Chapter 12

Identifying \(p=2\) we have $$\begin{aligned} c_{n} &=\frac{1}{4} \int_{-2}^{2} f(x) e^{-i n \pi x / 2} d x=\frac{1}{4}\left[\int_{-2}^{0}(-1) e^{-i n \pi x / 2} d x+\int_{0}^{2} e^{-i n \pi x / 2} d x\right] \\ &=\frac{i}{2 n \pi}\left[-1+e^{i n \pi}+e^{-i n \pi}-1\right]=\frac{i}{2 n \pi}\left[-1+(-1)^{n}+(-1)^{n}-1\right]=\frac{1-(-1)^{n}}{n \pi i} \end{aligned}$$ and $$c_{0}=\frac{1}{4} \int_{-2}^{2} f(x) d x=0$$ Thus $$f(x)=\sum_{n=-\infty \atop n \neq 0}^{\infty} \frac{1-(-1)^{n}}{i n \pi} e^{i n \pi x / 2}$$

For \(\lambda \leq 0\) the only solution of the boundary-value problem is \(y=0 .\) For \(\lambda=\alpha^{2}>0\) we have $$y=c_{1} \cos \alpha x+c_{2} \sin \alpha x.$$ Now $$y^{\prime}(x)=-c_{1} \alpha \sin \alpha x+c_{2} \alpha \cos \alpha x$$ and \(y^{\prime}(0)=0\) implies \(c_{2}=0,\) so $$y(1)+y^{\prime}(1)=c_{1}(\cos \alpha-\alpha \sin \alpha)=0 \quad \text { or } \quad \cot \alpha=\alpha.$$ The eigenvalues are \(\lambda_{n}=\alpha_{n}^{2}\) where \(\alpha_{1}, \alpha_{2}, \alpha_{3}, \ldots\) are the consecutive positive solutions of \(\cot \alpha=\alpha .\) The corresponding eigenfunctions are \(\cos \alpha_{n} x\) for \(n=1,2,3, \ldots .\) Using a CAS we find that the first four eigenvalues are approximately \(0.7402,11.7349,41.4388,\) and 90.8082 with corresponding approximate eigenfunctions \(\cos 0.8603 x, \cos 3.4256 x, \cos 6.4373 x,\) and \(\cos 9.5293 x\).

Since \(f(-x)=\sin (-3 x)=-\sin 3 x=-f(x), f(x)\) is an odd function.

$$\int_{-2}^{2} x x^{2} d x=\left.\frac{1}{4} x^{4}\right|_{-2} ^{2}=0$$

Since \(f(-x)=-x \cos (-x)=-x \cos x=-f(x), f(x)\) is an odd function.

$$\begin{array}{l} a_{0}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) d x=\frac{1}{\pi} \int_{-\pi}^{0}(-1) d x+\frac{1}{\pi} \int_{0}^{\pi} 2 d x=1 \\ a_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos n x d x=\frac{1}{\pi} \int_{-\pi}^{0}-\cos n x d x+\frac{1}{\pi} \int_{0}^{\pi} 2 \cos n x d x=0 \\ b_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin n x d x=\frac{1}{\pi} \int_{-\pi}^{0}-\sin n x d x+\frac{1}{\pi} \int_{0}^{\pi} 2 \sin n x d x=\frac{3}{n \pi}\left[1-(-1)^{n}\right] \\ f(x)=\frac{1}{2}+\frac{3}{\pi} \sum_{n=1}^{\infty} \frac{1-(-1)^{n}}{n} \sin n x \end{array}$$

$$\int_{-1}^{1} x^{3}\left(x^{2}+1\right) d x=\left.\frac{1}{6} x^{6}\right|_{-1} ^{1}+\left.\frac{1}{4} x^{4}\right|_{-1} ^{1}=0$$

For \(\lambda=0\) the solution of \(y^{\prime \prime}=0\) is \(y=c_{1} x+c_{2} .\) The condition \(y^{\prime}(0)=0\) implies \(c_{1}=0,\) so \(\lambda=0\) is an eigenvalue with corresponding eigenfunction 1. For \(\lambda=\alpha^{2} > 0\) we have \(y=c_{1} \cos \alpha x+c_{2} \sin \alpha x\) and \(y^{\prime}=-c_{1} \alpha \sin \alpha x+c_{2} \alpha \cos \alpha x .\) The condition \(y^{\prime}(0)=0\) implies \(c_{2}=0\) and so \(y=c_{1} \cos \alpha x .\) Now the condition \(y^{\prime}(L)=0\) implies \(-c_{1} \alpha \sin \alpha L=0 .\) For \(c_{1} \neq 0\) this condition will hold when \(\alpha L=n \pi\) or \(\lambda=\alpha^{2}=n^{2} \pi^{2} / L^{2},\) where \(n=1,2,3, \ldots .\) These are the positive eigenvalues with corresponding eigenfunctions \(\cos (n \pi x / L), n=1,2,3, \ldots\)

$$\begin{array}{l} a_{0}=\int_{-1}^{1} f(x) d x=\int_{-1}^{0} 1 d x+\int_{0}^{1} x d x=\frac{3}{2} \\ a_{n}=\int_{-1}^{1} f(x) \cos n \pi x d x=\int_{-1}^{0} \cos n \pi x d x+\int_{0}^{1} x \cos n \pi x d x=\frac{1}{n^{2} \pi^{2}}\left[(-1)^{n}-1\right] \\ b_{n}=\int_{-1}^{1} f(x) \sin n \pi x d x=\int_{-1}^{0} \sin n \pi x d x+\int_{0}^{1} x \sin n \pi x d x=-\frac{1}{n \pi} \\ f(x)=\frac{3}{4}+\sum_{n=1}^{\infty}\left[\frac{(-1)^{n}-1}{n^{2} \pi^{2}} \cos n \pi x-\frac{1}{n \pi} \sin n \pi x\right] \end{array}$$

$$\int_{0}^{2} e^{x}\left(x e^{-x}-e^{-x}\right) d x=\int_{0}^{2}(x-1) d x=\left.\left(\frac{1}{2} x^{2}-x\right)\right|_{0} ^{2}=0$$

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

$$\int_{0}^{\pi} \cos x \sin ^{2} x d x=\left.\frac{1}{3} \sin ^{3} x\right|_{0} ^{\pi}=0$$

$$\begin{aligned} c_{i} &=\frac{2 \alpha_{i}^{2}}{\left(4 \alpha_{i}^{2}+1\right) J_{0}^{2}\left(2 \alpha_{i}\right)} \int_{0}^{2} x J_{0}\left(\alpha_{i} x\right) d x \\ &=\frac{2 \alpha_{i}^{2}}{\left(4 \alpha_{i}^{2}+1\right) J_{0}^{2}\left(2 \alpha_{i}\right)} \cdot \frac{1}{\alpha_{i}^{2}} \int_{0}^{2 \alpha_{i}} t J_{0}(t) d t \\ &=\frac{2}{\left(4 \alpha_{i}^{2}+1\right) J_{0}^{2}\left(2 \alpha_{i}\right)} \int_{0}^{2 \alpha_{i}} \frac{d}{d t}\left[t J_{1}(t)\right] d t \\ &=\left.\frac{2}{\left(4 \alpha_{i}^{2}+1\right) J_{0}^{2}\left(2 \alpha_{i}\right)} t J_{1}(t)\right|_{0} ^{2 \alpha_{i}} \\ &=\frac{4 \alpha_{i} J_{1}\left(2 \alpha_{i}\right)}{\left(4 \alpha_{i}^{2}+1\right) J_{0}^{2}\left(2 \alpha_{i}\right)} \end{aligned}.$$ Thus $$f(x)=4 \sum_{i=1}^{\infty} \frac{\alpha_{i} J_{1}\left(2 \alpha_{i}\right)}{\left(4 \alpha_{i}^{2}+1\right) J_{0}^{2}\left(2 \alpha_{i}\right)} J_{0}\left(\alpha_{i} x\right).$$

Since \(f(-x)=e^{|-x|}=e^{|x|}=f(x), f(x)\) is an even function.

$$\int_{-\pi / 2}^{\pi / 2} x \cos 2 x d x=\left.\frac{1}{2}\left(\frac{1}{2} \cos 2 x+x \sin 2 x\right)\right|_{-\pi / 2} ^{\pi / 2}=0$$

The eigenfunctions are \(\sin \alpha_{n} x\) where \(\tan \alpha_{n}=-\alpha_{n} .\) Thus $$\begin{aligned} \left\|\sin \alpha_{n} x\right\|^{2} &=\int_{0}^{1} \sin ^{2} \alpha_{n} x d x=\frac{1}{2} \int_{0}^{1}\left(1-\cos 2 \alpha_{n} x\right) d x \\ &=\left.\frac{1}{2}\left(x-\frac{1}{2 \alpha_{n}} \sin 2 \alpha_{n} x\right)\right|_{0} ^{1}=\frac{1}{2}\left(1-\frac{1}{2 \alpha_{n}} \sin 2 \alpha_{n}\right) \\ &=\frac{1}{2}\left[1-\frac{1}{2 \alpha_{n}}\left(2 \sin \alpha_{n} \cos \alpha_{n}\right)\right] \\ &=\frac{1}{2}\left[1-\frac{1}{\alpha_{n}} \tan \alpha_{n} \cos \alpha_{n} \cos \alpha_{n}\right] \\ &=\frac{1}{2}\left[1-\frac{1}{\alpha_{n}}\left(-\alpha_{n} \cos ^{2} \alpha_{n}\right)\right]=\frac{1}{2}\left(1+\cos ^{2} \alpha_{n}\right). \end{aligned}$$

Since \(f(-x)=e^{-x}-e^{x}=-f(x), f(x)\) is an odd function.

$$\int_{\pi / 4}^{5 \pi / 4} e^{x} \sin x d x=\left.\left(\frac{1}{2} e^{x} \sin x-\frac{1}{2} e^{x} \cos x\right)\right|_{\pi / 4} ^{5 \pi / 4}=0$$

(a) If \(\lambda \leq 0\) the initial conditions imply \(y=0 .\) For \(\lambda=\alpha^{2}>0\) the general solution of the Cauchy-Euler differential equation is \(y=c_{1} \cos (\alpha \ln x)+c_{2} \sin (\alpha \ln x) .\) The condition \(y(1)=0\) implies \(c_{1}=0,\) so that \(y=c_{2} \sin (\alpha \ln x) .\) The condition \(y(5)=0\) implies \(\alpha \ln 5=n \pi, n=1,2,3, \ldots .\) Thus, the eigenvalues are \(n^{2} \pi^{2} /(\ln 5)^{2}\) for \(n=1,2,3, \ldots,\) with corresponding eigenfunctions \(\sin [(n \pi / \ln 5) \ln x]\). (b) The self-adjoint form is $$\frac{d}{d x}\left[x y^{\prime}\right]+\frac{\lambda}{x} y=0.$$ (c) An orthogonality relation is $$\int_{1}^{5} \frac{1}{x} \sin \left(\frac{m \pi}{\ln 5} \ln x\right) \sin \left(\frac{n \pi}{\ln 5} \ln x\right) d x=0, \quad m \neq n.$$

The fundamental period is \(T=4,\) so \(\omega=2 \pi / 4=\pi / 2\) and the values of \(n \omega\) are \(0, \pm \pi / 2, \pm \pi, \pm 3 \pi / 2, \ldots .\) From Problem \(1, c_{0}=0\) and \(\left|c_{n}\right|=\left(1-(-1)^{n}\right) / n \pi .\) The table shows some values of \(n\) with corresponding values of \(\left|c_{n}\right|\) The graph is a portion of the frequency spectrum. $$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \mathrm{n} & -5 & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \mathrm{c}_{\mathrm{n}} & 0.1273 & 0.0000 & 0.2122 & 0.0000 & 0.6366 & 0.0000 & 0.6366 & 0.0000 & 0.2122 & 0.0000 & 0.1273 \\ \hline \end{array}$$

$$\begin{array}{l} a_{0}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) d x=\frac{1}{\pi} \int_{-\pi}^{\pi}(x+\pi) d x=2 \pi \\ a_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos n x d x=\frac{1}{\pi} \int_{-\pi}^{\pi}(x+\pi) \cos n x d x=0 \\ b_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin n x d x=\frac{2}{n}(-1)^{n+1} \\\ f(x)=\pi+\sum_{n=1}^{\infty} \frac{2}{n}(-1)^{n+1} \sin n x \end{array}$$

For \(m \neq n\) $$\begin{array}{rl} \int_{0}^{\pi / 2} \sin (2 n+1) x & \sin (2 m+1) x d x \\ & =\frac{1}{2} \int_{0}^{\pi / 2}(\cos 2(n-m) x-\cos 2(n+m+1) x) d x \\ & =\left.\frac{1}{4(n-m)} \sin 2(n-m) x\right|_{0} ^{\pi / 2}-\left.\frac{1}{4(n+m+1)} \sin 2(n+m+1) x\right|_{0} ^{\pi / 2}=0. \end{array}$$ For \(m=n\) $$\begin{aligned} \int_{0}^{\pi / 2} \sin ^{2}(2 n+1) x d x &=\int_{0}^{\pi / 2}\left(\frac{1}{2}-\frac{1}{2} \cos 2(2 n+1) x\right) d x \\ &=\left.\frac{1}{2} x\right|_{0} ^{\pi / 2}-\left.\frac{1}{4(2 n+1)} \sin 2(2 n+1) x\right|_{0} ^{\pi / 2}=\frac{\pi}{4} \end{aligned}$$ so that $$\|\sin (2 n+1) x\|=\frac{1}{2} \sqrt{\pi}.$$

(a) The roots of the auxiliary equation \(m^{2}+m+\lambda=0\) are \(\frac{1}{2}(-1 \pm \sqrt{1-4 \lambda}) .\) When \(\lambda=0\) the general solution of the differential equation is \(c_{1}+c_{2} e^{-x} .\) The boundary conditions imply \(c_{1}+c_{2}=0\) and \(c_{1}+c_{2} c^{-2}=0 .\) since the determinant of the coefficients is not \(0,\) the only solution of this homogeneous system is \(c_{1}=c_{2}=0\) in which case \(y=0 .\) When \(\lambda=\frac{1}{4},\) the general solution of the differential equation is \(c_{1} e^{-x / 2}+c_{2} x e^{-x / 2}\) The boundary conditions imply \(c_{1}=0\) and \(c_{1}+2 c_{2}=0,\) so \(c_{1}=c_{2}=0\) and \(y=0 .\) Similarly, if \(0 < \lambda < \frac{1}{4},\) the general solution is $$y=c_{1} e^{\frac{1}{2}(-1+\sqrt{1-4 \lambda}) x}+c_{2} c^{\frac{1}{2}(-1-\sqrt{1-4 \lambda}) x}.$$ In this case the boundary conditions again imply \(c_{1}=c_{2}=0,\) and so \(y=0 .\) Now, for \(\lambda>\frac{1}{4},\) the general solution of the differential equation is $$y=c_{1} e^{-x / 2} \cos \sqrt{4 \lambda-1} x+c_{2} e^{-x / 2} \sin \sqrt{4 \lambda-1} x.$$ The condition \(y(0)=0\) implies \(c_{1}=0\) so \(y=c_{2} e^{-x / 2} \sin \sqrt{4 \lambda-1} x .\) From $$y(2)=c_{2} e^{-1} \sin 2 \sqrt{4 \lambda-1}=0$$ we see that the eigenvalues are determined by \(2 \sqrt{4 \lambda-1}=n \pi\) for \(n=1,2,3, \ldots .\) Thus, the eigenvalues are \(n^{2} \pi^{2} / 4^{2}+1 / 4\) for \(n=1,2,3, \ldots,\) with corresponding eigenfunctions \(e^{-x / 2} \sin (n \pi x / 2)\). (b) The self-adjoint form is $$\frac{d}{d x}\left[e^{x} y^{\prime}\right]+\lambda e^{x} y=0.$$ (c) An orthogonality relation is $$\int_{0}^{2} e^{x}\left(e^{-x / 2} \sin \frac{m \pi}{2} x\right)\left(e^{-x / 2} \cos \frac{n \pi}{2} x\right) d x=\int_{0}^{2} \sin \frac{m \pi}{2} x \cos \frac{n \pi}{2} x d x=0.$$

$$\begin{array}{l} a_{0}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) d x=\frac{1}{\pi} \int_{-\pi}^{\pi}(3-2 x) d x=6 \\ a_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos n x d x=\frac{1}{\pi} \int_{-\pi}^{\pi}(3-2 x) \cos n x d x=0 \\ b_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi}(3-2 x) \sin n x d x=\frac{4}{n}(-1)^{n} \\\ f(x)=3+4 \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \sin n x \end{array}$$

For \(m \neq n\) $$\begin{array}{rl} \int_{0}^{\pi / 2} \cos (2 n+1) x & \cos (2 m+1) x d x \\ & =\frac{1}{2} \int_{0}^{\pi / 2}(\cos 2(n-m) x+\cos 2(n+m+1) x) d x \\ & =\left.\frac{1}{4(n-m)} \sin 2(n-m) x\right|_{0} ^{\pi / 2}+\left.\frac{1}{4(n+m+1)} \sin 2(n+m+1) x\right|_{0} ^{\pi / 2}=0. \end{array}$$ For \(m=n\) $$\begin{aligned} \int_{0}^{\pi / 2} \cos ^{2}(2 n+1) x d x &=\int_{0}^{\pi / 2}\left(\frac{1}{2}+\frac{1}{2} \cos 2(2 n+1) x\right) d x \\ &=\left.\frac{1}{2} x\right|_{0} ^{\pi / 2}+\left.\frac{1}{4(2 n+1)} \sin 2(2 n+1) x\right|_{0} ^{\pi / 2}=\frac{\pi}{4} \end{aligned}$$ so that $$\|\cos (2 n+1) x\|=\frac{1}{2} \sqrt{\pi}.$$

Since \(f(x)\) is not defined for \(x<0\), it is neither even nor odd.

$$\begin{aligned} &a_{0}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) d x=\frac{1}{\pi} \int_{0}^{\pi} \sin x d x=\frac{2}{\pi}\\\ &a_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos n x d x=\frac{1}{\pi} \int_{0}^{\pi} \sin x \cos n x d x=\frac{1}{2 \pi} \int_{0}^{\pi}(\sin (1+n) x+\sin (1-n) x) d x \end{aligned}$$ $$\begin{aligned} &=\frac{1+(-1)^{n}}{\pi\left(1-n^{2}\right)} \quad \text { for } n=2,3,4, \ldots \\ a_{1} &=\frac{1}{2 \pi} \int_{0}^{\pi} \sin 2 x d x=0 \\ b_{n} &=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin n x d x=\frac{1}{\pi} \int_{0}^{\pi} \sin x \sin n x d x \\ &=\frac{1}{2 \pi} \int_{0}^{\pi}(\cos (1-n) x-\cos (1+n) x) d x=0 \quad \text { for } n=2,3,4, \ldots \\ b_{1} &=\frac{1}{2 \pi} \int_{0}^{\pi}(1-\cos 2 x) d x=\frac{1}{2} \\ f(x) &=\frac{1}{\pi}+\frac{1}{2} \sin x+\sum_{n=2}^{\infty} \frac{1+(-1)^{n}}{\pi\left(1-n^{2}\right)} \cos n x \end{aligned}$$

For \(m \neq n\) $$\begin{aligned} \int_{0}^{\pi} \sin n x \sin m x d x &=\frac{1}{2} \int_{0}^{\pi}(\cos (n-m) x-\cos (n+m) x) d x \\ &=\left.\frac{1}{2(n-m)} \sin (n-m) x\right|_{0} ^{\pi}-\left.\frac{1}{2(n+m)} \sin (n+m) x\right|_{0} ^{\pi}=0. \end{aligned}$$ For \(m=n\) $$\int_{0}^{\pi} \sin ^{2} n x d x=\int_{0}^{\pi}\left(\frac{1}{2}-\frac{1}{2} \cos 2 n x\right) d x=\left.\frac{1}{2} x\right|_{0} ^{\pi}-\left.\frac{1}{4 n} \sin 2 n x\right|_{0} ^{\pi}=\frac{\pi}{2}$$ so that $$\|\sin n x\|=\sqrt{\frac{\pi}{2}}.$$

Identifying \(2 p=\pi\) or \(p=\pi / 2,\) and using \(\cos x=\) \(\left(e^{i x}-e^{-i x}\right) / 2,\) we have $$\begin{aligned} c_{n}=& \frac{1}{\pi} \int_{0}^{\pi} f(x) e^{-2 i n x / \pi} d x \\ =& \frac{1}{\pi} \int_{0}^{\pi / 2}(\cos x) e^{-2 i n x / \pi} d x \\ =& \frac{1}{\pi} \int_{0}^{\pi / 2} \frac{1}{2}\left(e^{i x}-e^{-i x}\right) e^{-2 i n x / \pi} d x \\ =& \frac{1}{2 \pi} \int_{0}^{\pi / 2}\left(e^{(1-2 n / \pi) i x}-e^{-(1+2 n / \pi) i x}\right) d x \\ =& \frac{1}{2 \pi}\left[\frac{1}{i(1-2 n / \pi)} e^{(1-2 n / \pi) i x}\right.\\\ &\left.+\frac{1}{i(1+2 n / \pi)} e^{-(1+2 n / \pi) i x}\right]_{0}^{\pi / 2} \\\ =& \frac{2 n e^{-i n}+i \pi}{\pi^{2}-4 n^{2}} \end{aligned}$$ The fundamental period is \(T=\pi,\) so \(\omega=2 \pi / \pi=2\) and the values of \(n \omega\) are \(0,\pm 2,\pm 4,\pm 6, \ldots .\) Values of \(\left|c_{n}\right|\) for \(n=0,\pm 1,\pm 2,\pm 3,\pm 4,\) and ±5 are shown in the table. The bottom graph is a portion of the frequency spectrum.

Since \(f(-x)=\left|(-x)^{5}\right|=\left|x^{5}\right|=f(x), f(x)\) is an even function.

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

For \(m \neq n\) $$\begin{aligned} \int_{0}^{p} \sin \frac{n \pi}{p} x \sin \frac{m \pi}{p} x d x &=\frac{1}{2} \int_{0}^{p}\left(\cos \frac{(n-m) \pi}{p} x-\cos \frac{(n+m) \pi}{p} x\right) d x \\ &=\left.\frac{p}{2(n-m) \pi} \sin \frac{(n-m) \pi}{p} x\right|_{0} ^{p}-\left.\frac{p}{2(n+m) \pi} \sin \frac{(n+m) \pi}{p} x\right|_{0} ^{p}=0. \end{aligned}$$ For \(m=n\) $$\int_{0}^{p} \sin ^{2} \frac{n \pi}{p} x d x=\int_{0}^{p}\left(\frac{1}{2}-\frac{1}{2} \cos \frac{2 n \pi}{p} x\right) d x=\left.\frac{1}{2} x\right|_{0} ^{p}-\left.\frac{p}{4 n \pi} \sin \frac{2 n \pi}{p} x\right|_{0} ^{p}=\frac{p}{2}$$ so that $$\left\|\sin \frac{n \pi}{p} x\right\|=\sqrt{\frac{p}{2}}.$$

(a) The differential equation is $$\left(1+x^{2}\right) y^{\prime \prime}+2 x y^{\prime}+\frac{\lambda}{1+x^{2}} y=0.$$ Letting \(x=\tan \theta\) we have \(\theta=\tan ^{-1} x\) and $$\begin{aligned} \frac{d y}{d x} &=\frac{d y}{d \theta} \frac{d \theta}{d x}=\frac{1}{1+x^{2}} \frac{d y}{d \theta} \\ \frac{d^{2} y}{d x^{2}} &=\frac{d}{d x}\left[\frac{1}{1+x^{2}} \frac{d y}{d \theta}\right]=\frac{1}{1+x^{2}}\left(\frac{d^{2} y}{d \theta^{2}} \frac{d \theta}{d x}\right)-\frac{2 x}{\left(1+x^{2}\right)^{2}} \frac{d y}{d \theta} \\\ &=\frac{1}{\left(1+x^{2}\right)^{2}} \frac{d^{2} y}{d \theta^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}} \frac{d y}{d \theta}. \end{aligned}$$ The differential equation can then be written in terms of \(y(\theta)\) as $$\begin{array}{c} \left(1+x^{2}\right)\left[\frac{1}{\left(1+x^{2}\right)^{2}} \frac{d^{2} y}{d \theta^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}} \frac{d y}{d \theta}\right]+2 x\left[\frac{1}{1+x^{2}} \frac{d y}{d \theta}\right]+\frac{\lambda}{1+x^{2}} y \\\ =\frac{1}{1+x^{2}} \frac{d^{2} y}{d \theta^{2}}+\frac{\lambda}{1+x^{2}} y=0 \end{array}$$ or $$\frac{d^{2} y}{d \theta^{2}}+\lambda y=0.$$ The boundary conditions become \(y(0)=y(\pi / 4)=0 .\) For \(\lambda \leq 0\) the only solution of the boundary-value problem is \(y=0 .\) For \(\lambda=\alpha^{2}>0\) the general solution of the differential equation is \(y=c_{1} \cos \alpha \theta+c_{2} \sin \alpha \theta\) The condition \(y(0)=0\) implies \(c_{1}=0\) so \(y=c_{2} \sin \alpha \theta .\) Now the condition \(y(\pi / 4)=0\) implies \(c_{2} \sin \alpha \pi / 4=\) 0\. For \(c_{2} \neq 0\) this condition will hold when \(\alpha \pi / 4=n \pi\) or \(\lambda=\alpha^{2}=16 n^{2},\) where \(n=1,2,3, \ldots .\) These are the eigenvalues with corresponding eigenfunctions \(\sin 4 n \theta=\sin \left(4 n \tan ^{-1} x\right),\) for \(n=1,2,3, \dots\) (b) An orthogonality relation is $$\int_{0}^{1} \frac{1}{x^{2}+1} \sin \left(4 m \tan ^{-1} x\right) \sin \left(4 n \tan ^{-1} x\right) d x=0, \quad m \neq n.$$

(a) Adding \(c_{n}=\frac{1}{2}\left(a_{n}-i b_{n}\right)\) and \(c_{-n}=\frac{1}{2}\left(a_{n}+i b_{n}\right)\) we get \(c_{n}+c_{-n}=a_{n} .\) Subtracting, we get \(c_{n}-c_{-n}=-i b_{n}\) Multiplying both sides by \(i\) we obtain \(i\left(c_{n}-c_{-n}\right)=b_{n}\) (b) From $$a_{n}=c_{n}+c_{-n}=(-1)^{n} \frac{\sinh \pi}{\pi}\left[\frac{1-i n}{n^{2}+1}+\frac{1+i n}{n^{2}+1}\right]=\frac{2(-1)^{n} \sinh \pi}{\pi\left(n^{2}+1\right)}, \quad n=0,1,2, \ldots$$ and $$b_{n}=i\left(c_{n}-c_{-n}\right)=i(-1)^{n} \frac{\sinh \pi}{\pi}\left[\frac{1-i n}{n^{2}+1}-\frac{1+i n}{n^{2}+1}\right]=i(-1)^{n} \frac{\sinh \pi}{\pi}\left[-\frac{2 i n}{n^{2}+1}\right]=\frac{2(-1)^{n} n \sinh \pi}{\pi\left(n^{2}+1\right)}$$ the Fourier series of \(f\) is $$f(x)=\frac{\sinh \pi}{\pi}+\frac{2 \sinh \pi}{\pi} \sum_{n=1}^{\infty}\left[\frac{(-1)^{n}}{n^{2}+1} \cos n x+\frac{n(-1)^{n}}{n^{2}+1} \sin n x\right]$$

$$\begin{array}{l} a_{0}=\frac{1}{2} \int_{-2}^{2} f(x) d x=\frac{1}{2}\left(\int_{-1}^{0}-2 d x+\int_{0}^{1} 1 d x\right)=-\frac{1}{2} \\ a_{n}=\frac{1}{2} \int_{-2}^{2} f(x) \cos \frac{n \pi}{2} x d x=\frac{1}{2}\left(\int_{-1}^{0}(-2) \cos \frac{n \pi}{2} x d x+\int_{0}^{1} \cos \frac{n \pi}{2} x d x\right)=-\frac{1}{n \pi} \sin \frac{n \pi}{2} \\ b_{n}=\frac{1}{2} \int_{-2}^{2} f(x) \sin \frac{n \pi}{2} x d x=\frac{1}{2}\left(\int_{-1}^{0}(-2) \sin \frac{n \pi}{2} x d x+\int_{0}^{1} \sin \frac{n \pi}{2} x d x\right)=\frac{3}{n \pi}\left(1-\cos \frac{n \pi}{2}\right) \\ f(x)=-\frac{1}{4}+\sum_{n=1}^{\infty}\left[-\frac{1}{n \pi} \sin \frac{n \pi}{2} \cos \frac{n \pi}{2} x+\frac{3}{n \pi}\left(1-\cos \frac{n \pi}{2}\right) \sin \frac{n \pi}{2} x\right] \end{array}$$

For \(m \neq n\) $$\begin{aligned} \int_{0}^{p} \cos \frac{n \pi}{p} x \cos \frac{m \pi}{p} x d x &=\frac{1}{2} \int_{0}^{p}\left(\cos \frac{(n-m) \pi}{p} x+\cos \frac{(n+m) \pi}{p} x\right) d x \\ &=\left.\frac{p}{2(n-m) \pi} \sin \frac{(n-m) \pi}{p} x\right|_{0} ^{p}+\left.\frac{p}{2(n+m) \pi} \sin \frac{(n+m) \pi}{p} x\right|_{0} ^{p}=0. \end{aligned}$$ For \(m=n\) $$\int_{0}^{p} \cos ^{2} \frac{n \pi}{p} x d x=\int_{0}^{p}\left(\frac{1}{2}+\frac{1}{2} \cos \frac{2 n \pi}{p} x\right) d x=\left.\frac{1}{2} x\right|_{0} ^{p}+\left.\frac{p}{4 n \pi} \sin \frac{2 n \pi}{p} x\right|_{0} ^{p}=\frac{p}{2}.$$ Also $$\int_{0}^{p} 1 \cdot \cos \frac{n \pi}{p} x d x=\left.\frac{p}{n \pi} \sin \frac{n \pi}{p} x\right|_{0} ^{p}=0 \quad \text { and } \quad \int_{0}^{p} 1^{2} d x=p$$ so that $$\|1\|=\sqrt{p} \text { and }\left\|\cos \frac{n \pi}{p} x\right\|=\sqrt{\frac{p}{2}}.$$

(a) Letting \(\lambda=\alpha^{2}\) the differential equation becomes \(x^{2} y^{\prime \prime}+x y^{\prime}+\left(\alpha^{2} x^{2}-1\right) y=0 .\) This is the parametric Bessel equation with \(\nu=1 .\) The general solution is $$y=c_{1} J_{1}(\alpha x)+c_{2} Y_{1}(\alpha x).$$ since \(Y\) is unbounded at 0 we must have \(c_{2}=0,\) so that \(y=c_{1} J_{1}(\alpha x) .\) The condition \(J_{1}(3 \alpha)=0\) defines the eigenvalues \(\lambda_{n}=\alpha_{n}^{2}\) for \(n=1,2,3, \ldots .\) The corresponding eigenfunctions are \(J_{1}\left(\alpha_{n} x\right).\) (b) Using a CAS or Table 5.1 in the text to solve \(J_{1}(3 \alpha)=0\) we find \(3 \alpha_{1}=3.8317,3 \alpha_{2}=7.0156,3 \alpha_{3}=\) \(10.1735,\) and \(3 \alpha_{4}=13.3237 .\) The corresponding eigenvalues are \(\lambda_{1}=\alpha_{1}^{2}=1.6313, \lambda_{2}=\alpha_{2}^{2}=5.4687\), \(\lambda_{3}=\alpha_{3}^{2}=11.4999,\) and \(\lambda_{4}=\alpha_{4}^{2}=19.7245\).

Since \(f(x)\) is an even function, we expand in a cosine series: $$\begin{aligned} &a_{0}=\int_{1}^{2} 1 d x=1\\\&a_{n}=\int_{1}^{2} \cos \frac{n \pi}{2} x d x=-\frac{2}{n \pi} \sin \frac{n \pi}{2}.\end{aligned}$$ Thus $$f(x)=\frac{1}{2}+\sum_{n=1}^{\infty} \frac{-2}{n \pi} \sin \frac{n \pi}{2} \cos \frac{n \pi}{2} x.$$

When \(\lambda=0\) the differential equation is \(r(x) y^{\prime \prime}+r^{\prime}(x) y^{\prime}=0 .\) By inspection we see that \(y=1\) is a solution of the boundary-value problem. Thus, \(\lambda=0\) is an eigenvalue.

Since \(f(x)\) is an even function, we expand in a cosine series: $$\begin{aligned} &a_{0}=\frac{2}{\pi} \int_{0}^{\pi} x d x=\pi\\\ &a_{n}=\frac{2}{\pi} \int_{0}^{\pi} x \cos n x d x=\frac{2}{n^{2} \pi}\left[(-1)^{n}-1\right].\end{aligned}$$ Thus $$f(x)=\frac{\pi}{2}+\sum_{n=1}^{\infty} \frac{2}{n^{2} \pi}\left[(-1)^{n}-1\right] \cos n x.$$

$$\begin{array}{l} a_{0}=\frac{1}{5} \int_{-5}^{5} f(x) d x=\frac{1}{5}\left(\int_{-5}^{0} 1 d x+\int_{0}^{5}(1+x) d x\right)=\frac{9}{2} \\ a_{n}=\frac{1}{5} \int_{-5}^{5} f(x) \cos \frac{n \pi}{5} x d x=\frac{1}{5}\left(\int_{-5}^{0} \cos \frac{n \pi}{5} x d x+\int_{0}^{5}(1+x) \cos \frac{n \pi}{5} x d x\right)=\frac{5}{n^{2} \pi^{2}}\left[(-1)^{n}-1\right] \\ b_{n}=\frac{1}{5} \int_{-5}^{5} f(x) \sin \frac{n \pi}{5} x d x=\frac{1}{5}\left(\int_{-5}^{0} \sin \frac{n \pi}{5} x d x+\int_{0}^{5}(1+x) \cos \frac{n \pi}{5} x d x\right)=\frac{5}{n \pi}(-1)^{n+1} \\ f(x)=\frac{9}{4}+\sum_{n=1}^{\infty}\left[\frac{5}{n^{2} \pi^{2}}\left[(-1)^{n}-1\right] \cos \frac{n \pi}{5} x+\frac{5}{n \pi}(-1)^{n+1} \sin \frac{n \pi}{5} x\right] \end{array}$$

Since \(f(x)\) is an odd function, we expand in a sine series: $$b_{n}=\frac{2}{\pi} \int_{0}^{\pi} x \sin n x d x=\frac{2}{n}(-1)^{n+1}.$$ Thus $$f(x)=\sum_{n=1}^{\infty} \frac{2}{n}(-1)^{n+1} \sin n x.$$

$$\begin{array}{l} a_{0}=\frac{1}{2} \int_{-2}^{2} f(x) d x=\frac{1}{2}\left(\int_{-2}^{0}(2+x) d x+\int_{0}^{2} 2 d x\right)=3 \\ a_{n}=\frac{1}{2} \int_{-2}^{2} f(x) \cos \frac{n \pi}{2} x d x=\frac{1}{2}\left(\int_{-2}^{0}(2+x) \cos \frac{n \pi}{2} x d x+\int_{0}^{2} 2 \cos \frac{n \pi}{2} x d x\right)=\frac{2}{n^{2} \pi^{2}}\left[1-(-1)^{n}\right] \\ b_{n}=\frac{1}{2} \int_{-2}^{2} f(x) \sin \frac{n \pi}{2} x d x=\frac{1}{2}\left(\int_{-2}^{0}(2+x) \sin \frac{n \pi}{2} x d x+\int_{0}^{2} 2 \sin \frac{n \pi}{2} x d x\right)=\frac{2}{n \pi}(-1)^{n+1} \\ f(x)=\frac{3}{2}+\sum_{n=1}^{\infty}\left[\frac{2}{n^{2} \pi^{2}}\left[1-(-1)^{n}\right] \cos \frac{n \pi}{2} x+\frac{2}{n \pi}(-1)^{n+1} \sin \frac{n \pi}{2} x\right] \end{array}$$

We compute $$\begin{array}{l} c_{0}=\frac{1}{2} \int_{0}^{1} x P_{0}(x) d x=\frac{1}{2} \int_{0}^{1} x d x=\frac{1}{4} \\ c_{1}=\frac{3}{2} \int_{0}^{1} x P_{1}(x) d x=\frac{3}{2} \int_{0}^{1} x^{2} d x=\frac{1}{2} \\ c_{2}=\frac{5}{2} \int_{0}^{1} x P_{2}(x) d x=\frac{5}{2} \int_{0}^{1} \frac{1}{2}\left(3 x^{3}-x\right) d x=\frac{5}{16} \\ c_{3}=\frac{7}{2} \int_{0}^{1} x P_{3}(x) d x=\frac{7}{2} \int_{0}^{1} \frac{1}{2}\left(5 x^{4}-3 x^{2}\right) d x=0 \end{array}$$ $$\begin{array}{l} c_{4}=\frac{9}{2} \int_{0}^{1} x P_{4}(x) d x=\frac{9}{2} \int_{0}^{1} \frac{1}{8}\left(35 x^{5}-30 x^{3}+3 x\right) d x=-\frac{3}{32} \\ c_{5}=\frac{11}{2} \int_{0}^{1} x P_{5}(x) d x=\frac{11}{2} \int_{0}^{1} \frac{1}{8}\left(63 x^{6}-70 x^{4}+15 x^{2}\right) d x=0 \\ c_{6}=\frac{13}{2} \int_{0}^{1} x P_{6}(x) d x=\frac{13}{2} \int_{0}^{1} \frac{1}{16}\left(231 x^{7}-315 x^{5}+105 x^{3}-5 x\right) d x=\frac{13}{256}. \end{array}$$ Thus $$f(x)=\frac{1}{4} P_{0}(x)+\frac{1}{2} P_{1}(x)+\frac{5}{16} P_{2}(x)-\frac{3}{32} P_{4}(x)+\frac{13}{256} P_{6}(x)+\cdots.$$ The figure above is the graph of \(S_{5}(x)=\frac{1}{4} P_{0}(x)+\frac{1}{2} P_{1}(x)+\frac{5}{16} P_{2}(x)-\frac{3}{32} P_{4}(x)+\frac{13}{256} P_{6}(x)\).

By orthogonality \(\int_{a}^{b} \phi_{0}(x) \phi_{n}(x) d x=0\) for \(n=1,2,3, \ldots ;\) that is, \(\int_{a}^{b} \phi_{n}(x) d x=0\) for \(n=1,2,3, \ldots\)

$$\begin{array}{l} a_{0}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) d x=\frac{1}{\pi} \int_{0}^{\pi}\left(e^{x}-1\right) d x=\frac{1}{\pi}\left(e^{\pi}-\pi-1\right) \\\ a_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos n x d x=\frac{1}{\pi} \int_{0}^{\pi}\left(e^{x}-1\right) \cos n x d x=\frac{\left[e^{\pi}(-1)^{n}-1\right]}{\pi\left(1+n^{2}\right)} \\ b_{n}=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin n x d x=\frac{1}{\pi} \int_{0}^{\pi}\left(e^{x}-1\right) \sin n x d x=\frac{1}{\pi}\left(\frac{n e^{\pi}(-1)^{n+1}}{1+n^{2}}+\frac{n}{1+n^{2}}+\frac{(-1)^{n}}{n}-\frac{1}{n}\right) \\\ f(x)=\frac{e^{\pi}-\pi-1}{2 \pi}+\sum_{n=1}^{\infty}\left[\frac{e^{\pi}(-1)^{n}-1}{\pi\left(1+n^{2}\right)} \cos n x+\left(\frac{n}{1+n^{2}}\left[e^{\pi}(-1)^{n+1}+1\right]+\frac{(-1)^{n}-1}{n}\right) \sin n x\right] \end{array}$$

Using the facts that \(\phi_{0}\) and \(\phi_{1}\) are orthogonal to \(\phi_{n}\) for \(n>1\), we have $$\begin{aligned} \int_{a}^{b}(\alpha x+\beta) \phi_{n}(x) d x &=\alpha \int_{a}^{b} x \phi_{n}(x) d x+\beta \int_{a}^{b} 1 \cdot \phi_{n}(x) d x \\ &=\alpha \int_{a}^{b} \phi_{1}(x) \phi_{n}(x) d x+\beta \int_{a}^{b} \phi_{0}(x) \phi_{n}(x) d x \\ &=\alpha \cdot 0+\beta \cdot 0=0 \end{aligned}$$ for \(n=2,3,4, \dots\)

\(\mathrm{U} \operatorname{sing} \cos ^{2} \theta=\frac{1}{2}(\cos 2 \theta+1)\) we have $$\begin{aligned} P_{2}(\cos \theta) &=\frac{1}{2}\left(3 \cos ^{2} \theta-1\right)=\frac{3}{2} \cos ^{2} \theta-\frac{1}{2} \\ &=\frac{3}{4}(\cos 2 \theta+1)-\frac{1}{2}=\frac{3}{4} \cos 2 \theta+\frac{1}{4}=\frac{1}{4}(3 \cos 2 \theta+1), \end{aligned}$$

Using the fact that \(\phi_{n}\) and \(\phi_{m}\) are orthogonal for \(n \neq m\) we have $$\begin{aligned} \left\|\phi_{m}(x)+\phi_{n}(x)\right\|^{2} &=\int_{a}^{b}\left[\phi_{m}(x)+\phi_{n}(x)\right]^{2} d x=\int_{a}^{b}\left[\phi_{m}^{2}(x)+2 \phi_{m}(x) \phi_{n}(x)+\phi_{n}^{2}(x)\right] d x \\ &=\int_{a}^{b} \phi_{m}^{2}(x) d x+2 \int_{a}^{b} \phi_{m}(x) \phi_{n}(x) d x+\int_{a}^{b} \phi_{n}^{2}(x) d x \\ &=\left\|\phi_{m}(x)\right\|^{2}+\left\|\phi_{n}(x)\right\|^{2}. \end{aligned}$$

Since \(f(x)\) is an odd function, we expand in a sine series: $$b_{n}=\frac{2}{\pi} \int_{0}^{\pi} x^{3} \sin n x d x=\frac{2}{\pi}\left(-\left.\frac{x^{3}}{n} \cos n x\right|_{0} ^{\pi}+\frac{3}{n} \int_{0}^{\pi} x^{2} \cos n x d x\right)=\frac{2 \pi^{2}}{n}(-1)^{n+1}-\frac{12}{n^{2} \pi} \int_{0}^{\pi} x \sin n x d x$$ $$=\frac{2 \pi^{2}}{n}(-1)^{n+1}-\frac{12}{n^{2} \pi}\left(-\left.\frac{x}{n} \cos n x\right|_{0} ^{\pi}+\frac{1}{n} \int_{0}^{\pi} \cos n x d x\right)=\frac{2 \pi^{2}}{n}(-1)^{n+1}+\frac{12}{n^{3}}(-1)^{n}.$$ Thus $$f(x)=\sum_{n=1}^{\infty}\left(\frac{2 \pi^{2}}{n}(-1)^{n+1}+\frac{12}{n^{3}}(-1)^{n}\right) \sin n x.$$