Can Composition Increase a Power Series’ Radius of Convergence?

Yes. The radius of convergence of a composed power series can be larger than the radius of the inner series. This can happen because algebraic cancellation removes a singularity that limits the inner series. It is not automatic: depending on the functions, composition can also leave the radius unchanged or make it smaller.

What composition guarantees

Suppose the inner series is $g(z)=\sum_{n=1}^{\infty}b_nz^n$, with radius of convergence $R_g$, and the outer series is $f(w)=\sum_{n=0}^{\infty}a_nw^n$, with radius $R_f$. The condition $g(0)=0$ makes the formal composition $f(g(z))$ well defined as a power series: each coefficient in the result is determined by only finitely many terms.

For direct substitution of the convergent series, you need $|z|<R_g$ and $|g(z)|<R_f$. Wherever these hold, the composition is analytic. This gives a region where the composition is represented by the substituted series, but it does not always determine the full radius of the Taylor series of the composition. The resulting function may simplify or extend analytically beyond that region.

In particular, there is no general rule that the composition’s radius is the smaller of $R_g$ and $R_f$. The outer series’ radius applies to its input $w=g(z)$, not directly to $z$.

Worked example: the radius becomes infinite

Take the inner function

$$g(z)=\sqrt{1-z}-1.$$

Its Taylor series about $z=0$ begins

$$g(z)=-\frac{z}{2}-\frac{z^2}{8}-\frac{z^3}{16}-\cdots.$$

The binomial series for $\sqrt{1-z}$ has radius $1$, because the function has a branch point at $z=1$. Subtracting $1$ does not remove that singularity, so $g$ also has radius $1$.

Now choose the outer function $f(w)=(1+w)^2$. This is a polynomial, so its Taylor series has infinite radius. Since $g(0)=0$, the composition is an ordinary power-series composition. Simplifying it gives

$$\begin{aligned} f(g(z))&=(1+g(z))^2\\ &=\bigl(\sqrt{1-z}\bigr)^2\\ &=1-z. \end{aligned}$$

The Taylor series of $1-z$ is a polynomial, and therefore converges for every complex $z$. Its radius is infinite, even though the inner series has radius $1$. The square removes the branch-point behavior that limited the inner function.

This example shows why you must distinguish the radius of the inner series from the radius of the Taylor series of the simplified composition. The original substitution is directly justified for $|z|<1$, but the resulting function $1-z$ has an entire extension, and its Taylor series has infinite radius.

Composition can also reduce the radius

An increase is possible, but not guaranteed. For example, let $f(w)=1/(1-w)$ and $g(z)=2z$. The inner series has infinite radius, while

$$f(g(z))=\frac{1}{1-2z}$$

has a Taylor series about $0$ with radius $1/2$, determined by its singularity at $z=1/2$. Thus composition can produce a smaller radius than the inner series had.

The outer function’s singularities matter too. As $z$ varies, the values of $g(z)$ may reach a singularity of $f$ even while $g$ itself is analytic. Conversely, a composition may cancel a singularity of $g$, as in the square-root example.

Common mistakes

• Assuming the radii simply combine by taking a minimum. The outer radius concerns values of $g(z)$, and the composition’s Taylor radius depends on the resulting function of $z$.
• Treating the direct substitution region as the full answer. The condition $|g(z)|<R_f$ ensures convergence of the outer series after substitution, but simplification or analytic continuation may give a larger Taylor radius.
• Ignoring the composition condition. For two power series centered at $0$, the usual formal composition requires $g(0)=0$. If the inner series has a nonzero constant term, the outer series must instead be centered appropriately, or the composition must be handled as an analytic function where it is defined.
• Assuming cancellation always occurs. Singularities may persist, or the composition may encounter a singularity of the outer function sooner than the inner series encounters its own boundary.