Finding Taylor or Maclaurin series for a function

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

1. Core Definitions of Taylor and Maclaurin Series★★☆☆☆⏱ 2 min

Finding a Taylor or Maclaurin series means constructing an infinite power series representation of a given infinitely differentiable function. Coefficients are defined to match the function's value and all orders of derivative at a chosen center point $a$ .

📘 Definition

Taylor Series

A Taylor series centered at $a$ for a function $f (x)$ is an infinite power series built to match $f (x)$ and all its derivatives at $x = a$ . It converges to $f (x)$ on its interval of convergence.

Example:

A Taylor series centered at $a = 2$ will have terms of the form $c_{n} (x - 2)^{n}$

📘 Definition

Maclaurin Series

A Maclaurin series is a special case of a Taylor series that is always centered at $a = 0$ , the most common center tested on the AP Calculus BC exam.

Example:

All pre-memorized standard series for common functions are Maclaurin series

This topic is a core skill in Unit 10, which makes up 17-18% of the AP Calculus BC exam. Questions on finding series appear in both multiple-choice and free-response sections, and are often combined with other series skills like finding convergence or approximation.

2. Direct Computation via the Definition Method★★★☆☆⏱ 3 min

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The formal definition of a Taylor series centered at $a$ gives the general formula:

f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n

Where $f^{(n)} (a)$ is the $n$ -th derivative of $f$ evaluated at $a$ , and $f^{(0)} (a) = f (a)$ by convention. For Maclaurin series ( $a = 0$ ), this simplifies to:

f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n

The direct method requires computing the first few derivatives, identifying a pattern in coefficients, and writing the general term. It is most often used for functions that do not fit a known standard series, or for non-zero centers.

📐 Worked Example

Find the first four non-zero terms of the Maclaurin series for $f (x) = sin (3 x)$ using the definition.

1
Compute the first few derivatives of $f (x)$ :
2
$f^{(0)}(x) = \sin(3x),\ f^{(1)}(x) = 3\cos(3x),\ f^{(2)}(x) = -9\sin(3x),\ f^{(3)}(x) = -27\cos(3x),\ f^{(4)}(x) = 81\sin(3x),\ f^{(5)}(x) = 243\cos(3x),\ f^{(7)}(x) = -3^7 \cos(3x)$
3
Evaluate all derivatives at the center $a = 0$ :
4
$f^{(0)}(0) = 0,\ f^{(1)}(0) = 3,\ f^{(2)}(0) = 0,\ f^{(3)}(0) = -27,\ f^{(4)}(0) = 0,\ f^{(5)}(0) = 243,\ f^{(7)}(0) = -2187$
5
Calculate coefficients as $\frac{f ^{(n)} ( 0 )}{n !}$ for non-zero values:
6
$n=1: \frac{3}{1!} = 3,\ n=3: \frac{-27}{3!} = -\frac{9}{2},\ n=5: \frac{243}{5!} = \frac{81}{40},\ n=7: \frac{-2187}{7!} = -\frac{81}{186}$
7
Write the first four non-zero terms of the series:
8
$\sin(3x) = 3x - \frac{9}{2}x^3 + \frac{81}{40}x^5 - \frac{81}{186}x^7 + \dots$

Exam tip:

Use this method when the question explicitly asks you to use the definition of Taylor series, or for non-zero centers.

3. Constructing Series by Substitution★★★☆☆⏱ 3 min

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Once you memorize standard Maclaurin series for common functions, you do not need to recompute coefficients from scratch for most exam problems. Substitution is the most common shortcut: if you have a known series $f (u) = \sum_{n = 0}^{\infty} c_{n} u^{n}$ , then $f (g (x)) = \sum_{n = 0}^{\infty} c_{n} (g (x))^{n}$ within the new interval of convergence.

This method is tested more frequently than the direct definition method on the AP exam, because it tests your ability to manipulate known series rather than just compute derivatives.

📐 Worked Example

Find the Maclaurin series for $g (x) = \frac{1}{1 + 5 x ^{2}}$ , including the general term.

1
Recall the standard geometric Maclaurin series:
2
$\frac{1}{1 - u} = \sum_{n=0}^{\infty} u^n, \quad |u| < 1$
3
Rewrite $g (x)$ to match the geometric series form: $\frac{1}{1 + 5 x ^{2}} = \frac{1}{1 - ( - 5 x ^{2} )}$ , so substitute $u = - 5 x^{2}$ .
4
Apply the exponent to the entire substituted expression to get the general term:
5
$u^n = (-5x^2)^n = (-1)^n 5^n x^{2n}$
6
Write the full series:
7
$\frac{1}{1 + 5x^2} = \sum_{n=0}^{\infty} (-1)^n 5^n x^{2n}, \quad |x| < \frac{1}{\sqrt{5}}$

4. Constructing Series by Differentiation and Integration★★★★☆⏱ 3 min

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Power series can be differentiated and integrated term-by-term within their interval of convergence, so you can use this property to derive a new series from a known one. 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}

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

Term-by-term differentiation and integration do not change the radius of convergence of the original series; only convergence at the endpoints may change. This method is used when your target function is the derivative or integral of a function with a known series.

📐 Worked Example

Find the Maclaurin series for $h (x) = arctan (2 x)$ for $∣ x ∣ < 1/2$ , including the general term.

1
First note that the derivative of $h (x)$ is:
2
$h'(x) = \frac{2}{1 + (2x)^2} = \frac{2}{1 + 4x^2}$
3
Use substitution to get the series for $h^{'} (x)$ :
4
$h'(x) = 2 \sum_{n=0}^{\infty} (-1)^n 4^n x^{2n} = \sum_{n=0}^{\infty} (-1)^n 2^{2n+1} x^{2n}$
5
Integrate term-by-term from $0$ to $x$ to recover $h (x) = arctan (2 x) = \int_{0}^{x} h^{'} (t) d t$ :
6
$\sum_{n=0}^{\infty} (-1)^n 2^{2n+1} \int_0^x t^{2n} dt = \sum_{n=0}^{\infty} (-1)^n 2^{2n+1} \frac{x^{2n+1}}{2n+1}$
7
Check the constant of integration: $arctan (0) = 0$ , so $C = 0$ . Rewrite the final series:
8
$\arctan(2x) = \sum_{n=0}^{\infty} \frac{(-1)^n (2x)^{2n+1}}{2n+1}$

5. Adjusting Series by Multiplying by a Power of $x$★★★☆☆⏱ 3 min

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A common AP exam problem requires finding the series for a function that is a known function multiplied by $x^{k}$ (or $(x - a)^{k}$ for a general center). This simple manipulation checks your understanding of how power series terms work: when you multiply a power series by $(x - a)^{k}$ , every exponent increases by $k$ , and coefficients do not change.

📐 Worked Example

Write the first three non-zero terms of the Maclaurin series for $f (x) = x^{3} e^{- 2 x}$ .

1
Recall the standard Maclaurin series for $e^{u}$ :
2
$e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \dots$
3
Substitute $u = - 2 x$ into the series:
4
$e^{-2x} = 1 - 2x + \frac{4x^2}{2} - \frac{8x^3}{6} + \dots = 1 - 2x + 2x^2 - \frac{4}{3}x^3 + \dots$
5
Multiply the entire series by $x^{3}$ , which adds 3 to every exponent:
6
$x^3 e^{-2x} = x^3 \left(1 - 2x + 2x^2 - \frac{4}{3}x^3 + \dots\right)$
7
Simplify to get the first three non-zero terms:
8
$x^3 - 2x^4 + 2x^5 + \dots$

6. Common Pitfalls

Wrong move:

When substituting $u = - x^{2}$ into $e^{u} = \sum \frac{u ^{n}}{n !}$ , writing the general term as $\frac{- x ^{2 n}}{n !}$ instead of $\frac{( - 1 ) ^{n} x ^{2 n}}{n !}$

Why:

Students forget to apply the exponent to the negative sign, only applying it to the $x^{2}$ term.

Correct move:

Always wrap the entire substituted expression in parentheses before applying the exponent: $(- x^{2})^{n} = (- 1)^{n} (x^{2})^{n} = (- 1)^{n} x^{2 n}$

Wrong move:

When computing a Taylor series centered at $a = 2$ , using memorized Maclaurin coefficients and just replacing $x$ with $x - 2$ without re-evaluating derivatives.

Why:

Students confuse the general form of a Taylor series with pre-memorized Maclaurin coefficients, assuming shifting $x$ is enough for a non-zero center.

Correct move:

For a non-zero center, either recompute coefficients using the definition at $a$ , or rewrite the function in terms of $u = x - a$ and build the series around $u$ .

Wrong move:

When integrating term-by-term to find the series for $ln (1 + x)$ , leaving a non-zero constant of integration as the first term of the series.

Why:

Students forget that $ln (1 + 0) = 0$ , so the constant of integration evaluated at the center is zero.

Correct move:

Always evaluate the integrated function at the center $a$ to solve for $C$ ; for Maclaurin series, plug in $x = 0$ to the target function to get $C$ .

Wrong move:

When computing the n-th derivative of $e^{k x}$ for direct coefficient calculation, writing $f^{(n)} (0) = 1$ instead of $k^{n}$ .

Why:

Students get used to derivatives of $e^{x}$ and forget the chain rule gives an extra constant factor for the linear inner function.

Correct move:

For composite functions $f (k x)$ , always apply chain rule $n$ times, which gives an extra factor of $k^{n}$ for the n-th derivative.

Wrong move:

When asked for the coefficient of $x^{5}$ in $x^{2} sin (x)$ , taking the coefficient of $x^{5}$ from the $sin (x)$ series as the answer.

Why:

Students forget that multiplying by $x^{2}$ shifts all exponents up by 2.

Correct move:

If multiplying by $x^{k}$ , the coefficient of $x^{m}$ in the new series is equal to the coefficient of $x^{m - k}$ in the original series. For this example, the coefficient of $x^{5}$ in $x^{2} sin (x)$ is the coefficient of $x^{3}$ in $sin (x)$ .

7. Quick Reference Cheatsheet

Category	Formula	Notes
General Taylor Series (center $a$ )	$\sum_{n = 0}^{\infty} \frac{f ^{(n)} ( a )}{n !} (x - a)^{n}$	Use for non-zero centers or definition questions
General Maclaurin Series	$\sum_{n = 0}^{\infty} \frac{f ^{(n)} ( 0 )}{n !} x^{n}$	Special case of Taylor with $a = 0$ , most common on AP
Substitution Rule	If $f (u) = \sum c_{n} u^{n}$ , then $f (g (x)) = \sum c_{n} (g (x))^{n}$	Radius adjusts, coefficients follow substitution directly
Term-by-Term Differentiation	$f^{'} (x) = \sum_{n = 1}^{\infty} n c_{n} (x - a)^{n - 1}$	Same radius of convergence as original
Term-by-Term Integration	$\int f (x) d x = C + \sum_{n = 0}^{\infty} \frac{c _{n}}{n + 1} (x - a)^{n + 1}$	Solve for $C$ at $x = a$ ; usually $C = 0$ for Maclaurin
Multiply by Power of $x$	If $f (x) = \sum c_{n} x^{n}$ , then $x^{k} f (x) = \sum c_{n} x^{n + k}$	Only exponents shift; coefficients stay the same
Standard: $e^{x}$	$\sum_{n = 0}^{\infty} \frac{x ^{n}}{n !}$ , all real $x$	All terms non-zero, all exponents
Standard: $sin x$	$\sum_{n = 0}^{\infty} \frac{( - 1 ) ^{n} x ^{2 n + 1}}{( 2 n + 1 )!}$ , all real $x$	Only odd powers, alternating signs
Standard: $cos x$	$\sum_{n = 0}^{\infty} \frac{( - 1 ) ^{n} x ^{2 n}}{( 2 n )!}$ , all real $x$	Only even powers, alternating signs
Standard: $\frac{1}{1 - u}$	$\sum_{n = 0}^{\infty} u^{n}$ , $∣ u ∣ < 1$	Base geometric series for substitution

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 $x^{k}$ in modified series
2022 · FRQ
Derive series via differentiation of geometric series
2021 · MCQ
Maclaurin series via substitution of standard function

What's Next

Now that you can construct Taylor and Maclaurin series using all common methods, you are ready to move on to more advanced series topics that are heavily tested on the AP Calculus BC exam. Next, you will learn how to find the radius and interval of convergence for any power series, test for convergence at the endpoints of the interval, and then use the Lagrange error bound to estimate the error of Taylor polynomial approximations. Mastery of constructing series is the foundation for all these topics, so the methods you learned here will be used in almost every series question on the actual exam.