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 !
... 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 !