Use partial fractions to write $\frac{1}{x^2 + x – 1}$ as $\frac{A_1}{x – α_1} + \frac{A_2}{x – α_2}$ and write it as a power series

Question

Find the roots $α_1$, $α_2$ of $x^2 + x – 1$ and use partial fractions to write $\frac{1}{x^2 + x – 1}$ as $\frac{A_1}{x – α_1} + \frac{A_2}{x – α_2}$ , for suitable $A_1, A_2$. Using the power series for $\frac{1}{x – α}$ , find another power series for f(x). First of all, this problem is for the Fibonacci sequence. I have already showed that one of the power series f(x) = $\frac{-1}{x^2+x-1}$ and now I am trying to find another one. This is what I have done so far: I know that the roots $α_1=-(\frac{1+\sqrt5}{2})$ and $α_2=-(\frac{1-\sqrt5}{2})$, so $$\frac{A_1}{x + \frac{1+\sqrt5}{2}} + \frac{A_2}{x + \frac{1-\sqrt5}{2}}$$ $$1=A_1(\frac{1-\sqrt5}{2}) + A_2(\frac{1+\sqrt5}{2})$$ $$A_1=-A_2$$ $$1=-A_2(\frac{1-\sqrt5}{2}) +A_2(\frac{1+\sqrt5}{2})$$ $$1=\sqrt5A_2$$ $$A_2=\frac{1}{\sqrt5}$$$$A_1=-\frac{1}{\sqrt5}$$ This would make the partial fraction be $$\frac{-\frac{1}{\sqrt5}}{x +\frac{1+\sqrt5}{2}} + \frac{\frac{1}{\sqrt5}}{x +\frac{1-\sqrt5}{2}}$$. I'm not sure however if this would be correct. If it isn't, where did I go wrong? If it is, how can I write this as a power series?

Answer 1

HINT: You know the standard power series

$$\frac1{1-x}=\sum_{n\ge 0}x^n\tag{1}$$

for a simple geometric series, and you have functions of the form $\frac1{x+a}$. The trick is to manipulate what you have to replace $a$ by $1$:

$$\frac1{x+a}=\frac{\frac1a}{1+\frac{x}a}=\frac1a\cdot\frac1{1-\left(-\frac{x}a\right)}\;,$$

and you can now use $(1)$.

To get the power series I find it a bit easier to use partial fractions to write

$$\frac{-1}{1-x-x^2}=\frac{A}{1-\alpha x}+\frac{B}{1-\beta x}\;,$$

so that the denominators already have the desired $1$. Since $1-x-x^2=(1-\alpha x)(1-\beta x)$, we have $\alpha\beta=-1$ and $\alpha+\beta=1$, and solving the system for $\alpha$ and $\beta$ yields

$$\alpha=\frac{1+\sqrt5}2\qquad\text{and}\qquad\beta=\frac{1-\sqrt5}2\;.$$

I’ll leave it to you to solve for $A$ and $B$. Once you have them, you simply write

$$\begin{align*} \frac1{x^2+x-1}&=\frac{-1}{1-x-x^2}\\ &=\frac{A}{1-\alpha x}+\frac{B}{1-\beta x}\\ &=A\sum_{n\ge 0}(\alpha x)^n+B\sum_{n\ge 0}(\beta x)^n\\ &=\sum_{n\ge 0}\left(A\alpha^n+B\beta^n\right)x^n\;. \end{align*}$$