Beautiful proofs(#2): Euler’s Sum

$1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \hdots = \frac{\pi^2}{6}$

Say what? This one blew my mind when I first encountered it. But it turns out Euler was the one who came up with it and it’s proof is just beautiful!

Prerequisite
Say you have a quadratic equation $f(x)$ whose roots are $r_1,r_2$, then you can write $f(x)$ as follows:

$f(x) = (x-r_1)(x-r_2) = 0$  (or)

$f(x) = (r_1-x)(r_2-x) = 0$  (or)

$f(x) = (1- \frac{x}{r_1})(1- \frac{x}{r_2}) = 0$

$f(x) = 1 - (\frac{1}{r_1} + \frac{1}{r_2}) + \frac{x^2}{r_1 r_2} = 0$

As for as this proof is concerned we are only worried about the coefficient of $x$, which you can prove that for a n-degree polynomial is:

$a_1 = - (\frac{1}{r_1} + \frac{1}{r_2} + \hdots + + \frac{1}{r_n})$

where $r_1,r_2 \hdots r_n$ are the n-roots of the polynomial.

Now begins the proof

It was known to Euler that

$f(y) = \frac{sin(\sqrt{y})}{\sqrt{y}} = 1 - \frac{1}{3!}y + \hdots$

But this could also be written in terms of the roots of the equation as:

$f(y) = \frac{sin(\sqrt{y})}{\sqrt{y}} = 1 - (\frac{1}{r_1} + \frac{1}{r_2} + \hdots + + \frac{1}{r_n})y + \hdots$

Now what are the roots of $f(y)$ ?. Well, $f(y) = 0$ when $\sqrt{y} = n \pi$ i.e $y = n^2 \pi^2$ *

The roots of the equation are $y = \pi^2, 4 \pi^2, 9 \pi^2, \hdots$

Therefore,

$f(y) = \frac{sin(\sqrt{y})}{\sqrt{y}} = 1 - \frac{1}{3!}y + \hdots = 1 -( \frac{1}{\pi^2} + \frac{1}{4 \pi^2} + \hdots )y + \hdots$

Comparing the coefficient of y on both sides of the equation we get that:

$\frac{1}{6} = \frac{1}{\pi^2} + \frac{1}{4 \pi^2} + \frac{1}{ 9 \pi^2} + \hdots$

$\zeta(2) = \frac{\pi^2}{6} = 1 + \frac{1}{4} + \frac{1}{9} + \hdots$

Q.E.D

* n=0 is not a root since
$\frac{sin(\sqrt{y})}{\sqrt{y}} = 1$ at y = 0

** Now if all that made sense but you are still thinking : Why on earth did Euler use this particular form of the polynomial for this problem, read the first three pages of this article. (It has to do with convergence)

Legendre Differential Equation(#2): A friendly introduction

Now there is something about the Legendre differential equation that drove me crazy. What is up with the l(l+1) !!!

$(1-x^2)y^{''} -2xy^{'} + l(l+1)y = 0$

To understand why let’s take this form of the LDE and arrive at the above:

$(1-x^2)y^{''} -2xy^{'} + \lambda y = 0$

$y = \sum\limits_{n=0}^{\infty} a_n x^n$

If we do a power series expansion and following the same steps as the previous post, we end up with the following recursion relation.

$(n+2)(n+1)a_{n+2} = (\lambda -n(n+1))a_n$

or

$a_{n+2} = a_n \frac{\lambda - n(n+1)}{(n+1)(n+2)}$

Here’s the deal: We want a convergent solution for our differential solution. This means that as $n \rightarrow l , a_{n+2} \rightarrow 0$.

Hence we obtain that

$\lambda = l(l+1)$

Beautiful proofs (#1) : Divergence of the harmonic series

The harmonic series are as follows:

$\sum\limits_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \hdots$

And it has been known since as early as 1350 that this series diverges. Oresme’s proof to it is just so beautiful.

$S_1 = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \hdots$

$S_1 = 1 + \left(\frac{1}{2}\right) + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) \hdots$

Now replace ever term in the bracket with the lowest term that is present in it. This will give a lower bound on $S_1$.

$S_1 > 1 + \left(\frac{1}{2}\right) + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} \right) \hdots$

$S_1 > 1 + \left(\frac{1}{2}\right) + \left(\frac{1}{2}\right) + \left(\frac{1}{2}\right) + \left(\frac{1}{2}\right) + \hdots$

Clearly the lower bound of $S_1$ diverges and therefore $S_1$ also diverges. 😀
But it interesting to note that of divergence is incredibly small: 10 billion terms in the series only adds up to around 23.6 !