How is evaluated iteratively?

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- Mar 24th 2011, 06:21 AMCairoGamma Function
How is evaluated iteratively?

- Mar 24th 2011, 06:27 AMchisigma
- Mar 24th 2011, 08:23 AMCairo
Sorry.

Yes it is real - as in . - Mar 24th 2011, 01:05 PMDrexel28
- Mar 24th 2011, 01:14 PMchisigma
- Mar 24th 2011, 01:14 PMCairo
- Mar 24th 2011, 01:15 PMCairo
Thanks chisigma.

But is that the same expression that Drexel gets? - Mar 24th 2011, 09:40 PMDrexel28
Certainly. But, my original proof didn't work out. So I found an alternate...but it's fairly technical. You might find 's suggestion more useful. Also, I did find an alternate proof on page 22 of this book (a very good book by the wayP may be found for a more general result. Regardless...

For define . Note firstly then that . Moreover,

Finally, note that . So that where is the digamma function. But, a quick check (using digamma's series expression) shows then that and thus is convex, and so is log convex. The result then follows from the Bohr-Mollerup theorem.