Can the the radius of convergence increase due to composition of two power series?

When composing power series, is the radius of convergence the minimum of that of the individual series, or is it like for multiplication and addition of power series where the resultant radius of convergence may be larger than for the individual series?

If the radius of convergence is the minimum as described, then is the interval of convergence the intersection of the two individual intervals of convergence, or is behaviour at the end-points not guaranteed (just like it isn't when differentiating a power series to obtain another)?

Solution 1:

(Edited. The handling of the arising cases was not correct.)

A composition $h(z):=f\bigl(g(z)\bigr)$ of analytic functions makes sense whenever $g$ is analytic in a neighborhood of $z_0$ and $f$ is analytic in a neighborhood of $g(z_0)$.

Contrasting this the composition of the power series $g(z):=\sum_{k=0}^\infty a_k z^k$ with the series $f(w):=\sum_{j=0}^\infty b_j w^j$ only makes sense when $a_0=0$, i.e., when $g(0)=0$. Only in this case the Taylor coefficients of the composite function $h(z):=f\bigl(g(z)\bigr)$ can be calculated in finite terms.

Therefore let's assume $g(0)=0$. Let $0<R\leq\infty$ be the radius of convergence of the $g$-series, and put $$M_g(r):=\max_{0\leq \phi\leq 2\pi} |g(re^{i\phi}|\qquad(r<R)\ .$$ Then $M_g(0)=0$, and the function $r\mapsto M_g(r)$ is strictly increasing and continuous. Put $\lim_{r\to R-} M_g(r)=:\rho'\leq\infty$.

Finally, let $0<\rho\leq\infty$ be the radius of convergence of the $f$-series.

If $\rho\geq\rho'$ then $|g(z)|<\rho$ for all $z\in D_R$, and it follows that $h$ is analytic in $D_R$. Therefore the radius of convergence of the $h$-series is $\geq R$.

If $\rho<\rho'$ then there is an $R'<R$ such that $M_g(R')=\rho$. It follows that $|g(z)|<\rho$ for $z\in D_{R'}$, and it is easy to see that in this case the $h$-series has radius of convergence $\geq R'$.

Solution 2:

How about an example: $$ \frac x{1-x}\circ\frac x{1+x} = x\quad? $$

Solution 3:

Think what happens when you compose $f$ and $g$ into $f(g(z))$. If $z$ is in the convergence region of $g$, and $g(z)$ is in the convergence region of $f$, things should work out nicely. In particular, if $f$ converges everywhere, $f(g(z))$ will converge whenever $g(z)$ does. But $f(g(z))$ might fail to converge even if $g(z)$ does.