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The Riemann zeta function is the function of one complex variable $s$ defined by the series $\zeta(s) = \sum_{n \geq 1} \frac{1}{n^s}$ when $\operatorname{Re}(s)>1$. It admits a meromorphic continuation to $\mathbb{C}$ with only a simple pole at $1$. This function satisfies a functional equation relating the values at $s$ and $1-s$. This is the most simple example of an $L$-function and a central object of number theory.

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How is "large" defined in an equality for the modulus of Riemann zeta?

You could look at Chandee's paper (in Proc AMS) and on arxiv at http://arxiv.org/pdf/0906.4177v1.pdf where explicit bounds are worked out for zeta and L-function, and it is specified when they start …
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7 votes
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The horizontal distribution of zeros of $\zeta^\prime(s)$

The zeta function will have $\gg \log T$ zeros with ordinates in the interval $[t_2+1,t_2+2]$. The contribution of these zeros to the sum is of absolute value $\gg \log T$.
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8 votes
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Zeta zeros standard normal distribution about $\vartheta (\gamma_n)$

What you're observing is a remarkable theorem of Selberg. The usual notation is to let $N(T)$ denote the number of zeros of $\zeta(s)$ with ordinates between $0$ and $T$. Then the argument principle …
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5 votes

Picking a new set of primes

As Greg Martin has pointed out this topic of Beurling numbers has been extensively investigated. Usually the Beurling numbers are studied as semigroups contained in ${\Bbb R}$. In the particular con …
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13 votes
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Bound on $L^2$ norm of $1/\zeta(1+i t)$?

Problems like this are classical (as noted in Terry's answer), and have been considered more recently with attention to uniformity in the moments. To give a quick indication, one can show that $$ \z …
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Is it possible to show that $\sum_{n=1}^{\infty} \frac{\mu(n)}{\sqrt{n}}$ diverges?

One can show that $\sum_{n=1}^{\infty} \mu(n)/\sqrt{n}$ diverges. Suppose to the contrary that it converges, which as you note implies RH. Put $M_0(x)=\sum_{n\le x} \mu(n)/\sqrt{n}$, and our assumpti …
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5 votes

On extended Riemann Hypothesis and coefficients of Selberg Class L-functions

Here's an argument that I think demonstrates why no such tweak is possible if the degree is strictly bigger than $1$. For simplicity let us just consider the case when $L(s,f)$ has no pole. Suppose …
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12 votes

Objections to and arguments for the simplicity of all Riemann zeros

Quite a lot is known in the literature about zeros of $\zeta(s)$ and $\zeta^{\prime}(s)$, and in particular one can show that zeros of $\zeta^{\prime}(s)$ do get close to zeros of $\zeta(s)$. However …
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Sharpest bound on the zero free region of $\zeta^{\prime}$?

If $s=\sigma+it$ with $\sigma >1$ then note that $$ \Big| \frac{\zeta^{\prime}}{\zeta}(s) \Big| =\Big| \sum_{p} \frac{\log p}{p^s-1} \Big| \ge \frac{\log 2}{|2^s-1|} - \sum_{p\ge 3} \frac{\log p}{| …
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If $\zeta(s)=0$ with $\Re(s)=\frac{1}{2}$, is then $|\hat{\zeta}(s,3)|^2=\frac{1}{2}$?

This is just a simple calculation. Note that $$ \sum_{n=k}^{\infty} \frac{1}{2^{n+1}} \binom{n}{k} =1 $$ for any non-negative integer $k$, which immediately gives $$ \frac{1}{1-2^{1-s}} \sum_{n= …
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Does Littlewood's bound on $\zeta(1+it)$ extend to all the partial sums?

The short answer is yes, and this is treated explicitly for characters in the work of Granville and Soundararajan (the paper appeared in J. Amer. Math. Soc.). Their Theorem 2 gives that on GRH for $x …
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Generalization of the The Liouville Lambda function

Let's just consider the case $k=2$; you can try to generalize this argument for larger $k$. For $k=2$, $$ \sum_{n\le x} \lambda_2(n) = \sum_{\substack{ n\le x \\ \Omega(n) = 0,1 \mod 4}} 1 - \sum_{\ …
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Reference for Lindelöf Hypothesis implying finitely many zeros off critical line?

The Lindelof hypothesis (LH) does not seem to give such precise information about the zeros of $\zeta(s)$. Here are three known implications of Lindelof on the zeros, but they will be seen to fall fa …
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Is this theorem on $L$-functions known?

Closely related problems have been extensively studied; in particular much stronger versions of the corollaries are already known. Here are some references: Fujii was the first to show that a positi …
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