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

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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  
 
If $|q|>1$ or the pair $0 < |q| <1$ and $|z| < \dfrac{1}{|1-q|}$ hold, then the $q$-exponential $E_q$ is  
 
$$E_q(z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{[k]_q!},$$
 
$$E_q(z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{[k]_q!},$$
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=Properties=
 
=Properties=
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[[Meromorphic continuation of q-exponential E sub q]]<br />
<strong>Theorem:</strong> The following [[meromorphic continuation]] of $E_q$ holds:
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[[Q-difference equation for q-exponential E sub q]]<br />
$$E_q(z)=\dfrac{1}{(z(1-q);q)_{\infty}},$$
 
where $(z(1-q);q)_{\infty}$ denotes the [[q-Pochhammer symbol]].
 
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<strong>Proof:</strong> █
 
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=See also=
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[[q-Cos]]<br />
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[[q-exponential E sub 1/q]]<br />
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[[q-Sin]]<br />
  
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=References=
<strong>Theorem:</strong> The following formula holds:
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* {{PaperReference|q-exponential and q-gamma functions. I. q-exponential functions|1994|D.S. McAnally|prev=findme|next=q-Factorial}} $(3.2)$ (calls $E_q$ $\exp_q$)
$$D_q E_q(z) = aE_q(az),$$
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* {{BookReference|Quantum Calculus|2002|Victor Kac|author2=Pokman Cheung||prev=findme|next=findme}} $(9.5)$ (calls $E_q(x)$ $e_q^x$)
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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* {{BookReference|A Comprehensive Treatment of q-Calculus|2012|Thomas Ernst|prev=findme|next=Meromorphic continuation of q-exponential E sub q}}: ($6.150$)
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<strong>Proof:</strong> █
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[[Category:SpecialFunction]]
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Latest revision as of 04:27, 26 December 2016

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

Meromorphic continuation of q-exponential E sub q
Q-difference equation for q-exponential E sub q

See also

q-Cos
q-exponential E sub 1/q
q-Sin

References