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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?
(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?
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