Difference between revisions of "Q-exponential E sub q"
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The $q$-exponential $E_q$ is | The $q$-exponential $E_q$ is | ||
$$E_q(z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{[k]_q!} = \displaystyle\sum_{k=0}^{\infty} \dfrac{z^k(1-q)^k}{(q;q)_k}=\displaystyle\sum_{k=0}^{\infty} z^k \dfrac{(1-q)^k}{(1-q^k)(1-q^{k-1})\ldots(1-q)},$$ | $$E_q(z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{[k]_q!} = \displaystyle\sum_{k=0}^{\infty} \dfrac{z^k(1-q)^k}{(q;q)_k}=\displaystyle\sum_{k=0}^{\infty} z^k \dfrac{(1-q)^k}{(1-q^k)(1-q^{k-1})\ldots(1-q)},$$ | ||
| − | where $[k]_q!$ denotes the [[q-factorial|$q$-factorial]] and $(q;q)_k$ denotes the [[q-Pochhammer symbol|$q$-Pochhammer symbol]]. | + | where $[k]_q!$ denotes the [[q-factorial|$q$-factorial]] and $(q;q)_k$ denotes the [[q-Pochhammer symbol|$q$-Pochhammer symbol]]. Note that this function is different than the [[q-exponential e|$q$-exponential $e$]]. |
Revision as of 17:57, 20 May 2015
The $q$-exponential $E_q$ is $$E_q(z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{[k]_q!} = \displaystyle\sum_{k=0}^{\infty} \dfrac{z^k(1-q)^k}{(q;q)_k}=\displaystyle\sum_{k=0}^{\infty} z^k \dfrac{(1-q)^k}{(1-q^k)(1-q^{k-1})\ldots(1-q)},$$ where $[k]_q!$ denotes the $q$-factorial and $(q;q)_k$ denotes the $q$-Pochhammer symbol. Note that this function is different than the $q$-exponential $e$.
