6.3 解答

6.3

1.

(a)

$$f_{e}(x) = \left\{\begin{array}{ll} x+1, & -2 < x < -1\\ -x-1,... ...-1 < x < 0\\ x - 1, & 0 < x < 1\\ 1 - x, & 1 < x < 2 \end{array} \right .$$

(b)

$$f_{e}(x) = \left\{\begin{array}{ll} e^{-x}, & -\pi < x <0\\ e^... ... & -\pi < x <0\\ e^{x}, & 0 < x < \pi \end{array} \right . \ \ \framebox{終}$$

2.

(a) まず$f(x)$のフーリエ余弦級数を求める．

$$f_{e}(x) \sim \frac{a_{0}}{2} + \sum_{n=1}^{\infty}a_{n}\cos{n\pi x}$$

$$a_{0} = \frac{2}{1}\int_{0}^{1}f(x)dx = 2 \int_{0}^{1}x^2 dx = \frac{2}{3}$$

$$a_{n} = \frac{2}{1}\int_{0}^{1}f(x)\cos{n\pi x}dx$$

$$= 2\int_{0}^{1}x^2 \cos{n\pi x} dx \ \ \ \left(\begin{array}{ll} u ... ...= \cos{n\pi x}\\ du = 2x dx & v = \frac{\sin{n\pi x}}{n\pi} \end{array}\right)$$

$$= 2[\frac{x^2 \sin{n\pi x}}{n\pi} \mid_{0}^{1} - \frac{2}{n\pi}\int... ...v = \sin{n\pi x}\\ du = dx & v = -\frac{\cos{n\pi x}}{n\pi} \end{array}\right)$$

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

$$= -\frac{4}{n\pi}(\frac{-\cos{n\pi}}{n\pi} + \frac{\sin{n\pi x}}{(n\pi)^{2}}\mid_{0}^{1})$$

$$= -\frac{4}{n\pi}(\frac{-\cos{n\pi}}{n\pi} + \frac{\sin{n\pi}}{(n\pi)^{2}})$$

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

より

$$f_{e}(x) \sim \frac{1}{3} + \frac{4}{\pi^2}\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos{n\pi x}.$$

次に$f(x)$のフーリエ正弦級数を求める．

$$f_{o}(x) \sim \sum_{n=1}^{\infty}b_{n}\sin{n\pi x}$$

$$b_{n} = \frac{2}{1}\int_{0}^{1}f(x)\sin{n\pi x}dx$$

$$= 2\int_{0}^{1}x^2 \sin{n\pi x} dx \ \ \ \left(\begin{array}{ll} u ... ...in{n\pi x} dx\\ du = 2x dx & v = -\frac{\cos{n\pi x}}{n\pi} \end{array}\right.$$

$$= 2[\frac{-x^2 \cos{n\pi x}}{n\pi} \mid_{0}^{1} + \frac{2}{n\pi}\in... ... = \cos{n\pi x}dx\\ du = dx & v = \frac{\sin{n\pi x}}{n\pi} \end{array}\right.$$

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

$$= 2[\frac{-\cos{n\pi}}{n\pi} + \frac{2}{n\pi}(\frac{\sin{n\pi }}{n\pi}+ \frac{\cos{n\pi x}}{(n\pi)^2}\mid_{0}^{1})]$$

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

$$= 2[\frac{-(-1)^n}{n\pi} + \frac{2((-1)^n - 1)}{(n\pi)^3}]$$

より

$$f_{o}(x) \sim 2\sum_{n=1}^{\infty}[\frac{(-1)^{n+1}}{n\pi} + \frac{2((-1)^{n} - 1)}{(n\pi)^3}]\sin{n\pi x}$$

(b) まず$f(x)$のフーリエ余弦級数を求める．

$$f_{e}(x) \sim \frac{a_{0}}{2} + \sum_{n=1}^{\infty}a_{n}\cos{n x}$$

$$a_{0} = \frac{2}{\pi}\int_{0}^{\pi}\sin{\frac{x}{2}} dx = \frac{2}{\pi}(-2\cos{\frac{x}{2}}\mid_{0}^{\pi})$$

$$= \frac{2}{\pi}(-2\cos{\frac{\pi}{2}} + 2) = \frac{4}{\pi}.$$

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

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

$$= \frac{1}{\pi}[\frac{-\cos{(n+\frac{1}{2})x}}{n + \frac{1}{2}} + \frac{\cos{(n-\frac{1}{2})x}}{n - \frac{1}{2}}\mid_{0}^{\pi}]$$

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

$$= \frac{1}{\pi}[\frac{\sin{n\pi} + 1}{n + \frac{1}{2}} + \frac{-\sin{n\pi } - 1}{n - \frac{1}{2}}]$$

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

より

$$f_{e}(x) \sim \frac{2}{\pi} - \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{1}{4n^2 - 1}\cos{nx}$$

次に$f(x)$のフーリエ正弦級数を求める．

$$f_{o}(x) \sim \sum_{n=1}^{\infty}b_{n}\sin{nx}$$

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

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

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

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

$$= \frac{1}{\pi}[\frac{-2n\cos{n\pi}}{n^2 - \frac{1}{4} } ]$$

$$= \frac{8}{\pi}[\frac{n(-1)^{n+1}}{4n^2 - 1}]$$

より

$$f_{o}(x) \sim \frac{8}{\pi}\sum_{n=1}^{\infty}\frac{n(-1)^{n+1}}{4n^2 - 1}\sin{nx} \$$

(c) まず$f(x)$のフーリエ余弦級数を求める．

$$f_{e}(x) \sim \frac{a_{0}}{2} + \sum_{n=1}^{\infty}a_{n}\cos{n x}$$

$$a_{0} = \frac{2}{\pi}\int_{0}^{\pi}\cos{x} dx = \frac{2}{\pi}(-\sin{x}\mid_{0}^{\pi}) = 0.$$

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

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

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

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

より

$$f_{e}(x) \sim \cos{x}$$

次に$f(x)$のフーリエ正弦級数を求める．

$$f_{o}(x) \sim \sum_{n=1}^{\infty}b_{n}\sin{nx}$$

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

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

$$= \left\{\begin{array}{ll} \frac{1}{\pi}[-\frac{\cos{(n+1)\pi + 1}}... ...rac{1}{\pi}[-\frac{\cos{2\pi}}{2} + \frac{1}{2}] & \ \ n = 1 \end{array}\right.$$

$$= \left\{\begin{array}{cl} \frac{1}{\pi}[-\frac{2n\cos{(n+1)\pi + 2n}}{n^2 - 1}] & \ \ n \neq 1\\ 0& \ \ n = 1 \end{array}\right.$$

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

より

$$f_{0}(x) \sim \frac{2}{\pi}\sum_{n=2}^{\infty}\frac{n(1-n(-1)^{n+1})}{n^2 - 1}\sin{nx} \$$

3.

(a) $f(x)$の複素系フーリエ級数を求める．

$$f(x) \sim \sum_{n = -\infty}^{\infty}c_{n}e^{inx}$$

$$c_{n} = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}dx$$

より

$$c_{0} = \frac{1}{2\pi}\int_{-\pi}^{\pi}\vert x\vert dx = \frac{1}{\pi}\int_{0}^{\pi}x dx = \frac{\pi}{2}$$

$$c_{n} = \frac{1}{2\pi}\int_{-\pi}^{\pi}\vert x\vert e^{-inx}dx$$

$$= \frac{1}{2\pi}[\int_{-\pi}^{\pi}\vert x\vert(\cos{nx} + i\sin{nx})dx ]$$

$$= \frac{1}{2\pi}[\int_{-\pi}^{\pi}\vert x\vert\cos{nx}dx] = \frac{1}{\pi}\int_{0}^{\pi}x\cos{nx} dx ]$$

$$= \frac{1}{\pi}[\frac{x\sin{nx}}{n}\mid_{0}^{\pi} - \frac{1}{n} \int_{0}^{\infty}\sin{nx}dx]$$

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

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

より

$$f(x) \sim \frac{\pi}{2} + \frac{1}{\pi}\sum_{n = -\infty}^{\infty}\frac{(-1)^{n} - 1}{n^2}e^{inx} \ \framebox{終}$$

(b) $f(x)$の複素系フーリエ級数を求める．

$$c_{0} = \frac{1}{2\pi}\int_{-\pi}^{\pi}x^{2}dx = \frac{1}{\pi}\int_{0}^{\pi}x^{2} dx = \frac{\pi^{2}}{3}$$

$$c_{n} = \frac{1}{2\pi}\int_{-\pi}^{\pi}x^{2}e^{-inx}dx \ \left(\begin{arr... ... & dv = e^{-inx}dx\\ du = 2x dx & v = - \frac{e^{-inx}}{in} \end{array}\right.$$

$$= \frac{1}{2\pi}[\frac{-x^{2}e^{-inx}}{in}\mid_{-\pi}^{\pi} + \frac{2}{in}\int_{-\pi}^{\pi}xe^{-inx}dx$$

$$= \frac{1}{2\pi}[\frac{-\pi^{2}e^{-in\pi} + (-\pi)^{2}e^{in\pi}}{in... ...ac{-xe^{-inx}}{in}\mid_{-\pi}^{\pi} + \frac{1}{in}\int_{-\pi}^{\pi}e^{-inx}dx)]$$

$$= \frac{1}{2\pi}[\frac{2(-\pi e^{-in\pi} -\pi e^{in\pi})}{(in)^{2}}... ...ac{1}{(in)^{3}}e^{-inx}\mid_{-\pi}^{\pi}] \ (e^{in\pi} = e^{-in\pi} = (-1)^{n})$$

$$= \frac{1}{2\pi}[\frac{-4\pi (-1)^{n}}{(in)^{2}} - \frac{1}{(in)^{3}}(e^{-in\pi} - e^{in\pi})]$$

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

より

$$f(x) \sim \frac{\pi^2}{3} + \sum_{n=-\infty}^{\infty}\frac{(-1)^{n}}{n^2}e^{-in x} \ \framebox{終}$$