Chapter 5, Problem 5.48P

To determine

Prove the formulas given.

Answer to Problem 5.48P

The equations have been proven.

Explanation of Solution

The expression for Fourier series is

    $\begin{array}{c}f\left(t\right)={\displaystyle \sum _{n=0}^{\infty }\left[a_n\cos \left(n\omega t\right)+b_n\sin \left(n\omega t\right)\right]}\\ =\left\{\begin{array}{l}{a}_0+a_1\cos \omega t+a_1\cos \omega t+\mathrm{...}+a_n\cos n\omega t+\mathrm{...}+b_1\sin \omega t\\ +b_2\sin 2\omega t+\mathrm{...}+b_n\sin n\omega t\end{array}\right\}\end{array}$        (I)

Multiply (I) with $\cos m\omega t$ and integrate within the limit $-\frac{\tau }{2}$ to $\frac{\tau }{2}$ with respect to $dt$

    $\begin{array}{c}f\left(t\right)\cos m\omega tdt={\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{a}_0\cos \left(m\omega t\right)}dt+{\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{a}_1\cos \omega t\cos \left(m\omega t\right)}dt+{\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{a}_2\cos 2\omega t\cos \left(m\omega t\right)}dt+\mathrm{...}\\ +{\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{a}_n\cos m\omega t\cos \left(m\omega t\right)}dt+\mathrm{...}+{\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{b}_1\cos \omega t\cos \left(m\omega t\right)}dt+{\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{b}_2\cos 2\omega t\cos \left(m\omega t\right)}dt+\mathrm{...}\\ +{\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{b}_n\cos n\omega t\cos \left(m\omega t\right)}dt+\mathrm{...}\end{array}$

But,

    ${\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}\cos n\omega t\cos \left(m\omega t\right)}dt=\{\begin{array}{l}0\text{if}m\equiv n\\ \frac{\tau }{2}\text{if}m=n\ne 0\end{array}$

And

  ${\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}\cos n\omega t\cos \left(m\omega t\right)}dt=0\text{for}\text{all}n\text{and}m$

Therefore all the $b$ terms in the integral will become zero all $a$ terms become zero for except for $m=n$

Therefore,

    $\begin{array}{c}f\left(t\right)\cos m\omega tdt={\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{a}_n\cos n\omega t\cos \left(n\omega t\right)}dt\\ ={\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{a}_n{\cos }^{2}n\omega t}dt\\ ={\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{a}_n\left(\frac{1+\cos \left(2n\omega t\right)}{2}\right)}dt\\ =\frac{1}{2}{\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{a}_{n}d}t+{\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}\left(\frac{1+\cos \left(2n\omega t\right)}{2}\right)dt}\end{array}$        (II)

Here $\frac{\cos \left(2n\omega t\right)}{2}$ is an even function from definite integral properties.

Thus ${\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}\left(\frac{\cos \left(2n\omega t\right)}{2}\right)dt}=0$

Therefore,

    $\begin{array}{c}f\left(t\right)\cos m\omega tdt=\frac{1}{2}{\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{a}_{n}d}t+0\\ =\frac{a_n}{2}{\left(t\right)}_{-\frac{\tau }{2}}^{\frac{\tau }{2}}\\ a_n\left(\frac{\tau }{2}\right)\end{array}$

Rearrange the equation for $a_n$

    $a_n=\left(\frac{2}{\tau }\right)f\left(t\right)\cos m\omega tdt\text{for}n\ge 1$

For $a_0\ne 0$ substitute zero for $n$ in (II) and find $a_0$

    $\begin{array}{c}f\left(t\right)\cos m\omega tdt={\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{a}_0\left(\frac{1+\cos 2\left(0\right)\omega t}{2}\right)}dt\\ {\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}f\left(t\right)}dt={\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{a}_0\left(\frac{1+1}{2}\right)}dt\\ =a_0\tau \\ a_0=\frac{1}{\tau }{\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}f\left(t\right)}dt\end{array}$

Now, again multiply (I) with $\sin m\omega t$ and integrate within the limit $-\frac{\tau }{2}$ to $\frac{\tau }{2}$ with respect to $dt$

$\begin{array}{c}f\left(t\right)\cos m\omega tdt={\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{a}_0\cos \left(m\omega t\right)}dt+{\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{a}_1\cos \omega t\sin \left(m\omega t\right)}dt+{\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{a}_2\cos 2\omega t\sin \left(m\omega t\right)}dt+\mathrm{...}\\ +{\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{a}_n\cos m\omega t\sin \left(m\omega t\right)}dt+\mathrm{...}+{\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{b}_1\cos \omega t\sin \left(m\omega t\right)}dt+{\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{b}_2\cos 2\omega t\sin \left(m\omega t\right)}dt+\mathrm{...}\\ +{\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{b}_n\cos n\omega t\sin \left(m\omega t\right)}dt+\mathrm{...}\end{array}$

But,

    ${\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}\cos n\omega t\sin \left(m\omega t\right)}dt=\{\begin{array}{l}0\text{if}m\equiv n\\ \frac{\tau }{2}\text{if}m=n\ne 0\end{array}$

And

  ${\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}\cos n\omega t\sin \left(m\omega t\right)}dt=0\text{for}\text{all}n\text{and}m$

Therefore all the $b$ terms in the integral will become zero all $a$ terms become zero for except for $m=n$

Therefore,

    $\begin{array}{c}f\left(t\right)\sin m\omega tdt={\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{b}_n\sin n\omega t\sin \left(n\omega t\right)}dt\\ ={\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{b}_n{\sin }^{2}n\omega t}dt\\ ={\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{b}_n\left(\frac{1-\cos \left(2n\omega t\right)}{2}\right)}dt\\ =\frac{1}{2}{\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{b}_{n}d}t-{\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}\left(\frac{\cos \left(2n\omega t\right)}{2}\right)dt}\end{array}$

But ${\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}\left(\frac{\cos \left(2n\omega t\right)}{2}\right)dt}=0$

Equate and rearrange for $b_n$

    $\begin{array}{c}{\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}f\left(t\right)\sin m\omega tdt}=\frac{1}{2}{\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}{b}_{n}d}t-0\\ =\frac{b_n}{2}{\left[t\right]}_{-\frac{\tau }{2}}^{\frac{\tau }{2}}\\ =b_n\frac{\tau }{2}\\ b_n=\frac{2}{\tau }{\displaystyle {\int }_{-\frac{\tau }{2}}^{\frac{\tau }{2}}f\left(t\right)\sin m\omega tdt}\text{for}n\ge 1\end{array}$

And $b_0=0$

Hence proved.

Conclusion:

The equations have been proven.