# Representing functions as power series

> AP Calculus BC · Infinite Sequences and Series (Unit 10)
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u10-representing-functions-as-power-series/

This subtopic covers representing rational and transcendental functions as power series, using geometric series and term-by-term differentiation/integration. You will also learn how to find intervals of convergence for new representations.

**Prerequisites:** [Power series interval of convergence](https://www.owlsprep.com/study/ap-calculus-bc-u10-power-series-interval-convergence/); Testing series convergence; Basic differentiation and integration of polynomials; Infinite geometric series convergence rules

## Learning objectives

- Represent rational functions as power series using the geometric series formula
- Differentiate and integrate power series term-by-term to find new representations
- Find intervals of convergence for new power series representations
- Apply these techniques to common transcendental functions

## Geometric Power Series Representations

The foundation for all power series representations is the infinite geometric series formula you already know. For $|r| < 1$, the sum of the infinite series is $\sum_{n=0}^\infty r^n = \frac{1}{1-r}$. If we let $r = g(x)$, we can write any rational function rearranged to the form $\frac{1}{1-g(x)}$ directly as a power series, convergent when $|g(x)| < 1$.

**Geometric Power Series Representation** — A power series constructed from the infinite geometric series formula, used for functions that can be written in the form $\frac{c}{1 - g(x)}$ centered at $a$, where $g(x)$ is linear in $(x-a)$.

*Example:* $f(x) = \frac{4}{3-x}$ centered at 0

**Worked example:** Find the power series representation of $f(x) = \frac{4}{3 - x}$ centered at $a=0$, and state its interval of convergence.

1. Rewrite the denominator to match the $\frac{1}{1-r}$ form by factoring out the constant from the denominator:

   $$frac{4}{3left(1 - frac{x}{3}right)} = frac{4}{3} cdot frac{1}{1 - frac{x}{3}}$$
2. Here $r = \frac{x}{3}$. Apply the geometric series formula, which converges when $\left|\frac{x}{3}\right| < 1$:

   $$frac{1}{1 - frac{x}{3}} = sum_{n=0}^infty left(frac{x}{3}right)^n$$
3. Pull the constant factor out and simplify the general term:

   $$f(x) = frac{4}{3} sum_{n=0}^infty frac{x^n}{3^n} = sum_{n=0}^infty frac{4}{3^{n+1}} x^n$$
4. Check convergence at endpoints: at $x=3$, the series becomes $\sum \frac{4}{3}$, which diverges by the nth term test. At $x=-3$, the series becomes $\sum \frac{4(-1)^n}{3}$, which also diverges. The final interval of convergence is:

   $$(-3, 3)$$

> **Exam tip:** Always start by rewriting the denominator to get 1 as the constant term, matching the $1-r$ form. If you factor incorrectly, you will get wrong coefficients for every term of the series.

*Calculator:* forbidden

## Term-by-Term Differentiation of Power Series

Within the interval of convergence (excluding possibly endpoints), power series can be differentiated term-by-term just like finite polynomials. This property lets us build new power series from known existing representations. If $f(x) = \sum_{n=0}^\infty c_n (x-a)^n$ has radius of convergence $R$, the derivative has the same radius of convergence, but you must recheck convergence at endpoints.

$$f'(x) = sum_{n=1}^infty n c_n (x-a)^{n-1}$$

**Worked example:** Find the power series representation of $f(x) = \frac{2x}{(1 - x^2)^2}$ centered at $a=0$, using differentiation.

1. Start with the known geometric series for $\frac{1}{1 - x^2}$:

   $$frac{1}{1 - x^2} = sum_{n=0}^infty (x^2)^n = sum_{n=0}^infty x^{2n}, quad |x| < 1$$
2. Notice that the derivative of $\frac{1}{1-x^2}$ is exactly our target function. Differentiate term-by-term:

   $$frac{d}{dx}left(frac{1}{1-x^2}right) = sum_{n=0}^infty frac{d}{dx}left(x^{2n}right) = sum_{n=1}^infty 2n x^{2n - 1}$$
3. The radius of convergence remains $R=1$, same as the original. Check endpoints: at both $x=1$ and $x=-1$, the nth term does not approach zero, so the series diverges at both endpoints. Final result:

   $$f(x) = sum_{n=1}^infty 2n x^{2n-1}, quad (-1, 1)$$

> **Exam tip:** When you differentiate a power series, the constant term disappears, so the starting index always shifts from $n=0$ to $n=1$. Forgetting to adjust the starting index is a common multiple-choice trap answer.

*Calculator:* forbidden

## Term-by-Term Integration of Power Series

Like differentiation, power series can be integrated term-by-term within their interval of convergence. This is an extremely useful technique that lets us find power series for transcendental functions that are integrals of rational functions, such as $\ln(1+x)$ and $\arctan x$.

$$int f(x) dx = C + sum_{n=0}^infty frac{c_n}{n+1} (x-a)^{n+1}$$

The radius of convergence stays the same as the original series, but you must recheck convergence at endpoints. The constant of integration $C$ is found by substituting $x=a$ (the center of the series) into the function.

**Worked example:** Find the Maclaurin power series for $f(x) = \ln(1 + 3x)$, and state its interval of convergence.

1. First, note that $f'(x) = \frac{3}{1 + 3x}$, which can be written as a geometric series:

   $$frac{3}{1 + 3x} = 3 cdot frac{1}{1 - (-3x)} = sum_{n=0}^infty (-1)^n 3^{n+1} x^n$$
2. This converges when $|-3x| < 1$, so $|x| < \frac{1}{3}$. Integrate term-by-term:

   $$ln(1+3x) = int frac{3}{1+3x} dx = C + sum_{n=0}^infty (-1)^n 3^{n+1} cdot frac{x^{n+1}}{n+1}$$
3. Find $C$ by substituting $x=0$: $\ln(1 + 0) = 0 = C$, so $C=0$. Reindex to simplify:

   $$ln(1+3x) = sum_{n=1}^infty frac{(-1)^{n+1} 3^n x^n}{n}$$
4. Check endpoints: At $x=\frac{1}{3}$, the alternating harmonic series converges. At $x=-\frac{1}{3}$, the series becomes the negative harmonic series, which diverges. Final interval:

   $$left(-frac{1}{3}, frac{1}{3}right]$$

> **Exam tip:** Don't forget to solve for the constant of integration $C$ when integrating a power series. For Maclaurin series centered at 0, $C = f(0)$, which is almost always 0 for common functions like $\ln(1+x)$ or $\arctan x$.

*Calculator:* forbidden

## AP-Style Concept Check

**Check your understanding**

Answer this AP-style multiple-choice question to test your understanding:

1. Which of the following is the coefficient of $x^5$ in the Maclaurin series for $f(x) = \frac{x^2}{1 + 4x}$?

   - -64
   - -16
   - 16
   - 64

   *Why:* Rewrite $\frac{1}{1+4x}$ as $\sum_{n=0}^\infty (-4)^n x^n$, multiply by $x^2$ to get $\sum_{n=0}^\infty (-4)^n x^{n+2}$. Set $n+2=5$, so $n=3$, giving $(-4)^3 = -64$.

*Calculator:* forbidden

## Common pitfalls

- **Wrong:** Forgetting to check convergence at endpoints after differentiating or integrating a power series
  - Why it fails: Students know the radius of convergence stays the same, so they incorrectly assume the entire interval stays the same, but convergence behavior can change at endpoints.
  - Correct: After finding the radius of convergence, always test both endpoints with a convergence test, and add them to the interval if they converge.
- **Wrong:** Writing the power series for $\frac{1}{1+x}$ as $\sum_{n=0}^\infty x^n$ instead of $\sum_{n=0}^\infty (-1)^n x^n$
  - Why it fails: Students misidentify $r$ in the geometric series formula, forgetting the denominator is $1-r$, so $+x$ means $r=-x$.
  - Correct: Always explicitly rewrite the denominator as $1 - (\text{something})$ before writing the series, so $\frac{1}{1+x} = \frac{1}{1 - (-x)}$.
- **Wrong:** Reindexing incorrectly after differentiation or integration, leading to division by zero at $n=0$ (e.g. writing $\int \sum_{n=0}^\infty x^n dx = \sum_{n=0}^\infty \frac{x^n}{n}$)
  - Why it fails: Students rush the substitution when shifting the starting index of the series.
  - Correct: After reindexing, write out the first 2-3 terms of both the original and new series to confirm they match.
- **Wrong:** Pulling constants out of the series incorrectly when factoring the denominator (e.g. rewriting $\frac{3}{2-x}$ as $3\sum \left(\frac{x}{2}\right)^n$ instead of $\frac{3}{2}\sum \left(\frac{x}{2}\right)^n$)
  - Why it fails: Students forget that factoring a constant from the denominator moves its reciprocal to the outside of the fraction.
  - Correct: Always factor step-by-step: $\frac{A}{B - Cx} = \frac{A}{B\left(1 - \frac{Cx}{B}\right)} = \frac{A}{B} \cdot \frac{1}{1 - \frac{Cx}{B}}$.
- **Wrong:** Differentiating or integrating the coefficients $c_n$ along with the $x^n$ term
  - Why it fails: Students confuse the variable of differentiation $x$ with the index $n$, so they unnecessarily modify the coefficients.
  - Correct: Remember coefficients $c_n$ are constants with respect to $x$, so they stay unchanged when differentiating or integrating.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Geometric power series | $\frac{1}{1 - r} = \sum_{n=0}^\infty r^n$ | Converges only when $\|r\| < 1$; $r$ can be any function of $x$ |
| Term-by-term differentiation | If $f(x) = \sum_{n=0}^\infty c_n (x-a)^n$, then $f'(x) = \sum_{n=1}^\infty n c_n (x-a)^{n-1}$ | Same radius of convergence as original; always check endpoints |
| Term-by-term integration | If $f(x) = \sum_{n=0}^\infty c_n (x-a)^n$, then $\int f(x) dx = C + \sum_{n=0}^\infty \frac{c_n}{n+1} (x-a)^{n+1}$ | Same radius of convergence as original; $C = f(a)$ for center $a$ |
| $\frac{1}{1+x}$ (Maclaurin) | $\sum_{n=0}^\infty (-1)^n x^n$ | Interval of convergence $(-1, 1)$ |
| $\frac{1}{(1-x)^2}$ (Maclaurin) | $\sum_{n=0}^\infty (n+1) x^n$ | Derivative of $\frac{1}{1-x}$, interval $(-1, 1)$ |
| $\ln(1+x)$ (Maclaurin) | $\sum_{n=1}^\infty \frac{(-1)^{n+1} x^n}{n}$ | Integral of $\frac{1}{1+x}$, interval $(-1, 1]$ |
| $\arctan x$ (Maclaurin) | $\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}$ | Integral of $\frac{1}{1+x^2}$, interval $[-1, 1]$ |

## What's next

This topic gives you the core foundation for all work with Taylor and Maclaurin series, the next major topic in AP Calculus BC Unit 10. Representing functions as power series via geometric series, term-by-term differentiation, and integration lets you construct series for common functions without computing every derivative from scratch, which saves significant time on the AP exam and avoids computational errors. Without mastering the techniques here, you will struggle to construct Taylor series for composite or related functions, and will find approximation problems that commonly appear in free-response questions much harder than necessary. This topic is also the basis for integrating non-elementary functions, a frequent AP exam application.

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