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The general Mittag-Leffler function $$E_{a,b}(z)=\sum_{h=0}^{\infty}\frac{z^h}{\Gamma(ha+b)}$$ satifies the recurrence $$E_{a,b}(z)=zE_{a,b+a}(z)+\frac1{\Gamma(b)}.$$ I am having a hard time in proving this recurrence. Does it follow immediately from definition. I am not getting it.I think the property of Gamma function is used here. Any easy proofs? Thanks beforehand.

metamorphy
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vidyarthi
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1 Answers1

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Well $\;\displaystyle zE_{a,b+a}(z)=\sum_{h=0}^{\infty}\frac{z^{h+1}}{\Gamma((h+1)a+b)}$ so yes it is immediate.

For other interesting properties of the Mittag-Leffler function see this thread.

Raymond Manzoni
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  • Thanks!, yes now its immediate didnt get that one though Just the work of paranthesis – vidyarthi Apr 23 '22 at 15:22
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    Mittag-Leffler's function allows to make very nice mathematics starting with simple things like the "class 3" from this answer to very deep mathematics. Excellent excursions! – Raymond Manzoni Apr 23 '22 at 15:33
  • So your answer in that thread was inspired by Mittag-Leffler function? – vidyarthi Apr 23 '22 at 16:13
  • When I updated my answer with the "class 3" I noticed that it could be generalized to non integer values of the denominators and further linked to Mittag-Leffler (the relation between integrals and series is very interesting too). – Raymond Manzoni Apr 23 '22 at 16:52