Guest wrote:Equation: [tex]\sum_{k=1}^{\infty }\frac{sin(2 \pi k \sqrt{I + \lambda^{2}})}{k} = \frac{\pi}{2}[/tex].
That equation is quite beautiful!
And it may be quite important too!
Go Blue!
Guest wrote:Tentative Remark: Yes! The power series expansion is not appropriate here. But the Fourier series expansion is appropriate here.
[tex]\triangle n(I, \lambda) = \sqrt{I + \lambda^{2}} - Floor(\sqrt{I + \lambda^{2}})[/tex].
Moreover, [tex]\triangle n(I, \lambda) = \frac{1}{2} - \frac{1}{\pi} * \sum_{k=1}^{\infty }\frac{sin( 2 \pi k \sqrt{I + \lambda^{2}} )}{k}[/tex]
where [tex]1 \le \lambda < \frac{\sqrt{I}}{2}[/tex] for a given I such that I = p * q.
Tentative Remark: If our explicit definition (Fourier series expansion) of [tex]\triangle n(I, \lambda)[/tex] holds, then we have almost solved our integer factorization problem!
Question: What is the solution of [tex]\triangle n(I, \lambda) = 0[/tex] ?
Relevant Reference Link:
'Continuity and series expansions',
https://en.wikipedia.org/wiki/Floor_and_ceiling_functions#Continuity_and_series_expansions.
Dave.
Solving the equation, [tex]\triangle n(I, \lambda) = 0[/tex], may be quite technical.
Please refer to the paper, 'Fast algorithms for multiple evaluations of the Riemann zeta function' by Prof. A. M. Odlyzko and Prof. A. Schönhage for some ideas/algorithms on how to solve [tex]\triangle n(I, \lambda) = 0[/tex].
Relevant Reference Link:
http://www.dtc.umn.edu/~odlyzko/doc/arch/fast.zeta.eval.pdf;
Davet wrote:Here's more homework!
[tex]q = \frac{-\delta + \sqrt{\delta^{2} + 4 * I}}{2} = - \lambda+ \sqrt{\lambda^{2} + I}[/tex]
where [tex]\delta[/tex] is a positive even integer such that [tex]p - q = \delta = 2 * \lambda[/tex] and such that
[tex]2 \le \delta < 6.018 * 10^{498}[/tex]... where the known I = p * q for some unknown primes, p and q.
Good luck!
Guest wrote:We do not know!
Now suppose [tex]\sum_{n=1}^{\infty }\frac{sin( 2 \pi (2n-1) \sqrt{I + \lambda^{2}} )}{2n-1} = \frac{\pi}{a}[/tex].
And suppose, [tex]\sum_{n=1}^{\infty }\frac{sin( 4 \pi n \sqrt{I + \lambda^{2}} )}{2n} = \sum_{n=1}^{\infty }\frac{sin( 4 \pi n \sqrt{I + \lambda^{2}} )}{n} = \frac{\pi}{b}[/tex].
We assume a and b are nonzero rationals.
Hmm. What's next?
Guest wrote:An Update: We correct the previous post.Guest wrote:We do not know!
Now suppose [tex]\sum_{n=1}^{\infty }\frac{sin( 2 \pi (2n-1) \sqrt{I + \lambda^{2}} )}{2n-1} = \frac{\pi}{a}[/tex].
And suppose, [tex]\sum_{n=1}^{\infty }\frac{sin( 4 \pi n \sqrt{I + \lambda^{2}} )}{2n} = \frac{\pi}{b}[/tex].
We assume a and b are nonzero rationals.
Hmm. What's next?
Guest wrote:An Update: We correct the previous post.Guest wrote:We do not know!
Now suppose (1) [tex]\sum_{n=1}^{\infty }\frac{sin( 2 \pi (2n-1) \sqrt{I + \lambda^{2}} )}{2n-1} = \frac{\pi}{a}[/tex].
And suppose, (2) [tex]\sum_{n=1}^{\infty }\frac{sin( 4 \pi n \sqrt{I + \lambda^{2}} )}{2n} = \frac{\pi}{b}[/tex].
We assume a and b are nonzero rationals.
Hmm. What's next?
Guest wrote:Guest wrote:An Update: We correct the previous post.Guest wrote:We do not know!
Now suppose (1) [tex]\sum_{n=1}^{\infty }\frac{sin( 2 \pi (2n-1) \sqrt{I + \lambda^{2}} )}{2n-1} = \frac{\pi}{a}[/tex].
And suppose, (2) [tex]\sum_{n=1}^{\infty }\frac{sin( 4 \pi n \sqrt{I + \lambda^{2}} )}{2n} = \frac{\pi}{b}[/tex].
We assume a and b are nonzero rationals.
Hmm. What's next?
Let [tex]I = 257 * 127[/tex]. We compute [tex]\lambda = 65[/tex]. Let's compute equations one and two. What are the values for a and b?
Dave wrote:An Update: We correct the previous post.
"We do not know!"
Now suppose (1) [tex]\sum_{n=1}^{\infty }\frac{sin( 2 \pi (2n-1) \sqrt{I + \lambda^{2}} )}{2n-1} = \frac{\pi}{a}[/tex].
And suppose, (2) [tex]\sum_{n=1}^{\infty }\frac{sin( 4 \pi n \sqrt{I + \lambda^{2}} )}{2n} = \frac{\pi}{b}[/tex].
We assume a and b are nonzero real numbers.
Hmm. What's next?
Guest wrote:Remark: (1) [tex]\triangle n(I, \lambda) = \frac{1}{2} - \frac{1}{\pi} * \sum_{k=1}^{\infty }\frac{sin( 2 \pi k \sqrt{I + \lambda^{2}} )}{k} = 0[/tex].
Please solve that equation for [tex]\lambda[/tex] with our known value, [tex]I[/tex]. Please see the previous post. Thank you!
Guest wrote:Remark: Montgomery's Pair Correlation Conjecture is true!
Go Blue!
Guest wrote:Guest wrote:Remark: Montgomery's Pair Correlation Conjecture is true!
Go Blue!
Hah! Please prove Montgomery's Pair Correlation Conjecture. A claim is not proof
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