A strange operator

March 9, 2017 / 153armstrong / 1 Comment

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 !

Beautiful proofs(#2): Euler’s Sum

March 5, 2017October 27, 2017 / 153armstrong / Leave a comment

$1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \hdots = \frac{\pi^2}{6}$

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)

On the origins of Taylor/Maclaurin Series

March 2, 2017 / 153armstrong / Leave a comment

Many a times it is not discussed as to How the Taylor/Maclaurin series came to be in its current form. This short snippet is all about it.

Let us assume that some function $f(x)$ can be written as a power series expansion. i.e

$f(x) = a_0 + a_1 x + a_2 x^2 + \hdots$ .

We are left with the task of finding out the coefficients of the power series expansion.

Substitution x = 0, we obtain the value of $a_0$ .

$a_0 = f(0)$ .

Lets differentiate $f(x)$ wrt x.

$\frac{d}{dx} f(x) = a_1 + 2a_2 x + \hdots$

Evaluating at x =0 , we get

$\frac{d}{dx} f(0) = a_1$

And likewise:

$\frac{d^2}{dx^2} f(0) = 2.1.a_2 = 2! \space a_2$

$\frac{d^3}{dx^3} f(0) = 3.2.1.a_3 = 3! \space a_3$

$\vdots$

$\frac{d^n}{dx^n} f(0) = n.n-1...3.2.1.a_n = n! a_n$

That’s it we have found all the coefficient values, the only thing left to do is to plug it back into the power series expression:

$f(x) = f(0) + \frac{d}{dx}f(0) \frac{x}{1!} + \frac{d^2}{dx^2}f(0) \frac{x^2}{2!} + \frac{d^3}{dx^3} f(0) \frac{x^3}{3!} \hdots$ .

The above series expanded about the point x = 0 is called as the ‘Maclaurin Series’. The same underlying principle can be extended for expanding about any other point as well i.e ‘Taylor Series’.

Legendre Differential Equation(#2): A friendly introduction

February 26, 2017 / 153armstrong / 1 Comment

Now there is something about the Legendre differential equation that drove me crazy. What is up with the l(l+1) !!!

$(1-x^2)y^{''} -2xy^{'} + l(l+1)y = 0$

To understand why let’s take this form of the LDE and arrive at the above:

$(1-x^2)y^{''} -2xy^{'} + \lambda y = 0$

$y = \sum\limits_{n=0}^{\infty} a_n x^n$

If we do a power series expansion and following the same steps as the previous post, we end up with the following recursion relation.

$(n+2)(n+1)a_{n+2} = (\lambda -n(n+1))a_n$

$a_{n+2} = a_n \frac{\lambda - n(n+1)}{(n+1)(n+2)}$

Here’s the deal: We want a convergent solution for our differential solution. This means that as $n \rightarrow l , a_{n+2} \rightarrow 0$ .

Hence we obtain that

$\lambda = l(l+1)$

Legendre Differential equation (#1) : A friendly introduction

February 26, 2017February 26, 2017 / 153armstrong / 2 Comments

In this series of posts about Legendre differential equation, I would like to de-construct the differential equation down to the very bones. The motivation for this series is to put all that I know about the LDE in one place and also maybe help someone as a result.

The Legendre differential equation is the following:

$(1-x^2)y^{''} -2xy^{'} + l(l+1)y = 0$

where $y^{'} = \frac{dy}{dx}$ and $y^{''} = \frac{d^{2}y}{dx}$

We will find solutions for this differential equation using the power series expansion i.e
$y = \sum\limits_{n=0}^{\infty} a_n x^n$

$y^{'} = \sum\limits_{n=0}^{\infty} na_n x^{n-1}$

$y^{''} = \sum\limits_{n=0}^{\infty} n(n-1)a_n x^{n-2}$

We will plug in these expressions for the derivatives into the differential equation.

$l(l+1)y = l(l+1)\sum\limits_{n=0}^{\infty} a_n x^n$ – (i)

$-2xy^{'} = -2\sum\limits_{n=0}^{\infty} na_n x^{n}$ – (ii)

$(1-x^2)y^{''} = (1-x^2)\sum\limits_{n=0}^{\infty} n(n-1)a_n x^{n-2}$

$= \sum\limits_{n=0}^{\infty} n(n-1)a_n x^{n-2} - \sum\limits_{n=0}^{\infty} n(n-1)a_n x^{n}$ – (iii)

** Note: Begin

$\sum\limits_{n=0}^{\infty} n(n-1)a_n x^{n-2}$

Let’s take $\lambda = n-2$ .
As n -> $0$ . , $\lambda$ -> $-2$ .
As n -> $\infty$ , $\lambda$ -> $\infty$ .

$\sum\limits_{\lambda = -2}^{\infty} (\lambda+2)(\lambda+1)a_n x^{\lambda}$

$= 0 + 0 + \sum\limits_{\lambda = 0}^{\infty} (\lambda+2)(\lambda+1)a_n x^{\lambda}$

Again performing a change of variables from $\lambda$ to n.

$= \sum\limits_{n= 0}^{\infty} (n+2)(n+1)a_n x^{n}$

** Note: End

(iii) can now be written as follows.

$\sum\limits_{n=0}^{\infty} x^n \left((n+1)(n+2)a_{n+2} - n(n-1)a_n \right)$ – (iv)

(i)+(ii)+(iv).

$\sum\limits_{n=0}^{\infty} x^n \left((n+2)(n+1)a_{n+2} - (l(l+1)-n(n+1))a_n \right)$

x = 0 is a trivial solution and therefore we get the indicial equation:

$(n+2)(n+1)a_{n+2} - (l(l+1)-n(n+1))a_n = 0$

$(n+2)(n+1)a_{n+2} = (l^2 - n^2 + l - n)a_n = 0$

$(n+2)(n+1)a_{n+2} = ((l-n)(l+n)+ l - n)a_n = 0$

$(n+2)(n+1)a_{n+2} = (l-n)(l+n+1)a_n = 0$

We get the following recursion relation on the coefficients of the power series expansion.

$a_{n+2} = a_n \frac{(l+n+1)(l-n)}{(n+1)(n+2)}$

Next post: What do these coefficients mean ?

Beautiful proofs (#1) : Divergence of the harmonic series

February 26, 2017 / 153armstrong / Leave a comment

The harmonic series are as follows:

And it has been known since as early as 1350 that this series diverges. Oresme’s proof to it is just so beautiful.

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.

Clearly the lower bound of S₁ diverges and therefore S₁ 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 !

Beautiful proofs (#1) : Divergence of the harmonic series

February 22, 2017February 22, 2017 / 153armstrong / Leave a comment

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 !

Ecstasy Shots

A poetry of logical ideas

series

A strange operator

Beautiful proofs(#2): Euler’s Sum

On the origins of Taylor/Maclaurin Series

Legendre Differential Equation(#2): A friendly introduction

Legendre Differential equation (#1) : A friendly introduction

Beautiful proofs (#1) : Divergence of the harmonic series

Beautiful proofs (#1) : Divergence of the harmonic series