inverse of polynomial as sum of two power series

Question

given polynomial $$g(x) =ax^2+bx+c$$
I try to find $g(x)^{-1}$ as a sum of two power series.

I wrote $g(x) = \alpha\beta(1-\frac{x}{\alpha})(1-\frac{x}{\alpha})$ when $ \alpha,\beta = \frac{ - b\pm\sqrt{b^2 - 4ac}}{2a} \in C$
then $$g(x)^{-1} = \frac{1}{\alpha\beta} \frac{1}{(1-\frac{x}{\alpha})(1-\frac{x}{\beta})} = \frac{1}{\alpha\beta} \left[\frac{A}{(1-\frac{x}{\alpha})} + \frac{B}{(1-\frac{x}{\beta})} \right]$$
then I found $$A = \frac{-\beta}{\alpha - \beta} ,B = \frac{\alpha }{\alpha - \beta}$$
so I can get $$g(x)^{-1} = \frac{1}{\alpha\beta}\left[\frac{-\beta}{\alpha - \beta} \sum_{n=0}^{n=\infty} {\left(\frac{x}{\alpha }\right)}^n + \frac{\alpha }{\alpha - \beta} \sum_{n=0}^{n=\infty} {\left(\frac{x}{\beta}\right)}^n\right]$$

my question is, is there any way to assure that the each power series is Real or change it into sum of two Real power series?

Answer 1

The coefficient of $x^n$ will be $$\dfrac1{\alpha\beta(\alpha-\beta)}\left(\dfrac\alpha{\beta^n}-\dfrac\beta{\alpha^n}\right)=\dfrac{\alpha^{n+1}-\beta^{n+1}}{(\alpha\beta)^{n+1}(\alpha-\beta)}$$

As

$$(\alpha+\beta)(\alpha^n-\beta^n)=\alpha^{n+1}-\beta^{n+1}+\alpha\beta(\alpha^{n-1}-\beta^{n-1})$$

If $f(m)=\dfrac{\alpha^m-\beta^m}{\alpha-\beta},$ $$f(n+1)=(\alpha+\beta)f(n)-\alpha\beta f(n-1)$$

Use strong induction to establish that if $f(m)$ is real for $m\le n,f(n+1)$ will be real

Assumption: $\alpha+\beta,\alpha\beta$ are real