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September 29th, 2014, 09:58 PM   #1
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Riemann Zeta Function proof

i have my proof in a Microsoft word document download link:
Proof of The Riemann Zeta function
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September 30th, 2014, 07:14 AM   #2
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If I understand correctly, you're claiming that $\zeta(-52)=0$ and $\zeta(2)\approx1.0191193511008111.$

$\zeta(-n)=0$ for all even integers n > 1, but $\zeta(2)=\pi^2/6\approx1.6449340668$ has been known for a long time (the Basel problem).

I'll note that neither of these values are related to the Riemann hypothesis.
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September 30th, 2014, 09:03 PM   #3
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Quote:
Originally Posted by CRGreathouse View Post
If I understand correctly, you're claiming that $\zeta(-52)=0$ and $\zeta(2)\approx1.0191193511008111.$

$\zeta(-n)=0$ for all even integers n > 1, but $\zeta(2)=\pi^2/6\approx1.6449340668$ has been known for a long time (the Basel problem).

I'll note that neither of these values are related to the Riemann hypothesis.
Well did you check -52,-42,-32,-22,-12 (all my solutions) or just assumed that they were wrong (likely) because I got zeta at 2 wrong.

also how was zeta 2 calculated because what I got was

zeta(n) ≈ 1 + n+31/n-(12^n+1)

n = 2

so

zeta(2) ≈ 1 + 33/1728 = 1.0190972222222222

zeta(2) should be roughly = 1.0190972222222222

so I'm confused now.

Thanks in advance if anyone can explain this to me.

Last edited by skipjack; October 4th, 2014 at 04:03 AM.
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October 1st, 2014, 06:22 AM   #4
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$\zeta(2)=1+\frac{1}{2^2}+\frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \cdots$ which already shows that $\zeta(2)>1.5.$ If you know a little calculus you can see that
$$
1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \int_8^{\infty}\frac{dx}{x^2} < \zeta(2) < 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \int_7^{\infty}\frac{dx}{x^2}
$$
where the lower and upper bounds are 1.636797 and 1.6546542, respectively. Of course you can choose any cutoff you like to get more or less precise.
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October 1st, 2014, 08:43 AM   #5
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$$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$
follows from the Fourier Series expansion of $$f(x) = \frac{3x^2 - 6\pi x + 2\pi^2}{12} = \sum_{n=1}^\infty \frac{\cos nx}{n^2} \qquad x \in (0, 2\pi) $$by setting $x=0$.
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