# Leading term of a power-series solution

1. Jan 25, 2015

### StephvsEinst

Hey! I have a question: how can you find the power-series solution and its leading term of the following ODE around the point x=+-(1-b) (where b is near x):

$$-(1-x^2)\frac{\partial^2 f^m_l}{\partial x^2}+2x\frac{f^m_l}{\partial x}+\frac{m^2}{1-x^2}f^m_l=l(l+1)f^m_l ,$$

and m and l are constants.

EDIT: Let me ask a better question - how can you find the solution to the following ODE:

$$sin(\theta)\frac{\partial}{\partial \theta}(sin(\theta)\frac{\partial f(\theta)}{\partial \theta})+[l(l+1)sin^2(\theta)-m^2)]f(\theta)=0$$

Thank you.

Last edited: Jan 25, 2015
2. Jan 26, 2015

### Staff: Mentor

I didn't solve it, but defining f(θ)=g(sin(θ)) makes the equation easier. Maybe the more general f(θ)=g(sin(θ)) + h(cos(θ)) is even better.

3. Jan 26, 2015

### StephvsEinst

I already solved it: it isn't an homonegeous ODE, so to calculate its power-series solution you substitute x=cos(θ)=sqrt(1-(sin(θ))^2). The solution is:

$$\Theta(\theta)=A_l^mP_l^m(x)$$

where
$$P_l^m(x)\equiv (1-x^2)^{\vert m \vert/2}\left(\frac{d}{dx}\right)P_l(x)$$.
This is its associated Legendre function.

4. Jan 27, 2015

### StephvsEinst

What I meant was that the solution is:
$$\Theta(\theta)=A_l^mP_l^m(cos\theta)$$

where

$$P_l^m(x)\equiv (1-x^2)^{\vert m \vert/2}\left(\frac{d}{dx}\right)P_l(x),$$

where

$$x=cos\theta.$$