0
$\begingroup$

I am trying to express this equation in terms of theta:

$$(1-x^2)y''(x)-2xy'(x)+n(n+1)y(x)=0$$

where

$$x=cos\theta$$

I know I can begin with Legendre's equation:

$$(1-x^2)\frac{d^2P_n}{dx^2}-2x\frac{dP_n}{dx}+n(n+1)P_n(x)=0$$

For some reason,

$$\frac{dP_n}{dx}=\frac{d\Theta_n}{d\theta}\frac{d\theta}{dx}=-\frac{1}{sin\theta}\frac{d\Theta_n}{d\theta}$$

I do not understand where the

$$\frac{1}{sin\theta} $$

comes from.

Can someone explain this to me?

$\endgroup$
1
$\begingroup$

I may have figured it out actually, very simple.

If

$$x=cos\theta$$

then

$$dx/d\theta=-sin\theta$$

then just inverse

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.