Theory

Power series and Taylor series

Where a power series converges, how to compute the radius and the set of convergence, how to differentiate and integrate term by term, and how Taylor expansions represent the elementary functions.

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Power series and radius of convergence
Computing the radius

Power series and radius of convergence

1. Power series

A power series centered at $x_{0} \in R$ is a series of the form

n = 0 \sum \infty a_{n} (x - x_{0})^{n} = a_{0} + a_{1} (x - x_{0}) + a_{2} (x - x_{0})^{2} + \dots

where the numbers $a_{n}$ are the coefficients and $x_{0}$ is the center; by convention $(x - x_{0})^{0} = 1$ also at $x = x_{0}$ . For each fixed value of $x$ the series is a numerical series, so the study of a power series is the study of the set of points $x$ at which that numerical series converges. At $x = x_{0}$ all the terms with $n \geq 1$ are zero, therefore every power series converges at least at its own center, with sum $a_{0}$ . The convergence tests for numerical series are in the chapter on series in Mathematical Analysis 1.

The geometric series as a power series

The series $n = 0 \sum \infty x^{n}$ is a power series centered at $0$ with all coefficients equal to $1$ . For $∣ x ∣ < 1$ it is a convergent geometric series with sum $\frac{1}{1 - x}$ ; for $∣ x ∣ \geq 1$ the general term $x^{n}$ does not tend to $0$ , so the series does not converge. The set of convergence is the interval $(- 1, 1)$ , symmetric with respect to the center. The following theorem shows that this shape of the set of convergence is general.

y = \dfrac{1}{1 - x}

y = 1 + x + x^{2}

y = \dfrac{1 - x^{7}}{1 - x}

y = \frac{1}{1 - x}

with the partial sums

1 + x + x^{2}

(dashed) and

1 + x + \dots + x^{6}

(red): inside

(- 1, 1)

they approach the sum, outside they move away from it

2. Radius of convergence theorem

For every power series $\sum a_{n} (x - x_{0})^{n}$ there exists a unique $R \in [0, + \infty]$ , called the radius of convergence, such that

if $∣ x - x_{0} ∣ < R$ the series converges absolutely;
if $∣ x - x_{0} ∣ > R$ the series does not converge.

In the extreme cases: if $R = 0$ the series converges only at $x_{0}$ ; if $R = + \infty$ the series converges absolutely for every $x \in R$ . The theorem reduces the study of convergence to the computation of the number $R$ ; at the points with $∣ x - x_{0} ∣ = R$ it states nothing, so the endpoints require a separate study.

Proof

We first prove the following fact: if the series converges at a point $x_{1} \neq = x_{0}$ , then it converges absolutely at every $x$ with $∣ x - x_{0} ∣ < ∣ x_{1} - x_{0} ∣$ . Since the numerical series $\sum a_{n} (x_{1} - x_{0})^{n}$ converges, by the necessary condition for convergence the general term tends to $0$ , so it is bounded: there exists $M > 0$ such that $∣ a_{n} (x_{1} - x_{0})^{n} ∣ \leq M$ for every $n$ . Fixing $x$ with $∣ x - x_{0} ∣ < ∣ x_{1} - x_{0} ∣$ , we set $q = \frac{∣ x - x _{0} ∣}{∣ x _{1} - x _{0} ∣} < 1$ and estimate:

∣ a_{n} (x - x_{0})^{n} ∣ = ∣ a_{n} (x_{1} - x_{0})^{n} ∣ \cdot q^{n} \leq M q^{n} .

The geometric series $\sum M q^{n}$ converges because $0 \leq q < 1$ , so by the comparison test the series $\sum ∣ a_{n} (x - x_{0})^{n} ∣$ converges.

Now set

R = sup {∣ x - x_{0} ∣ : the series converges at x},

the supremum of a nonempty set, since it contains $0$ . If $∣ x - x_{0} ∣ < R$ , by definition of supremum there exists a point of convergence $x_{1}$ with $∣ x - x_{0} ∣ < ∣ x_{1} - x_{0} ∣$ , so by the initial fact the series converges absolutely at $x$ . If $∣ x - x_{0} ∣ > R$ , the series does not converge at $x$ , because otherwise the number $∣ x - x_{0} ∣$ would belong to the set and would be greater than its supremum, which is absurd.

3. Set of convergence

The set of convergence $E$ of a power series is the set of points $x$ at which the series converges. For $0 < R < + \infty$ , the radius theorem gives

(x_{0} - R, x_{0} + R) \subseteq E \subseteq [x_{0} - R, x_{0} + R],

so $E$ is an interval centered at $x_{0}$ that contains the open interval and may include or exclude each of the two endpoints. The complete study of a power series therefore consists of two steps: the computation of $R$ and the study of the numerical series at each of the points $x_{0} - R$ and $x_{0} + R$ .

Computing the radius

1. Ratio test for the radius

If $a_{n} \neq = 0$ eventually and there exists

n \to \infty lim \frac{∣ a _{n + 1} ∣}{∣ a _{n} ∣} = ℓ \in [0, + \infty],

then $R = \frac{1}{ℓ}$ , with the conventions $R = + \infty$ for $ℓ = 0$ and $R = 0$ for $ℓ = + \infty$ . The statement follows by applying the ratio test to the series of absolute values: for $x \neq = x_{0}$ the ratio between two consecutive terms tends to $ℓ ∣ x - x_{0} ∣$ , which is less than $1$ if and only if $∣ x - x_{0} ∣ < \frac{1}{ℓ}$ , and greater than $1$ (with a general term that does not tend to $0$ ) if $∣ x - x_{0} ∣ > \frac{1}{ℓ}$ . For the series $n = 0 \sum \infty \frac{x ^{n}}{n !}$ the ratio is $\frac{n !}{( n + 1 )!} = \frac{1}{n + 1} \to 0$ , so $R = + \infty$ .

2. Root test for the radius

If there exists

n \to \infty lim n ∣ a_{n} ∣ = ℓ \in [0, + \infty],

then $R = \frac{1}{ℓ}$ , with the same conventions as in the ratio test. The justification is identical, with the root test in place of the ratio test. This test is suited to the case where the coefficient $a_{n}$ is an $n$ -th power: for $n = 1 \sum \infty (\frac{n}{2 n + 1})^{n} x^{n}$ we have $n ∣ a_{n} ∣ = \frac{n}{2 n + 1} \to \frac{1}{2}$ , so $R = 2$ .

Lacunary series

A power series is lacunary if it contains only some powers, for example only the even ones ( $x^{2 n}$ ), only the odd ones ( $x^{2 n + 1}$ ) or only the multiples of three ( $x^{3 n}$ ). Infinitely many zeros appear in the sequence of coefficients, so the formulas based on the ratio and on the root of the coefficients do not apply. In these cases the series is studied with a substitution: for $n = 0 \sum \infty \frac{x ^{2 n}}{4 ^{n}}$ we set $t = x^{2}$ and obtain the geometric series $\sum (\frac{t}{4})^{n}$ , which converges if and only if $∣ t ∣ < 4$ , that is, $x^{2} < 4$ , that is, $∣ x ∣ < 2$ . The radius is therefore $R = 2$ and not $4$ . Alternatively, the ratio or root test is applied directly to the terms of the series, with $x$ as a parameter.

The theory continues with 7 more sections (behavior at the endpoints, total convergence and continuity of the sum, term by term differentiation and integration…), plus all worked examples, the solutions and the PDF.

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