- #1

Mentallic

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I'd like to know how to prove this summation. And if possible, what is the significance of having [itex]\pi[/itex] in the answer?

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- Thread starter Mentallic
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- #1

Mentallic

Homework Helper

- 3,798

- 94

I'd like to know how to prove this summation. And if possible, what is the significance of having [itex]\pi[/itex] in the answer?

- #2

uart

Science Advisor

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If you're familar with Fourier series then one nice method is to consider the expansion of a "saw-tooth" wave as follows.

Let [itex]y(x) = \pi x[/itex] : [itex]-0.5 \leq x \leq 0.5[/itex]

Now make it periodic as per [itex]y(x) = y(x-k)[/itex] : [itex]-0.5+k \leq x \leq 0.5+k[/itex], for all integer k.

It's fairly easy to show that the Fourier series expansion is,

[tex]y = \sin(2 \pi x) - \frac{1}{2} \sin(4 \pi x) \, ... \, + \frac{(-1)^{k+1}}{k} \sin(2k \pi x) + \, ...[/tex]

Consider the mean squared value of y, calculated two different ways. Firstly calculate directly from y(x),

[tex]MS(y) = \int_{x=-0.5}^{+0.5} (\pi x)^2 dx = \frac {\pi^2}{12}[/tex]

Now repeating the calculation but this time using the Fourier series (and making use of the fact that the terms are orthagonal) we get,

[tex]MS(y) =0.5 ( 1 + 1/4 + 1/9 + ... 1/k^2 + ... )[/tex]

Equating these two expressions for the mean squared value gives the required sum.

Let [itex]y(x) = \pi x[/itex] : [itex]-0.5 \leq x \leq 0.5[/itex]

Now make it periodic as per [itex]y(x) = y(x-k)[/itex] : [itex]-0.5+k \leq x \leq 0.5+k[/itex], for all integer k.

It's fairly easy to show that the Fourier series expansion is,

[tex]y = \sin(2 \pi x) - \frac{1}{2} \sin(4 \pi x) \, ... \, + \frac{(-1)^{k+1}}{k} \sin(2k \pi x) + \, ...[/tex]

Consider the mean squared value of y, calculated two different ways. Firstly calculate directly from y(x),

[tex]MS(y) = \int_{x=-0.5}^{+0.5} (\pi x)^2 dx = \frac {\pi^2}{12}[/tex]

Now repeating the calculation but this time using the Fourier series (and making use of the fact that the terms are orthagonal) we get,

[tex]MS(y) =0.5 ( 1 + 1/4 + 1/9 + ... 1/k^2 + ... )[/tex]

Equating these two expressions for the mean squared value gives the required sum.

Last edited:

- #3

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http://www.secamlocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf

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