스털링 근사 공식의 엄밀한 증명
📂함수 스털링 근사 공식의 엄밀한 증명 정리 lim n → ∞ n ! e n ln n − n 2 π n = 1
\lim_{n \to \infty} {{n!} \over {e^{n \ln n - n} \sqrt{ 2 \pi n} }} = 1
n → ∞ lim e n l n n − n 2 πn n ! = 1
설명 스털링 근사 혹은 스털링 공식 stirling formula 은 통계학이나 물리학 등 여러 곳에서 유용하게 쓰인다. 또 다른 표현으로는 감마 함수 를 사용해 다음과 같이 적을 수 있다.
Γ ( n ) ≈ e n ln n − n 2 π n
\Gamma ( n ) \approx {e^{n \ln n - n} \sqrt{ 2 \pi n}}
Γ ( n ) ≈ e n l n n − n 2 πn
본 증명은 ‘제타함수의 비밀’이라는 책의 부록에 실려있는 것으로, 저자인 쿠로카와 노부시게 黑川信繁 라는 일본의 수학자가 고등학교 때 발견하고 몇 년 뒤 수학세미나數學セミナ- 1972년 6월호 에 게재했다고 밝혔다. 그러나 직접 찾아본 결과 생략이 많아서 그닥 참고할 것은 못 되며, 사실 그 논지도 증명이 아니었다. 마찬가지로 ‘제타함수의 비밀’을 직접 찾아보는 것도 별로 도움이 되지 않을 것이다. 자신있게 단언하건대 국내에서 쿠로카와의 증명을 가장 쉽고 친절하게 설명한 것은 이 포스트다.
다른 해석학 교재에 나오는 증명도 있지만 굳이 이 증명을 선택한 이유는 엄밀성을 포기하지 않는 선에서 지금까지 본 어떤 증명보다 쉬웠기 때문이다. 물론 보조정리를 증명하는 과정을 다 포함하면 더 길어지겠지만, 적어도 스털링 근사만을 위한 보조정리를 증명하는 삽질은 피할 수 있었다. 그나마 이 증명법에서 등장하는 보조정리는 상당히 짜임새 있는 맥락에서 나온 것으로, 그 과정을 따라가는 것 또한 공부하는 입장에서 상당히 즐거운 일이다.
같이보기 증명 함수 f ( x ) : = ln ( 1 + x ) f( x ) := \ln (1 + x) f ( x ) := ln ( 1 + x )
Part 1. lim n → ∞ [ ∑ k = 1 n f ( k n ) − n ∫ 0 1 f ( x ) d x ] = f ( 1 ) − f ( 0 ) 2 \displaystyle \lim_{n \to \infty} \left[ \sum_{k=1}^{n} f \left( {{k} \over {n}} \right) - n \int_{0}^{1} f(x) dx \right] = {{f(1) - f(0) } \over {2}} n → ∞ lim [ k = 1 ∑ n f ( n k ) − n ∫ 0 1 f ( x ) d x ] = 2 f ( 1 ) − f ( 0 )
f ′ ( x ) = 1 1 + x \displaystyle f '(x) = {{1} \over {1 + x}} f ′ ( x ) = 1 + x 1 x ∈ [ k − 1 n , k n ] \displaystyle x \in \left[ {{k-1} \over {n}} , {{k} \over {n}} \right] x ∈ [ n k − 1 , n k ] f ′ ( X ) f ' (X) f ′ ( X ) 1 1 + k n \displaystyle {{1} \over {1 + {{k} \over {n}} }} 1 + n k 1 1 1 + k − 1 n \displaystyle {{1} \over {1 + {{k-1} \over {n}} }} 1 + n k − 1 1
1 1 + k n ≤ f ( k n ) − f ( x ) k n − x ≤ 1 1 + k − 1 n
{{1} \over {1 + {{k} \over {n}} }} \le {{ f \left( {{k} \over {n}} \right) - f(x) } \over { {{k} \over {n}} - x }} \le {{1} \over {1 + {{k-1} \over {n}} }}
1 + n k 1 ≤ n k − x f ( n k ) − f ( x ) ≤ 1 + n k − 1 1
각 변에 ( k n − x ) \displaystyle \left( {{k} \over {n}} - x \right) ( n k − x )
1 1 + k n ( k n − x ) ≤ f ( k n ) − f ( x ) ≤ 1 1 + k − 1 n ( k n − x )
{{1} \over {1 + {{k} \over {n}} }} \left( {{k} \over {n}} - x \right) \le f \left( {{k} \over {n}} \right) - f(x) \le {{1} \over {1 + {{k-1} \over {n}} }} \left( {{k} \over {n}} - x \right)
1 + n k 1 ( n k − x ) ≤ f ( n k ) − f ( x ) ≤ 1 + n k − 1 1 ( n k − x )
각 변을 k − 1 n \displaystyle {{k-1} \over {n}} n k − 1 k n \displaystyle {{k} \over {n}} n k
∫ k − 1 n k n ( k n − x ) d x = k n 1 n − [ 1 2 x 2 ] k − 1 n k n = 2 k 2 n 2 − k 2 − k 2 + 2 k − 1 2 n 2 = 1 2 n 2
\begin{align*}
\int_{ {{k-1} \over {n}} }^{ {{k} \over {n}} } \left( {{k} \over {n}} - x \right) dx =& {{k} \over {n}} {{1} \over {n}} - \left[ {{1} \over {2}} x^2 \right]_{ {{k-1} \over {n}} }^{ {{k} \over {n}} }
\\ =& {{2k} \over {2n^2}} - {{k^2 - k^2 + 2k - 1} \over {2n^2}}
\\ =& {{1} \over {2n^2}}
\end{align*}
∫ n k − 1 n k ( n k − x ) d x = = = n k n 1 − [ 2 1 x 2 ] n k − 1 n k 2 n 2 2 k − 2 n 2 k 2 − k 2 + 2 k − 1 2 n 2 1
이므로
1 2 n 2 1 1 + k n ≤ ∫ k − 1 n k n [ f ( k n ) − f ( x ) ] d x ≤ 1 2 n 2 1 1 + k − 1 n
{{1} \over {2n^2}} {{1} \over {1 + {{k} \over {n}} }} \le \int_{ {{k-1} \over {n}} }^{ {{k} \over {n}} } \left[ f \left( {{k} \over {n}} \right) - f(x) \right] dx \le {{1} \over {2n^2}} {{1} \over {1 + {{k-1} \over {n}} }}
2 n 2 1 1 + n k 1 ≤ ∫ n k − 1 n k [ f ( n k ) − f ( x ) ] d x ≤ 2 n 2 1 1 + n k − 1 1
각 변에 n n n
1 2 n 1 1 + k n ≤ n ∫ k − 1 n k n [ f ( k n ) − f ( x ) ] d x ≤ 1 2 n 1 1 + k − 1 n
{{1} \over {2n}} {{1} \over {1 + {{k} \over {n}} }} \le n \int_{ {{k-1} \over {n}} }^{ {{k} \over {n}} } \left[ f \left( {{k} \over {n}} \right) - f(x) \right] dx \le {{1} \over {2n}} {{1} \over {1 + {{k-1} \over {n}} }}
2 n 1 1 + n k 1 ≤ n ∫ n k − 1 n k [ f ( n k ) − f ( x ) ] d x ≤ 2 n 1 1 + n k − 1 1
각 변을 k = 1 k=1 k = 1 n n n
1 2 ∑ k = 1 n 1 n 1 1 + k n ≤ n ∑ k = 1 n ∫ k − 1 n k n [ f ( k n ) − f ( x ) ] d x ≤ 1 2 ∑ k = 1 n 1 n 1 1 + k − 1 n
{{1} \over {2}} \sum_{k=1}^{n} {{1} \over {n}} {{1} \over {1 + {{k} \over {n}} }} \le n \sum_{k=1}^{n} \int_{ {{k-1} \over {n}} }^{ {{k} \over {n}} } \left[ f \left( {{k} \over {n}} \right) - f(x) \right] dx \le {{1} \over {2}} \sum_{k=1}^{n} {{1} \over {n}} {{1} \over {1 + {{k-1} \over {n}} }}
2 1 k = 1 ∑ n n 1 1 + n k 1 ≤ n k = 1 ∑ n ∫ n k − 1 n k [ f ( n k ) − f ( x ) ] d x ≤ 2 1 k = 1 ∑ n n 1 1 + n k − 1 1
가운데 변은
n ∑ k = 1 n ∫ k − 1 n k n [ f ( k n ) − f ( x ) ] d x = n [ k n − k − 1 n ] ∑ k = 1 n f ( k n ) − n ∫ 0 1 f ( x ) d x = ∑ k = 1 n f ( k n ) − n ∫ 0 1 f ( x ) d x
\begin{align*}
n \sum_{k=1}^{n} \int_{ {{k-1} \over {n}} }^{ {{k} \over {n}} } \left[ f \left( {{k} \over {n}} \right) - f(x) \right] dx =& n \left[ {{k} \over {n}} - {{k-1} \over {n}} \right] \sum_{k=1}^{n} f \left( {{k} \over {n}} \right) - n \int_{0}^{1} f(x) dx
\\ =& \sum_{k=1}^{n} f \left( {{k} \over {n}} \right) - n \int_{0}^{1} f(x) dx
\end{align*}
n k = 1 ∑ n ∫ n k − 1 n k [ f ( n k ) − f ( x ) ] d x = = n [ n k − n k − 1 ] k = 1 ∑ n f ( n k ) − n ∫ 0 1 f ( x ) d x k = 1 ∑ n f ( n k ) − n ∫ 0 1 f ( x ) d x
구분구적법과 정적분 : lim n → ∞ ∑ k = 1 n f ( a + p n k ) p n = ∫ 0 1 p f ( a + p x ) d x \lim _{ n\to \infty }{ \sum _{ k=1 }^{ n }{ f\left( a+\frac { p }{ n }k \right) \frac { p }{ n } } } =\int _{ 0 }^{ 1 }{ pf(a+px)dx } n → ∞ lim k = 1 ∑ n f ( a + n p k ) n p = ∫ 0 1 p f ( a + p x ) d x
좌변과 우변은 n → ∞ n \to \infty n → ∞
1 2 ∑ k = 1 ∞ 1 n 1 1 + k n = 1 2 ∑ k = 1 ∞ 1 n 1 1 + k − 1 n = 1 2 ∫ 0 1 1 1 + x d x = 1 2 [ ln ∣ 1 + x ∣ ] 0 1 = f ( 1 ) − f ( 0 ) 2
\begin{align*}
{{1} \over {2}} \sum_{k=1}^{\infty} {{1} \over {n}} {{1} \over {1 + {{k} \over {n}} }} =& {{1} \over {2}} \sum_{k=1}^{\infty} {{1} \over {n}} {{1} \over {1 + {{k-1} \over {n}} }}
\\ =& {{1} \over {2}} \int_{0}^{1} {{1} \over {1 + x}} dx
\\ =& {{1} \over {2}}\left[ \ln | 1 + x | \right]_{0}^{1}
\\ =& {{f(1) - f(0) } \over {2}}
\end{align*}
2 1 k = 1 ∑ ∞ n 1 1 + n k 1 = = = = 2 1 k = 1 ∑ ∞ n 1 1 + n k − 1 1 2 1 ∫ 0 1 1 + x 1 d x 2 1 [ ln ∣1 + x ∣ ] 0 1 2 f ( 1 ) − f ( 0 )
샌드위치 정리 에 따라
lim n → ∞ [ ∑ k = 1 n f ( k n ) − n ∫ 0 1 f ( x ) d x ] = f ( 1 ) − f ( 0 ) 2
\lim_{n \to \infty} \left[ \sum_{k=1}^{n} f \left( {{k} \over {n}} \right) - n \int_{0}^{1} f(x) dx \right] = {{f(1) - f(0) } \over {2}}
n → ∞ lim [ k = 1 ∑ n f ( n k ) − n ∫ 0 1 f ( x ) d x ] = 2 f ( 1 ) − f ( 0 )
Part 2. lim n → ∞ [ ( 2 n ) ! n ! n n ( e 4 ) n ] = 2 \displaystyle \lim_{n \to \infty} \left[ {{(2n)!} \over {n! n^n}} \left( {{e} \over {4}} \right)^{n} \right] = \sqrt{2} n → ∞ lim [ n ! n n ( 2 n )! ( 4 e ) n ] = 2
위의 Part 1에서 얻은 결과의 각 항을 살펴보자.
첫 항 ∑ k = 1 n f ( k n ) \displaystyle \sum_{k=1}^{n} f \left( {{k} \over {n}} \right) k = 1 ∑ n f ( n k )
∑ k = 1 n f ( k n ) = ∑ k = 1 n ln ( n + k n ) = ln ( n + 1 n n + 2 n ⋯ n + n n ) = ln ( ( 2 n ) ! n ! n n )
\begin{align*}
\sum_{k=1}^{n} f \left( {{k} \over {n}} \right) =& \sum_{k=1}^{n} \ln \left( {{n+k} \over {n}} \right)
\\ =& \ln \left( {{n+1} \over {n}} {{n+2} \over {n}} \cdots {{n+n} \over {n}} \right)
\\ =& \ln \left( {{ (2n)! } \over { n! n^n }} \right)
\end{align*}
k = 1 ∑ n f ( n k ) = = = k = 1 ∑ n ln ( n n + k ) ln ( n n + 1 n n + 2 ⋯ n n + n ) ln ( n ! n n ( 2 n )! )
두번째 항 n ∫ 0 1 f ( x ) d x \displaystyle n \int_{0}^{1} f(x) dx n ∫ 0 1 f ( x ) d x
n ∫ 0 1 f ( x ) d x = n [ ( 1 + x ) ln ( 1 + x ) − x ] 0 1 = 2 n ln 2 − n = ln ( 2 2 n e n ) = ln ( 4 e ) n
\begin{align*}
n \int_{0}^{1} f(x) dx =& n \left[ (1+x) \ln (1+x) -x \right]_{0}^{1}
\\ =& 2n \ln 2 - n
\\ =& \ln \left( {{2^{2n}} \over {e^{n}}} \right)
\\ =& \ln \left( {{4} \over {e}} \right)^{n}
\end{align*}
n ∫ 0 1 f ( x ) d x = = = = n [ ( 1 + x ) ln ( 1 + x ) − x ] 0 1 2 n ln 2 − n ln ( e n 2 2 n ) ln ( e 4 ) n
따라서
lim n → ∞ [ ∑ k = 1 n f ( k n ) − n ∫ 0 1 f ( x ) d x ] = lim n → ∞ ln [ ( 2 n ) ! n ! n n ( e 4 ) n ] = ln 2 2 = ln 2
\begin{align*}
\lim_{n \to \infty} \left[ \sum_{k=1}^{n} f \left( {{k} \over {n}} \right) - n \int_{0}^{1} f(x) dx \right] =& \lim_{n \to \infty} \ln \left[ {{(2n)!} \over {n! n^n}} \left( {{e} \over {4}} \right)^{n} \right]
\\ =& {{ \ln 2 } \over {2}}
\\ =& \ln \sqrt{2}
\end{align*}
n → ∞ lim [ k = 1 ∑ n f ( n k ) − n ∫ 0 1 f ( x ) d x ] = = = n → ∞ lim ln [ n ! n n ( 2 n )! ( 4 e ) n ] 2 ln 2 ln 2
이고, 정리하면
lim n → ∞ [ ( 2 n ) ! n ! n n ( e 4 ) n ] = 2
\lim_{n \to \infty} \left[ {{(2n)!} \over {n! n^n}} \left( {{e} \over {4}} \right)^{n} \right] = \sqrt{2}
n → ∞ lim [ n ! n n ( 2 n )! ( 4 e ) n ] = 2
Part 3. lim n → ∞ 4 n ( n ! ) 2 n ( 2 n ) ! = π \displaystyle \lim_{n \to \infty} {{4^n (n!)^2 } \over {\sqrt{n} (2n)! } } = \sqrt{ \pi } n → ∞ lim n ( 2 n )! 4 n ( n ! ) 2 = π
월리스 곱 :
∏ n = 1 ∞ 4 n 2 4 n 2 − 1 = lim n → ∞ 2 ⋅ 2 1 ⋅ 3 ⋅ 4 ⋅ 4 3 ⋅ 5 ⋅ ⋯ ⋅ 2 n ⋅ 2 n ( 2 n − 1 ) ⋅ ( 2 n + 1 ) = π 2
\prod_{n=1}^{\infty} {{4n^2} \over {4n^2 - 1}} = \lim_{n \to \infty} {{2 \cdot 2 } \over { 1 \cdot 3 } } \cdot {{4 \cdot 4 } \over { 3 \cdot 5 } } \cdot \cdots \cdot {{2n \cdot 2n } \over { (2n-1) \cdot (2n+1) } } = {{ \pi } \over {2}}
n = 1 ∏ ∞ 4 n 2 − 1 4 n 2 = n → ∞ lim 1 ⋅ 3 2 ⋅ 2 ⋅ 3 ⋅ 5 4 ⋅ 4 ⋅ ⋯ ⋅ ( 2 n − 1 ) ⋅ ( 2 n + 1 ) 2 n ⋅ 2 n = 2 π
π 2 = lim n → ∞ 2 2 ⋅ 4 2 ⋅ ⋯ ⋅ ( 2 n ) 2 1 ⋅ 3 2 ⋅ ⋯ ⋅ ( 2 n − 1 ) 2 ⋅ ( 2 n + 1 ) = lim n → ∞ 2 2 ⋅ 4 2 ⋅ ⋯ ⋅ ( 2 n ) 2 1 ⋅ 3 2 ⋅ ⋯ ⋅ ( 2 n − 1 ) 2 ⋅ 2 n
\begin{align*}
{{\pi} \over {2}} =& \lim_{n \to \infty} {{2^2 \cdot 4^2 \cdot \cdots \cdot (2n)^2} \over {1 \cdot 3^2 \cdot \cdots \cdot (2n-1)^2 \cdot (2n+1)}}
\\ =& \lim_{n \to \infty} {{2^2 \cdot 4^2 \cdot \cdots \cdot (2n)^2} \over {1 \cdot 3^2 \cdot \cdots \cdot (2n-1)^2 \cdot 2n }}
\end{align*}
2 π = = n → ∞ lim 1 ⋅ 3 2 ⋅ ⋯ ⋅ ( 2 n − 1 ) 2 ⋅ ( 2 n + 1 ) 2 2 ⋅ 4 2 ⋅ ⋯ ⋅ ( 2 n ) 2 n → ∞ lim 1 ⋅ 3 2 ⋅ ⋯ ⋅ ( 2 n − 1 ) 2 ⋅ 2 n 2 2 ⋅ 4 2 ⋅ ⋯ ⋅ ( 2 n ) 2
양변에 2 2 2
π = lim n → ∞ 2 2 ⋅ 4 2 ⋅ ⋯ ⋅ ( 2 n ) 2 1 ⋅ 3 2 ⋅ ⋯ ⋅ ( 2 n − 1 ) 2 ⋅ n
\pi = \lim_{n \to \infty} {{2^2 \cdot 4^2 \cdot \cdots \cdot (2n)^2} \over {1 \cdot 3^2 \cdot \cdots \cdot (2n-1)^2 \cdot n }}
π = n → ∞ lim 1 ⋅ 3 2 ⋅ ⋯ ⋅ ( 2 n − 1 ) 2 ⋅ n 2 2 ⋅ 4 2 ⋅ ⋯ ⋅ ( 2 n ) 2
양변에 루트를 취하면
π = lim n → ∞ 1 n 2 ⋅ 4 ⋅ ⋯ ⋅ ( 2 n ) 1 ⋅ 3 ⋅ ⋯ ⋅ ( 2 n − 1 ) = lim n → ∞ 1 n 2 ⋅ 4 ⋅ ⋯ ⋅ ( 2 n ) ⋅ 2 ⋅ 4 ⋅ ⋯ ⋅ ( 2 n ) 1 ⋅ 2 ⋅ 3 ⋅ ⋯ ⋅ ( 2 n − 1 ) ⋅ 2 n = lim n → ∞ 4 n ( n ! ) 2 n ( 2 n ) !
\begin{align*}
\sqrt{ \pi } =& \lim_{n \to \infty} {{1} \over {\sqrt{n}}} {{2 \cdot 4 \cdot \cdots \cdot (2n)} \over {1 \cdot 3 \cdot \cdots \cdot (2n-1) }}
\\ =& \lim_{ n \to \infty} {{1} \over {\sqrt{n}}} {{2 \cdot 4 \cdot \cdots \cdot (2n) \cdot 2 \cdot 4 \cdot \cdots \cdot (2n)} \over {1 \cdot 2 \cdot 3 \cdot \cdots \cdot (2n-1) \cdot 2n }}
\\ =& \lim_{n \to \infty} {{4^n (n!)^2 } \over {\sqrt{n} (2n)! }}
\end{align*}
π = = = n → ∞ lim n 1 1 ⋅ 3 ⋅ ⋯ ⋅ ( 2 n − 1 ) 2 ⋅ 4 ⋅ ⋯ ⋅ ( 2 n ) n → ∞ lim n 1 1 ⋅ 2 ⋅ 3 ⋅ ⋯ ⋅ ( 2 n − 1 ) ⋅ 2 n 2 ⋅ 4 ⋅ ⋯ ⋅ ( 2 n ) ⋅ 2 ⋅ 4 ⋅ ⋯ ⋅ ( 2 n ) n → ∞ lim n ( 2 n )! 4 n ( n ! ) 2
정리하면
lim n → ∞ 4 n ( n ! ) 2 n ( 2 n ) ! = π
\lim_{n \to \infty} {{4^n (n!)^2 } \over {\sqrt{n} (2n)! } } = \sqrt{ \pi }
n → ∞ lim n ( 2 n )! 4 n ( n ! ) 2 = π
Part 4. 취합
lim n → ∞ n ! e n ln n − n n = lim n → ∞ n ! n n e − n n = lim n → ∞ ( 2 n ) ! 4 n n ! ( e n ) n 4 n ( n ! ) 2 n ( 2 n ) !
\begin{align*}
\lim_{n \to \infty} {{n!} \over {e^{n \ln n - n }\sqrt{n}} } =& \lim_{n \to \infty} {{n!} \over {n^{n} e^{-n} \sqrt{n} }}
\\ =& \lim_{n \to \infty} { { (2n)!} \over { 4^{n} n! } } \left( {{e} \over {n}} \right)^{n} {{4^n (n!)^2} \over {\sqrt{ n} (2n)! }}
\end{align*}
n → ∞ lim e n l n n − n n n ! = = n → ∞ lim n n e − n n n ! n → ∞ lim 4 n n ! ( 2 n )! ( n e ) n n ( 2 n )! 4 n ( n ! ) 2
Part 2, 3에 따라
lim n → ∞ ( 2 n ) ! 4 n n ! ( e n ) n 4 n ( n ! ) 2 n ( 2 n ) ! = 2 π = 2 π
\lim_{n \to \infty} { { (2n)!} \over { 4^{n} n! } } \left( {{e} \over {n}} \right)^{n} {{4^n (n!)^2} \over {\sqrt{ n} (2n)! }} = \sqrt{2} \sqrt{\pi} = \sqrt{2 \pi}
n → ∞ lim 4 n n ! ( 2 n )! ( n e ) n n ( 2 n )! 4 n ( n ! ) 2 = 2 π = 2 π lim n → ∞ n ! e n ln n − n 2 π n = 1
\lim_{n \to \infty} {{n!} \over {e^{n \ln n - n} \sqrt{ 2 \pi n} }} = 1
n → ∞ lim e n l n n − n 2 πn n ! = 1
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