"트리감마 함수(trigamma function)"의 두 판 사이의 차이

2010년 7월 17일 (토) 03:24 판

\(\sum_{n=1}^\infty \frac{1}{n^3\binom{2n}{n}}=\frac{\pi\sqrt{3}}{18}(\psi^{(1)}(\frac{1}{3})-\psi^{(1)}(\frac{2}{3}))-\frac{4}{3}\zeta(3)\)

여기서 \(\psi^{(1)}\)는 트리감마(trigamma)함수.

http://www.wolframalpha.com/input/?i=(polylog[2,exp(-i*2pi/3)]-polylog[2,exp(i*2pi/3)])*i/2

@@ 6번째 줄: / 6번째 줄: @@
 <h5 style="line-height: 3.428em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.166em; background-image: ; background-color: initial; background-position: 0px 100%;">개요</h5>
+[[중심이항계수(central binomial coefficient)]]
+<math>\sum_{n=1}^\infty \frac{1}{n^3\binom{2n}{n}}=\frac{\pi\sqrt{3}}{18}(\psi^{(1)}(\frac{1}{3})-\psi^{(1)}(\frac{2}{3}))-\frac{4}{3}\zeta(3)</math>
+여기서 <math>\psi^{(1)}</math>는 트리감마(trigamma)함수.
@@ 39번째 줄: / 55번째 줄: @@
 http://www.wolframalpha.com/input/?i=i*0.5*(-(i+trigamma(1/6))/(12+sqrt(3))-(i+trigamma(1/3))/(12+sqrt(3))%2B(i+trigamma(2/3))/(12+sqrt(3))%2B(i+trigamma(5/6))/(12+sqrt(3)))
+http://mathworld.wolfram.com/GiesekingsConstant.html