# [SOLVED] Von-Mangoldt Formula

• Jun 7th 2010, 06:59 AM
raheel88
[SOLVED] Von-Mangoldt Formula
Consider the Von Mangoldt explicit formula, namely,

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

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

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

where $\displaystyle \gamma_i$ 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 $\displaystyle \sqrt x$

• Jun 7th 2010, 07:03 AM
chiph588@
Quote:

Originally Posted by raheel88
Consider the Von Mangoldt explicit formula, namely,

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

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

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

where $\displaystyle \gamma_i$ 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 $\displaystyle \sqrt x$

Try this.
• Jun 7th 2010, 07:31 AM
chiph588@
Quote:

Originally Posted by chiph588@
Try this.

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

Originally Posted by chiph588@
Are you familiar with the formula $\displaystyle \frac{x^\rho}{\rho}+\frac{x^{\overline{\rho}}}{\ov erline\rho}=\frac{2x^{\beta}}{\beta^2+\gamma^2}(\b eta\cos(\gamma\log(x))+\gamma\sin(\gamma\log(x)))$?

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 $\displaystyle O (1)$ term, surely you can also absorb sine as well? isnt sine just a transformation of cosine?
• Jun 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 $\displaystyle O (1)$ term, surely you can also absorb sine as well? isnt sine just a transformation of cosine?

You cant assume its $\displaystyle O(1)$ though for sine since the term in the series only has $\displaystyle \gamma_n$ in the denominator, not $\displaystyle \gamma_n^2$ like the other series.
• Jun 7th 2010, 08:10 AM
raheel88
that's plausible...thanks for your help chiph588.

you saved the day once again!
• Jun 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 $\displaystyle \gamma_n\sim\frac{2\pi n}{\log(n)}$.

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

Originally Posted by chiph588@
I should also mention this works because $\displaystyle \gamma_n\sim\frac{2\pi n}{\log(n)}$.

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
• Jun 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 $\displaystyle N(t)$ where $\displaystyle N(t)$ is the number of zeros of $\displaystyle \zeta(s)$ with ordinate less than $\displaystyle t$?

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.
• Jun 7th 2010, 10:57 AM
raheel88
Quote:

Originally Posted by chiph588@
Are you familiar with the estimation theorem for $\displaystyle N(t)$ where $\displaystyle N(t)$ is the number of zeros of $\displaystyle \zeta(s)$ with ordinate less than $\displaystyle t$?

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 $\displaystyle N(t) \sim \frac{t\ln t}{2\pi}$ ?
Sure!
• Jun 7th 2010, 11:09 AM
chiph588@
Quote:

Originally Posted by raheel88
You mean $\displaystyle N(t) \sim \frac{t\ln t}{2\pi}$ ?
Sure!

Ok cool!

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

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

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

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

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