M344: Calculus III
Section 11.8
Power Series¶
A power series is a series of the form
$$
p(x) = \sum_{n=0}^{\infty} c_n (x-x_0)^n = c_0 + c_1 (x-x_0) + c_2(x-x_0)^2 + c_3(x-x_0)^3 + \cdots
$$
where $x$ is a variable, $c_n$ are constants called the coefficients of the series, and $x_0$ is a fixed constant called the center of the series.
For any fixed $x$, the series becomes a usual series that we can test for convergence or divergence using any of the tests that we have studied.
It is possible that a power series could converge for some values $x$ but diverge for other values of $x$.
Consider a power series
$$
p(x) = \sum_{n=0}^\infty c_n (x-x_0)^n.
$$
If we apply the ratio test to the terms $a_n = c_n(x-x_0)^n$ of this series, then we obtain
$$
\left\vert \frac{a_{n+1}}{a_n} \right\vert = \left\vert \frac{c_{n+1}}{c_n} \right\vert\, \vert x - x_0 \vert \longrightarrow L \vert x - x_0 \vert,
$$
as $n \to \infty$, where $L$ is the limit obtained by applying the ratio test to the sequence of coefficients, $\left\{ c_n \right\}$. The ratio test tells us that the series $p(x)$ will be absolutely convergent if
$$
L \vert x - x_0 \vert < 1.
$$
If $L > 0$ is a number, then the power series will converge absolutely for all $x$ that satisfy
$$
\vert x - x_0 \vert < \frac{1}{L}.
$$
In this case, the number $R = \tfrac{1}{L}$ is called the radius of convergence of the power series $p$. By convention, if $L = 0$, then we set $R = +\infty$. If $L = +\infty$, or $L$ does not exist, then we set $R = 0$.
Now suppose that the radius of covergence of the power series $p$ is $R$, so that $p$ is absolutely convergent on the interval
$$
x_0 - R < x < x_0 + R. \tag{$*$}
$$
Recall that the ratio test is inconclusive when $\vert x - x_0 \vert = R$. It is possible that series could converge or diverge for $(x-x_0) = \pm R$. The convergence of the series at these two endpoints must be checked individually. Regardless, the interval on which the power series converges is at least $(*)$. This is the domain of the power series when regarded as a function. It is referred to as the interval of convergence of the power series.
1. Example¶
If we let $x_0 = 0$ and $c_n = 1$ for all $n$, we get a geometric series,
$$
p(x) = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + \cdots + x^n + \cdots.
$$
Now, let's consider plugging in some different values of $x$ into this function. If we let $x = \tfrac{1}{2}$, we get
$$
p\big(\tfrac{1}{2}\big) = \sum_{n=0}^{\infty} \left(\dfrac{1}{2}\right)^n = 1 + \dfrac{1}{2} + \left(\dfrac{1}{2}\right)^2 + \cdots,
$$
which we know converges. In fact, its value is $p\big(\tfrac{1}{2}\big) = 2$.
However, if we plug in an $x$ value greater than 1, like $x = 2$, we get
$$
p(2) = \sum_{n=1}^{\infty} 2^n = 1 + 2 + 2^2 + \cdots,
$$
which diverges. This analysis tells us that $x = \tfrac{1}{2}$ is in the domain of this power series, while $x = 2$ is not in the domain. In fact, we know from our study of geometric series that this power series converges for $\vert x \vert < 1$. This is the domain, or interval of convergence, of the power series. The radius of convergence of this series is $R = 1$.
2. Example¶
Determine the radius and interval of convergence of the power series.
$$
\sum_{n=1}^\infty 2^nn^2x^n
$$
Video solution
3. Example¶
Determine the radius and interval of convergence of the power series.
$$
\sum_{n=1}^\infty \frac{\sqrt{n}}{8^n}(x +6)^n
$$
Video solution
4. Example¶
Determine the radius and interval of convergence of the power series.
$$
\sum_{n=1}^\infty n! (2x-1)^n
$$
Video solution
5. Example¶
Determine the radius and interval of convergence of the power series.
$$
\sum_{n=1}^\infty \frac{n}{2^n (n^2 + 1)}\, x^n
$$
Video solution
6. Example¶
Determine the radius and interval of convergence of the power series.
$$
\sum_{n=1}^\infty \frac{n!\, x^n}{1\cdot 3\cdot 5 \cdots (2n-1)}
$$
Video solution
Partial Sums of Power Series¶
Consider a power series
$$
p(x) = \sum_{n=0}^\infty c_n (x-x_0)^n,
$$
and suppose that the radius of convergence of $p$ is greater than $0$. The partial sums of $p$ are polynomials that approximate the graph of $p$ near $x_0$.
$$
p_k(x) = \sum_{n=0}^k c_n(x-x_0)^n = c_0 + c_1(x-x_0) + \cdots + c_k (x-x_0)^k.
$$
The degree of the polynomial $p_k$ is at most $k$, and it is equal to $k$ if $c_k \neq 0$. Moreover, for any fixed $x$ in the interval of convergence of $p$, the value of $p(x)$ is equal to the limit
$$
p(x) = \lim_{k\to\infty} p_k(x).
$$
We consider the following example without proof.
7. Example¶
We will soon learn that the power series representation of $\sin x$ about $x_0=0$ is given by
$$
\sin x = \sum\limits_{n=0}^{\infty} \dfrac{(-1)^n}{(2n+1)!}\, x^{2n+1}.
$$
This is known as the Taylor-Maclaurin series for sine.
The first $3$ partial sums of this series are the polynomials,
$$ \displaystyle
\begin{aligned}
s_0(x) & = x, \\
s_1(x) & = x- \dfrac{x^3}{3!}, \\
s_2(x) & = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!}. \\
\end{aligned}
$$
When plotted on the same set of axes as $y = \sin x$, we see that the graphs of these polynomials closely approximate the graph of $\sin x$ near the origin, and that the closeness of the approximation extends further away from the origin as $k$ increases.
See the graph
We are not yet ready to prove that this power series indeed equals the sine function, but you can (and should!) verify that the radius of convergence of this series is $R = +\infty$. One consquence of this fact, provided the series really does equal the sine function, is that for any $x$-value, and any fixed $\varepsilon > 0$, one can find a $k > 0$ such that $\vert p_k(x) - \sin(x) \vert < \varepsilon$.
In practice, this means that if one is working on a bounded interval with a pre-determined acceptable error tolerance level of $\varepsilon > 0$, then $\sin(x)$ can always be replaced by a polynomial function on that interval.
Some of you may have seen this swindle pulled in a physics class. When working very close to the origin with a tolerance of $\varepsilon \sim 10^{-3}$, physicist often replace $\sin(x)$ by its first-order partial sum $s_0(x) = x$.
8. Example¶
The Bessel function of order 1 is the function defined by the power series
$$
J_1(x) = \sum_{n=0}^\infty \frac{(-1)^n\ x^{2n+1}}{n!(n+1)!2^{2n+1}}
$$
The graph of $J_1$ is plotted below.
We plot the function together with 11 of its partial sums on the inteval $[-20,20]$. The Bessel function itself is the thick blue curve, and the partial sums are the thinner curves.
As the degree $k$ increases, the thin curves approximate more and more of the actual curve. You can imagine how as $k \to \infty$, the entire curve will be approximated by the partial sums.
9. Example¶
Suppose $k$ is a positive integer. Find the radius of convergence of the series.
$$
\sum_{n=0}^\infty \frac{(n!)^k}{(kn)!}\, x^n
$$
Video solution
10. Example¶
Given the power series
$$
f(x) = 1 + 2x + x^2 + 2x^3 + x^4 + 2x^5 + x^6 + \cdots,
$$
determine the radius and interval of convergence, and find a representation of this function as a rational function. Clearly state the domain on which the rational function corresponds with the power series.
Video solution
Discussion¶
Questions? You can ask them here.