Operations with Power Series: Learn It 1

Giselle Miller

Operations with Power Series: Learn It 1

Add and subtract power series
Multiply two power series together
Find derivatives of power series term by term
Integrate power series term by term

Combining Power Series

In the previous section, we learned how to represent certain functions using power series. Now we’ll explore how to manipulate these power series—combining, modifying, and transforming them to create new power series representations.

This ability to work with power series is incredibly valuable for two key reasons:

First, it lets us find power series for functions that might otherwise be difficult to work with directly. For example, if we know the power series for [latex]f(x) = \frac{1}{1-x}[/latex], we can use calculus operations to find the power series for [latex]f'(x) = \frac{1}{(1-x)^2}[/latex].

Second, we can define entirely new functions that can’t be expressed using elementary functions. This becomes especially important when solving differential equations that have no elementary solutions.

Basic Operations with Power Series

When you have two power series that converge on the same interval, you can perform several operations to create new power series.

theorem: combining power series

Suppose the power series [latex]\sum_{n=0}^{\infty} c_n x^n[/latex] and [latex]\sum_{n=0}^{\infty} d_n x^n[/latex] both converge to functions [latex]f[/latex] and [latex]g[/latex] respectively on a common interval [latex]I[/latex].

Addition/Subtraction: [latex]\sum_{n=0}^{\infty} (c_n \pm d_n) x^n[/latex] converges to [latex]f \pm g[/latex] on [latex]I[/latex]
Scalar and Power Multiplication: [latex]\sum_{n=0}^{\infty} bx^m c_n x^n[/latex] converges to [latex]bx^m f(x)[/latex] on [latex]I[/latex] for any integer [latex]m \geq 0[/latex] and real number [latex]b[/latex]
Composition: [latex]\sum_{n=0}^{\infty} c_n (bx^m)^n[/latex] converges to [latex]f(bx^m)[/latex] for all [latex]x[/latex] where [latex]bx^m[/latex] is in [latex]I[/latex]

Proof

We prove i. in the case of the series [latex]\displaystyle\sum _{n=0}^{\infty }\left({c}_{n}{x}^{n}+{d}_{n}{x}^{n}\right)[/latex]. Suppose that [latex]\displaystyle\sum _{n=0}^{\infty }{c}_{n}{x}^{n}[/latex] and [latex]\displaystyle\sum _{n=0}^{\infty }{d}_{n}{x}^{n}[/latex] converge to the functions f and g, respectively, on the interval I. Let x be a point in I and let [latex]{S}_{N}\left(x\right)[/latex] and [latex]{T}_{N}\left(x\right)[/latex] denote the Nth partial sums of the series [latex]\displaystyle\sum _{n=0}^{\infty }{c}_{n}{x}^{n}[/latex] and [latex]\displaystyle\sum _{n=0}^{\infty }{d}_{n}{x}^{n}[/latex], respectively. Then the sequence [latex]\left\{{S}_{N}\left(x\right)\right\}[/latex] converges to [latex]f\left(x\right)[/latex] and the sequence [latex]\left\{{T}_{N}\left(x\right)\right\}[/latex] converges to [latex]g\left(x\right)[/latex]. Furthermore, the Nth partial sum of [latex]\displaystyle\sum _{n=0}^{\infty }\left({c}_{n}{x}^{n}+{d}_{n}{x}^{n}\right)[/latex] is

[latex]\begin{array}{cc}\hfill {\displaystyle\sum _{n=0}^{N}} \left({c}_{n}{x}^{n}+{d}_{n}{x}^{n}\right)& ={\displaystyle\sum _{n=0}^{N}} {c}_{n}{x}^{n}+{\displaystyle\sum _{n=0}^{N}}{d}_{n}{x}^{n}\hfill \\ & ={S}_{N}\left(x\right)+{T}_{N}\left(x\right).\hfill \end{array}[/latex]

Because

[latex]\begin{array}{cc} \hfill {\underset{N\to\infty}\lim} \left({S}_{N}\left(x\right)+{T}_{N}\left(x\right)\right) & ={\underset{N\to\infty}\lim} {S}_{N}\left(x\right)+{\underset{N\to\infty}\lim} {T}_{N}\left(x\right)\hfill \\ & =f\left(x\right)+g\left(x\right),\hfill \end{array}[/latex]

we conclude that the series [latex]\displaystyle\sum _{n=0}^{\infty }\left({c}_{n}{x}^{n}+{d}_{n}{x}^{n}\right)[/latex] converges to [latex]f\left(x\right)+g\left(x\right)[/latex].

[latex]_\blacksquare[/latex]

These operations open up powerful possibilities. Since we know the power series representation for [latex]f(x) = \frac{1}{1-x}[/latex], we can now find power series representations for related functions like:

[latex]y = \frac{3x}{1-x^2}[/latex]
[latex]y = \frac{1}{(x-1)(x-3)}[/latex]

Why the Same Interval? The interval of convergence restriction is crucial. Power series behave nicely only within their convergence intervals, so combining series requires working within the intersection of their individual convergence intervals.

We’ll examine products of power series in a later theorem. For now, let’s focus on applying these combining techniques and determining intervals of convergence for the new power series we create.

When you combine power series using the operations above, the new series will generally have the same interval of convergence as the original series. However, there are some important considerations to keep in mind when working with compositions and transformations.

Suppose that [latex]\displaystyle\sum _{n=0}^{\infty }{a}_{n}{x}^{n}[/latex] is a power series whose interval of convergence is [latex]\left(-1,1\right)[/latex], and suppose that [latex]\displaystyle\sum _{n=0}^{\infty }{b}_{n}{x}^{n}[/latex] is a power series whose interval of convergence is [latex]\left(-2,2\right)[/latex].

Find the interval of convergence of the series [latex]\displaystyle\sum _{n=0}^{\infty }\left({a}_{n}{x}^{n}+{b}_{n}{x}^{n}\right)[/latex].
Find the interval of convergence of the series [latex]\displaystyle\sum _{n=0}^{\infty }{a}_{n}{3}^{n}{x}^{n}[/latex].

Watch the following video to see the worked solution to the above example.

You can view the transcript for “6.2.1” here (opens in new window).

In the next example, we show how to use combining power series and the power series for a function [latex]f[/latex] to construct power series for functions related to [latex]f[/latex]. Specifically, we consider functions related to the function [latex]f\left(x\right)=\frac{1}{1-x}[/latex] and we use the fact that [latex]\frac{1}{1-x}=\displaystyle\sum _{n=0}^{\infty }{x}^{n}=1+x+{x}^{2}+{x}^{3}+\cdots[/latex] for [latex]|x|<1[/latex].

Use the power series representation for [latex]f\left(x\right)=\frac{1}{1-x}[/latex] combined with Combining Power Series to construct a power series for each of the following functions. Find the interval of convergence of the power series.

[latex]f\left(x\right)=\frac{3x}{1+{x}^{2}}[/latex]
[latex]f\left(x\right)=\frac{1}{\left(x - 1\right)\left(x - 3\right)}[/latex]

First write [latex]f\left(x\right)[/latex] as
[latex]f\left(x\right)=3x\left(\frac{1}{1-\left(\text{-}{x}^{2}\right)}\right)[/latex].

Using the power series representation for [latex]f\left(x\right)=\frac{1}{1-x}[/latex] and parts ii. and iii. of Combining Power Series, we find that a power series representation for f is given by

[latex]\displaystyle\sum _{n=0}^{\infty }3x{\left(\text{-}{x}^{2}\right)}^{n}=\displaystyle\sum _{n=0}^{\infty }3{\left(-1\right)}^{n}{x}^{2n+1}[/latex].

Since the interval of convergence of the series for [latex]\frac{1}{1-x}[/latex] is [latex]\left(-1,1\right)[/latex], the interval of convergence for this new series is the set of real numbers x such that [latex]|{x}^{2}|<1[/latex]. Therefore, the interval of convergence is [latex]\left(-1,1\right)[/latex].
- To find the power series representation, use partial fractions to write [latex]f\left(x\right)=\frac{1}{\left(1-x\right)\left(x - 3\right)}[/latex] as the sum of two fractions. We have
[latex]\begin{array}{cc}\hfill \frac{1}{\left(x - 1\right)\left(x - 3\right)}& =\frac{\text{-}\frac{1}{2}}{x - 1}+\frac{\frac{1}{2}}{x - 3}\hfill \\ & =\frac{\frac{1}{2}}{1-x}-\frac{\frac{1}{2}}{3-x}\hfill \\ & =\frac{\frac{1}{2}}{1-x}-\frac{\frac{1}{6}}{1-\frac{x}{3}}.\hfill \end{array}[/latex]

First, using part ii. of Combining Power Series, we obtain

[latex]\frac{\frac{1}{2}}{1-x}=\displaystyle\sum _{n=0}^{\infty }\frac{1}{2}{x}^{n}\text{for}|x|<1[/latex].

Then, using parts ii. and iii. of Combining Power Series, we have

[latex]\frac{\frac{1}{6}}{1-\frac{x}{3}}=\displaystyle\sum _{n=0}^{\infty }\frac{1}{6}{\left(\frac{x}{3}\right)}^{n}\text{for}|x|<3[/latex].

Since we are combining these two power series, the interval of convergence of the difference must be the smaller of these two intervals. Using this fact and part i. of Combining Power Series, we have

[latex]\frac{1}{\left(x - 1\right)\left(x - 3\right)}=\displaystyle\sum _{n=0}^{\infty }\left(\frac{1}{2}-\frac{1}{6\cdot {3}^{n}}\right){x}^{n}[/latex]

where the interval of convergence is [latex]\left(-1,1\right)[/latex].

In the previous example, we showed how to find power series for certain functions. In the next example we show how to do the opposite: given a power series, determine which function it represents.

Consider the power series [latex]\displaystyle\sum _{n=0}^{\infty }{2}^{n}{x}^{n}[/latex]. Find the function f represented by this series. Determine the interval of convergence of the series.