式展開: 11月 2022

準備1．$2nA_n=\left(2n-1\right)A_{n-1}$

$$\begin{eqnarray} A_n&=&\int_0^{\frac{\pi}{2}}\cos^{2n}{\left(x\right)}\mathrm{d}x \\&=&\int_0^{\frac{\pi}{2}}\cos{\left(x\right)}\cos^{2n-1}{\left(x\right)}\mathrm{d}x \\&=&\left[\sin{\left(x\right)\cdot\cos^{2n-1}{\left(x\right)}}\right]_0^{\frac{\pi}{2}} -\int_0^{\frac{\pi}{2}}\sin{\left(x\right)}\left\{-\left(2n-1\right)\cos^{2n-2}{\left(x\right)}\sin{\left(x\right)}\right\}\mathrm{d}x \\&&\;\dots\;(fg)^\prime=f^\prime g+fg^\prime,f^\prime g=(fg)^\prime-fg^\prime,\int f^\prime g=\int(fg)^\prime-\int fg^\prime,\int f^\prime g=\left[fg\right]-\int fg^\prime \\&&\;\dots\;f^\prime(x)=\cos{\left(x\right)} \\&&\;\dots\;f(x)=\sin{\left(x\right)} \\&&\;\dots\;g(x)=\cos^{2n-1}{\left(x\right)}=u^{2n-1}(x)\;\cdots\;u=\cos{\left(x\right)} \\&&\;\dots\;g^\prime(x)=\frac{\mathrm{d}g}{\mathrm{d}x}=\frac{\mathrm{d}g}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x} =\left(2n-1\right)u^{2n-2}(x)\cdot\left(-\sin{\left(x\right)}\right) =-\left(2n-1\right)\cos^{2n-2}{\left(x\right)}\sin{\left(x\right)} \\&=&\left[0-0\right]+\left(2n-1\right)\int_0^{\frac{\pi}{2}}\color{red}{\sin^2{\left(x\right)}}\color{black}{}\cos^{2n-2}{\left(x\right)}\mathrm{d}x \\&=&\left(2n-1\right)\int_0^{\frac{\pi}{2}}\color{red}{\left(1-\cos^2{\left(x\right)}\right)}\color{black}{}\cos^{2n-2}{\left(x\right)}\mathrm{d}x \\&=&\left(2n-1\right)\int_0^{\frac{\pi}{2}}\left(\cos^{2n-2}{\left(x\right)}-\cos^{2n}{\left(x\right)}\right)\mathrm{d}x \\&=&\left(2n-1\right)\int_0^{\frac{\pi}{2}}\cos^{2(n-1)}{\left(x\right)}\mathrm{d}x-\left(2n-1\right)\int_0^{\frac{\pi}{2}}\cos^{2n}{\left(x\right)}\mathrm{d}x \\&=&\left(2n-1\right)A_{n-1}-\left(2n-1\right)A_n \\A_n+\left(2n-1\right)A_n&=&\left(2n-1\right)A_{n-1} \\2nA_n&=&\left(2n-1\right)A_{n-1} \end{eqnarray}$$

準備2．$A_n=(2n-1)nB_{n-1}-2n^2B_{n}$

$$\begin{eqnarray} A_n&=&\int_0^{\frac{\pi}{2}}\cos^{2n}{\left(x\right)}\mathrm{d}x \\&=&\left[x\cdot\cos^{2n}{\left(x\right)}\right]_0^{\frac{\pi}{2}} -\int_0^{\frac{\pi}{2}}x\left\{2n\cos^{2n-1}{\left(x\right)\left(-\sin{\left(x\right)}\right)}\right\}\mathrm{d}x \\&&\;\dots\;(fg)^\prime=f^\prime g+fg^\prime,f^\prime g=(fg)^\prime-fg^\prime,\int f^\prime g=\int(fg)^\prime-\int fg^\prime,\int f^\prime g=\left[fg\right]-\int fg^\prime \\&&\;\dots\;f^\prime(x)=1 \\&&\;\dots\;f(x)=x \\&&\;\dots\;g(x)=\cos^{2n}{\left(x\right)}=u^{2n}\;\cdots\;u=\cos{\left(x\right)} \\&&\;\dots\;g^\prime(x)=\frac{\mathrm{d}g}{\mathrm{d}x}=\frac{\mathrm{d}g}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x} =2nu^{2n-1}\cdot\left(-\sin{\left(x\right)}\right) =2n\cos^{2n-1}{\left(x\right)}\left(-\sin{\left(x\right)}\right) \\&=&\left[0-0\right]+2n\int_0^{\frac{\pi}{2}}x\sin{\left(x\right)}\cos^{2n-1}{\left(x\right)}\mathrm{d}x \\&=&2n\int_0^{\frac{\pi}{2}}x\sin{\left(x\right)}\cos^{2n-1}{\left(x\right)}\mathrm{d}x \\&=&2n\left[\left[\frac{1}{2}x^2\cdot\sin{\left(x\right)}\cos^{2n-1}{\left(x\right)}\right]_0^{\frac{\pi}{2}} -\int_0^{\frac{\pi}{2}}\frac{1}{2}x^2\left\{ \cos{\left(x\right)}\cdot\cos^{2n-1}{\left(x\right)} +\sin{\left(x\right)}\cdot(2n-1)\cos^{2n-2}{\left(x\right)}\left(-\sin{\left(x\right)}\right) \right\}\mathrm{d}x \right] \\&=&-2n\frac{1}{2}\int_0^{\frac{\pi}{2}}x^2\left\{ \cos^{2n}{\left(x\right)}-(2n-1)\color{red}{\sin^2{\left(x\right)}}\color{black}{}\cos^{2n-2}{\left(x\right)} \right\}\mathrm{d}x \\&=&-n\int_0^{\frac{\pi}{2}}\left[ x^2\cos^{2n}{\left(x\right)}-(2n-1)x^2\color{red}{\left\{1-\cos^2{\left(x\right)}\right\}}\color{black}{}\cos^{2n-2}{\left(x\right)} \right]\mathrm{d}x \\&=&-n\int_0^{\frac{\pi}{2}}\left[ x^2\cos^{2n}{\left(x\right)}-(2n-1)x^2\left\{\cos^{2n-2}{\left(x\right)}-\cos^2{\left(x\right)}\cos^{2n-2}{\left(x\right)}\right\} \right]\mathrm{d}x \\&=&-n\int_0^{\frac{\pi}{2}}\left[ x^2\cos^{2n}{\left(x\right)}-(2n-1)x^2\left\{\cos^{2n-2}{\left(x\right)}-\cos^{2n}{\left(x\right)}\right\} \right]\mathrm{d}x \\&=&-n\int_0^{\frac{\pi}{2}}\left\{ x^2\cos^{2n}{\left(x\right)}-(2n-1)x^2\cos^{2n-2}{\left(x\right)}+(2n-1)x^2\cos^{2n}{\left(x\right)} \right\}\mathrm{d}x \\&=&-n\int_0^{\frac{\pi}{2}}\left\{ -(2n-1)x^2\cos^{2n-2}{\left(x\right)}+2nx^2\cos^{2n}{\left(x\right)} \right\}\mathrm{d}x \\&=&-n\int_0^{\frac{\pi}{2}} -(2n-1)x^2\cos^{2(n-1)}{\left(x\right)} \mathrm{d}x -n\int_0^{\frac{\pi}{2}} 2nx^2\cos^{2n}{\left(x\right)} \mathrm{d}x \\&=&(2n-1)n\int_0^{\frac{\pi}{2}} x^2\cos^{2(n-1)}{\left(x\right)} \mathrm{d}x -2n^2\int_0^{\frac{\pi}{2}} x^2\cos^{2n}{\left(x\right)} \mathrm{d}x \\&=&(2n-1)nB_{n-1}-2n^2B_{n}\;\cdots\;B_n=\int_0^{\frac{\pi}{2}}x^2\cos^{2n}{\left(x\right)}\mathrm{d}x \end{eqnarray}$$

準備3．$\lim_{n\rightarrow\infty}\frac{B_n}{A_n}=0$

$$\begin{eqnarray} B_n&=&\int_0^{\frac{\pi}{2}}x^2\cos^{2n}{\left(x\right)}\mathrm{d}x \\&\leq&\int_0^{\frac{\pi}{2}}\left\{\frac{\pi}{2} \sin{\left(x\right)}\right\}^2\cos^{2n}{\left(x\right)}\mathrm{d}x \\&&\;\cdots\;\frac{2}{\pi}x\leq\sin{\left(x\right)}\;\cdots\;x\in\left[0,\frac{\pi}{2}\right] \\&&\;\cdots\;x\leq\frac{\pi}{2}\sin{\left(x\right)} \\&&\;\cdots\;(0,0)と(\frac{\pi}{2},1)を通る直線\frac{2}{\pi}xは，区間\left[0,\frac{\pi}{2}\right]では常に\sin{\left(x\right)}以下である(図を参照)． \\&=&\frac{\pi^2}{4}\int_0^{\frac{\pi}{2}}\sin^{2}{\left(x\right)}\cos^{2n}{\left(x\right)}\mathrm{d}x \\&=&\frac{\pi^2}{4}\int_0^{\frac{\pi}{2}}\color{red}{\sin^{2}{\left(x\right)}}\color{black}{}\cos^{2n}{\left(x\right)}\mathrm{d}x \\&=&\frac{\pi^2}{4}\int_0^{\frac{\pi}{2}}\color{red}{\left\{1-\cos^{2}{\left(x\right)}\right\}}\color{black}{}\cos^{2n}{\left(x\right)}\mathrm{d}x \\&=&\frac{\pi^2}{4}\int_0^{\frac{\pi}{2}}\left\{1-\cos^{2}{\left(x\right)}\right\}\cos^{2n}{\left(x\right)}\mathrm{d}x \\&=&\frac{\pi^2}{4}\int_0^{\frac{\pi}{2}}\left\{\cos^{2n}{\left(x\right)}-\cos^{2n+2}{\left(x\right)}\right\}\mathrm{d}x \\&=&\frac{\pi^2}{4}\left[\int_0^{\frac{\pi}{2}}\cos^{2n}{\left(x\right)}\mathrm{d}x-\int_0^{\frac{\pi}{2}}\cos^{2(n+1)}{\left(x\right)}\mathrm{d}x\right] \\&=&\frac{\pi^2}{4}\left(A_n-A_{n+1}\right) \\&=&\frac{\pi^2}{4}\left(A_n-\frac{2(n+1)-1}{2(n+1)}A_{n}\right) \\&=&\frac{\pi^2}{4}\left(\frac{2(n+1)-\left\{2(n+1)-1\right\}}{2(n+1)}A_{n}\right) \\&=&\frac{\pi^2}{4}\left(\frac{2(n+1)-2(n+1)+1}{2(n+1)}A_{n}\right) \\&=&\frac{\pi^2}{4}\left(\frac{1}{2(n+1)}A_{n}\right) \\&=&\frac{\pi^2}{4}\frac{A_{n}}{2(n+1)} \\B_n&\leq&\frac{\pi^2}{4}\frac{A_n}{2(n+1)} \\B_n\frac{1}{A_n}&\leq&\frac{\pi^2}{4}\frac{A_n}{2(n+1)}\frac{1}{A_n}\;\cdots\;A_n\geq0 \\\frac{B_n}{A_n}&\leq&\frac{\pi^2}{4}\frac{1}{2(n+1)} \\\lim_{n\rightarrow\infty}\frac{B_n}{A_n}&\leq&\lim_{n\rightarrow\infty}\frac{\pi^2}{4}\frac{1}{2(n+1)}=0 \\\lim_{n\rightarrow\infty}\frac{B_n}{A_n}&\geq&0 \;\cdots\;A_n,B_nは共に被積分凾数\left(\cos^{2n}{\left(x\right)}, x^2\cos^{2n}{\left(x\right)}\right)が偶数乗のため，常に値は0以上． \\&&\;\cdots\;よって，その積分結果も常に0以上となる． \\\lim_{n\rightarrow\infty}\frac{B_n}{A_n}&=&0\;\cdots\;前述の2つから，はさみうちの原理より極限は0． \end{eqnarray}$$

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

$$\begin{eqnarray} \\\sum_{n=1}^{\infty}\frac{1}{n^2}&=&\lim_{N\rightarrow\infty} \sum_{n=1}^{N}\frac{1}{n^2} \\&=&\lim_{N\rightarrow\infty} \sum_{n=1}^{N}\left\{\frac{(2n-1)B_{n-1}}{nA_n}-\frac{2B_{n}}{A_n}\right\} \\&&\;\cdots\;A_n=(2n-1)nB_{n-1}-2n^2B_{n}\;(準備2．) \\&&\;\cdots\;\frac{A_n}{n^2}=\frac{(2n-1)nB_{n-1}}{n^2}-\frac{2n^2B_{n}}{n^2} \\&&\;\cdots\;\frac{A_n}{n^2}=\frac{(2n-1)B_{n-1}}{n}-2B_{n} \\&&\;\cdots\;\frac{1}{n^2}=\frac{(2n-1)B_{n-1}}{nA_n}-\frac{2B_{n}}{A_n} \\&=&\lim_{N\rightarrow\infty} \sum_{n=1}^{N}\frac{2B_{n-1}}{A_{n-1}}-\frac{2B_{n}}{A_n} \\&&\;\cdots\;2nA_n=\left(2n-1\right)A_{n-1}\;(準備1．) \\&&\;\cdots\;\frac{2}{A_{n-1}}nA_n=\left(2n-1\right) \\&&\;\cdots\;\frac{2}{A_{n-1}}=\frac{\left(2n-1\right)}{nA_n} \\&=&2\lim_{N\rightarrow\infty} \sum_{n=1}^{N}\frac{B_{n-1}}{A_{n-1}}-\frac{B_{n}}{A_n} \\&=&2\lim_{N\rightarrow\infty} \left\{ \left(\frac{B_{1-1}}{A_{1-1}}-\frac{B_{1}}{A_1}\right) +\left(\frac{B_{2-1}}{A_{2-1}}-\frac{B_{2}}{A_2}\right) +\left(\frac{B_{3-1}}{A_{3-1}}-\frac{B_{3}}{A_3}\right) +\cdots+\left(\frac{B_{N-1}}{A_{N-1}}-\frac{B_{N}}{A_N}\right) \right\} \\&=&2\lim_{N\rightarrow\infty} \left\{ \frac{B_{0}}{A_{0}}\cancel{-\frac{B_{1}}{A_1} +\frac{B_{1}}{A_{1}}}\cancel{-\frac{B_{2}}{A_2} +\frac{B_{2}}{A_{2}}}\cancel{-\frac{B_{3}}{A_3} +\cdots}+\cancel{\cdots+\frac{B_{N-1}}{A_{N-1}}}-\frac{B_{N}}{A_N} \right\} \;\cdots\;Telescoping Series \\&=&2\left(\frac{B_{0}}{A_{0}}-\lim_{N\rightarrow\infty}\frac{B_{N}}{A_N}\right) \\&=&2\left(\frac{B_{0}}{A_{0}}-0\right) \;\cdots\;\lim_{n\rightarrow\infty}\frac{B_n}{A_n}=0\;(準備3．) \\&=&2\frac{B_{0}}{A_{0}} \\&=&\left.2\frac{\int_0^{\frac{\pi}{2}}x^2\cos^{2\cdot n}{\left(x\right)}\mathrm{d}x}{\int_0^{\frac{\pi}{2}}\cos^{2\cdot n}{\left(x\right)}\mathrm{d}x}\right|_{n=0} \\&=&2\frac{\int_0^{\frac{\pi}{2}}x^2\cos^{2\cdot0}{\left(x\right)}\mathrm{d}x}{\int_0^{\frac{\pi}{2}}\cos^{2\cdot0}{\left(x\right)}\mathrm{d}x} \\&=&2\frac{\int_0^{\frac{\pi}{2}}x^2\mathrm{d}x}{\int_0^{\frac{\pi}{2}}\mathrm{d}x} \\&=&2\frac{\left[\frac{1}{3}x^3\right]_0^{\frac{\pi}{2}}}{\left[x\right]_0^{\frac{\pi}{2}}} \\&=&2\frac{\left[\frac{1}{3}\left(\frac{\pi}{2}\right)^3-\frac{1}{3}0^3\right]}{\left[\frac{\pi}{2}-0\right]} \\&=&2\frac{\frac{\pi^3}{24}}{\frac{\pi}{2}} \\&=&2\frac{\pi^2}{12} \\&=&\frac{\pi^2}{6} \end{eqnarray}$$

式展開

バーゼル問題

準備1．\(2nA_n=\left(2n-1\right)A_{n-1}\)

準備2．\(A_n=(2n-1)nB_{n-1}-2n^2B_{n}\)

準備3．\(\lim_{n\rightarrow\infty}\frac{B_n}{A_n}=0\)

\(\sum_{n=1}^{\infty}\frac{1}{n^2}\)