# The generalized product rule ( Leibniz Formula )

If f and g are n-times differentiable functions, then : $(fg)^{'} = fg^{'} + gf^{'}$

Now, we would like to find out a generalized expression for the n-th derivative of fg. In order to arrive at that formulation lets calculate a few derivatives to see whether we can find a pattern: $(fg)^{'} = fg^{'} + gf^{'}$ $(fg)^{''} = \left(fg^{'} + gf^{'}\right)^{'} = fg^{''} + 2 f^{'}g^{'} + gf^{''}$ $(fg)^{'''} = \left(fg^{''} + 2 f^{'}g^{'} + gf^{''} \right)^{'} = fg^{'''} + 3 f^{''}g^{'} + 3 f^{'}g^{''} + gf^{'''}$ $(fg)^{''''} = fg^{''''} + 4 f^{'''}g^{'} + 6f^{''}g^{''} + 4 f^{'}g^{'''} + gf^{''''}$ $\vdots$

You must have noticed a pattern in the above expressions. The coefficients seem are the one in the binomial expansion of $(x+y)^n$ Therefore we can write the expression for the n-derivative of fg as the following expression: $(fg)^n = \sum\limits_{i=0}^{n} \binom{n}{i} f^{(i)}g^{(n-i)}$
where (i) means to differentiate i-times.

This is also known as Leibniz Formula.

** This plays an important role when we start discussing about the Associated Legendre Differential Equation.

# Legendre Differential equation (#1) : A friendly introduction

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 ?