Power Series and Taylor Series

In this lesson you’ll learn that analyticity and having a convergent power series are the same thing over $\mathbb{C}$, that the radius of convergence is always the distance to the nearest singularity, and how to build new series from old ones quickly.

A power series about $z_0$ is

$$\sum_{n=0}^{\infty}a_n(z-z_0)^n$$

Every such series has a radius of convergence $R \in [0,\infty]$ with the following behaviour:

• Absolute convergence for $|z-z_0| < R$, and uniform convergence on any closed disc strictly inside.
• Divergence for $|z-z_0| > R$.
• Anything can happen on the circle $|z-z_0| = R$ itself.

Note the shape of the convergence set: a disc, not an interval. In real analysis the interval of convergence had two endpoints to check separately; here the boundary is a whole circle, and boundary behaviour is genuinely subtle.

Within the disc, a power series is analytic and may be differentiated and integrated term by term, with the derived series having the same radius.

This is the result that makes the word analytic unambiguous.

Taylor’s theorem for analytic functions. If $f$ is analytic on the disc $|z-z_0|<R$, then

$$f(z) = \sum_{n=0}^{\infty}\frac{f^{(n)}(z_0)}{n!}(z-z_0)^n$$

for every $z$ in that disc. Conversely, any convergent power series defines an analytic function inside its disc.

So differentiable on an open set and locally a convergent power series are equivalent conditions over $\mathbb{C}$.

In real analysis these are different, and dramatically so: $e^{-1/x^2}$ is infinitely differentiable at 0 with every derivative zero, so its Taylor series is identically 0 and does not represent the function. No complex analytic function can do that. If you met that counterexample in real analysis, this is where it gets ruled out.

The proof is Cauchy’s integral formula plus a geometric series expansion of $\frac{1}{z-w}$, which is why this lesson comes after that one.

The radius of convergence of the Taylor series of $f$ about $z_0$ equals the distance from $z_0$ to the nearest singularity of $f$.

This is the single most useful fact in the lesson. It means you can read off the radius by looking at a picture, without computing a single coefficient.

It also resolves the puzzle from lesson 1. The function $\frac{1}{1+z^2}$ has poles at $\pm i$, so about 0 the radius is 1, about 1 it is $\sqrt2$, and about 2 it is $\sqrt5$. On the real line nothing distinguishes $x = \pm1$, and the real graph offers no explanation whatsoever.

The classical tests still work when you want to verify:

$$R = \lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right| \qquad\text{or}\qquad \frac1R = \limsup_{n\to\infty}|a_n|^{1/n}$$

but the geometric statement is faster and explains why.

Each of these is the real series with $x$ replaced by $z$, valid on the stated disc:

$$e^z = \sum_{n=0}^\infty\frac{z^n}{n!}, \qquad R = \infty$$ $$\sin z = \sum_{n=0}^\infty\frac{(-1)^nz^{2n+1}}{(2n+1)!}, \qquad \cos z = \sum_{n=0}^\infty\frac{(-1)^nz^{2n}}{(2n)!}, \qquad R = \infty$$ $$\frac{1}{1-z} = \sum_{n=0}^\infty z^n, \qquad R = 1$$ $$\operatorname{Log}(1+z) = \sum_{n=1}^\infty\frac{(-1)^{n+1}z^n}{n}, \qquad R = 1$$

The three entire functions have $R=\infty$ because they have no singularities at all. The last two have $R=1$ because of a singularity at exactly distance 1: a pole at $z=1$ for the geometric series, and the branch point at $z=-1$ for the logarithm.

Substitution and algebra beat differentiation. Almost no one computes Taylor coefficients from $f^{(n)}(z_0)/n!$ in practice; you manipulate known series instead.

If $f$ is analytic at $z_0$ with $f(z_0)=0$, the series there begins at some term:

$$f(z) = a_m(z-z_0)^m + a_{m+1}(z-z_0)^{m+1}+\cdots, \qquad a_m \ne 0$$

The integer $m$ is the order of the zero. Equivalently $f = (z-z_0)^mg(z)$ with $g$ analytic and $g(z_0)\ne0$.

Two consequences: zeros of a non-zero analytic function are isolated, since $g$ stays nonzero nearby, and that isolation is what proves the identity theorem. The exception is the zero function, which is why the identity theorem’s conclusion is “identically equal” rather than “equal somewhere.”

Example 1: Find the radius of convergence of the Taylor series of $\frac{1}{z-3}$ about $z_0=0$.

Solution. The only singularity is at $z=3$, at distance 3 from the origin.

$$R = 3$$

Verifying by expansion, $\frac{1}{z-3} = -\frac13\cdot\frac{1}{1-z/3} = -\frac13\sum(z/3)^n$, which converges for $|z/3|<1$. ✓

Example 2: Same function, about $z_0 = 1$.

Solution. The distance from 1 to 3 is 2, so $R = 2$. ∎

The function did not change; the centre did. The radius is a property of the pair.

Example 3: Find the series for $\frac{1}{1+z^2}$ about 0 by substitution.

Solution. Start from the geometric series and substitute $-z^2$ for $z$:

$$\frac{1}{1-(-z^2)} = \sum_{n=0}^\infty(-z^2)^n = 1-z^2+z^4-z^6+\cdots$$

valid when $|-z^2|<1$, that is $|z|<1$, so $R=1$. Consistent with the poles at $\pm i$ being at distance 1. ✓

Example 4: Find the first few terms of the series for $\frac{e^z}{1-z}$ about 0.

Solution. Multiply the two known series:

$$\left(1+z+\frac{z^2}{2}+\cdots\right)\left(1+z+z^2+\cdots\right)$$

Collecting by degree:

• constant: $1$
• $z$: $1+1 = 2$
• $z^2$: $\frac12+1+1 = \frac52$

$$\frac{e^z}{1-z} = 1+2z+\frac52z^2+\cdots$$

The radius is 1, set by the pole at $z=1$; the exponential factor contributes no singularity. ∎

Example 5: Find the order of the zero of $f(z)=z^2\sin z$ at 0.

Solution. Since $\sin z = z - \frac{z^3}{6}+\cdots$ has a simple zero at 0,

$$z^2\sin z = z^3 - \frac{z^5}{6}+\cdots$$

The first nonzero term is $z^3$, so the zero has order 3. ∎

Orders add under multiplication, which is the quick way to see this: order 2 from $z^2$ plus order 1 from $\sin z$.

Example 6: Explain why $e^{-1/x^2}$ has no complex analogue of its bad behaviour.

Solution. As a real function it is smooth at 0 with all derivatives 0, so its Taylor series is identically 0 while the function is not.

Now consider $g(z)=e^{-1/z^2}$ in the plane. Along the imaginary axis, $z=iy$ gives $-1/z^2 = 1/y^2$, so

$$|g(iy)| = e^{1/y^2}\to\infty \text{ as } y\to0$$

So $g$ has a genuine essential singularity at 0 and is not analytic there. The real function’s strange behaviour was the shadow of a complex singularity, and once you can see the singularity there is nothing left to explain.

Example 7: Use the identity theorem via zeros.

$f$ is analytic on $\mathbb{C}$ and $f(1/n)=0$ for every positive integer $n$. Show $f\equiv0$.

Solution. The points $1/n$ accumulate at 0, which is in the domain. If $f$ were not identically zero, its zeros would be isolated, contradicting the accumulation.

$$f\equiv0$$

∎

Compare $\sin(\pi/z)$, which vanishes at every $1/n$ and is not identically zero - but it is not analytic at 0, so there is no contradiction. The hypothesis that the accumulation point is in the domain is essential.

Function evaluation in software. Library routines for $\exp$, $\sin$, and $\log$ reduce the argument to a small range and then sum a truncated series, with the truncation point chosen from a remainder bound. The radius of convergence is what tells you the reduction is necessary.

Generating functions. In combinatorics and algorithm analysis, a counting sequence is packed into a power series, and the radius of convergence gives the exponential growth rate of the coefficients. Locating the nearest singularity is the standard technique for asymptotics.

Perturbation theory. Physical quantities are expanded in a small parameter, and whether the expansion converges is a question about singularities in the complex parameter plane. Many important series in quantum field theory have zero radius of convergence, which is a fact about complex singularities.

Control system approximation. Padé and Taylor approximations of transfer functions are valid inside a disc bounded by the nearest pole, which is why models degrade near resonance.

Numerical analysis of ODE solvers. The stability region of a method is described by a power series in the step size, and its radius controls the maximum usable step.

Signal interpolation. Analytic signals are reconstructed from samples using series expansions whose convergence depends on the location of singularities of the signal’s analytic extension.