"Basic hypergeometric series"의 두 판 사이의 차이
둘러보기로 가기
검색하러 가기
| 1번째 줄: | 1번째 줄: | ||
| − | Series[QPochhammer[q, q], {q, 0, 100}]<br> Series[\!\(<br> \*UnderoverscriptBox[\(\[Product]\), \(k = 1\), \(100\)]\((1 -<br> q^k)\)\), {q, 0, 100}]<br> f[q_] := \!\(<br> \*UnderoverscriptBox[\(\[Sum]\), \(k = 0\), \(100\)]\(PartitionsP[<br> k] q^k\)\)<br> Series[1/QPochhammer[q, q], {q, 0, 100}]<br> Series[f[q], {q, 0, 100}]<br> d[n_] := DivisorSigma[1, n]<br> g[q_] := \!\(<br> \*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(100\)]\(d[k] q^k\)\)<br> Expand[f[q]*g[q]] | + | # Series[QPochhammer[q, q], {q, 0, 100}]<br> Series[\!\(<br> \*UnderoverscriptBox[\(\[Product]\), \(k = 1\), \(100\)]\((1 -<br> q^k)\)\), {q, 0, 100}]<br> f[q_] := \!\(<br> \*UnderoverscriptBox[\(\[Sum]\), \(k = 0\), \(100\)]\(PartitionsP[<br> k] q^k\)\)<br> Series[1/QPochhammer[q, q], {q, 0, 100}]<br> Series[f[q], {q, 0, 100}]<br> d[n_] := DivisorSigma[1, n]<br> g[q_] := \!\(<br> \*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(100\)]\(d[k] q^k\)\)<br> Expand[f[q]*g[q]] |
| − | + | # | |
| − | |||
| 9번째 줄: | 8번째 줄: | ||
<math>\sum_{n\geq 0}^{\infty}\frac{q^{n^2/2}}{(q)_n}\sim \exp(\frac{\pi^2}{12t}-\frac{t}{48})</math> | <math>\sum_{n\geq 0}^{\infty}\frac{q^{n^2/2}}{(q)_n}\sim \exp(\frac{\pi^2}{12t}-\frac{t}{48})</math> | ||
| − | + | ||
| − | + | # f[q_] := QHypergeometricPFQ[{}, {}, q, -q^(1/2)]<br> g[q_] := Exp[-(Pi^2/(12 Log[q]))]<br> Table[N[f[2^(-i)]/g[2^(-i)], 10], {i, 5, 1000}] | |
2010년 3월 21일 (일) 09:22 판
- Series[QPochhammer[q, q], {q, 0, 100}]
Series[\!\(<br> \*UnderoverscriptBox[\(\[Product]\), \(k = 1\), \(100\)]\((1 -<br> q^k)\)\), {q, 0, 100}]
f[q_] := \!\(<br> \*UnderoverscriptBox[\(\[Sum]\), \(k = 0\), \(100\)]\(PartitionsP[<br> k] q^k\)\)
Series[1/QPochhammer[q, q], {q, 0, 100}]
Series[f[q], {q, 0, 100}]
d[n_] := DivisorSigma[1, n]
g[q_] := \!\(<br> \*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(100\)]\(d[k] q^k\)\)
Expand[f[q]*g[q]]
q-hypergeometric series
\(\sum_{n\geq 0}^{\infty}\frac{q^{n^2/2}}{(q)_n}\sim \exp(\frac{\pi^2}{12t}-\frac{t}{48})\)
- f[q_] := QHypergeometricPFQ[{}, {}, q, -q^(1/2)]
g[q_] := Exp[-(Pi^2/(12 Log[q]))]
Table[N[f[2^(-i)]/g[2^(-i)], 10], {i, 5, 1000}]