Legendre Polynomials - how to find P0(u) and P2(u)?

[rwooduk]

Pl(u) is normalized such that Pl(1) = 1. Find P0(u) and P2(u)

note: l, 0 and 2 are subscript



recursion relation

an+2 = [n(n+1) - l (l+1) / (n+2)(n+1)] an

n is subscript

substituted λ = l(l+1) and put n=0 for P0(u) and n=2 for P2(u), didnt get very far

please could someone give me a point in the right direction or a youtube video to explain the process used to solve this? really lost.

thanks for any advice

 

[Ray Vickson]

Pl(u) is normalized such that Pl(1) = 1. Find P0(u) and P2(u)

note: l, 0 and 2 are subscript



recursion relation

an+2 = [n(n+1) - l (l+1) / (n+2)(n+1)] an

n is subscript

substituted λ = l(l+1) and put n=0 for P0(u) and n=2 for P2(u), didnt get very far

please could someone give me a point in the right direction or a youtube video to explain the process used to solve this? really lost.

thanks for any advice

You wrote
[tex] a_n + 2 = \left[n(n+1) - \frac{l(l+1)}{(n+2)(n+2)} \right] a_n [/tex]
Is that what you really mean, or did you want
[tex] a_{n+2} = \left[ \frac{n(n+1) - l(l+1)}{(n+2)(n+1)}\right] a_n ?[/tex]
If so, please use parentheses properly.

Anyway, I don't see what this approach will get you, since you have the recursion for a fixed value of ##l##---that is, the recursion that tells you how to get ##P_l(x)## for fixed ##l##, given some initial conditions. There is nothing there that tells you how to relate ##P_{l-1}(x)## and ##P_{l+1}(x)## to ##P_l(x)##. Normalization comes from outside the differential equation and outside the recursion. Look up Legendre functions in a textbook or via Google.

Note added in edit: I see that you might have meant ##P_l##; the font you used made it look like ##P_1##. Anyway, you might find it useful to look at http://www.phys.ufl.edu/~fry/6346/legendre.pdf .

 

[rwooduk]

apologies, yes it's this:

[tex] a_{n+2} = \left[ \frac{n(n+1) - l(l+1)}{(n+2)(n+1)}\right] a_n[/tex]

i think the idea is to put l=0 for P0(u) and l=2 for P2(u)

but not really sure what to do after that.

will have a look at the link, thank-you.