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Fourier Series

suvadip

Member
Feb 21, 2013
69
If \(\displaystyle f(x)=x+1\), expand \(\displaystyle f(x)\) in Fourier series and hence show that

\(\displaystyle \sum_{n=0}^\infty \frac{1}{(2n-1)^2}=\frac{\pi^2}{8}\)


This question was set in an exam. I am in a position to try it if there is some interval say \(\displaystyle [-\pi \quad \pi]\) or like that.

But there is no interval in the question. Please give me some suggestion how to proceed.
 

chisigma

Well-known member
Feb 13, 2012
1,704
If \(\displaystyle f(x)=x+1\), expand \(\displaystyle f(x)\) in Fourier series and hence show that

\(\displaystyle \sum_{n=0}^\infty \frac{1}{(2n-1)^2}=\frac{\pi^2}{8}\)


This question was set in an exam. I am in a position to try it if there is some interval say \(\displaystyle [-\pi \quad \pi]\) or like that.

But there is no interval in the question. Please give me some suggestion how to proceed.
The problem is, in my opinion, badly defined. A possible improvement consists in the Fourier series expansion of the function...


$\displaystyle f(x)= \begin{cases} \pi - x &\text{if}\ 0 < x < \pi\\ \pi + x &\text{if}\ - \pi < x < 0\end{cases}$ (1)

In that case, writing...

$\displaystyle f(x)= \frac{a_{0}}{2} + \sum_{n=1}^{\infty} (a_{n}\ \cos n x + b_{n}\ \sin n x)$ (2)

... we can compute the coeffcients $a_{n}$ [the $b_{n}$ are zero because the function is even...] as follows...

$\displaystyle a_{0} = \frac{2}{\pi} \int_{0}^{\pi} (\pi - x)\ dx = \pi$ (3)


$\displaystyle a_{n} = \frac{2}{\pi} \int_{0}^{\pi} (\pi - x)\ \cos n x\ dx = \frac{2}{\pi}\ \frac{\cos n\ \pi -1}{n^{2}}$ (4)

... so that is...

$\displaystyle f(x) = \frac{\pi}{2} + \frac{4}{\pi}\ (\cos x + \frac {\cos 3x}{9} + \frac{\cos 5 x}{25} + ...)$ (5)

Now setting in (5) $x=0 \implies f(x)= \pi$ we obtain with simple steps...


$\displaystyle 1 + \frac {1}{9} + \frac{1} {25} + ... = \frac{\pi^{2}}{8}$ (6)


Kind regards


$\chi$ $\sigma$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
If \(\displaystyle f(x)=x+1\), expand \(\displaystyle f(x)\) in Fourier series and hence show that

\(\displaystyle \sum_{n=0}^\infty \frac{1}{(2n-1)^2}=\frac{\pi^2}{8}\)


This question was set in an exam. I am in a position to try it if there is some interval say \(\displaystyle [-\pi \quad \pi]\) or like that.

But there is no interval in the question. Please give me some suggestion how to proceed.
Hi suvadip! :)

The idea of the Fourier series is that you limit your domain to [$-\pi,\pi$], and then extend it again while repeating it.
This is a saw tooth wave.

Anyway, for the calculation of the Fourier series you only look at [$-\pi,\pi$].
The resulting series will be equal to the sawtooth wave.


However, with f(x)=x+1, you won't get the result that is required.
What you do get is:
$$f(x)=1+\sum \frac {-2 \cdot (-1)^n}{n} \sin nx$$
$$x + 1 = 1 + 2(\sin x - \frac 1 2 \sin 2x + \frac 1 3 \sin 3x - ...)$$
Filling in $x=\frac \pi 2$ leads to another result:
$$1 - \frac 1 3 + \frac 1 5 - ... = \frac \pi 4$$


I suspect that the intended function is $f(x)=|x|+1$, which is similar to the function $\chi$ $\sigma$ suggested and yields the same result.
 
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