# Math Help - Sum over Complex Zeros

1. ## Sum over Complex Zeros

I am given $\psi(x)=x-\sum{\frac{x^\rho}{\rho}}+O(1)$ where $\rho$ denotes complex zeros.

And also given $\frac{x^\rho}{\rho}+\frac{x^{\overline{\rho}}}{\ov erline\rho}=\frac{2x^{\beta}}{\beta^2+\gamma^2}(\b eta\cos(\gamma\log(x))+\gamma\sin(\gamma\log(x)))$ where $\rho=\beta+i\gamma$

Assume the Riemann Hypothesis is true, and write the complex zeros as $\frac{1}{2}+i\gamma_{1},\frac{1}{2}+i\gamma_{2} . . . . . . . .$.

Then show that $\frac{\psi(x)-x}{\sqrt{x}}= -2\sum\limits_{n = 1}^\infty \frac{sin(\gamma_{n}logx)}{\gamma_{n}}+O(1)$

Any help is greatly appreciated.

2. If, as You have written,...

$\frac{x^\rho}{\rho}+\frac{x^{\overline{\rho}}}{\ov erline\rho}=\frac{2x^{\beta}}{\beta^2+\gamma^2}(\b eta\cos(\gamma\log(x))+\gamma\sin(\gamma\log(x)))$ (1)

... and for all complex zeroes is $\beta=\frac{1}{2}$, then all the terms of the sum $\sum (\frac{x^{\rho}}{\rho}+ \frac{x^{\overline{\rho}}}{\overline\rho})$ contain the term $x^{\beta}= \sqrt{x}$...

Kind regards

$\chi$ $\sigma$

3. Originally Posted by skamoni
I am given $\psi(x)=x-\sum{\frac{x^\rho}{\rho}}+O(1)$ where $\rho$ denotes complex zeros.

And also given $\frac{x^\rho}{\rho}+\frac{x^{\overline{\rho}}}{\ov erline\rho}=\frac{2x^{\beta}}{\beta^2+\gamma^2}(\b eta\cos(\gamma\log(x))+\gamma\sin(\gamma\log(x)))$ where $\rho=\beta+i\gamma$

Assume the Riemann Hypothesis is true, and write the complex zeros as $\frac{1}{2}+i\gamma_{1},\frac{1}{2}+i\gamma_{2} . . . . . . . .$.

Then show that $\frac{\psi(x)-x}{\sqrt{x}}= -2\sum\limits_{n = 1}^\infty \frac{sin(\gamma_{n}logx)}{\gamma_{n}}+O(1)$

Any help is greatly appreciated.
Lemma: If $\rho$ is a root of $\zeta(s)$, $\overline{\rho}$ is too. (This is not too hard to verify.)

So we have $\psi(x) = x-\sum{\frac{x^\rho}{\rho}}+O(1) = x-\sum_{\Im{\rho}>0}{\left(\frac{x^\rho}{\rho}+\frac {x^{\overline{\rho}}}{\overline{\rho}}\right)}+O(1 )$
(I'm a little scared about being able to change order of summation up above... maybe someone else should take a look at this line to confirm its validity.)

$\psi(x)-x = -\sum_{\gamma>0}\frac{2\sqrt{x}}{\frac{1}{4}+\gamma ^2}(\frac{1}{2}\cos(\gamma\log(x))+\gamma\sin(\gam ma\log(x)))+O(1)$
$= -2\sqrt{x} \sum_{n=1}^{\infty} \frac{1}{\frac{1}{4}+\gamma_n^2}(\frac{1}{2}\cos(\ gamma_n\log(x))+\gamma_n\sin(\gamma_n\log(x)))+O(1 )$

Since we have $O(1)$ in our equation, we can drop the $\frac{1}{4}$ in the denominator.
Now split up the sum.

$\frac{\psi(x)-x}{\sqrt{x}} = -2 \sum_{n=1}^{\infty} \frac{\cos(\gamma_n\log(x))}{2\gamma_n^2}-2\sum_{n=1}^{\infty}\frac{\gamma_n\sin(\gamma_n\lo g(x))}{\gamma_n^2}+O(1)$

Since $|\cos(x)|\leq 1 \; \forall \; x\in\mathbb{R}$ and we have $\gamma_n^2$ in our denominator, the first sum gets absorbed into the $O(1)$ term.

Thus we get $\frac{\psi(x)-x}{\sqrt{x}}= -2\sum\limits_{n = 1}^\infty \frac{sin(\gamma_{n}\log(x))}{\gamma_{n}}+O(1)$.

4. Originally Posted by chiph588@

So we have $\psi(x) = x-\sum{\frac{x^\rho}{\rho}}+O(1) = x-\sum_{\Im{\rho}>0}{\left(\frac{x^\rho}{\rho}+\frac {x^{\overline{\rho}}}{\overline{\rho}}\right)}+O(1 )$
(I'm a little scared about being able to change order of summation up above... maybe someone else should take a look at this line to confirm its validity.)
There's a chance I may be able to help...but more for my own curiosity, what did you do in the above? I don't actually see what happened

5. Originally Posted by Drexel28
There's a chance I may be able to help...but more for my own curiosity, what did you do in the above? I don't actually see what happened
The first sum is summing over all the roots of $\zeta(s)$. So since we if $\rho$ is a root then $\overline{\rho}$ is a root too, we can just look in the upper half plane (i.e. when $\Im{\rho}>0$) for roots and sum the conjugate pairs together.

6. Originally Posted by chiph588@
The first sum is summing over all the roots of $\zeta(s)$. So since we if $\rho$ is a root then $\overline{\rho}$ is a root too, we can just look in the upper half plane (i.e. when $\Im{\rho}>0$) for roots and sum the conjugate pairs together.
Ok, I think I get it. So you're saying that since $\zeta(\rho)=0\implies \zeta(\overline{\rho})=0$ it is obvious that every zero comes in pairs $\{\rho,\overline{\rho}\}$ where one is such that $\text{Im}\text{ }<0$ and the other positive. So $\sum_{\rho}f(\rho)=\sum_{\rho,\text{ }\Im \rho<0}f(\rho)+\sum_{\rho,\text{ }\Im \rho>0}f(\rho)$ $=\sum_{\rho,\text{ }\rho>0}f(\rho)+\sum_{\overline{\rho},\text{ }\Im \rho>0}f(\rho)=\sum_{\rho,\text{ }\Im \rho>0}+\sum_{\rho,\text{ }\rho>0}f(\overline{\rho})=\sum_{\rho,\text{ }\Im\rho>0}\left\{f(\rho)+f(\overline{\rho})\right \}$?

7. Originally Posted by Drexel28
Ok, I think I get it. So you're saying that since $\zeta(\rho)=0\implies \zeta(\overline{\rho})=0$ it is obvious that every zero comes in pairs $\{\rho,\overline{\rho}\}$ where one is such that $\text{Im}\text{ }<0$ and the other positive. So $\sum_{\rho}f(\rho)=\sum_{\rho,\text{ }\Im \rho<0}f(\rho)+\sum_{\rho,\text{ }\Im \rho>0}f(\rho)$ $=\sum_{\rho,\text{ }\rho>0}f(\rho)+\sum_{\overline{\rho},\text{ }\Im \rho>0}f(\rho)=\sum_{\rho,\text{ }\Im \rho>0}+\sum_{\rho,\text{ }\rho>0}f(\overline{\rho})=\sum_{\rho,\text{ }\Im\rho>0}\left\{f(\rho)+f(\overline{\rho})\right \}$?
I don't understand some of your intermediate steps, but $\sum_{\rho}f(\rho)=\sum_{\rho,\text{ }\Im\rho>0}\left\{f(\rho)+f(\overline{\rho})\right \}$ is correct.

$\rho$ is complex, so $\rho>0$ doesn't make sense.

8. Originally Posted by chiph588@
I don't understand some of your intermediate steps, but $\sum_{\rho}f(\rho)=\sum_{\rho,\text{ }\Im\rho>0}\left\{f(\rho)+f(\overline{\rho})\right \}$ is correct.
$\sum_{\rho}f(\rho)=\sum_{\rho,\text{ }\Im \rho>0}f(\rho)+\sum_{\rho,\text{ }\Im \rho<0}f(\rho)$- The sum over all equals the sum over the ones in the upper plane plus those in the lower plane.

All the ones in the lower plane have a unique one in the upper plane (it's conjugate). So, to hit all the ones in the lower plane we sum over all the conjugates of the ones in the upper.

So $\sum_{\rho,\text{ }\Im \rho}f(\rho)+\sum_{\rho,\text{ }\Im \rho<0}f(\rho)=\sum_{\rho,\text{ }\rho>0}f(\rho)+\sum_{\overline{\rho},\text{ }\Im \rho>0}f(\rho)$

But clearly this is equal to

$\sum_{\rho,\text{ }\Im \rho>0}f(\rho)+\sum_{\rho,\text{ }\Im\rho>0}f(\overline{\rho})$- We just changed in the second sum where the conjugation occurs.

But, we can add these two together to get $\sum_{\rho,\text{ }\Im \rho>0}\left\{f(\rho)+f(\overline{\rho})\right\}$.

This is assuming everything converged.

9. Originally Posted by Drexel28
$\sum_{\rho}f(\rho)=\sum_{\rho,\text{ }\Im \rho>0}f(\rho)+\sum_{\rho,\text{ }\Im \rho<0}f(\rho)$- The sum over all equals the sum over the ones in the upper plane plus those in the lower plane.

All the ones in the lower plane have a unique one in the upper plane (it's conjugate). So, to hit all the ones in the lower plane we sum over all the conjugates of the ones in the upper.

So $\sum_{\rho,\text{ }\Im \rho}f(\rho)+\sum_{\rho,\text{ }\Im \rho<0}f(\rho)=\sum_{\rho,\text{ }\rho>0}f(\rho)+\sum_{\overline{\rho},\text{ }\Im \rho>0}f(\rho)$

But clearly this is equal to

$\sum_{\rho,\text{ }\Im \rho>0}f(\rho)+\sum_{\rho,\text{ }\Im\rho>0}f(\overline{\rho})$- We just changed in the second sum where the conjugation occurs.

But, we can add these two together to get $\sum_{\rho,\text{ }\Im \rho>0}\left\{f(\rho)+f(\overline{\rho})\right\}$.

This is assuming everything converged.
Yes that's correct.

10. Originally Posted by chiph588@
Yes that's correct.
Well, I don't see anything wrong with that. That said, I am NOT very well-versed in advanced CA.

11. Originally Posted by skamoni
Assume the Riemann Hypothesis is true, and write the complex zeros as $\frac{1}{2}+i\gamma_{1},\frac{1}{2}+i\gamma_{2} . . . . . . . .$.
I think it would be interesting to work it through assuming the Riemann Hypothesis is false. Recall if it's false, then instead of a conjugate pair of zeros, we have a pair of conjugate pair: $\sigma\pm \gamma i$ and $1/2+\sigma\pm \gamma i$ with $0<\sigma<1/2$.

. . . guess I need to check to make sure Von Mangoldt's expression for $\psi(x)$ is valid independent of the placement of zeros in the critical strip. I think it is since it can be derived using the Residue Theorem (all the terms are sums of residues).