1. ## [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 $\sqrt x$

2. 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 $\sqrt x$

Try this.

3. 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)))
$
?

4. 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)))
$
?
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?

5. 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.

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

you saved the day once again!

7. 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!

8. 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!
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

9. 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.

10. 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!

11. 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 $2\pi n\sim\gamma_n\log(\gamma_n)\implies 2\pi n\sim\gamma_n\log(n) \implies \gamma_n\sim\frac{2\pi n}{\log(n)}$.

12. Nice!