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

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$$e_q(z) = \displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{[k]_q!} = \displaystyle\sum_{k=0}^{\infty} \dfrac{z^k(1-q)^k}{(q;q)_k}=\displaystyle\sum_{k=0}^{\infty} z^k \dfrac{(1-q)^k}{(1-q^k)(1-q^{k-1})\ldots(1-q)},$$
 
$$e_q(z) = \displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{[k]_q!} = \displaystyle\sum_{k=0}^{\infty} \dfrac{z^k(1-q)^k}{(q;q)_k}=\displaystyle\sum_{k=0}^{\infty} z^k \dfrac{(1-q)^k}{(1-q^k)(1-q^{k-1})\ldots(1-q)},$$
 
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]].
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=Properties=
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{{:Q-Euler formula for e sub q}}

Revision as of 23:46, 3 May 2015

The $q$-exponential $e_q$ is defined by the formula $$e_q(z) = \displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{[k]_q!} = \displaystyle\sum_{k=0}^{\infty} \dfrac{z^k(1-q)^k}{(q;q)_k}=\displaystyle\sum_{k=0}^{\infty} z^k \dfrac{(1-q)^k}{(1-q^k)(1-q^{k-1})\ldots(1-q)},$$ where $[k]_q!$ denotes the $q$-factorial and $(q;q)_k$ denotes the $q$-Pochhammer symbol.

Properties

Theorem

The following formula holds: $$e_q(iz)=\cos_q(z)+i\sin_q(z),$$ where $e_q$ is the $q$-exponential $e_q$, $\cos_q$ is the $q$-$\cos$ function and $\sin_q$ is the $q$-$\sin$ function.

Proof

References