The generalized product rule ( Leibniz Formula )

March 1, 2017March 1, 2017 / 153armstrong / Leave a comment

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 $(x+y)^n$

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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(#4) : A friendly introduction

March 1, 2017 / 153armstrong / Leave a comment

When you are working with Spherical harmonics, then the Legendre Differential Equation does not appear in its natural form i.e

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

Instead, it appears in this form:

$\frac{d^2 y}{d\theta^2} + \frac{cos\theta}{sin\theta} \frac{dy}{d\theta} + l(l+1)y = 0$

It seems daunting but the above is the same as the LDE. We can arrive at it by taking $x = cos(\theta)$ and proceeding as follows:

$\frac{dy}{dx} = \frac{dy}{d(cos\theta)} = \frac{-1}{sin\theta}\frac{dy}{d \theta}$

$\frac{d^2 y}{dx^2} = \frac{d}{d(cos\theta)}\left( \frac{-1}{sin\theta}\frac{dy}{d \theta} \right) = \frac{-1}{sin\theta}\frac{d}{d \theta}\left( \frac{-1}{sin\theta}\frac{dy}{d \theta} \right)$

Now, applying chain rule, we obtain that

$\frac{d^2 y}{dx^2} = \frac{-1}{sin\theta} \left( \frac{-1}{sin\theta} \frac{d^2 y}{d\theta^2} - \frac{cos\theta}{sin^2 \theta} \frac{dy}{d\theta} \right)$

Now simplifying the above expression, we obtain that:

$\frac{d^2 y}{dx^2} = \frac{1}{sin^2\theta} \left( \frac{d^2 y}{d\theta^2} + \frac{cos\theta}{sin\theta} \frac{dy}{d\theta} \right)$

Plugging in the values of $\frac{dy}{dx}$ and $\frac{d^2 y}{dx^2}$ into the Legendre Differential Equation,

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

$(1-cos^2 \theta)y^{''} -2cos\theta y^{'} + l(l+1)y = 0$

$\frac{1- cos^2 \theta}{sin^2 \theta} \left( \frac{d^2 y}{d\theta^2} + \frac{cos\theta}{sin\theta}\frac{dy}{d\theta} \right) +\frac{2cos\theta}{sin\theta} \frac{dy}{d\theta} + l(l+1)y = 0$

Now if we do some algebra and simplify the trigonometric identities, we will arrive at the following expression for the Legendre Differential Equation:

$\frac{d^2 y}{d\theta^2} + \frac{cos\theta}{sin\theta} \frac{dy}{d\theta} + l(l+1)y = 0$

If we take the solution for the LDE as $f(x)$ , then the solution to the LDE in the above form is merely $f(cos\theta)$ .

Legendre Differential Equation(#3): A friendly introduction

February 27, 2017 / 153armstrong / Leave a comment

This post is just a note on the notation that is used across internet sources and books while referring to the LDE.

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

If one takes $p(x) = 1-x^2$ , then it follows that $p^{'}(x) = -2x$ . The differential equation can be rewritten as follows:

$p(x)y^{''} + p^{'}(x)y^{'} + l(l+1) y = 0$

Now the first two terms must seem familiar to you from the chain rule. ( $(py)^{'} = py^{'} + yp^{'}$ ). Ergo,

$(py^{'})^{'} + l(l+1)y = 0$

$\frac{d}{dx}(p \frac{dy}{dx}) + l(l+1)y = 0$

Now, putting back the value of p :

$\frac{d}{dx}\left((1-x^2) \frac{dy}{dx} \right) + l(l+1)y = 0$

And you will see this form of the LDE also in many places and I thought it was worth mentioning how one ended up in that form.

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