Decompose the function to power series $f(x) = \frac{1}{1-x-x^2},x_o=0$

Question

We have the $f(x)$ function to represent it as power series with center at defined point $x_0$:

$$f(x) = \frac{1}{1-x-x^2},x_o=0$$

Also we should define the radius of convergence.

It seems like we should use the formula:

$$\frac{1}{1-t}=\sum_{n=0}^\infty t^n,0<t<1$$

I tried to use it this way:

$$f(x)=\{t=x+x^2\}=\frac{1}{1-t}=\sum_{n=0}^\infty t^n = \sum_{n=0}^\infty (x+x^2)^n$$

Then we can found radius of convergence using following formula: $$ \mathbb {C}=\lim_{n\rightarrow\infty}\sqrt[n]{|b_n|}=\lim_{n\rightarrow\infty}\sqrt[n]{|x+x^2|^n}=|x+x^2|$$

Convergence is true, while: $|x+x^2|<1$

Solving this, we get: $$-\frac12-\frac{\sqrt{5}}2<x<-\frac12+\frac{\sqrt{5}}2$$

So, it seems that we got $x_0=- \frac12$, but this is not $0$ ($x_0=0$)

I got this problem from "Collection of problems in mathematical analysis, Kudryavtsev. Book 2.". Under the theme "Power series".

Answer 1

Rewrite,

$$f(x) = \frac{1}{1-x-x^2}=\frac{1}{(1-\frac xa)(1-\frac xb)}$$

where $a=\frac{-1+\sqrt 5}{2}$ and $b=\frac{-1-\sqrt 5}{2}$ are the roots of $1-x-x^2 = 0$. Then, decompose the denominator,

$$f(x) = \frac{p}{1-\frac xa} + \frac{q}{1-\frac xb}$$

where $p =\frac{5+\sqrt 5}{10}$ and $q= \frac{5-\sqrt 5}{10}$. Now, expand the two terms separately,

$$\frac{1}{1-\frac xa}=\sum_{n=0}^\infty \frac {x^n}{a^n}, \>\>\> \frac{1}{1-\frac xb}=\sum_{n=0}^\infty \frac {x^n}{b^n} $$

The first expansion converges if $|x| < a = \frac{\sqrt 5 -1}{2}$ and the second expansion converges if $|\frac xb| < 1$ and then $ |x| < |b| = \frac{\sqrt 5 +1}{2}$.

$ \frac{\sqrt 5 -1}{2}<\frac{\sqrt 5 +1}{2}$, thus, both converge if

$$|x| < \frac{\sqrt 5 -1}{2}$$

Add the two expansions together to get the power series of $f(x)$,

$$f(x) = p\sum_{n=0}^\infty \frac {x^n}{a^n} + q\sum_{n=0}^\infty \frac {x^n}{b^n} = \sum_{n=0}^\infty \left( \frac {p}{a^n} + \frac{q}{b^n}\right)x^n $$

where the values of $p$, $q$, $a$ and $b$ have been derived earlier.

Answer 2

Some hints and additional info referring to comments:

• If there is any convergent power series expansion of $f$ around $x=0$, then it must satisfy $$f(x) - xf(x) - x^2 f(x) \stackrel{!}{=} 1 \Leftrightarrow \sum_{n=0}a_nx^n - \sum_{n=0}a_nx^{n+1}- \sum_{n=0}a_nx^{n+2} \stackrel{!}{=} 1$$
• Writing as a single power series gives $$a_0 + (a_1 -a_0)x + \sum_{n=2}^{\infty}(a_n - a_{n-1}-a_{n-2})x^n \stackrel{!}{=}1$$
• This gives a recursion for $a_n$ which turns out to be the recursion for the Fibonacci-sequence $F_n$: $$a_0=1, a_1 = a_0, a_n = a_{n-1}+a_{n-2} \; (n\geq 2)\Rightarrow \boxed{a_n = F_n}$$
• The explicit formula for $F_n$ (which can be found, for example, by solving the recursion) is $$F_n = \frac{1}{\sqrt{5}}\left( \left( \frac{1+\sqrt{5}}{2}\right)^n - \left( \frac{1-\sqrt{5}}{2}\right)^n \right)$$
• Since $\boxed{\lim_{n\to \infty}\frac{F_{n+1}}{F_n} = \phi = \frac{1+\sqrt{5}}{2}}$ (the golden ratio), you get the radius of convergence $$\boxed{R = \frac{1}{\phi}}$$