## Why is the area under one hump of a sine curve exactly 2?

I was talking with a student recently who told me that he always found the fact that $latex int_0^{pi} sin x , dx = 2$ amazing. “How is it that the area under one hump of the sine curve comes out exactly 2?” He asked me if there is an easy way to see that, or is it something you just have to discover by doing the computation.

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## 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.

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

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:

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

## An infinite joke

$\frac{1}{0} = \infty$

Now flip this over by 90 degree counter clockwise :

$- 10 = 8$

$- 18 = 0$

Now flip this over again by 90 degree clockwise :

$\frac{1}{\infty} = 0$

## 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

## 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)

(Shearing)(Rotation)

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.)