# Polynomial solution to Legendre's equation

1. Nov 1, 2015

### Mark Brewer

• Member warned about posting with no effort
1. The problem statement, all variables and given/known data
Starting from the recurrence relation, show that, when l is an integer, the polynomial solution to Legendre's equation is
yl(x) = Kl ∑ from k = 0 to (l/2) of (((-1)k) / k!) (((2l - 2k)!) / (l-k)! (l - 2k)!) (xl-2k)

where Kl is an arbitrary constant (depending on l) and x (floor) stands for the largest integer less than or equal to x. Hint. THis is a tricky exercise in algebraic manipulation. Rewrite the recurrence relation in the form al-2k = f(l,k) al-2k+2, (where you are to evalute f(l,k) explicitly), then run out the recurrence to obtain the desired result. You will need to be clever in manipulating factorial expression. For example, somewhere along the line you will have to use the formula (l)(l-1)...(l-2k+1) = ((l!) / (l-2k)!), and a bunch more like it.
Mod note: Fixed. One of your closing subscript tags -- [/sub]) was actually [sub].
2. Relevant equations

recurrence equation: am+2 = [((l-m)(l+m+1) / (m+2)(m+1))

3. The attempt at a solution

I'm lost to where to start. Any help to get me started would be much appreciated.]

Last edited by a moderator: Nov 1, 2015
2. Nov 1, 2015

### Mark Brewer

I'm not sure why the second half of my problem is in a subscript. The double checked the problem by previewing it, but I could get the text out of being in a subscript. My apologies for this.

3. Nov 1, 2015

### Staff: Mentor

You need to try something -- that's the rule here at this forum. You could start with the hints given in the problem.

Also, this doesn't look like a recurrence equation to me. The right side should have $a_m$ and $a_{m+1}$ in it.

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