6.2 解答

6.2

(a) 1.

$$a_{0} = \frac{1}{\pi}[\int_{-\pi}^{0}1dx + \int_{0}^{\pi}2dx] = \frac{1}{\pi}[\pi + 2\pi] = 3$$

次に， $\int_{0}^{\pi}\cos{nx}dx = 0$より

$$a_{n} = \frac{1}{\pi}[\int_{-\pi}^{0}\cos{nx}dx + \int_{0}^{\pi}2\cos{nx}dx] = 0$$

$$b_{n} = \frac{1}{\pi}[\int_{-\pi}^{0}\sin{nx}dx + \int_{0}^{\pi}2\sin{nx}dx]$$

$$= \frac{1}{\pi}[-\frac{\cos{nx}}{n}\mid_{-\pi}^{0} - \frac{2\cos{nx}}{n}\mid_{0}^{\pi}]$$

$$= \frac{1}{\pi}[\frac{-1 + \cos{n\pi}}{n} - \frac{2(\cos{n\pi} - 1)}{n}]$$

$$= \frac{1}{\pi}[\frac{1 - \cos{n\pi}}{n}] = \frac{1 - (-1)^{n}}{n\pi}$$

したがって，$f(x)$のフーリエ級数は

$$f(x) \sim \frac{3}{2} + \sum_{n=1}^{\infty} \frac{1 - (-1)^{n}}{n\pi}\sin{nx}$$

またフーリエ級数のグラフは

(b) $f(x) = \vert\sin{x}\vert$は偶関数より$b_{n} = 0$．

$$a_{0} = \frac{1}{\pi}\int_{-\pi}^{\pi}\vert\sin{x}\vert dx = \fra... ...int_{0}^{\pi}\sin{x}dx = \frac{2}{\pi}[-\cos{x}\mid_{0}^{\pi}] = \frac{4}{\pi}$$

$$a_{n} = \frac{2}{\pi}\int_{0}^{\pi}\sin{x}\cos{nx}dx$$

$$= \frac{2}{\pi}\int_{0}^{\pi}[\frac{\sin{(n + 1)x} - \sin{(n-1)x}}{2}dx$$

$$= \left\{\begin{array}{ll} \frac{1}{\pi}[-\frac{\cos{(n+1)x}}{n+1} ... ...{\cos{(n-1)x}}{n-1} \mid_{0}^{\pi}] & (n \neq 1)\\ 0 & n =1 \end{array}\right.$$

$$= \frac{1}{\pi}[-\frac{\cos{(n+1)\pi}}{n+1} +\frac{\cos{(n-1)\pi}}{n-1} + \frac{1}{n+1} - \frac{1}{n-1}]$$

$$= \frac{1}{\pi}[-\frac{(-1)^{n+1}}{n+1} +\frac{(-1)^{n-1}}{n-1} + \frac{-2}{n^2 - 1} ]$$

$$= \frac{1}{\pi}[\frac{n((-1)^{n-1} - (-1)^{n+1}) + (-1)^{n+1} + (-1)^{n-1} -2}{n^2 - 1} ]$$

$$= \frac{1}{\pi}[\frac{2(-1)^{n+1} -2}{n^2 - 1} ]$$

$$= \frac{2}{\pi}[\frac{(-1)^{n+1} -1}{n^2 - 1} ]$$

$$= \frac{2}{\pi}[\frac{-2}{4m^2 - 1}] \ \ (n = 2m)$$

よって

$$f(x) \sim \frac{2}{\pi} - \frac{4}{\pi}\sum_{m=1}^{\infty}\frac{1}{4m^2 - 1}\cos{2mx}.$$

またフーリエ級数のグラフは

(c) $f(x) = x$は奇関数より $a_{n} = 0 \ (n \geq 0)$.

$$b_{n} = \frac{1}{\pi}\int_{-\pi}^{\pi}x \sin{nx}dx$$

$$= \frac{2}{\pi}[-\frac{x\cos{nx}}{n}\mid_{0}^{\pi} + \frac{1}{n}\int_{0}^{\pi}\cos{nx}dx$$

$$= \frac{2}{\pi}[\frac{1}{n^2}\sin{nx}\mid_{0}^{\pi}$$

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

よって

$$f(x) \sim 2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin{nx}.$$

またフーリエ級数のグラフは

2. $f(x) = x^2$より$f(x)$は偶関数．よって$b_{n} = 0$．また

$$a_{0} = \frac{2}{\pi}\int_{0}^{\pi}x^2 dx = \frac{2}{\pi}(\frac{x^3}{3}\mid_{0}^{\pi} = \frac{2 \pi^2}{3},$$

$$a_{n} = \frac{2}{\pi}\int_{0}^{\pi}x^2 \cos{nx}dx \ \ \left(\begin{array}... ...x^2 & dv = \cos{nx}dx\\ du = 2xdx & v = \frac{\sin{nx}}{n} \end{array}\right .$$

$$= \frac{2}{\pi}[\frac{x^2 \sin{nx}}{n}\mid_{0}^{\pi} - \frac{2}{n}\... ... = x & dv = \sin{nx}dx\\ du = dx & v = -\frac{\cos{nx}}{n} \end{array}\right .$$

$$= \frac{-4}{n\pi}[\frac{-x \cos{nx}}{n}\mid_{0}^{\pi} + \frac{1}{n}\int_{0}^{\pi}\cos{nx}dx ]$$

$$= \frac{-4}{n\pi}[\frac{-\pi \cos{n\pi}}{n}$$

$$= \frac{4 (-1)^{n}}{n^2}$$

ここでDirichlet条件より

$$\frac{f(x-0) + f(x+0)}{2} = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^2} \cos{nx}$$

となるので$x = \pi$とおくと， $\cos{n\pi} = (-1)^{n}$より

$$\pi^2 = f(\pi) = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{1}{n^2}$$

よって

$$\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6} .$$

また$x = 0$とおくと，

$$0 = f(0) = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^2}$$

よって

$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2} = \frac{\pi^2}{12} \ \framebox{終}.$$

3. $f(x) = x$は奇関数より $a_{n} = 0 \ (n \geq 0)$.

$$b_{n} = \frac{1}{\pi}\int_{-\pi}^{\pi}x \sin{nx}dx$$

$$= \frac{2}{\pi}[-\frac{x\cos{nx}}{n}\mid_{0}^{\pi} + \frac{1}{n}\int_{0}^{\pi}\cos{nx}dx$$

$$= \frac{2}{\pi}[\frac{1}{n^2}\sin{nx}\mid_{0}^{\pi}$$

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

よって

$$f(x) \sim 2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin{nx}$$

また$f(x)$はDirichlet条件を満たすので

$$\frac{\pi}{2} = f(\frac{\pi}{2}) = 2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin{\frac{n \pi}{2}}$$

ここで

$$\sin{\frac{n\pi}{2}} = \left\{\begin{array}{ll} (-1)^{m} & n = 2m+1\\ 0 & n =2m \end{array} \right .$$

より

$$\frac{\pi}{2} = 2\sum_{m=0}^{\infty}\frac{(-1)^{2m+2}}{2m+1}(-1)^{m}$$

$$= 2\sum_{m=0}^{\infty}\frac{(-1)^{m}}{2m+1}$$

よって

$$\sum_{m=0}^{\infty}\frac{(-1)^{m}}{2m+1} = \frac{\pi}{4} \ \framebox{終}.$$