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An analytic function equals its power series on the largest disc about the centre containing no singularity, so the radius of convergence is a distance in the plane — which is why a perfectly smooth real function can have a series that stops converging for no visible reason.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to state Taylor's theorem for an analytic function, find a radius of convergence by locating the nearest singularity rather than by testing coefficients, obtain a series by bending a function into a geometric one, say which real interval a complex disc of convergence cuts out, and explain why a real power series' radius is decided by the plane.
Cauchy's integral formula and the estimate that follows from it, and power series from the second calculus course with their radius of convergence. What is new is that the radius stops being something computed from the coefficients and becomes something read off a picture.
A power series about $z_0$ is $\sum a_n (z - z_0)^{n}$. Its radius of convergence $R$ is the number such that it converges for $|z - z_0| < R$ and diverges for $|z - z_0| > R$; the disc of convergence is that open disc. A function is represented by a series on a set when the series converges to it at every point of the set. A singularity of $f$ is a point at which $f$ fails to be analytic.
Taylor's theorem. If $f$ is analytic on the disc $|z - z_0| < R$, then for every $z$ in that disc
$$f(z) = \sum_{n=0}^{\infty} a_n (z - z_0)^{n}, \qquad a_n = \frac{f^{(n)}(z_0)}{n!} = \frac{1}{2\pi i}\oint_C \frac{f(z)}{(z - z_0)^{n+1}}\,dz.$$
And the radius is a distance. The series converges on the largest open disc about $z_0$ containing no singularity of $f$. So
$$R = \text{the distance from } z_0 \text{ to the nearest singularity of } f,$$
and if there are none at all, the series converges on the whole plane.
This is worth comparing with the real case carefully, because it is where complex analysis pays a debt that real analysis could not. Over the reals, $\dfrac{1}{1 + x^{2}}$ is infinitely differentiable on the whole line, is bounded by $1$, and its Taylor series about $0$ nevertheless diverges for $|x| > 1$. Nothing on the real line explains why. In the plane the explanation is immediate: the function has singularities at $\pm i$, both at distance $1$ from the origin, and the disc of convergence cannot reach past them. The real interval was the shadow of a complex disc all along.
Consequences that get used. A power series may be differentiated and integrated term by term inside its disc, and the result has the same radius. Two analytic functions with the same series about a point agree near it. And since every analytic function has a series, analytic and locally a convergent power series are the same property — which is why the word is used for both.
Another way: picture
Put a pin at the centre of the expansion and mark every singularity of the function. Now grow a circle from the pin until it touches the first mark. That circle is the boundary of convergence, and it does not care whether the offending singularity is on the real axis, off it, or invisible from wherever you happen to be standing.
Another way: steps
The formula $a_n = f^{(n)}(z_0)/n!$ is the definition and almost never the method. In practice a series is obtained by manipulating one of four known ones, and knowing that is most of the skill.
| Series | Coefficients | Radius about $0$ |
|---|---|---|
| $\dfrac{1}{1 - w} = \sum w^{n}$ | all one | $|w| < 1$ |
| $e^{z} = \sum \dfrac{z^{n}}{n!}$ | reciprocal factorials | the whole plane |
| $\sin z = \sum \dfrac{(-1)^{n}z^{2n+1}}{(2n+1)!}$ | odd powers only | the whole plane |
| $\cos z = \sum \dfrac{(-1)^{n}z^{2n}}{(2n)!}$ | even powers only | the whole plane |
The geometric series does most of the work, because any denominator can be bent into the shape $1 - w$. For $\dfrac{1}{3 - z}$ about $0$, factor out the $3$: $\dfrac{1}{3}\cdot\dfrac{1}{1 - z/3}$, so $w = z/3$ and the radius is $3$. For the same function about $1$, write $3 - z = 2 - (z - 1)$ and factor out the $2$: the radius is $2$, which is the distance from $1$ to $3$, exactly as the theorem promises.
Multiplying, substituting and differentiating known series are all legitimate inside the disc, and each is faster than computing a single derivative of the original. The one thing to keep track of is the radius: substituting $w = z/3$ into a series valid for $|w| < 1$ gives one valid for $|z| < 3$.
A real power series that stops converging for no visible reason is the classic puzzle of a second calculus course, and the classic wrong explanations are that the function must misbehave somewhere real, or that the coefficients simply grow too fast for no reason. Both are attempts to explain a two-dimensional fact one-dimensionally.
$\dfrac{1}{1 + x^{2}}$ is the standard case: smooth and bounded on the whole real line, series diverging past $|x| = 1$, and the explanation sitting at $\pm i$ where no amount of looking along the real axis will find it.
Two smaller errors follow from the same habit. The nearest singularity, not the most obvious one: a function with singularities at $2$ and at $-3i$ has radius $2$ about the origin, and a reader scanning the real axis will find the first and stop. And the boundary is a separate question: the radius tells you the series converges inside and diverges outside, and says nothing whatever about the circle itself, where it may converge everywhere, nowhere, or on part of it.
Find the radius of convergence about $0$ of $\dfrac{1}{(z - 2)(z + 3i)}$.
Locate the singularities.
The singularities are $2$ and $-3i$, at distances $2$ and $3$ from the origin.
Measure each distance.
The radius is the smaller, $2$. No coefficient was computed and none was needed.
The nearest one decides.
Expand $\dfrac{1}{2 - z}$ about $0$. Factor: $\dfrac{1}{2}\cdot\dfrac{1}{1 - z/2}$.
Make the denominator look like one minus something.
The geometric series gives $\dfrac{1}{2}\sum (z/2)^{n} = \sum \dfrac{z^{n}}{2^{n+1}}$, valid for $|z| < 2$.
Substitute and keep track of the radius.
And $2$ is the distance from $0$ to the only singularity. The two routes agree, as the theorem says they must.
A check, for free.
The singularities are $\pm i$; the distances from $2$ are both $\sqrt{4 + 1} = \sqrt{5}$.
Distances in the plane, not along the axis.
So the radius is $\sqrt{5}$.
Expand $f(z) = \dfrac{1}{1 - 5z}$ as a Taylor series about $0$ and fill in the table.
| Value | |
|---|---|
| The coefficient of the constant term | |
| The coefficient of the first power | |
| The coefficient of the second power | |
| The coefficient of the third power | |
| The reciprocal of the radius of convergence |
What is the radius of convergence of the Taylor series about $0$ of $f(z) = \dfrac{1}{z - \left(8 + 6i\right)}$?
Answer:
The Taylor series about $0$ of $\dfrac{1}{z - \left(20 + 21i\right)}$ converges on a disc. Which real numbers $x$ lie strictly inside that disc?
This task has no paper form; do it on a device.
Each series is the Taylor series about $0$. Match each function to the radius of convergence.
| $1$ | $3$ | $7$ | No limit: the whole plane | |
|---|---|---|---|---|
| $\dfrac{1}{1 - z}$ | ||||
| $\dfrac{1}{z - 3}$ | ||||
| $\dfrac{1}{z^{2} + 49}$ | ||||
| $e^{z}$ |
Put the steps of finding the radius of convergence of a Taylor series in order.
Number the steps in order (write the number in the box):
$f$ is analytic on and inside the circle $|z| = 5$, with $|f| \le 3$ on it. Its Taylor series about $0$ is $\sum a_j z^{j}$. What bound does the Cauchy estimate give for $\left|a_{2}\right|$? Give a fraction.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Expand $f(z) = \dfrac{1}{1 - 4z}$ as a Taylor series about $0$ and fill in the table.
| Value | |
|---|---|
| The coefficient of the constant term | |
| The coefficient of the first power | |
| The coefficient of the second power | |
| The coefficient of the third power | |
| The reciprocal of the radius of convergence |
You can find a radius of convergence by looking for singularities and can expand a quotient as a geometric series. Say in your own words why the series for one over one plus the square stops at radius one. Next: what happens when the centre of the expansion is itself a singularity.
9. Your turn: the radius of the Taylor series of $\dfrac{1}{1 + z^{2}}$ about $z = 2$, step 2