- #1
ehrenfest
- 2,020
- 1
The even solutions of an SHO are:
[tex]h^+(y) = \sum_{s = 0}^{\infty}a_s y^{2s} [/tex]
where a is given by the recursion
[tex] a_{s+1} = a_s \left( \frac{4s + 1 - \epsilon}{2(s+1)(2s+1)} \right)[/tex]
The solutions are square integrable iff
a_n = 0 so that the polynomial is finite.
What I do not understand is why my book (Robinett) says
h_0(y) = 1 and not 0 when a_0 = 0 for n = 0?
[tex]h^+(y) = \sum_{s = 0}^{\infty}a_s y^{2s} [/tex]
where a is given by the recursion
[tex] a_{s+1} = a_s \left( \frac{4s + 1 - \epsilon}{2(s+1)(2s+1)} \right)[/tex]
The solutions are square integrable iff
a_n = 0 so that the polynomial is finite.
What I do not understand is why my book (Robinett) says
h_0(y) = 1 and not 0 when a_0 = 0 for n = 0?