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)

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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 !