Difference between revisions of "Q-exponential E sub q"

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The $q$-exponential $E_q$ is  
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If $|q|>1$ or the pair $0 < |q| <1$ and $|z| < \dfrac{1}{|1-q|}$ hold, then the $q$-exponential $E_q$ is  
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$$E_q(z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{[k]_q!} = \displaystyle\sum_{k=0}^{\infty} \dfrac{z^k(1-q)^k}{(q;q)_k}=\displaystyle\sum_{k=0}^{\infty} z^k \dfrac{(1-q)^k}{(1-q^k)(1-q^{k-1})\ldots(1-q)},$$
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$$E_q(z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{[k]_q!},$$
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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 $e$]].
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where $[k]_q!$ denotes the [[q-factorial|$q$-factorial]].
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=Properties=
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<div class="toccolours mw-collapsible mw-collapsed">
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<strong>Theorem:</strong> The following [[meromorphic continuation]] of $E_q$ holds:
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$$E_q(z)=\dfrac{1}{(z(1-q);q)_{\infty}},$$
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where $(z(1-q);q)_{\infty}$ denotes the [[q-Pochhammer symbol]].
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<div class="mw-collapsible-content">
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<strong>Proof:</strong> █
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</div>
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</div>
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<div class="toccolours mw-collapsible mw-collapsed">
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<strong>Theorem:</strong> The following formula holds:
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$$D_q E_q(z) = aE_q(az),$$
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where $D_q$ is the [[q-difference operator|$q$-difference operator]] and $E_q$ is the [[Q-exponential E sub q|$q$-exponential $E_q$]].
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<div class="mw-collapsible-content">
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<strong>Proof:</strong> █
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</div>
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</div>

Revision as of 21:39, 5 September 2015

If $|q|>1$ or the pair $0 < |q| <1$ and $|z| < \dfrac{1}{|1-q|}$ hold, then the $q$-exponential $E_q$ is $$E_q(z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{[k]_q!},$$ where $[k]_q!$ denotes the $q$-factorial.

Properties

Theorem: The following meromorphic continuation of $E_q$ holds: $$E_q(z)=\dfrac{1}{(z(1-q);q)_{\infty}},$$ where $(z(1-q);q)_{\infty}$ denotes the q-Pochhammer symbol.

Proof: █


Theorem: The following formula holds: $$D_q E_q(z) = aE_q(az),$$ where $D_q$ is the $q$-difference operator and $E_q$ is the $q$-exponential $E_q$.

Proof: █

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