# Chebyshev function

1. Jul 27, 2006

### eljose

let be the Chebyshev function:

$$\psi(x)= x- \sum_{\rho} \frac{ x^{\rho}}{\rho}+C-log(1-x^{2})$$ (1)

Where the sum is over the Non trivial zeros of $$\zeta(s)$$

then we have that $$\psi(n) -\psi(n-1)=\Lambda (n)$$ where the Lambda function is only nonzero witha value of log(p) for n=p^{k} with k a positive integer...my question is.. if we use (1) to calculate Chebyshev function..couldn't we get an expression for the log(p)?..thanks.

2. Jul 27, 2006

### eljose

Another question if we have that:

$$\psi(x)-x=\Omega (x^{1/2})$$ then setting x=exp(u) so

$$\psi(e^u )-e^u =\Omega (e^{u/2})$$ now if we put $$\psi(e^u ) =g(u)$$ my question is if for g(u) we have:

$$g(u)-e^u =\Omega (e^{u/2})$$

3. Jul 27, 2006

### matt grime

It's called the von Mangoldt Function (the thing you called the Lambda Function).

4. Jul 27, 2006

### shmoe

Not sure what you want here, given a p you can find log(p) with a calculator.

If you are hoping to use the explicit formula to determine if a specific number is prime by using it to find psi(n) and psi(n-1) exactly (or at least accurately enough to determine if lambda(n)>0) then you are out of luck. You are better off with the usual primality testing algorithms.

If you just want to express lambda(n) as a big fat sum over the zeros of the zeta function, then of course this will work.

Sure, but why? This change of variables that you seem to love x->e^u does nothing profound at all.

5. Jul 28, 2006

### eljose

But why not?..in fact if you knew:

$$\psi(n)-\psi(n-1)=\Lambda (n)$$ then you could use a computer to represent n=2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,................. and when you have a "peak" of the function different from 0 you know you're dealing with a prime..

the problem of $$g(u)-e^u = \Omega (e^{u/2})$$ is in my opinion to find if the derivative of the function above will satisfy:

$$g' (u)-e^u =\Omega (e^{u/2})$$ and what can we say about the quotient $$\frac{f(x)}{x}$$ if f(x) satisfies $$f(x)= \Omega (x)$$

6. Jul 28, 2006

### matt grime

because you've not proven it allows us to do anything more easily than we already can. It is trivial to come up with methods that allow you to find primes and factorize integers, but they are almost always worse than just doing trial division, unless you happen to be Pollard.

and how are you going to find these peaks?

7. Jul 28, 2006

### shmoe

As always, do it. Of course you can use the explicit formula to find psi(n) as accurately as you need by taking enough terms in the sum. The problem is the 'enough' required to get 'accurately as you need' will be 'so many this is a terrible way of doing things'.

What you should do is go to odlyzko's website and download a bunch of zeros. Take a look at the explicit formula with the sum truncated to some finite number of terms, 10, 100, 1000, etc, try different values. Plot a graph of psi(x) and what this truncated version of the left hand side gives. See how far out this is close enough to determine psi(x). (there's free software you can use, like pari/gp, so no complaining about not being able to do this).

Failing that there used to be websites that had java animations of this. Go find one at look at it (I've given links before, not going to bother finding them again).

The derivative of g is zero almost everywhere, psi is a step function with jumps at the prime powers, so no, the derivative of g does not satisfy what you've claimed.

Derivatives don't work this way with asymptotics. if f(x)=x*sin(x^2) then clearly f(x)=O(x) but does f'(x)=O(1)?

8. Aug 23, 2006

### saltydog

Can I be picky? You guys get annoyed by that? That's not quite right up there and I just wish to be precise as best as I can at my meager level of familiarity with the subject.

Chebyshev's function is:

$$\psi(x)=\sum_{p^n\leq x} ln(p)$$

Von Mongoldt's explicit expression is:

$$\psi_o(x)=x-\sum_{\rho}\frac{x^{\rho}}{\rho}-C-1/2 ln(1-\frac{1}{x^2})$$

The two functions are related however:

$$\psi(x)=\left\{ \begin{array}{rcl} \psi_o(x) & \mbox{for} & x\neq p^n \\ \psi_o(x)+1/2 ln(p) & \mbox{for} & x=p^n \end{array}\right$$

Pretty sure that's right. Kindly correct if not.

Thanks for the Ingham reference Shmoe. It's a beautiful problem for me.

Last edited: Aug 23, 2006
9. Aug 23, 2006

### saltydog

Just though I'd mention how to read into Mathematica, the first 100000 imaginary components of the zeta zeros from Odlyzko's site:

http://www.dtc.umn.edu/~odlyzko/zeta_tables/zeros1

First cut and paste the list to a *.txt file. Then use the code below, changing of course the location of where you put the file:

Code (Text):
(* set up path to read in zeta zeros from file *)

$Path = Append[$Path, "c:\Math\pf forum 4"];

(* now read in zeros *)

Keep in mind the list is only the imaginary components of $\rho$ and not $\overline{\rho}$. Anyway, I did this and computed the zero sum of von Mangoldt's formula in order to exhibit the discontinuity of the series at powers of primes. Note the plot below. The sum needs to be done symmetrically (both $\rho$ and $\overline{\rho}$ together before going to the next one.

#### Attached Files:

• ###### zerosum.JPG
File size:
12.9 KB
Views:
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Last edited: Aug 23, 2006
10. Aug 23, 2006

### shmoe

Looks fine. Dealing with this $$\psi_{0}$$ is often skipped by making a statement about $$\psi(x)$$ that's valid when "x is half an odd integer", or some such dodge.

I'm sure you've also plotted the truncated explicit formula vs. the von mangoldt function itself? It's kind of neat to see just how well it follows the step shape for a given number of zeros before being drowned out by the main term.

11. Oct 5, 2006

A surprising article relating "Chebyshev function" and Riemann hypothesis"

http://arxiv.org/abs/math.GM/0607095

Author claims that he has found an operator $$H=p^2 + V(x)$$ so RH is satisfied $$\zeta (1/2+iH)=0$$

12. Oct 5, 2006

### matt grime

Yes, very surprising.... written by the OP eljose, in fact..., who changed his login name to, erm, lokofer. His style of posting is very similar to yours in fact, Karlisbad (syntax, spelling, grammar, use of lots of .... and smilies for no reason). Here for instance: