What's the radius of convergence for power series of $1/(1-(x+x^2))$? Is it symmetrical? When reading the section on composition of power series in Book Calculus With Analytic Geometry (George F. Simmons) (2nd edition, pp. 517), the author claimed you can replace $x$ in a power series $f(x):=\sum_{j=0}^\infty b_j x^j$ with another power series $g(z):=\sum_{k=0}^\infty a_k z^k$, and write the composition of two series as $h(z):=f\bigl(g(z)\bigr) = \sum_{k=0}^\infty c_k z^k$. However, I have problem with radius of convergence of this new series.
To be concrete, suppose I want to know the power series of $1/(1-x-x^2)$. I start with $1/(1-x) = \sum_{i=0}^\infty x^i, |x|<1$, and then replace $x$ with $x+x^2$. Since the constant term of $x+x^2$ ($g(z)$ in the above paragraph) is zero, I can certainly rearrange $\sum_{i=0}^\infty (x+x^2)^i$ in terms of $\sum_{k=0}^\infty c_k x^k$. When does this new series converge? Suppose I compute $ |x+x^2|<1$, then I get $(-1-\sqrt{5})/2 < x < (-1+\sqrt{5})/2 $. But, is this inconsistent with the theorem that the radius of convergence for a power series is symmetrical (say $(-R,R)$)?
 A: A start: An approach that is useful in the analysis of series like yours is to find the roots $\alpha$ and $\beta$ of $1-x-x^2=0$, and then imagine using  the method of partial fractions to express $\frac{1}{x^2+x-1}$ as $\frac{A}{x-\alpha}+\frac{B}{x-\beta}$. To imagine is enough, for our purposes we  don't need to know $A$ and $B$
The interval of convergence can now be read off. 
A: The reciprocals of the poles of the denominator of this rational function are the roots of $X^2-X-1$, which are the golden ratio $\phi$ and its "conjugate" $-\phi^{-1}$. The largest in absolute value of these two is $\phi$, so the coefficients $c_n$ of your series are asymptotically $\sim c\phi^n$ for some $c\neq0$, and the radius of convergence is $\frac1\phi$.
You make an interesting observation that this convergence condition is not equivalent to $|x+x^2|<1$. Let's see what happens when you expand $\sum_{i=0}^\infty (x+x^2)^i$. You get $1+x+2x^2+3x^3+5x^4-8x^5+\cdots$, which looks familiar; indeed it is $\sum_iF_{i+1}x^i$ where $F_i$ are the Fibonacci numbers (which indeed grow asymptotically as $\frac1{\sqrt5}\phi^n$). Now what happens if one sets $x=-1$, a value outside the radius of convergence, but which conveniently satisfies $|x+x^2|=0<1$? On one hand you have been expanding $\sum_{i=0}^\infty 0^i$ which should give $1$. On the other hand substituting $x=-1$ into the series gives $1-1+2-3+5-8+\cdots$ which diverges. What gives? Well if you try to find your powers of $x+x^2$ back in the power series, you can group together parts of several terms so that each group of terms (except the initial $1$) evaluates to $0$ for $x=-1$. But in order to do so, you have been, at $x=-1$, grouping together terms in an absolutely divergent series, and that is not allowed (and may produce a convergent series). Indeed an easier form of this swindle is
$$
  \begin{align}
  1-1+2-3+5-8-\cdots &= 1 - (0+1)+(1+1)-(1+2)+(2+3)-(3+5) \\
  &= 1+(-1+1)+(1-1)+(-2+2)+(3-3)+(-5+\cdots \\ &=1.
  \end{align}
$$
Your power series is divergent at$~x=-1$ (and indeed everywhere outside the interval $[-\frac1\phi,\frac1\phi]$).
A: Hint:
the new series is convergent when $|x+x^2|<1$ anddivergent when $|x+x^2|>1$.
