# 감마함수의 비와 라마누잔의 연분수

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## 개요

• 감마함수의 비를 다음과 같이 연분수로 표현가능

$\frac{\Gamma \left(\frac{1}{4} (-n+x+1)\right) \Gamma \left(\frac{1}{4} (n+x+1)\right)}{\Gamma \left(\frac{1}{4} (-n+x+3)\right) \Gamma \left(\frac{1}{4} (n+x+3)\right)}=\cfrac{4}{x-\cfrac{n^2-1}{2 x-\cfrac{n^2-9}{2 x-\cfrac{n^2-25}{2 x-\cfrac{n^2-49}{2 x-\cfrac{n^2-81}{2 x-\cfrac{n^2-121}{2 x-\cfrac{n^2-169}{2 x-\cfrac{n^2-225}{2 x-\cfrac{n^2-289}{2 x-\cdots}}}}}}}}}}$

## 감마함수와 무한곱

정리

$\frac{\Gamma(\frac{1}{4})\Gamma(\frac{3}{4})}{\Gamma(\frac{2}{4})\Gamma(\frac{2}{4})}=\sqrt{2}$

증명

감마함수의 다음 표현 $\Gamma(z) = \lim_{n \to \infty} \frac{n! \; n^z}{z \; (z+1)\cdots(z+n)}$과 다음 무한곱의 연분수 표현을 사용하여 얻을 수 있다 $1+{q \over 1+q + } {q^2 \over 1+q^2+} {q^3 \over 1+q^3} \cdots=\frac{(q^2;q^4)_{\infty}^2}{(q^1;q^4)_{\infty}(q^3;q^4)_{\infty}}$ ■

## 메모

$$R(a,b)=\frac{f(a,a^{-1}b)}{f(aq,a^{-1}b)}-a=\frac{R^{N}(a,b)}{R^{D}(a,b)}=1+\frac{bq}{1+aq} {\ \atop+} \frac{bq^2}{1+aq^2}{\ \atop+} \frac{bq^3}{1} {\ \atop+\dots}$$

$$R(1,1)=\frac{R^{N}(1,1)}{R^{D}(1,1)}=1+{q \over 1+q + } {q^2 \over 1+q^2+} {q^3 \over 1+q^3} \cdots=\frac{(q^2;q^4)_{\infty}^2}{(q^1;q^4)_{\infty}(q^3;q^4)_{\infty}}$$

## 관련논문

• Cao, Xiaodong, and Cristinel Mortici. “Multiple-Correction and Summation of the Rational Series.” arXiv:1511.00198 [math], October 31, 2015. http://arxiv.org/abs/1511.00198.
• Cao, Xiaodong, Yoshio Tanigawa, and Wenguang Zhai. “The Fastest Possible Continued Fraction Approximations of a Class of Functions.” arXiv:1508.00176 [math], August 1, 2015. http://arxiv.org/abs/1508.00176.
• Ramanathan, K. G. 1987. “Hypergeometric Series and Continued Fractions.” Indian Academy of Sciences. Proceedings. Mathematical Sciences 97 (1-3): 277–296 (1988). doi:10.1007/BF02837830.