq(DijkstraFusc(k)/DijkstraFusc(k+1),k=2^(n-1)..2^n-1) end:
```
## The case of rational numbers
Now we apply the Schinzel-Sierpinski encoding to the Calkin-Wilf tree.
```SchinzelSierpinski :=
proc(a,b) local p,q,P; P := select(isprime,[$1..4000]):
for p in P do for q in P do
if (p+1)/(q+1) = a/b then RETURN(`if`(a<b,-p*q,p*q)) fi
od od; print("More primes needed!") end:
SchinzelSierpinskiEncodedCalkinWilfTree := proc(n) local l;
seq(SchinzelSierpinski(numer(l),denom(l)),l=CalkinWilfTree(n)) end:
seq(print(SchinzelSierpinskiEncodedCalkinWilfTree(i)),i=0..5);
```
### The case r = (p + 1) / ( q + 1)
```// The encoded tree [form r=(p+1)/(q+1)] in breadth-first traversal.
A182914 = 4, -10, 10, -33, 15, -15, 33, -22, 6, -209, 133,
-133, 209, -6, 22, -57, 667, -91, 65, -14, 589, -1189, 39,
// The rightmost branch of the encoded tree,
// i.e. the codes for the natural numbers.
seq(SchinzelSierpinski(n, 1), n = 1..42);
A182915 = 4, 10, 33, 22, 57, 34, 205, 46, 265, 58, 129,
141, 721, 82, 177, 94, 201, 106, 565, 118, 249, 655, 685,
```
There is a third interesting sequence here: the numbers which do not appear in the encoded tree (apart from the sign). This is the complement of A182914 relative to the semiprimes A001358. For instance the squares of odd primes are not members of A182915 (they all map to '1' but do not meet the minimality condition).
### The case r = (p − 1) / ( q − 1)
Schinzel and Sierpinski considered in their conjecture besides r = (p + 1)/(q + 1) also the form r = (p − 1)/(q − 1); in fact both forms are easy consequences of a more general conjecture. So let us look also at this case. The encoded tree starts
```4,
-6,6,
-21,35,-35,21,
-10,221,-77,55,-55,77,-221,10,
```
This leads to another set of sequences.
```// The encoded tree [form r=(p-1)/(q-1)] in breadth-first traversal.
4, -6, 6, -21, 35, -35, 21, -10, 221, -77, 55, -55, 77, -221, 10,
-33, 46513, -493, 377, -119, 187, -1333, 559, -559, 1333, -187, 119,
// The rightmost branch of this encoded tree,
// i.e. the codes for the natural numbers.
4, 6, 21, 10, 33, 14, 145, 51, 57, 22, 69, 26, 265, 87, 93, 34, 721,
38, 2101, 123, 129, 46, 141, 485, 505, 159, 545, 58, 177, 62,
```
## Is there a connection with supersingular primes?
Is there any pattern in the encodings of the CW-tree? If we look only at the positive values of the first five lines
```4,
10,
33,15,
22,6,209,133,
57,(667),91,65,14,(589),(1189),39,
```
we see 13 values which are products of exactly two supersingular primes (from the sequence A108764 by Jonathan Vos Post). This is not a very significant concordance; however, it is remarkable how it points in the right direction: to group theory. It is known that the set of shifted primes generates a subgroup of the multiplicative group of rational numbers (see the references).
And the conjecture of Schinzel and Sierpinski can be formulated in terms of group theory. Let Q* denote the multiplicative group of positive rationals and let G be the subgroup generated by shifted primes p+1 then the conjecture of Schinzel and Sierpinski can be stated as: Q* = G.
The supersingular primes are the set of primes that divide the group order of a big sporadic group, the so called Monster group. In what way the group orders of other (sporadic) groups are possibly reflected in the encoded Calkin-Wilf tree has not yet been studied to my knowledge.
## Appendix
```CalkinWilfTree_level := proc(n)
local k, DijkstraFusc;
DijkstraFusc := proc(m) option remember;
local a, b, n; a := 1; b := 0; n := m;
while n > 0 do
if type(n, odd) then b := a+b else a := a+b fi;
n := iquo(n,2)
od;
b end:
seq(DijkstraFusc(k)/DijkstraFusc(k+1),
k=2^(n-1)..2^n-1) end:
for i from 1 to 6 do CalkinWilfTree_level(i) od;
```
```#-- A fast and save implementation.
SchinzelSierpinski := proc(l, primetype, outtype)
local a, b, r, p, q, uno, sgn, SearchLimit;
a := numer(l); b := denom(l);
SearchLimit := 40000; q := 2;
sgn := `if`(a < b, -1, 1);
uno := `if`(primetype = "plus", 1, -1);
do r := a*(q + uno);
if r mod b = 0 then p := r/b - uno;
if isprime(p) then
if outtype = "pri" then RETURN(p/q)
elif outtype = "cod" then RETURN(sgn*p*q)
elif outtype = "num" then RETURN(p)
elif outtype = "den" then RETURN(q)
else ERROR("Wrong type specification!")
fi fi fi;
q := nextprime(q);
if q > SearchLimit then
ERROR("SearchLimit reached! Input was: ",a, b)
fi
od end:
for i from 1 to 3 do
seq(SchinzelSierpinski(l,"plus","pri"),
l=CalkinWilfTree_level(i)) od;
```
The parameters in the call SchinzelSierpinski(l,para1,para2) are for para1 'plus' for the type r = (p+1)/(q+1) and 'minus' for the type r = (p-1)/(q-1). For para2: 'pri' stands for the encoding to a rational number p/q, p, q, primes; 'cod' for the encoding to semiprimes; 'num' and 'den' returns the numerator and the denominator of the rational number respectively.
## References
Calkin, Neil; Wilf, Herbert (2000), "Recounting the rationals", Amer. Math. Monthly 107 (4): 360–363
E. Dijkstra, Selected Writings on Computing, Springer, 1982, p. 232.
Elliott, P. D. T. A. "The multiplicative group of rationals generated by the shifted primes. I." J. Reine Angew. Math. 463 (1995), 169–216.
Elliott, P. D. T. A. "The multiplicative group of rationals generated by the shifted primes. II." J. Reine Angew. Math. 519 (2000), 59–71.
Matthew M. Conroy, A sequence related to a conjecture of Schinzel, J. Integ. Seqs. Vol. 4 (2001), #01.1.7.
Schinzel, A. and Sierpinski, W. "Sur certaines hypothèses concernant les nombres premiers." Acta Arith. 4, 185-208, 1958. erratum 5 (1958) p. 259.
## See also
Stern's Diatomic Array
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|
http://math.stackexchange.com/questions/171974/application-of-the-chebyshev-inequality?answertab=votes
|
Application of the Chebyshev inequality
was revising my stuffs for my stochastics exams and came across this question that I couldn't figure my way around..
Let $X_1,\ldots,X_n$ be independent, identically distributed random variables with $E(X_1) =a$ and
$$S_n = \frac{1}{n} \sum_{i = 1}^n X_i$$
Using the Chebychev inequality, give the smallest possible value of $x$, where $\mathbb{P}(\left | S_{100} - a \right | \geq x) \leq 0.01$, in the case where $X_{1}$ with the parameters $p \in [0,1]$ is:
a) binomially distributed ($B(10,p)$)
b) geometrically distributed
I gathered that $0.01 =\frac{Var(X))}{x^2}$ owing to the Chebychev inequality and figured that if I could derive $Var(X)$, I could then get $x$. And using the given distributions (binomial or geometric), I attempted to solve for $Var(X) = E(X^2) - (E(X))^2$, although I couldn't proceed any further with the finding of $E(X^2)$.
Although first and foremost, am I heading off into the right track? If so, how should I go about finding $E(X^2)$? Thanks for the help, as always!
-
– karakfa Jul 17 '12 at 15:46
1 Answer
I think this is the right track. Since $S_{n}$ is a sum of iid r.v.s we have
\begin{equation} \text{E}(S_n) = \frac{1}{n} \text{E}(X_1+\ldots+X_n) =\frac{1}{n}\left[ \text{E}(X_1)+\ldots + \text{E}(X_n) \right] = \text{E}(X_1) = a \end{equation} This can be used to relate $a$ to the mean of the distributions you want. Similarly,
\begin{equation} \text{Var}(S_n) = \text{Var}\left(\frac{X_1+\ldots+X_n}{n}\right) = \frac{1}{n^2} \text{Var}(X_1+\ldots+X_n) = \frac{n}{n^2} \text{Var}(X_1) = \frac{\text{Var}(X_1)}{n} \end{equation} This follows since, being iid, the $X_i$ are uncorrelated. Now, noting that $0.01 = 1/n$ the result becomes \begin{equation} x^2 = \text{Var}(X_1) \end{equation} Thus,
1) For $X_i \sim Bin(10,p)$, $\text{Var}(X_1) = 10 p(1-p)$
2) For $X_i \sim Geom(p)$, $\text{Var}(X_1) = (1-p)/p^2$
-
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|
http://math.stackexchange.com/questions/38292/factoring-methods-tricks
|
# Factoring Methods/Tricks
One of the things I've struggled with most in algebra/calculus is all the "factoring tricks". When I take time away from doing math I inevitably forget most if not all of them. The old proverb "use it or lose it" definitely comes into play.
I'm a huge fan of Khan Academy and have started to keep a list of tutorials which demonstrate factoring, but I haven't found them all:
Was wondering if anyone could help me fill in the gaps by providing their own list of factoring tips be them in the thread itself, or to online resources like Khan Academy, this site, or elsewhere
-
It is not possible to answer your question without reading through all your links and lessons, because who knows what is covered in "Factoring Special Products". So it would be fine to indicate shortly what kind of expressions are addressed. – Phira May 10 '11 at 18:28
Also you have to be aware that you are also asking about applications of factoring methods in your examples. – Phira May 10 '11 at 18:36
"flagged" for community wiki. – Rasmus May 10 '11 at 20:22
## 4 Answers
Remembering the geometric series helps for factoring things like $a^n-b^n$:
$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1})$$
A similar factorization applies for $a^\ell+b^\ell$ for $\ell$ odd:
$$\begin{align*} a^3+b^3&=(a+b)(a^2-ab+b^2)\\ a^5+b^5&=(a+b)(a^4-a^3 b+a^2 b^2-ab^3+b^4) \end{align*}$$
whose general pattern I'll leave for you to tease out.
-
Completing the rectangle:
$$ab+7a+11b=ab+7a+11b+77-77=(a+7)(b+11)-77.$$
-
Completing a square with a middle term that is a square itself:
$$x^4+4=x^4+4+4x^2-4x^2=(x^2+2)^2-(2x)^2=(x^2+2-2x)(x^2+2+2x).$$
-
2
Also $x^4-x^2+16=x^4+8x^2+16-9x^2=(x^2+4)^2-(3x)^2=(x^2+4-3x)(x^2+4+3x)$. – Isaac May 10 '11 at 19:59
As indicated in my comment above, you did not really say what material is covered in your links, but in any case I suggest you look at the algebra section of alcumus:
http://www.artofproblemsolving.com/Alcumus/Introduction.php
-
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|
http://mathoverflow.net/questions/24754?sort=oldest
|
## (Co)homology of the Eilenberg-MacLane spaces K(G.n)
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $(G, n)$ be a pair such that $n$ is a natural number, $G$ is a finite group which is abelian if $n \geq 1$. It is well-known that $\pi_n((K,(G,n)) = G$ and $\pi_i (K(G,n)) = 0$ if $i \neq n$.
Also it is known that these spaces $K(G,n)$ play a very important role for cohomology. For any abelian group $G$, and any CW-complex $X$, the set $[X, K(G,n)]$ of homotopy classes of maps from $X$ to $K(G,n)$ is in natural bijection with the $n^{\mathrm{th}}$ singular cohomology group $H^n(X; G)$ with coefficients in $G$.
But what is known about the cohomology of the $K(G,n)$ themselves? It is interesting in the light of the above. Here I mean the singular cohomology with integral coefficients.
-
1
The (co)homology of $K(G,1)$ is well-known to equal the group (co)homology of $G$ with integer coefficients. I don't know what happens for $n>1$. – Robin Chapman May 15 2010 at 14:23
7
See an introductory algebraic topology text like Hatcher or May. The (co)homology of Eilenberg-Maclane spaces are heavily studied. In a "stable range" this cohomology is called the Steenrod Algebra. – Ryan Budney May 15 2010 at 14:29
3
Hatcher's spectral sequence text computes only cohomologies of Eilenberg-MacLane spaces with Z/2-coefficients, but her refers to J. P. May, A general approach to Steenrod operations, Springer Lecture Notes 168 (1970), 153–231 for an integral computation. – Lennart Meier May 15 2010 at 21:18
## 4 Answers
Computing the integral cohomology of $K(\pi,n)$'s is feasible but a bit tricky. In fact the only reference I know is expos\'e 11 of H. Cartan's seminar, year 7 (available on Numdam). I'd be interested if there are other sources that cover that.
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I just noticed this answer. I asked for other sources as a question: mathoverflow.net/questions/50417/… I'm trying to get hold of Schafer's thesis, to see if it fills this niche. – Daniel Moskovich Jan 11 2011 at 13:37
@algori: Cartan seems to have also written up his calculations in some papers: "Sur les groupes d'Eilenberg-Mac Lane. I,II." (French) Proc. Nat. Acad. Sci. U. S. A. 40, (1954). I haven't looked at them yet, but presumably the presentation is cleaner. – Mark Grant Aug 18 2011 at 14:25
### 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.
If you don't mind a little notational pain, you can look at the original papers where Eilenberg and Mac Lane worked many cases out: On the homology of groups $H(\pi,n)$ (I, II, III). Annals of Mathematics ~1953.
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Scott, is the ring structure in the cohomology computed there as well? – algori May 15 2010 at 15:26
I don't have the papers in front of me, but I will tentatively say "no". – S. Carnahan♦ May 15 2010 at 16:13
Expanding slightly on Ryan's comment: it's an easy fact (often attributed to Serre) that the set of cohomology operations $H^k(-;G)\to H^r(-;H)$ (i.e. natural transformations) is in 1-1 correspondence with $[K(G,k),K(H,r)]=H^r(K(G,k);H)$ (for any abelian groups $G,H$). There are tons of these, some easy, some not so easy to understand, corresponding to how easy the calculation of $H^r(K(G,k);H)$ is. A nice subset are the stable operations (compatible with a certain suspension $[K(G,k),K(H,r)]\to [K(G,k+1),K(H,r+1)]$ which come in families, the most common family being the Steenrod algebra, corresponding to $G=H=Z/p$. There are non-stable operations also, eg the Pontryagin square $H^k(-;Z/2)\to H^{2k}(-;Z/4)$.
-
You might find some useful information in the Lausanne thesis of Alain Clément:
http://doc.rero.ch/record/482/files/Clement_these.pdf
In particular, he gives an account of Cartan's results in Chapter 2, then describes a C++ program for computing integral (co)homology of certain ($2$-local) Eilenberg-Mac Lane spaces in Chapter 3. An appendix lists the integral (co)homology groups of $K(\mathbb{Z}_2,2)$, $K(\mathbb{Z}_2,3)$, $K(\mathbb{Z}_4,2)$ and $K(\mathbb{Z}_4,3)$ up to degree $200$.
-
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|
http://mathoverflow.net/questions/64828?sort=oldest
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## Asymptotic Formula for a Mertens Style Sum
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hello,
I am wondering if there is a simple asymptotic formula for
`$$\sum_{p\leq x}\frac{\left(\log p\right)^{k}}{p},$$`
where $k\geq0$ is some integer. If $k$ is $0,$ by using the Prime Number Theorem we have
`$$\sum_{p\leq x}\frac{1}{p}=\log \log x+b+O\left(e^{-c\sqrt{\log x}}\right).$$`
Similarly, the prime number theorem and integration by parts solves the case $k=1$ and gives
`$$\sum_{p\leq x}\frac{\left(\log p\right)}{p}=\log x+C+O\left(e^{-c\sqrt{\log x}}\right).$$`
My question is do these integrals have a nice asymptotic formula for every $k$? Specifically, I mean with an error term of the form $O\left(e^{-c\sqrt{\log x}}\right).$
Thanks!
Remark: This question is related, and in particular if it is solved with a nice enough asymptotic, then so is this. (but not vice versa)
-
## 3 Answers
Expanding on Frank's answer: by partial summation, we have that $$\sum_{p \leq x} \frac{(\log p)^k}{p} = \frac{(\log x)^k}{x} \pi(x) - \int_{2}^{x} \pi(t) \left(\frac{kt (\log t)^{k-1} - (\log t)^k}{t^2}\right) dt.$$ Using the fact that $\pi(x) = \mathrm{Li}(x) + E(x)$, where $E(x) = O(e^{-c\sqrt{\log x}})$, we have that the first term is equal to $$\frac{(\log x)^k}{x} \mathrm{Li}(x) + O((\log x)^k e^{-c\sqrt{\log x}}) = \frac{(\log x)^k}{x} \mathrm{Li}(x) + O(e^{-c_k \sqrt{\log x}}).$$ For the second term, $$- \int_{2}^{x} \pi(t) \left(\frac{kt (\log t)^{k-1} - (\log t)^k}{t^2}\right) dt = R_1(x) + R_2 + R_3(x),$$ where $$R_1(x) = - \int_{2}^{x} \mathrm{Li}(t) \left(\frac{kt (\log t)^{k-1} - (\log t)^k}{t^2}\right) dt,$$ $$R_2 = - \int_{2}^{\infty} E(t) \left(\frac{kt (\log t)^{k-1} - (\log t)^k}{t^2}\right) dt,$$ $$R_3(x) = \int_{x}^{\infty} E(t) \left(\frac{kt (\log t)^{k-1} - (\log t)^k}{t^2}\right) dt.$$ Let's look at $R_1(x)$ first. Using the fact that $\mathrm{Li}(t) = \int_{2}^{s} \frac{ds}{\log s}$ and interchanging the order of integration, then evaluating the integral with respect to $t$, we obtain $$R_1(x) = - \frac{(\log x)^k}{x} \mathrm{Li}(x) + \frac{(\log x)^k}{k} - \frac{(\log 2)^k}{k}.$$ For $R_3(x)$, a simple calculation shows that $$R_3(x) \ll - \int_{x}^{\infty}{(\log t)^{k-1} e^{-c\sqrt{\log t}} \left(\frac{kt - \log t}{t^2}\right) dt} = - \int_{\log x}^{\infty}{u^{k-1} e^{-c\sqrt{u}} \left(\frac{ke^{u} - u}{e^u}\right) du}.$$ We can rewrite this integral as $$- \int_{\log x}^{\infty}{\left(ku^{k-1} e^{-c\sqrt{u}} - \frac{c}{2} u^{k - 1/2} e^{-c\sqrt{u}}\right) + \left(\frac{c}{2} u^{k - 1/2} e^{-c\sqrt{u}} - u^k e^{-c\sqrt{u} - u}\right) du}$$ and from this it is clear that $$R_3(x) \ll_k - \int_{\log x}^{\infty}{\left(ku^{k-1} e^{-c\sqrt{u}} - \frac{c}{2} u^{k - 1/2} e^{-c\sqrt{u}}\right) du} = (\log x)^k e^{-c \sqrt{\log x}} \ll e^{-c_k \sqrt{\log x}}.$$ Finally, it's clear from the estimate for $R_3(x)$ that the integral defining $R_2$ converges, and so we obtain $$\sum_{p \leq x} \frac{(\log p)^k}{p} = \frac{(\log x)^k}{k} - \frac{(\log 2)^k}{k} + R_2 + O_k(e^{-c\sqrt{\log x}}).$$
Note that it's possible to make this valid for much more than just nonnegative integers $k$ - you could also have $k$ complex.
-
### 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.
Sure. You can write the expression as a Stieltjes integral
$$\int_2^x \frac{(\log t)^k}{t} d\pi(t).$$
Integrate by parts to get
$$\frac{(\log t)^k}{t} \pi(t) - \int_2^t \pi(t) \frac{d}{dt} \Big( \frac{(\log t)^k}{t} \Big).$$
Now apply the prime number theorem and replace $\pi(t)$ by $\frac{t}{\log t} + E(t)$ where the error bound $E(t)$ (assumed increasing in $t$) is given by the expression you gave above. You now have an asymptotic formula of the type you were looking for, involving an elementary integral which you could clean up further if you like. The total error is $\ll (\log t)^k E(t).$
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The first line cannot be correct, I think you mean $d(s(t))$ rather than $d(\pi(t))$ where $$s(t)=\sum_{p\leq t} \frac{1}{p}.$$ Then using $s(t)=\log\log t+E(t)$, and the fact that $d(\log \log t)=\frac{dt}{t\log t}$ we get a main term of $$\int_2^x \frac{1}{t}(\log t)^{k-1}dt.$$ Fortunately this directly has an antiderivative of $\frac{1}{k}(\log t)^k.$ The error term is also of the form $e^{-c\sqrt{\log x}}$ by integration by parts. (A constant also jumps out) – Eric Naslund May 17 2011 at 18:45
@Eric: Oops, I did indeed make a mistake, thank you, corrected. (My correction is a little bit different than you suggest however.) – Frank Thorne May 17 2011 at 19:56
Here is an answer that is similar in spirit to Frank and Peter's answers, but possibly simpler.
Summing by parts, we see that
$$\sum_{p\le x} \frac{\log^k p}{p} = (\log x)^{k-1} \sum_{p\leq x} \frac{\log p}{p} -(k-1)\int_{2^-}^x (\log u)^{k-2}\sum_{p\le u} \frac{\log p}{p} \frac{du}{u}.$$
Now use the formula $$\sum_{p\leq x} \frac{\log p}{p} = \log x + c_1 + O(\exp(-c_2\sqrt{\log x}))$$ and it is not hard to derive that $$\sum_{p\le x} \frac{\log^k p}{p} = \frac{\log^k x}{k} + c_3 + O(\exp(-c_4\sqrt{\log x})).$$
This seems easier than dealing with $Li(x)$.
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Just a small quibble - the sum in the integrand in the first line shouldn't have a $k$-th power in it. – Peter Humphries May 13 2011 at 7:41
@Peter: I fixed it. – Micah Milinovich May 13 2011 at 14:06
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http://math.stackexchange.com/questions/206552/convergence-in-l-infty-and-l-1-even-if-infinite-measure-space
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Convergence in $L_\infty$ and $L_1$ even if infinite measure space
Let $\{f_n\}_{n\in \mathbb{N}}$ be a sequence of measurable functions on a measure space and $f$ measurable. In the literature, assuming the measure space $X$ has finite measure, if $f_n$ converges to $f$ in $L^{\infty}$-norm , then $f_n$ converges to $f$ in $L^{1}$-norm.
Even if $X$ has infinite measure, does it converge to $f$ in $L^{1}$-norm?
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1 Answer
No. Try $f_n$ the constant function such that $f_n(x)=\frac1n$ for every $x$ in $X$.
For an example where each $g_n$ is in $L^1\cap L^\infty$, $g_n\to0$ in $L^\infty$ and not in $L^1$, try $g_n=\frac1n\mathbf 1_{[0,n^2]}$ on $X=\mathbb R$ with Lebesgue measure.
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Here the $f_n$ are not even in $L^1$. Maybe the question can be adjusted: if $f_n\in L^1\cap L^\infty$ and $f_n$ converges in $L^\infty$, does it converge in $L^1$? – Hagen von Eitzen Oct 3 '12 at 13:11
@HagenvonEitzen See Edit. – Did Oct 3 '12 at 13:19
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http://math.stackexchange.com/questions/42115/when-linear-combinations-of-independent-random-variables-are-still-independent
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When linear combinations of independent random variables are still independent?
We have a random vector $X=(X_j)_{j=1,...,n}$, whose components $X_j$ are mutually independent. We build a new random vector $Y=A X+b$, with $Y=(Y_k)_{k=1,...,m}$, $A=(a_{k,j})_{k=1,...,m;j=1,...,n}$, $b=(b_k)_{k=1,...,m}$.
When the components of $Y$ are still mutually independent random variables? How to prove the result? I suppose this can be related with the rank of $A$ but I'm not sure and I cannot find a way to prove if I don't make some special assumption about the law of $X$.
I know $U$ and $V$ are independent iff $f(U)$ and $g(V)$ are independent for each and every pair of measurable function $f$ and $g$. This can help, I think, when, for example, $Y_1=X_1+X_2$ and $Y_2=X_3+X_4$.
But when does $Y_1 = X_1 + X_2$ and $Y_2 = X_1 - X_2$? Or when does $Y_1=X_1 + X_2$ and $Y_2=X_5$?
None of the texts I searched discuss seriously this topic, but leave it in a certain way to "intuition"?
Thanks
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I don't think there's a general result here unless the $X_i$ are Gaussian. – Qiaochu Yuan May 30 '11 at 13:14
As you've noted, the general result is true if $A = a_1 \oplus a_2 \oplus \cdots \oplus a_p$ for some $p$ where the $a_i$ are row vectors. – cardinal May 30 '11 at 13:48
3 Answers
If $A = a_1 \oplus a_2 \oplus \cdots \oplus a_m$, for $m \leq n$, where $a_i$ are row vectors of dimension $n_i$ such that $\sum_{i=1}^m n_i = n$ and $\oplus$ denotes the direct sum, then the random vector $Y$ has independent coordinates.
This is not hard to see since $Y_1$ is measurable with respect to $\sigma(X_1, \ldots X_{n_1})$, $Y_2$ is measurable with respect to $\sigma(X_{n_1+1}, \ldots, X_{n_1+n_2})$, etc., and these $\sigma$-algebras are independent since the $X_i$ are independent (essentially, by definition).
Obviously, this result still holds if we consider matrices that are column permutations of the matrix $A$ described above. Indeed, as we see below, in the case where the distribution of each $X_i$ is non-normal (though perhaps depending on the index $i$), this is essentially the only form that $A$ can take for the desired result to hold.
In the normal-distribution case, as long as $A A^T = D$ for some diagonal matrix $D$, then the coordinates of $Y$ are independent. This is easily checked with the moment-generating function.
Suppose $X_1$ and $X_2$ are iid with finite variance. If $X_1 + X_2$ is independent of $X_1 - X_2$, then $X_1$ and $X_2$ are normal distributed random variables. See here. This result is known as Bernstein's theorem and can be generalized (see below). A proof can be found in Feller or here (Chapter 5).
In the case where $A$ cannot be written as a direct sum of row vectors, you can always cook up a distribution for $X$ such that $Y$ does not have independent coordinates. Indeed, we have
Theorem (Lukacs and King, 1954): Let $X_1, X_2, \cdots, X_n$ be $n$ independently (but not necessarily identically) distributed random variables with variances $\sigma_i^2$, and assume that the $n$th moment of each $X_i(i = 1, 2, \cdots, n)$ exists. The necessary and sufficient conditions for the existence of two statistically independent linear forms $Y_1 = \sum^n_{i=1} a_i X_i$ and $Y_2 = \sum^n_{i=1} b_i X_i$ are
1. Each random variable which has a nonzero coefficient in both forms is normally distributed, and
2. $\sum^n_{i=1} a_i b_i \sigma^2_i = 0$.
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Great answer. (And thanks for the Lukacs and King result.) – Did Jun 5 '11 at 8:21
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I think little can be said in general, apart from the rather trivial case in which each of the resulting $y_i$ depends on non-intersecting subssets of $X$ (what would correspond to each column of A having no more than one non-null value).
Some weaker result can be obtained by considering non-correlation instead of independence (or, what would equivalent, restricting to gaussian variables), because the covariances matrices are simply related
$C_Y = A \; C_X A^t$
By assumption, $C_X$ is diagonal, and we want fo find out for which $A$ matrices $C_Y$ is also diagonal. Again, little can be said in general, more than just that. If we restrict further the assumptions, and assume that $x_i$ are iid (or just equal variances), then we get that orthogonality of $A$ in sufficient (but not necessary).
In particular, for the case $Y_1 = X_1 + X_2$, $Y_2 = X_1 - X_2$, $Y$ is not guaranteed to be uncorrelated, unless we assume $x_i$ are iid (or have same variance).
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Something relatively strong can be said. See my updated answer. – cardinal May 30 '11 at 14:46
@cardinal: At first sight, that seems practically equivalent to what I stated. The normal assumption is equivalent to use correlation in place of independence. And the summation seems as the non-diagonal terms in my $C_y$ to me. – leonbloy May 30 '11 at 16:38
It appears much stronger to me. – cardinal May 30 '11 at 16:42
Ah, yes, I see it now. – leonbloy May 30 '11 at 17:23
The Skitovich-Darmois theorem A (Skitovich(1953), Darmois(1953), see also A. Kagan, Yu. Linnik, and C.R. Rao (1973, Ch.3)). Let $\xi_j$, where $j=1, 2,\dots, n,$ and $n\geq 2$, be independent random variables. Let $\alpha_j, \beta_j$ be nonzero constants. If the linear statistics $L_1=\alpha_1\xi_1+\cdots+\alpha_n\xi_n$ and $L_2=\beta_1\xi_1+\cdots+\beta_n\xi_n$ are independent, then all random variables $\xi_j$ are Gaussian.
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http://math.stackexchange.com/questions/153681/absolute-value-on-a-number-line
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# Absolute value on a number line
Solve : |x-4|>a if case1:a>0 and case2:a<0
I am getting answers which look similar in both cases. please i wish to know why it is so and how different both answers are when plotted on a number line
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can u show your work? – Bhargav Jun 4 '12 at 10:12
let a>0 so x>4+a or x<4-a , let a<0 so x>4+a or x<4-a .Though i know that both answer's meaning is different i am unable to find out how the points included in both cases are different – meg_1997 Jun 4 '12 at 10:14
– TMM Jun 4 '12 at 11:07
## 3 Answers
If $a \lt 0$, all $x$ will satisfy it as all absolute values are $\ge 0$. If $a \gt 0$ you need the points more than $a$ from $4$.
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:i was having problem to solve:|x-2|+|x-5|=3 . I was trying to post this question but could not(i was being said that it does not meet our quality standards).So posted it here. – meg_1997 Jun 4 '12 at 15:43
@meg_1997: when you have two absolute value signs, it is easiest to consider each region of $x$ and resolve the signs. So for $x\le 2$ both expressions are negative and need to be inverted. You are left with $7-2x=3$ AND $x \le 2$. You solve the equality and see if it meets the inequality. Then there are two more sections of the real line to consider the same way. – Ross Millikan Jun 4 '12 at 21:15
The work in your comment was a good start. Another thing to notice is that if $a>0$, then $4-a<4+a$, so your solution consists of two disjoint "rays" corresponding to the inequalities you wrote. However, when $a<0$, then $4+a<4-a$, and in particular each $x$ such that $x\geq 4-a$ satisfies $x>4+a$. That is one way to see that your method leads to the same answer mentioned in Ross's answer. But Ross's method of observing that $|x-4|\geq 0>a$ for all $x$ if $a<0$ is a little easier.
To make things a little more concrete, consider what happens when $a=5$: Your method says that $x>9$ or $x<-1$, which is correct. Now when $a=-5$, your method says that $x>-1$ or $x<9$, which is true for all real numbers.
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– Ross Millikan Jun 4 '12 at 10:56
Consider multiple cases
Case 1: $a = 0$
Then, any number other than $4$ will satisfy your inequality.
Case 2: $a < 0$
Then, any $x$ will satisfy your inequality since absolute values are $\ge 0$
Case 3: $a > 0$
Then $|x - 4| > a$ if and only if $x$ is farther than $a$ units from $4$. Hence, $x - 4 > a$ or $x - 4 < -a$. So the set of all real numbers that satisfy your inequality is $$(-\infty, 4 - a) \cup (4+a, \infty)$$
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http://physics.stackexchange.com/questions/23173/what-makes-the-stars-that-are-farther-from-the-nucleus-of-the-galaxy-go-faster-t
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# What makes the stars that are farther from the nucleus of the galaxy go faster than those in the middle?
It has no sense that stars that have a bigger radius and apparently less angular speed($\omega$) goes faster than the ones near the center.
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## 4 Answers
The dark matter or the dark energy.
We don't know at all what are they. We just know that they exist, but as I say, we don't have idea.
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3
well dark energy isn't required to explain the flat rotation curves. And while we don't what DM is, we know what DM isn't, cf MACHOs – Nic Apr 2 '12 at 14:58
DM isnt that mysterious. Also, this doesn't really answer the question--you need to include the explanation as well. – Manishearth♦ Apr 2 '12 at 15:02
1
– Frédéric Grosshans Apr 2 '12 at 15:12
# Short answer
The question is a bit ambiguous. If the question is
why do star velocity increase with distance close to the galactic centre ?
the answer is
because their orbit encompass more mass, and this corresponds to a stronger gravity pull.
If the question is
why does their velocity stays constant and does not decrease at big radii, where the star density decreases ?
the current consensus answer is
We have to add dark matter to observed stars to explain that.
# Long answer
The curve B below (taken from wikipedia) plots the the observed star velocity as function of the distance from the galactic centre.
The curve A corresponds to the expected curve without dark matter. As you can see, beyond a given distance, the velocity is expected to decrease, but it actually stays roughly constant. Dark matter has been initially postulated as a solution to this discrepancy.
The increase of velocity close to the centre is independent to the presence of dark-matter or not. The velocity of a star on a circular orbit of radius $r$ in the galactic plane is given by a balance of the centrifugal acceleration and the gravity it feels: $$\begin{gather} \frac{v^2}r= G \frac{m(r)}{r^2}\\ v=\sqrt{ G\frac{m(r)}{r}} \end{gather}$$ where $m(r)$ is the mass of stars contained in a spheroid centred on the galactic centre of radius $r$ (see e.g. here for more details. And then adapt it to the geometry of a galaxy). If $r$ is smaller than the galaxy thickness, the number of star is proportional to the volume of the sphere, and we expect $$\begin{align} m(r)&\propto r^3& v&\propto r \end{align}$$ which is consistent with the initial increase.
When $r$ is bigger than the thickness, if the star density is constant, we have then $$\begin{align} m(r)&\propto r^3& v&\propto \sqrt r, \end{align}$$ and this still corresponds to a velocity increase.
When $r$ is big enough, the density of star decreases with $r$ upto a point where $m(r)< C r$ and this should give the decreasing curve A. On the other hand, the observed curve B is essentially constant, and this can bee seen as a measurement of $m(r)\propto r$. This is not consistent with the observed star repartition, but it is consistent with the presence of dark (i.e. not seen) matter with a radial density $\propto\frac1{r^2}$.
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+1 for using mathcal :D. And for the interesting, detailed, answer. – Manishearth♦ Apr 2 '12 at 16:30
@Manishearth : Apparently, \mathcal was not standard (even if I learned it this way ;-) ), so Garmen1778 removed it. I hope the +1 stays anyway ! – Frédéric Grosshans Jun 4 '12 at 15:36
since no one mentioned it, i think its only fair to provide at least one answer about MOND (Modified Newtonian Dynamics).
Basically the galaxy rotation curve is the reason dark matter was proposed in the first instance. However, the dark matter explanation, putting aside for a moment other considerations as supersymmetric weakly coupled partners, is not very satisfying from a scientific point of view, since its an instance of adjusting parameters (i.e: unseen matter) in order to preserve a model. Think of Ptolomeus model of the solar system, with the epicyclic orbits postulated in order to preserve the earth in the center of the model.
MOND as an alternative explanation to dark matter is widely discredited, specially after the observation of gravitational lensing in the middle regions of the Bullet cluster, which suggests to some that a transparent source of gravity is causing the lensing effect.
However, despite this, the MOND hypothesis, at least as an heuristic to retrodict the galaxy rotation curves works extremely well. The hypothesis basically says that the gravitational mass coupling becomes weaker when accelerations drop below $a_0 \approx 10^{-10} m/s^2$. From that simple assumption, it is able to predict most of the galaxy rotation curves that are currently observed
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`postdict` The word you're looking for is `retrodict`--queer word; only really comes to use in physics :) – Manishearth♦ Apr 2 '12 at 16:34
thanks i fixed it – lurscher Apr 2 '12 at 16:35
– Qmechanic♦ Apr 3 '12 at 22:08
The striking similarity of a galaxy and an hurricane .
Do you need DM or SMBH in the hurricane ? Their force field is similar. The hurricane is governed by a vortex, then how can a galaxy be subjected to such a field that is more important than the central force at the centre of the galaxy ? I'm sure that there is a way.
Edit add
I will try to explain why there is a vortex :
All images were taken from outra Física CC.
In this PSE-link (anti gravity) I show with graphs and equations how the voids have to grow due to the gravitational field. All the mass that was previously inside them is now in the exterior shell of the voids.
When two voids intersect a we have :
and the field is like
To see the field of a vortex you need to remove the mean field
As you see there is no need of DM neither SMBH. The only think that is needed is to follow the rules of physics (and a lot of imagination of my friend Alfredo that allow s me to see only physics as the true nature of Nature, when everybody else can only explain with a magical Dark Matter).
There are recent news (arxiv and cosmiclog) that DM probably has to have a two different composition/properties and the astronomers dont now how to justify the observed behaviour.
Of course one can continue the pursue of DM quest. I dont.
The resistence that the readers have to this kind of model, that only follows the laws of gravity, has to have with the fact that this asks for a different kind of beginning of the Universe. Why not? Ahh the BB and the Dark Energy, you say!
Then, the beginning of the Universe can be an infinite universe (or almost), homogeneous, and with temperature = 0. Temperature will grow, contradicting our deepest convictions that this is impossible.
If you want to understand clearer read that blog from the first to the last post (it is written, in portugueese, in such a way that any person can understand), or to focus only the present question you should start at O nascimento de uma Bolha until you arrive at As espirais Galácticas.
And the Dark Energy? You got rid of it, again with the help of my friend Alfredo, with his paper: A self-similar model of the Universe unveils the nature of dark energy. I posted a short math proof of his argument in this PSE link (are-the-rest-masses-of-fundamental-particles-certainly-constants). Why sould the particles be shrinking giving us the impression of a space expansion ? Their associated fields expand in space, the fields have energy, and are sourced by the particles. To preserve the total energy budget they must 'shrink'. Hey, in the lab I dont see that shrinking, how come? It is impossible to detect the phenomena locally because the lab, the instrumentation and yourself are shrinking, and above all no one likes to be shrinking.
I do realize that when DM was 'invented' was only a data fit. They have no model to justify the observations and instead of thinking harder they postulated its existence. Do we need GR in the cosmology if the space is flat arxiv-BAO-survey? No. That is why I invite the readers to pay attention to my friends paper, and be the first to say 'I spot an error'.
When I went to school I already 'knew' that the space is expanding, and the same with you. It is shocking if it is 'on the contrary'.
## The bigest enemy we face in the search of the knowledge is our 'rooted beliefs'.
Now, that I took my precious time to write this long answer I kindly ask the downvoters to say in what particular points they beleive that I am mistaken.
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-1 : It is not an answer about the velocity distribution. If you explicitely link you hurricane intuition with the increase of velocity, I may remove the -1 – Frédéric Grosshans Apr 3 '12 at 12:39
@Frédéric - the galacic vortex is in place. The outer regions have bigger speeds as you can see. – Helder Velez Apr 3 '12 at 19:14
I don't think this place is for original research. Your answer seems to be far from the scientific consensus. Maybe the consensus is wrong, but if you manage to convince a significant part of the astrophysics community that your explanation is the good one, I promise to remove my downvote, even if you will probably not care at this step. – Frédéric Grosshans Apr 10 '12 at 13:25
And I will not read your explanation on your blog beacause : 1. Even if I'm a professional physicist, I do not have enough knowledge in astrophysics to judge the relevance of arguments about the dark matter model. 2. I'm not any person, since I cannot understand Portuguese. Essayez en Français, pour que je puisse comprendre ;-) – Frédéric Grosshans Apr 10 '12 at 13:29
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http://mathematica.stackexchange.com/questions/2583/rearranging-a-polynomial?answertab=oldest
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Rearranging a Polynomial
In Mathematica 8.04 on Windows, I want to display a formula in standard textbook format. The formula is the variance of an $N$-security portfolio. For two securities it is:
$$w_1^2\sigma_1^2+w_2^2\sigma_2^2+2\sigma_{1,2}w_1 w_2$$
To generate this, I have created the following function:
````psd[n_Integer] :=
Sum[Sum[Subscript["w", i] Subscript["w", j] If[i == j, Subsuperscript["σ", i, 2],
Subscript["σ", Min[i, j], Max[i, j]]], {i, 1, n}], {j, 1, n}];
````
This works, but the terms are out of order as shown in this image of the output of `psd[2]`:
I'd like to take the last two terms (with the exponents) and place them in front of the first two. Suppose that I call the output "a", then I can get the parts of a using `Take[a,1]` and `Take[a,-2]`.
What I would like to be able to do is to put those two parts back together in reverse order as shown in the original equation. Ideally, I'd be able to specify a sort order, but I can't seem to figure that out. I've tried all of the orders for `MonomialList` with no luck, and when I add the list parts it gets resorted anyway.
The order of the terms doesn't matter for calculation, but this is for presentation and I'd like it to look like it does in every textbook. The point is to demonstrate how the formula grows longer as more securities are added to the portfolio. Except for the ordering, it works perfectly in a `Manipulate`. Any ideas?
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Sorry Tim, on my screen the first formula looked like it had a minus not a plus. I have deleted my comment. You know you can use LaTeX markup on the site? – Verbeia♦ Mar 4 '12 at 21:31
1 Answer
You were definitely on the right track with `MonomialList`. Here is a solution. Others will probably find nicer ways. Using the trick found here, we first define a Format that looks like "Plus" but doesn't rearrange things:
````Format[myPlus[expr__]] := Row[Riffle[{expr}, "+"]]
````
With this format in hand, we can wrap your original function in the following:
````myPolynomial[n_Integer] := myPlus @@ RotateRight[Sort@MonomialList[psd[n]], n]
````
The `Sort` places things in what MMA considers canonical order. Lucky for us, it happens to put all the $\sigma_{1,2} w_1w_2$-like terms together first. Then we rotate the list using `RotateRight`. Since we know there will be `n` functions of the type $w_1^2\sigma_1^2$, we rotate it that many places. Using the non-rearranging form of plus, we get the desired result!
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Thank you! This works perfectly. I didn't realize that I could use `Format` to define my own formats. I think that I'll find other uses for that. – Tim Mayes Mar 4 '12 at 20:05
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You could also write `myPlus` this way: `myPlus[expr__] := Plus @@@ HoldForm[{expr}]` – Mr.Wizard♦ Mar 4 '12 at 22:56
Ahhh, that's the combination of Plus and HoldForm that I couldn't quite figure out. Thanks @Mr.Wizard – tkott Mar 4 '12 at 23:30
@Mr.Wizard, thanks for the simplification. I kept trying to test `HoldForm`, but I couldn't get it to work. – Tim Mayes Mar 4 '12 at 23:41
@Tim both methods have benefits; the `Format` method is good because the head `myPlus` remains and can later be replaced (`/. myPlus -> Plus`). My direct method can be used if you need staged release using `ReleaseHold`, or simply prefer `ReleaseHold` as more generic. – Mr.Wizard♦ Mar 5 '12 at 7:00
show 1 more comment
lang-mma
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http://mathhelpforum.com/calculus/193272-find-surface-area-donut.html
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# Thread:
1. ## find surface area of a donut
I have a problem where i dont even know where to begin. Usually, when i post problems here i have something to show but this time i dont. Im sorry for that.
here is the problem
:
FInd the surface area of the torus (or donut) obtained by rotating the circle (x - b)2 + y2 = a2; (a < b) about the
y-axis.
would i do something like the Area of a surface of revolution theorem?:
$S = 2 \pi \int_a^b f(t) \sqrt {(dx/dt)^2 + (dy/dt)^2} dt$
if so i am not even sure what to do to get it into the form to put it into that formula?
Any help is appreciated since i am lost!!
Thank you
2. ## Re: find surface area of a donut
Originally Posted by icelated
FInd the surface area of the torus (or donut) obtained by rotating the circle (x - b)2 + y2 = a2; (a < b) about the y-axis.
Use the First Guldin Theorem: the surface area $S$ of a surface of revolution generated by the revolution of a curve about an external axis is equal to the product of the arc length $s$ of the generating curve and the distance $d$ traveled by the curve's geometric centroid $\bar{x}$ .
3. ## Re: find surface area of a donut
Originally Posted by FernandoRevilla
Use the First Guldin Theorem:
Im not suppose to use pappus's theorem. When i googled First Guldin Theorem everything came back to pappus. What is First Guldin Theorem? =(
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http://en.wikipedia.org/wiki/Value_(mathematics)
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# Value (mathematics)
In mathematics, value commonly refers to the output of a function. In the most basic case, that of unary, single-valued functions, there is one input (the argument) and one output (the value of the function).
Example: If the function $f$ is defined by prescribing that $f(x) = 2x^2-3x+1$ for each real number $x$, then the input 3 will yield the function value 10 (since indeed 2 · 32 – 3 · 3 + 1 = 10).
The function $f$ of the example is real-valued, since each and every possible function value is real. On the other hand, it is not injective, since different inputs may yield the same value; e.g., $f(-1.5) = 10$, too.
In some contexts, for convenience, functions may be considered to have several arguments and/or several values; also cf. the discussion in the article function. However, strictly seen, this is not an extension, since such functions may be considered as having single families and/or sets as input or output.
Value is also used in other senses, e.g., to specify a certain instance of a variable.
Example: $f(x) = 0$ for two separate values of $x$, namely, for $x=0.5$ and for $x=1$.
## See also
This mathematics-related article is a stub. You can help Wikipedia by expanding it.
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http://mathoverflow.net/questions/101629?sort=votes
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## Around Banach isomorphism theorem
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $E$ be a normed (real) space which is not complete. Is it always possible to find $f$ a continuous bijective linear function from $E$ to $E$ such that $f^{-1}$ is not continuous?
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## 1 Answer
No. If $E$ has codimension less than the continuum in a Banach space then such an $f$ must be open. This was proved by Saxon and Levin; see the proposition on page 95 of
this paper
which is
Saxon, Stephen; Levin, Mark, Every countable-codimensional subspace of a barrelled space is barrelled. Proc. Amer. Math. Soc. 29 1971 91–96.
So for an example, take any discontinuous linear functional on a Banach space and let $E$ be its kernel.
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Thanks Bill. What would you recommend as books to go between "initial functional analysis" (Banach isomorphism theorem, Banach-Steinhaus theorem...) and the research articles you mention? I have to bridge the gap! – Jean-Pierre M. Jul 8 at 19:52
Sorry, I don't. This is TVS theory rather than Banach space theory. I would look at the books referenced in the Levin-Saxon paper; i.e., Edwards' book and Bourbaki's book on TVS (v. 5). Probably one of them contains the result that finite codimensional subspaces of Banach spaces are barrelled. IIRC, Edwards' book is reader friendly, while Bourbaki's has the advantage of being in French. – Bill Johnson Jul 8 at 20:05
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http://nrich.maths.org/34/note
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### Tangles
A personal investigation of Conway's Rational Tangles. What were the interesting questions that needed to be asked, and where did they lead?
### Bryony's Triangle
Watch the video to see how to fold a square of paper to create a flower. What fraction of the piece of paper is the small triangle?
# Chocolate
## Chocolate
This challenge is about chocolate. You have to imagine (if necessary!) that everyone involved in this challenge enjoys chocolate and wants to have as much as possible.
There's a room in your school that has three tables in it with plenty of space for chairs to go round. Table $1$ has one block of chocolate on it, table $2$ has two blocks of chocolate on it and, guess what, table $3$ has three blocks of chocolate on it.
Now ... outside the room is a class of children. Thirty of them all lined up ready to go in and eat the chocolate. These children are allowed to come in one at a time and can enter when the person in front of them has sat down. When a child enters the room they ask themself this question:
"If the chocolate on the table I sit at is to be shared out equally when I sit down, which would be the best table to sit at?"
However, the chocolate is not shared out until all the children are in the room so as each one enters they have to ask themselves the same question.
It is fairly easy for the first few children to decide where to sit, but the question gets harder to answer, e.g.
It maybe that when child $9$ comes into the room they see:
• $2$ people at table $1$
• $3$ people at table $2$
• $3$ people at table $3$
So, child $9$ might think:
"If I go to:
• table $1$ there will be $3$ people altogether, so one block of chocolate would be shared among three and I'll get one third.
• table $2$ there will be $4$ people altogether, so two blocks of chocolate would be shared among four and I'll get one half.
• table $3$, there will be $4$ people altogether, so three blocks of chocolate would be shared among four and I'll get three quarters.
Three quarters is the biggest share, so I'll go to table $3$."
Go ahead and find out how much each child receives as they go to the "best table for them". As you write, draw and suggest ideas, try to keep a note of the different ideas, even if you get rid of some along the way.
THEN when a number of you have done this, talk to each other about what you have done, for example:
A. Compare different methods and say which you think was best.
B. Explain why it was the best.
C. If you were to do another similar challenge, how would you go about it?
### Why do this problem?
This is an excellent problem for helping youngsters to develop their concepts of fractions. It's not so much to do with arithmetical manipulation of fractions but more to do with youngsters exploring and developing their ideas. By encouraging learners to share their methods, there is an opportunity to discuss which might be the 'best' (this might depend on the individual's preference too).
### Possible approach
Children will need plenty of (the same sized) paper available for folding and tearing in order to explore sizes of fraction.
To begin the activity, you could act out the problem using large sheets of paper to stand for the chocolate bars (or real bars!), placed on tables. The acting could go through the situation with the first six children coming up one by one to the tables. Encourage them to justify their decisions and make sure the whole group agrees with their choice.
Then learners could work in pairs on what happens when further children come to the tables. By listening to their conversations you can get a good insight into the ways that those youngsters think about and visualise fractions. Some surprises are very likely!
After some time, bring everyone together again to talk about their ways of working. Invite comments about each method and then once all the different ways have been explained, ask pairs to discuss which method they would use now they have seen so many. You can then suggest they continue working on the problem, choosing any approach (or see the extension below). It would be interesting to talk to those pairs who have changed the way they tackle the task to find out what it is about their new method that they preferred to the original one. Some of their reflections could be recorded for display.
### Key questions
What size do you think this is ...?
Why? (In response to the answer to the above.)
### Possible extension
Challenge children to come up with a system or pattern that would help them to solve similar challenges.
### Possible support
You could start off with just two tables and a total of three bars of chocolate.
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.
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http://physics.stackexchange.com/questions/55297/fictitious-forces-confusion/55299
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Fictitious forces confusion
I have a hard time understanding the subject of fictitious forces. Let's discuss a few examples:
1) I'm sitting inside a vehicle which is accelerating in a straight line. I feel like someone is pushing me to the seat. So, on the one hand, I'm told that this happens according to the third Newton's law: this pressure is the result of me pushing the seat as a reaction to the seat pushing me (because it is accelerating with the same acceleration as the car). On the other hand, these forces are acting on different objects and I'm told that there is another fictitious force acting on me in an opposite direction to the acceleration. So what is right? And if there is a fictitious force, then why some call it a "math trick" when they are real and I can feel them?
2) I do not understand why some call centrifugal force a fictitious force. The earth is pulling the earth with its "invisible" string called gravitation. That's why the moon is still there. And this is the centripetal force. However the moon is also pulling the earth according to the Newton's third law, and that's why we have tides. This is the centrifugal force. So why it is fictitious? What it has to do with frame of reference? When we observe this in non-inertial frame of reference (such as the moon), does it simply mean that we can't call it anymore a reaction force according to the Newton's third law? But why if it is virtually the reaction force?
3) Accelerating elevator - similar to the first example - let's say the elevator is accelerating upwards. So we get that $\vec{N}=m(g+a)$, and that's the same as me pushing the floor. That is why I feel heavier. Then why some add to here the fictitious force?
I will appreciate any answer.
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2 Answers
1) You surely feel the pressure when you accelerate. Whether you attribute it to fictitious forces or other forces depends on your choice of the "reference frame" (vantage point). From the viewpoint of your body's reference frame, which is not an inertial frame, there exist fictitious forces (inertia and/or centrifugal and/or Coriolis' force) that are pushing your body towards the seat. In an inertial reference frame, such as the vantage point of people who stand on the sidewalk and watch you, the pressure is exerted by the seat because it's accelerating i.e. pushing (you are pushing on the seat as well, by the third Newton's law) and there are no additional fictitious forces. Both of these descriptions are OK but the description from the inertial systems (e.g. the sidewalk system) is described by simpler, more universal equations. Without a loss of generality, we may describe all of physics from these frames and these frames never force us (and never allow us) to add any fictitious forces. The frame of your (accelerating) body may be considered "unnatural" and therefore all the forces that appear in that frame are artifacts of the frame's being unnatural, and therefore they are called "fictitious". They may be avoided.
2) Centrifugal forces are the textbook examples of fictitious forces; they have to be added if you describe the reality from the viewpoint of rotating systems. They are avoided if you use non-rotating frames. However, the tides have nothing to do with centrifugal forces. The tidal forces appear because the the side of the Earth that is further from the Moon is less strongly attracted to the Moon than the side that is closer to the Moon. In other words, the tidal forces totally depend on the non-uniformity of the gravitational field around the Moon – the force decreases with the distance. You could create the same attractive force as the Moon exerts by using a heavier body that is further than the Moon. The attractive i.e. "centripetal" force would be the same but the tidal forces would be weaker!
3) In an inertial system – connected with the Earth's surface, for example – the force acting on you is $mg$ downwards from the Earth's attraction plus $ma$ from the extra upwards accelerating elevator. The part $ma$ has a clear new source, object that causes it, namely the elevator. However, in a freely falling frame, for example, the gravitational downward $mg$ force cancels against the fictitious inertial force $mg$ upwards. However, the material of the elevator is now accelerating by the acceleration $g+a$ upwards so the total force is $m(g+a)$ again.
As you can see, whether there are fictitious forces depends on the reference frame. What I feel is your trouble is that you're not used to describe processes from the viewpoint of inertial reference frames. Take a spinning carousel. There is a centripetal force acting on the children and this force, $F=mr\omega^2$, is the reason why the children aren't moving along straight paths with the uniform velocity (as Newton's first law would suggest). Instead, they're deviating from the uniform straight motion and move along circles. The centripetal, inwards directed force $mr\omega^2$ from the pressure from the seats is the reason. (For planets, the centripetal force is the gravitational one.) There are no fictitious forces, in particular no centrifugal force, in the description using the inertial system (from the viewpoint of the sidewalk). However, from your rotating viewpoint, there is a centrifugal force $mr\omega^2$ acting outwards that's always there because the frame is rotating. This force is cancelled against the pressure from the seat, a centripetal force $mr\omega^2$, and the result is zero which implies that in the rotating frame, the coordinates stay constant in this case, especially the distance $r$ from the axis of the spinning carousel. Both frames are possible: one of them forces you to add fictitious forces, the other one (inertial frame) doesn't contain any such forces.
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1
Thank you very much, sir. I understand it now better. – apdmn6072 Feb 27 at 17:41
By the way, you stated that "There are no fictitious forces, in particular no centrifugal force, in the description using the inertial system (from the viewpoint of the sidewalk).". However, there must be some force which will act against the centripetal force (otherwise the object will just move towards the center) - reactive centrifugal force which is described by third Newton's law - it is the reaction to the string or whatever wants to pull you to the center. Am I wrong? Thanks in advance. – apdmn6072 Feb 27 at 20:32
Hi, no, as I tried to emphasize, the centripetal force is exactly what's needed for the object to move along the circle in the inertial frame's language (not towards the center!). If no centripetal force were acting, the object would move along a straight line, away from the circle (outside to infinity), as declared by the first Newton's law (one that you seem unfamiliar with but it's the most important one)! – Luboš Motl Feb 28 at 6:21
So you basically say that there is no reaction force acting on the string? How it could possibly be? Then what is "reactive centrifugal force" and why it is described in Landau's books (as well as in other decent books)? And it is described from inertial frame of reference. And I am familiar with Newton's laws. You right when you said that my trouble was with reference frames. The whole issue of fictitious forces is just a way to describe things in non-inertial frames. – apdmn6072 Feb 28 at 11:05
Dear apdmn, there may also be something wrong about your understanding of the third law and the meaning of the reaction force. On a carousel, the seat acts on the baby with a centripetal force that keeps the baby on the circular path. Third Newton's law implies that there is also the opposite force by which the baby acts on the seat or the whole carousel. But this is a force acting on the seat or carousel, not the baby, so it doesn't contribute to the motion of the baby! If it did, all forces would just cancel which they surely don't. Reaction means that A acts on B and B acts on A. – Luboš Motl Mar 1 at 12:41
show 1 more comment
Luboš Motl already answered your subquestions very well, but I thought I'd work out an example in detail for your understanding.
Some people like to work with fictitious forces, most people don't. The thing is that Newton's laws don't work in reference frames that have a non-zero (finite) acceleration. In those reference frames we need to resort to mysterious fictitious forces to make things add up. In inertial frames (frames with zero acceleration), Newton's laws work fine and we need no hocus-pocus.
Now for my example, let's look at the situation of a satellite of mass $m$ orbiting the earth. In the reference frame $O'$ of the satellite, there's only one force exerted on it: the gravitational pull of the earth. In its own reference frame, the satellite is also stationary, so its acceleration with respect to $O'$ is zero. Now, we know this can't be the whole story, otherwise Newton's second law $m\mathbf{a} = \mathbf{F}$ would yield
$$\mathbf{a} = G\frac{M}{r^2}\mathbf{\hat{r}'}$$
where $G$ is the gravitational constant, $M$ is the mass of the earth and $\mathbf{r'}$ is the vector from the satellite to the earth's center. This would mean the acceleration is non-zero, which we know is not the case. So we have to add a fictitious centrifugal force to get $a$ to be zero:
$$m\mathbf{a} = G\frac{mM}{r'^2}\mathbf{\hat{r}'} + \mathbf{F}_{cf}$$
where the centrifugal force $\mathbf{F}_{cf}$ is equal to $-m\frac{v^2}{r'}\mathbf{\hat{r}'}$ where $v$ is the orbital velocity and to get $\mathbf{a}=\mathbf{0}$ we demand that this force cancels out with gravity (a condition from which we can derive the orbital speed!).
A different way of looking at this is by saying: well, we know the reference frame $O'$ isn't inertial, because the second law of Newton gives us nonsense (this doesn't necessarily mean you're not in an inertial frame but let's assume it does). So let's take the reference frame fixed to the center of the earth $O$, which is a fair approximation of an inertial frame for these purposes. From this reference frame we can clearly observe that the satellite's reference frame $O'$ moves with some finite acceleration $\mathbf{a}_{O'}$. Moreover, since it's moving (approximately) on a circle, we know an expression for this acceleration: $$\mathbf{a}_{O'} = -\frac{v^2}{r}\mathbf{\hat{r}}$$ where $v$ is the orbital velocity and $\mathbf{\hat{r}}$ is the vector from the center of the earth to the satellite. So let's insert this acceleration into the equation:
$$m\mathbf{a}_{O'} = -G\frac{mM}{r^2}\mathbf{\hat{r}}$$
(notice the sign change because $\mathbf{\hat{r}} = -\mathbf{\hat{r}'}$) We can rewrite this as follows:
$$\begin{align} \mathbf{0} &= -G\frac{mM}{r^2}\mathbf{\hat{r}} - m\mathbf{a}_{O'} \\ \mathbf{0} &= -G\frac{mM}{r^2}\mathbf{\hat{r}} + m\frac{v^2}{r}\mathbf{\hat{r}} \end{align}$$
This is the same equation as before. So we end up with the same physics, but the handling of the problem is much more natural in inertial frames of reference. We needed some artificial construct to get things right in the non-inertial frame, while we had no issues at all in the inertial frame.
Another way of looking at fictitious forces is with fictitious accelerations: the maths are all the same, but the mindset is different. Instead of adding a fictitious force, which you know isn't real, you add a "fictitious" acceleration, which is more true to the actual physics since it's the acceleration of the reference frame if you were to look at it from an inertial frame. If you add the fictitious acceleration $\mathbf{\hat{r}'}v^2/r'$ in the left hand side of the equation $m\mathbf{a} = \mathbf{F}$ for the reference frame $O'$ attached to the satellite (where $\mathbf{a}$ was zero), you get:
$$m\left(\mathbf{0}+\frac{v^2}{r'}\mathbf{\hat{r}'}\right) = G\frac{mM}{r^2}\mathbf{\hat{r}'}$$
This is again the same equation.
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I appreciate your answer, sir. Thank you. It is indeed a great addition to the another answer. – apdmn6072 Feb 27 at 17:42
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http://math.stackexchange.com/questions/tagged/ring-theory+category-theory
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# Tagged Questions
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### $M\cong_{R\operatorname{-Mod}}N$ if and only if $M\cong_{\operatorname{Ab}}N$ (Warning: This statement is FALSE!)
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### Inherited Morita similar rings
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### Is the functor $R \mapsto \mathbb{M}_n(R)$ a right adjoint?
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### On limits, schemes and Spec functor
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### Equivalent module categories
Let $A$ and $B$ be rings and let $A\text{-mod}$ and $B\text{-mod}$ be their abelian module categories. Let $F:A\text{-mod}\to M\text{-mod}$ and $F':B\text{-mod}\to A\text{-mod}$ be functors which ...
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### Some isomorphism conditions
Here I want to ask some true or false problems regarding isomorphisms (and if false, is there some extra condition to make it true). I do not what is the correct title for these problem. And I also ...
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http://mathoverflow.net/questions/10991/can-premodular-be-relaxed-as-a-condition-for-uniqueness-of-bruguieres-mueger-mo
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## Can “premodular” be relaxed as a condition for uniqueness of Bruguieres/Mueger modularization?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Suppose that C is a ribbon monoidal category with dominant ribbon functors F_1: C->D_1 and F_2: C->D_2 such that D_1 and D_2 are modular tensor categories, does it follow that D_1 and D_2 are equivalent as MTCs? Here dominant means that every object in the target is a summand of an object in the image of the functor.
This is certainly true if C is premodular (semisimple with finitely many simple objects) as was proved by Bruguieres. What if C is not premodular? I haven't been able to locate a more general statement in the literature.
The particular case I have in mind is where C is the Kuperberg G_2-spider specialized to q a particular root of unity. After semisimplification C is in fact premodular, but actually proving that is likely to be a lot of work (it would require writing down inductive formulas for simples, etc.).
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## 2 Answers
Noah, this is a comment to your answer to your question: unfortunately your functor is not braided. Indeed the braiding on the square V_2 has 2 eigenvalues and the braiding on
the square of F_2(V_2) has 4 different eigenvalues.. Also, I think that TL_{-1} is related with third (or sixth) root of 1.
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Excellent point! – Noah Snyder Jan 12 2010 at 18:17
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I think this is a counterexample to the result I was looking for. Let C be the idempotent completion of TL_{-1}, the Temperley-Lieb category with loop value -1. (Or equivalently, the category of finite dimensional representations of U_q(sl_2) where q is an third root of unity.)
Let D_1 be the quotient of C by negligibles. This category is just Rep(Z/2) but with the dimension of the nontrivial rep being -1.
Let B be the Fibonacci category and let B' be its Galois conjugate, let x and x' be the nontrivial objects. Let D_2 be the Deligne tensor product of B and B'. Let F_2 be the functor sending V_2 to $x \boxtimes x'$, this exists by the universal property of Temperley-Lieb (i.e. $x \boxtimes x'$ is symmetrically self-dual and has quantum dimension -1). This functor is dominant because $F_2(V_2 \otimes V_2)$ has every object of D_2 as a summand.
As Victor Ostrik points out F_2 is not a ribbon functor, so this is not a counterexample.
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Despite this not being a counterexample to my question, it still suggests that the result is unlikely to be true. What this counterexample shows is that there are functors of pivotal categories C->D with D semisimple which don't factor through the "semisimplification" (quotient by all negligibles) of C. This stops the most obvious route to a positive answer to my question. – Noah Snyder Jan 12 2010 at 18:48
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http://physics.stackexchange.com/questions/35431/is-the-law-of-conservation-of-energy-still-valid/35464
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# Is the law of conservation of energy still valid?
Is the law of conservation of energy still valid or have there been experiments showing that energy could be created or lost?
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It is absolutely valid in our dimensions. When one goes to big bang models and general relativity, conjectures abound. – anna v Sep 2 '12 at 4:10
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Richart, is there some reason you have to suspect it isn't valid? It will help us give a better answer if we can address that specifically. – David Zaslavsky♦ Sep 2 '12 at 4:32
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– Richart Bremer Sep 2 '12 at 6:10
@annav Well in greece, a lot of things are "valid". – Richart Bremer Sep 2 '12 at 6:23
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There is no global energy conservation in cosmology, if that is what you mean. Locally energy is still conserved ($\nabla_\mu T^{\mu\nu}=0$). – C.R. Sep 2 '12 at 6:45
show 1 more comment
## 5 Answers
The energy conservation becomes vacuous or invalid in the general theory of relativity and especially in cosmology. See
http://motls.blogspot.com/2010/08/why-and-how-energy-is-not-conserved-in.html
Why and what does it imply? First of all, Noether's theorem makes the energy conservation law equivalent to the time-translational symmetry. In general backgrounds in GR, the time-translational symmetry is broken (especially in cosmology), so the corresponding energy conservation law is broken, too, despite the fact that the energy conservation law (and the corresponding time-translational symmetry) is an unassailable principle in all of pre-general-relativistic physics.
One example of a possible subtlety we have to be careful about: $\nabla_\mu T^{\mu\nu}=0$ holds in GR but because it contains the covariant derivative $\nabla$, this law can't be brought to the equivalent integral form. The extra Christoffel symbol terms explicitly measure how much the energy conservation law is violated at the given point. There's no way to redefine $T_{\mu\nu}$ so that the conservation law would hold with partial derivatives $\partial_\mu$ but the energy would still retain a coordinate-independent value that actually constrains the final state in any way.
If one views the background as variable and appreciates that the underlying laws as being time-translational-invariant, it doesn't help because the time-translational symmetry is a subgroup of the diffeomorphism group which is a local (gauge) symmetry in GR, and all physical states must therefore be invariant under it. The invariance is the same thing as saying that the generator – the energy itself – identically vanishes. So we may declare that there's a conserved energy in GR but it's zero.
We may see the same point if we try to associate energy to gravitational waves. In general spacetimes, we will fail to find a good formula. It's not hard to see why. The total stress-energy tensor comes from the variation of the action with respect to the metric tensor. The variation of the "matter-field" part of the action gives us the matter part of the energy/momentum density. However, the variation of the gravitational part, the Einstein-Hilbert action, gives us an additional term, the Einstein curvature tensor. Of course, the sum of both vanishes – this condition is nothing else than Einstein's equations – because the metric tensor is a dynamical variable in GR and the action has to be stationary under variations of all dynamical fields.
We may also try to invent other definitions of the total energy in general spacetimes. They will either explicitly refuse to be conserved; or they will be identically zero; or they will depend on the chosen spacetime coordinates (in the latter case, it will actually be the case that the whole "beef" of the energy will be just an artifact of the choice of coordinates and there will be no "meaningful piece" that would actually depend on the matter distribution). There's no way to define "energy" in general (cosmological) situations that would be nonzero, coordinate-choice-independent, and conserved at the same moment.
For asymptotically flat or other asymptotically time-translationally-invariant spacetimes, we may again define the total energy, the ADM mass, but it is not possible to exactly say "where it is located" and the cleanest way to determine the ADM mass is from the asymptotic conditions of the spacetime.
Cosmology
In cosmology, the most explicit example of the text above is the FRW uniform and isotropic cosmology. In that case, the total energy stored in dust which has $p=0$, vanishing pressure, is conserved. However, the total energy stored in radiation is decreasing as $1/a$ where $a$ are the linear dimensions of the Universe simply because each photon (or particle of radiation) sees its wavelength grow as $a$ and energy goes like $1/\lambda$ i.e. $1/a$.
There are other states of matter I could discuss such as cosmic strings and cosmic domain walls which obey different power laws. But the most interesting example I will mention is the cosmological constant. It's an energy density of the vacuum. Because the cosmological constant is "constant", this energy density is always and everywhere the same. So because the density is constant and the volume of spacetime grows as $a^3$ in our spacetime dimension, the total energy stored in the Universe grows as $a^3$, too.
Cosmic inflation is driven by a "temporary cosmological constant" so the total energy of the Universe grows with the volume of the Universe, too. In Alan Guth's words, inflation (or the Universe) is the ultimate free lunch. Inflation explains why the mass/energy of the visible Universe is so much hugely larger than the mass scales of particle physics.
For different mixtures of matter obeying different equations of state (roughly speaking, with different ratios of pressure and energy density), one will see the total energy increase or decrease or be constant. Generally, the total energy of the Universe will tend to increase as the Universe expands if the Universe is filled with matter of increasingly negative pressure; the total energy will decrease if it is filled with matter of increasingly positive pressure.
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+1 Very clear answer!! I've just learned some nice stuff today! – J. C. Leitão Sep 2 '12 at 8:32
Shouldn't the total energy of the Universe be infinite ? – jjcale Sep 2 '12 at 17:07
I've already quoted the link. – Richart Bremer Sep 3 '12 at 1:29
Thanks, JCL. ... Dear Richart, I realize that. Still, I believe that I also have the right to quote my own text about the very same topic. ;-) @jjcale: the total energy of the visible Universe (whose current radius if 46 billion light years) is finite. Whether or not the total energy of the whole Universe is finite depends on whether the Universe (its volume) is finite. We don't know. We just know that if finite, it is very large - the curvature radius of the whole Universe is at least hundreds of billions of light years, way greater than the radius of the visible Universe. – Luboš Motl Sep 3 '12 at 6:04
– Raindrop Jan 31 at 22:09
It is still valid.
Only two hypothetical exclusions exist:
1) Quantum uncertainty principle. Energy can be uncertain if time is certain and vice versa. So, virtual particles can violate energy conservation law for a small amounts of time. These violations are averaged in normal scales.
2) General relativity model. Universe are not equal over time. Since energy conservation is a consequence of time uniformity, it is possible that it is violated in cosmological scale.
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Energy has been shown to be conserved in all circumstances where it is currently possible to test it experimentally. It is also conserved according to theory in any system with time translation invariance. This is the case in all known physics including general relativity.
Some people have tried to argue that energy is not conserved in general relativity, or that conservation is approximate, trivial or meaningless. This is not the case. A variety of falacious arguments are used to support non-conservation, e.g. some theorists say that general relativity does not have time-translation invariance because the gravitational field is not invariant. The obvious solution to this is to include the gravitational field as a dynmaical field with its own time-transaltion invariance. Despite this such incorrect arguments have even made their way into textbooks written by well known cosmologists. This is not something where you should rely on the word of authority. Check the maths and the logic yourself.
This subject has been discussed on physics stackexchange several time before so I wont expand on this answer. Suffice to say that energy is conserved in all established physical laws including general realtivity. It is not approximate, or trivial or true only in special cases. See also the discussions at my blog http://blog.vixra.org/category/energy-conservation/
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Dear @Phil, concerning the article you linked to – which is probably meant as evidence of your otherwise unjustified and invalid propositions – we have already discussed that but if your "conserved total energy" is always equal to zero, then it is trivial, by the very definition of trivial. In every system, you may invent infinitely many such new "conservation laws" – the number of Phil-Gibbs-owned-platinum-comets is also conserved and zero – and because the value is zero, the knowledge of $E_{\rm initial}$ provides us with no information about the final state. – Luboš Motl Sep 3 '12 at 6:10
– Luboš Motl Sep 3 '12 at 6:18
– Philip Gibbs May 6 at 12:26
I think Lubos said it very well. Phil Gibbs is, I believe, way off track in thinking energy exists in general relativity. It definitely does not, as there is no background space-time against which one can ask whether the laws are time translationally invariant.
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Can anyone here say there is research to back their claims,e.g. that physics' laws can be put aside at and before the start of our universes? The answer is no, all the math and theorems are simply a creation of scientists to support theories and suppositions that they have never observed,can not be observed if the laws that govern universes now where non-existent at the beginning. It all sounds very wise and learned but the truth is it boils down to this: Scientists can not explain the existence of a super small matter,energy or whatever you may call it supposedly existed at the time of the Big bang. Regardless of whether its value was zero (which seems to connotate in one small dot would imply a static state) and does not explain the unpredictable expansion that also added up to zero which we call our universe now which scientists have conveniently exempted all existing theory to fit into their limited knowledge of that universe. All this talk about zero values, a singularity and such as well as the numbers to back these suppositions does have some value, of course. There was a singularity when creation began; and there is more than likely a singularity when creation began before that and before that. The theories proposed are based on time as a straight line or perhaps even a curved line in the case of einstein's theories concerning the curvature of the universes. But a zero state at the time of the creation of our universe(s) implies that time also equaled zero. This means that any one point in time at the time of the big bang would have been equal to the beginning and the end at the same identical time. Time would have been circular, truly there was a singularity. The Bible calls that Creator as the ALPHA and the OMEGA, the beginning and the end, yes at the same instant, zero and everything that exists which add up to zero. Time that moves in an infinite circle truly will keep the laws of the conservation of matter and energy in place.Where the beginining can also be the end. Who can explain what amount of energy or what force could have re-arranged zero circular time into a straight line or perhaps even a curved phenom? No one on earth can imagine this kind of time, it is a singularity and since we are all bound by time that runs along a straight line, from birth to death and all points in between, no one can even begin to understand the alpha and the omega, God that is the beginning and the end. How is it possible to understand this with mathematical equations that can not begin to de scribe this circular time and all that really happened at the Beginning(s)? The wisdom of men is but foolishness to God. All the attempts to erase the Alpha and Omega from science only creates more unanswered questions and will require even more exceptions to the laws of physics in the future. For what, just to keep cherished theories that proclaim our wisdom to the rest of us on earth? For what does any of these cherished theories actually benefit anything but a few scientists' egos and their unwillingness to accept that the beginning of knowledge increases the chance that they may find out they don't know anyhting at all! It is all vanity. Time is circular and the furthest reaches of space are actually right next to us. If the magnetic field is removed or weakened, things have a hard time holding atoms together, but suppose if a person was able to weaken magnetic fields enough to separate atoms that enabled them to walk through a wall or other planes that exists next to us concurrently? In order to advance science it is necessary to open our minds and hearts leaving behind our egos, moving ahead like children with curiousity and joy, not caring if we win or lose but just exploring for the joy of it. Perhaps then the men will be able to see the Truth that has always been there and find out that we didn't have to create singularity math justifications to find it. It can not be found via the wisdom of men. Men are bound by linear time and therefore can not experience in anyway what happened on before and perhaps not even after creation. Many questions have not been answered. Why do some planets rotate in reverse? Why don't all galaxies rotate the same way? Why aren't they all shaped the same? Why don't they all move apart or together at the same speed?Why would have any part of the big bang cause some particles to be hurled further out than others if they were moving through what should have been empty space, a vacuum? Why aren't all solar systems set up about the same way following a uniform path, regardless of their locations? Some of these questions are certainly observable and even answerable, but moving from a circular system into a linear or curved one has to trace the question from its inception to where it exists presently.
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Hi Charles. Welcome to Physics.SE. Your answer is pretty hard to read. Could you format it somewhat, so that it doesn't look chatty..? – Ϛѓăʑɏ βµԂԃϔ Feb 1 at 1:08
## protected by Qmechanic♦Feb 1 at 1:37
This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.
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http://stats.stackexchange.com/questions/32863/how-does-a-multivariate-mixed-cdf-look-like
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# How does a multivariate mixed CDF look like?
If I have multiple CDFs $X_1,X_2,\cdots,X_n$, (for simplicity I assume 2 : $X,Y$) and if $X$ is discrete and $Y$ is continuous, how would the joint CDF look like?
I understand that:
• If they were both discrete, it would be a staircase in 2 dimensions. Similar to stacking books atop each other on a table.
• If they were both continuous, it would be a nice smooth graph coming to 1 at $(\infty,\infty)$
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It sounds like you are talking about the CDF rather than the PDF. – Neil G Jul 23 '12 at 21:01
@NeilG, isn't it the same? – Nunoxic Jul 23 '12 at 21:09
– whuber♦ Jul 23 '12 at 21:10
@whuber. I am using PDF (caps) for Cumulative and pdf (small) for the usual. – Nunoxic Jul 23 '12 at 21:15
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Inquest, because you are in a small minority who uses this convention, I have edited your question so it will be understood by others as you intended it. (+1 for an interesting question.) – whuber♦ Jul 23 '12 at 21:18
show 1 more comment
## 1 Answer
You've actually already put together all the right pieces in your question. The cdf is
• along one axis, say $x$, a staircase (similar to stacking books atop each other on a table)
• along the other axis, say $y$, continuous
So, imagine a staircase where each step has its own continuous curve, and each successive step's curve is greater than or equal to the last step's curve at every point.
Or, looking at it from the perpendicular direction, each part of a continuous curve has a corresponding staircase, each staircase greater than or equal to the last at every point.
As you say, the cdf goes to 1 as $x$ and $y$ go to $(\infty, \infty)$.
This plot of a multivariate CDF is for a Binomial($1/3$,$2$) variable $X$ and an independent Beta$(3,4)$ variable $Y$. Because the variables are independent this CDF is the product of the individual CDFs.
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@whuber: Wow, nice diagram!! (Thanks for adding it!) – Neil G Jul 24 '12 at 21:31
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http://mathoverflow.net/questions/6074?sort=newest
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## Kahler differentials and Ordinary Differentials
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What's the relationship between Kahler differentials and ordinary differential forms?
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I would have thought this got mentioned in some book on either commutative algebra or differential geometry... I'm not an expert, but the universal property of Kahler differentials should be suggestive. – Yemon Choi Nov 19 2009 at 9:00
I agree, but I just can't find where. – Abtan Massini Nov 19 2009 at 9:43
There is some discussion of this in section 8.1 of "Elements of noncommutative geometry" by Gracia-Bondia, Varilly, and Figueroa. See in particular the example starting on page 324. – Zach Conn Jun 20 2011 at 23:26
## 5 Answers
Let $M$ be a differentiable manifold, $A=C^\infty (M)$ its ring of global differentiable functions and $\Omega^1 (M)$ the A-module of global differential forms of class $C^\infty$. The A-module of Kähler differentials $\Omega_k(A)$ is the free A-module over the symbols $adb$ ($a,b \in A$) divided out by the relations
$d(a+b)=da+db,\quad d(ab)= adb+bda,\quad d\lambda=0 \quad(a,b\in A, \quad \lambda \in k)$
There is an obvious surjective map $\quad \Omega_k(A) \to \Omega^1 (M)$ because the relations displayed above are valid in the classical interpretation of the calculus (Leibniz rule). However, I do not believe at all that it is injective.
For example, if $\: M=\mathbb R$ , I see absolutely no reason why dsin(x)=cos(x)dx should be true in $\Omega_k(A)$ (Beware the sirens of calculus).Things would be worse if we considered $C^\infty$ functions which, contrary to the sinus, are not analytic . The same sort of reasoning applies to holomorphic manifolds and also to local rings of differentiable or holomorphic functions on manifolds.
To sum up: the differentials used in differentiable or holomorphic manifold theory are a quotient of the corresponding Kähler differentials but are not isomorphic to them. (And I think David's claim that they are isomorphic is mistaken)
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I think your definition of Kahler differentials differs from that in the other two entries. – John McCarthy Nov 19 2009 at 19:49
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Dear John, this is entirely possible. The definition I use is the same as in Weibel's book on Homological Algebra (8.8.1.),Qing Liu's (6.1.1.)and Hartshorne's (II,8) on Algebraic Geometry, Matsumura's on Commutative Ring Theory (9.25),Bourbaki's Algebra ( III,\$10,11) and actually in all the books on the subject that I know of. What other definition of Kähler differentials do you have in mind? Greetings, – Georges Elencwajg Nov 19 2009 at 20:36
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Georges is quite correct in that it is not trye that $d\mathrm{sin}=\mathrm{cos}dx$ in the module of Kähler differentials for $A$. – Mariano Suárez-Alvarez Nov 19 2009 at 20:40
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Hello everybody, I'm not a mathematician :) so sorry not to write the details here. Basically, I studied engineering and control theory. In nonlinear control theory we often use differentials, either Kahler or ordinary, to study the systems. There has been a long discussion whether those two notions are isomorphic or not. I believe, we have the answer finally. Together with my colleagues we have shown that differential one-froms are isomorphic to a quotient space (module) of Kahler differentials. These two modules coincide when they are modules over a ring of linear differential operators over the field of algebraic functions. It was published as G. Fu, M. Halás, Ü. Kotta, Z. Li: Some remarks on Kähler differentials and ordinary differentials in nonlinear control theory. In: Systems and control letters, article in press, 2011. Available online at http://www.sciencedirect.com/science/article/pii/S0167691111001198
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There is a discussion of this issue at the $n$-category cafe. I'd encourage people who were interested in this question to head over there and see if they can lend some insight.
Here is a sketch of a proof that $d (e^x) \neq e^x dx$ in the Kahler differentials of $C^{\infty}(\mathbb{R})$. More generally, we should be able to show that, if $f$ and $g$ are $C^{\infty}$ functions with no polynomial relation between them, then $df$ and $dg$ are algebraically independent, but I haven't thought through every detail.
Choose any sequence of points $x_1$, $x_2$, in $\mathbb{R}$, tending to $\infty$. Inside the ring `$\prod_{i=0}^{\infty} \mathbb{R}$`, let $X$ and $e^X$ be the sequences $(x_i)$ and `$(e^{x_i})$`. Choose a nonprincipal ultrafilter on the $x_i$ and let $K$ be the corresponding quotient of `$\prod_{i=0}^{\infty} \mathbb{R}$`.
$K$ is a field. Within $K$, the elements $X$ and $e^X$ do not obey any polynomial relation with real coefficients. (Because, for any nonzero polynomial $f$, $f(x,e^x)$ only has finitely many zeroes.) Choose a transcendence basis, `$\{ z_a \}$`, for $K$ over $\mathbb{R}$ and let `$L$` be the field `$\mathbb{R}(z_a)$`.
Any function `$\{ z_a \} \to L$` extends to a unique derivation $L \to L$, trivial on $\mathbb{R}$. In particular, we can find $D:L \to L$ so that $D(X)=0$ and $D(e^X) =1$. Since $K/L$ is algebraic and characteristic zero, $D$ extends to a unique derivation $K \to K$. Taking the composition $C^{\infty} \to K \to K$, we have a derivation $C^{\infty}(\mathbb{R}) \to K$ with $D(X)=0$ and $D(e^X)=1$. By the universal property of the Kahler differentials, this derivation factors through the Kahler differentials. So there is a quotient of the Kahler differentials where $dx$ becomes $0$, and $d(e^x)$ does not, so $dx$ does not divide $d(e^x)$.
I'm traveling and can't provide references for most of the facts I am using aobut derivations of fields, but I think this is all in the appendix to Eisenbud's Commutative Algebra.
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It is a standard fact that given a field extension $L/K$ with $K$ of characteristc zero, a subset $S\subseteq L$ is algebraically independent over $K$ iff $\{ds:s\in S\}$ is linearly independent in $\Omega_{L/K}$ over $L$. If $\mathcal M$ is the field of meromorphic functions on $\mathbb C$, one can use this to show that $e^x$ and $x$ are linearly independent in the $\mathcal M$-vector space $\Omega_{\mathcal M/\mathbb C}$. – Mariano Suárez-Alvarez Dec 25 2009 at 16:24
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Don;t you hate when people say "it is a standard fact that..." without giving references? The above claim is stated and proved, for example, in Matsumura's Commutative Algebra as theorem 87. – Mariano Suárez-Alvarez Dec 25 2009 at 16:30
UPDATE The previous answer that was here was pretty much completely wrong. Thanks to Georges for several corrections. I encourage everybody to vote my answer down and his up.
If $M$ is a $C^{\infty}$ manifold and $A$ is the ring $C^{\infty}(M)$ then there is a natural map from the Kahler differentials $\Omega_{A/\mathbb{R}}$ to the $C^{\infty}$ one-forms. This is surjective but far from injective. For example, $d \sin x \neq \cos x dx$ in the Kahler differentials. The basic problem is that Kahler differentials are only linear for finite sums, so they can't "see" nonpolynomial relations.
If you replace $C^{\infty}$ by complex analytic and $M$ is an open simply connected set in $\mathbb{C}^n$, then the Kahler differentials map to the holomorphic $(1,0)$-forms. This is still true for any complex manifold if interpreted as a statement about sheaves; let $\mathcal{H}$ be the sheaf of holomorphic functions and define the sheaf of Kahler differentials by sheafifying the presheaf $U \mapsto \Omega_{\mathcal{H}(U)/\mathbb{C}}$; then we again have a map from Kahler differentials to holomorphic $(1,0)$ forms.
In algebraic geometry, if $M$ is a smooth complex algebraic variety, one usually considers the sheaf gotten by sheafifiying the presheaf $U \mapsto \Omega_{\mathcal{O}(U)/\mathbb{C}}$. (We are now using the Zariski topology.) These are, by definition, the algebraic $(1,0)$ forms. They are not isomorphic to the holomorphic $(1,0)$ forms but, if $M$ is projective, they have the same cohomology by GAGA.
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Dear David, unless I misunderstand you I think the statements in the differentiable and holomorphic case are not quite correct. Would you care to have a look at my answer below? Friendly, – Georges Elencwajg Nov 19 2009 at 17:58
I think I'm right. I was a little confused by your answer because you spent a lot of time doing constructions with local rings, which I don't think you need to do. I'd be glad to hear about any error. Is it not true that any 1-form on a C^{\infty} manifold can be written as sum f_i d g_i, with f_i and g_i C^{infty} functions? I think this follows from the local fact and partitions of unity. Or am I wrong that two such expressions are equal if and only if their equality follows from the Liebnitz rule and linearity? Or is there some other error? – David Speyer Nov 19 2009 at 20:41
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I think you are confusing me with Alberto (just below), since I didn't spend any time at all on constructions with local rings! MY post is the last one. It is perfectly true that 1-forms on M are, as you say, sums of global forms of the type fdg ( Kähler differentials surject on classical 1-forms as I mention in my post). The problem is, as you suggest yourself, that the equality of honest 1-forms cannot be proved just by linearity and Leibniz. To repeat my example, how would you prove in the case of M=R(the reals) that d(sin x)= (cos x)dx (true as one-forms) is true in the Kähler module ? – Georges Elencwajg Nov 19 2009 at 22:16
Dear David, your gentlemanly reaction to my comment is incredibly gracious and generous. Let me use this opportunity to thank you and your colleagues for this (addictive!) wonderful site, whose lightning-fast success (1600 users in a little over a month) is more than deserved. – Georges Elencwajg Nov 20 2009 at 7:42
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Dear David, I see some people (yourself maybe?!) have voted you down. This is ridiculous: I am voting you up ! – Georges Elencwajg Nov 20 2009 at 19:50
show 1 more comment
UPDATE: My answer essentially just gives the definition of Kahler differentials and differential forms and misses the point of the question. Georges' answer addresses the relationship between the two. As David before me, I also encourage you to vote Georges' answer up and mine down.
Let $M$ be a smooth manifold and $p$ a point in $M$. The usual definition of the tangent space to $M$ at $p$ is as the vector space of linear maps $D: C^{\infty}(M) \to \mathbb{R}$ satisfying the Leibniz rule $$D(fg) = D(f)g(p) + f(p)D(g)$$ Equivalently, let $I$ be the ideal of $C^{\infty}(M)$ consisting of all functions vanishing at $p$. Then $T_p M$ is the dual of the vector space $I/I^2$ (which you hence call the cotangent space to $M$ at $p$). Indeed, $D(f) = 0$ for every $f \in I^2$, and conversely any linear map $r: I/I^2 \to \mathbb{R}$ gives rise to a derivation $D(f) := r(f-f(p))$.
Now let $X$ be a scheme over a field $k$ (you can generalize this to a morphism of schemes) and $x$ a closed point. Consider the local ring $\mathcal{O}_{x, X}$ of functions regular at $X$. Then the stalk at $x$ of the sheaf of Kahler differentials $\Omega^1_X$ corepresents the functor taking an $\mathcal{O}_{x,X}$-module $\mathcal{F}_x$ to `$\mathrm{Der}(\mathcal{O}_{x,X}, \mathcal{F}_x)$`. In particular, `$$\mathrm{Der}(\mathcal{O}_{x,X}, k) \cong \mathrm{Hom}(\Omega^1_{X,x}, k)$$`
It is in this sense that you think of $\Omega^1_{X,x}$ as the cotangent space to $X$ at $x$. Indeed, in this case $\Omega^1_{X,x} \cong m/m^2$ where $m \subset \mathcal{O}_{x, X}$ is the ideal of functions vanishing at $x$.
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I think "represents" is not to be taken in it's literal sense in the sentence "the stalk at x of the sheaf of Kähler differentials represents derivations". But it can easily be rewritten to eventually get a representation of $Der(\mathcal{O}_{X,x}, k)$. – Konrad Voelkel Nov 19 2009 at 16:15
You're right: that choice of words was unfortunate. I have clarified it now. – Alberto García-Raboso Nov 19 2009 at 16:51
5
Dear Alberto, thank you for your update but of course I strongly discourage anybody to vote you down! I am incredibly impressed by the intellectual honesty and fair-play that I am witnessing from David and you: this is MUCH more admirable than any result on Kähler differentials. – Georges Elencwajg Nov 20 2009 at 19:42
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http://mathhelpforum.com/advanced-statistics/51839-solved-density-function.html
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# Thread:
1. ## [SOLVED] Density function
I've got this problem:
f(y) = $(1/\alpha)my^{m-1}e^{{-y}^{m}/\alpha}$, y>0
0, elsewhere
Now I'm supposed to find $E(Y^k)$ for any positive integer k.
I start out like this:
$E(Y^k)$ = $\int_0^\infty y^{k}f(y) dy$ = $\int_0^\infty y^{k} (1/\alpha)my^{m-1}e^{{-y}^{m}/\alpha} dy$
I then move out $(1/\alpha)$ and m outside the integral =
$(1/\alpha)m$ $\int_0^\infty y^{k} y^{m-1}e^{{-y}^{m}/\alpha} dy$ = $(1/\alpha)m$ $\int_0^\infty y^{k+m-1}e^{{-y}^{m}/\alpha} dy$
and from here I don't know how to go on. The correct answer is $\Gamma (k/m+1)\alpha ^{k/m}$
and I know that $\Gamma (\alpha) = \int_0^\infty y^{\alpha-1}e^{-y} dy$
but I don't see how $(1/\alpha)m$ $\int_0^\infty y^{k+m-1}e^{{-y}^{m}/\alpha} dy$ can be rewritten to the correct answer. Thanks in advance for your help.
2. Originally Posted by approx
I've got this problem:
f(y) = $(1/\alpha)my^{m-1}e^{{-y}^{m}/\alpha}$, y>0
0, elsewhere
Now I'm supposed to find $E(Y^k)$ for any positive integer k.
I start out like this:
$E(Y^k)$ = $\int_0^\infty y^{k}f(y) dy$ = $\int_0^\infty y^{k} (1/\alpha)my^{m-1}e^{{-y}^{m}/\alpha} dy$
I then move out $(1/\alpha)$ and m outside the integral =
$(1/\alpha)m$ $\int_0^\infty y^{k} y^{m-1}e^{{-y}^{m}/\alpha} dy$ = $(1/\alpha)m$ $\int_0^\infty y^{k+m-1}e^{{-y}^{m}/\alpha} dy$
and from here I don't know how to go on. The correct answer is $\Gamma (k/m+1)\alpha ^{k/m}$
and I know that $\Gamma (\alpha) = \int_0^\infty y^{\alpha-1}e^{-y} dy$
but I don't see how $(1/\alpha)m$ $\int_0^\infty y^{k+m-1}e^{{-y}^{m}/\alpha} dy$ can be rewritten to the correct answer. Thanks in advance for your help.
Read post #5 of this thread: http://www.mathhelpforum.com/math-he...-function.html
3. mr fantastic: I'm sorry to say that I still don't understand how to rewrite that expression into the right answer.
4. Originally Posted by approx
mr fantastic: I'm sorry to say that I still don't understand how to rewrite that expression into the right answer.
The first three lines of the reference I've given you are crystal clear I would have thought. What exactly don't you understand?
5. Thanks for your fast answer. I don't understand which substitutions I'm supposed to do. Should I let:
$u = y^m/\alpha$ ?
which gives $(1/\alpha)m$ $\int_0^\infty y^{k+m-1}e^{{-u}} du$ ?
6. Originally Posted by approx
Thanks for your fast answer. I don't understand which substitutions I'm supposed to do. Should I let:
$u = y^m/\alpha$ ?
which gives $(1/\alpha)m$ $\int_0^\infty y^{k+m-1}e^{{-u}} du$ ?
Your substitution is incorrect for a number of reasons:
1. You still have y in the integral, there should only be u's.
2. You have not substituted the correct exprssion for dy. $dy \neq du$ !
Note that $u = \frac{y^m}{\alpha} \Rightarrow dy = \frac{\alpha}{m \, y^{m-1}}\, du$ and $y = (\alpha \, u)^{1/m}$.
It's expected that at this level you can correctly substitute a change of variable in an integral.
7. Ok. So I've done some thinking and came up with this substitution:
$u = y^m$, which gives $y = u^{1/m}$ and $dy = (1/m) u^{{(1/m)}-1} du$. Right?
And then I go on:
$(m/\alpha) \int_0^\infty u^{({1/m})^{k+m-1}}e^{-u/\alpha}(1/m) u^{(1/m)-1}du$
I move $1/m$ outside which gives $(1/m)(m/\alpha) = (1/\alpha)$ outside the integral.
$(1/\alpha) \int_0^\infty u^{({1/m})^{k+m-1}}e^{-u/\alpha}(1/m) u^{(1/m)-1}du$
=
$(1/\alpha) \int_0^\infty u^{(({(k-1)}/m)+1)}e^{-u/\alpha} u^{(1/m)-1}du$
I put together the u:s and receives:
$(1/\alpha) \int_0^\infty u^{k/m}e^{-u/\alpha} du$
Am I right so far?
8. Originally Posted by approx
Ok. So I've done some thinking and came up with this substitution:
$u = y^m$, which gives $y = u^{1/m}$ and $dy = (1/m) u^{{(1/m)}-1} du$. Right?
And then I go on:
$(m/\alpha) \int_0^\infty u^{({1/m})^{k+m-1}}e^{-u/\alpha}(1/m) u^{(1/m)-1}du$
I move $1/m$ outside which gives $(1/m)(m/\alpha) = (1/\alpha)$ outside the integral.
$(1/\alpha) \int_0^\infty u^{({1/m})^{k+m-1}}e^{-u/\alpha}(1/m) u^{(1/m)-1}du$
=
$(1/\alpha) \int_0^\infty u^{(({(k-1)}/m)+1)}e^{-u/\alpha} u^{(1/m)-1}du$
I put together the u:s and receives:
$(1/\alpha) \int_0^\infty u^{k/m}e^{-u/\alpha} du$
Am I right so far?
Yes. Now substitute $w = \frac{u}{\alpha} \Rightarrow u = \alpha w$.
9. Thank you! I got the right answer after the last substitution.
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http://dsp.stackexchange.com/questions/6214/probability-of-random-numbers
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# Probability of random numbers
This is the question i found on internet under DSP section so thats why i am posting it here. Help me understand it please.
"A computer adds 1000 random numbers that have each been rounded off to the nearest 10$^{th}$. Find the probability that the total round-off error for the sum is $\ge$1."
How probability of any system can be greater than 1?
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@DilipSarwate so it means i have two numbers one come up from the summation of 1000 random numbers without rounding off and another number is the summation of those random numbers after rounding off and have to find the probability of Sum2-Sum1 is greater or equal to 1? – heavenhitman Dec 8 '12 at 20:04
@DilipSarwate ok thanks for the help. – heavenhitman Dec 8 '12 at 21:19
## 1 Answer
The total roundoff error for the sum of $N$ numbers is:
$$S = \sum_{i=0}^{N-1} E_i$$
The roundoff error for the $i$-th number is represented by the random variable $E_i$. If we assume that the random number generator used by the computer yields numbers $X_i$ taken from a uniform distribution, then the difference between each $X_i$ and the nearest tenth (which is the roundoff error $E_i$) is uniformly distributed on the interval $(-\frac{0.1}{2}, \frac{0.1}{2}) = (-0.05, 0.05)$.
What we're concerned with, though, is the distribution of $S$. Since $S$ is the sum of $N$ independent, identically distributed (iid) random variables, then via the central limit theorem, as $N \to \infty$, $S$ will tend to a Gaussian distribution. If we assume that your case of $N=1000$ is "large enough" for the Gaussian assumption to hold, we can easily estimate the probability that you seek. It's certainly possible to exactly calculate the distribution of $S$, but the Gaussian assumption is likely close enough for most applications with such large $N$.
A Gaussian distribution is characterized by its first two moments, so if we can find those for $S$, then we have all the information we need. These are easy to calculate for a sum of iid random variables. The mean of $S$ is equal to:
$$\mathbb{E}(S) = \sum_{i=0}^{N-1} \mathbb{E}(E_i) = 0$$
The variance of $S$ is equal to:
$$\mathbb{E}\left((S - \mathbb{E}(S))^2\right) = \sum_{i=0}^{N-1} \mathbb{E}\left((E_i - \mathbb{E}(E_i))^2\right)$$
Recall that the random variables $E_i$ are distributed uniformly. It is well known that the uniform distribution over the interval $(a,b)$ has variance $\frac{1}{12}(b-a)^2$. For this case, that yields a variance $\sigma_{E_i}^2 = \frac{0.01}{12}$. Therefore, the variance of the total roundoff error $S$ is $\sigma_{S}^2 = \frac{0.01N}{12}$.
So in summary, we can approximate $S$'s distribution as Gaussian with mean zero and variance $\sigma_{S}^2 = \frac{0.01N}{12}$. Based on those parameters, you can easily calculate the estimated probability distribution function (pdf), then integrate that result to arrive at whatever probability you seek. The probability that there is a total roundoff error with magnitude greater than one would be:
$$\begin{align} P(|S| > 1) &= P(S>1 \lor S < -1) \\ &= 1 - P(-1 < S < 1) \\ &= 1 - \int_{-1}^{1}f_S(s)ds \end{align}$$
where $f_S(s)$ is the Gaussian distribution's pdf that we arrived at before.
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thanks for detailed answer sir. @DilipSarwate thanks for the explanation and calculator it really made it easy to calculate :) – heavenhitman Dec 8 '12 at 22:23
@JasonR Very nice answer, however I have a minor quibble. Does the RHS of your variance equation reduce to what is shown for any $\mathbb{E}\{S\}$? In this case it happens to be 0, so it would. – Mohammad Dec 9 '12 at 0:20
@JasonR (contd) However it threw me off because $\mathbb{E}\{(S - \mathbb{E}(S))^2\} = \mathbb{E}\{(\sum E_i - \mathbb{E}\{\sum E_i\})^2\} = \mathbb{E}\{(\sum E_i - \sum\mathbb{E}\{E_i\})^2 \}$. Factoring, this then leads to: $\mathbb{E}\{(\sum(E_i - \mathbb{E}\{E_i\}))^2\}$, which is not the same as shown on RHS. Your RHS is correct for this case of zero-mean and iid, but I am not sure it holds for any arbitrary mean, as the LHS would seem to indicate. – Mohammad Dec 9 '12 at 0:20
The expectation operator and summation operator are linear and therefore the expectation operator can be distributed over the summation. – Bryan Dec 10 '12 at 14:05
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http://math.stackexchange.com/questions/262269/find-the-sum-of-series
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# Find the sum of series
Does there exist an explicit formula for the sum of the series $$\sum_{n=1}^\infty \frac{1}{n^2-z^2}?$$
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– Raymond Manzoni Dec 19 '12 at 20:42
That's very nice, many thanks! (The answer is accepted, so I hope it contains a correct proof.) – deltuva Dec 19 '12 at 20:45
– Raymond Manzoni Dec 19 '12 at 20:46
## 3 Answers
Equation $(18)$ on this page states that $$\pi \cot(\pi z)=\frac{1}{z}+2z\sum_{n=1}^\infty \frac{1}{z^2-n^2}.$$
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This answer handles precisely this question by showing that the Cauchy Principal Value of $$\sum_{k\in\mathbb{Z}}\frac1{k+z}=\pi\cot(\pi z)$$ and derives an explicit value for $\displaystyle\sum_{k=1}^\infty\frac{1}{k^2-z^2}$ in $(9)$.
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Yes, there does. Of course the sum is not defined if $z$ is a nonzero integer.
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http://math.stackexchange.com/questions/271497/puzzle-stars-in-the-universe-probability-of-mutual-nearest-neighbors?answertab=oldest
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# Puzzle: stars in the universe - probability of mutual nearest neighbors
If the stars are distributed randomly within the universe, what is the probability for a star to be the nearest neighbor of a star that is its nearest neighbor? What if the number of spatial dimensions is higher than 3 or even grows without limit?
PS. I am interested in the asymptotic behavior with the number of stars growing without bound.
PPS. It's a well-known problem (the "birds on a wire problem" in higher dimensions) and I know the answers. However, no idea how to solve this analytically.
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1
Are you assuming something about the geometry of the universe? Might not an unexpected shape change the answer? – Geoff Robinson Jan 6 at 12:49
What I had in mind is the asymptotic behavior in the limit of the number of stars N growing without limit. Should make the problem geometry independent, isn't it? – Johannes Jan 6 at 14:23
Johannes. are you convinced that the answer is the same for a ball as for a solid torus? – Gerry Myerson Jan 6 at 14:58
I was thinking that with the number of stars growing without limit the difference in probability between a ball or a solid torus to vanish. You guys seem to suggest this is wrong. Need to think about this. If what you suggest is correct, the 1D birds problem should give different answers depending on the landing geometry being a line segment or a circle. – Johannes Jan 6 at 15:09
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"Randomly" isn't usually used by probabilists or statisticians as a synonym of "uniformly" or the like, but it seems widespread among mathematicians generally. – Michael Hardy Jan 6 at 18:08
## 1 Answer
Let's assume an infinite $n$-dimensional universe with stars distributed according to a Poisson process with density $\lambda$. The probability that there is no nearest neighbour within a hypherspherical volume $V=\alpha_nr^n$ of some star $S$ is $\exp(-\lambda V)=\exp(-\lambda\alpha_nr^n)$. Thus the density for finding the nearest neighbour at distance $r$ is $\lambda\alpha_nnr^{n-1}\exp(-\lambda\alpha_nr^n)$.
Now conditional on the nearest neighbour of $S$ being at distance $r$, there is no star in the sphere around $S$ with radius $r$, and the sphere around $S$'s nearest neighbour $T$ with radius $r$ overlaps this sphere in a symmetrical lune, whose volume is proportional to $r^n$, say, $\beta_nr^n$, with $\beta_1=1$ and $\beta_2=2\pi/3-\sqrt3/2\approx1.2284$ (see MathWorld). The probability that there is no other star within distance $r$ of $T$, and hence $S$ is the nearest neighbour of $T$, is $\exp(-\lambda(\alpha_n-\beta_n)r^n)$. Thus the total probability of $S$ being the nearest neighbour of $T$ is
$$\int_0^\infty\lambda\alpha_nnr^{n-1}\exp(-\lambda\alpha_nr^n)\exp(-\lambda(\alpha_n-\beta_n)r^n)\mathrm dr=\frac{\alpha_n}{2\alpha_n-\beta_n}\;.$$
For $n=1$, we have $\alpha_1=2$ and $\beta_1=1$, so the probability is $p_1=2/3$. For $n=2$, we have $\alpha_2=\pi$ and $\beta_2=2\pi/3-\sqrt3/2$, so the probability is
$$p_2=\frac\pi{2\pi-2\pi/3+\sqrt3/2}=\frac1{4/3+\sqrt3/(2\pi)}\approx0.6215\;.$$
We can write the probability for $n$ dimensions as
$$p_n=\frac{\displaystyle\int_{-\pi/2}^{\pi/2}\cos^nx\,\mathrm dx}{2\displaystyle\int_{-\pi/2}^{\pi/6}\cos^nx\,\mathrm dx}\;,$$
the ratio of the volume of a sphere to the volume of the union of two spheres with centres on each other's surfaces. Here's a Sage worksheet that calculates these probabilities for $n$ up to $10$:
````sage: t=var('t');
sage: def f(n,x): return integral (cos (t)^n,t).substitute (t=x);
sage: def g(n): return (f (n,pi/2) - f (n,-pi/2)) / (2 * (f (n,pi/6) - f(n,-pi/2)));
sage: [g(n) for n in range (1,11)]
[2/3, 6*pi/(8*pi + 3*sqrt(3)), 16/27, 12*pi/(16*pi + 9*sqrt(3)), 256/459, 30*pi/(40*pi + 27*sqrt(3)), 2048/3807, 840*pi/(1120*pi + 837*sqrt(3)), 65536/124659, 840*pi/(1120*pi + 891*sqrt(3))]
sage: [g(n).n () for n in range (1,11)]
[0.666666666666667, 0.621504896887431, 0.592592592592593, 0.572465550279709, 0.557734204793028, 0.546588665492621, 0.537956396112425, 0.531153988127765, 0.525722170079978, 0.521339529895594]
````
The probability for odd $n$ is rational, and the probability for even $n$ is of the form $3/(4+q_n\sqrt3/\pi)$ with $q_n$ rational. (You can show this analytically by deriving a recurrence relation for $\int_0^x\cos^nx\,\mathrm dx$ by integrating by parts.) Here are tables for odd $n$
$$\begin{array}{r|c|c} n&p_n&p_n\\\hline 1&\frac23&0.66667\\\hline 3&\frac{16}{27}&0.59259\\\hline 5&\frac{256}{459}&0.55773\\\hline 7&\frac{2048}{3807}&0.53796\\\hline 9&\frac{65536}{124659}&0.52572\\ \end{array}$$
and even $n$:
$$\begin{array}{r|c|c} n&q_n&p_n\\\hline 2&\frac12&0.62150\\\hline 4&\frac34&0.57247\\\hline 6&\frac9{10}&0.54659\\\hline 8&\frac{279}{280}&0.53115\\\hline 10&\frac{297}{280}&0.52134\\ \end{array}$$
Thus, for stars in our three-dimensional universe the probability is $16/27\approx0.59259$. Our star, the sun, is not its nearest neighbour's nearest neighbour: Its nearest neighbour is Proxima Centauri at a distance of about $4.2$ light-years, whose nearest neighbour is the binary star system Alpha Centauri.
The ratio of the volume of the lune to the volume of the sphere goes to zero with $n\to\infty$, so the probability for a star to be the nearest neighbour of its nearest neighbour goes to $1/2$.
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Thanks for the follow-up, joriki. – Johannes Feb 20 at 18:13
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http://math.stackexchange.com/questions/37989/what-is-the-discriminant-of-a-quadratic-extension-over-a-number-field
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# What is the discriminant of a quadratic extension over a number field?
Let $K$ be a number field and $d \in \mathcal{O}_K \setminus \mathcal{O}_K^2$. What is the discriminant of the extension $K[\sqrt{d}]/K$ ? Do we know its ring of integers and which primes are split or not ? (I know that these questions are solved when $K=\mathbb{Q}$, but I can't find any result otherwise).
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I don't have answer, but here is an example of an extreme situation. Let $K$ be an number field of class number 2. Then its Hilbert class field is a quadratic extension of $K$ where the relative discriminant $D_{H/K}$ is the unit ideal. A search on Wikipedia about Hilbert class field gives the example $K=\mathbb{Q}(\sqrt{-15})$ and $H=\mathbb{Q}(\sqrt{-3}, \sqrt{5})$. – Jiangwei Xue May 9 '11 at 10:41
@Arturo : Thank you for the fix. @Jiangwei : so it might be hopeless to find an answer depending on $K$ and $d$... Can we say something if we suppose that $O_K$ is factorial ? – user10676 May 9 '11 at 15:08
@user10676: Second pings don't usually work; I would suggest moving the second part of your comment to a new comment if you want to catch Jianwei's attention. – Arturo Magidin May 9 '11 at 15:09
@user01676, assuming $\mathcal{O}_K$ will certainly make things easier. However, please know that algorithms that computes the ring of integers and discriminants are well established. You may check out chapter 6 of "A course in computational number theory" by H. Cohen. (I must admit that I am not familiar with the algorithm). Such algorithms are implemented in Pari/Gp, which you can download for free. – Jiangwei Xue May 9 '11 at 16:57
1
The title is "A Course in Computational Algebraic Number Theory" – Plop May 9 '11 at 17:56
## 1 Answer
At least when working theoretically, or by hand, it is probably easiest to solve this sort of question locally. (I don't have any feeling for how computationally implemented algorithms for this sort of question work, and so won't comment any more on them.)
In short, fix a prime $v$ of $K$, and consider the completion $K_v$ and the local extension $K_v[\sqrt{d}]/K_v$. The discriminant of $K[\sqrt{d}]/K$ will be a product of the local discriminants, and so we are reduced to computing these.
How do we analyze $K_v[\sqrt{d}]/K_v$? Well, you could first try to compute the finite index subgroup $(K_v^{\times})^2$ of $K_v^{\times}$. Note that $K_v^{\times} = \mathcal O_{K_v}^{\times} \times \pi^{\mathbb Z}$, if $\pi$ is some choice of uniformizer, and so $(K_v^{\times})^2 = (\mathcal O_{K_v}^{\times})^2\times \pi^{2\mathbb Z}$.
Computing $(\mathcal O_{K_v}^{\times})^2\subset \mathcal O_{K_v}^{\times}$ is straightforward. E.g. if $v$ has odd residue characteristic, then any element in $1 + \pi \mathcal O_{K_v}$ is a square, and so it is just a question of computing the subgroup of squares in the residue field at $v$. If $v$ has even residue characteristic, then it is still pretty easy in any particular case to compute the square units, say using the fact that any unit congruent to $1 \bmod 4\pi$ is a square (as one sees using the binomial expansion for $(1+x)^{1/2}$).
Once you've done this step, you can easily check if $d$ is a square in $K_v$, which tells you whether or not $K[\sqrt{d}]/K$ is split at $v$.
To compute the discriminant is not that much harder. If $v$ has odd residue characteristic, then we may divide $d$ through by even powers of $\pi$ so that it is either a unit, or else is exactly divisible by $\pi$. In the first case, $K_v[\sqrt{d}]/K_v$ will be unramified at $v$, and hence the contribution to the discriminant from $v$ will be $1$. In the second case, $K_v[\sqrt{d}]/K_v$ will be tamely ramified at $v$, and the contribution to the discriminant from $v$ will be one power of $v$.
If $v$ is of even residue characteristic, then the computations are more involved, because even if $d$ (perhaps after dividing through by even powers of $\pi$) is a unit, it can still happen that $K_v[\sqrt{d}]/K_v$ is ramified (consider the case $\mathbb Q_2(i)/\mathbb Q_2$), although it need not be (consider the case $\mathbb Q_2(\sqrt{5})/\mathbb Q_2$).
If you would like more details, I can provide them.
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So what are the more details? I am rather intrigued. Thanks for any information. – awllower Feb 2 at 4:34
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http://mathoverflow.net/questions/118831/subspaces-of-l-p-and-banach-mazur-distance/118835
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## Subspaces of $l_p$ and Banach-Mazur distance
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
This is a question I posted on SE, and I have been advised to post it here.
http://math.stackexchange.com/questions/146427/subspaces-of-l-p-and-banach-mazur-distance
It is well-known that every subspace of $l_2$ is isometric to $l_2$. When $p\neq 2$, $l_p$ has subspaces that are not even isomorphic, let alone isometric, to $l_p$. Suppose $X$ is a subspace of $l_p$ with $p\neq 2$ such that $X$ is isomorphic to $l_p$. What can one say about the Banach-Mazur distance between $X$ and $l_p$? More precisely, which one of the following mutually exclusive options holds true:
1) Given $K$, there exists a subspace $X$ of $l_p$, isomorphic to $l_p$, such that for any isomorphism $T:X\to l_p$ one has $||T||\cdot||T^{-1}||>K$.
OR
2) There exist a constant $K$ (possibly depending on $p$), such that for any subspace $X$ of $l_p$, isomorphic to $l_p$, there exist an isomorphism $T:X\to l_p$ such that $||T||\cdot||T^{-1}||\leq K$.
Intuitively, I very strongly suspect it is 1) but I do not have an argument to exclude 2) and, if it is indeed 1), I would like to see a concrete example of a subspace having that property.
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## 1 Answer
(1) is correct. It follows from the fact that there is a sequence $(E_n)$ of finite dimensional subspaces of $\ell_p$ s.t. $\gamma_p(E_n) \to \infty$. Here $\gamma_p(X)$ is the factorization constant of the identity on $X$ through an $L_p$ space; that is, $\gamma_p(X)= \inf \|T\|\cdot \|S\|$, where the infimum is over all $T:X\to Y$, $S:Y:\to X$, and $Y$ an $L_p$ space.
For fixed $n$, set $X_n=\ell_p(E_n)$. Since $E_n$ is finite dimensional, the space $X_n$ is isomorphic to $\ell_p$ (and embeds isometrically into $\ell_p$) but the Banach-Mazur distance of $X_n$ to $\ell_p$ is at least $\gamma_p(E_n)$.
There remains the sticky point of producing a sequence $(E_n)$ as above. One way is to take a subspace $X$ of $\ell_p$ that fails the approximation property and let $E_n$ be a sequence of finite dimensional subspaces of $X$ with $E_1\subset E_2 \subset \dots$ and $\cup E_n$ dense in $X$. Or, for $p<2$, $E_n$ can be (Banach-Mazur arbitrarily close to) $\ell_r^n$ with `$p<r<2$`, because $\ell_r$ embeds isometrically into $L_p(0,1)$ and there are various ways of checking that $\gamma_p(\ell_r^n)\to \infty$. Since I do not know your background (you have, unfortunately, chosen to remain anonymous), I do not know which approach of producing $(E_n)$ is best for you.
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Thank you, I am modestly familiar with Szankowski's construction, so I think I understand what the choice of $(E_n)$ can be. But it is not clear to me why $\gamma_p(E_n)\to\infty$, when $X$ fails AP. I can see that by looking at Szankowski's concrete definition (LT(II), Theorem 1.g.4 ) and taking $E_n=\span\{z_1, z_2,\dots z_n\}$ as in the proof, but I suppose there exists a simpler, abstract argument why any $(E_n)$ as above have the factorization constant going to infinity. – Theo Jan 13 at 22:17
Yes, it is not obvious. What is clear is that $(\sum_{n=1}^\infty E_p)_p$ cannot have the uniform approximation property, while $\ell_p$ does. This does not require the specifics of Szankowski's construction but uses another concept, the uniform approximation property. AFAIK, there is no really simple proof using only elementary facts about $\ell_p$. Another approach is to show that there are finite dimensional subspaces $F_n$ of $\ell_p$ s.t. the GL-constants of $F_n$ tend to infinity, but that also uses a non elementary concept and requires more arguments. – Bill Johnson Jan 13 at 23:38
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http://www.sciforums.com/showthread.php?114948-STRENGTHS-and-WEAKNESSES-of-the-THEORY-OF-RELATIVITY&p=2965652
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# Thread:
1. James, Are you well versed in every crank theory that has been proposed? No? Then how do you know the theories are false if you haven't studied them for years? What? What's that you say? You say that you don't need to bother wasting your time learning every aspect of a crank theory in order to know that it's BS? Well then, there you have it. Why should I waste my time trying to learn every aspect of relativity when I've already shown that it's BS? At the start, the second postulate is WRONG! If I have a theory that is based on two postulates of the earth being flat, do you really need to go any further??
2. Originally Posted by James R
Are you able to answer the following question yet?
Is an object always at rest in its own reference frame?
Silly you, James. Do you propose that an object can be in motion relative to itself?
3. Motor Daddy:
Originally Posted by Motor Daddy
James, Are you well versed in every crank theory that has been proposed? No?
Of course not. I take them as I find them, one crank at a time. Mostly they come to me, not the other way round.
Then how do you know the theories are false if you haven't studied them for years?
I don't know they are false until I have examined them.
What? What's that you say? You say that you don't need to bother wasting your time learning every aspect of a crank theory in order to know that it's BS? Well then, there you have it.
Obviously, if there's a fundamental mistake in page 1 of a 100 page theory, there's no need to read further...
Why should I waste my time trying to learn every aspect of relativity when I've already shown that it's BS? At the start, the second postulate is WRONG! If I have a theory that is based on two postulates of the earth being flat, do you really need to go any further??
You don't even know the basics of relativity, as is well evidenced by your posts here. Despite my taking considerable time in a futile attempt to bring you a little up to speed, you still haven't grasped the concept of a reference frame, on which the whole of relativity is based. And it's been years, hasn't it? Years of no progress at all.
Where have you shown that relativity is BS? Not on sciforums. Anybody can review the thread I've linked above, where your many errors and misconceptions were laid bare for all to see.
The second postulate of relativity (speed of light postulate) is supported by experimental evidence. Your proposed alternative is supported by nothing, as I clearly showed in the above-linked thread.
Originally Posted by Motor Daddy
Silly you, James. Do you propose that an object can be in motion relative to itself?
It's ok. I'm happy to wait another 6 months for you to try to grapple with coming up with an answer.
4. Originally Posted by Motor Daddy
James, Are you well versed in every crank theory that has been proposed? No? Then how do you know the theories are false if you haven't studied them for years? What? What's that you say? You say that you don't need to bother wasting your time learning every aspect of a crank theory in order to know that it's BS? Well then, there you have it. Why should I waste my time trying to learn every aspect of relativity when I've already shown that it's BS? At the start, the second postulate is WRONG! If I have a theory that is based on two postulates of the earth being flat, do you really need to go any further??
Crank theory is really easy to round file. They 'almost always' make a claim or prediction that has already been empirically falsified. Makes it really easy to ignore. Just as James said. You've shown that relativity is wrong? How can that be when the theory of relativity hasn't been empirically falsified? You don't falsify theory with 'chicken scratch' nonsense you post in a public science forum. Especially when the learned folks in the public science forum conclude your efforts are clueless.
5. Originally Posted by Sibilia
WEAKNESSES
1 - All reference frames are equally valid. (Only applies to the speed of light in vacuum).
2 - Time is a dimension of space. (The time is flowing, the space is static, the bodies move in).
3 - It is possible the time travel.
4 - Gravity is not a force.
5 - The cosmological constant.
Points 1, 2 and 5 aren't weaknesses, they are strengths, especially 1. Any kind of symmetry or invariance or independence is always a help when formulating mathematical models of something physical. Phase invariance in electromagnetics leads to charge conservation for example. In the case of the reference frames etc, the two postulates of special relativity lead to Lorentz and Poincare transforms, which lead to momentum, angular momentum and energy conservation in relativity. Point 3 is not a weakness if time travel is physically possible, not to mention the fact relativity doesn't demand time travel to be possible, it just has the possibility to describe it. And Point 4 is a matter of semantics and interpretation and ultimately irrelevant.
Originally Posted by Motor Daddy
James, Are you well versed in every crank theory that has been proposed? No? Then how do you know the theories are false if you haven't studied them for years? What? What's that you say? You say that you don't need to bother wasting your time learning every aspect of a crank theory in order to know that it's BS? Well then, there you have it. Why should I waste my time trying to learn every aspect of relativity when I've already shown that it's BS? At the start, the second postulate is WRONG! If I have a theory that is based on two postulates of the earth being flat, do you really need to go any further??
It's a matter of the scientific method and Occam's razor.
General relativity is presently the most tested, accurate and developed model of gravity known to science. It has a century (or rather 97 years) of work examining it and even had technologies come about because of it (GPS network). As such it has demonstrably utility. This doesn't mean it is perfect but even if it weren't perfect it has demonstrable utility. It's for that reason physics students still learn the work of Maxwell and Newton, even though pretty much everything they did has been surpassed since then.
GR is thus the benchmark. A contender to replace it would need to demonstrate it can model just as much just as well. No crank's work has ever managed that. Few even manage to make testable predictions or formalise anything mathematicially. If you can give an example of one which works as well as GR please do. But for this reason JamesR is justified in not having to sift through every single nut idea on gravity. This is not the case for you and GR. You claim it's wrong but you cannot provide an experiment which demonstrates it. You have attempted to argue against it on the grounds of logical consistency but you've shown you don't know the relevant details. You are saying "Relativity is wrong in what it says about....." but you then demonstrate you don't know what relativity actually says. Your 'light sphere in a box' nonsense illustrates that. There's plenty of books and lecture notes and websites freely available which will help you learn the specifics of what relativity says about such systems but you don't seem to bother to look at them. As such you're whining about something you haven't actually understood. Someone like myself dismissing the claims of people like Farsight or QQ or Martillo when it comes to gravity is different, in that those people cannot demonstrate their assertions lead to anything quantitative and thus there's nothing of any substance to dismiss. General relativity has demonstrated it has things of substance and a great many people have learnt it over the last century. You're claiming to have reason to dismiss it but when you demonstrate you haven't even looked into how relativity constructs it's predictions you completely undermine your position.
Trying to excuse this by saying JamesR is being just as badly when he doesn't go through every crank claim in detail is just laughable. It's laughable for the reason I just explained and it's laughable because it amounts to saying "Well if I'm doing something in the wrong way then so are you!", which is an admission your approach is flawed. Two wrongs don't make a right and in your case you usually clock up about 10 wrongs in short order.
6. Brucep days:
You're delusional. You're explanations are delusional. You're relativity illiterate and certifiably blind if you think it's clear to you.
If I agree with the Relativity, I understand it, and if not, I am delusional.
I prefer to be delusional than to be in mistake.
James R says:
Please give an example of one of the laws of nature that this contradicts.
S = d/t, S is speed, d = distance, t = time
How do you calculate S using the concept of spacetime?
7. To whoever wanted to answer:
Why Albert Einstein not received the Nobel Prize for Relativity?
8. Originally Posted by Sibilia
To whoever wanted to answer:
Why Albert Einstein not received the Nobel Prize for Relativity?
This is turning into a full-blown crackpot thread.
9. Originally Posted by Sibilia
If I agree with the Relativity, I understand it, and if not, I am delusional.
I prefer to be delusional than to be in mistake.
S = d/t, S is speed, d = distance, t = time
How do you calculate S using the concept of spacetime?
The very fact you're asking that question suggests you're insufficiently familiar with relativity to be making the assertions about it you are.
The computation of position and motion through space-time is the central principles of general relativity, as they follow from the main mathematical object, the metric. This encodes notions of distance and action, from which dynamics can be extracted. The definition of a line element is $\textrm{d}s^{2} = g_{ab}\textrm{d}x^{a}\textrm{d}x^{b}$ for the metric g in some coordinates $\{x^{a}\}$. The length of some curve C is then computed via integration, $s(C) = \int_{C} \textrm{d}s = \int_{C} \sqrt{g_{ab}\textrm{d}x^{a}\textrm{d}x^{b}}$. We parametrise the curve by the proper time of something moving along the curve via $\mathbf{x}(\tau) \sim x^{a}(\tau)$ to give $s(C) = \int_{0}^{1} \sqrt{g_{ab}\dot{x}^{a}\dot{x}^{b}} d\tau$. An object which is not experiencing an external force will follow a geodesic. The geodesic equation follows from the length expression I've just given via the Euler-Lagrange equations and gives $\ddot{x}^{a} + \Gamma_{bc}^{a}\dot{x}^{b}\dot{x}^{c} = 0$ for $\Gamma_{bc}^{a} = \frac{1}{2}g^{ad}(g_{bd,c} + g_{dc,b} - g_{bc,d})$. This is essentially the generalisation of Newton's 2nd Law $\ddot{x} = F$ (for unit mass). The geodesic equation is 2nd order so it requires a pair of initial conditions, for instance an initial position and an initial velocity but then you just plug them into the equation and solve. This gives you the motion of something moving along a geodesic, ie the position as a function of proper time and thus velocity.
The expression you gave is what you get when you make pretty much every possible simplifying assumption you can possibly do so it isn't surprising it doesn't seem to work properly for more elaborate scenarios. s=d/t is only valid if the speed was constant and in a straight line and you have a universal notion of t, which isn't the case in relativity. You're making the same mistake Motor Daddy and chinglu do, which is to dismiss relativity without having any understanding of how it works and what it's predictions are. Given it's demonstrable success and the complete lack of anything even remotely close to the level of predictive power and verification GR has dismissing it in the manner you are is a tad unwise.
10. Originally Posted by AlphaNumeric
The very fact you're asking that question suggests you're insufficiently familiar with relativity to be making the assertions about it you are.
The computation of position and motion through space-time is the central principles of general relativity, as they follow from the main mathematical object, the metric. This encodes notions of distance and action, from which dynamics can be extracted. The definition of a line element is $\textrm{d}s^{2} = g_{ab}\textrm{d}x^{a}\textrm{d}x^{b}$ for the metric g in some coordinates $\{x^{a}\}$. The length of some curve C is then computed via integration, $s(C) = \int_{C} \textrm{d}s = \int_{C} \sqrt{g_{ab}\textrm{d}x^{a}\textrm{d}x^{b}}$. We parametrise the curve by the proper time of something moving along the curve via $\mathbf{x}(\tau) \sim x^{a}(\tau)$ to give $s(C) = \int_{0}^{1} \sqrt{g_{ab}\dot{x}^{a}\dot{x}^{b}} d\tau$. An object which is not experiencing an external force will follow a geodesic. The geodesic equation follows from the length expression I've just given via the Euler-Lagrange equations and gives $\ddot{x}^{a} + \Gamma_{bc}^{a}\dot{x}^{b}\dot{x}^{c} = 0$ for $\Gamma_{bc}^{a} = \frac{1}{2}g^{ad}(g_{bd,c} + g_{dc,b} - g_{bc,d})$. This is essentially the generalisation of Newton's 2nd Law $\ddot{x} = F$ (for unit mass). The geodesic equation is 2nd order so it requires a pair of initial conditions, for instance an initial position and an initial velocity but then you just plug them into the equation and solve. This gives you the motion of something moving along a geodesic, ie the position as a function of proper time and thus velocity.
The expression you gave is what you get when you make pretty much every possible simplifying assumption you can possibly do so it isn't surprising it doesn't seem to work properly for more elaborate scenarios. s=d/t is only valid if the speed was constant and in a straight line and you have a universal notion of t, which isn't the case in relativity. You're making the same mistake Motor Daddy and chinglu do, which is to dismiss relativity without having any understanding of how it works and what it's predictions are. Given it's demonstrable success and the complete lack of anything even remotely close to the level of predictive power and verification GR has dismissing it in the manner you are is a tad unwise.
I was being nice saying delusional. It's much worse being intellectually dishonest. That's what happens when you make pronouncements without making a real effort to understand why the pronouncements are bs.
11. Originally Posted by prometheus
I'm going to give you a chance to explain this in a bit more detail. Please define "local background."
Again, please explain what you mean by "planar" in this context.
I'm not entirely sure I can define local background adequately without derailing this thread, but seeing as it's going in the bin soon, i'll make a quick attempt.
Background :- the (natural) constraint(s) imposed upon an object by it's surroundings.
Local background :- additional constraint(s) imposed by local conditions.
The assumption is that the periodicity (of elements and subatomic particles) is universal.
However, if ALL experimentation is performed using apparatus comprising of materials with an observed periodicity, then how can we detect other periodicities?
How can we make such an assumption that one particular periodicity is universal?
When we impose immense electric fields on gasses in vacuum at 2 kelvin, are we changing the local background?
Why shouldn't particles travel faster than light if we impose different conditions?
Can we accurately predict the periodicity of subatomic particles if we go changing the machine and the energies we use? (answer is no)
It's a bit more complicated than that but you get the drift I hope, even if you disagree.
"Planar" :- geometrically flat, two-dimensional Euclidean space, suffering from the same problems that date back to Copernicus (ie r=0 r=∞)
I know we have had this argument before, and I'll agree that even though your simply rotating the planar function to suit, it is still a good approximation by any measure, it doesn't however make it anything more than a simple planar model.
("The non-planar model is so complicated however that it remains impossible for you puny humans to even simulate with your current technology" - Dave the alien)
12. Tach says:
This is turning into a full-blown crackpot thread.
Tach, you have to psichoanalyze yourself. Nobody understood Relativity, even Einstein.
Einstein was not able to unify all forces. He was a genius but he was human.
Alphanumeric, thanks for your contribution, but I think Calculus is a tool to the service of our
personal thought.
13. DUALITY BECOMING-TIME
We have suggested that the duration is a duality: becoming-time. The duration is the stay or course of events in reality. Becoming is the inherent property of matter to experience changes. no a cause for the becoming, nor be deprived the matter of this property. Therefore the becoming is absolute. Time is the rhythm or speed, in one sense, that phenomena and events occur. We refer to a steady and conventional rhythm (clocks) to measure the duration of events. The rhythm of time can be variable, this is regarded as relative.
14. Nobody understood Relativity, even Einstein.
Einstein understood it quite well.
I think the only thing we know for sure is that you don't understand Relativity.
Tach says:
This is turning into a full-blown crackpot thread.
Oh yeah.
15. Einstein understood it quite well.
I think the only thing we know for sure is that you don't understand Relativity.
I will continue despite your criticism, that does not affect me. With all due respect.
16. Originally Posted by Sibilia
DUALITY BECOMING-TIME
We have suggested that the duration is a duality: becoming-time. The duration is the stay or course of events in reality. Becoming is the inherent property of matter to experience changes. no a cause for the becoming, nor be deprived the matter of this property. Therefore the becoming is absolute. Time is the rhythm or speed, in one sense, that phenomena and events occur. We refer to a steady and conventional rhythm (clocks) to measure the duration of events. The rhythm of time can be variable, this is regarded as relative.
Whose we? Write it down so we know how it works. Duality B-S.
17. Originally Posted by Sibilia
Alphanumeric, thanks for your contribution, but I think Calculus is a tool to the service of our
personal thought.
Calculus is a methodical, logically sound, way of constricting implications of postulates. Given the postulates of special relativity we can derive the mass-energy relationship, Lorentz transforms and make quantitative predictions. It helps remove our tendencies to make unsound choices.
The post of yours which followed what I just quoted is a hair's breath away from getting the thread kicked to pseudo-science. It's still here because it has prompted sound discussion. Don't ruin that by posting random nonsense you just pulled from nowhere.
18. Originally Posted by AlphaNumeric
The post of yours which followed what I just quoted is a hair's breath away from getting the thread kicked to pseudo-science. It's still here because it has prompted sound discussion. Don't ruin that by posting random nonsense you just pulled from nowhere.
That, or philosophy, as becoming has no concise meaning in physics, only philosophy (or as concise as philosophy ever gets anyway).
19. Originally Posted by AlphaNumeric
The post of yours which followed what I just quoted is a hair's breath away from getting the thread kicked to pseudo-science. It's still here because it has prompted sound discussion. Don't ruin that by posting random nonsense you just pulled from nowhere.
When possible try spitting a thread rather than condemning the whole thing to the "aether". (pun intended)
20. Originally Posted by brucep
No, it's not a force.
Huh? C'mon ... learn some basic physics! The Universal Law of Gravitation
Originally Posted by brucep
Objects follow the natural path, geodesic path, through spacetime curvature unless acted upon by a FORCE. The geodetic effect was measured in the proper frame of the GPB experimental apparatus. Spacetime curvature is an experimentally confirmed fact of nature.
Delirious .... the trajectory of objects depends on their speed !
There is for each speed, another curvature of space ?
Originally Posted by brucep
How about learning something rather than making incoherent comments about stuff you don't understand. So far you're just trolling anybody who does understand this stuff.
Look in the mirror!
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http://physics.stackexchange.com/questions/44959/in-classical-mechanics-are-complex-numbers-unphysical/44961
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# In classical mechanics, are complex numbers unphysical? [duplicate]
Possible Duplicate:
Physics math without $\sqrt{-1}$
When I produce a complex final solution to a problem that began without complex coefficients at all, I have so far (with my limited expertise) tended to discard it as unphysical, unless I had to tinker some more, in which case the $i$s may square and real-ity be restored. Is this acceptable?
I understand that the Wessel plane is useful in modelling planetary motion, and many other things, but the fact (?) that you could do it all (albeit more awkwardly) without them seems to be indicative that they are a little contrived.
In short, is there a counter-example?; a physical theory (not a particular, neat, mathematical model of a physical theory) in classical mechanics that could not be formulated without complex numbers?
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– Qmechanic♦ Nov 23 '12 at 22:51
## marked as duplicate by Qmechanic♦, Manishearth♦, Emilio PisantyDec 10 '12 at 10:17
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
## 2 Answers
The answer is "It depends how you use complex numbers"!
If you are modelling motion on the plane, you could use $\mathbb{C}$ instead of $\mathbb{R}^{2}$ perfectly fine. In this case $1$ and $i$ are "unit vectors".
If you are working with, e.g., waves in a plane...complex numbers are quite natural.
This is markedly different from quantum theory, though.
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Also complex numbers show up in problems that result in harmonic functions (sine, cosine, etc) depending on how they are formulated. Mass-spring systems are a classic example. – tpg2114 Nov 23 '12 at 22:40
yes but those examples can be formulated without complex numbers, which is what I think constitutes the main point of the question – Nivalth Nov 23 '12 at 22:42
1
@Burzum: But classical mechanics can be formulated without calculus...does this mean calculus is contrived? – Alex Nelson Nov 23 '12 at 23:05
@Alex good question I obviously need to think more about it. – Nivalth Nov 23 '12 at 23:57
3
@Burzum The question poses "Are complex numbers unphysical" and the answer is a definitive no. Granted at the end it does ask if solutions are not possible without complex numbers, but I recall doing very complicated mass-spring systems that would not be possible to solve without Laplace transforms and imaginary numbers. Also, control theory requires complex numbers for things like stability analysis, which still falls under classical mechanics (sort of, it's analysis of classical mechanical systems). – tpg2114 Nov 24 '12 at 1:48
show 1 more comment
No. For example, complex numbers are used in an essential way in classical electromagnetism to study networks of conductors, capacitors, and inductors.
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http://mathhelpforum.com/differential-equations/167192-exponential-method-3-variables.html
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# Thread:
1. ## Exponential method 3 variables
Solve using exponential method.
My book as a good example for separation of variables but neglected one for the exponential method.
$u_{xx}+u_{yy}+u_{zz}-u=0$
$\exp{(rx+sy+tz)}$
$\exp{(rx+sy+tz)}[r^2+s^2+t^2-1]=0$
I usually solve for r. Do I just continue working as I would with 2 variables?
Thanks.
Does the fact that $r^2+s^2+t^2-1=0$ is a sphere of radius help or make solving this easier?
$r=\pm\sqrt{1-s^2-t^2}$
$\exp{(x\pm\sqrt{1-s^2-t^2}+sy+tz)}\Rightarrow$ $\exp{(x\sqrt{1-s^2-t^2}+sy+tz)}+\exp{(-x\sqrt{1-s^2-t^2}+sy+tz)}$
I haven't the slightest on how to proceed now.
2. Originally Posted by dwsmith
Solve using exponential method.
My book as a good example for separation of variables but neglected one for the exponential method.
$u_{xx}+u_{yy}+u_{zz}-u=0$
$\exp{(rx+sy+tz)}$
$\exp{(rx+sy+tz)}[r^2+s^2+t^2-1]=0$
I usually solve for r. Do I just continue working as I would with 2 variables?
Thanks.
Does the fact that $r^2+s^2+t^2-1=0$ is a sphere of radius help or make solving this easier?
$r=\pm\sqrt{1-s^2-t^2}$
$\exp{(x\pm\sqrt{1-s^2-t^2}+sy+tz)}\Rightarrow$ $\exp{(x\sqrt{1-s^2-t^2}+sy+tz)}+\exp{(-x\sqrt{1-s^2-t^2}+sy+tz)}$
I haven't the slightest on how to proceed now.
Dear dwsmith,
Yes. You can solve for r, proceed the same way you have done with 2 variable problems. In this case the exponent of the exponentials will contain s and t. Remember when we deal with two variable PDEs' the exponential terms only contained s.
The solution would be, $u(x,y,z)=A\exp{(x\sqrt{1-s^2-t^2}+sy+tz)}+B\exp{(-x\sqrt{1-s^2-t^2}+sy+tz)}$
By the way can you please tell me the book you are refering? I am having trouble finding a good book for PDEs'.
3. I am using Elementary Partial Differential Equations by Berg and McGregor. It is an older book but my professor likes it.
Amazon.com: Elementary Partial Differential Equations (9780070048508): P. W. Berg: Books
Since it is out of print, it is really cheap. Mine, surprisingly, is in mint condition for being from the 1966 and only cost 25 dollars.
4. The question now becomes what is the correct substitution for the exponential method so that it is equal to the separation of variables method (see below).
$u(x,y,z)=\varphi(x)\psi(y)\omega(z)$
$\varphi''(x)\psi(y)\omega(z)+\varphi(x)\psi''(y)\o mega(z)+\varphi(x)\psi(y)\omega''(z)-\varphi(x)\psi(y)\omega(z)=0$
$\displaystyle \frac{\varphi''(x)\psi(y)\omega(z)+\varphi(x)\psi' '(y)\omega(z)+\varphi(x)\psi(y)\omega''(z)-\varphi(x)\psi(y)\omega(z)}{\varphi(x)\psi(y)\omeg a(z)}=0$
$\displaystyle \frac{\varphi''(x)}{\varphi(x)}+\frac{\psi''(y)}{\ psi(y)}+\frac{\omega''(z)}{\omega(z)}-1=0$
$\displaystyle \frac{\varphi''(x)}{\varphi(x)}=1-\frac{\psi''(y)}{\psi(y)}-\frac{\omega''(z)}{\omega(z)}$
$\displaystyle \frac{\varphi''(x)}{\varphi(x)}=\lambda\Rightarrow \varphi''(x)-\lambda\varphi(x)=0\Rightarrow m^2-\lambda\Rightarrow m=\pm\sqrt{\lambda}$
$\displaystyle \lambda=1-\frac{\psi''(y)}{\psi(y)}-\frac{\omega''(z)}{\omega(z)}$
$\displaystyle \frac{\psi''(y)}{\psi(y)}=\mu\Rightarrow \psi''(y)-\mu\psi(y)=0\Rightarrow n^2-\mu\Rightarrow n=\pm\sqrt{\mu}$
$\displaystyle \mu=1-\lambda-\frac{\omega''(z)}{\omega(z)}$
$\displaystyle \omega''(z)-\omega(z)+\lambda\omega(z)+\mu\omega(z)=0\Rightarr ow a^2-1+\lambda+\mu\Rightarrow a=\pm\sqrt{1-\lambda-\mu}$
$u(x,y,z;\lambda,\mu)=\exp{(x\pm\sqrt{\lambda})}*\e xp{(y\pm\sqrt{\mu})}*\exp{(z\pm\sqrt{1-\lambda-\mu})}$
I can't figure out the correct substitution for the exponential method to match the separations method.
5. Originally Posted by dwsmith
The question now becomes what is the correct substitution for the exponential method so that it is equal to the separation of variables method (see below).
$u(x,y,z)=\varphi(x)\psi(y)\omega(z)$
$\varphi''(x)\psi(y)\omega(z)+\varphi(x)\psi''(y)\o mega(z)+\varphi(x)\psi(y)\omega''(z)-\varphi(x)\psi(y)\omega(z)=0$
$\displaystyle \frac{\varphi''(x)\psi(y)\omega(z)+\varphi(x)\psi' '(y)\omega(z)+\varphi(x)\psi(y)\omega''(z)-\varphi(x)\psi(y)\omega(z)}{\varphi(x)\psi(y)\omeg a(z)}=0$
$\displaystyle \frac{\varphi''(x)}{\varphi(x)}+\frac{\psi''(y)}{\ psi(y)}+\frac{\omega''(z)}{\omega(z)}-1=0$
$\displaystyle \frac{\varphi''(x)}{\varphi(x)}=1-\frac{\psi''(y)}{\psi(y)}-\frac{\omega''(z)}{\omega(z)}$
$\displaystyle \frac{\varphi''(x)}{\varphi(x)}=\lambda\Rightarrow \varphi''(x)-\lambda\varphi(x)=0\Rightarrow m^2-\lambda\Rightarrow m=\pm\sqrt{\lambda}$
$\displaystyle \lambda=1-\frac{\psi''(y)}{\psi(y)}-\frac{\omega''(z)}{\omega(z)}$
$\displaystyle \frac{\psi''(y)}{\psi(y)}=\mu\Rightarrow \psi''(y)-\mu\psi(y)=0\Rightarrow n^2-\mu\Rightarrow n=\pm\sqrt{\mu}$
$\displaystyle \mu=1-\lambda-\frac{\omega''(z)}{\omega(z)}$
$\displaystyle \omega''(z)-\omega(z)+\lambda\omega(z)+\mu\omega(z)=0\Rightarr ow a^2-1+\lambda+\mu\Rightarrow a=\pm\sqrt{1-\lambda-\mu}$
$u(x,y,z;\lambda,\mu)=\exp{(x\pm\sqrt{\lambda})}*\e xp{(y\pm\sqrt{\mu})}*\exp{(z\pm\sqrt{1-\lambda-\mu})}$
I can't figure out the correct substitution for the exponential method to match the separations method.
Dear dwsmith,
Substitute, $\mu=s^2~and~\lambda=1-s^2-t^2$. Hope you can continue.
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http://mathhelpforum.com/calculus/67525-integral-test.html
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# Thread:
1. ## integral test
I just want to quickly check....
I used the integral test to check if $\sum_{n=2}^{\infty}\frac{1}{nln(n)}$converges.... and i found that it does. Can someone verify this? Thanks
2. Hi,
Originally Posted by tsal15
I used the integral test to check if $\sum_{n=2}^{\infty}\frac{1}{nln(n)}$converges.... and i found that it does.
No it does not, this series is divergent. (it can be shown using the integral test )
3. Originally Posted by tsal15
I just want to quickly check....
I used the integral test to check if $\sum_{n=2}^{\infty}\frac{1}{nln(n)}$converges.... and i found that it does. Can someone verify this? Thanks
$\sum_{n=2}^{\infty}\frac{1}{n \ln(n)} > \int_2^{+\infty} \frac{1}{x \ln x} \, dx$ and I'm pretty confident that the integral does not have a finite value .....
4. Originally Posted by mr fantastic
$\sum_{n=2}^{\infty}\frac{1}{n \ln(n)} > \int_2^{+\infty} \frac{1}{x \ln x} \, dx$ and I'm pretty confident that the integral does not have a finite value .....
Are you sure that sign is around the right way?
5. Originally Posted by Prove It
Are you sure that sign is around the right way?
All the left hand rectangles of width 1 (starting from x = 2) have areas lying above the corresponding area under the curve so I'm pretty sure I'm sure.
But since I'm about to go to bed I will leave it to others to point out my mistakes if there are any.
6. Originally Posted by mr fantastic
All the left hand rectangles of width 1 (starting from x = 2) have areas lying above the corresponding area under the curve so I'm pretty sure I'm sure.
But since I'm about to go to bed I will leave it to others to point out my mistakes if there are any.
Yes that would make sense.
Plus, to show that it's divergent it WOULD have to be greater than something tending to infinity wouldn't it...
7. Always verify that $f$ must be positive, continuous and decreasing on $[1,\infty[.$
8. Originally Posted by Krizalid
Always verify that $f$ must be positive, continuous and decreasing on $[1,\infty[.$
Yes when using the integral test
Thanks for the reminder
9. Originally Posted by mr fantastic
$\sum_{n=2}^{\infty}\frac{1}{n \ln(n)} > \int_2^{+\infty} \frac{1}{x \ln x} \, dx$ and I'm pretty confident that the integral does not have a finite value .....
Ok now, I've got a problem...
the integral equals => $ln(ln(x)$ with infinity and 2 being the bounds
therefore, $ln(ln(\infty)) - ln(ln(2))$
ln(ln(2)) is a value. Thus even if $ln(ln(\infty))$ does diverge, ln(ln(2)) will converge..... so does this change things?
10. Originally Posted by tsal15
Ok now, I've got a problem...
the integral equals => $ln(ln(x)$ with infinity and 2 being the bounds
therefore, $ln(ln(\infty)) - ln(ln(2))$
ln(ln(2)) is a value. Thus even if $ln(ln(\infty))$ does diverge, ln(ln(2)) will converge..... so does this change things?
Nope, the integral diverges.
11. Originally Posted by tsal15
Ok now, I've got a problem...
the integral equals => $ln(ln(x)$ with infinity and 2 being the bounds
therefore, $ln(ln(\infty)) - ln(ln(2))$
ln(ln(2)) is a value. Thus even if $ln(ln(\infty))$ does diverge, ln(ln(2)) will converge..... so does this change things?
You're interested in the final value of the integral, not the individual bits that contribute to calculating this value.
12. Originally Posted by tsal15
Ok now, I've got a problem...
the integral equals => $ln(ln(x)$ with infinity and 2 being the bounds
therefore, $ln(ln(\infty)) - ln(ln(2))$
ln(ln(2)) is a value. Thus even if $ln(ln(\infty))$ does diverge, ln(ln(2)) will converge..... so does this change things?
In the grand scheme of things, $\ln{\ln{2}}$ is very small when compared to infinity.
So even though this value is being subtracted, it's still close enough to infinity to say that the integral tends to infinity.
Therefore it diverges.
Make sense?
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http://regularize.wordpress.com/2011/08/26/second-news-from-ilas-11/
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regularize
Trying to keep track of what I stumble upon
August 26, 2011
Second news from ILAS 11
Posted by Dirk under Conference, Math | Tags: conference, ILAS |
On Thursday and Friday, a minisymposium on “Compressed Sensing and Sparse Approximation Algorithms” takes place at ILAS 11.
Holger Rauhut talked about Compressive Sensing and Structured Random Matrices and his abstract was
Compressive sensing is a recent paradigm in signal processing and sampling theory that predicts that sparse signals can be recovered from a small number of linear and non- adaptive measurements using convex optimization or greedy algorithms. Quite remark- ably, all good constructions of the so called measurement matrix known so far are based on randomness. While Gaussian random matrices provide optimal recovery guarantees, such unstructured matrices are of limited use in applications. Indeed, structure often allows to have fast matrix vector multiplies. This is crucial in order to speed up recovery algorithms and to deal with large scale problems. The talk discusses models of structured random matrices that are useful in certain applications, and presents corresponding recovery guarantees. An important type of structured random matrix arises in connection with sampling sparse expansions in terms of bounded orthogonal systems (such as the Fourier system). The second type of structured random matrices to be discussed are partial random circulant matrices, that is, from convolution. In particular, we present recent results with J. Romberg and J. Tropp on the restricted isometry property of such matrices.
Mark A. Iwen talked about Compressed Sensing for Manifold Data and his abstract was
Preliminary work involving the recovery of manifold data using compressive measurements will be discussed, along with related sampling bounds for the approximation of manifold data via a recent multiscale piecewise linear approximation method known as “Geometric Wavelets” [Allard, Chen, Maggioni, Multiscale Geometric Methods for Data Sets II: Geometric Wavelets, Submitted, 2011].
John Wright talked about Local Correctness of Dictionary Learning Algorithms and his abstract was
The idea that many important classes of signals can be well-represented by linear com- binations of a small set of atoms selected from a given dictionary has had dramatic impact on the theory and practice of signal processing. For practical problems in which an appropriate sparsifying dictionary is not known ahead of time, a very popular and successful heuristic is to search for a dictionary that minimizes an appropriate sparsity surrogate over a given set of sample data. In this talk, we describe steps towards un- derstanding when this problem can be solved by efficient algorithms. We show that under mild hypotheses, the dictionary learning problem is locally well-posed: the desired solution is a local minimum of the 1 norm. Namely, if ${A \in \mathbf{R}^{m\times n}}$ is an incoherent (and possibly overcomplete) dictionary, and the coefficients ${X \in \mathbf{R}^{n\times p}}$ follow a random sparse model, then with high probability ${(A, X)}$ is a local minimum of the 1 norm over the manifold of factorizations ${(A', X')}$ satisfying ${A' X' = Y}$ , provided the number of samples ${p = \Omega(n^3 k)}$. For overcomplete A, this is the first result showing that the dictionary learning problem is locally solvable.
The last speaker of the first session was Jakob Lemvig, talking on Sparse Dual Frames and his abstract was
Frames are generalizations of bases which lead to redundant signal expansions, and they play an important role in many applications, e.g., in the theory of nonuniform sampling, wireless communications, and Sigma-Delta quantization. Sparsity of frame vectors (in some fixed orthonormal basis) is a new paradigm in frame theory that among other things allows for simple representation of the frame vectors and fast decomposi- tion and reconstruction procedures. Recently, a general construction method for sparse tight frames was proposed in [Casazza, Heinecke, Krahmer, Kutyniok, Optimally sparse frames, preprint]. In this talk, however, we will initiate the study of sparse dual frames of a given (not necessarily tight nor sparse) frame. We present theoretical and experimental results showing that sparse duals can have many desirable properties as well as providing fast reconstruction.
Today we have to following speakers:
Gitta Kutyniok, Separation of Data by Sparse Approximations
Modern data is customarily of multimodal nature, and analysis tasks typically require separation into the single components. Although a highly ill-posed problem, the morphological difference of these components often allows a precise separation. A novel methodology consists in applying 1 minimization to a composed dictionary consisting of tight frames each sparsifying one of the components. In this talk, we will first discuss a very general approach to derive estimates on the accuracy of separation using cluster coherence and clustered/geometric sparsity. Then we will use these results to analyze performance of this methodology for separation of image components.
Götz E. Pfander, From the Bourgain Tzafriri Restricted Invertibility Theorem to restricted isometries
The Bourgain-Tzafriri Restricted Invertibility Theorem states conditions under which a Riesz bases (for its span) can be extracted from a given system of norm one vectors in finite dimensional spaces. The challenge is to choose a large subset while making sure that the resulting lower Riesz bound remains large. An upper Riesz bound of the selected Riesz bases is inherited from the frame bound of the original system of vectors. In this talk, we shall present an algorithm that allows us to control both, the lower and the upper, Riesz bounds.
Sadegh Jokar, Compressed Sensing and Sparse Solution of PDEs
Compressed sensing says that a high dimensional but sparse signal can be reconstructed from only a small number of (random) measurements. We consider a related but different problem. We are interested in the numerical solution of partial differential equations ${Lu = f}$ with a differential operator L using some ideas from compressed sensing. Considering a classical Galerkin or Petrov-Galerkin finite ele- ment approach, one seeks a solution u in some function space ${U}$ (which is spanned by ${\{\varphi_1 , \dots,\varphi_n\}}$), represented as ${u = \sum_{i=1}^n u_i \varphi_i}$. Here we would like to find the sparsest representation of the solution ${u}$ in the space ${U = span\{u_1 , \dots , u_n \}}$ using discretized finite elements via ${\ell^1}$-minimization. Specially in adaptive methods, the usual approach to achieve this goal is to use local a posteriori error estimation to determine where a refinement is needed, but we use 1 -minimization and linear programming to perform the adaptive refinement in the finite element method in such a way that the solution is sparsely represented by a linear combination of basis functions. We also perform the analysis of solutions of underdetermined linear systems that have Kronecker product structure and show some theoretical results in this direction.
In fact, Sadegh also talked about his work on Compressed Sensing with matrices which are Kronecker product where he investigated how several interesting constants behave under the formation of Kronecker products. He (I guess) invented the term “sparkity” for the smallness of the spark of a matrix, and then the result sound funny: The sparkity of a Kronecker product is equal to the minimum of the sparkity of the products.
Felix Krahmer, New and Improved Johnson-Lindenstrauss Embeddings via the Restricted Isometry Property
The Johnson-Lindenstrauss (JL) Lemma states that any set of p points in high dimensional Euclidean space can be embedded into ${O(\delta^{-2} \log(p))}$ dimensions, without distorting the distance between any two points by more than a factor between ${1-\delta}$ and ${1 + \delta}$ . We establish a new connection between the JL Lemma and the Restricted Isometry Property (RIP), a well-known concept in the theory of sparse recovery often used for showing the success of ${\ell^1}$-minimization. Consider an ${m \times N}$ matrix satisfying the ${(k, \delta_k )}$-RIP and an arbitrary set ${E}$ of ${O(e^k }$points in ${R^N}$ . We show that with high probability, such a matrix with randomized column signs maps ${E}$ into ${R^m}$ without distorting the distance between any two points by more than a factor of ${1 \pm 4\delta k}$ . Consequently, matrices satisfying the Restricted Isometry of optimal order provide optimal Johnson-Lindenstrauss embeddings up to a logarithmic factor in ${N}$. Moreover, our results yield the best known bounds on the necessary embedding dimension m for a wide class of structured random matrices. Our results also have a direct application in the area of compressed sensing for redundant dictionaries.
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http://mathhelpforum.com/calculus/168667-how-do-i-find-relative-extrema-points-inflection-function.html
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# Thread:
1. ## How do I find the relative extrema and points of inflection of this function?
y = (x)/(x^2+9)
2. Local Maxima: The points where $\displaystyle \frac{dy}{dx} = 0$ and $\displaystyle \frac{d^2y}{dx^2} < 0$.
Local Minima: The points where $\displaystyle \frac{dy}{dx} = 0$ and $\displaystyle \frac{d^2y}{dx^2} > 0$.
Inflection Points may or may not appear where $\displaystyle \frac{d^2y}{dx^2} = 0$. It helps to evaluate these points and compare them to the graph of the function.
3. Originally Posted by Prove It
Local Maxima: The points where $\displaystyle \frac{dy}{dx} = 0$ and $\displaystyle \frac{d^2y}{dx^2} < 0$.
Local Minima: The points where $\displaystyle \frac{dy}{dx} = 0$ and $\displaystyle \frac{d^2y}{dx^2} > 0$.
Inflection Points may or may not appear where $\displaystyle \frac{d^2y}{dx^2} = 0$. It helps to evaluate these points and compare them to the graph of the function.
"Inflections Points" are defined as points where the first derivative changes sign. Where that happens the second derivative must be 0 but the second derivative may be 0 where the first derivative does not change sign (example, $f(x)= x^3$ at x= 0). Determine where the second derivative is 0, then look at the first derivative on either side of those points.
(Surely you knew this basic information, onanyc, so what is your difficulty with this particular problem?)
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http://unapologetic.wordpress.com/2009/10/22/multilinear-functionals/?like=1&source=post_flair&_wpnonce=39170bc4ce
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# The Unapologetic Mathematician
## Multilinear Functionals
Okay, time for a diversion from all this calculus. Don’t worry, there’s tons more ahead.
We’re going to need some geometric concepts tied to linear algebra, and before we get into that we need to revisit an old topic: tensor powers and the subspaces of symmetric and antisymmetric tensors. Specifically, how do all of these interact with duals. Through these post we’ll be working with a vector space $V$ over a field $\mathbb{F}$, which at times will be assumed to be finite-dimensional, but will not always be.
First, we remember that elements of the dual space $V^*=\hom_\mathbb{F}(V,\mathbb{F})$ are called “linear functionals”. These are $\mathbb{F}$-linear functions from the vector space $V$ to the base field $\mathbb{F}$. Similarly, a “$n$-multilinear functional” is a function $f$ that takes $n$ vectors from $V$ and gives back a field element in $\mathbb{F}$ in a way that’s $\mathbb{F}$-linear in each variable. That is,
$\displaystyle f(v_1,\dots,av_i+bw_i,\dots,v_n)=af(v_1,\dots,v_i,\dots,v_n)+bf(v_1,\dots,w_i,\dots,v_n)$
for scalars $a$ and $b$, and for any index $i$. Equivalently, by the defining universal property of tensor products, this is equivalent to a linear function $f:V^{\otimes n}\rightarrow\mathbb{F}$ — a linear functional on $V^{\otimes n}$. That is, the space of $n$-multilinear functionals is the dual space $\left(V^{\otimes n}\right)^*$.
There’s a good way to come up with $n$-multilinear functionals. Just take $n$ linear functionals and sew them together. That is, if we have an $n$-tuple of functionals $(\lambda^1,\dots,\lambda^n)\in\left(V^*\right)^{\times n}$ we can define an $n$-multilinear functional by the formula
$\displaystyle\left[m(\lambda^1,\dots,\lambda^n)\right](v_1\otimes\dots\otimes v_n)=\prod\limits_{i=1}^n\lambda^i(v_i)$
We just feed the $i$th tensorand $v_i$ into the $i$th functional $\lambda^i$ and multiply all the resulting field elements together. Since field multiplication is multilinear, so is this operation. Then the universal property of tensor products tells us that this mapping from $n$-tuples of linear functionals to $n$-multilinear functionals is equivalent to a unique linear map from the $n$th tensor power $\left(V^*\right)^{\otimes n}\rightarrow\left(V^{\otimes n}\right)^*$. It’s also easy to show that this map has a trivial kernel.
This is not to say that dualization and tensor powers commute. Indeed, in general this map is a proper monomorphism. But it turns out that if $V$ is finite-dimensional, then it’s actually an isomorphism. Just count the dimensions — if $V$ has dimension $d$ then each space has dimension $d^n$ — and use the rank-nullity theorem to see that they must be isomorphic. That is, every $n$-multilinear functional is a linear combination of the ones we can construct from $n$-tuples of linear functionals.
Now we can specialize this result. We define a multilinear functional to be symmetric if its value is unchanged when we swap two of its inputs. Equivalently, it commutes with the symmetrizer. That is, it must kill everything that the symmetrizer kills, and so must really define a linear functional on the subspace of symmetric tensors. That is, the space of symmetric $n$-multilinear functionals is the dual space $\left(S^nV\right)^*$. We can construct such symmetric multilinear functionals by taking $n$ linear functionals as before and symmetrizing them. This gives a monomorphism $S^n\left(V^*\right)\rightarrow\left(S^nV\right)^*$, which is an isomorphism if $V$ is finite-dimensional.
Similarly, we define a multilinear functional to be asymmetric or “alternating” if its value changes sign when we swap two of its inputs. Then it commutes with the antisymmetrizer, must kill everything the antisymmetrizer kills, and descends to a linear functional on the subspace of antisymmetric tensors. As before, we can construct just such an antisymmetric $n$-multilinear functional by antisymmetrizing $n$ linear functionals, and get a monomorphism $A^n\left(V^*\right)\rightarrow\left(A^nV\right)^*$. And yet again, this map is an isomorphism if $V$ is finite-dimensional.
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Posted by John Armstrong | Algebra, Linear Algebra
## 5 Comments »
1. [...] and at least one of them will be useful. Remember that we also have symmetric and alternating multilinear functionals in play, so the same constructions will give rise to even more [...]
Pingback by | October 26, 2009 | Reply
2. [...] dual to the exterior algebra is the algebra of all alternating multilinear functionals on , providing a counterpart to the algebra of polynomial functions on . But where the variables in [...]
Pingback by | October 27, 2009 | Reply
3. [...] of degree . This is the antisymmetrization of the tensor product of all these linear functionals. We’ve seen that we can consider this as a linear functional on the space of degree tensors by applying the [...]
Pingback by | October 29, 2009 | Reply
4. [...] all these together, we find that the -dimensional volume of a parallelepiped with sides is an alternating multilinear functional, with the sides as variables, and so it lives somewhere in the exterior algebra . We’ll have [...]
Pingback by | November 2, 2009 | Reply
5. [...] Yesterday we established that the -dimensional volume of a parallelepiped with sides should be an alternating multilinear functional of those sides. But now we want to investigate which [...]
Pingback by | November 3, 2009 | Reply
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