Question

expand f(x)=xsinx as fourier series in (0,2π)​

Answer 1

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\blue{ \boxed{ \sf \: 2sinx \: cosy = sin(x + y) + sin(x - y)}}$

$\blue{ \boxed{ \sf \: 2sinx \: siny = cos(x - y) - cos(x + y)}}$

$\blue{ \boxed{ \sf \: {2sin}^{2}x = 1 - cos2x}}$

$\blue{ \boxed{ \sf \: cos2n\pi = 1}}$

$\blue{ \boxed{ \sf \: sin2n\pi = 0}}$

$\blue{ \boxed{ \sf \: sin0 = 0}}$

$\blue{ \boxed{ \sf \: cos0 = 1}}$

$\blue{ \boxed{ \sf \: Fourier \: series \: of \: f(x) \: in \: [ 0, 2\pi] \: is \: given \: by}}$

$\sf \: f(x) = \dfrac{a_0}{2} + \sum_{n=1}^{\infty}[a_n \cos(nx) + b_n \sin(nx)]$

where,

$\blue{ \boxed{ \sf \: a_n = \frac{1}{\pi}\int_{0}^{2\pi}f(x) \cos(nx)\ dx}}$

$\blue{ \boxed{ \sf \: b_n = \frac{1}{\pi}\int_{0}^{2\pi}f(x) \sin(nx)\ dx}}$

$\blue{ \boxed{ \sf \: a_0 = \frac{1}{\pi}\int_{0}^{2\pi}f(x)\ dx}}$

Please see the attachment for solution :-

After that

$\sf \: b_n = \dfrac{1}{2\pi} [\dfrac{cos2n\pi}{{(n-1)}^{2}} -\dfrac{cos0}{{(n-1)}^{2}}]-\dfrac{1}{2\pi}[\dfrac{cos2n\pi}{{(n+1)}^{2}} -\dfrac{cos0}{ {(n+1)}^{2}}]$

$\sf \: b_n = \dfrac{1}{2\pi} [1 - 1]-\dfrac{1}{2\pi}[1 - 1]$

$\sf \: b_n = \dfrac{1}{2\pi} [0]-\dfrac{1}{2\pi}[0]$

$\bf\implies \:b_n = 0 \: \: \: \: \: \: n \ne \: 1$

Hence,

$\sf \: Fourier \: series \: of \: f(x) = xsinx \: on \: [ 0, 2\pi] \: is$

$\sf \: f(x) = \dfrac{a_0}{2} +a_1 + b_1 \sum_{n=2}^{\infty}[a_n \cos(nx) + b_n \sin(nx)]$

On substituting the values, we get

$\sf \: f(x) = \dfrac{ - 2}{2}-\dfrac{1}{2} +\pi + \sum_{n=2}^{\infty}[ \dfrac{2}{ {n}^{2} - 1 } \cos(nx) +0]$

$\sf \: f(x) = - 1-\dfrac{1}{2} +\pi + \sum_{n=2}^{\infty}[ \dfrac{2}{ {n}^{2} - 1 } \cos(nx)]$

$\sf \: f(x) = -\dfrac{3}{2} +\pi + \sum_{n=2}^{\infty}[ \dfrac{2}{ {n}^{2} - 1 } \cos(nx)]$

Answer 2

Answer:

Expansion of $f(x)=xsinx$ as Fourier series in $(0,2\pi)$: $f(x)=-\frac{3}{2}+\pi+\sum_{n=2}^{\infty}\left[\frac{2}{n^{2}-1} \cos (n x)\right]$

Step-by-step explanation:

Fourier series- A Fourier series may be a sum using only basic waves chosen to mathematically represent the waveform for nearly any periodic function. These basic waves are sine and cosine waves whose frequency is an integer multiple of the fundamental of the periodic function.

Fourier series of $f(x)$ in $[0,2 \pi]$ is given by:

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

where,

$\begin{aligned}&a_{n}=\frac{1}{\pi} \int_{0}^{2 \pi} f(x) \cos (n x) d x \\&b_{n}=\frac{1}{\pi} \int_{0}^{2 \pi} f(x) \sin (n x) d x \\&a_{0}=\frac{1}{\pi} \int_{0}^{2 \pi} f(x) d x\end{aligned}$

Step 1-

$\begin{aligned}a_{0} &=\frac{1}{\pi} \int_{0}^{2 \pi} f(x) d x \\&=\frac{1}{\pi} \int_{0}^{2 \pi} x \sin x d x \\&=\frac{1}{\pi}|-x \cos x+\left.\sin x\right|_{0} ^{2 \pi}\end{aligned}$

Since,

$\int x \sin x d x\right.&=x \int \sin x d x-\int\left(\frac{d}{d x} x \int \sin x d x\right) d x \\&\left.=-x \cos x-\int 1 \cdot \cos x\right) d x \\&=-x \cos x+\int \cos x d x \\&=-x \cos x+\sin x\\end{aligned}$

$\begin{aligned}&=\frac{1}{\pi}\{[-2 \pi \cos 2 \pi+\sin 2 \pi]-[0+\sin 0]\} \\&=\frac{1}{\pi}(-2 \pi+0) \\&=-2\end{aligned}$

$\therefore a_{0}=-2$

Step 2-

$\begin{aligned}a_{n} &=\frac{1}{\pi} \int_{0}^{2 \pi} f(x) \cos n x d x \\&=\frac{1}{\pi} \int_{0}^{2 \pi} \cdot(x \sin x \cos n x) d x \\&=\frac{1}{2 \pi} \int_{0}^{2 \pi} x(2 {sin} x \cos n x) d x \\&=\frac{1}{2 \pi} \int_{0}^{2 \pi} x(\sin (1+n) x+\sin (1-n) x) d x \\&=\frac{1}{2 \pi} \int_{0}^{2 \pi} x(\sin (n+1) x-\sin (n-1) x) d x \\&=\frac{1}{2 \pi} \int_{0}^{2 \pi}[x \sin (n+1) x-x \sin (n-1) x] d x\end{aligned}$

Now, by using By parts:

$\begin{gathered}=\frac{1}{2 \pi}\left\{\left[\frac{-x \cos (n+1) x+\sin (n+1) x}{n+1}\right]_{0}^{2 \pi}-\left[\frac{-x \cos (n-1) x}{n-1}\right.\right\left.\left.+\frac{\sin (n-1) x}{(n-1)^{2}}\right]_{0}^{2 \pi}\right\}\end{gathered}$

$\begin{aligned}&=\frac{1}{2 \pi}\left[-2 \pi \frac{\cos (n+1) 2 \pi+0+2 \pi \cos (n-1) 2 \pi}{(n+1)}\right] \\&=\frac{2 \pi}{2 \pi}\left[-\frac{(-1)^{2}}{n+1}+\frac{(-1)^{2}}{n-1}\right]\end{aligned}$

$\begin{aligned}&=\left[-\frac{1}{n+1}+\frac{1}{n-1}\right] \\&=\left[-\frac{n+1+n+1}{(n+1)(n-1)}\right] \\&=\frac{2}{n^{2}-1} ; n \neq 1\end{aligned}$

Calculate $a_1$:

$\begin{aligned}a_{1} &=\frac{1}{\pi} \int_{0}^{2 \pi} f(x) \cdot \cos x d x \\&=\frac{1}{\pi} \int_{0}^{2 \pi} x \sin x \cos x d x \\&=\frac{1}{2 \pi} \int_{0}^{2 \pi} x \sin 2 x d x \\&=\frac{1}{2 \pi}\left[-x \cos 2 x+\frac{\sin 2 x}{4}\right]_{0}^{2 \pi}\end{aligned}$

$\begin{aligned}&=\frac{1}{2 \pi}\left[-2 \pi \frac{\cos 4 \pi}{2}+0-0-0\right] \\&=\frac{1}{2 \pi}\left[-\pi(-1)^{4}\right] \\&=-\frac{1}{2}\end{aligned}$

For $n =1$:

$\begin{aligned}b_{1} &=\frac{1}{\pi} \int_{0}^{2 \pi} f(x) \cdot \sin x d x \\&=\frac{1}{\pi} \int_{0}^{2 \pi} x \sin x \cdot \sin x d x \\&=\frac{1}{\pi} \int_{0}^{2 \pi} x \sin ^{2} x d x \\&=\frac{1}{\pi} \int_{0}^{2 \pi} x\left(\frac{1-\cos 2 x}{2}\right) d x \\&=\frac{1}{2 \pi} \int_{0}^{2 \pi}(x-x \cos 2 x) d x\end{aligned}$

$\begin{aligned}&=\frac{1}{2 \pi}\left[\frac{x^{2}}{2}-\left(x \sin 2 x+\frac{\cos 2 x}{4}\right)\right]_{0}^{2 \pi} \\&=\frac{1}{2 \pi}\left[\frac{4 \pi^{2}}{2}-0-\frac{1}{4}-0+0+\frac{1}{4}\right] \\&=\pi\end{aligned}$

Step 3-

$\begin{aligned}b_{n} &=\frac{1}{\pi} \int_{0}^{2 \pi} f(x) \sin 2 \pi x \\b_{n} &=\frac{1}{\pi} \int_{0} x \sin x \sin n x d x \\&=\frac{1}{2 \pi} \int_{0}^{2 \pi} x \cdot(2 \sin x \sin x n) d x \\&=\frac{1}{2 \pi} \int_{0}^{2 \pi} x(\cos (n-1) x-\cos (n+1) x) d x\end{aligned}$

$=\frac{1}{2 \pi} \int_{0}^{2 \pi}(x \cos (n-1) x-x \cos (n+1) x) d x$

$\begin{aligned}&=\frac{1}{2 \pi}\left\{\left[\frac{x \sin (n-1) x}{n-1}+\frac{\cos (n-1) x}{(n-1)^{2}}\right]_{0}^{2 \pi}-\right&\left.\left[x \frac{\sin (n+1) x}{(n+1)}+\frac{{cos}(n+13 x}{(n+1)^{2}}\right]_{0}^{2 \pi}\right\}\end{aligned}$

Now solve as shown in the image below: