Undergrad Understanding Legendre differential equation

[chwala]

TL;DR

Analysis of $(1-x^2)y^{"}-2x y'+ n(n+1)y=0$.

I am looking at this now, $(1-x^2)y^{"}-2x y' + n(n+1)y=0$, my interest being on $P_n(x)$, i will go straight to the point,

... steps leading to this are clear, i.e

we have the recursive relationship,

$a_{k+2} = \dfrac{(k-n)(k+n+1)}{(k+2)(k+1)}$

For even solutions, we set $a_1 =0$ and $a_0=1$ Is this always the case?, of course i get the even and odd part... using symmetry,

Now in my understanding, we shall end up with,

$k=0, a_2=-3$

$k=2, a_4 = 0$

...

$y(x)=a_0x^0 + a_2x^2 + a_4x^4$ (which terminates at $k=2$).

$y(x)=1-3x^2= P_2(x)$ my step here is clear, now my confusion is here or misunderstanding, arises from the following lines,

$y(x)=a_0(1-3x^2)$ where is this $a_0$ coming from where? or its set again for conformity? and then further, $P_2(x)= \dfrac{1}{2}(3x^2-1)$, how?

...i read scale factor or something, i may need a clear explanation...cheers !

 

[anuttarasammyak]

https://en.wikipedia.org/wiki/Legendre_polynomials

Does your sequence coincide with this ?

 

[pasmith]

TL;DR Summary: Analysis of $(1-x^2)y^{"}-2x y'+ n(n+1)y=0$.

I am looking at this now, $(1-x^2)y^{"}-2x y' + n(n+1)y=0$, my interest being on $P_n(x)$, i will go straight to the point,

... steps leading to this are clear, i.e

we have the recursive relationship,

$a_{k+2} = \dfrac{(k-n)(k+n+1)}{(k+2)(k+1)}$

This is the correct recurrence. If you se l = k - 2 you will get it in the form posted by @anuttarasammyak.

For even solutions, we set $a_1 =0$ and $a_0=1$ Is this always the case?

Note that the ODE is linear, so a constant multiple of a solution is a solution. The same is true of the recurrence relation for the a_k. But the recurrence has to start somewhere, so a_0 and a_1 are arbitrary constants.

The recurrence is of the form a_{k+2} = f(k)a_k. It follows that if a_1 = 0 and a_0 \neq 0 then y will be even, and if a_0 = 0 and a_1 \neq 0 then y will be odd.

The Legendre polynomials are normalized such that P_n(1) = 1. This means that we don't get P_{2m} by setting (a_0,a_1) = (1,0); instead we have to leave a_0 arbitrary until we know the value of y(1) in terms of a_0.