Representing functions as power series — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Geometric power series representations, term-by-term differentiation and integration of power series, finding intervals of convergence for new representations, and representing rational and transcendental functions as power series.

You should already know: How to test series convergence and find intervals of convergence for general power series. Basic differentiation and integration of polynomial functions. The convergence rule for infinite geometric series.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Representing functions as power series?

A power series is an infinite series of the form ${\sum }_{n=0}^{\infty }{c}_n(x-a{)}^n$, centered at $x=a$, that can represent a function $f(x)$ for all $x$ within its interval of convergence. The core insight here is that many common functions can be written as infinite polynomials, which lets us perform calculus operations that are difficult or impossible on the original function. Synonyms for this topic include power series expansions and power series representations.

Per the AP Calculus BC CED, the full unit of Infinite Sequences and Series makes up 17–18% of the total exam score, and this subtopic accounts for approximately one-quarter of the unit’s exam questions. It appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often combined with other topics like approximation or integration of non-elementary functions. On the AP exam, you will almost always be asked to find a representation centered at 0 (a Maclaurin series) unless stated otherwise.

2. Geometric Power Series Representations

The most basic and commonly tested power series representation comes from the infinite geometric series formula you already know: for $|r|<1$, the sum of the infinite geometric series is $$\sum _{n=0}^{\infty }{r}^n=\frac{1}{1-r}$$ If we let $r=g(x)$, a function of $x$, we can rewrite any rational function that can be manipulated into the form $\frac{1}{1-g(x)}$ directly as a power series: $$\frac{1}{1-g(x)}=\sum _{n=0}^{\infty }[g(x){]}^n$$ This series converges if and only if $|g(x)|<1$, which gives us the bounds for the interval of convergence. To center the series at a point $x=a$ instead of 0, we simply rearrange the denominator to get a term of $(x-a)$ matching the structure above. This technique is the starting point for almost all other power series constructions on the AP exam.

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: Factor 3 out of the denominator to get a constant term of 1: $$\frac{4}{3\left(1-\frac{x}{3}\right)}=\frac{4}{3}\cdot \frac{1}{1-\frac{x}{3}}$$ Here, $r=\frac{x}{3}$.
2. Apply the geometric series formula: $$\frac{1}{1-\frac{x}{3}}=\sum _{n=0}^{\infty }{\left(\frac{x}{3}\right)}^n,\text{for }\left|\frac{x}{3}\right|<1$$
3. Simplify the series by pulling out the constant factor: $$f(x)=\frac{4}{3}\sum _{n=0}^{\infty }\frac{x^n}{3^n}=\sum _{n=0}^{\infty }\frac{4}{3^{n+1}}{x}^n$$
4. Find the full interval of convergence: Solve $\left|\frac{x}{3}\right|<1⟹|x|<3$. Check endpoints: at $x=3$, the series becomes $\sum \frac{4}{3}$, which diverges by the nth term test; at $x=-3$, it becomes $\sum \frac{4(-1{)}^n}{3}$, which also diverges. The 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.

3. Term-by-Term Differentiation of Power Series

Within the interval of convergence (excluding possibly the endpoints), power series can be differentiated term-by-term just like finite polynomials. This property lets us construct new power series from existing representations we already know. If $f(x)={\sum }_{n=0}^{\infty }{c}_n(x-a{)}^n$ has a radius of convergence $R$, then its derivative is: $$f^{\prime }(x)=\sum _{n=1}^{\infty }n{c}_n(x-a{)}^{n-1}$$ The radius of convergence $R$ stays the same for the differentiated series, but you must recheck convergence at the endpoints, since convergence behavior can change after differentiation. This technique is especially useful for finding series of derivatives of known functions, like $\frac{1}{(1-x{)}^2}$ from the geometric series of $\frac{1}{1-x}$.

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 $\frac{1}{1-x^2}$: $$\frac{1}{1-x^2}=\sum _{n=0}^{\infty }(x^2{)}^n=\sum _{n=0}^{\infty }{x}^{2n},\text{ converges for }|x|<1$$
2. Notice that the derivative of $\frac{1}{1-x^2}$ is $\frac{2x}{(1-x^2{)}^2}$, which is exactly our 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 starting index shifts to $n=1$ because the $n=0$ term is a constant ($x^0=1$), whose derivative is 0. Check radius of convergence: it matches the original series, $R=1$.
4. Check endpoints: At $x=1$ and $x=-1$, the nth term $2n(1{)}^{2n-1}=2n$ does not approach 0, so the series diverges at both endpoints. The final series is $f(x)={\sum }_{n=1}^{\infty }2n{x}^{2n-1}$ with interval of convergence $(-1,1)$.

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

4. Term-by-Term Integration of Power Series

Just like differentiation, power series can be integrated term-by-term within their interval of convergence. This is one of the most useful applications of power series, because it lets us find representations for transcendental functions that are integrals of rational functions (like $\ln (1+x)$, $\arctan x$). If $f(x)={\sum }_{n=0}^{\infty }{c}_n(x-a{)}^n$ has radius of convergence $R$, then the indefinite integral is: $$\int f(x)dx=C+\sum _{n=0}^{\infty }\frac{c_n}{n+1}(x-a{)}^{n+1}$$ Again, the radius of convergence stays the same, but you must recheck convergence at the endpoints. The constant of integration $C$ is almost always found by substituting $x=a$ (the center of the series) into the original function.

Worked Example

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

1. First, note that $f^{\prime }(x)=\frac{3}{1+3x}$, which we can write as a geometric power series: $$\frac{3}{1+3x}=3\cdot \frac{1}{1-(-3x)}=3\sum _{n=0}^{\infty }(-3x{)}^n=\sum _{n=0}^{\infty }(-1{)}^n{3}^{n+1}{x}^n$$ This converges when $|-3x|<1⟹|x|<\frac{1}{3}$.
2. Integrate term-by-term to get $\ln (1+3x)$: $$\ln (1+3x)=\int \frac{3}{1+3x}dx=C+\sum _{n=0}^{\infty }(-1{)}^n{3}^{n+1}\cdot \frac{x^{n+1}}{n+1}$$
3. Find the constant of integration: substitute $x=0$, we get $\ln (1+0)=0=C$, so $C=0$.
4. Check endpoints: At $x=\frac{1}{3}$, the series becomes ${\sum }_{n=0}^{\infty }\frac{(-1{)}^n{3}^{n+1}(1/3{)}^{n+1}}{n+1}={\sum }_{n=1}^{\infty }\frac{(-1{)}^{n+1}}{n}$, which converges by the alternating series test. At $x=-\frac{1}{3}$, the series becomes ${\sum }_{n=1}^{\infty }-\frac{1}{n}$, which diverges. The final series is $\ln (1+3x)={\sum }_{n=1}^{\infty }\frac{(-1{)}^{n+1}{3}^n{x}^n}{n}$ with interval of convergence $\left(-\frac{1}{3},\frac{1}{3}\right]$.

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 $\arctan x$.

5. Common Pitfalls (and how to avoid them)

• 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 at endpoints can change. 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 $1-(\text{something})$ before writing the series, so $\frac{1}{1+x}=\frac{1}{1-(-x)}$.
• Wrong move: Reindexing incorrectly after differentiation or integration, leading to a division by zero at $n=0$. For example, writing $\int {\sum }_{n=0}^{\infty }{x}^{n}dx={\sum }_{n=0}^{\infty }\frac{x^n}{n}$. Why: Students rush the substitution when shifting the starting index. 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. For example, 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: 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: $\frac{A}{B-Cx}=\frac{A}{B\left(1-\frac{Cx}{B}\right)}=\frac{A}{B}\cdot \frac{1}{1-\frac{Cx}{B}}$.
• 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. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Which of the following is the coefficient of $x^5$ in the Maclaurin series for $f(x)=\frac{x^2}{1+4x}$? (A) $-64$ (B) $-16$ (C) $16$ (D) $64$

Worked Solution: Start by rewriting the function as a geometric power series: $\frac{1}{1+4x}=\frac{1}{1-(-4x)}={\sum }_{n=0}^{\infty }(-4x{)}^n={\sum }_{n=0}^{\infty }(-4{)}^n{x}^n$, which converges for $|x|<1/4$. Multiply by $x^2$ to get $f(x)=x^2{\sum }_{n=0}^{\infty }(-4{)}^n{x}^n={\sum }_{n=0}^{\infty }(-4{)}^n{x}^{n+2}$. To find the coefficient of $x^5$, set $n+2=5$, so $n=3$. Substitute $n=3$ to get the coefficient: $(-4{)}^3=-64$. The correct answer is A.

Question 2 (Free Response)

Let $f(x)=\frac{1}{1+x^2}$. (a) Find the Maclaurin series for $f(x)$, and state the radius of convergence $R$. (b) Use term-by-term integration to find the Maclaurin series for $g(x)=\arctan x$. (c) Identify the exact value of the infinite series ${\sum }_{n=0}^{\infty }\frac{(-1{)}^n}{(2n+1)2^{2n+1}}$. Justify your answer.

Worked Solution: (a) Rewrite $f(x)=\frac{1}{1-(-x^2)}$. By the geometric series formula, this becomes: $$f(x)=\sum _{n=0}^{\infty }(-x^2{)}^n=\sum _{n=0}^{\infty }(-1{)}^n{x}^{2n}$$ Convergence requires $|-x^2|<1⟹|x|<1$, so radius of convergence $R=1$. (b) We know $\int \frac{1}{1+x^2}dx=\arctan x+C$. Integrate term-by-term: $$\arctan x=C+\int \sum _{n=0}^{\infty }(-1{)}^n{x}^{2n}dx=C+\sum _{n=0}^{\infty }(-1{)}^n\frac{x^{2n+1}}{2n+1}$$ Substitute $x=0$: $\arctan (0)=0=C$, so the series is $\arctan x={\sum }_{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{2n+1}$. (c) Comparing the given series to the series for $\arctan x$, the general term matches when $x=\frac{1}{2}$. Since $|\frac{1}{2}|<1=R$, the series converges, so the exact sum is $\arctan \left(\frac{1}{2}\right)$.

Question 3 (Application / Real-World Style)

The electric potential at a distance $z$ along the central axis of a uniformly charged ring of radius $R$ (with $z>R$) is given by: $$V(z)=\frac{kQ}{\sqrt{z^2+R^2}}$$ where $k=8.99\times 10^9{\text{ Ncdotpm}}^2/{\text{C}}^2$ is Coulomb's constant, and $Q$ is total charge on the ring in coulombs. Find the first two non-zero terms of the power series representation of $V(z)$ for $z>R$, written as a series in powers of $\frac{R}{z}$. Then compute the approximate potential for $R=0.1\text{ m}$, $z=1\text{ m}$, $Q=1\times 10^{-6}\text{ C}$.

Worked Solution: First, factor $z^2$ out of the square root: $$V(z)=\frac{kQ}{z\sqrt{1+{\left(\frac{R}{z}\right)}^2}}=\frac{kQ}{z}{\left(1+{\left(\frac{R}{z}\right)}^2\right)}^{-1/2}$$ Using the binomial series expansion for $(1+u{)}^k$ with $|u|<1$ (here $u=(R/z{)}^2<1$), the first two terms are $1-\frac{1}{2}u$, so the two-term approximation is: $$V(z)\approx \frac{kQ}{z}\left(1-\frac{1}{2}{\left(\frac{R}{z}\right)}^2\right)$$ Substitute the given values: $\frac{kQ}{z}=\frac{(8.99\times 10^9)(1\times 10^{-6})}{1}=8990\text{ V}$, and $\frac{R}{z}=0.1$. The approximation becomes $V\approx 8990\left(1-\frac{1}{2}(0.1{)}^2\right)=8990-44.95=8945.05\text{ V}$. In context, this shows that for large $z$, the potential of a charged ring is very close to the potential of a point charge ($\frac{kQ}{z}$), with a small correction for the finite size of the ring.

7. Quick Reference Cheatsheet

8. What's Next

This chapter gives you the core foundation for all work with Taylor and Maclaurin series, the next major topic in 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 FRQ much harder than necessary. This topic is also the basis for integrating non-elementary functions, a frequent AP exam application.

Taylor and Maclaurin series Power series interval of convergence Taylor approximation error bounds