Finding Power Series Representations
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 6.1: “Power Series and Functions” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/6-1-power-series-and-functions |
| Textbook used in class | Stewart, Calculus, Section 11.9: “Representations of Functions as Power Series” (Examples 1, 2, 3) |
Opening Scenario
The geometric series sum $\dfrac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$ for $|x| < 1$ is the universal starting point for representing functions as power series. By substituting a function of $x$ for $x$, multiplying both sides by a constant, or shifting the variable, you turn one formula into representations of many related functions. This skill is building a toolkit from a single formula.
Quick Reference
Base formula: \[ \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n, \qquad |x| < 1. \]
Common substitutions:
| Substitution in $1/(1-u)$ | Result | Interval |
|---|---|---|
| $u = -x$ | $\dfrac{1}{1+x} = \sum_{n=0}^\infty (-1)^n x^n$ | $\|x\| < 1$ |
| $u = x^2$ | $\dfrac{1}{1-x^2} = \sum_{n=0}^\infty x^{2n}$ | $\|x\| < 1$ |
| $u = -x^2$ | $\dfrac{1}{1+x^2} = \sum_{n=0}^\infty (-1)^n x^{2n}$ | $\|x\| < 1$ |
| $u = x/3$ | $\dfrac{1}{1-x/3} = \sum_{n=0}^\infty x^n/3^n$ | $\|x\| < 3$ |
Key Concepts
1. Substitution Into the Base Formula
Example 1. Find a power series for $\dfrac{1}{1 + x^2}$ and state its interval of convergence. (Stewart 11.9, Example 1.)
Substitute $u = -x^2$ into $1/(1-u) = \sum u^n$: \[ \frac{1}{1 + x^2} = \frac{1}{1 - (-x^2)} = \sum_{n=0}^{\infty} (-x^2)^n = \sum_{n=0}^{\infty} (-1)^n x^{2n}. \]
The substitution $u = -x^2$ requires $|u| < 1$, i.e., $x^2 < 1$, i.e., $|x| < 1$.
Boxed answer: $\dfrac{1}{1+x^2} = \displaystyle\sum_{n=0}^{\infty} (-1)^n x^{2n}$, for $|x| < 1$.
Example 2. Find a power series for $\dfrac{x^3}{x + 2}$ centered at $0$. (Stewart 11.9, Example 2.)
Factor out to match the base form: \[ \frac{x^3}{x+2} = x^3 \cdot \frac{1}{x+2} = x^3 \cdot \frac{1}{2(1 + x/2)} = \frac{x^3}{2} \cdot \frac{1}{1 - (-x/2)}. \]
Substitute $u = -x/2$: \[ \frac{1}{1+x/2} = \sum_{n=0}^{\infty} \left(-\frac{x}{2}\right)^n = \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{2^n}. \]
So: \[ \frac{x^3}{x+2} = \frac{x^3}{2}\sum_{n=0}^{\infty} \frac{(-1)^n x^n}{2^n} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+3}}{2^{n+1}}. \]
Converges for $|x/2| < 1$, i.e., $|x| < 2$.
Boxed answer: $\displaystyle\sum_{n=0}^{\infty} \dfrac{(-1)^n x^{n+3}}{2^{n+1}}$, for $|x| < 2$.
2. Why Substitution Preserves Convergence Proofs
When you substitute $u = g(x)$ into a convergent power series, the new series converges wherever $|g(x)| < 1$ (the convergence condition for the base series). The interval of convergence for the new series is exactly the set of $x$ satisfying that condition.
thinking the interval of convergence is always $(-1, 1)$ after substitution. The interval for $\sum u^n$ is $|u| < 1$. When $u = x/3$, this becomes $|x| < 3$: the interval is $(-3, 3)$. The substitution scales the interval. Always translate the condition $|u| < 1$ back into a condition on $x$.
Common Errors Summary
| Error | Correction |
|---|---|
| Using the formula without factoring to get the $1/(1-u)$ form | Rewrite the function so the denominator is $1$ minus something; factor out a constant if needed |
| Keeping $u$ instead of substituting back to $x$ | Replace $u$ with the expression in $x$ before writing the final answer |
| Giving the interval as $|u| < 1$ instead of the $x$-interval | Solve $|g(x)| < 1$ to get the interval in $x$ |
Leveled Practice
Problem 1. Find a power series for $\dfrac{1}{5 - x}$ centered at $0$ and give its interval of convergence.
Show answer
$\dfrac{1}{5-x} = \dfrac{1}{5} \cdot \dfrac{1}{1 - x/5} = \dfrac{1}{5}\sum_{n=0}^\infty \left(\dfrac{x}{5}\right)^n = \sum_{n=0}^\infty \dfrac{x^n}{5^{n+1}}$.
Converges for $|x/5| < 1$, i.e., $|x| < 5$. Interval: $(-5, 5)$.
Problem 2. Use the series for $1/(1+x^2)$ to find a power series for $\arctan x$. (You will need differentiation/integration from the next lesson; state the strategy.)
Show answer
Since $\dfrac{d}{dx}\arctan x = \dfrac{1}{1+x^2} = \sum_{n=0}^\infty (-1)^n x^{2n}$, integrating term by term gives \[ \arctan x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1} + C. \] At $x = 0$: $\arctan 0 = 0$ and the series is $0$, so $C = 0$.
$\arctan x = \displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{2n+1}$, for $|x| \leq 1$.
Mastery Checklist
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