# Associated Legendre polynomials

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1. Nov 22, 2014

### bvol

m=1 and l=1

x = cos(θ)

What would be the solution to this?

Thanks.

Last edited: Nov 22, 2014
2. Nov 22, 2014

### ZetaOfThree

We won't do the calculation for you. Why can't you do it yourself? What part of it doesn't make sense to you?

3. Nov 22, 2014

### bvol

I've tried a lot of things, but I don't get it. (NOT a physics/math student)

I get this:

4cos(θ)^3 - 4cos(θ)

But it should be sin(θ) since I'm applying this formula:

and the solution for 1,1 is Sqrt(3/8π) e^((+/-)iφ) * sin(θ)
I get how the part in italics is derived, but not how the part in bold has been derived from the associated legendre polynomial..

4. Nov 22, 2014

### ZetaOfThree

I think you've made an error in your calculation. If we plug in $\ell = 1$ and $m=1$, we have $$P^{1}_{1}={(-1)^1 \over 2 \cdot 1!} (1-x^2)^{1/2} \cdot {d^2 \over dx^2}(x^2-1)$$
Do you have that much? It should be easy to simplify that, then plug in $x=\cos{\theta}$.

5. Nov 22, 2014

### bvol

Great I have it now. -(1-cos()^2)^(1/2), which is equal to sin(θ).

Right I have it now, like you said I made a small error in not applying l+m, but instead just l and some errors on other places as well.

Thanks for the quick reply! :)