On the origins of Taylor/Maclaurin Series

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


\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

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)