Way to find exact value of Riemann Zeta Function

Question

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?

Answer 1

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.

Answer 2

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}]