어려운 수학문제 쉽게 만들기 (편집) (1)

(편집 필터 규칙)

==# 첫 번째 예제 #==
우선 간단한 식을 하나 생각한다. 아래 식을 보자.
><math>1 \times 1 = 1</math>
조작을 시작한다! 우선 <math>1=\ln e=\frac{\pi}{\pi}=\sin^2 \theta+\cos^2 \theta</math>이므로
><math>\ln e \times  \frac{\pi}{\pi}= \sin^2 \theta+\cos^2 \theta</math>
<math>e=\lim_{n \to \infty} (1+\frac{1}{n})^n</math>이므로
><math>\ln \{ \lim_{n \to \infty} (1+\frac{1}{n})^n \} \times  \frac{\pi}{\pi}= \sin^2 \theta+\cos^2 \theta</math>
한편 [[바젤 문제]]에 의해 <math>\pi = \sqrt{\zeta ( 2 ) \times 6} = \sqrt{\sum_{k=1}^{\infty} \frac {6}{k^2}}</math>이므로
><math>\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</math>
또한 그레고리 급수에 의해 <math>\pi=4\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}</math>이므로
><math>\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</math>
[[오일러 파이 함수]]의 성질 <math>n=\sum_{d|n}\phi(d)</math>와 <math>\phi(n)=n\prod_{p_i|n}\left(1-\frac{1}{p_i}\right)</math>를 이용하면
><math>\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}}}</math><math>= \sin^2 \theta+\cos^2 \theta</math>
윌런스의 공식 <math>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]</math>를 이용하면
><math>\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}}}</math><math>= \sin^2 \theta+\cos^2 \theta</math>

분모는 충분히 더러워졌으므로, 이제 분자를 더럽히자. <math>6=1+2+3</math>이고, <math>1=-\sum_{d \mid n,d>1}\mu(d)\ (n>1)</math>, <math>2=[e]=[\sum_{n=0}^{\infty} \frac{1}{n!}]</math>, <math>3=[\pi]</math>이므로
><math>\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}}} </math><math>=\sin^2 \theta+\cos^2 \theta</math>
우변도 더럽혀 보자. 테일러 전개를 이용한다. <math>\sin x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1}</math>이고 <math>\cos x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} x^{2n}</math>이므로
><math>\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}}}</math>
><math>={\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} \theta^{2n+1}}^2+{\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} \theta^{2n}}^2</math>
>----
><math>\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}}} </math><math>={\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} \theta^{2n+1}}^2+{\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} \theta^{2n}}^2</math>
<math>e=\lim_{n \to \infty} (1+\frac{1}{n})^n</math>이므로
><math>\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}}} </math><math>={\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} \theta^{2n+1}}^2+{\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} \theta^{2n}}^2</math>
>----
><math>\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}}} </math><math>={\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} \theta^{2n+1}}^2+{\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} \theta^{2n}}^2</math>
<math>\lim_{x \to 0} {\frac{\sin x}{x}}=1</math>이므로
><math>\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}}} </math><math>={\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} \theta^{2n+1}}^2+{\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} \theta^{2n}}^2</math>
<math>[1/\gamma]=1</math>이므로
><math>\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}}} </math><math>={\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} \theta^{2n+1}}^2+{\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} \theta^{2n}}^2</math>
<math>\gamma=\lim_{n\to\infty} \sum_{k=1}^n \frac{1}{k} - \ln n</math> 이므로
><math>\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}}} </math><math>={\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} \theta^{2n+1}}^2+{\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} \theta^{2n}}^2</math>
<math>{\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</math>이므로
>[math(\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)]

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