Talk:Hurwitz zeta function

Multiplication theorem

There is a useful generalization of the multiplication theorem for the Hurwitz zeta function. In Maple notation:

   sum(Zeta(s,a+p/q),p=1..q-1) = q^s*Zeta(s,q*a)

where (q-1) is a natural number. It reduces to the simpler form on writing a=1/q (and recalling that Zeta(s,1) is the Riemann zeta function). [The proof is trivial ...]

So put it into the article. The above equation in LaTeX is

$\sum _{p=1}^{q-1}\zeta (s,a+p/q)=q^{s}\,\zeta (s,qa)$

PAR 20:21, 3 August 2006 (UTC)

Sorry - I made an error. The lower bound of the sum (over p) should be zero, not unity! So -

$\sum _{p=0}^{q-1}\zeta (s,a+p/q)=q^{s}\,\zeta (s,qa)$

Hair Commodore 22:00, 30 October 2006 (UTC)

region of convergence

The series, and other formulas, should include a discussion of the regions of absolute and conditional convergence. I slapped an expert tag on the section introducing this idea. linas (talk) 20:36, 18 November 2007 (UTC)

Well my markup foo is lacking, but ..... The series always locally converges. At a series of small points arranged around a singularity pole, the radius converges upto the singularity, so the singularities are surrounded by convergence from all sides. Thye singular points of divergence are then not divergent ranges, but singular points which are singular, and so the function converges to the true value of the function in an infinite radius of convergence. The function is thus fully described in it's entirety, it is an or has an entire description. How do? Same man who added entire gamma formula to bottom of this page 188.29.164.25 (talk) 12:24, 27 January 2015 (UTC)

The Hurwitz zeta function generalizes the polygamma function:

How does it generalize PolyGamma if the identity in fact holds only for integers? Usually PolyGamma is generalized in a different way.--MathFacts (talk) 13:04, 14 November 2009 (UTC)

A representation of Gamma

Gamma[z] = constant Exp[D[(s-1) HurwitzZeta[s,p],s]] /. p -> z /. s-> 1

I think this proves Gamma is entire. It requires the use of L'Hopitals rule when in the limit, and using the differential in the second argument need integrating. And stuff the whole lot into the 0th polygamma definition. 188.29.165.77 (talk) 15:36, 26 January 2015 (UTC)

https://sites.google.com/site/thebohrmagneton/the-entire-gamma-function pdf bottom of page, final page has formula. just need the constant calculating to fit positives.

Gamma[z] = Const Exp[Sum[1/(n+1) Sum[(-1)^(k+1) Binomial[n, k] Log[k+z], {k, 0, n}], {n, 0, Infinity}]]

Assuming the binomial n!/((n-k)!k!) and z in Complex 188.29.164.25 (talk) 12:14, 27 January 2015 (UTC)

The nature of the entire description means the negative axis is correct, and the Euler graph is wrong. This is to do with the termination of the Euler definition of factorial ending as a count down to 1, and not continuing to minus infinity. So this ENTIRE gamma is based on an infinite product of the value of this ENTIRE gamma down to minus infinity.

The Euler gamma makes a mistake of including an absolute magnitude operation in the algorithm by carrying out an inverse plus one iteration on negative parameter values. This prevents entirity, and also using the Risch algorithm on the Euler gamma.

Why do differential equations need like the Bessel, or Bessel-Clifford need alternate solution functions? The negative gamma values? Another note on the multiplicative inverse of entire functions. Think about it! — Preceding unsigned comment added by 188.29.164.25 (talk) 12:38, 27 January 2015 (UTC)

You have discovered an identity for the (log of the) gamma function that (you claim) converges everywhere on the complex plane (except at the poles, of course). But that does not mean that the gamma function is "entire" -- its the same as it always was. The word "entire" does not mean what you think it means. It means entire function. The claim that it converges everywhere is dubious. You might want to do some numerical exploration. I see two issues: binomial coefficients get huge before they get small, so you risk having a sum that has immense, giant terms in the middle that have to delicately cancel each other out. The other issue is that you have an infinite number of logarithms in sum. Each logarithm has a cut. You can put the cut wherever you want, but managing them all is a trick. Understanding the monodromy is tricky. You should compare your formula to the very first formula given in the section called "the log-gamma function" here: gamma function. Interesting work, but you are not done yet! 67.198.37.16 (talk) 18:30, 25 January 2019 (UTC)