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Sharing a cool series 
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#1
Aug2913, 08:43 AM

P: 38

example: 1+(1/2)+2+(1/4)+4+(1/8)+8+... = 0
What does this imply? 


#2
Aug2913, 09:50 AM

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This turns out to be a subset of a remarkably elegant result in combinatorics, namely counting integer points inside of polyhedra.
Given a polyhedron P, [tex] S[P](x) = \sum_{m\in P\cap \mathbb{Z}^d} x^m [/tex] where for [itex] x = (\xi_1,....,\xi_d)[/itex] and [itex] m = (\eta_1,....,\eta_d)[/itex], [itex] x^m = \xi_1^{\eta_1} \xi_2^{\eta_2}.... \xi_d^{\eta_d} [/itex] is a map from polyhedra which do not contain lines to functions of x. Every such function can be rewritten as a rational function of x which is welldefined everywhere except for a couple of singular points. You can extend this map linearly on the set of linear combinations of indicator functions of polyhedra  [P](x) is the function which equals 1 if x is in P, and 0 if x is not in P. Then you can define S[P](x) for any indicator function of a polyhedron as above which does not contain a line, and this extends linearly to linear combinations of polyhedra and is welldefined. For example, in R^{2} let P be the top right quadrant (x >= 0, y >= 0), A the set (x>=0, y>= x), B the set (y>=0, x>=y) and C the set (x>=0, x=y). Then [P] = [A] + [B]  [C] and you can check (it's trivial in this case) that S[P](x) = S[A](x) + S[B](x)  S[C](x) The remarkble part is that the extension of this function to linear combinations of polyhedra without lines means that you have extended it to include polyhedra that contain lines, for example the real line can be written as [x>0=] + [x<= 0]  [x=0] and it turns out that when you take the rational functions involved you always the corresponding function for a polyhedron containing a line is zero. This turns out to be important because identities amongst polyhedra that do no contain lines are typically trivial (like that [P] = [A]+[B][C]), but when you include lines you can get identities which are geometrically meaningful quantities, plus a bunch of garbage polyhedra containing lines that turn out not to matter 


#3
Aug2913, 11:26 AM

P: 38

How did you connect the polyhedron idea with the original one? (I'm not familiar with the ideas behind polyhedra.)
I just thought of it as some kind of a "period" p of the real numbers. (x+p=x) where p is not necessarily 0. Period as in how much do you have to add to make something recur. 


#4
Aug2913, 05:04 PM

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Sharing a cool series
OK a simple example. Let's say I want to know how many integers there are between 2 and 5 (inclusive, and yes I realize this is a stupid question). I can write the interval [2,5] as
[tex] [2,5] = [2,\infty) + (\infty,5]  (\infty,\infty)[/tex] My objective is to write down a polynomial which has a term of x^{k} for each k in [2,5], i.e. I want to find the polynomial [itex] x^2 + x^3 + x^4 + x^5[/itex]. For each guy on the right hand side I can write down a rational function that corresponds to the integer points inside of it [tex] \sum_{n \geq 2} x^n = x^2 \frac{1}{1x} [/tex] at least for some values of x [tex] \sum_{n \leq 5} x^n = x^5 \frac{x}{x1} [/tex] at least for some values of x, and because the interval [itex] (\infty, \infty)[/itex] contains a line I will assign the value [tex] \sum_{\infty}^{\infty} x^n = 0 [/tex]. If you add these all up, you do in fact get that [tex] x^2 + x^3 + x^4 + x^5 = \frac{x^2}{1x} + \frac{x^6}{x1} [/tex] now if I want to know how many integer points are in [2,5] I can clearly plug in x=1 to the left hand side and get an answer. I can also take the right hand side and take the limit as x goes to 1 and get the correct answer as well. This seems like a stupid example, but now consider a triangle in 2dimensions. You can actually do the same procedure. The tangent cone of a vertex of a triangle is simply take the vertex, and extend the edges to infinity to get a cone. It turns out that your triangle is equal to the sum of the three tangent cones, plus some half spaces which contain lines and therefore we can ignore them (the fact that they can be ignored is a deep theory of which your one dimensional sum equaling zero is a simple example of). So if you can write down the rational function corresponding to those three cones you have found the corresponding polynomial for the triangle. You can repeat this process in higher dimensions and this turns out to be algorithmically the best known way (that I'm aware of) to count integer points in convex objects like this 


#5
Aug2913, 07:47 PM

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#6
Aug2913, 08:32 PM

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[tex] \sum_{n\geq 0} x^n = \frac{1}{1x} [/tex] [tex] \sum_{n\leq 1} x^n = \frac{1}{x1} [/tex] Of course these converge for different values of x but suggest (and it turns out this is the correct way to think about it as I detail in my posts) that [tex] \sum_{n=\infty}^{\infty} x^n = \frac{1}{1x} + \frac{1}{x1} = 0[/tex] 


#7
Aug3013, 08:12 AM

P: 38

I'm wondering if this would work with n=1 and n=0.
(Though it probably won't.) 


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