"Basic hypergeometric series"의 두 판 사이의 차이

2020년 11월 13일 (금) 17:22 판

오일러의 오각수정리(pentagonal number theorem)\((1-x)(1-x^2)(1-x^3) \cdots = 1 - x - x^2 + x^5 + x^7 - x^{12} - x^{15} + x^{22} + x^{26} + \cdots\)
오일러공식\(\prod_{n=0}^{\infty}(1+zq^n)=\sum_{n\geq 0}\frac{q^{n(n-1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)} z^n\)

\(\sum_{n\geq 0}^{\infty}\frac{q^{n^2/2}}{(q)_n}\sim \exp(\frac{\pi^2}{12t}-\frac{t}{48})\)

f[q_] := QHypergeometricPFQ[{}, {}, q, -q^(1/2)] g[q_] := Exp[-(Pi^2/(12 Log[q])) + Log[q]/48] Table[N[f[1 - 1/10^(i)]/g[1 - 1/10^(i)], 50], {i, 1, 5}] // TableForm

@@ 1번째 줄: / 1번째 줄: @@
 ==theory==
-* [http://pythagoras0.springnote.com/pages/4145675 오일러의 오각수정리(pentagonal number theorem)]<br><math>(1-x)(1-x^2)(1-x^3) \cdots = 1 - x - x^2 + x^5 + x^7 - x^{12} - x^{15} + x^{22} + x^{26} + \cdots</math><br>
+* [http://pythagoras0.springnote.com/pages/4145675 오일러의 오각수정리(pentagonal number theorem)]<math>(1-x)(1-x^2)(1-x^3) \cdots = 1 - x - x^2 + x^5 + x^7 - x^{12} - x^{15} + x^{22} + x^{26} + \cdots</math>
-*  오일러공식<br><math>\prod_{n=0}^{\infty}(1+zq^n)=\sum_{n\geq 0}\frac{q^{n(n-1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)} z^n</math><br>
+*  오일러공식<math>\prod_{n=0}^{\infty}(1+zq^n)=\sum_{n\geq 0}\frac{q^{n(n-1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)} z^n</math>
@@ 28번째 줄: / 28번째 줄: @@
-# f[q_] := QHypergeometricPFQ[{}, {}, q, -q^(1/2)]<br> g[q_] := Exp[-(Pi^2/(12 Log[q])) + Log[q]/48]<br> Table[N[f[1 - 1/10^(i)]/g[1 - 1/10^(i)], 50], {i, 1, 5}] // TableForm
+# f[q_] := QHypergeometricPFQ[{}, {}, q, -q^(1/2)] g[q_] := Exp[-(Pi^2/(12 Log[q])) + Log[q]/48] Table[N[f[1 - 1/10^(i)]/g[1 - 1/10^(i)], 50], {i, 1, 5}] // TableForm