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 :

Screenshot from 2017-06-07 00:19:11

\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.

Circle_cos_sin.gif

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

 

nth roots of unity : A geometric approach

When one is dealing with complex numbers, it is many a times useful to think of them as transformations. The problem at hand is to find the nth roots of unity. i.e

z^n = 1

Multiplication as a Transformation

Multiplication in the complex plane is mere rotation and scaling. i.e

z_{1} = r_{1}e^{i\theta_{1}}, z_{2} = r_{2}e^{i\theta_{2}} 

z_{1}z_{2} = \underbrace{r_{1} r_{2}}_{scaling} \underbrace{e^{i(\theta_{1} + \theta_{2})}}_{rotation}

Now what does finding the n roots of unity mean?

If you start at 1 and perform n equal rotations( because multiplication is nothing but rotation + scaling ), you should again end up at 1.

We just need to find the complex numbers that do this.i.e

z^n = 1

\underbrace{zz \hdots z}_{n} = 1

z = re^{i\theta}

r^{n}e^{i(\theta + \theta + \hdots \theta)} = 1e^{2\pi k i}

r^{n}e^{in\theta} =1e^{2\pi k i}

This implies that :

\theta = \frac{2\pi k}{n}, r = 1

And therefore :

z = e^{\frac{2\pi k i}{n}}

Take a circle, slice it into n equal parts and voila you have your n roots of unity.

main-qimg-6da134e3e5735a9fb92355d53f95e4ed

Okay, but what does this imply ?

Multiplication by 1 is a 360^o/0^o rotation.

drawing

When you say that you are multiplying a positive real number(say 1) with 1 , we get a number(1) that is on the same positive real axis.

Multiplication by (-1) is a 180^o rotation.

drawing-1

When you multiply a positive real number (say 1) with -1, then we get a number (-1) that is on the negative real axis

The act of multiplying 1 by (-1) has resulted in a 180o transformation. And doing it again gets us back to 1.

Multiplication by i is a 90^o rotation.

drawing-2

Similarly multiplying by i takes 1 from real axis to the imaginary axis, which is a 90o rotation.

This applies to -i as well.

That’s about it! – That’s what the nth roots of unity mean geometrically. Have a good one!

 

Solving the Laplacian in Spherical Coordinates (#1)

In this post, let’s derive a general solution for the Laplacian in Spherical Coordinates. In future posts, we shall look at the application of this equation in the context of Fluids and Quantum Mechanics.

sph_coor

x = rsin\theta cos\phi
y = rsin\theta cos\phi
z = rcos\theta

where

0 \leq r < \infty
0 \leq \theta \leq \pi
0 \leq \phi < 2\pi

The Laplacian in Spherical coordinates in its ultimate glory is written as follows:

\nabla ^{2}f ={\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial f}{\partial r}}\right)+{\frac {1}{r^{2}\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial f}{\partial \theta }}\right)+{\frac {1}{r^{2}\sin ^{2}\theta }}{\frac {\partial ^{2}f}{\partial \phi ^{2}}} = 0

To solve it we use the method of separation of variables.

f = R(r)\Theta(\theta)\Phi(\phi)

Plugging in the value of f into the Laplacian, we get that :

\frac{\Theta \Phi}{r^2} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right) + \frac{R \Phi}{r^2 sin \theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{d\theta} \right) + \frac{\Theta R}{r^2 sin^2 \theta} \frac{d^2 \Phi}{d\phi^2} = 0

Dividing throughout by R\Theta\Phi and multiplying throughout by r^2, further simplifies into:

\underbrace{ \frac{1}{R} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right)}_{h(r)} + \underbrace{\frac{1}{\Theta sin \theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{d\theta} \right) + \frac{1}{\Phi sin^2 \theta} \frac{d^2 \Phi}{d\phi^2}}_{g(\theta,\phi)} = 0

It can be observed that the first expression in the differential equation is merely a function of r and the remaining a function of \theta and \phi only. Therefore, we equate the first expression to be \lambda = l(l+1) and the second to be -\lambda = -l(l+1). The reason for choosing the peculiar value of l(l+1) is explained in another post.

\underbrace{ \frac{1}{R} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right)}_{l(l+1)} + \underbrace{\frac{1}{\Theta sin \theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{d\theta} \right) + \frac{1}{\Phi sin^2 \theta} \frac{d^2 \Phi}{d\phi^2}}_{-l(l+1)} = 0 (1)

 

The first expression in (1) the Euler-Cauchy equation in r.

\frac{d}{dr} \left( r^2\frac{dR}{dr} \right) = l(l+1)R

The general solution of this has been in discussed in a previous post and it can be written as:

R(r) = C_1 r^l + \frac{C_2}{r^{l+1}}

 

The second expression in (1) takes the form as follows:

\frac{sin \theta}{\Theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ l(l+1)sin^2 \theta + \frac{1}{\Phi} \frac{d^2 \Phi}{d\phi^2} = 0

The following observation can be made similar to the previous analysis

\underbrace{\frac{sin \theta}{\Theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ l(l+1)sin^2 \theta }_{m^2} + \underbrace{\frac{1}{\Phi} \frac{d^2 \Phi}{d\phi^2}}_{-m^2} = 0 (2)

 

The first expression in the above equation (2) is the Associated Legendre Differential equation.

\frac{sin \theta}{\Theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ l(l+1)sin^2 \theta = m^2

sin \theta \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ \Theta \left( l(l+1)sin^2 \theta - m^2 \right) = 0

The general solution to this differential equation can be given as:
\Theta(\theta) = C_3 P_l^m(cos\theta) + C_4 Q_l^m(cos\theta)

 

The solution to the second term in the equation (2) is a trivial one:

\frac{d^2 \Phi}{d\phi^2} = m^2 \Phi
\Phi(\phi) = C_5 e^{im\phi} + C_6 e^{-im\phi}

 

Therefore the general solution to the Laplacian in Spherical coordinates is given by:

R\Theta\Phi = \left(C_1 r^l + \frac{C_2}{r^{l+1}} \right) \left(C_3 P_l^m(cos\theta) + C_4 Q_l^m(cos\theta \right) \left(C_5 e^{im\phi} + C_6 e^{-im\phi}\right)

A strange operator

In a previous post on using the Feynman’s trick for Discrete calculus, I used a very strange operator ( \triangledown ). And whose function is the following :

\triangledown n^{\underline{k}} = \frac{n^{\underline{k+1}}}{k+1}

What is this operator? Well, to be quite frank I am not sure of the name, but I used it as an analogy to Integration. i.e

\int x^{n} = \frac{x^{n+1}}{n+1} + C

What are the properties of this operator ? Let’s use the known fact that n^{\underline{k+1}} = (n-k) n^{\underline{k}}

\triangledown n^{\underline{k}} = \frac{n^{\underline{k+1}}}{k+1}

\triangledown n^{\underline{k}} = \frac{(n-k) n^{\underline{k}}}{k+1}

And applying the operator twice yields:

\triangledown^2 n^{\underline{k}} = \frac{n^{\underline{k+2}}}{(k+1)(k+2)}

\triangledown^2 n^{\underline{k}} = \frac{(n-k-1) n^{\underline{k+1}}}{(k+1)(k+2)}

\triangledown^2 n^{\underline{k}} = \frac{(n-k-1)(n-k) n^{\underline{k}}}{(k+1)(k+2)}

We can clearly see a pattern emerging from this already, applying the operator once more :

\triangledown^3 n^{\underline{k}} = \frac{(n-k-2)(n-k-1)(n-k) n^{\underline{k}}}{(k+1)(k+2)(k+3)}

\vdots

Or in general, the operator that has the characteristic prescribed in the previous post is the following:

\triangledown^m n^{\underline{k}} = \frac{n^{\underline{k+m}}}{(k+m)^{\underline{m}}} n^{\underline{k}}

If you guys are aware of the name of this operator, do ping me !

Matrix Multiplication and Heisenberg Uncertainty Principle

We now understand that Matrix multiplication is not commutative (Why?). What has this have to do anything with Quantum Mechanics ?

Behold the commutator operator:
[\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A}

where \hat{A},\hat{B} are operators that are acting on the wavefunction \psi . This is equal to 0 if they commute and something else if they don’t.

One of the most important formulations in Quantum mechanics is the Heisenberg’s Uncertainty principle and it can be written as the commutation of the momentum operator (p) and the position operator (x):

[\hat{p}, \hat{x}] = \hat{p}\hat{x} - \hat{x}\hat{p} = i\hbar

If you think of p and x as some Linear transformations. (just for the sake of simplicity).

This means that measuring distance and then momentum is not the same thing as measuring momentum and then distance. Those two operators do not commute! You can sort of visualize them in the same way as in the post.

But in Quantum Mechanics, the matrices that are associated with \hat{p} and \hat{x} are infinite dimensional. ( The harmonic oscillator being the simple example to this )

\hat{x} = \sqrt{\frac{\hbar}{2m \omega}} \begin{bmatrix} 0 & \sqrt{1} & 0 & 0 & \hdots \\ \sqrt{1} &  0 &\sqrt{2} & 0 & \hdots \\ 0 & \sqrt{2} &  0 &\sqrt{3}  & \hdots \\  0 & 0 & \sqrt{3} &  0  & \hdots \\  \vdots & \vdots & \vdots & \vdots \end{bmatrix}

\hat{p} = \sqrt{\frac{\hbar m \omega}{2}} \begin{bmatrix} 0 & -i & 0 & 0 & \hdots \\ i &  0 & -i \sqrt{2} & 0 & \hdots \\ 0 & i\sqrt{2} &  0 &\-i \sqrt{3}  & \hdots \\  0 & 0 & i\sqrt{3} &  0  & \hdots \\  \vdots & \vdots & \vdots & \vdots \end{bmatrix}

 

 

 

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