I Understanding Legendre differential equation

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chwala
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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)} ##

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 !
 
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https://en.wikipedia.org/wiki/Legendre_polynomials
1744630981099.png


Does your sequence coincide with this ?
 
chwala said:
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.
 
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