From the following:

$$\sum_0^\infty [ n(n-1)a_nx^{n-2} - n(n-1)a_n x^n -2na_nx^n + l(l+1)a_n x^n ] = 0$$ (a)

I'm trying to get to:

$$\sum_0^\infty [ (n+2)(n+1)a_{n+2} - [n(n+1) + l(l+1)]a_n]x^n = 0$$ (b)

Unfortunately, I'm not being very successful.

I can either get the first term ($(n+2)(n+1)a_{n+2} x^n$) by saying that $n=n+2$ or get the 2nd and 3rd terms $[n(n+1) + l(l+1)]a_nx^n$ just by simple calculations.

So, how do I get from a) to b)?


IMO this is not your problem, but the sign of the $l(l+1)$ term is positive in (a) but negative in (b). You can split and re-combine the sums with index shift because the $n(n-1)$ term is zero for $n=0,1.\;$ Here my manipulations $$\sum_0^\infty \Big( n(n-1)a_nx^{n-2} - n(n-1)a_n x^n -2na_nx^n + l(l+1)a_n x^n \Big)=$$ $$\sum_0^\infty n(n-1)a_nx^{n-2} + \sum_0^\infty \left(- n(n-1) -2n + l(l+1) \right)a_n x^n=$$ $$\sum_2^\infty n(n-1)a_nx^{n-2} + \sum_0^\infty \left(- n(n+1) + l(l+1) \right)a_n x^n=$$ $$\sum_0^\infty (n+2)(n+1)a_{n+2}x^n + \sum_0^\infty \left(- n(n+1) + l(l+1) \right)a_n x^n=$$ $$\sum_0^\infty \Big( (n+2)(n+1)a_{n+2} - (n(n+1) - l(l+1))a_n \Big) x^n$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.