# Tag Info

Accepted

### Is the Riemann zeta function surjective?

The Riemann zeta function is surjective. First, $\zeta(1/z)$ is holomorphic in the punctured disk $0<|z|<1$. Looking at $z=(1/2+it)^{-1}$ with $t\to\infty$ reveals that $\zeta(1/z)$ has an ...
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• 535

### A new way of approaching the pole of the Riemann zeta function - and a new conjectured formula

The paper entitled Euler constant as a renormalized value of Riemann zeta function at its pole by Andrei Vieru contains a derivation of the first formula in the OP (Benoȋt Cloitre's formula), and ...
• 176k

### Could the Riemann zeta function be a solution for a known differential equation?

The fact that $\zeta$ satisfy no algebraic differential equation is due to its famous relation with the Gamma function which was proved by Hölder not to satisfy such an equation. Detailed answer can ...
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### Could the Riemann zeta function be a solution for a known differential equation?

The Riemann zeta function is "hypertranscendental" in the sense shown HERE It is not the solution $y(x)$ of a differential equation of the form $$F(x,y,y',y'',\dots,y^{(n)})=0$$ where $F$ is a ...
• 40.1k
Accepted

### Has Apéry's proof of the irrationality of $\zeta(3)$ ever been used to prove the irrationality of other constants?

The proof of irrationality of $\displaystyle\sum_{n=0}^{+\infty}\frac1{F_n}$ (where $F_n$ is the $n$-th Fibonacci number) by RIchard André-Jeannin is an adaptation of the original Apery's proof of the ...
• 3,627
Accepted

### values of $\zeta$ function are linearly independent?

People have been exerting steady effort to prove/disprove irrationality of the Riemann zeta values in your list. Of course, $\zeta(3)$ is known to be irrational due to Roger Apery. Such investigations,...
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### Is there a closed form for $\int_0^\infty\frac{\tanh^3(x)}{x^2}dx$?

Rewrite the integrand and apply Taylor expansion to $\frac1{(1+e^{-2x})^3}$ so that $$\frac{\tanh^3x}{x^2}=\sum_{j\geq0}(-1)^j\binom{j+2}2 \frac{(1-e^{-2x})^3}{x^2}\binom{j+2}2e^{-2jx}.$$ Integrate ...
• 41.7k
Accepted

### $P(s)=1-\sqrt{\frac{2}{\zeta(s)}-\sqrt{\frac{2}{\zeta(2s)}-\sqrt{\frac{2}{\zeta(4s)}-\sqrt{\frac{2}{\zeta(8s)}-...}}}}$

This has been answered in the comments by Lucia. The identity $$P(s)=1-\sqrt{\frac{2}{\zeta(s)}-\sqrt{\frac{2}{\zeta(2s)}-\sqrt{\frac{2}{\zeta(4s)}-\sqrt{\frac{2}{\zeta(8s)}-\cdots}}}}$$ is false. By ...

### Algebraic independence of shifts of the Riemann zeta function

Hmm, it was more difficult than I expected to leverage universality to establish the claim. But one can proceed by probabilistic reasoning instead, basically exploiting the phase transition in the ...
• 107k
Accepted

### Algebraic independence of shifts of the Riemann zeta function

$\zeta(s - z)$ has an Euler product $\prod_p \frac{1}{1 - p^{z-s}}$, and so a monomial $\prod_i \zeta(s - z_i)$ (with the $z_i$ not necessarily distinct) has an Euler product \prod_i \zeta(s - z_i) =...
• 115k
There is a connection! (Though see the edit below.) Keating and Snaith make their conjecture by modeling the distribution of $\zeta(s)$ by the distribution of the characteristic polynomial of a random ...