Difference between revisions of "Product representation of q-exponential E sub 1/q"
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The following formula holds for $0 \leq |q| \leq 1$: | The following formula holds for $0 \leq |q| \leq 1$: | ||
$$E_{\frac{1}{q}}(z) = \displaystyle\prod_{k=0}^{\infty} \left[ 1 + (1-q)zq^k \right],$$ | $$E_{\frac{1}{q}}(z) = \displaystyle\prod_{k=0}^{\infty} \left[ 1 + (1-q)zq^k \right],$$ | ||
| − | where $E_{\frac{1}{q}}$ denotes the [[Q-exponential E sub 1/q]]. | + | where $E_{\frac{1}{q}}$ denotes the [[Q-exponential E sub 1/q|$q$-exponential $E_{\frac{1}{q}}$]]. |
==Proof== | ==Proof== | ||
==References== | ==References== | ||
| − | * {{BookReference|A Comprehensive Treatment of q-Calculus|2012|Thomas Ernst|prev=Q-difference equation for q-exponential E sub 1/q|next=}}: (6.155) | + | * {{BookReference|A Comprehensive Treatment of q-Calculus|2012|Thomas Ernst|prev=Q-difference equation for q-exponential E sub 1/q|next=Limit of q-exponential E sub 1/q for 0 less than q less than 1}}: ($6.155$) |
[[Category:Theorem]] | [[Category:Theorem]] | ||
[[Category:Unproven]] | [[Category:Unproven]] | ||
Latest revision as of 07:41, 18 December 2016
Theorem
The following formula holds for $0 \leq |q| \leq 1$: $$E_{\frac{1}{q}}(z) = \displaystyle\prod_{k=0}^{\infty} \left[ 1 + (1-q)zq^k \right],$$ where $E_{\frac{1}{q}}$ denotes the $q$-exponential $E_{\frac{1}{q}}$.
Proof
References
- 2012: Thomas Ernst: A Comprehensive Treatment of q-Calculus ... (previous) ... (next): ($6.155$)
