# Remembering Leonhard Euler's birthday...

#### chisigma

##### Well-known member

Kind regards

$\chi$ $\sigma$

Last edited:

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Nothing significant just a small list

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

#### chisigma

##### Well-known member
In the year 1720, when he was only thirteen, Leonhard Euler entered in the University of Basel where he graduated in Philosophy. Leonhard's father, a protestant pastor, wanted him to become theologian but Johann Bernoulli, professor in Basel, convinced him that Math was his future...

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

$\chi$ $\sigma$