Question

In each of the following problems you are given a function on the interval$\mathbf{-}\mathbf{\pi }\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{\pi }$.

Sketch several periods of the corresponding periodic function of period $\mathbf{2}\mathbf{\pi }$ . Expand the periodic function in a sine-cosine Fourier series.

$\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{ll}\mathbf{0}\mathbf{,}& \mathbf{-}\mathbf{\pi }\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{0}\mathbf{,}\\ \mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}\mathbf{,}& \mathbf{}\mathbf{}\mathbf{0}\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{\pi }\end{array}\mathbf{.}$.

Short answer

The expansion is $f\left(x\right)=\frac{1}{\mathrm{\pi }}+\frac{1}{2}\sin x-\frac{2}{\mathrm{\pi }}\sum _{n=2m}^{\infty }\frac{\cos \left(nx\right)}{n^2-1}$where $m=1,2,3,......$.

Step-by-step solution

Given

The given function is $f\left(x\right)=\left\{\begin{array}{l}0,-\mathrm{\pi }<\mathrm{x}<0\\ \sin x,0<x<\mathrm{\pi }\end{array}\right.$.

The given points are $x=0,\pm \frac{\mathrm{\pi }}{2},\pm \mathrm{\pi },\pm 2\mathrm{\pi }$.

Definition of Fourier series and L’Hospital’s rule

The Fourier series for the function f(x) :

$$\begin{gathered} \mathit{f}\left(x\right)\mathbf{=}\frac{{\mathbf{a}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{+}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\left(a_{n}cosnx+b_{n}sinnx\right) \\ \mathbf{}\mathbf{}{\mathit{a}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }}{\mathbf{\int }}_{\mathbf{-}\mathbf{\pi }}^{\mathbf{x}}\mathit{f}\left(x\right)\mathit{d}\mathit{x} \\ \mathbf{}\mathbf{}{\mathit{a}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }}{\mathbf{\int }}_{\mathbf{-}\mathbf{\pi }}^{\mathbf{x}}\mathit{f}\left(x\right)\mathit{c}\mathit{o}\mathit{s}\mathit{n}\mathit{x}\mathit{d}\mathit{x} \\ \mathbf{}\mathbf{}{\mathit{a}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }}{\mathbf{\int }}_{\mathbf{-}\mathbf{\pi }}^{\mathbf{x}}\mathit{f}\left(x\right)\mathit{c}\mathit{o}\mathit{s}\mathit{n}\mathit{x}\mathit{d}\mathit{x} \end{gathered}$$

Iff(x)is an even function:

$$\begin{gathered} \mathbf{}\mathbf{}{\mathit{b}}_{\mathbf{n}}\mathbf{=}\mathbf{0} \\ \mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{a}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{+}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}{\mathit{a}}_{\mathbf{n}}\mathit{c}\mathit{o}\mathit{s}\mathit{n}\mathit{x} \end{gathered}$$

Iff(x)is an odd function:

role="math" localid="1660884845189" $$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}{\mathit{a}}_{\mathbf{0}}\mathbf{=}{\mathit{a}}_{\mathbf{n}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0} \\ \mathit{f}\left(x\right)\mathbf{=}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}{\mathit{b}}_{\mathbf{n}}\mathit{s}\mathit{i}\mathit{n}\mathit{n}\mathit{x} \end{gathered}$$

L'hospital's rule states that if $\underset{\mathbf{x}\mathbf{\to }\mathbf{c}}{\mathbf{l}\mathbf{i}\mathbf{m}}\mathit{f}\left(x\right)\mathbf{=}\underset{\mathbf{x}\mathbf{\to }\mathbf{c}}{\mathbf{l}\mathbf{i}\mathbf{m}}\mathit{g}\left(x\right)\mathbf{=}\mathbf{0}\mathbf{}\mathit{o}\mathit{r}\mathbf{}\mathbf{\pm }\mathbf{\infty }\mathbf{,}\mathit{g}\left(x\right)\mathbf{\ne }\mathbf{0}$for all xinlwith$\mathit{x}\mathbf{\ne }\mathit{c}$ and$\underset{\mathbf{x}\mathbf{\to }\mathbf{c}}{\mathbf{l}\mathbf{i}\mathbf{m}}\frac{\mathbf{f}\mathbf{\text{'}}\left(x\right)}{\mathbf{g}\mathbf{\text{'}}\left(x\right)}$exists, then$\underset{\mathbf{x}\mathbf{\to }\mathbf{c}}{\mathbf{l}\mathbf{i}\mathbf{m}}\frac{\mathbf{f}\left(x\right)}{\mathbf{g}\left(x\right)}\mathbf{=}\underset{\mathbf{x}\mathbf{\to }\mathbf{c}}{\mathbf{l}\mathbf{i}\mathbf{m}}\frac{\mathbf{f}\mathbf{\text{'}}\left(x\right)}{\mathbf{g}\mathbf{\text{'}}\left(x\right)}$.

Sketch the function

The sketch for the given function is shown below.

In the sketch the given function is shifted to the left (blue line) by $\frac{\mathrm{\tau \tau }}{2}$.

Determine the type of function

New function is $g\left(x\right)=\left\{\begin{array}{l}\cos x,-\frac{\mathrm{\pi }}{2}<x<\frac{\mathrm{\pi }}{2}\\ 0,x\in \left(-\mathrm{\pi },-\frac{\mathrm{\pi }}{2}\right)\cup \left(\frac{\mathrm{\pi }}{2},\mathrm{\pi }\right)\end{array}\right.$.

The function is an even function $\left(b_n=0\right)$.

Find the Coefficients and the convergence points

Coefficients of $a_0$:

$$\begin{gathered} a_0=\frac{1}{\mathrm{\pi }}{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}f\left(x\right)dx \\ =\frac{1}{\mathrm{\pi }}{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\cos xdx \\ =\frac{1}{\mathrm{\pi }}{\left[\sin x\right]}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}} \\ =\frac{2}{\mathrm{\pi }} \end{gathered}$$

Coefficients of $a_n$:

$$\begin{gathered} a_n=\frac{1}{\mathrm{\pi }}{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}f\left(x\right)\cos nxdx \\ =\frac{1}{\mathrm{\pi }}{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\left(\cos x\right)\cos nxdx \\ =\frac{1}{\mathrm{\pi }}{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\left(\frac{e^{inx}+e^{-inx}}{2}\right)\left(\frac{e^{ix}+e^{-ix}}{2}\right)dx \\ =\frac{1}{\mathrm{\pi }}{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\left(e^{inx}+e^{-inx}\right)\left(e^{ix}+e^{-ix}\right)dx \end{gathered}$$

Calculate further as shown below.

$$\begin{gathered} a_n=\frac{1}{4\mathrm{\pi }}{\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\left(e^{ix\left(n+1\right)}+e^{ix\left(n-1\right)}+e^{ix\left(1-n\right)}+e^{-ix\left(n+1\right)}\right)dx \\ =\frac{1}{4\mathrm{\pi }}{\left[\frac{e^{ix\left(n+1\right)}}{i\left(n+1\right)}+\frac{e^{ix\left(n-1\right)}}{i\left(n+1\right)}+\frac{e^{ix\left(1-n\right)}}{i\left(n+1\right)}+\frac{e^{-ix\left(n+1\right)}}{i\left(n+1\right)}\right]}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}} \\ =\frac{1}{4\mathrm{\pi }}{\left[\left\{\frac{e^{ix\left(n+1\right)}}{i\left(n+1\right)}-\frac{e^{-ix\left(n+1\right)}}{i\left(n+1\right)}\right\}+\left\{\frac{e^{ix\left(n-1\right)}}{i\left(n+1\right)}-\frac{e^{ix\left(1-n\right)}}{i\left(n+1\right)}\right\}\right]}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}} \\ =\frac{1}{2\mathrm{\pi }}{\left[\frac{\sin \left\{x\left(n+1\right)\right\}}{n+1}+\frac{\sin \left\{x\left(n-1\right)\right\}}{n-1}\right]}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}} \\ forn=1,\underset{x\to 1}{\lim }\frac{\sin \left\{x\left(n-1\right)\right\}}{n-1}. \end{gathered}$$

Use L’Hospital’s rule

$$\begin{gathered} \underset{n\to 1}{\lim }\frac{\cos \left\{x\left(nx-x\right)\right\}}{1}=\underset{x\to 1}{\lim }\cos \left\{x\left(nx-x\right)\right\} \\ =x\cos \left(x-x\right) \\ =x\cos (0 \end{gathered}$$Use L'Hospital's rule as follows:

Now, $$\begin{gathered} a_1=\underset{1\to 1}{\lim }\frac{1}{2\mathrm{\tau \tau }}{\left[\frac{\sin \left\{\mathrm{x}\left(\mathrm{n}-1\right)\right\}}{\mathrm{n}-1}\right]}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}} \\ =\frac{1}{2\mathrm{\tau \tau }}{\left[x\right]}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}} \\ =\frac{1}{2} \end{gathered}$$.

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

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

Then, $g\left(x\right)=\frac{1}{\mathrm{\pi }}+\frac{1}{2}\cos x-\frac{2}{\mathrm{\pi }}\sum _{n=1}^{\infty }\cos \frac{n\mathrm{\pi }}{2}\frac{\cos \left(nx\right)}{n^2-1}$.

For $f\left(x\right)$, $f\left(x\right)=g\left(x-\mathrm{\tau \tau }/2\right)$.

Find the expansion of the function

The expansion of the given function is given below.

$$\begin{gathered} f\left(x\right)=\frac{1}{\mathrm{\pi }}+\frac{1}{2}\cos \left(x-\mathrm{\pi }/2\right)-\frac{2}{\mathrm{\pi }}\sum _{n=1}^{\infty }\cos \frac{n\mathrm{\pi }}{2}\frac{\cos \left\{n\left(x-\mathrm{\pi }/2\right)\right\}}{n^2-1} \\ =\frac{1}{\mathrm{\pi }}+\frac{1}{2}\sin x-\frac{2}{\mathrm{\pi }}\sum _{n=1}^{\infty }\cos \frac{n\mathrm{\pi }}{2}\frac{\cos \left(nx\right)\cos {\displaystyle \frac{n\mathrm{\pi }}{2}}+\sin \left(nx\right)\sin \frac{n\mathrm{\pi }}{2}}{n^2-1} \\ =\frac{1}{\mathrm{\pi }}+\frac{1}{2}\sin x-\frac{2}{\mathrm{\pi }}\sum _{n=1}^{\infty }\frac{\cos \left(nx\right){\cos }^2\frac{n\mathrm{\pi }}{2}+\sin \left(nx\right)\cos \frac{n\mathrm{\pi }}{2}\sin \frac{n\mathrm{\pi }}{2}}{n^2-1} \\ =\frac{1}{\mathrm{\pi }}+\frac{1}{2}\sin x-\frac{2}{\mathrm{\pi }}\sum _{n=1}^{\infty }\frac{\cos \left(nx\right){\cos }^2\frac{n\mathrm{\pi }}{2}+{\displaystyle \frac{1}{2}}\sin \left(nx\right)\sin \left(n\mathrm{\pi }\right)}{n^2-1} \end{gathered}$$

So, $=\frac{1}{\mathrm{\pi }}+\frac{1}{2}\sin x-\frac{2}{\mathrm{\pi }}\sum _{n=1}^{\infty }\frac{\cos \left(nx\right){\cos }^2\frac{n\mathrm{\pi }}{2}}{n^2-1}$where $\left[\sin \left(n\mathrm{\tau \tau }=0\right)\right]$.

$f\left(x\right)=\frac{1}{\mathrm{\pi }}+\frac{1}{2}\sin x-\frac{2}{\mathrm{\pi }}\sum _{n=2m}^{\infty }\frac{\cos \left(nx\right)}{n^2-1}$ where $m=1,2,3,..........$.