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#### chisigma

##### Well-known member

- Feb 13, 2012

- 1,704

That's the Google's home page today... how many Euler's milestones are missing?...

Kind regards

$\chi$ $\sigma$

Last edited:

- Thread starter chisigma
- Start date

- Thread starter
- #1

- Feb 13, 2012

- 1,704

That's the Google's home page today... how many Euler's milestones are missing?...

Kind regards

$\chi$ $\sigma$

Last edited:

- Jan 17, 2013

- 1,667

1-\(\displaystyle \sum^{\infty}_{n=1}\frac{1}{n^s}= \prod^{\infty}_{p} (1-p^{-1})^{-1}\)

2-\(\displaystyle \frac{\sin(\pi x) }{\pi} = x\prod^{\infty}_{n=1} \left( 1-\frac{x^2}{n^2} \right) \)

3-\(\displaystyle \sum^{\infty}_{n=1}\frac{1}{n^2}= \frac{\pi^2}{6}\)

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- Feb 13, 2012

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At the time the most relevant unsolved problem in Basel was to evaluate the 'infinite sum'...

$\displaystyle S= \sum_{n=1}^{\infty} \frac{1}{n^{2}}$ (1)

The convergence of (2) had been demonstrated but nobody had any idea on how to evaluate the sum. Leonhard arrived to solve the 'Basel problem' avoiding the 'brute force attak' but with simple considerations about the well known function $\sin x$. It was well known the 'infinite sum'...

$\displaystyle \sin x = \sum_{n=0}^{\infty} \frac {(-1)^{n}}{(2n +1)!}\ x^{2n+1} = x - \frac {x^{3}}{6} + ... $ (2)

... and the first step of Leonhard was to devide the (2) by x obtaining...

$\displaystyle \frac{\sin x}{x} = \sum_{n=0}^{\infty} \frac {(-1)^{n}}{(2n +1)!}\ x^{2n} = 1 - \frac {x^{2}}{6} + ... $ (3)

The first step is not extraordinary but succesive steps are very exciting and reveal the genious. The (3) is, althogh with infinite terms, a 'polynomial' and we know that a polynomial of degree k can be written as...

$\displaystyle P(x) = a\ \prod_{n=1}^{k} (1-\frac{x}{x_{n}})$ (4)

... where $a=P(0)$ and the $\displaystyle x_{n}, n=1,2,...,k$ are the 'zeroes' of P(x). Very well!... what are the ‘zeroes’ of $\displaystyle P(x)= \frac{\sin x} {x}$ ?... not a very difficult task, they are $\displaystyle x_{n}= \pm \pi\ n,\ n=1,2,...$, so that we can write...

$\displaystyle \frac{\sin x}{x} = \prod_{n=1}^{\infty} (1- \frac{x^{2}}{\pi^{2}\ n^{2}})$ (5)

Now we can imagine how happy was Leonhard when hi saw that, comparing (3) and (5), he arrived to solve the Basel's problem finding that is...

$\displaystyle \frac{1}{\pi^{2}} \sum_{n=1}^{\infty} \frac{1}{n^{2}} = \frac{1}{6}$ (6)

All that is of course well known... a little less undestable for me is that, almost three centuries after, neither the 'infinite sum' (3) nor the 'infinite product' (5) are present in most of the 'Holybooks'...

Kind regards

$\chi$ $\sigma$