So far we studied series of numbers. The question we will be answering today is: Can we get some variables involved and use series to represent functions?
Tianran Chen
Department of Mathematics
Auburn University at Montgomery
We studied geometric series of the form \[ \sum_{k=0}^\infty r^k = 1 + r + r^2 + r^3 + r^4 + \cdots \] and we now know exactly for which $r$ this series will converge.
A natural question to ask is then: Can we treat $r$ as a variable? And will that produce a function? This is what power series is all about.
They are expressions that look like polynomials with infinitely many terms.
A power series centered at $x=0$ (or about $x=0$) is an expression \[ \sum_{k=0}^\infty c_k x^k = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \cdots \] where $c_0,c_1,c_2,\ldots$ are real constants known as coefficients.
For example, the following expressions are power series (about $x=0$): \begin{align*} &1 + x + x^2 + x^3 + x^4 + \cdots \\ &1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots \\ &1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \cdots \end{align*}
In a power series in the variable $x$ (about $x=0$), we only allow terms to have nonnegative integer powers of $x$.
Therefore the following expressions are not series (about $x=0$): \begin{align*} &\frac{1}{x} + 1 + x + x^2 + x^3 + x^4 + \cdots \\ &1 + \sqrt{x} + \frac{x}{2} + \frac{\sqrt{x^3}}{6} + \frac{x^2}{24} + \cdots \\ &1 - \ln x + 2 \ln x - 3 \ln x + \cdots \end{align*}
In general, a power series centered at $x = a$ is an expression \[ \sum_{k=0}^\infty c_k (x-a)^k = c_0 + c_1 (x-a) + c_2 (x-a)^2 + c_3 (x-a)^3 + \cdots \] for some real constants (the coefficients) $c_0,c_1,c_2,\ldots$.
Simply put, a power series is an infinite series in which each term has a nonnegative integer power of a variable (or a variable minus a number).
For example, the expression \[ 1 + (x - 3) + (x-3)^2 + (x-3)^3 + (x-3)^4 + \cdots \] is a power series centered at $x=3$.
Keep in mind that we may not be able to assign meaningful value to it once we "plug in" a value for $x$. This is the question of convergence.
It is clear that a power series $\sum_{k=0}^\infty c_k(x-a)^k$ always converges at its center $x=a$: If we "plug in" $x=a$, we get \[ c_0 + c_1 (a-a) + c_2 (a-a)^2 + \cdots = c_0 + 0 + 0 + \cdots = c_0. \]
An important question we need to study is that given a power series centered at $x=a$, are there any other values we can plug in for $x$ and get a convergent series (and hence a meaningful value)?
Each power series centered at $x=a$ has a radius of convergence such that for any $x$ within this radius around its center, the series converges.
Theorem. A power series $\sum_{k=0}^\infty c_k (x-a)^k$ has three possibilities:
Consider the power series \[ 1+ x+ x^2 + x^3 + x^4 + x^5 + \cdots \] centered at $x=0$.
From what we know about geometric series, we can see this series converges if and only if $|x| < 1$. Since the center is $x=0$, we can conclude that the radius of convergence is 1.
Consider the power series \[ \sum_{k=0}^\infty \frac{2^k(x-1)^k}{3} = \frac{1}{3} + \frac{2(x-1)}{3} + \frac{2^2(x-1)^2}{3} + \frac{2^3(x-1)^3}{3} + \cdots \] centered at $x=1$.
This is a geometric series $\sum_{k=0}^\infty a r^k$ with $a = \frac{1}{3}$ and $r = 2(x-1)$.
We know such a series converges if and only if $|r| < 1$, i.e., $|2(x-1)| < 1$. This is equivalent to \[ |x-1| < 1/2. \] Since the center is $x=1$, this inequality states that the radius of convergence is $1/2$.
Consider the power series \[ \sum_{k=0}^\infty \frac{x^k}{k!} = 1 + \frac{x}{1} + \frac{x^2}{2} + \frac{x^3}{6} + \cdots \] centered at $x=0$.
We will apply the ratio test here by computing the limit of the ratio between two consecutive terms: \[ \lim_{k \to \infty} \frac{ \frac{x^{k+1}}{(k+1)!} }{ \frac{x^k}{k!} } = \lim_{k \to \infty} \frac{ \frac{x}{(k+1)!} }{ \frac{1}{k!} } = \lim_{k \to \infty} \frac{ x }{ k } = 0 \] for any given $x$.
By the ratio test, we can conclude that this series converges for any given $x$ value. Therefore the radius of convergence is $\infty$.
The set of real numbers for which the series converge form a interval, and it is called the interval of convergence of the power series.
Within this interval of convergence, power series having the same center can be manipulated algebraically, e.g., they can be added, and multiplied.
For example, \begin{align*} \sum_{k=0} b_k (x-a)^k + \sum_{k=0} c_k (x-a)^k &= \sum_{k=0} (b_k + c_k) (x-a)^k \\ b \cdot \sum_{k=0} c_k (x-a)^k &= \sum_{k=0} b c_k (x-a)^k \end{align*}
When the interval of convergence consists of more than one number, within this interval, the power series defines a function that is continuous and differentiable. Indeed, we can show that all derivatives of this function exists and are continuous in that interval.
This also means, within their interval of convergence, power series can be differentiated and integrated term by term.
\[ \frac{d}{dx} \sum_{k=0}^\infty c_k (x-a)^k = \sum_{k=0}^\infty \frac{d}{dx} \left[ c_k (x-a)^k \right] = \sum_{k=1}^\infty c_k k (x-a)^{k-1} \]
\[ \int \sum_{k=0}^\infty c_k (x-a)^k = \sum_{k=0}^\infty \int \left[ c_k (x-a)^k \right] = \sum_{k=0}^\infty c_k \frac{(x-a)^{k+1}}{k+1} \]
Functions given by power series are called analytic functions.
Many familiar functions in calculus are analytic. For Example, \begin{align*} e^x &= \sum_{k=0}^\infty \frac{x^k}{k!} & \ln(1+x) &= \sum_{k=1}^\infty \frac{ (-1)^{k-1} }{k} x^k & \\ \cos(x) &= \sum_{k=0}^\infty \frac{(-1)^k x^{2k}}{(2k)!} & \sin(x) &= \sum_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{(2k+1)!} \end{align*} (we will learn how to compute them soon.)
Historically, power series are the main weapons for mathematicians for a long time. It is likely that mathematicians thought that all "reasonable" functions are analytic. This assumption turn out to be quite wrong. Nonetheless, power series remain extremely useful in applied mathematics.