# Sine Fourier series and coefficient of Fourier series

#### Markov

##### Member
1) Find the sine Fourier series of $f(x)=x(\pi-x),\,0\le x\le\pi$ and show that $\displaystyle\sum_{k\ge1}\frac{(-1)^k}{(2k-1)^3},\,\sum_{k\ge1}\frac1{(2k-1)^6}$ and $\displaystyle\sum_{k\ge1}\frac{\sin\big((2k-1)\sqrt5\pi\big)}{(2k-1)^3}$ and to show that $\displaystyle\sum_{k\ge0}\frac{\sin(2n-1)x}{(2n-1)^3}>0$ for all $x\in(0,\pi).$

2) Let $f$ be an odd piecewise continuous function of period $4L$ and which is even with respect $x=L.$ Show that the Fourier series of $f$ is $\displaystyle\sum_{n\ge1}b_{2n-1}\sin\frac{(2n-1)\pi}{2L}x,$ where $b_{2n-1}=\displaystyle\frac2L\int_0^L f(x)\sin\frac{(2n-1)\pi}{2L}x\,dx.$

Attempts:

1) I don't know how to find the Fourier series here, how to work with a non-symmetric interval?

2) No ideas here, how to start?

#### chisigma

##### Well-known member
1) Find the sine Fourierseries of $f(x)=x(\pi-x),\,0\le x\le\pi$...
If f(x) is $2 \pi$ periodic then is...

$\displaystyle a_{n}= \frac{1}{\pi}\ \int_{0}^{\pi} x\ (\pi-x) \cos nx\ dx = \left\{\begin{array}{cl}\frac{\pi^{2}}{6},&n=0\\ \frac{(-1)^{n}}{\pi\ n^{2}},&n > 0\end{array}\right.$

$\displaystyle b_{n}= \frac{1}{\pi}\ \int_{0}^{\pi} x\ (\pi-x) \sin nx\ dx = \frac{2}{\pi}\ \frac{1 - (-1)^{n}}{n^{3}}$

Kind regards

$\chi$ $\sigma$

Last edited:

#### Markov

##### Member
Okay, makes sense! Could you help me with second part?

#### chisigma

##### Well-known member
First of all I need to correct my previous post because the values of the $a_{n}$ aren't correct ... Here the Fourier coefficients for...

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

$\displaystyle a_{n}= \frac{1}{\pi}\ \int_{0}^{\pi} x\ (\pi-x) \cos nx\ dx = \left\{\begin{array}{cl}\frac{\pi^{2}}{6},&n=0\\ \frac{(-1)^{n}}{n^{2}},&n > 0\end{array}\right.$

$\displaystyle b_{n}= \frac{1}{\pi}\ \int_{0}^{\pi} x\ (\pi-x) \sin nx\ dx = \frac{2}{\pi}\ \frac{1 - (-1)^{n}}{n^{3}}$

Very sorry!...

Kind regards

$\chi$ $\sigma$

#### chisigma

##### Well-known member
The results obtained in the last post permits us to write...

$\displaystyle f(x)= \frac{\pi^{2}}{12} - (\cos x - \frac{1}{4}\ \cos 2x +\frac{1}{9}\ \cos 3x - \frac{1}{16}\ \cos 4x +...) + \frac{4}{\pi}\ (\sin x + \frac{1}{27}\ \sin 3x + \frac{1}{125}\ \sin 5x +...)$ (1)

... where for $0 \le x < \pi$ is $f(x)=x\ (\pi-x)$. S
etting in (2) $x=0$ we obtain...

$\displaystyle \frac{\pi^{2}}{12}= 1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16}+...$ (2)

... and setting in (2) $x=\frac{\pi}{2}$ we obtain...

$\displaystyle \frac{\pi^{3}}{32} = 1 - \frac{1}{27} + \frac{1}{125} -...$ (3)

Other values of
$0 \le x < \pi$ can be inserted in (2) obtaining more interesting results and the task is left to the reader...

Kind regards

$\chi$ $\sigma$

#### chisigma

##### Well-known member
The task is greatly more simple using the 'double identity', easy obtainable from the Fourier expansion just described...

$\displaystyle \frac{\pi^{2}}{12} - (\cos x - \frac{1}{4}\ \cos 2x +\frac{1}{9}\ \cos 3x - \frac{1}{16}\ \cos 4x +...) = \frac{4}{\pi}\ (\sin x + \frac{1}{27}\ \sin 3x + \frac{1}{125}\ \sin 5x +...)= \frac{x\ (\pi-x)}{2}$ (1)

... where for $0 \le x \le \pi$. For example setting in (1) $x=\frac{\pi}{4}$ You obtain...

$\displaystyle 1 + \frac{1}{27} - \frac{1}{125} - \frac{1}{343} +... = \pi^{3}\ \frac{3}{64\ \sqrt{2}}= 1.02772258593...$ (2)

Kind regards

$\chi$ $\sigma$

#### chisigma

##### Well-known member
In the last post we have seen that the'identity'...

$\displaystyle \frac{4}{\pi}\ (\sin x +\frac{1}{27}\ \sin 3x + \frac{1}{125}\ \sin 5x +...)= \frac{x\ (\pi-x)}{2}$ (1)

... is true for $0 \le x \le \pi$ and in that interval the 'infinite sum' is greater or equal to 0. A simple question: whatabout $-\pi \le x \le 0$?... the answer is easy: in that case change the sign of second term of (1)...

In the starting post it was requested the value of $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{3}}\ \sin (2n+1) x$ for $x=\pi\ \sqrt{5}$... because is $\displaystyle \pi \sqrt{5}= \pi (\sqrt{5}-2)\ \text{mod}\ 2 \pi$ is...

$\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{3}}\ \sin (2n+1)\ \pi \sqrt{5} = \frac{11 - 5 \sqrt{5}}{8}\ \pi^{3} = -.6989585560358...$ (2)

Kind regards

$\chi$ $\sigma$

#### Markov

##### Member
Thanks a lot! That helped me a lot! Could you help me with second problem please?

#### chisigma

##### Well-known member
For the second problem is essential for me to understand better that...

... let $f$ be an odd piecewise continuous function of period $4L$ and which is even with respect $x=L$...
... what does it mean that?... may be that is even in the two half periods and odd in the full period?...

Kind regards

$\chi$ $\sigma$

#### Markov

##### Member
Sorry for the delay, well, I wouldn't know really what exactly that means.