라마누잔 타우 함수 라마누잔 타우 함수(Ramanujan Tau Function) τ ( n ) {\displaystyle \;\;\tau (n)} 는 타우 함수(Tau Function)로도 불린다.[1] n {\displaystyle n} 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 τ ( n ) {\displaystyle \tau (n)} 1 −24 252 −1472 4830 −6048 −16744 84480 −113643 −115920 534612 −370944 −577738 401856 1217160 987136 = 1 x 1 − 24 x 2 + 252 x 3 − 1472 x 4 + 4830 x 5 − 6048 x 6 + . . . . {\displaystyle \qquad =1x^{1}-24x^{2}+252x^{3}-1472x^{4}+4830x^{5}-6048x^{6}+....} = x ( 1 − 3 x 1 + 5 x 3 − 7 x 6 + . . . . ) 8 {\displaystyle \qquad =x(1-3x^{1}+5x^{3}-7x^{6}+....)^{8}} = x ∏ n = 1 ∞ ( 1 − x n ) 24 {\displaystyle \qquad =x\prod _{n=1}^{\infty }(1-x^{n})^{24}} = ∑ n = 1 ∞ τ ( n ) x n {\displaystyle \qquad =\sum _{n=1}^{\infty }\tau (n)x^{n}} 라마누잔의 계산식[편집] τ ( n ) ( n − 1 ) = ∑ k = 1 b n ( − 1 ) k + 1 ( 2 k + 1 ) ( n − 1 − 9 K ) τ ( n − K ) {\displaystyle \tau (n)(n-1)=\sum _{k=1}^{b_{n}}(-1)^{k+1}(2k+1)\left(n-1-9K\right)\tau \left(n-K\right)} b n = 8 n + 1 − 1 2 {\displaystyle b_{n}={{{\sqrt {8n+1}}-1} \over 2}} K = k ( k + 1 ) 2 {\displaystyle K={{k(k+1)} \over 2}} 따라서, τ ( n ) ( n − 1 ) = ∑ k = 1 b n ( − 1 ) k + 1 ( 2 k + 1 ) ( n − 1 − 9 K ) τ ( n − K ) {\displaystyle \tau (n)(n-1)=\sum _{k=1}^{b_{n}}(-1)^{k+1}(2k+1)\left(n-1-9K\right)\tau \left(n-K\right)} 데데킨트 에타 함수와의 관계[편집] ∑ n = 1 ∞ τ ( n ) x n = η ( τ ) 24 = Δ ( τ ) {\displaystyle \sum _{n=1}^{\infty }\tau (n)x^{n}=\eta (\tau )^{24}=\Delta (\tau )} x = e 2 π i τ {\displaystyle x=e^{2\pi i\tau }} η ( τ ) {\displaystyle \eta (\tau )} 데데킨트 에타 함수 같이 보기[편집] 약수함수 L-함수 각주[편집] ↑ http://mathworld.wolfram.com/TauFunction.html