Chapter 16, Problem 16.10P

To determine

Show that the Fourier coefficients of the triangular wave of Figure 16.7 are zero for $n$ even and ${\left(-1\right)}^m\frac{32}{{\left(2m+1\right)}^2{\pi }^2}\left(1-\cos \frac{\left(2m+1\right)\pi }{8}\right)$ for $n$ odd.

Answer to Problem 16.10P

Showed that the Fourier coefficients of the triangular wave of Figure 16.7 are zero for $n$ even and ${\left(-1\right)}^m\frac{32}{{\left(2m+1\right)}^2{\pi }^2}\left(1-\cos \frac{\left(2m+1\right)\pi }{8}\right)$ for $n$ odd.

Explanation of Solution

Consider the Figure shown below.

From the Figure, it is clear that, $u_0\left(x\right)=0$ for $0<x<3\&5<x<8$, $u_0\left(x\right)$ is increasing in the interval $x=3\text{to}4$ from $0\text{to}1$, and decreasing from $1\text{to}0$ in the interval $x=4\text{to}5$.

Write the equation 16.33.

    $B_n=\frac{2}{L}{\displaystyle \underset{0}{\overset{L}{\int }}{u}_0\left(x\right)}\sin \frac{n\pi x}{L}dx$        (I)

Divide the integral into four parts, use $8$ for $L$, and solve.

    $\begin{array}{c}{B}_n=\frac{2}{8}\left[{\displaystyle \underset{0}{\overset{3}{\int }}{u}_0\left(x\right)}\sin \frac{n\pi x}{8}dx+{\displaystyle \underset{3}{\overset{4}{\int }}{u}_0\left(x\right)}\sin \frac{n\pi x}{8}dx+{\displaystyle \underset{4}{\overset{5}{\int }}{u}_0\left(x\right)}\sin \frac{n\pi x}{8}dx+{\displaystyle \underset{5}{\overset{8}{\int }}{u}_0\left(x\right)}\sin \frac{n\pi x}{8}dx\right]\\ =\frac{1}{4}\left[0+{\displaystyle \underset{3}{\overset{4}{\int }}{u}_0\left(x\right)}\sin \frac{n\pi x}{8}dx+{\displaystyle \underset{4}{\overset{5}{\int }}{u}_0\left(x\right)}\sin \frac{n\pi x}{8}dx+0\right]\\ =\frac{1}{4}\left[{\displaystyle \underset{3}{\overset{4}{\int }}{u}_0\left(x\right)}\sin \frac{n\pi x}{8}dx+{\displaystyle \underset{4}{\overset{5}{\int }}{u}_0\left(x\right)}\sin \frac{n\pi x}{8}dx\right]\end{array}$        (II)

Consider the enlarged view of triangle in the Figure 1 is shown below.

Let D is a point on straight line CA, the coordinate of this point is $\left(x,u_o\left(x\right)\right)$, E is the foot of the perpendicular of D on CB, then the triangle CAH of CDE are similar triangles. So the ratio of corresponding sides of the two triangles are equal.

    $\frac{DE}{AH}=\frac{CE}{CH}$        (III)

From the Figure 2, use $u_0\left(x\right)$ for $DE$, $1$ for $AH$, $x-3$ for $CE$, and $1$ for $CH$.

    $\begin{array}{c}\frac{u_0\left(x\right)}{1}=\frac{x-3}{1}\\ u_0\left(x\right)=x-3\text{ for }3<x<4\end{array}$        (IV)

Let F is a point on straight line AB, the coordinate of this point is $\left(x,u_o\left(x\right)\right)$, G is the foot of the perpendicular of F on CB, then the triangle AHB of FGB are similar triangles. So the ratio of corresponding sides of the two triangles are equal.

    $\frac{FG}{AH}=\frac{GB}{HB}$        (V)

From the Figure 2, use $u_0\left(x\right)$ for $FG$, $1$ for $AH$, $5-x$ for $CE$, and $1$ for $CH$.

    $\begin{array}{c}\frac{u_0\left(x\right)}{1}=\frac{5-x}{1}\\ u_0\left(x\right)=5-x\text{ for 4}<x<5\end{array}$        (VI)

Use equation (IV) and (VI) in the equation (II).

    $B_n=\frac{1}{4}\left[{\displaystyle \underset{3}{\overset{4}{\int }}\left(x-3\right)}\sin \frac{n\pi x}{8}dx+{\displaystyle \underset{4}{\overset{5}{\int }}\left(5-x\right)}\sin \frac{n\pi x}{8}dx\right]$        (VII)

Integrate the first part of the equation (VII).

    $\begin{array}{c}{\displaystyle \underset{3}{\overset{4}{\int }}\left(x-3\right)}\sin \frac{n\pi x}{8}dx={\displaystyle \underset{3}{\overset{4}{\int }}x}\sin \frac{n\pi x}{8}dx-3{\displaystyle \underset{3}{\overset{4}{\int }}\sin \frac{n\pi x}{8}d}x\\ =\left\{{\left[x\frac{\left(-\cos \frac{n\pi x}{8}\right)}{\frac{n\pi }{8}}\right]}_3^4-{\displaystyle \underset{3}{\overset{4}{\int }}1\cdot \frac{\left(-\cos \frac{n\pi x}{8}\right)}{\frac{n\pi }{8}}}\right\}-3{\left[\frac{\left(-\cos \frac{n\pi x}{8}\right)}{\frac{n\pi }{8}}\right]}_3^4\\ =\frac{-8}{n\pi }\left(4\cos \frac{4n\pi }{8}-3\cos \frac{3n\pi }{8}\right)+\frac{8}{n\pi }{\displaystyle \underset{3}{\overset{4}{\int }}\cos \frac{n\pi x}{8}dx}+3\frac{8}{n\pi }\left(\cos \frac{4n\pi }{8}-\cos \frac{3n\pi }{8}\right)\end{array}$        (VIII)

Solve the equation (VIII).

    $\begin{array}{c}{\displaystyle \underset{3}{\overset{4}{\int }}\left(x-3\right)}\sin \frac{n\pi x}{8}dx=\frac{-32}{n\pi }\cos \frac{n\pi }{2}+\frac{24}{n\pi }\cos \frac{3n\pi }{8}+\frac{8}{n\pi }{\left(\frac{\sin \left(\frac{n\pi x}{8}\right)}{\frac{n\pi }{8}}\right)}_3^4+\frac{24}{n\pi }\cos \frac{n\pi }{2}-\frac{24}{n\pi }\cos \frac{3n\pi }{8}\\ =-\frac{8}{n\pi }\cos \frac{n\pi }{2}+{\left(\frac{8}{n\pi }\right)}^2\left(\sin \frac{4n\pi }{8}-\sin \frac{3n\pi }{8}\right)\\ =-\frac{8}{n\pi }\cos \frac{n\pi }{2}+{\left(\frac{8}{n\pi }\right)}^2\left(\sin \frac{n\pi }{2}-\sin \frac{3n\pi }{8}\right)\end{array}$        (IX)

Integrate the second part of the equation (VII).

    $\begin{array}{c}{\displaystyle \underset{3}{\overset{4}{\int }}\left(5-x\right)}\sin \frac{n\pi x}{8}dx={\displaystyle \underset{4}{\overset{5}{\int }}5}\sin \frac{n\pi x}{8}dx-{\displaystyle \underset{4}{\overset{5}{\int }}x\sin \frac{n\pi x}{8}d}x\\ ={\left[\frac{\left(-\cos \frac{n\pi x}{8}\right)}{\frac{n\pi }{8}}\right]}_4^5-\left\{{\left[x\cdot \frac{\left(-\cos \frac{n\pi x}{8}\right)}{\frac{n\pi }{8}}\right]}_4^5-{\displaystyle \underset{4}{\overset{5}{\int }}1\cdot \frac{\left(-\cos \frac{n\pi x}{8}\right)}{\frac{n\pi }{8}}dx}\right\}\\ =\frac{-40}{n\pi }\left(\cos \frac{5n\pi }{8}-\cos \frac{4n\pi }{8}\right)+\frac{8}{n\pi }\left(5\cos \frac{5n\pi }{8}-4\cos \frac{4n\pi }{8}\right)-\frac{8}{n\pi }{\displaystyle \underset{4}{\overset{5}{\int }}\cos \frac{n\pi x}{8}dx}\end{array}$        (X)

Solve the equation (X).

    $\begin{array}{c}{\displaystyle \underset{3}{\overset{4}{\int }}\left(5-x\right)}\sin \frac{n\pi x}{8}dx=\frac{-40}{n\pi }\cos \frac{5n\pi }{2}+\frac{40}{n\pi }\cos \frac{n\pi }{2}+\frac{40}{n\pi }\cos \frac{5n\pi }{8}-\frac{32}{n\pi }\cos \frac{n\pi }{2}-\frac{8}{n\pi }{\left(\frac{\sin \left(\frac{n\pi x}{8}\right)}{\frac{n\pi }{8}}\right)}_4^5\\ =\frac{8}{n\pi }\cos \frac{n\pi }{2}-{\left(\frac{8}{n\pi }\right)}^2\left(\sin \frac{5n\pi }{8}-\sin \frac{4n\pi }{8}\right)\\ =\frac{8}{n\pi }\cos \frac{n\pi }{2}-{\left(\frac{8}{n\pi }\right)}^2\left(\sin \frac{5n\pi }{2}-\sin \frac{n\pi }{2}\right)\end{array}$        (XI)

Use equation (X) and (XI) in (VII).

    $\begin{array}{c}{B}_n=\frac{1}{4}\left[-\frac{8}{n\pi }\cos \frac{n\pi }{2}+{\left(\frac{8}{n\pi }\right)}^2\left(\sin \frac{n\pi }{2}-\sin \frac{3n\pi }{8}\right)+\frac{8}{n\pi }\cos \frac{n\pi }{2}-{\left(\frac{8}{n\pi }\right)}^2\left(\sin \frac{5n\pi }{2}-\sin \frac{n\pi }{2}\right)\right]\\ =\frac{1}{4}\left[{\left(\frac{8}{n\pi }\right)}^2\left(\sin \frac{n\pi }{2}-\sin \frac{3n\pi }{8}-\sin \frac{5n\pi }{2}+\sin \frac{n\pi }{2}\right)\right]\\ =\frac{16}{n^2{\pi }^2}\left(2\sin \frac{n\pi }{2}-\sin \frac{3n\pi }{8}-\sin \frac{5n\pi }{2}\right)\\ =\frac{16}{n^2{\pi }^2}\left(2\sin \frac{n\pi }{2}-\left[\sin \frac{3n\pi }{8}+\sin \frac{5n\pi }{2}\right]\right)\end{array}$        (XII)

Solve equation (XII).

    $\begin{array}{c}{B}_n=\frac{16}{n^2{\pi }^2}\left(2\sin \frac{n\pi }{2}-\left[\sin \frac{3n\pi }{8}+\sin \frac{5n\pi }{2}\right]\right)\\ =\frac{16}{n^2{\pi }^2}\left(2\sin \frac{n\pi }{2}-2\sin \left(\frac{\frac{5n\pi }{2}+\frac{3n\pi }{8}}{2}\right)\cos \left(\frac{\frac{5n\pi }{2}-\frac{3n\pi }{8}}{2}\right)\right)\\ =\frac{16}{n^2{\pi }^2}\left(2\sin \frac{n\pi }{2}-2\sin \frac{n\pi }{2}\cos \frac{n\pi }{8}\right)\\ =\frac{16}{n^2{\pi }^2}2\sin \frac{n\pi }{2}\left(1-\cos \frac{n\pi }{8}\right)\\ =\frac{32}{n^2{\pi }^2}\sin \frac{n\pi }{2}\left(1-\cos \frac{n\pi }{8}\right)\end{array}$        (XIII)

If $n$ is even, $\sin \frac{n\pi x}{2}=0$, and if $n$ is odd, $\sin \frac{n\pi x}{2}={\left(-1\right)}^m\text{ where}m=0,1,\mathrm{2..}$. So equation (XIII) becomes zero for even $n$. Then the equation (XIII) can be written as for odd $n$

    $B_n={\left(-1\right)}^m\frac{32}{{\left(2m+1\right)}^2{\pi }^2}\left[1-\cos \frac{\left(2m+1\right)\pi }{8}\right]$.

Conclusion:

Therefore, the Fourier coefficients of the triangular wave of Figure 16.7 are zero for $n$ even and ${\left(-1\right)}^m\frac{32}{{\left(2m+1\right)}^2{\pi }^2}\left(1-\cos \frac{\left(2m+1\right)\pi }{8}\right)$ for $n$ odd.