[SOLVED] Von-Mangoldt Formula

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June 7th 2010, 06:59 AM
raheel88

[SOLVED] Von-Mangoldt Formula

Consider the Von Mangoldt explicit formula, namely,

$\Psi (x) = x - \sum \frac{x^\rho}{\rho} + O (1)$

How can I prove that, assuming the Riemann Hypothesis to be true, then,

$\frac{\Psi (x) - x}{\sqrt x} = -2\sum \frac{sin(\gamma_n t)}{\gamma}$

where $\gamma_i$ is the imaginary part of the i-th non-trivial zeta zero.

I've had a decent stab at it but I get stuck in trying to manipulate the sum once I've divided by $\sqrt x$

Thanks in advance!
June 7th 2010, 07:03 AM
chiph588@

Quote:

Originally Posted by raheel88

Consider the Von Mangoldt explicit formula, namely,

$\Psi (x) = x - \sum \frac{x^\rho}{\rho} + O (1)$

How can I prove that, assuming the Riemann Hypothesis to be true, then,

$\frac{\Psi (x) - x}{\sqrt x} = -2\sum \frac{sin(\gamma_n t)}{\gamma}$

where $\gamma_i$ is the imaginary part of the i-th non-trivial zeta zero.

I've had a decent stab at it but I get stuck in trying to manipulate the sum once I've divided by $\sqrt x$

Thanks in advance!

Try this.
June 7th 2010, 07:31 AM
chiph588@

Quote:

Originally Posted by chiph588@

Try this.

Are you familiar with the formula $\frac{x^\rho}{\rho}+\frac{x^{\overline{\rho}}}{\ov erline\rho}=\frac{2x^{\beta}}{\beta^2+\gamma^2}(\b eta\cos(\gamma\log(x))+\gamma\sin(\gamma\log(x)))<br />$ ?
June 7th 2010, 07:39 AM
raheel88

Quote:

Originally Posted by chiph588@

Are you familiar with the formula $\frac{x^\rho}{\rho}+\frac{x^{\overline{\rho}}}{\ov erline\rho}=\frac{2x^{\beta}}{\beta^2+\gamma^2}(\b eta\cos(\gamma\log(x))+\gamma\sin(\gamma\log(x)))<br />$ ?

yes, that's easy to prove

the step i'm trying to get my head around is when you absorb the cosine term into the $O (1)$ term, surely you can also absorb sine as well? isnt sine just a transformation of cosine?
June 7th 2010, 07:47 AM
chiph588@

Quote:

Originally Posted by raheel88

yes, that's easy to prove

the step i'm trying to get my head around is when you absorb the cosine term into the $O (1)$ term, surely you can also absorb sine as well? isnt sine just a transformation of cosine?

You cant assume its $O(1)$ though for sine since the term in the series only has $\gamma_n$ in the denominator, not $\gamma_n^2$ like the other series.
June 7th 2010, 08:10 AM
raheel88

that's plausible...thanks for your help chiph588.

you saved the day once again!
June 7th 2010, 09:20 AM
chiph588@

Quote:

Originally Posted by raheel88

that's plausible...thanks for your help chiph588.

you saved the day once again!

I should also mention this works because $\gamma_n\sim\frac{2\pi n}{\log(n)}$ .

This is my signature below! (Cool)
June 7th 2010, 09:42 AM
raheel88

Quote:

Originally Posted by chiph588@

I should also mention this works because $\gamma_n\sim\frac{2\pi n}{\log(n)}$ .

This is my signature below! (Cool)

yeh i spotted that...very intersting result! I've not come across it before in my course, so a proof would be nice whenever you're not too busy.

regards
June 7th 2010, 10:50 AM
chiph588@

Quote:

Originally Posted by raheel88

yeh i spotted that...very intersting result! I've not come across it before in my course, so a proof would be nice whenever you're not too busy.

regards

Are you familiar with the estimation theorem for $N(t)$ where $N(t)$ is the number of zeros of $\zeta(s)$ with ordinate less than $t$ ?

The proof follows from that. I'll be glad to show you if you have seen this. I can't think of any other way.
June 7th 2010, 10:57 AM
raheel88

Quote:

Originally Posted by chiph588@

Are you familiar with the estimation theorem for $N(t)$ where $N(t)$ is the number of zeros of $\zeta(s)$ with ordinate less than $t$ ?

The proof follows from that. I'll be glad to show you if you have seen this. I can't think of any other way.

You mean $N(t) \sim \frac{t\ln t}{2\pi}$ ?
Sure!
June 7th 2010, 11:09 AM
chiph588@

Quote:

Originally Posted by raheel88

You mean $N(t) \sim \frac{t\ln t}{2\pi}$ ?
Sure!

Ok cool!

First note that $N(\gamma_n-1)\leq n \leq N(\gamma_n+1)$ .

From the theorem you just mentioned, we get $2\pi N(\gamma_n\pm1)\sim(\gamma_n\pm1)\log(\gamma_n\pm1 )\sim\gamma_n\log(\gamma_n)$ .

Thus by the squeeze theorem we have $2\pi n\sim\gamma_n\log(\gamma_n)$ .

So $\log(2\pi n)\sim\log(\gamma_n\log(\gamma_n))\implies \log(2\pi)+\log(n)\sim\log(\gamma_n)+\log(\log(\ga mma_n))$ $\implies \log(n)\sim\log(\gamma_n)$ .

Finally we see $2\pi n\sim\gamma_n\log(\gamma_n)\implies 2\pi n\sim\gamma_n\log(n) \implies \gamma_n\sim\frac{2\pi n}{\log(n)}$ .
June 7th 2010, 11:13 AM
raheel88

Nice!