Some more on last stage of Riemann Proof

Aron Palmer

Abstract:

 

More detail on the last stages of my earlier proof.

 

1. General

 

If we have

 

then we also have

 

and therefore if

we get

 

2. Specific


Therefore from theory of elliptic functions,

and

 

So in our case

 and

 

3. Derivative

 

If we have

 then we have

 

which is a very small positive number.

 

We also have

which has roots at

 

Two of these are imaginary and two are real.

The real ones are only at t less than zero when

or

 

which is roughly 1.00227

Therefore the only term which can be decreasing at t == 0 is the first term and since the derivative is zero at t == 0 (even function) this shows that there cannot be a derivative of zero before hand since the negative gradient of the first term must equal all the others at the origin only (all the others are increasing, the first term is decreasing). 

 

 

 

4. Product

We have

and

from the Hadamard product form of the zeta function where the roots, pi are the roots of the Zeta(s+1/2) function.

Since

we have

or

Therefore the two sided Laplace transform of our function does indeed represent a product form of the Zeta function.

 

 

5. Graphs

A graph of

Shows that function being even and having only a derivative of 0 at t=0, and being positive decreasing function are plausible.

 

And a graph of the first 4 terms.

Shows that only the first term (its the one on the left) having a negative gradient for t<0 is plausible.

Contact: aron.palmer@ukonline.co.uk

Aron Palmer 20 February 2006