# Fourier Series

##### Member
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
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
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 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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