# Fibonacci sequence in the hiding

What ?!! There exists such an elegant decimal representation of the Fibonacci sequence? Well yes! and the only thing that you need to know to prove this is that if the Fibonacci numbers were the coefficients to a power series expansion, then the Fibonacci generating function is given as follows:

$1x + 1x^{2} + 2x^{3} + 3x^{4} + 5x^{5} + \hdots = \frac{x}{1-x-x^{2}}$

Subsituting the value of $x = \frac{1}{10}$, we get :

$\frac{1}{10} + \frac{1}{10}^{2} + 2(\frac{1}{10})^{3} + 3(\frac{1}{10})^{4} + 5(\frac{1}{10})^{5} + \hdots = \frac{\frac{1}{10}}{1-\frac{1}{10}-\frac{1}{10}^{2}}$

$0.1 + 0.01 + 0.002 + 0.0003 + 0.00005 + \hdots = \frac{10}{89}$

$0.01 + 0.001 + 0.0002 + 0.00003 + 0.000005 + \hdots = \frac{1}{89}$

Proved. 😀

# 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:

$f(x) = (x-r_1)(x-r_2) = 0$  (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)

# 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 !