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- Jun 15th 2014, 12:25 PMrobinlrandallRe: Help with Riemann's Hypothesis
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RR - Jun 15th 2014, 12:44 PMShakarriRe: Help with Riemann's Hypothesis
- Jun 15th 2014, 12:58 PMrobinlrandallRe: Help with Riemann's Hypothesis
Of course all Riemann Hypothesis constraints are in place:

0 < s < 1 the "critical strip"

Am I forgetting any other "normal" R.H. assumptions?

Robin - Jun 15th 2014, 01:48 PMrobinlrandallRe: Help with Riemann's Hypothesis
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So I'm saying what if I have 0 = Sum 0 x an

n=1

therefore every element is 0

Then if that means s=1/2 and it checks out they are true zeroes

and "an" are constants for all n "as defined by the equation", where are there

any more zeroes to be found? Maybe there's y1 - y123456 = 0 buried somewhere in the equation but I don't see it yet. I'm still looking. Let's see:

(n^(3/4) - n^(1/4))/n - (m^(3/4) - m^(1/4))/m + ... = 0

Yes, I have some work to do. But I chipping away.

Robin - Jun 16th 2014, 12:26 AMShakarriRe: Help with Riemann's Hypothesis
But you don't have 0 x an, you have bn x an. You are assuming that bn is equal to zero and it doesn't matter if that assumption leads to RH being true, it is still an assumption.

Just take a simple example to test $\displaystyle n^s=n^{1-s}$ and you will see that it is false.

For example, n=2, s=0.4+0.2i leads to $\displaystyle 2^{0.4+0.2i}= 1.31+0.182i$ and $\displaystyle 2^{0.6-0.2i}=1.5-0.209i$