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

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Why on earth is matrix multiplication NOT commutative ? – Intuition

One is commonly asked to prove in college as part of a linear algebra problem set that matrix multiplication is not commutative. i.e If A and B are two matrices then :

AB \neq BA

But without getting into the Algebra part of it, why should this even be true ? Let’s use linear transformations to get a feel for it.

If A and B are two Linear Transformations namely Rotation and Shear. Then it means that.

(Rotation)(Shearing) \neq (Shearing)(Rotation)

Is that true? Well, lets perform these linear operations on a unit square and find out:

(Rotation)(Shearing)

2

(Shearing)(Rotation)

1

You can clearly see that the resultant shape is not the same upon the two transformations. This means that the order of matrix multiplication matters a lot ! ( or matrix multiplication is not commutative.)

Basis Vectors are instructions !

Basis vectors are best thought of in the context of roads.

Imagine you are in a city – X which has only roads that are perpendicular to one another.

0003

You can reach any part of the city but the only constraint is that you need to move along these perpendicular roads to get there.

lon6gm

 

Now lets say you go to another city-Y which has a different structure of roads.

0001

In this case as well you can get from one part of the city to any other, but you have to travel these ‘Sheared cubic’ pathways to get there.

xgwzgl

Just like these roads determine how you move about in the city, Basis Vectors encode information on how you move about on a plane. What do I mean by that ?

The basis vector of City-X is given as:

screenshot-from-2017-02-19-021951

\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

This to be read as – ” If you would like to move in City-X you can only do so by taking 1 step in the x-direction or 1 step in the y-direction ”

The basis vector of City-Y is given as:

screenshot-from-2017-02-19-021644

\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}

This to be read as – ” If you would like to move in City-Y you can only do so by taking 1 step in the x-direction or  1 step along the diagonal OB ”

 

Conclusion:

By having the knowledge about the Basis Vectors of any city, you can travel to any destination by merely scaling these basis vectors.

As an example, lets say need to get to the point (3,2), then in City-X,  you would take 2 steps in the x-direction and 3 steps in the y-direction

\begin{bmatrix} 3 \\  2 \end{bmatrix}  =  3* \begin{bmatrix} 1 \\  0 \end{bmatrix}  +  2 * \begin{bmatrix} 0 \\  1 \end{bmatrix}

And similarly in City-Y, you would take 1 step along the x -direction and 2 steps along the diagonal OB.

\begin{bmatrix} 3 \\  2 \end{bmatrix}  =  1* \begin{bmatrix} 1 \\  0 \end{bmatrix}  +  2 * \begin{bmatrix} 1 \\  1 \end{bmatrix}

Destination Arrived 😀

On the travelling wave: An intuition

The aim of this post is to understand the travelling wave solution. It is sometimes not explained in textbook as to why the solution “travels”.

We all know about our friend – ‘The sinusoid’.

y = sin(x)

y becomes 0 whenever sin(x) = 0 i.e x = n \pi

Now the form of the travelling sine wave is as follows:

y= sin(x - \omega t)

When does the value for y become 0 ? Well, it is when

x - \omega t = n \pi

x = n \pi + \omega t

As you can see this value of x is dependent on the value of time ‘t’, which means as time ticks, the value of x is pushed forward/backward by a \omega .

drawing

When the value of \omega > 0 , the wave moves forward and when \omega < 0 , the wave moves backward.

Here is a slowly moving forward sine wave.

nxg3p