Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm wondering if there is any information on the inverse of the Riemann zeta function (not it's reciprocal, but its functional inverse). This would obviously be a multi-valued function.

share|improve this question
1  
What kind of information? –  Qiaochu Yuan Aug 9 '11 at 23:39
27  
Supposedly it takes some constrained values at $0$. –  Moosbrugger Aug 9 '11 at 23:39
2  
There are several books on the Riemann zeta function. Tell us which ones you've looked through, so we don't duplicate the effort you've already put in. –  Gerry Myerson Aug 9 '11 at 23:40
    
@Gerry: I haven't looked at any. I was just curious about this topic and I googled it, not finding anything even remotely related (when people say inverse, they usually mean reciprocal). I'm looking for papers or books that have considered constructing the branches of the inverse function. –  Victor Liu Aug 10 '11 at 0:33
add comment

3 Answers

I fiddled with this recently; however I did not yet arrive at an interesting result for the inverse of the $\zeta$. But one could use the alternating-zeta (or Dirichlet's eta- $\eta$) function. It is not too difficult to construct an invertible powerseries for the related eta-function; one can simply consider the sequence of formal powerseries for $ {1 \over (1+1)^x }, {1 \over (1+2)^x}, ... $ and adds the coefficients at the same powers of x. These produce non-convergent series, but which can be Euler-summed. In Pari/GP one does simply:

ps_eta = sumalt(k=0,(-1)^k*taylor(1/(1+k)^x,x)) 

and gets

$ \small pseta(x)= 0.500000 + 0.225791 x - 0.0305146 x^2 - 0.00391245 x^3 + 0.00208483 x^4 - 0.000312274 x^5 + O(x^{6}) $
From this a powerseries for zeta is also constructible:

ps_zeta = ps_eta /(1-2*2^-x) 
printtex(ps_zeta+taylor(1/(1-x),x))     

$ \small \begin{array} {lll} pszeta(x) &=& - 0.500000 + (0.0810615-1) x - (0.00317823+1) x^2 - (0.000785194+1) x^3 \\\ & & + (0.000120700-1) x^4 - (0.00000194090+1) x^5 + O(x^{6}) \end{array} $

(That powerseries is related to the power series using the Stieltjes-constant by replacing x by x+1 )


The power series for $\eta$ can be recentered at the fixpoint $ \small fp \sim 0.629334 $ to get a powerseries without a constant term which can then be inverted. Let's call this $\eta_{fp} = \eta (x+fp)-fp $ then the powerseries begins like

$ \small \begin{array} {lll} pseta_{fp}(x) &=& 0.184574 x - 0.0337023 x^2 + 0.000152965 x^3 + 0.00117594 x^4 \\\ & & - 0.000254950 x^5 + 0.0000216757 x^6 + 0.00000147274 x^7 - 0.000000714222 x^8 + O(x^{9}) \end{array} $

From this we can generate a powerseries for the inverse of $ \eta_{fp}$. The range of convergence is small; but using eulersummation one can compute values for the inverse of $ \eta_{fp}$. Even fractional iterates are accessible; here is a plot which shows the fractional iteration of $\eta(x,h)$, beginning at x=1 where h is the iteration-parameter (all is computed using the centered version $\eta_{fp}$ ).

The plot has to be read that at h=0 we have $\eta(x,0)=x=1 $, at h=1 we have $\eta(x,1)=\eta(x)= \log(2)$, at h=2 we have $\eta(x,2)=\eta(\eta(x))$ at h->inf we get the fixpoint fp and the inverse is at h=-1: $ \eta(x,-1)=\eta^{-1}(x) \to \infty $

fractional iteration of eta

I'm not yet ready with a small script/sketch of an article where I explore this in a bit more detail.
However, as I said, the inverse of zeta does not behave so nicely - the eigenvalue for the iteration is negative and one cannot uniquely get fractional roots out of this. Also the according powerseries may be too bad configured/has too small range of convergence.


[update] Here is a plot for a range of the inverse alternating zeta; for the "extreme" values at the borders I used Eulersummation because the power series has very small range of convergnce.

inverse eta

share|improve this answer
add comment

The question is about the value distribution of $\zeta(s)$; it is considered (without speaking of inverse) in some detail in Chapter XI of Titchmarsh's book.

share|improve this answer
add comment

The following Mathematica program gives the first few zeta zeros to good accuracy by applying series reversion or the inverse as you call it, twice:

Clear[n]
Do[
Clear[x, a, b, c, d];
 b = 10;
 c = N[2*Pi*Exp[1]*(n - 11/8)/Exp[1]/LambertW[(n - 11/8)/Exp[1]], 10];
 a = Normal[InverseSeries[Series[(Zeta[1/2 + I*x]), {x, c, b}], x]];
 x = 0;
 d = N[Re[a]];
 Clear[a, x];
 a = Normal[InverseSeries[Series[(Zeta[1/2 + I*x]), {x, d, b}], x]];
 x = 0;
 Print[N[Re[a]]], {n, 1, 20}]

14.1347, 21.022, 25.0109, 30.4249, 32.9351, 37.5862, 40.9187, 43.3271, 48.0052, 49.7738, 52.9703, 56.4462, 59.347, 60.8318, 65.1125, 67.0798, 69.5464, 72.0672, 75.7047, 77.1448

Of course by giving not the approximation of the zeros as a starting point, but the zeros them selves it converges even faster for $x=0$. Expanding for other values not close to zeros the function appears to be much less orderly when looking at a few values.

share|improve this answer
    
Is your Mathematica program based on the rather recent preprint of LeClair and França? –  Sylvain JULIEN Dec 18 '13 at 19:39
    
Yes it is, I hesitated to cite it for some reason, but here is the link to the OEIS sequence where I first found it in a comment by Peter Bala, and the link to the arxiv paper by LeClair: Reference 1: oeis.org/A177885 Reference 2: arxiv.org/abs/1305.2613 –  matsgranvik Dec 18 '13 at 19:48
    
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.