View source for Q-difference equation for q-exponential E sub 1/q

==Theorem==
The following formula holds:
$$D_q E_{\frac{1}{q}}(az)=aE_{\frac{1}{q}}(qaz),$$
where $D_q$ denotes the [[q-difference operator]] and $E_{\frac{1}{q}}$ denotes the [[Q-exponential E sub 1/q|$q$-exponential $E_{\frac{1}{q}}$]].

==Proof==

==References==
* {{BookReference|A Comprehensive Treatment of q-Calculus|2012|Thomas Ernst|prev=q-exponential E sub 1/q|next=Product representation of q-exponential E sub 1/q}}: ($6.154$)

[[Category:Theorem]]
[[Category:Unproven]]