Talk:Reciprocal times Derivative of Gamma Function

Should I replace a simpler proof with the existing proof?

I discover a simpler proof. The starting point is also Weierstrass form, but using logarithm and derivative to prove it.

This will yield to same result but can be more understood by people.

So should I replace it with the existing proof or just add a new proof?

And sorry for my bad English. --Henry kong (talk) 05:22, 10 January 2016 (UTC)

See Help:FAQ/Questions about contributions/You undid my corrections. You can name the existing proof "Proof 1" and put your proof below, as "Proof 2".

Please be sure to follow the house style as close as possible. — Lord_Farin (talk) 10:41, 10 January 2016 (UTC)

I see you removed the Digamma Function category tag from this page. Would you be amenable to reconsidering that edit? --Robkahn131 (talk) 01:09, 11 January 2024 (UTC)

Well no, because the digamma function is not referenced anywhere on that page.

While we "know" that $\dfrac {\map {\Gamma'} z} {\map \Gamma z}$ actually is the digamma function, this can only confuse the reader who knows nothing about this and doesn't know what the digamma function is.

If we decide to refactor this and rebuild the structure of all this material so as to remove $\dfrac {\map {\Gamma'} z} {\map \Gamma z}$ and replace it with $\map \psi z$ then that would be appropriate, but I am trying to reduce the incidence of material being in categories that are not directly relevant. --prime mover (talk) 07:30, 11 January 2024 (UTC)