# Recurrence relation for the Legendre functions

1. May 31, 2012

### aaaa202

My book wants to find solutions to Legendre's equation:

(1-x2)y'' - 2xy' 0 l(l+1)y = 0 (1)

By assuming a solution of the form:

y = Ʃanxn , the sum going from 0->∞ (2)

Now by plugging (2) into (1) one finds:

Ʃ[n(n-1)anxn-2-n(n-1)anxn - 2nanxn +
l(l+1)anxn = 0

Rearranging my book writes this as:

Ʃ[(n+2)(n+1)an+2 - (n(n+1)-l(l+1))an]xn = 0

I can't understand the emphasized bit. How did we go from xn-2 to xn using an+2?

Last edited: May 31, 2012
2. May 31, 2012

### Ray Vickson

Until you get the hang of these things the best policy is to write it out in a bit more detail: from $y = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n + \cdots,$ we have
$$y' = a_1 + 2 a_2 x + 3a_3 x^2 + \cdots + (n+1) a_{n+1} x^n + \cdots, \\ y'' = 2a_2 + 3\cdot 2 a_3 x + 4 \cdot 3 a_4 x^2 + \cdots + (n+2)(n+1) a_{n+2} x^n + \cdots.$$ So now you can write out the terms of
$$(1-x^2) y'' - 2xy' + l(l+1) y,$$ collect all the terms in $x^0, \: x^2, \: x^2, \ldots, x^n, \ldots$ and require them all to be zero.

RGV