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

==Theorem==
The following formula holds:
$$D_q E_q(az) = aE_q(az),$$
where $D_q$ is the [[q-difference operator|$q$-difference operator]] and $E_q$ is the [[Q-exponential E sub q|$q$-exponential $E_q$]].

==Proof==

==References==
* {{BookReference|A Comprehensive Treatment of q-Calculus|2012|Thomas Ernst|prev=Meromorphic continuation of q-exponential E sub q|next=Q-exponential E sub 1/q}}: ($6.152$)

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