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Feb 8, 2019 at 4:04 vote accept Richard Diagram
Feb 7, 2019 at 3:34 comment added Richard Diagram I think you're on to something here. The more I mess around with this the more I'm doubting its truth. I think I need further conditions on $f$ to get this to work; or I need to finesse the asymptotics more. Frankly, I could only prove $\mathcal{I}_n$ is bounded as $n \to \infty$. I can't even prove it converges to $0$. I may have to do a work around, and take a more complicated path. Thanks for your answer though, I don't know why I didn't think of using the Schroder function more explicitly. I was approaching this using the Mellin transform, and the identity $I_{n+1}(z) - I_n(z) = I_n(f(z))$
Feb 7, 2019 at 2:55 history answered Joe Silverman CC BY-SA 4.0