式展開: 11月 2022

準備1． $2 n A_{n} = (2 n - 1) A_{n - 1}$

\begin{array}{rcl} A_{n} & = & \int_{0}^{\frac{π}{2}} \cos^{2 n} (x) d x \\ = & \int_{0}^{\frac{π}{2}} \cos (x) \cos^{2 n - 1} (x) d x \\ = & {[\sin (x) \cdot \cos^{2 n - 1} (x)]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} \sin (x) {- (2 n - 1) \cos^{2 n - 2} (x) \sin (x)} d x \\ \dots (f g)^{'} = f^{'} g + f g^{'}, f^{'} g = (f g)^{'} - f g^{'}, \int f^{'} g = \int (f g)^{'} - \int f g^{'}, \int f^{'} g = [f g] - \int f g^{'} \\ \dots f^{'} (x) = \cos (x) \\ \dots f (x) = \sin (x) \\ \dots g (x) = \cos^{2 n - 1} (x) = u^{2 n - 1} (x) \dots u = \cos (x) \\ \dots g^{'} (x) = \frac{d g}{d x} = \frac{d g}{d u} \frac{d u}{d x} = (2 n - 1) u^{2 n - 2} (x) \cdot (- \sin (x)) = - (2 n - 1) \cos^{2 n - 2} (x) \sin (x) \\ = & [0 - 0] + (2 n - 1) \int_{0}^{\frac{π}{2}} \sin^{2} (x) \cos^{2 n - 2} (x) d x \\ = & (2 n - 1) \int_{0}^{\frac{π}{2}} (1 - \cos^{2} (x)) \cos^{2 n - 2} (x) d x \\ = & (2 n - 1) \int_{0}^{\frac{π}{2}} (\cos^{2 n - 2} (x) - \cos^{2 n} (x)) d x \\ = & (2 n - 1) \int_{0}^{\frac{π}{2}} \cos^{2 (n - 1)} (x) d x - (2 n - 1) \int_{0}^{\frac{π}{2}} \cos^{2 n} (x) d x \\ = & (2 n - 1) A_{n - 1} - (2 n - 1) A_{n} \\ A_{n} + (2 n - 1) A_{n} & = & (2 n - 1) A_{n - 1} \\ 2 n A_{n} & = & (2 n - 1) A_{n - 1} \end{array}

準備2． $A_{n} = (2 n - 1) n B_{n - 1} - 2 n^{2} B_{n}$

\begin{array}{rcl} A_{n} & = & \int_{0}^{\frac{π}{2}} \cos^{2 n} (x) d x \\ = & {[x \cdot \cos^{2 n} (x)]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} x {2 n \cos^{2 n - 1} (x) (- \sin (x))} d x \\ \dots (f g)^{'} = f^{'} g + f g^{'}, f^{'} g = (f g)^{'} - f g^{'}, \int f^{'} g = \int (f g)^{'} - \int f g^{'}, \int f^{'} g = [f g] - \int f g^{'} \\ \dots f^{'} (x) = 1 \\ \dots f (x) = x \\ \dots g (x) = \cos^{2 n} (x) = u^{2 n} \dots u = \cos (x) \\ \dots g^{'} (x) = \frac{d g}{d x} = \frac{d g}{d u} \frac{d u}{d x} = 2 n u^{2 n - 1} \cdot (- \sin (x)) = 2 n \cos^{2 n - 1} (x) (- \sin (x)) \\ = & [0 - 0] + 2 n \int_{0}^{\frac{π}{2}} x \sin (x) \cos^{2 n - 1} (x) d x \\ = & 2 n \int_{0}^{\frac{π}{2}} x \sin (x) \cos^{2 n - 1} (x) d x \\ = & 2 n [{[\frac{1}{2} x^{2} \cdot \sin (x) \cos^{2 n - 1} (x)]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} \frac{1}{2} x^{2} {\cos (x) \cdot \cos^{2 n - 1} (x) + \sin (x) \cdot (2 n - 1) \cos^{2 n - 2} (x) (- \sin (x))} d x] \\ = & - 2 n \frac{1}{2} \int_{0}^{\frac{π}{2}} x^{2} {\cos^{2 n} (x) - (2 n - 1) \sin^{2} (x) \cos^{2 n - 2} (x)} d x \\ = & - n \int_{0}^{\frac{π}{2}} [x^{2} \cos^{2 n} (x) - (2 n - 1) x^{2} {1 - \cos^{2} (x)} \cos^{2 n - 2} (x)] d x \\ = & - n \int_{0}^{\frac{π}{2}} [x^{2} \cos^{2 n} (x) - (2 n - 1) x^{2} {\cos^{2 n - 2} (x) - \cos^{2} (x) \cos^{2 n - 2} (x)}] d x \\ = & - n \int_{0}^{\frac{π}{2}} [x^{2} \cos^{2 n} (x) - (2 n - 1) x^{2} {\cos^{2 n - 2} (x) - \cos^{2 n} (x)}] d x \\ = & - n \int_{0}^{\frac{π}{2}} {x^{2} \cos^{2 n} (x) - (2 n - 1) x^{2} \cos^{2 n - 2} (x) + (2 n - 1) x^{2} \cos^{2 n} (x)} d x \\ = & - n \int_{0}^{\frac{π}{2}} {- (2 n - 1) x^{2} \cos^{2 n - 2} (x) + 2 n x^{2} \cos^{2 n} (x)} d x \\ = & - n \int_{0}^{\frac{π}{2}} - (2 n - 1) x^{2} \cos^{2 (n - 1)} (x) d x - n \int_{0}^{\frac{π}{2}} 2 n x^{2} \cos^{2 n} (x) d x \\ = & (2 n - 1) n \int_{0}^{\frac{π}{2}} x^{2} \cos^{2 (n - 1)} (x) d x - 2 n^{2} \int_{0}^{\frac{π}{2}} x^{2} \cos^{2 n} (x) d x \\ = & (2 n - 1) n B_{n - 1} - 2 n^{2} B_{n} \dots B_{n} = \int_{0}^{\frac{π}{2}} x^{2} \cos^{2 n} (x) d x \end{array}

準備3． $lim_{n \to \infty} \frac{B_{n}}{A_{n}} = 0$

\begin{array}{rcl} B_{n} & = & \int_{0}^{\frac{π}{2}} x^{2} \cos^{2 n} (x) d x \\ \leq & \int_{0}^{\frac{π}{2}} {\frac{π}{2} \sin (x)}^{2} \cos^{2 n} (x) d x \\ \dots \frac{2}{π} x \leq \sin (x) \dots x \in [0, \frac{π}{2}] \\ \dots x \leq \frac{π}{2} \sin (x) \\ \dots (0, 0) と (\frac{π}{2}, 1) を 通 る 直 線 \frac{2}{π} x は ， 区 間 [0, \frac{π}{2}] で は 常 に \sin (x) 以 下 で あ る (図 を 参 照) ． \\ = & \frac{π^{2}}{4} \int_{0}^{\frac{π}{2}} \sin^{2} (x) \cos^{2 n} (x) d x \\ = & \frac{π^{2}}{4} \int_{0}^{\frac{π}{2}} \sin^{2} (x) \cos^{2 n} (x) d x \\ = & \frac{π^{2}}{4} \int_{0}^{\frac{π}{2}} {1 - \cos^{2} (x)} \cos^{2 n} (x) d x \\ = & \frac{π^{2}}{4} \int_{0}^{\frac{π}{2}} {1 - \cos^{2} (x)} \cos^{2 n} (x) d x \\ = & \frac{π^{2}}{4} \int_{0}^{\frac{π}{2}} {\cos^{2 n} (x) - \cos^{2 n + 2} (x)} d x \\ = & \frac{π^{2}}{4} [\int_{0}^{\frac{π}{2}} \cos^{2 n} (x) d x - \int_{0}^{\frac{π}{2}} \cos^{2 (n + 1)} (x) d x] \\ = & \frac{π^{2}}{4} (A_{n} - A_{n + 1}) \\ = & \frac{π^{2}}{4} (A_{n} - \frac{2 (n + 1) - 1}{2 (n + 1)} A_{n}) \\ = & \frac{π^{2}}{4} (\frac{2 (n + 1) - {2 (n + 1) - 1}}{2 (n + 1)} A_{n}) \\ = & \frac{π^{2}}{4} (\frac{2 (n + 1) - 2 (n + 1) + 1}{2 (n + 1)} A_{n}) \\ = & \frac{π^{2}}{4} (\frac{1}{2 (n + 1)} A_{n}) \\ = & \frac{π^{2}}{4} \frac{A_{n}}{2 (n + 1)} \\ B_{n} & \leq & \frac{π^{2}}{4} \frac{A_{n}}{2 (n + 1)} \\ B_{n} \frac{1}{A_{n}} & \leq & \frac{π^{2}}{4} \frac{A_{n}}{2 (n + 1)} \frac{1}{A_{n}} \dots A_{n} \geq 0 \\ \frac{B_{n}}{A_{n}} & \leq & \frac{π^{2}}{4} \frac{1}{2 (n + 1)} \\ lim_{n \to \infty} \frac{B_{n}}{A_{n}} & \leq & lim_{n \to \infty} \frac{π^{2}}{4} \frac{1}{2 (n + 1)} = 0 \\ lim_{n \to \infty} \frac{B_{n}}{A_{n}} & \geq & 0 \dots A_{n}, B_{n} は 共 に 被 積 分 凾 数 (\cos^{2 n} (x), x^{2} \cos^{2 n} (x)) が 偶 数 乗 の た め ， 常 に 値 は 0 以 上 ． \\ \dots よ っ て ， そ の 積 分 結 果 も 常 に 0 以 上 と な る ． \\ lim_{n \to \infty} \frac{B_{n}}{A_{n}} & = & 0 \dots 前 述 の 2 つ か ら ， は さ み う ち の 原 理 よ り 極 限 は 0 ． \end{array}

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

\begin{array}{rcl} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} & = & lim_{N \to \infty} \sum_{n = 1}^{N} \frac{1}{n^{2}} \\ = & lim_{N \to \infty} \sum_{n = 1}^{N} {\frac{(2 n - 1) B_{n - 1}}{n A_{n}} - \frac{2 B_{n}}{A_{n}}} \\ \dots A_{n} = (2 n - 1) n B_{n - 1} - 2 n^{2} B_{n} (準 備 2 ．) \\ \dots \frac{A_{n}}{n^{2}} = \frac{(2 n - 1) n B_{n - 1}}{n^{2}} - \frac{2 n^{2} B_{n}}{n^{2}} \\ \dots \frac{A_{n}}{n^{2}} = \frac{(2 n - 1) B_{n - 1}}{n} - 2 B_{n} \\ \dots \frac{1}{n^{2}} = \frac{(2 n - 1) B_{n - 1}}{n A_{n}} - \frac{2 B_{n}}{A_{n}} \\ = & lim_{N \to \infty} \sum_{n = 1}^{N} \frac{2 B_{n - 1}}{A_{n - 1}} - \frac{2 B_{n}}{A_{n}} \\ \dots 2 n A_{n} = (2 n - 1) A_{n - 1} (準 備 1 ．) \\ \dots \frac{2}{A_{n - 1}} n A_{n} = (2 n - 1) \\ \dots \frac{2}{A_{n - 1}} = \frac{(2 n - 1)}{n A_{n}} \\ = & 2 lim_{N \to \infty} \sum_{n = 1}^{N} \frac{B_{n - 1}}{A_{n - 1}} - \frac{B_{n}}{A_{n}} \\ = & 2 lim_{N \to \infty} {(\frac{B_{1 - 1}}{A_{1 - 1}} - \frac{B_{1}}{A_{1}}) + (\frac{B_{2 - 1}}{A_{2 - 1}} - \frac{B_{2}}{A_{2}}) + (\frac{B_{3 - 1}}{A_{3 - 1}} - \frac{B_{3}}{A_{3}}) + \dots + (\frac{B_{N - 1}}{A_{N - 1}} - \frac{B_{N}}{A_{N}})} \\ = & 2 lim_{N \to \infty} {\frac{B_{0}}{A_{0}} - \frac{B_{1}}{A_{1}} + \frac{B_{1}}{A_{1}} - \frac{B_{2}}{A_{2}} + \frac{B_{2}}{A_{2}} - \frac{B_{3}}{A_{3}} + \dots + \dots + \frac{B_{N - 1}}{A_{N - 1}} - \frac{B_{N}}{A_{N}}} \dots T e l e s c o p i n g S e r i e s \\ = & 2 (\frac{B_{0}}{A_{0}} - lim_{N \to \infty} \frac{B_{N}}{A_{N}}) \\ = & 2 (\frac{B_{0}}{A_{0}} - 0) \dots lim_{n \to \infty} \frac{B_{n}}{A_{n}} = 0 (準 備 3 ．) \\ = & 2 \frac{B_{0}}{A_{0}} \\ = & {2 \frac{\int_{0}^{\frac{π}{2}} x^{2} \cos^{2 \cdot n} (x) d x}{\int_{0}^{\frac{π}{2}} \cos^{2 \cdot n} (x) d x} |}_{n = 0} \\ = & 2 \frac{\int_{0}^{\frac{π}{2}} x^{2} \cos^{2 \cdot 0} (x) d x}{\int_{0}^{\frac{π}{2}} \cos^{2 \cdot 0} (x) d x} \\ = & 2 \frac{\int_{0}^{\frac{π}{2}} x^{2} d x}{\int_{0}^{\frac{π}{2}} d x} \\ = & 2 \frac{{[\frac{1}{3} x^{3}]}_{0}^{\frac{π}{2}}}{{[x]}_{0}^{\frac{π}{2}}} \\ = & 2 \frac{[\frac{1}{3} {(\frac{π}{2})}^{3} - \frac{1}{3} 0^{3}]}{[\frac{π}{2} - 0]} \\ = & 2 \frac{\frac{π^{3}}{24}}{\frac{π}{2}} \\ = & 2 \frac{π^{2}}{12} \\ = & \frac{π^{2}}{6} \end{array}

式展開

バーゼル問題

準備1．2nAn=(2n−1)An−1

準備2．An=(2n−1)nBn−1−2n2Bn

準備3．limn→∞BnAn=0

∑n=1∞1n2

準備1． $2 n A_{n} = (2 n - 1) A_{n - 1}$

準備2． $A_{n} = (2 n - 1) n B_{n - 1} - 2 n^{2} B_{n}$

準備3． $lim_{n \to \infty} \frac{B_{n}}{A_{n}} = 0$

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