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I have been considering the Riemann Zeta function, but with a limited domain. In this case I have been considering $$f(x)=\sum _{n=1}^{\infty }\:\frac{1}{n^x}, x>1$$ Considering that this is well define for every element in the domain, and the fact that this would have a horizontal asymptote at y=1 as the first term in the series will always be 1, is it possible to set this sum equal to some $n>1$ and find a closed form of the x that will give this result? I believe that this definition is continuous and is decreasing everywhere in the domain. What I mean by this question is a way to find $x$ in an equation such as $$3=\sum _{n=1}^{\infty }\:\frac{1}{n^x}, x>1$$ If not in closed form, perhaps a method to find the x to any level of accuracy as wanted, without just plugging this into wolphram alpha and solving for. Also, is the inverse of the Riemann Zeta function well defined and has it been studied?

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  • $\begingroup$ Well, we know $\zeta$ is a bijection between $(1,\infty)$ and itself, and is specifically decreasing. Thus in principle this equation is solvable. But we have very limited analytic information, beyond knowing $\lim_{x \to 1^+} \zeta(x)=+\infty$ and knowing the values at even integers. $\endgroup$ – Ian Dec 16 '17 at 21:09
  • $\begingroup$ $\sum_{n=N+1}^\infty n^{-x} < \int_N^\infty t^{-x}dt = \frac{N^{1-x}}{x-1}$. Thus for $N$ large enough, you can approximate $\zeta(x) = 3$ by $\sum_{n=1}^N n^{-x} = 3$. $\endgroup$ – reuns Dec 16 '17 at 23:04
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As noted by Ian, $\zeta$ is bijective on $(1,\infty)$, and thus, a quick and dirty bisection approach can approximate $\zeta^{-1}(x)$ for any $x\in(1,\infty)$.

The approach I'm taking is this:

Relation to the Dirichlet eta function.

$$\zeta(s)=\frac{\eta(s)}{1-2^{1-s}}$$

Euler summing the Dirichlet eta function.

$$\eta(s)=\sum_{n=0}^\infty\frac1{2^{n+1}}\sum_{k=0}^n\binom nk\frac{(-1)^k}{(k+1)^s}$$

Then bisection:

$$a_0=1,b_0=1-\log_2(1-1/x)$$

Let

$$c_n=\frac{a_n+b_n}2$$

If $\zeta(c_n)>x$, then

$$a_{n+1}=c_n,b_{n+1}=b_n$$

Otherwise,

$$a_{n+1}=a_n,b_{n+1}=c_n$$

Where

$$\zeta^{-1}(x)=\lim_{n\to\infty}c_n$$

This program applies the above approach, truncating the Euler sum on the Dirichlet eta function to about 10 terms, and returning $c_n$ when $b_n-a_n<0.00001$.

For $x=3$, my program returns $1.4178443839456096$, and according to WolframAlpha, $\zeta(1.4178443839456096)\approx3.0000087817793577$, which is pretty decent.

The program is adjustable as well now. Inputting larger values on "accuracy" will raise the accuracy, but be warned the program gets exponentially slower.

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  • $\begingroup$ Why don't you use $\eta(s) = \sum_{n=1}^\infty (-1)^{n+1} n^{-s} =\sum_{n=1}^N (-1)^{n+1} n^{-s}+ \mathcal{O}(\frac{N^{1-s}}{s-1})$ ? $\endgroup$ – reuns Dec 16 '17 at 23:31
  • $\begingroup$ @reuns That has slower convergence (sub-linear) $\endgroup$ – Simply Beautiful Art Dec 16 '17 at 23:39
  • $\begingroup$ Yes but the goal is to make things as simple as possible. $\endgroup$ – reuns Dec 16 '17 at 23:55
  • $\begingroup$ @reuns How do you know the OP doesn't prefer a more accurate method over a simpler one, especially since they've already accepted my answer? $\endgroup$ – Simply Beautiful Art Dec 17 '17 at 0:00
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This can be calculated using series expansion: $$\zeta(s)=\frac{1}{s-1} + \sum_{n=0}^\infty \frac{(-1)^n\gamma_n}{n!}(s-1)^n$$

where $\gamma_n$ are Stiletjes constants: https://en.wikipedia.org/wiki/Stieltjes_constants

The approximate values are given in the article. This equation can be solved for 3: and the $$s=1.417845935787357292962112538319040923439322566473290579706$$

The Mathematica will solve this as well:

Solve[Zeta[s] == 3, {s}]

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  • $\begingroup$ How would one go about solving for $s$ from the given expression? $\endgroup$ – Simply Beautiful Art Dec 16 '17 at 23:46
  • $\begingroup$ We can take as many elements of the sum as needed and solve polynomial equation for $s$. $\endgroup$ – Gevorg Hmayakyan Dec 16 '17 at 23:49
  • $\begingroup$ Solving for $\sum_{k=0}^\infty c_k z^k = C$ isn't especially easier than solving for $\sum_{n=1}^\infty a_n n^{-s} = C$. $\endgroup$ – reuns Dec 17 '17 at 2:19
  • $\begingroup$ I understand your point, but for at least for $n=5$ it is solvable in terms of roots. And as the $\gamma_n$ is decreasing enough quickly so it may give good analytical result for some small values. $\endgroup$ – Gevorg Hmayakyan Dec 17 '17 at 7:00

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