ヘロンの公式
余弦定理
$$\begin{eqnarray} |BH|^2&=&c^2-|AH|^2 \\&=&c^2-c^2 \cos^2(A) \\&=&a^2-|CH|^2 \\&=&a^2-\{b-c \cos(A)\}^2 \\&=&a^2-b^2+2bc\cos(A)-c^2\cos^2(A) \end{eqnarray}$$ $$\begin{eqnarray} c^2\cancel{-c^2 \cos^2(A)}&=&a^2-b^2+2bc\cos(A)\cancel{-c^2\cos^2(A)} \\c^2&=&a^2-b^2+2bc\cos(A) \\b^2+c^2-a^2&=&2bc\cos(A) \\\cos(A)&=&\frac{b^2+c^2-a^2}{2bc} \end{eqnarray}$$ヘロンの公式
$$\begin{eqnarray} S^2&=&\left(\frac{1}{2}|AC||BH|\right)^2 \\&=&\left\{\frac{1}{2}bc\sin{(A)}\right\}^2 \\&=&\frac{1}{4}b^2c^2\sin^2{(A)} \\&=&\frac{1}{4}b^2c^2\left\{1-\cos^2{(A)}\right\}\;\cdots\;\sin^2{(x )}+\cos^2{(x)}=1 \\&=&\frac{1}{4}b^2c^2-\frac{1}{4}b^2c^2\cos^2{(A)} \\&=&\frac{1}{4}b^2c^2-\frac{1}{4}b^2c^2\left(\frac{b^2+c^2-a^2}{2bc}\right)^2\;\cdots\;\cos{(A)}=\frac{b^2+c^2-a^2}{2bc} \\&=&\frac{1}{4}b^2c^2-\frac{1}{4}\cancel{b^2c^2}\frac{\left(b^2+c^2-a^2\right)^2}{4\cancel{b^2c^2}} \\&=&\frac{1}{4}b^2c^2-\frac{1}{16}\left(b^2+c^2-a^2\right)^2 \\&=&\frac{1}{16}\left\{4b^2c^2-\left(b^2+c^2-a^2\right)^2\right\} \\&=&\frac{1}{16}\left\{\left(2bc\right)^2-\left(b^2+c^2-a^2\right)^2\right\} \\&=&\frac{1}{16}\left[\left\{2bc+\left(b^2+c^2-a^2\right)\right\}\left\{2bc-\left(b^2+c^2-a^2\right)\right\}\right] \\&=&\frac{1}{16}\left\{\left(2bc+b^2+c^2-a^2\right)\left(2bc-b^2-c^2+a^2\right)\right\} \\&=&\frac{1}{16}\left[\left\{\left(b^2+2bc+c^2\right)-a^2\right\}\left\{a^2-\left(b^2-2bc+c^2\right)\right\}\right] \\&=&\frac{1}{16}\left[\left\{\left(b+c\right)^2-a^2\right\}\left\{a^2-\left(b-c\right)^2\right\}\right]\;\cdots\;(a \pm b)^2=a^2\pm2ab+b^2 \\&=&\frac{1}{16}\left[\left\{\left(b+c+a\right)\left(b+c-a\right)\right\}\left\{\left(a+b-c\right)\left(a-\left(b-c\right)\right)\right\}\right] \\&=&\frac{1}{16}\left[\left\{\left(b+c+a\right)\left(b+c-a\right)\right\}\left\{\left(a+b-c\right)\left(a-b+c\right)\right\}\right] \\&=&\frac{1}{16}\left(a+b+c\right)\left(-a+b+c\right)\left(a-b+c\right)\left(a+b-c\right) \\&=&\frac{a+b+c}{2}\frac{-a+b+c}{2}\frac{a-b+c}{2}\frac{a+b-c}{2} \\&=&\frac{a+b+c}{2}\frac{-a+b+c\color{red}{+a-a}}{2}\frac{a-b+c\color{red}{+b-b}}{2}\frac{a+b-c\color{red}{+c-c}}{2} \\&=&\frac{a+b+c}{2}\frac{a+b+c-2a}{2}\frac{a+b+c-2b}{2}\frac{a+b+c-2c}{2} \\&=&\frac{a+b+c}{2}\left(\frac{a+b+c}{2}-a\right)\left(\frac{a+b+c}{2}-b\right)\left(\frac{a+b+c}{2}-c\right) \\&=&s\left(s-a\right)\left(s-b\right)\left(s-c\right)\;\cdots\;s=\frac{a+b+c}{2} \end{eqnarray}$$ $$\begin{eqnarray} S&=&\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\;\cdots\;s=\frac{a+b+c}{2} \end{eqnarray}$$バーゼル問題
オリジナル: https://youtu.be/6olRd_dDS80
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．\(\lim_{n\rightarrow\infty}\frac{B_n}{A_n}=0\)
\(\sum_{n=1}^{\infty}\frac{1}{n^2}\)
$$\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}$$
登録:
投稿 (Atom)