Do I really have to memorize all six Maclaurin series?

Yes. The AP exam expects instant recall of e^x, x, x, 1/1-x, (1+x), and x. Deriving them from scratch on a timed test wastes minutes you do not have, and manipulation problems assume you already know them.

How does substitution change the interval of convergence?

Apply the original convergence condition to the substituted expression. If 1/1-x converges for |x|<1 and you substitute u=4x, then you need |4x|<1, so |x|<14. The radius shrinks or grows depending on the substitution.

Why do differentiation and integration keep the same radius of convergence?

It is a theorem about power series: within the open interval of convergence, a power series behaves like a polynomial, and term-by-term differentiation or integration produces a series with the identical radius. Only the endpoints can change, so recheck those separately if the problem asks.

When would I integrate a series instead of finding an antiderivative directly?

When the function has no elementary antiderivative, such as e^-x^2, x/x, or (x^2). You write the series, integrate term by term, and use the result to approximate a definite integral, often pairing it with the alternating series error bound.

AP-CALCBC-10.14-10.15

U10.14 Common Maclaurin Series and Manipulation

Q: Why do differentiation and integration keep the same radius of convergence?

It is a theorem about power series: within the open interval of convergence, a power series behaves like a polynomial, and term-by-term differentiation or integration produces a series with the identical radius. Only the endpoints can change, so recheck those separately if the problem asks.

Q: When would I integrate a series instead of finding an antiderivative directly?

When the function has no elementary antiderivative, such as e^-x^2, x/x, or (x^2). You write the series, integrate term by term, and use the result to approximate a definite integral, often pairing it with the alternating series error bound.

Master the six common Maclaurin series for AP Calculus BC and learn to derive new series fast using substitution, differentiation, and integration.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U10.14 Common Maclaurin Series and Manipulation, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start U10.14 Common Maclaurin Series and Manipulation free

What this lesson covers

By this point you can build Taylor polynomials and find intervals of convergence. Now comes the payoff: instead of computing derivatives one at a time, you memorize a small library of Maclaurin series and manipulate them algebraically to get almost anything the exam throws at you. Need the series for

x^2 e^{-x}

\int \sin(x^2)\,dx

? You will build it in seconds.

This lesson locks in the six series you must know cold —

e^x

\sin x

\cos x

\frac{1}{1-x}

\ln(1+x)

, and

\arctan x

— and then teaches the three legal moves (substitute, differentiate, integrate) that turn one known series into many.

The Six Series You Must Memorize

A Maclaurin series is just a Taylor series centered at

x=0

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

. Rather than re-derive these on test day, commit the following to memory along with their intervals of convergence.

Function	Series	Converges for
$e^x$	$\sum_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\cdots$	all $x$
$\sin x$	$\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\cdots$	all $x$
$\cos x$	$\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\cdots$	all $x$
$\frac{1}{1-x}$	$\sum_{n=0}^{\infty}x^n=1+x+x^2+\cdots$	$-1<x<1$
$\ln(1+x)$	$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}=x-\frac{x^2}{2}+\cdots$	$-1<x\le 1$
$\arctan x$	$\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1}=x-\frac{x^3}{3}+\cdots$	$-1\le x\le 1$

Notice the patterns:

\sin

uses odd powers,

\cos

uses even powers, and both have factorials. The geometric series

\frac{1}{1-x}

is the parent of

\ln(1+x)

and

\arctan x

, which is why those two have no factorials — just

n

2n+1

in the denominator.

Manipulation Move 1: Substitution

The fastest way to generate a new series is to substitute an expression in for

x

in a known series. Wherever you see

x

, replace it — including inside every power and every coefficient position.

For example, to find the series for

e^{-x^2}

, start with

e^{u}=\sum \frac{u^n}{n!}

and let

u=-x^2

e^{-x^2}=\sum_{n=0}^{\infty}\frac{(-x^2)^n}{n!}=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{n!}=1-x^2+\frac{x^4}{2!}-\cdots

Substitution also transforms the interval of convergence. Since

e^u

converges for all

u

e^{-x^2}

converges for all

x

. But for

\frac{1}{1-x}

, which needs

|x|<1

, substituting

u=2x

gives

\frac{1}{1-2x}=\sum(2x)^n

, valid only when

|2x|<1

, i.e.

|x|<\tfrac12

.

A common misconception is forgetting to raise the substituted quantity to the full power. If you plug

2x

into

x^n

, the term is

(2x)^n=2^n x^n

, not

2x^n

. The exam loves this trap. Always wrap your substitution in parentheses before simplifying, then apply exponent rules carefully.

Manipulation Moves 2 and 3: Differentiate and Integrate

A power series can be differentiated or integrated term by term within its open interval of convergence, and the radius of convergence stays the same (endpoints may change).

Differentiating

\frac{1}{1-x}=\sum x^n

term by term gives

\frac{1}{(1-x)^2}=\sum_{n=1}^{\infty} n x^{n-1}

. Integrating

\frac{1}{1+x}=\sum(-x)^n

gives

\ln(1+x)

— this is exactly how the

\ln(1+x)

series is derived. Similarly, integrating

\frac{1}{1+x^2}=\sum(-1)^n x^{2n}

produces

\arctan x

.

When you integrate, do not forget the constant of integration

+C

. Evaluate at a convenient point (usually

x=0

) to solve for

C

. For

\ln(1+x)

, at

x=0

the function is

\ln 1=0

, and the series with no constant is also

0

, so

C=0

Move	Effect on term $c_n x^n$	Radius $R$
Substitute $u=g(x)$	replace $x$ with $g(x)$	changes with $g$
Differentiate	$\to n c_n x^{n-1}$	unchanged
Integrate	$\to \frac{c_n}{n+1}x^{n+1}+C$	unchanged

These moves also let you evaluate hard integrals like

\int_0^1 \frac{\sin x}{x}\,dx

\int e^{-x^2}\,dx

, which have no elementary antiderivative — you integrate the series instead.

How the Exam Tests This

AP questions rarely ask you to recite a series in isolation. Instead they chain the moves. A typical multiple-choice item gives

f(x)=x^2\cos(3x)

and asks for the coefficient of

x^4

, or for the first three nonzero terms. A free-response part might ask you to write a series for an integrand, integrate term by term, and then approximate a definite integral, often finishing with an alternating series error bound (connecting back to U10.7).

Strategy: identify the parent series, perform the substitution, then multiply or differentiate or integrate as needed, and only keep the terms you need. If asked for the coefficient of a specific power, figure out which

n

produces that power rather than expanding everything.

A frequent misconception is multiplying a whole series by

x^k

incorrectly. Multiplying

\sin x=x-\frac{x^3}{6}+\cdots

x^2

simply raises every exponent by 2:

x^3-\frac{x^5}{6}+\cdots

. Another pitfall: mixing up factorial denominators from

\sin/\cos/e^x

with the non-factorial denominators from

\ln

and

\arctan

. Keep the two families straight. Finally, always state the interval of convergence when asked — substitution can shrink it, and endpoints require separate checking.

Key terms

Maclaurin series.: A Taylor series centered at $x=0$ : $\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$ .
Substitution method.: Replacing $x$ in a known series with another expression (like $-x^2$ or $2x$ ) to build a new series.
Term-by-term differentiation.: Differentiating each term of a power series individually; valid inside the interval of convergence with the same radius.
Term-by-term integration.: Integrating each term of a power series individually; requires adding a constant $C$ determined by a known value.
Radius of convergence.: Half the length of the interval where a power series converges; preserved under differentiation and integration.
Nonzero terms.: The terms of a series with nonzero coefficients; AP prompts often ask for the first three nonzero terms, skipping vanished ones.
Geometric series parent.: $\frac{1}{1-x}=\sum x^n$ , the source series from which $\ln(1+x)$ and $\arctan x$ are derived by integration.

Worked example

Find the first three nonzero terms of the Maclaurin series for

f(x)=x\,\ln(1+2x)

, and state its radius of convergence.

Start with the known series

\ln(1+u)=u-\frac{u^2}{2}+\frac{u^3}{3}-\cdots

, valid for

-1<u\le 1

.

Substitute

u=2x

. Wrap in parentheses:

\ln(1+2x)=(2x)-\frac{(2x)^2}{2}+\frac{(2x)^3}{3}-\cdots

.

Simplify each term:

(2x)=2x

;

\frac{(2x)^2}{2}=\frac{4x^2}{2}=2x^2

;

\frac{(2x)^3}{3}=\frac{8x^3}{3}

. So

\ln(1+2x)=2x-2x^2+\frac{8x^3}{3}-\cdots

.

Now multiply the entire series by

x

, raising every exponent by one:

x\,\ln(1+2x)=2x^2-2x^3+\frac{8x^4}{3}-\cdots

.

The first three nonzero terms are

2x^2-2x^3+\frac{8x^4}{3}

.

For the radius:

\ln(1+u)

needs

|u|<1

, so

|2x|<1

, giving

|x|<\tfrac12

. Multiplying by

x

does not change the radius, so

R=\tfrac12

Practice questions

What is the coefficient of

x^6

in the Maclaurin series for

\cos(x^2)

$-\frac{1}{6}$
$\frac{1}{24}$
$0$
$-\frac{1}{2}$

Answer: $0$

Start with

\cos u=1-\frac{u^2}{2!}+\frac{u^4}{4!}-\cdots

and substitute

u=x^2

\cos(x^2)=1-\frac{x^4}{2!}+\frac{x^8}{4!}-\cdots

. Every power of

x

is a multiple of 4 (0, 4, 8, ...), so there is no

x^6

term. Its coefficient is

0

Use the Maclaurin series for

\frac{1}{1-x}

to find a series for

\frac{x}{1-x^3}

, then give its first three nonzero terms and its interval of convergence.

Answer: $x + x^4 + x^7 + \cdots$ , converging for $-1<x<1$ .

Substitute

u=x^3

into

\frac{1}{1-u}=\sum u^n

to get

\frac{1}{1-x^3}=\sum x^{3n}=1+x^3+x^6+\cdots

. Multiply by

x

\frac{x}{1-x^3}=x+x^4+x^7+\cdots

. Convergence needs

|x^3|<1

, i.e.

|x|<1

; endpoints

x=\pm 1

diverge since terms do not shrink to zero, so the interval is

-1<x<1

Which expression gives the Maclaurin series for

e^{3x}

$\sum_{n=0}^{\infty}\frac{3x^n}{n!}$
$\sum_{n=0}^{\infty}\frac{3^n x^n}{n!}$
$\sum_{n=0}^{\infty}\frac{x^n}{3^n n!}$
$\sum_{n=0}^{\infty}\frac{(3x)^n}{n}$

Answer: $\sum_{n=0}^{\infty}\frac{3^n x^n}{n!}$

Substitute

u=3x

into

e^u=\sum\frac{u^n}{n!}

, wrapping in parentheses:

\frac{(3x)^n}{n!}=\frac{3^n x^n}{n!}

. The trap answer

\frac{3x^n}{n!}

fails to raise the 3 to the

n

th power; the factorial must remain because

e^x

is a factorial-family series.

FAQ

Do I really have to memorize all six Maclaurin series?: Yes. The AP exam expects instant recall of $e^x$ , $\sin x$ , $\cos x$ , $\frac{1}{1-x}$ , $\ln(1+x)$ , and $\arctan x$ . Deriving them from scratch on a timed test wastes minutes you do not have, and manipulation problems assume you already know them.
How does substitution change the interval of convergence?: Apply the original convergence condition to the substituted expression. If $\frac{1}{1-x}$ converges for $|x|<1$ and you substitute $u=4x$ , then you need $|4x|<1$ , so $|x|<\tfrac14$ . The radius shrinks or grows depending on the substitution.
Why do differentiation and integration keep the same radius of convergence?: It is a theorem about power series: within the open interval of convergence, a power series behaves like a polynomial, and term-by-term differentiation or integration produces a series with the identical radius. Only the endpoints can change, so recheck those separately if the problem asks.
When would I integrate a series instead of finding an antiderivative directly?: When the function has no elementary antiderivative, such as $e^{-x^2}$ , $\frac{\sin x}{x}$ , or $\cos(x^2)$ . You write the series, integrate term by term, and use the result to approximate a definite integral, often pairing it with the alternating series error bound.

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