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Given $P_k$, the truncated Prime $\zeta$ function, defined like $$ P_k(it):=\sum_{n=1}^k p_n^{it}, $$ where $p_n$ is the $n$th prime. What's the probability to find a value or range of $t$ less than $T$, where $|P_k(it)|<\epsilon$?

EDIT By that I mean, among all values of $t<T$, how large is the portion of values, where $|P_k(it)|<\epsilon$?

To give an example, I plotted $P_{500}(it)$, where $0<t<400$:

$\hskip0.7in$enter image description here

The lower plot shows the portion of values that lies below a certain threshold. Almost all values lie below $.1$. I interpret the lower plot as the sum over the distribution.

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It's not clear what you mean by "the probability to find...". That might depend on how you are looking for it. –  Robert Israel Nov 14 '12 at 23:14
    
There seems to be no probability at all in this question. –  Did Nov 15 '12 at 7:29
    
@did, why not? The probability to find a value $|P_k(it)|\le k$ is $100\%$ irrespective of $t$ and $T$... –  draks ... Nov 15 '12 at 8:00
    
(Not independent of T.) The only probability space involved is [0,T] endowed with the normalized Lebesgue measure... Unless the proof involves some probabilistic techniques such as conditioning, change of measure, independence, whatever (and I see no reason to suspect it does), this is purely (deterministic) number theory. –  Did Nov 15 '12 at 9:07
    
Ok, so what's the solution to this purely (deterministic) number theoretic problem? Sorry if I put it in the wrong box... –  draks ... Nov 15 '12 at 9:49

1 Answer 1

up vote 1 down vote accepted

If $t$ is chosen randomly (uniformly) from a large enough interval, $p_n^{it}$ for different $n$ should be good approximations to independent random variables with uniform distribution on the unit circle. This distribution has mean $0$ and covariance matrix $\pmatrix{1/2 & 0\cr 0 & 1/2\cr}$. The sum, for large $k$, has approximately a bivariate normal distribution with mean $0$ and covariance $\pmatrix{k/2 & 0\cr 0 & k/2\cr}$. Its absolute value has PDF $f(r) = \dfrac{2 r}{k} e^{-r^2/k}$ for $r > 0$,
CDF $F(r) = 1 - e^{-r^2/k}$ for $r > 0$, mean $\sqrt{k\pi}/2$ and standard deviation $\sqrt{(1-\pi/4) k}$.

Are you sure your $P_k$ doesn't have a factor $1/k$?

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it has, thanks... –  draks ... Mar 9 at 14:06

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