What is the convergence radius of this power series?
#4
(05/27/2011, 08:06 PM)JmsNxn Wrote: Well the next line must be true, because, using pari gp

\( \ln(x) = e^{-x} (\sum_{n=0}^{\infty} \frac{x^n}{n!} \psi_0(n+1)) \)
this series converges, at least for values like ln(20), ln(100), and numbers just greater than e they have very close convergence, maybe 4 to 6 decimal places.

This would mean that
\( \ln(x)e^x \) could be developed into a powerseries at 0.
But obviously it still has a singularity there.
This would imply that \( \ln(0)e^0 = \psi_0(1) \)
and that the first derivative of \( \ln(x)e^x \) at 0 is \( \psi_0(2) \).

And both is wrong AFAIK, the digamma function is finite on 1 and 2 while all derivatives of ln(x)e^x at 0 are infinite.
But then it seems there must some other error in the deriviations.
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Messages In This Thread
RE: What is the convergence radius of this power series? - by bo198214 - 05/28/2011, 09:35 AM

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