## Beautiful Proofs(#3): Area under a sine curve !

So, I read this post on the the area of the sine curve some time ago and in the bottom was this equally amazing comment :

$\sum sin(\theta)d\theta =$  Diameter of the circle/ The distance covered along the x axis starting from $0$ and ending up at $\pi$.

And therefore by the same logic, it is extremely intuitive to see why:

$\int\limits_{0}^{2\pi} sin/cos(x) dx = 0$

Because if a dude starts at $0$ and ends at $0/ 2\pi/ 4\pi \hdots$, the effective distance that he covers is 0.

If you still have trouble understanding, follow the blue point in the above gif and hopefully things become clearer.

## 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:

$f(x) = x^2 - (r_1 + r_2) x + r_1r_2$

You can also divide throughout by $r_1r_2$ and arrive at this form:

$f(x) = r_1r_2 \left( \frac{x^2}{r_1r_2} - (\frac{1}{r_1} + \frac{1}{r_2}) x + 1 \right)$

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,

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

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

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

$\frac{\pi^2}{6} = 1 + \frac{1}{4} + \frac{1}{9} + \hdots = S_2$

Q.E.D

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

## On the beauty of Parametric Integration and the Gamma function

Parametric integration is one such technique that once you are made aware of it, you will never for the love of god forget it. Let me demonstrate :

Now this integral might seem familiar to many of you and to evaluate it is rather simple as well.

$\int\limits_0^{\infty} e^{-sx} dx = \frac{1}{s}$

Knowing this you can do lots of crazy stuff. Lets differentiate this expression wrt to the parameter in the integral – s (Hence the name parametric integration ). i.e

$\frac{d}{ds}\int\limits_0^{\infty} e^{-sx} dx = \frac{d}{ds}\left(\frac{1}{s}\right)$

$\int\limits_0^{\infty} x e^{-sx} dx = \frac{1}{s^2}$

Look at that, by simple differentiation we have obtained the expression for another integral. How cool is that! It gets even better.
Lets differentiate it once more:

$\int\limits_0^{\infty} x^2 e^{-sx} dx = \frac{2*1}{s^3}$

$\int\limits_0^{\infty} x^3 e^{-sx} dx = \frac{3*2*1}{s^4}$

$\vdots$

If you keep on differentiating the expression n times, one gets this :

$\int\limits_0^{\infty} x^n e^{-sx} dx = \frac{n!}{s^{n+1}}$

Now substituting the value of s to be 1, we obtain the following integral expression for the factorial. This is known as the gamma function.

$\int\limits_0^{\infty} x^n e^{-x} dx = n! = \Gamma(n+1)$

There are lots of ways to derive the above expression for the gamma function, but parametric integration is in my opinion the most subtle way to arrive at it. 😀

## 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 !