My Math Forum Riemann Zeta Function proof

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 September 29th, 2014, 09:58 PM #1 Newbie   Joined: Sep 2014 From: place Posts: 16 Thanks: 0 Riemann Zeta Function proof i have my proof in a Microsoft word document download link: Proof of The Riemann Zeta function
 September 30th, 2014, 07:14 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 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.
September 30th, 2014, 09:03 PM   #3
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 Originally Posted by CRGreathouse 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.

 October 1st, 2014, 06:22 AM #4 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms $\zeta(2)=1+\frac{1}{2^2}+\frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \cdots$ 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. Thanks from topsquark
 October 1st, 2014, 08:43 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,031 Thanks: 2342 Math Focus: Mainly analysis and algebra $$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$ 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 $x=0$. Thanks from topsquark

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