# Legendre's Diff Eq using Frobenius

1. Dec 26, 2011

### brew_guru

1. The problem statement, all variables and given/known data

Show, by direct examination of the Frobenius series solution to Legendre's differential equation that;
$$P_n(x) = \sum_{k=0}^{N} \frac{(-1)^k(2n-k)!} {2^n k! (n-k)! (n-2k)!}x^{n-2k} ;\ N=\frac{n}{2}\ \mathrm{n\ even,}\ N=\frac{n-1}{2}\ \mathrm{n\ odd}$$

Write down the first five Legendre polynomials due to this formula. Show also that $$P_n(1)=1$$ and $$P_n(-1)=(-1)^n$$

2. Relevant equations
The Frobenius/Puiseux/Fuchs series solution is: $$y= \sum_{n=0}^{\infty} a_{n}x^{k+n}$$
Legendre's Diff Eq is: $$(1-x^{2})y''-2xy'+l(l+1)=0$$

3. The attempt at a solution
$$y= \sum_{n=0}^{\infty} a_{n}x^{k+n} y'= \sum_{n=0}^{\infty} (k+n)a_{n}x^{k+n-1} y''= \sum_{n=0}^{\infty} (k+n)(k+n-1)a_{n}x^{k+n-2}$$
Setting the $$x^k$$ power to zero yields:
$$a_2 = \frac{k(k+1)-l(l+1)}{(k+2)(k+1)} a_0$$
Setting the $$x^{k+1}$$ power to zero yields:
$$a_3 = \frac{(k+1)(k+2)+l(l+1)}{(k+3)(k+2)}a_1$$
I believe setting the $$x^{k-2}\ \mathrm{and}\ x^{k-1}$$ powers yield:
$$k(k-1)a_0 = 0 k(k+1)a_1 = 0$$
although this would seem to imply my first two constants are zero, which would make all higher order constants zero as well since each constant can be written in terms of an earlier constant. This is the first part I am stuck on, and the second is getting an answer in terms of $$P_n(x)$$ instead of $$y(x)$$. Thank you for any direct help or links to reading material!

Last edited: Dec 26, 2011
2. Dec 26, 2011

### LCKurtz

Use the forward slash to close tex tags. You should also preview your posts to confirm they are displaying correctly. I fixed your tags so your post is readable. I don't have time right now to look at it but at least now someone might read it.

Last edited: Dec 26, 2011
3. Dec 26, 2011

### vela

Staff Emeritus
Since you're expanding about x0=0, which isn't a singular point of the differential equation, you can expand the solution as a plain old Taylor series:
$$y = \sum_{n=0}^\infty a_n x^n$$You don't need to mess with the k+n nonsense. When you do this, you get the same recursion relation as before.

Note if you set k=0 in your work above, the conditions
\begin{align*}
k(k-1)a_0 &= 0 \\
k(k+1)a_1 &= 0
\end{align*} are satisfied automatically. There are no restrictions on a0 and a1.

Last edited: Dec 26, 2011