Representing functions as power series

AP Calculus BC· AP Calculus BC CED — Infinite Sequences and Series· 14 min read

1. Geometric Power Series Representations★★☆☆☆⏱ 4 min

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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 $∣ g (x) ∣ < 1$ .

📘 Definition

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 $\frac{x}{3} < 1$ :
$frac{1}{1 - frac{x}{3}} = sum_{n=0}^infty left(frac{x}{3}right)^n$
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.

2. Term-by-Term Differentiation of Power Series★★★☆☆⏱ 4 min

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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 $f (x) = \frac{2 x}{( 1 - x ^{2} ) ^{2}}$ 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.

3. Term-by-Term Integration of Power Series★★★☆☆⏱ 4 min

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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 $f (x) = ln (1 + 3 x)$ , and state its interval of convergence.

1
First, note that $f^{'} (x) = \frac{3}{1 + 3 x}$ , 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 $∣ - 3 x ∣ < 1$ , 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 $arctan x$ .

4. AP-Style Concept Check★★★★☆⏱ 2 min

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✓ Quick check

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

Which of the following is the coefficient of $x^{5}$ in the Maclaurin series for $f (x) = \frac{x ^{2}}{1 + 4 x}$ ?
- -64
- -16
- 16
- 64
Reveal answer
-64 —
Rewrite $\frac{1}{1 + 4 x}$ 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 $(- 4)^{3} = - 64$ .

5. Common Pitfalls

Wrong move:

Forgetting to check convergence at endpoints after differentiating or integrating a power series

Why:

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 move:

After finding the radius of convergence, always test both endpoints with a convergence test, and add them to the interval if they converge.

Wrong move:

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:

Students misidentify $r$ in the geometric series formula, forgetting the denominator is $1 - r$ , so $+ x$ means $r = - x$ .

Correct move:

Always explicitly rewrite the denominator as $1 - (something)$ before writing the series, so $\frac{1}{1 + x} = \frac{1}{1 - ( - x )}$ .

Wrong move:

Reindexing incorrectly after differentiation or integration, leading to division by zero at $n = 0$ (e.g. writing $\int \sum_{n = 0}^{\infty} x^{n} d x = \sum_{n = 0}^{\infty} \frac{x ^{n}}{n}$ )

Why:

Students rush the substitution when shifting the starting index of the series.

Correct move:

After reindexing, write out the first 2-3 terms of both the original and new series to confirm they match.

Wrong move:

Pulling constants out of the series incorrectly when factoring the denominator (e.g. rewriting $\frac{3}{2 - x}$ as $3 \sum (\frac{x}{2})^{n}$ instead of $\frac{3}{2} \sum (\frac{x}{2})^{n}$ )

Why:

Students forget that factoring a constant from the denominator moves its reciprocal to the outside of the fraction.

Correct move:

Always factor step-by-step: $\frac{A}{B - C x} = \frac{A}{B ( 1 - \frac{C x}{B} )} = \frac{A}{B} \cdot \frac{1}{1 - \frac{C x}{B}}$ .

Wrong move:

Differentiating or integrating the coefficients $c_{n}$ along with the $x^{n}$ term

Why:

Students confuse the variable of differentiation $x$ with the index $n$ , so they unnecessarily modify the coefficients.

Correct move:

Remember coefficients $c_{n}$ are constants with respect to $x$ , so they stay unchanged when differentiating or integrating.

6. Quick Reference 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 $\int f (x) d x = C + \sum_{n = 0}^{\infty} \frac{c _{n}}{n + 1} (x - a)^{n + 1}$	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)	$\sum_{n = 0}^{\infty} \frac{( - 1 ) ^{n} x ^{2 n + 1}}{2 n + 1}$	Integral of $\frac{1}{1 + x ^{2}}$ , interval $[- 1, 1]$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Find coefficient of x^5 in series
2022 · FRQ
Arctan series via integration

Going deeper

official syllabusAP Calculus BC CEDOfficial College Board course description

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.