piiiiiiiiiiii

[khansaheb]

 

[mario99]

[imath]\displaystyle \pi = \int_{-1}^{1}\frac{1}{\sqrt{1 - x^2}} \ dx[/imath]

 

[fresh_42]

I like
[math] \zeta(2)=\dfrac{\pi^2}{6}=\prod_{p\text{ prime}}\dfrac{1}{1-{p}^{-2}}=\prod_{p\text{ prime}} \dfrac{p^2}{(p-1)(p+1)} [/math]I mean, what in the world have prime numbers to do with a circle?

 

[mario99]

I like
[math] \zeta(2)=\dfrac{\pi^2}{6}=\prod_{p\text{ prime}}\dfrac{1}{1-{p}^{-2}}=\prod_{p\text{ prime}} \dfrac{p^2}{(p-1)(p+1)} [/math]I mean, what in the world have prime numbers to do with a circle?

It is fascinating to see that they are related but cannot tell the exact reason. This zeta function, [imath]\zeta(2)[/imath], is famous as the exact solution of [imath]\displaystyle \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}[/imath]. Would you be the first one to tell the world the exact solution of [imath]\displaystyle \zeta(3) = \sum_{k=1}^{\infty} \frac{1}{k^3} = \ ?[/imath]. I would not be surprised if it includes [imath]\pi[/imath] and prime numbers.

 

[topsquark]

Would you be the first one to tell the world the exact solution of [imath]\displaystyle \zeta(3) = \sum_{k=1}^{\infty} \frac{1}{k^3} = \ ?[/imath]. I would not be surprised if it includes [imath]\pi[/imath] and prime numbers.

It has been proven that there is no closed form solution for odd zeta function values.

-Dan

 

[topsquark]

Khaaaaaaaaaaaaaaaaaaaaansaheb!

-Dan

 

[fresh_42]

It is fascinating to see that they are related but cannot tell the exact reason.

That's astonishingly easy: Let [imath] \mathbb{P} [/imath] be the set of primes, and recall that the zeta-function is
[math] \zeta(s)=\displaystyle{\sum_{n=1}^\infty \dfrac{1}{n^s}}. [/math]
[math]\begin{array}{lll} \left(1-\dfrac{1}{p^s}\right)\zeta(s)&=\displaystyle{\sum_{n=1}^\infty \dfrac{1}{n^s}-\sum_{n=1}^\infty \dfrac{1}{(pn)^s} =\sum_{\stackrel{n=1}{n\not\in p\mathbb{Z}}}^\infty \dfrac{1}{n^s} =\sum_{\stackrel{n=1}{p\nmid n}}^\infty \dfrac{1}{n^s}}\\[24pt] \prod_{p\in\mathbb{P}}\left(1-\dfrac{1}{p^s}\right)\zeta(s)&=\displaystyle{\sum_{\stackrel{n=1}{n\not\equiv 0\mod p\,\forall \,p\in \mathbb{P}}}^\infty \dfrac{1}{n^s}=\sum_{n\in \{1\}}\dfrac{1}{n^s}=1}\\[24pt] \zeta(s)&=\displaystyle{1/\prod_{p\in\mathbb{P}}\left(1-\dfrac{1}{p^s}\right)= \prod_{p\in\mathbb{P}}\left(1/\left(1-\dfrac{1}{p^s}\right)\right)=\prod_{p\in\mathbb{P}}\dfrac{1}{1-p^{-s}}} \end{array}[/math]

This zeta function, [imath]\zeta(2)[/imath], is famous as the exact solution of [imath]\displaystyle \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}[/imath]. Would you be the first one to tell the world the exact solution of [imath]\displaystyle \zeta(3) = \sum_{k=1}^{\infty} \frac{1}{k^3} = \ ?[/imath]. I would not be surprised if it includes [imath]\pi[/imath] and prime numbers.

[imath] \zeta(3) [/imath] is called Apéry's constant and although there are quite a few funny formulas, some even involving [imath] \pi, [/imath] there doesn't seem to be a nice closed formula without infinite sums, products or nasty integrals like
[math] \zeta(3)=\displaystyle{\dfrac{2}{3}\pi^3\int_0^1 x\left(x-\dfrac{1}{2}\right)\left(x-1\right) \cot( \pi x) \,dx} [/math]

 

[mario99]

It has been proven that there is no closed form solution for odd zeta function values.

-Dan

Thank you Dan for telling me this new information.

That's astonishingly easy: Let [imath] \mathbb{P} [/imath] be the set of primes, and recall that the zeta-function is
[math] \zeta(s)=\displaystyle{\sum_{n=1}^\infty \dfrac{1}{n^s}}. [/math]

Oops, the multiplication trick did not come in my mind. Even with this trick it is still fascinating!

[imath] \zeta(3) [/imath] is called Apéry's constant and although there are quite a few funny formulas, some even involving [imath] \pi, [/imath] there doesn't seem to be a nice closed formula without infinite sums, products or nasty integrals like
[math] \zeta(3)=\displaystyle{\dfrac{2}{3}\pi^3\int_0^1 x\left(x-\dfrac{1}{2}\right)\left(x-1\right) \cot( \pi x) \,dx} [/math]

I would prefer to approximate the infinite series [imath]\displaystyle \sum_{k=1}^{\infty}\frac{1}{k^3}[/imath] rather than solving that nasty integral.