Power Series

A sequence of partial sums of polynomial functions.

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Concept	Content
Root Criterion	An important criterion to prove absolute convergence by means of the behavior of the n-th roots of the underlying sequence's terms.
Sequences of Bounded Functions	The concept of sequences but for functions instead of real numbers.
Polynomials	A basic class of functions that consists a linear combinations of monomials.

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Concept	Content
Examples of Differentiable Functions	Calculation of derivatives by example.
Taylor's Theorem	An approximation method for differentiable functions.
Uniform Convergence for Differentiable Functions	Compatibility of differentiability with uniform limits.

Study Power Series

Very roughly speaking, power series are “infinite polynomials”. A precise definition is the following:

Definition 1. Let a sequence $(a_{k})_{k \in N}$ in $K$ be given and let $x_{0} \in K$ . Then the function $f : D (f) \to K$ defined by the series $f (x) = \sum_{k = 0}^{\infty} a_{k} (x - x_{0})^{k}$ is called power series.
The set $D (f) := {x \in K : \sum_{k = 0}^{\infty} a_{k} (x - x_{0})^{k} is convergent}$ is called domain of convergence. The domain of convergence at least includes $x_{0}$ since $\sum_{k = 0}^{\infty} a_{k} (x_{0} - x_{0})^{k} = a_{0}$ .

We have already seen several examples of power series in this chapter.

Example 2.

The exponential function is defined via the power series $\exp (x) = \sum_{k = 0}^{\infty} \frac{x^{k}}{k!},$ i.e., $(a_{k})_{k \in N} = (\frac{1}{k!})_{k \in N}$ and $x_{0} = 0$ . Here $D (f) = C$ .
The sine function is defined via the power series $\sin (x) = \sum_{k = 0}^{\infty} \frac{(- 1)^{k} x^{2 k + 1}}{(2 k + 1)!},$ i.e., $(a_{k})_{k \in N} = (0, \frac{1}{1!}, 0, - \frac{1}{3!}, 0, \frac{1}{5!}, 0, - \frac{1}{7!}, \dots)_{k \in N}$ and $x_{0} = 0$ . Again $D (f) = C$ .
$\cos$ , $\cosh$ , $\sinh$ are defined via the power series...
The function $f (x) = \sum_{k = 1}^{\infty} \frac{(x - 1)^{k}}{k}$ is a power series.

Next we characterise the domain of convergence.

Theorem 3 (Theorem of Cauchy-Hadamard). Let a power series $f (x) = \sum_{k = 0}^{\infty} a_{k} (x - x_{0})^{k}$ be given. Let $r := \frac{1}{\underset{k \to \infty}{lim sup} \sqrt[k]{| a_{k} |}},$ where we formally define $1 / \infty := 0$ and $1 / 0 := \infty$ . Then for all $x \in K$ with $| x - x_{0} | < r$ holds $x \in D (f)$ . Furthermore, for all $x \in K$ with $| x - x_{0} | > r$ holds $x \notin D (f)$ .
The number $r$ as defined above is called the radius of convergence.

Proof: We have to show the following two statements:

For all $x \in K$ with $| x - x_{0} | \cdot \underset{k \to \infty}{lim sup} \sqrt[k]{| a_{k} |} < 1$ , the power series is convergent.
For all $x \in K$ with $| x - x_{0} | \cdot \underset{k \to \infty}{lim sup} \sqrt[k]{| a_{k} |} > 1$ , the power series is divergent.

Statement (i) just follows from the limit form of the root criterion, namely $\underset{k \to \infty}{lim sup} \sqrt[k]{| (x - x_{0})^{k} a_{k} |} = | x - x_{0} | \cdot \underset{k \to \infty}{lim sup} \sqrt[k]{| a_{k} |} < 1.$ For showing (ii), we also make use of the formula $\underset{k \to \infty}{lim sup} \sqrt[k]{| (x - x_{0})^{k} a_{k} |} = | x - x_{0} | \cdot \underset{k \to \infty}{lim sup} \sqrt[k]{| a_{k} |} > 1.$ This implies that the sequence $(x - x_{0})^{k} a_{k}$ does not converge to 0 and therefore, the power series cannot converge. $◻$

Geometrically, the above result implies that for all $x$ inside a circle with midpoint $x_{0}$ and radius $r$ , the series is convergent and outside this circle, we have divergence.
The Cauchy-Hadamard Theorem characterizes convergence/divergence of the power series in dependence of $x$ whether it is inside or outside the circle around $x_{0}$ with radius $r$ . In the case $| x - x_{0} | = r$ , this result does not tell us anything. Indeed, we may have points on the circle with $x \in D (f)$ and also points on the circle with $x \notin D (f)$ .

To see this, let us reconsider Example 2 d):
The radius of convergence is given by $r = \frac{1}{\underset{k \to \infty}{lim sup} \sqrt[k]{| \frac{1}{k} |}} = 1.$ So we have convergence for all $x \in (0, 2)$ and divergence for all $x \in (- \infty, 0) \cup (2, \infty)$ . The remaining real points which are not characterized by the Theorem of Cauchy-Hadamard are $x = 0$ and $x = 2$ . In the case $x = 0$ , we obtain the series $\sum_{k = 0}^{\infty} \frac{(- 1)^{k}}{k}$ which is convergent by the Leibniz criterion. Plugging in $x = 2$ , the power series becomes a harmonic series $\sum_{k = 0}^{\infty} \frac{1}{k}$ that is well-known to be divergent.

Example 4.

For the power series defined by the exponential function, we have $(a_{k})_{k \in N} = {(\frac{1}{k!})}_{k \in N}, x_{0} = 0.$ The radius of convergence is then given by $r = \frac{1}{\underset{k \to \infty}{lim sup} \sqrt[k]{| \frac{1}{k!} |}} = \frac{1}{0} = \infty .$ As a consequence, the series converges for every $x \in C$ . The same holds true for the series of $\sin$ , $\cos$ , $\sinh$ , $\cosh$ .
As we have already seen above, the radius of convergence of the power series $f (x) = \sum_{k = 0}^{\infty} \frac{(x - 1)^{k}}{k}$ is $r = 1$ .
Consider the power series $f (x) = \sum_{k = 0}^{\infty} k! x^{k} .$ The radius of convergence is given by $r = \frac{1}{\underset{k \to \infty}{lim sup} \sqrt[k]{| k! |}} = \frac{1}{\infty} = 0.$ So this series is divergent for any $x \neq 0$ .

Sometimes the computation of the radius of convergence $r = \underset{n \to \infty}{lim sup} \sqrt[n]{| a_{n} |}$ of a power series $\sum_{n = 1}^{\infty} a_{n} (x - x_{0})^{n}$ is quite difficult. In such cases the following theorem might be better suited, which follows from the quotient criterion and is stated without proof.

Theorem 5. Suppose that $\sum_{n = 1}^{\infty} a_{n} (x - x_{0})^{n}$ is a power series with coefficients $a_{n} \in K$ such that $a_{n} \neq 0$ for all $n \geq N$ with fixed $N \in N$ . If $lim_{n \to \infty} \frac{| a_{n} |}{| a_{n + 1} |}$ exists in $R \cup {+ \infty}$ than it is the radius of convergence of the power series.

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Power Series

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