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I invented a function, somewhat similar to that of the Riemann zeta, but replacing the reciprocal of powers by superexponentiation. Namely,

$\mathfrak{H}(z) = \sum_{n=1}^{\infty} \frac{1}{{}^z n}$ where by definition ${}^z 1 = 1$

This seems to diverge for z < 1.

I tried many methods of analytically continue it in the complex plane analogues to zeta but none of them seems to be that useful. But Cohen-Villegas-Zagier's acceleration (CVZ) method seems to be converging it in a unique way. But I haven't dared to prove whether CVZ gives the right or even an analytic continuation.

There are many interesting properties I found under CVZ like : H^[m](z) seems to be converging towards either -7.0744329865020 + i39.959874355410 or -7.0744329865020 - i39.959874355410. If we denote these values by z1 and z2, it seems that H(z1) = z2. And also, interestingly, z1 = z2* ! These are surely not fixed points but what are they?

I've also posted this in a forum : http://www.mymathforum.com/viewtopic.php?f=15&t=38901

I've added a little details there so please read it. (I am too lazy to include those detailed informations here right now)

Any comments and graphs would be appreciated,
Balarka
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(03/11/2013, 11:38 AM)Balarka Sen Wrote: [ -> ]I invented a function, somewhat similar to that of the Riemann zeta, but replacing the reciprocal of powers by superexponentiation. Namely,

$\mathfrak{H}(z) = \sum_{n=1}^{\infty} \frac{1}{{}^z n}$ where by definition ${}^z 1 = 1$
(...)
I've also posted this in a forum : http://www.mymathforum.com/viewtopic.php?f=15&t=38901

Hi Balarka,
I've looked at your computation of H(3), where your method seems to work incorrect. Look at the first five summands only, we get
Code:
vectorv(5,k,1/k^k^k)*1.0 %977 = [1.00000000000, 0.0625000000000, 1.31137265240 E-13, 7.45834073120 E-155, 5.23282787918 E-2185]~
and its pretty obvious, that the sum is near 1.0625000.. instead of your 1.0625508... value

Gottfried
Gottfried Helms Wrote:I've looked at your computation of H(3), where your method seems to work incorrect.

Oops, sorry. I computed the sum with the integrand 1/(n^n)^n instead of 1/n^(n^n); interesting that they are a very good approximation to each other, can this be a coincidence?

PS : I have edited my post on MMF (My Math Forum)
Nobody's interested? I thought it worthed a great amount of attention . . .
Im not sure if this function is analytic. Im not an expert for CVZ ; I do not know what it does for nonanalytic functions ?

Why is this function special ? It looks nice but Im not sure about having properties ...

Some intrest though

regards

tommy1729
tommy1729 Wrote:Why is this function special ? It looks nice but Im not sure about having properties ...

I can't answer any of these unless I have a strong base. As in my post in MyMathForum that I linked to, it has zeros at the negative integers and got 2 poles at z =1 and z = -1, respectively.

The most interesting behavior I found is that the H(z) = H(z*)* where '*' denotes the conjugate.

However I can't ensure any of these unless I get a good, well-formed analytic continuation. I request everyone here interested in this to help me at this issue. CVZ certainly isn't a bad idea but I still need to verify it.

The thing in which I am most interested is if H(z) has any non trivial zeros on the complex plane.

Balarka
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(03/15/2013, 09:42 AM)Balarka Sen Wrote: [ -> ]
tommy1729 Wrote:Why is this function special ? It looks nice but Im not sure about having properties ...

The most interesting behavior I found is that the H(z) = H(z*)* where '*' denotes the conjugate.

However I can't ensure any of these unless I get a good, well-formed analytic continuation.

The conjugate thing is just as i would expect. It holds for each term and hence also for its sum.

I repeat ; im not sure if this function is analytic.

regards

tommy1729
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.