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(almost) proof of TPID 13 - Printable Version

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(almost) proof of TPID 13 - fivexthethird - 05/06/2016

Actually, the statement I'm proving is more general:

Theorem: Let be holomorphic and bounded on the right half-plane for some . Then is equal to its newton series starting at 0 on that half-plane,

We need the following very simple lemma:
Lemma: Let be the analytic continuation of the mellin transform. Then if
1. The sum is absolutely convergent for all x
2. The are all holomorphic.
3. The derivative of the sum at 0 is equal to its term-wise derivative at 0
Proof: The sum and the integral are trivially interchanged. The other term is just

The inner sum is clearly absolutely convergent, so we can interchange the sums. Then we can add the two sums of the transform term-wise to get the result.
A more general result is most likely well-known but I haven't found any proof of it.

Now, satisfies the conditions for Ramanujan's master theorem to hold, so we have :



As the Mellin transform will converge when , the result follows.

Of course, this isn't quite what TPID 13 actually wants: this proves convergence of the newton series of starting at every , but not starting at the desired .


RE: (almost) proof of TPID 13 - JmsNxn - 05/06/2016

Cant we just take the limit as ? Namely







And therefore



Granted showing the limit can be pulled through is trivial. Maybe I'm missing something though.