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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)?

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  • $\begingroup$ What do you mean by "preserved"? You're combining two power series, each with its own radius of convergence. $\endgroup$ – Chris Eagle Mar 7 '13 at 13:04
  • $\begingroup$ @ChrisEagle Thanks for pointing out my bad phrasing, I've edited it away. $\endgroup$ – Ryan G Mar 7 '13 at 13:12
  • $\begingroup$ Do you know a proof of the radius of convergence for a power series that you can modify to answer this? $\endgroup$ – Loki Clock Mar 7 '13 at 13:17
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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.

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  • $\begingroup$ Yes, that's intuitively what I would expect. But I'm asking my question in light of the fact that when you add two power series, the new power series may have a bigger radius of convergence that that of either series. So does composition have this "bonus" behaviour too? Secondly, how about at the boundaries of convergence? $\endgroup$ – Ryan G Mar 7 '13 at 13:10
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    $\begingroup$ @Ryan, no, that behavior doesn't carry over to composition and divergence can happen even in the simplest cases. Try calculating the constant term of the power series for $\frac{1}{1-z}$ composed with itself. But the problem doesn't always come from the constant term, for example composing $\frac{z}{1-z}$ with itself leads to a series whose radius of convergence is only $1/2$. $\endgroup$ – Antonio Vargas Mar 7 '13 at 20:50
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(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 $\geq R'$.

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  • $\begingroup$ @user71551: I hope all is correct now. (I had erroneously assumed that $g(z)=\sum_{k\geq1} a_k z^k$ is unbounded in its circle of convergence.) $\endgroup$ – Christian Blatter Apr 24 '13 at 16:24
  • $\begingroup$ With $a_0\ne0$ the composition also makes sense if $|a_0|<\rho$. $\endgroup$ – user Feb 2 at 18:41
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How about an example: $$ \frac x{1-x}\circ\frac x{1+x} = x\quad? $$

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