[Joey_21]

Would anybody (namely CP5670 or others who have experience in this area ) happen to know an identity for the Hurwitz Zeta function to allow me to calculate Zeta(s, 1 / 4) and/or Zeta(s, 3 / 4)?

[CP5670]

lol, I was actually looking for a very similar thing just a few days ago (formulas for 1/3, 1/4 and 1/6 arguments) but it was for the zeta derivative; something of a coincidence there. Anyway, you could try Rademacher's formula:


(p and q being integers with 0 < p < q )

of course, this is really only useful for integer s (positive integers really, since the negatives reduce to simple bernoulli polys anyway). There is also the basic zeta addition formula but that can only express a 1/4 in terms of a 3/4 (along with 1/2 and 1) or the other way around.

[Joey_21]

Yea, that identity sorta really doesn't help because I've seen it already.

Here's my work so far:
4^s * Zeta(s) = Zeta(s, 1/4) + Zeta(s, 1/2) + Zeta(s, 3/4) + Zeta(s)

Zeta(s, 1/2) = (2^s-1) * Zeta(s)

4^s * Zeta(s) - Zeta(s) - (2^s-1) * Zeta(s) = Zeta(s, 1/4) + Zeta(s, 3/4)... looking for a way to either extract the two or find some kind of ratio that'll transform the Zeta(s, 1/4) + Zeta(s, 3/4) into a Zeta(s, 1/4) - Zeta(s, 3/4) using only the left-hand arguments.

[Petrarch of the VBB]

I am most definitely NOT taking Maths A-Level, after looking at this.

[Joey_21]

Originally posted by Petrarch of the VBB
I am most definitely NOT taking Maths A-Level, after looking at this.

I'm not doing this project for class, I'm doing it for fun.

(look in siggy for calculator)

Last thing we even talked about in Calculus I class was how to integrate and antiderive so you can see I've done a bit of recreational study of my own.

[CP5670]

lol, same case here...

yeah, that first thing is the addition formula I was talking about. You might be able to put it in terms of just z(s) though, if that's what you are looking for.

[edit] hmm, on second thought, it looks like that is not so easy after all; let me see if this other thing I have works...

[Joey_21]

Well, as you see by the last part of that post I'm trying to extract Zeta(s, 1/4) and Zeta(s, 3/4) or make it to where those two arguments are subtracted because I'm trying to put the Dirichlet Beta function into my calculator because:
Beta(x) = 1/4^s * [Zeta(s, 1/4) - Zeta(s, 3/4)]

The summation series for the Dirichlet Beta function just doesn't suffice.

Edit: Note, I already have a fully-working Riemann Zeta function in my calculator (although not uploaded at the mo) so if everything could be in terms of it that would work fine.

[CP5670]

Ah, I think I see what you want to do now; you want to put the z(s,1/4) - z(s,3/4) bit into a single zeta, right? I think the closest you can get to that is to eliminate just one of either the 1/4 or the 3/4 and replace it with a 1 (normal riemann zeta). Or you could turn it into a lerch/polylog function.

Unfortunately, I haven't seen any general formula for z(s,p/q) that does not involve other zetas except for s=1 with the pole subtracted out. There might be other such things for zeta derivatives though.

If you just need a numerical evaluation though, you could try using one of the several zeta integrals and applying some numerical technique on those.

[Joey_21]

Hmm, I suppose I could try the lerch transcendental. It looked a bit scary at first so I thought I would save it for last. I've been checking my work with a polygamma function I programmed since


but, unfortunately, the polygamma function can't have non-integer derivatives... or can it?

[CP5670]

The m there can actually be any complex number (using that liouville fraction integral), but such a polygamma can really only be expressed with zeta functions (or with a KDF two-variable hypergeometric, but it is far more messy that way). Almost all of the formulas for the lerch function are slightly more complicated versions of the zeta formulas and the proofs are nearly identical too, so it is pretty easy to work with if you have the zeta things.

hmm, now that I think of it, I think the 1/4 and 3/4 thing that you originally wanted is indeed possible (i.e. putting them in terms of 1), but only for odd integer s.

[Joey_21]

I tried the lerch transcendent... there wouldn't happen to be any recurrence identities for the Dirichlet Beta function, would there?

When 0 < n < 6, the summation diverges, and adding more terms doesn't help much either.

[CP5670]

0 to 6? The Lerch transcendant has a pole at z=1 so the power series only goes that far in z. There are some transformation formulas to take it beyond z=1, but its values there are not purely real anyway (a bit like the normal logarithm).

The Dirichlet function appears to have a variable first argument in its zeta functions, so there would probably not be any recurrence relation for it (a recurrence formula in s for z(s,x) would be completely revolutionary and solve almost all the zeta function problems ).

If you are using series evaluations though you can just use the sum and reflection equation for the dirichlet function given on the Mathworld site.

[Joey_21]

Originally posted by CP5670
The Dirichlet function appears to have a variable first argument in its zeta functions, so there would probably not be any recurrence relation for it (a recurrence formula in s for z(s,x) would be completely revolutionary and solve almost all the zeta function problems ).

Oops. I accidentally typed 'x' instead of 's'. The Dirichlet Beta function actually only involves one argument.

Here's what I've gathered so far:

[Zeta(s, 1/4) + Zeta(s, 3/4)] / 2^s = Zeta(s, 1/2)

Zeta(s, 1/2) / (2^s - 1) = Zeta(s)

Zeta(s) * (2^s - 1) * 2^s = Zeta(s, 1/4) + Zeta(s, 3/4)

And after all of that, it would appear that there's an identity to just seperate the right-hand pieces. Apparently this is not the case.

[diamondgeezer]

[Galemp]

I'll just stick with my Riemannian Manifolds, thank you very much.

[Joey_21]

Hmm... another identity I noticed:

-2 + [Zeta(s, 1/4) + Zeta(s, 3/4)] / Zeta(s, 3/4) = [Zeta(s, 1/4) - Zeta(s, 3/4)] / Zeta(s, 3/4)

Although it doesn't help me out just yet, this may help with finding a pattern.

[CP5670]

The problem there is sort of the similar to the one that arises in simplifying G(1/4); too many extra terms come up in the addition/multiplication formulas for it to be of any use. I think that in general it is only possible to get such a result for z(s,1/2). There are some such values known for specific values of s, but not for a completely general s. Here is the reflection equation I mentioned earlier:

e^pis/2 z(s, x) + e^-pis/2 z(s, 1-x) = (2p)^s e^2pix F(e^2pix, 1-s, 1) / G(s)

Although this holds for all s, notice that the e coefficients on the left only become ±1 or ±i for integer s and even then, the difference of the two zetas (i.e. opposite signs) can only be obtained for odd s.

[Joey_21]

Calculators don't really hold imaginary units very well, so meh.

[J.F.K.]

Originally posted by Joey_21
I'm not doing this project for class, I'm doing it for fun.

(look in siggy for calculator)

Dude, that's totally cool. I only understand about half of what you're saying, so extra kudos to you

[Arnav]

Originally posted by Petrarch of the VBB
I am most definitely NOT taking Maths A-Level, after looking at this.

Yeah, do that. Advanced Subsidiary Maths was bad enough... once I nearly cried, it was that bad.

stay away from maths, leave it to people who do it for "fun".