Series termination

Question

The successive terms in a power series are given by the recurrence relation $$\frac{a_{n+1}}{a_{n}}= \frac{n(n-1)+\lambda}{9(n+1)(n+2)}$$ where $\lambda=\text{const.}\in\mathbb{R}$. So our power series is \begin{align}\sum^{\infty}_{n=0}a_{n}x^{n} &= a_0 + a_1x + a_2x^2+a_3x^3+\cdots\\&=a_{0}\left(1 + \frac{\lambda}{18}x^{2} + \frac{\lambda(\lambda+2)}{1944}x^{4} + \frac{\lambda(\lambda+2)(\lambda+12)}{524880}x^{6} +\cdots\right) \\ &+ a_{1} x \left(1 + \frac{\lambda}{54}x^{2} + \frac{\lambda(\lambda+6)}{9760}x^{4} \frac{\lambda(\lambda+6)(\lambda+20)}{3674160}x^{6} + \cdots\right) \\ &= a_{0}\,f{(x)} + a_{1}x\,g{(x)}\end{align} Where \begin{align}f(x) &= 1 + \frac{\lambda}{18}x^{2} + \frac{\lambda(\lambda+2)}{1944}x^{4} + \frac{\lambda(\lambda+2)(\lambda+12)}{524880}x^{6} +\cdots \\ g(x) &= 1 + \frac{\lambda}{54}x^{2} + \frac{\lambda(\lambda+6)}{9760}x^{4} \frac{\lambda(\lambda+6)(\lambda+20)}{3674160}x^{6} + \cdots\end{align} I must find all the values $\lambda$ for which either $f(x)$ and $g(x)$ terminate (ie. form a polynomial). Although i'm not really sure what to do. If either one of these terminate then the general term $a_{n} = 0$ for some $n\ge0$. This happens if and only if $\lambda = n(1-n) \quad \forall n\ge0$. But this is using the recurrence relation for both of them together, how would I get a condition for them separately?

Answer 1

If lambda equals n(1-n) with n even, then your power series for f(x) terminates. You then get a polynomial for your original power series by choosing a_1 to be 0. (By "original power series" I mean your power series for a_0*f + a_1*x*g).

On the other hand, if lambda equals n(1-n) with n odd, then your power series for g(x) terminates. You then get a polynomial for your original power series by choosing a_0 to be 0.

Thus, assuming neither a_0 nor a_1 is equal to zero, there is no single value of lambda that will yield a polynomial for your original power series!

Lastly, Im not sure where you got this question from, but just so that you know its physical significance: these sort of recurrence relations are very common in solving the Schrodinger equation in quantum theory.

Hope this helps, and please feel free to ask any more questions if you need!