Difference between revisions of "Basic hypergeometric series psi"
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(Created page with "The bilateral basic hypergeometric series $\psi$ is defined by $${}_j\psi_{\ell}(a_1,\ldots,a_j;b_1,\ldots,b_k;q,z)=\displaystyle\sum_{k=-\infty}^{\infty} \dfrac{(a_1;q)_k\ldo...") |
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The bilateral basic hypergeometric series $\psi$ is defined by | The bilateral basic hypergeometric series $\psi$ is defined by | ||
$${}_j\psi_{\ell}(a_1,\ldots,a_j;b_1,\ldots,b_k;q,z)=\displaystyle\sum_{k=-\infty}^{\infty} \dfrac{(a_1;q)_k\ldots(a_j;q)_k}{(b_1;q)_k\ldots(b_{\ell};q)_k}\left( (-1)^k q^{ {k \choose 2} } \right)^{\ell-j}z^k.$$ | $${}_j\psi_{\ell}(a_1,\ldots,a_j;b_1,\ldots,b_k;q,z)=\displaystyle\sum_{k=-\infty}^{\infty} \dfrac{(a_1;q)_k\ldots(a_j;q)_k}{(b_1;q)_k\ldots(b_{\ell};q)_k}\left( (-1)^k q^{ {k \choose 2} } \right)^{\ell-j}z^k.$$ | ||
| + | |||
| + | =Properties= | ||
| + | <div class="toccolours mw-collapsible mw-collapsed"> | ||
| + | <strong>Theorem:</strong> The following formula holds: | ||
| + | $${}_1\psi_1(a,b;q,z) = \dfrac{\left(\frac{b}{a};q \right)_{\infty} (q,q)_{\infty} \left( \frac{q}{az};q \right)_{\infty} (az;q)_{\infty} }{(b;q)_{\infty} \left( \frac{b}{az};q \right)_{\infty} \left( \frac{q}{a};q \right)_{\infty} (z;q)_{\infty}}.$$ | ||
| + | <div class="mw-collapsible-content"> | ||
| + | <strong>Proof:</strong> █ | ||
| + | </div> | ||
| + | </div> | ||
| + | |||
| + | =See Also= | ||
| + | [[Basic hypergeometric phi]] | ||
| + | |||
| + | [[Category:SpecialFunction]] | ||
Latest revision as of 21:38, 17 June 2017
The bilateral basic hypergeometric series $\psi$ is defined by $${}_j\psi_{\ell}(a_1,\ldots,a_j;b_1,\ldots,b_k;q,z)=\displaystyle\sum_{k=-\infty}^{\infty} \dfrac{(a_1;q)_k\ldots(a_j;q)_k}{(b_1;q)_k\ldots(b_{\ell};q)_k}\left( (-1)^k q^{ {k \choose 2} } \right)^{\ell-j}z^k.$$
Properties
Theorem: The following formula holds: $${}_1\psi_1(a,b;q,z) = \dfrac{\left(\frac{b}{a};q \right)_{\infty} (q,q)_{\infty} \left( \frac{q}{az};q \right)_{\infty} (az;q)_{\infty} }{(b;q)_{\infty} \left( \frac{b}{az};q \right)_{\infty} \left( \frac{q}{a};q \right)_{\infty} (z;q)_{\infty}}.$$
Proof: █
