与えられた3つの関数をフーリエ級数展開する問題です。それぞれの関数は周期関数であると仮定します。関数は以下の通りです。 (1) $f(x) = 2x - 1$ ($-\pi \leq x \leq \pi$) (2) $f(x) = x + 1$ ($-1 \leq x \leq 1$) (3) $f(x) = \begin{cases} -2 & (-2 \leq x \leq 0) \\ 2 & (0 \leq x \leq 2) \end{cases}$

1. 問題の内容

与えられた3つの関数をフーリエ級数展開する問題です。それぞれの関数は周期関数であると仮定します。関数は以下の通りです。

(1) $f(x) = 2x - 1$ ($-\pi \leq x \leq \pi$)

(2) $f(x) = x + 1$ ($-1 \leq x \leq 1$)

(3) $f(x) = \begin{cases} -2 & (-2 \leq x \leq 0) \\ 2 & (0 \leq x \leq 2) \end{cases}$

2. 解き方の手順

(1) $f(x) = 2x - 1$ (周期 $2\pi$) の場合

フーリエ級数は次のように表されます。

$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx))$

ここで、$a_0$, $a_n$, $b_n$ はフーリエ係数であり、以下のように計算されます。

$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) dx$

$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) dx$

$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) dx$

まず、$a_0$ を計算します。

$a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} (2x - 1) dx = \frac{1}{\pi} [x^2 - x]_{-\pi}^{\pi} = \frac{1}{\pi} [(\pi^2 - \pi) - (\pi^2 + \pi)] = \frac{-2\pi}{\pi} = -2$

次に、$a_n$ を計算します。

$a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} (2x - 1) \cos(nx) dx = \frac{1}{\pi} \int_{-\pi}^{\pi} 2x \cos(nx) dx - \frac{1}{\pi} \int_{-\pi}^{\pi} \cos(nx) dx$

$2x \cos(nx)$ は奇関数なので、$\int_{-\pi}^{\pi} 2x \cos(nx) dx = 0$

$\int_{-\pi}^{\pi} \cos(nx) dx = [\frac{1}{n} \sin(nx)]_{-\pi}^{\pi} = \frac{1}{n}(\sin(n\pi) - \sin(-n\pi)) = 0$

よって、$a_n = 0$

最後に、$b_n$ を計算します。

$b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} (2x - 1) \sin(nx) dx = \frac{1}{\pi} \int_{-\pi}^{\pi} 2x \sin(nx) dx - \frac{1}{\pi} \int_{-\pi}^{\pi} \sin(nx) dx$

$\int_{-\pi}^{\pi} \sin(nx) dx = [-\frac{1}{n} \cos(nx)]_{-\pi}^{\pi} = -\frac{1}{n}(\cos(n\pi) - \cos(-n\pi)) = 0$

$\int_{-\pi}^{\pi} 2x \sin(nx) dx = 2 \int_{-\pi}^{\pi} x \sin(nx) dx = 2[-\frac{x}{n} \cos(nx) + \frac{1}{n^2} \sin(nx)]_{-\pi}^{\pi} = 2[-\frac{\pi}{n} \cos(n\pi) - \frac{\pi}{n} \cos(-n\pi)] = -\frac{4\pi}{n} \cos(n\pi) = -\frac{4\pi}{n} (-1)^n = \frac{4\pi}{n} (-1)^{n+1}$

$b_n = \frac{1}{\pi} (\frac{4\pi}{n} (-1)^{n+1}) = \frac{4}{n} (-1)^{n+1}$

したがって、$f(x) = -1 + \sum_{n=1}^{\infty} \frac{4}{n} (-1)^{n+1} \sin(nx)$

(2) $f(x) = x + 1$ (周期 $2$) の場合

フーリエ級数は次のように表されます。

$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos(\frac{n\pi x}{L}) + b_n \sin(\frac{n\pi x}{L}))$

ここで、$L = 1$ なので、

$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos(n\pi x) + b_n \sin(n\pi x))$

$a_0 = \frac{1}{1} \int_{-1}^{1} (x + 1) dx = [\frac{x^2}{2} + x]_{-1}^{1} = (\frac{1}{2} + 1) - (\frac{1}{2} - 1) = 2$

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

$\int_{-1}^{1} x \cos(n\pi x) dx = 0$ (奇関数)

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

よって、$a_n = 0$

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

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

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

よって、$b_n = \frac{2}{n\pi} (-1)^{n+1}$

したがって、$f(x) = 1 + \sum_{n=1}^{\infty} \frac{2}{n\pi} (-1)^{n+1} \sin(n\pi x)$

(3) $f(x) = \begin{cases} -2 & (-2 \leq x \leq 0) \\ 2 & (0 \leq x \leq 2) \end{cases}$ (周期 $4$)

$L = 2$ なので、

$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos(\frac{n\pi x}{2}) + b_n \sin(\frac{n\pi x}{2}))$

$a_0 = \frac{1}{2} \int_{-2}^{2} f(x) dx = \frac{1}{2} (\int_{-2}^{0} -2 dx + \int_{0}^{2} 2 dx) = \frac{1}{2} ([-2x]_{-2}^{0} + [2x]_{0}^{2}) = \frac{1}{2} ((0 - 4) + (4 - 0)) = 0$

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

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

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

よって、$a_n = 0$

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

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

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

よって、$b_n = \frac{2}{n\pi} (1 - (-1)^n) + \frac{2}{n\pi} (1 - (-1)^n) = \frac{4}{n\pi} (1 - (-1)^n)$

$b_n = \begin{cases} 0 & (n:even) \\ \frac{8}{n\pi} & (n:odd) \end{cases}$

したがって、$f(x) = \sum_{n=1,3,5,...}^{\infty} \frac{8}{n\pi} \sin(\frac{n\pi x}{2}) = \sum_{k=0}^{\infty} \frac{8}{(2k+1)\pi} \sin(\frac{(2k+1)\pi x}{2})$

3. 最終的な答え

(1) $f(x) = -1 + \sum_{n=1}^{\infty} \frac{4}{n} (-1)^{n+1} \sin(nx)$

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

(3) $f(x) = \sum_{k=0}^{\infty} \frac{8}{(2k+1)\pi} \sin(\frac{(2k+1)\pi x}{2})$