Have I found correct formula? $\zeta(3)$

Question

Have I found the correct formula? Or is this only numerical aproximation? $\zeta(3)=\frac{2{\pi}^2}{7}(\ln 2-\frac{4}{15})$

Reedited: I add another aproximation(may be better): $\zeta(3)=\frac{2{\pi}^2}{7}(\ln 2-\frac{e\;\pi}{32})$

Just for fun, new one: $$\zeta(3)\dot{=}4\pi^2\left(\ln\left(\frac{16665385931}{9990000}\right)-e^2\right)$$

Your approximation is $\approx 1.2026268$, $\zeta(3) \approx 1.2020569$ — Daniel Fischer, Dec 02 '13 at 16:59
Thank you, as I expected, my guess is only bad aproximation. But may be will inspire real mathematicians to find correct formula. — Marek, Dec 02 '13 at 17:06

score 5 · Answer 1 · answered Dec 02 '13 at 19:11

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If such a formula were to exist, AND depend on $\pi$, it would be a function of $\pi^3$. But it's unlikely, since the equation of the circle is NOT $x^3+y^3=r^3$. Which is why only even zetas depend on $\pi$.

answered Dec 02 '13 at 19:11

Lucian

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Can you expand on your last two sentences please? Where did the comparison to the circle equation come from? I would love to know! – zerosofthezeta Dec 05 '13 at 06:22
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All even Zetas are dependent upon the value of $\zeta(2)$ due to a combination between Euler's infinite product expression for the sine function, and Euler's formula, as shown here. At the same time, there are countless formulas linking the Zeta and Gamma functions. But then again, the Gamma function itself is linked to geometric shapes, as seen here. – Lucian Dec 05 '13 at 14:44

score 2 · Answer 2 · edited Mar 10 '17 at 19:47

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See Apéry's constant. I think that's just an approximation.

edited Mar 10 '17 at 19:47

Lucian

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answered Dec 02 '13 at 16:58

tenpercent

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Have I found correct formula? $\zeta(3)$

2 Answers2