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Hyperzeta function
#8
tommy1729 Wrote:im not sure if this function is analytic.

I have no idea how to show that. The function is infinitely differentiable on the region Re(z) > 1 but showing that the taylor series converges exactly to that function on the domain of interest could be a hard thing to prove. It seems computationally right though, since I have constructed the taylor series at z = 2 with 5 terms which seems to be working good around the neighborhood of z = 2.

BTW, as a matter of fact, I found a zero of the function which is situated at (approximately) z = 1.295 + i * 0.522. Interestingly enough, the zero is at Re(z) > 1 implying that it is computable by hand without even accelerating it. Here is the method I used compute the hyperzeta function at different points :

Code:
\r kneserquiet.gp
tet(x, y) = init(x); sexp(y)
H(n, z) = 1 + sum(k = 2, n, 1/tet(k, z))

You could use 1 + sumalt(k = 2, 1/tet(k, z)) to compute the value faster (it takes approximately 12 minutes) although one shouldn't trust the value at Re(z) < 1 since it analytically continues the function at that region.
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Messages In This Thread
Hyperzeta function - by Balarka Sen - 03/11/2013, 11:38 AM
RE: Hyperzeta function - by Gottfried - 03/11/2013, 03:16 PM
RE: Hyperzeta function - by Balarka Sen - 03/11/2013, 06:32 PM
RE: Hyperzeta function - by Balarka Sen - 03/12/2013, 07:41 PM
RE: Hyperzeta function - by tommy1729 - 03/12/2013, 11:57 PM
RE: Hyperzeta function - by Balarka Sen - 03/15/2013, 09:42 AM
RE: Hyperzeta function - by tommy1729 - 03/16/2013, 12:18 AM
RE: Hyperzeta function - by Balarka Sen - 05/08/2013, 01:36 PM

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