M344: Calculus III
Section 11.9
Representing functions as power series¶
In this section, we will investigate how we can use known power series, and calculus operations on these series, to represent functions.
We motivate this section by beginning with a celebration of the utility of geometric series.
1. Example¶
Consider the power series given by
$$
\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 \cdots.
$$
Recall that this series is geometric, and therefore has radius of convergence $R = 1$ and interval of convergence $\vert x\vert < 1$. Since it is a geometric series, we can immediately write down its sum on this interval.
$$
\sum_{n=0}^\infty x^n = \frac{1}{1-x}.
$$
These two functions do indeed coincide, but only on their shared domain. Despite the fact that the domain of the function $f(x) = \tfrac{1}{1-x}$ is $(-\infty,1)\cup(1,\infty)$, it is only equal to this power series representation on the interval of convergence, $(-1,1)$.
It is not a surprise that the point $x=1$ is an obstruction, but we might wonder why the common domain only extends to $x=-1$ in the other direction. We discussed this idea briefly in Section 11.8, but let's take another look at it now.
Consider the first five partial sums of the power series.
$$
\begin{align}{}
S_0(x) & = 1 \\
S_1(x) & = 1 + x \\
S_2(x) & = 1 + x + x^2 \\
S_3(x) & = 1 + x + x^2 + x^3 \\
S_4(x) & = 1 + x + x^2 + x^3 + x^4
\end{align}
$$
Let's plot these together with the function $f(x) = \tfrac{1}{1-x}$ on the same set of axes. Each $S_k$, as $k$ increases, should be a better approximation of the graph of $f$, centered at $x_0 = 0$.
See the graph
Figure. The thick black curve is the graph of the function $f(x) = \tfrac{1}{1-x}$. Recall that this function has a vertical asymptote at $x = 1$.
The thinner, multi-colored curves are the polynomial partial sum approximations. The horizontal line is $S_0(x) = 1$. As $k$ increases, the curves gradually provide better approximations of the portion of the graph near the $y$-axis. The higher degree approximations do a good job of approximating the asymptote as $k \to \infty$. However, the price that one pays is that the same polynomials do not do well at approximating the (very well behaved) portion of the curve on the interval $(-\infty,-1)$. In fact, they do a horrible job.
Recall that the domain, or interval of convergence, of a power series is always a symmetric interval about the center point. The graph above helps illustrate why this must be the case.
2. Example¶
Express the function
$$
f(x) = \dfrac{1}{1+x^2}
$$
as the sum of a power series and find the interval of convergence.
Video solution
3. Example¶
Find a power series representation of the function.
$$
f(x) = \dfrac{4}{2x + 3}
$$
What is its interval of convergence?
Video solution
Differentiation and Integration of Power Series¶
Since convergent power series are defined by limits of polynomial functions, it should not be too much of a surprise that we can perform our favorite calculus operations on them.
Theorem¶
Consider a power series
$$
f(x) = \sum\limits_{n=0}^{\infty} c_n (x-x_0)^n
$$
with radius of convergence $R>0$. Then $f$ is differentiable on the interval $(x_0-R, x_0+R)$ and
$$
f'(x) = c_1 + 2c_2(x-x_0) + 3c_3(x-x_0)^2 + \cdots = \sum\limits_{n=1}^{\infty} n c_n (x-x_0)^{n-1} = \sum\limits_{n=0}^{\infty} (n+1)c_{n+1} (x-x_0)^{n}. \tag{1}
$$
Moreover, since $f$ is differentiable on $I = \vert x - x_0\vert < R$, then it is also continuous on that interval. Therefore $f$ is integrable on $I$, and
$$
\displaystyle\int f(x) \, dx = C + c_0(x-x_0) + c_1\dfrac{(x-x_0)^2}{2} + c_2\dfrac{(x-x_0)^3}{3} + \cdots = C + \sum\limits_{n=0}^{\infty}\dfrac{c_n}{n+1}\,(x-x_0)^{n+1}, \tag{2}
$$
where $C$ is an unknown constant.
The radii of convergence of the power series in equations $(1)$ and $(2)$ both remain $R$. (Implicit exercise: show it!)
Note that to keep the index of equation $(1)$ starting at $n=0$, we had to shift the terms inside of the $\Sigma$. The two equations in the the thereom above can be written more succinctly as follows.
$$
\begin{array}{}
\displaystyle\dfrac{d}{dx} \left[ \sum\limits_{n=1}^{\infty} c_n (x-x_0)^n\right] & = \displaystyle\sum\limits_{n=0}^{\infty} \dfrac{d}{dx} \left[c_n (x-x_0)^{n}\right], \quad \text{and} \\[2ex]
\displaystyle\int \left[ \sum\limits_{n=1}^{\infty} c_n (x-x_0)^n\right]dx & = C + \displaystyle\sum\limits_{n=0}^{\infty} \left[ \displaystyle\int c_n (x-a)^{n} \, dx \right]
\end{array}
$$
4. Example¶
Find a power series representation for the function and find its radius of convergence.
$$
f(x) = \ln\big(5-x\big)
$$
Video solution
5. Example¶
Find a power series representation for the function. What is its radius of convergence?
$$
f(x) = \frac{x^2}{(1+x)^3}
$$
Video solution
Once we understand power series representations of functions, we can use power series to reframe and reimagine old problems from calculus.
6. Example¶
Evaluate the integral as a power series.
$$
\int \frac{\arctan x}{x}\,dx
$$
What is the radius of convergence of the solution?
Video solution
7. Example¶
Show that the power series
$$
f(x) = \sum_{n=0}^\infty \frac{(-1)^n\,x^{2n}}{(2n)!}
$$
is a solution to the differential equation $y'' + y = 0$. Based on your knowledge of calculus, can you guess what function this this power series represents?
Video solution
We end these notes by investigating a series representation of $\pi$.
8. Example¶
Use the power series for $\arctan x$ to show that
$$
\pi = 2\sqrt{3}\,\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)\, 3^n}.
$$
Video solution
Discussion¶
Questions? You can ask them here.