Power Series and Analytic Functions

Power Series

We will now look at an important type of series known as a power series which we define as follows.

Definition: A Power Series in Powers of $x-c$ is a series that takes on the form

$\sum _{n=0}^{\infty }a_{n}(x-c)^{n}=a_{0}+a_{1}(x-c)+a_{2}(x-c)^{2}+a_{3}(x-c)^{3}+...$ where $a_{0},a_{1},a_{2},a_{3},...$ are the Coefficients of The Power Series and $c$ is called the Center of Convergence of the power series.

We note that the power series is a function of a variable $x$ , and that the power series will either converge or diverge for each value of $x$ chosen. If the series converges for some $x$ , then the power series becomes a representation of some function for those values of $x$ . If the series diverges for some $x$ , then the power series is no longer a representation of a specific function for some $x$ .

For example, consider the geometric series $\sum _{n=0}^{\infty }x^{n}=1+x+x^{2}+...$ . Recall that if $\mid x\mid <1$ then $\sum _{n=0}^{\infty }x^{n}={\frac {1}{1-x}}$ . Therefore for $-1<x<1$ , the series $\sum _{n=0}^{\infty }x^{n}$ is a power series representation of the function $f(x)={\frac {1}{1-x}}$ on the interval $(-1,1)$ . We note that since all geometric series diverge if $\mid x\mid \geq 1$ , then the power series $\sum _{n=0}^{\infty }x^{n}$ is not a representation of $f$ on the interval $(-\infty ,-1]\cup [1,\infty )$ .

Now we note that the geometric series $\sum _{n=0}^{\infty }x^{n}$ has a center of convergence at $c=0$ . Notice that the interval to which this series convergences is for $(-1,1)=\{x\in \mathbb {R} :-1<x<1\}$ and that $c=0$ is indeed the center of this interval (hence the name center of convergence).

Now the following theorem will outline what sort of intervals of convergence we can expect from power series.

Theorem 1: Let $\sum _{n=0}^{\infty }a_{n}(x-c)^{n}$ be a power series. Then one of the following conclusions can be made regarding the points of convergence for the power series:

1) The series converges only at the center of convergence $c$ .

2) The series converges for every $x\in \mathbb {R}$ .

3) There exists a positive number $R\in \mathbb {R}$ such that the series converges if $\mid x-c\mid <R$ and diverges if $\mid x-c\mid >R$ where the converge and diverges of $x$ such that $\mid x-c\mid =R$ is undetermined.

Proof: We note that every power series converges at it's center of convergence, that is at the value $x=c$ since $\sum _{n=0}^{\infty }a_{n}(x-c)^{n}{\bigg |}_{x=c}=\sum _{n=0}^{\infty }a_{n}(c-c)^{n}=\sum _{n=0}^{\infty }0$ , and this sum converges to $0$ , specifically, this convergence is absolute convergence. We have thus shown that the validity of the first possibility.

We now need to show that if a series converges at any $x_{0}\neq c$ then it converges absolutely to every $x$ that is even closer to the center of convergence $c$ , that is $\mid x-c\mid <\mid x_{0}-c\mid$ . So the convergence of any $x_{0}\neq c$ will imply the absolute convergence for every $x$ such that $c-x_{0}<x<c+x_{0}$ . Now suppose that the power series $\sum _{n=0}^{\infty }a_{n}(x_{0}-c)^{n}$ is convergent. We know by the divergence theorem that then $\lim _{n\to \infty }a_{n}(x_{0}-c)^{n}=0$ . Therefore the sequence $\{a_{n}(x_{0}-c)^{n}\}$ converges and is bounded so $\forall n\in \mathbb {N}$ there exists a positive real number $K$ such that ${\bigg |}a_{n}(x_{0}-c)^{n}{\bigg |}<K$ .

Now suppose that $r={\frac {\mid x-c\mid }{\mid x_{0}-c\mid }}<1$ . This happens for all $x$ such that $\left|x-c\right|<\left|x_{0}-c\right|$ (the distance between $x$ and $c$ is less than the distance between $x_{0}$ and $c$ ), and so:

\quad \quad \sum _{n=0}^{\infty }{\bigg |}a_{n}(x-c)^{n}{\bigg |}=\sum _{n=0}^{\infty }{\bigg |}a_{n}{\bigg |}{\bigg |}(x-c)^{n}{\bigg |}=\sum _{n=0}^{\infty }{\bigg |}a_{n}{\bigg |}{\bigg |}(x_{0}-c)^{n}{\bigg |}{\bigg |}{\frac {x-c}{x_{0}-c}}{\bigg |}^{n}<K\sum _{n=0}^{\infty }{\bigg |}{\frac {x-c}{x_{0}-c}}{\bigg |}^{n}=K\sum _{n=0}^{\infty }r^{n}={\frac {K}{1-r}}<\infty

Therefore the power series $\sum _{n=0}^{\infty }a_{n}(x_{0}-c)^{n}$ is absolutely convergent which satisfies the second and third possibilities. $\blacksquare$

Definition: The Interval of Convergence $R$ of a power series $\sum _{n=0}^{\infty }a_{n}(x-c)^{n}$ is the interval of $x$ values centered at the the center of convergence $c$ for which the power series converges.

From the theorem above, we note that there are only 6 possibilities for the center of convergence summarized by the following corollary.

Corollary 1: Let $\sum _{n=0}^{\infty }a_{n}(x-c)^{n}$ be a power series. If this power series only converges at $x=c$ then the interval of convergence is the closed interval $[c,c]=\{c\}$ . If the power series converges for all $x\in \mathbb {R}$ then the interval of convergence is the entire real line $(-\infty ,\infty )=\{x\in \mathbb {R} \}$ . Otherwise, the interval of convergence of a power series is some interval centered at the center of convergence $c$ , that is either:

1) $[c-R,c+R]=\{x\in \mathbb {R} :c-R\leq x\leq c+R\}$ .

2) $(c-R,c+R)=\{x\in \mathbb {R} :c-R<x<c+R\}$ .

3) $[c-R,c+R)=\{x\in \mathbb {R} :c-R\leq x<c+R\}$ .

4) $(c-R,c+R]=\{x\in \mathbb {R} :c-R<x\leq c+R\}$ .

We note that on the interval $(c-R,c+R)$ the power series $\sum _{n=0}^{\infty }a_{n}(x-c)^{n}$ is absolutely convergent. We must check the end points of the power series when $x=c-R$ and $x=c+R$ to determine if the power series converges at $c-R$ and $c+R$ .

Operations on power series

Addition and subtraction

When two functions f and g are decomposed into power series around the same center c, the power series of the sum or difference of the functions can be obtained by termwise addition and subtraction. That is, if

f(x)=\sum _{n=0}^{\infty }a_{n}(x-c)^{n}

and

g(x)=\sum _{n=0}^{\infty }b_{n}(x-c)^{n}

then

f(x)\pm g(x)=\sum _{n=0}^{\infty }(a_{n}\pm b_{n})(x-c)^{n}.

It is not true that if two power series ${\textstyle \sum _{n=0}^{\infty }a_{n}x^{n}}$ and ${\textstyle \sum _{n=0}^{\infty }b_{n}x^{n}}$ have the same radius of convergence, then ${\textstyle \sum _{n=0}^{\infty }\left(a_{n}+b_{n}\right)x^{n}}$ also has this radius of convergence. If ${\textstyle a_{n}=(-1)^{n}}$ and ${\textstyle b_{n}=(-1)^{n+1}\left(1-{\frac {1}{3^{n}}}\right)}$ , then both series have the same radius of convergence of 1, but the series ${\textstyle \sum _{n=0}^{\infty }\left(a_{n}+b_{n}\right)x^{n}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3^{n}}}x^{n}}$ has a radius of convergence of 3.

Multiplication and division

With the same definitions for $f(x)$ and $g(x)$ , the power series of the product and quotient of the functions can be obtained as follows:

{\begin{aligned}f(x)g(x)&=\left(\sum _{n=0}^{\infty }a_{n}(x-c)^{n}\right)\left(\sum _{n=0}^{\infty }b_{n}(x-c)^{n}\right)\\&=\sum _{i=0}^{\infty }\sum _{j=0}^{\infty }a_{i}b_{j}(x-c)^{i+j}\\&=\sum _{n=0}^{\infty }\left(\sum _{i=0}^{n}a_{i}b_{n-i}\right)(x-c)^{n}.\end{aligned}}

The sequence ${\textstyle m_{n}=\sum _{i=0}^{n}a_{i}b_{n-i}}$ is known as the convolution of the sequences $a_{n}$ and $b_{n}$ .

For division, if one defines the sequence $d_{n}$ by

{\frac {f(x)}{g(x)}}={\frac {\sum _{n=0}^{\infty }a_{n}(x-c)^{n}}{\sum _{n=0}^{\infty }b_{n}(x-c)^{n}}}=\sum _{n=0}^{\infty }d_{n}(x-c)^{n}

then

f(x)=\left(\sum _{n=0}^{\infty }b_{n}(x-c)^{n}\right)\left(\sum _{n=0}^{\infty }d_{n}(x-c)^{n}\right)

and one can solve recursively for the terms $d_{n}$ by comparing coefficients.

Solving the corresponding equations yields the formulae based on determinants of certain matrices of the coefficients of $f(x)$ and $g(x)$

d_{0}={\frac {a_{0}}{b_{0}}}

d_{n}={\frac {1}{b_{0}^{n+1}}}{\begin{vmatrix}a_{n}&b_{1}&b_{2}&\cdots &b_{n}\\a_{n-1}&b_{0}&b_{1}&\cdots &b_{n-1}\\a_{n-2}&0&b_{0}&\cdots &b_{n-2}\\\vdots &\vdots &\vdots &\ddots &\vdots \\a_{0}&0&0&\cdots &b_{0}\end{vmatrix}}

Differentiation and integration

Once a function $f(x)$ is given as a power series as above, it is differentiable on the interior of the domain of convergence. It can be differentiated and integrated quite easily, by treating every term separately:

{\begin{aligned}f'(x)&=\sum _{n=1}^{\infty }a_{n}n(x-c)^{n-1}=\sum _{n=0}^{\infty }a_{n+1}(n+1)(x-c)^{n},\\\int f(x)\,dx&=\sum _{n=0}^{\infty }{\frac {a_{n}(x-c)^{n+1}}{n+1}}+k=\sum _{n=1}^{\infty }{\frac {a_{n-1}(x-c)^{n}}{n}}+k.\end{aligned}}

Both of these series have the same radius of convergence as the original one.

Analytic Functions

A function f defined on some open subset U of R or C is called analytic if it is locally given by a convergent power series. This means that every a ∈ U has an open neighborhood V ⊆ U, such that there exists a power series with center a that converges to f(x) for every x ∈ V.

Every power series with a positive radius of convergence is analytic on the interior of its region of convergence. All holomorphic functions are complex-analytic. Sums and products of analytic functions are analytic, as are quotients as long as the denominator is non-zero.

If a function is analytic, then it is infinitely differentiable, but in the real case the converse is not generally true. For an analytic function, the coefficients a_n can be computed as

a_{n}={\frac {f^{\left(n\right)}\left(c\right)}{n!}}

where $f^{(n)}(c)$ denotes the nth derivative of f at c, and $f^{(0)}(c)=f(c)$ . This means that every analytic function is locally represented by its Taylor series.

The global form of an analytic function is completely determined by its local behavior in the following sense: if f and g are two analytic functions defined on the same connected open set U, and if there exists an element c∈U such that f⁽ⁿ⁾(c) = g⁽ⁿ⁾(c) for all n ≥ 0, then f(x) = g(x) for all x ∈ U.

If a power series with radius of convergence r is given, one can consider analytic continuations of the series, i.e. analytic functions f which are defined on larger sets than { x : |x − c| < r } and agree with the given power series on this set. The number r is maximal in the following sense: there always exists a complex number x with |x − c| = r such that no analytic continuation of the series can be defined at x.

The power series expansion of the inverse function of an analytic function can be determined using the Lagrange inversion theorem.

Behavior near the boundary

The sum of a power series with a positive radius of convergence is an analytic function at every point in the interior of the disc of convergence. However, different behavior can occur at points on the boundary of that disc. For example:

Divergence while the sum extends to an analytic function: ${\textstyle \sum _{n=0}^{\infty }z^{n}}$ has radius of convergence equal to $1$ and diverges at every point of $|z|=1$ . Nevertheless, the sum in $|z|<1$ is ${\textstyle {\frac {1}{1-z}}}$ , which is analytic at every point of the plane except for $z=1$ .
Convergent at some points divergent at others: ${\textstyle \sum _{n=1}^{\infty }(-1)^{n+1}{\frac {z^{n}}{n}}}$ has radius of convergence $1$ . It converges for $z=1$ , while it diverges for $z=-1$
Absolute convergence at every point of the boundary: ${\textstyle \sum _{n=1}^{\infty }{\frac {z^{n}}{n^{2}}}}$ has radius of convergence $1$ , while it converges absolutely, and uniformly, at every point of $|z|=1$ due to Weierstrass M-test applied with the hyper-harmonic convergent series ${\textstyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}}$ .
Convergent on the closure of the disc of convergence but not continuous sum: Sierpiński gave an example of a power series with radius of convergence $1$ , convergent at all points with $|z|=1$ , but the sum is an unbounded function and, in particular, discontinuous. A sufficient condition for one-sided continuity at a boundary point is given by Abel's theorem.