우선 간단한 식을 하나 생각한다. 아래 식을 보자.
1 \times 1 = 1
조작을 시작한다! 우선
1=\ln e=\frac{\pi}{\pi}=\sin^2 \theta+\cos^2 \theta이므로
\ln e \times \frac{\pi}{\pi}= \sin^2 \theta+\cos^2 \theta
e=\lim_{n \to \infty} (1+\frac{1}{n})^n이므로
\ln \{ \lim_{n \to \infty} (1+\frac{1}{n})^n \} \times \frac{\pi}{\pi}= \sin^2 \theta+\cos^2 \theta
한편
바젤 문제에 의해
\pi = \sqrt{\zeta ( 2 ) \times 6} = \sqrt{\sum_{k=1}^{\infty} \frac {6}{k^2}}이므로
\ln \{ \lim_{n \to \infty} (1+\frac{1}{n})^n \} \times \frac{\pi}{\sqrt{\sum_{k=1}^{\infty} \frac {6}{k^2}}}= \sin^2 \theta+\cos^2 \theta
또한 그레고리 급수에 의해
\pi=4\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}이므로
\ln \{ \lim_{n \to \infty} (1+\frac{1}{n})^n \} \times \frac{4\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}}{\sqrt{\sum_{k=1}^{\infty} \frac {6}{k^2}}}= \sin^2 \theta+\cos^2 \theta
오일러 파이 함수의 성질
n=\sum_{d|n}\phi(d)와
\phi(n)=n\prod_{p_i|n}\left(1-\frac{1}{p_i}\right)를 이용하면
\ln \{ \lim_{n \to \infty} (1+\frac{1}{n})^n \} \times \frac{4\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}}{\sqrt{\sum_{k=1}^{\infty} \frac {6}{\left(\sum_{d|k}d\prod_{p_i|d}\left(1-\frac{1}{p_i}\right)\right)^2}}}= \sin^2 \theta+\cos^2 \theta
윌런스의 공식
p_n=1+\sum_{m=1}^{2^n}\left[\sqrt[n]{n}\left(\sum_{x=1}^m\left[\cos^2 \pi \frac{(x-1)!+1}{x}\right]\right)\right]를 이용하면
\ln \{ \lim_{n \to \infty} (1+\frac{1}{n})^n \} \times \frac{4\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}}{\sqrt{\sum_{k=1}^{\infty} \frac {6}{\left(\sum_{d|k}d\prod_{p_i|d}\left(1-\left(1+\sum_{m=1}^{2^i}\left[\sqrt[i]{i}\left(\sum_{x=1}^m\left[\cos^2 \pi \frac{(x-1)!+1}{x}\right]\right)\right]\right)^{-1}\right)\right)^2}}}= \sin^2 \theta+\cos^2 \theta
분모는 충분히 더러워졌으므로, 이제 분자를 더럽히자.
6=1+2+3이고,
1=-\sum_{d \mid n,d>1}\mu(d)\ (n>1),
2=[e]=[\sum_{n=0}^{\infty} \frac{1}{n!}],
3=[\pi]이므로
\ln \{ \lim_{n \to \infty} (1+\frac{1}{n})^n\} \times \frac{4\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}}{\sqrt{\sum_{k=1}^{\infty} \frac {-\sum_{d \mid 2k,d>1}\mu(d)+\left[\sum_{n=0}^{\infty} \frac{1}{n!}\right]+[\pi]}{\left(\sum_{d|k}d\prod_{p_i|d}\left(1-\left(1+\sum_{m=1}^{2^i}\left[\sqrt[i]{i}\left(\sum_{x=1}^m\left[\cos^2 \pi \frac{(x-1)!+1}{x}\right]\right)\right]\right)^{-1}\right)\right)^2}}}=\sin^2 \theta+\cos^2 \theta
우변도 더럽혀 보자. 테일러 전개를 이용한다.
\sin x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1}이고
\cos x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} x^{2n}이므로
\ln \left\{ \lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n \right\} \times \frac{4\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}}{\sqrt{\sum_{k=1}^{\infty} \frac {-\sum_{d \mid 2k,d>1}\mu(d)+\left[\sum_{n=0}^{\infty} \frac{1}{n!}\right]+[\pi]}{\left(\sum_{d|k}d\prod_{p_i|d}\left(1-\left(1+\sum_{m=1}^{2^i}\left[\sqrt[i]{i} \left(\sum_{x=1}^m\left[\cos^2 \pi \frac{(x-1)!+1}{x}\right]\right)\right]\right)^{-1}\right)\right)^2}}}
={\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} \theta^{2n+1}}^2+{\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} \theta^{2n}}^2
\ln \{ \lim_{n \to \infty} (\frac{e}{e}+\frac{1}{n})^n \} \times \frac{4\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}}{\sqrt{\sum_{k=1}^{\infty} \frac {-\sum_{d \mid 2k,d>1}\mu(d)+\left[\sum_{n=0}^{\infty} \frac{1}{n!}\right]+[\pi]}{\left(\sum_{d|k}d\prod_{p_i|d}\left(1-\left(1+\sum_{m=1}^{2^i}\left[\sqrt[i]{i}\left(\sum_{x=1}^m\left[\cos^2 \pi \frac{(x-1)!+1}{x}\right]\right)\right]\right)^{-1}\right)\right)^2}}}={\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} \theta^{2n+1}}^2+{\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} \theta^{2n}}^2
e=\lim_{n \to \infty} (1+\frac{1}{n})^n이므로
\ln \{ \lim_{n \to \infty} (\frac{e}{ (1+\frac{1}{n})^n}+\frac{1}{n})^n \}\times \frac{4\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}}{\sqrt{\sum_{k=1}^{\infty} \frac {-\sum_{d \mid 2k,d>1}\mu(d)+\left[\sum_{n=0}^{\infty} \frac{1}{n!}\right]+[\pi]}{\left(\sum_{d|k}d\prod_{p_i|d}\left(1-\left(1+\sum_{m=1}^{2^i}\left[\sqrt[i]{i}\left(\sum_{x=1}^m\left[\cos^2 \pi \frac{(x-1)!+1}{x}\right]\right)\right]\right)^{-1}\right)\right)^2}}}={\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} \theta^{2n+1}}^2+{\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} \theta^{2n}}^2
\ln \{ \lim_{n \to \infty} (\frac{1}{\frac{(1+\frac{1}{n})^n}{e}}+\frac{1}{n})^n \}\times \frac{4\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}}{\sqrt{\sum_{k=1}^{\infty} \frac {-\sum_{d \mid 2k,d>1}\mu(d)+\left[\sum_{n=0}^{\infty} \frac{1}{n!}\right]+[\pi]}{\left(\sum_{d|k}d\prod_{p_i|d}\left(1-\left(1+\sum_{m=1}^{2^i}\left[\sqrt[i]{i}\left(\sum_{x=1}^m\left[\cos^2 \pi \frac{(x-1)!+1}{x}\right]\right)\right]\right)^{-1}\right)\right)^2}}}={\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} \theta^{2n+1}}^2+{\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} \theta^{2n}}^2
\lim_{x \to 0} {\frac{\sin x}{x}}=1이므로
\ln \{ \lim_{n \to \infty} (\frac{\lim_{x \to 0} {\frac{\sin x}{x}}}{\frac{ (1+\frac{1}{n})^n}{e}}+\frac{1}{n})^n \}\times \frac{ 4\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1} }{\sqrt{\sum_{k=1}^{\infty} \frac {-\sum_{d \mid 2k,d>1}\mu(d)+\left[\sum_{n=0}^{\infty} \frac{1}{n!}\right]+[\pi]}{\left(\sum_{d|k}d\prod_{p_i|d}\left(1-\left(1+\sum_{m=1}^{2^i}\left[\sqrt[i]{i}\left(\sum_{x=1}^m\left[\cos^2 \pi \frac{(x-1)!+1}{x}\right]\right)\right]\right)^{-1}\right)\right)^2}}}={\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} \theta^{2n+1}}^2+{\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} \theta^{2n}}^2
[1/\gamma]=1이므로
\ln \{ \lim_{n \to \infty} (\frac{\lim_{x \to 0} {\frac{\sin x}{x}}}{\frac{ (1+\frac{1}{n})^n}{e}}+\frac{[\frac{1}{\gamma}]}{n})^n \}\times \frac{ 4\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1} }{\sqrt{\sum_{k=1}^{\infty} \frac {-\sum_{d \mid 2k,d>1}\mu(d)+\left[\sum_{n=0}^{\infty} \frac{1}{n!}\right]+[\pi]}{\left(\sum_{d|k}d\prod_{p_i|d}\left(1-\left(1+\sum_{m=1}^{2^i}\left[\sqrt[i]{i}\left(\sum_{x=1}^m\left[\cos^2 \pi \frac{(x-1)!+1}{x}\right]\right)\right]\right)^{-1}\right)\right)^2}}}={\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} \theta^{2n+1}}^2+{\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} \theta^{2n}}^2
\gamma=\lim_{n\to\infty} \sum_{k=1}^n \frac{1}{k} - \ln n 이므로
\ln \{ \lim_{n \to \infty} (\frac{\lim_{x \to 0} {\frac{\sin x}{x}}}{\frac{ (1+\frac{1}{n})^n}{e}}+\frac{[\frac{1}{ \sum_{k=1}^n \frac{1}{k} - \ln n}]}{n})^n \}\times \frac{ 4\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1} }{\sqrt{\sum_{k=1}^{\infty} \frac {-\sum_{d \mid 2k,d>1}\mu(d)+\left[\sum_{n=0}^{\infty} \frac{1}{n!}\right]+[\pi]}{\left(\sum_{d|k}d\prod_{p_i|d}\left(1-\left(1+\sum_{m=1}^{2^i}\left[\sqrt[i]{i}\left(\sum_{x=1}^m\left[\cos^2 \pi \frac{(x-1)!+1}{x}\right]\right)\right]\right)^{-1}\right)\right)^2}}}={\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} \theta^{2n+1}}^2+{\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} \theta^{2n}}^2
{\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} \theta^{2n+1}}^2+{\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} \theta^{2n}}^2=1이므로
\ln \{ \lim_{n \to \infty} (\frac{\lim_{x \to 0} {\frac{\sin x}{x}}}{\frac{ (1+\frac{1}{n})^n}{e}}+\frac{[\frac{1}{ \sum_{k=1}^n \frac{1}{k} - \ln n}]}{n})^n \}\times \frac{ 4\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1} }{\sqrt{\sum_{k=1}^{\infty} \frac {-\sum_{d \mid 2k,d>1}\mu(d)+\left[\sum_{n=0}^{\infty} \frac{1}{n!}\right]+[\pi]}{\left(\sum_{d|k}d\prod_{p_i|d}\left(1-\left(1+\sum_{m=1}^{2^i}\left[\sqrt[i]{i}\left(\sum_{x=1}^m\left[\cos^2 \pi \frac{(x-1)!+1}{x}\right]\right)\right]\right)^{-1}\right)\right)^2}}}={\sum^{\infty}_{n=0} \frac{(-({\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} \theta^{2n+1}}^2+{\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} \theta^{2n}}^2))^n}{(2n+{\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} \theta^{2n+1}}^2+{\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} \theta^{2n}}^2)!} \theta^{2n+{\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} \theta^{2n+1}}^2+{\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} \theta^{2n}}^2}}^2+{\sum^{\infty}_{n=0} \frac{(-({\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} \theta^{2n+1}}^2+{\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} \theta^{2n}}^2))^n}{(2n)!} \theta^{2n}}^2
