AP Calculus BC

Common Maclaurin Series Table (BC)

The four Maclaurin series to memorize are e^x = sum of x^n/n! (all x), sin x = sum of (-1)^n x^(2n+1)/(2n+1)! (all x), cos x = sum of (-1)^n x^(2n)/(2n)! (all x), and 1/(1-x) = sum of x^n (only for |x| < 1). Two common extensions, ln(1+x) and arctan x, hold for -1 < x <= 1 and |x| <= 1.

The four series the CED names

All of Unit 10 is BC-only, and these four Maclaurin series are the ones the College Board builds on directly in Topic 10.14: $e^x$ , $\sin x$ , and $\cos x$ are named as the foundation for constructing other series (LIM-8.F.2), and $\frac{1}{1-x}$ is the geometric building block (LIM-8.F.1). Memorize these four cold. Almost every other series on the BC exam is reached by reshaping one of them.

Function	Maclaurin series (general term)	Converges for	When to reach for it
$e^x$	$\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$	all real $x$	Any $e^{(\text{expression})}$ : substitute the exponent for $x$ . Converges everywhere, so there is no interval to check.
$\sin x$	$\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$	all real $x$	Odd powers only, matching that $\sin$ is an odd function. Use for $\sin$ of any expression; safe to substitute freely.
$\cos x$	$\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$	all real $x$	Even powers only, matching that $\cos$ is even. It is the term-by-term derivative of the $\sin x$ series, so you can recover one from the other.
$\frac{1}{1-x}$	$\displaystyle\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots$	$\|x\| < 1$	The geometric series with ratio $x$ (LIM-8.F.1). Base for $\frac{1}{1+x}$ , $\frac{1}{1-x^2}$ , and similar rationals. Diverges once $\|x\| \ge 1$ , so always carry the interval.

The three that converge for all $x$ ( $e^x$ , $\sin x$ , $\cos x$ ) are the ones you can substitute into without a second thought: any expression can replace $x$ and the result still converges everywhere. The geometric series is the exception. It collapses the moment $|x| \ge 1$ , so any substitution into it comes with an interval condition attached.

Two extensions worth memorizing

These two are not in the CED (Course and Exam Description) essential knowledge, so treat them as commonly-taught extras rather than required facts. You can derive each by integrating a geometric-type series term by term (Topic 10.15), but memorizing them saves time. They are labeled here as extensions so you know they sit outside the core four above.

Function (extension)	Maclaurin series (general term)	Converges for	Where it comes from
$\ln(1+x)$	$\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$	$-1 < x \le 1$	Integrate the $\frac{1}{1+x}$ series term by term. The sum starts at $n = 1$ : there is no constant term because $\ln 1 = 0$ .
$\arctan x$	$\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots$	$-1 \le x \le 1$	Integrate the $\frac{1}{1+x^2}$ series term by term. Odd powers like $\sin x$ , but the denominators are $2n+1$ , not factorials.

Both extensions come from integrating a known series, which is why their intervals pick up an endpoint the plain geometric series does not have. Term-by-term integration keeps the radius of convergence the same (LIM-8.D.6) but can change what happens at the endpoints, so $\ln(1+x)$ converges at $x = 1$ and $\arctan x$ converges at both $x = \pm 1$ . An interval question still needs each endpoint tested on its own (Topic 10.13).

Choosing which series to start from

Method selection is the real skill. Before computing anything, look at $f(x)$ and decide which base series it is a disguised version of. The derivative formula $\frac{f^{(n)}(a)}{n!}$ (Topic 10.11) always works but is slow, so save it for functions that match nothing on the list. This table maps a function's shape to the series and the single operation that builds it fastest.

If $f(x)$ looks like	Start from	Do this	Example
A standard function with $x$ replaced by an expression: $\cos(3x)$ , $e^{-x^2}$	the matching base series	substitute the expression in place of $x$	$e^{-x^2} = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}$
A base series times a power of $x$ : $x^2 \sin x$	the base series	multiply every term by that power	$x^2 \sin x = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+3}}{(2n+1)!}$
A rational form $\frac{1}{1 \pm ax}$ or $\frac{1}{1 \pm ax^2}$	$\frac{1}{1-x}$	substitute, factoring out constants first	$\frac{1}{1+x^2} = \displaystyle\sum_{n=0}^{\infty} (-1)^n x^{2n}$
A function whose derivative is a known series: $\ln(1+x)$ , $\arctan x$	the derivative's series (e.g. $\frac{1}{1+x}$ or $\frac{1}{1+x^2}$ )	integrate term by term, then fix the constant	$\int \frac{1}{1+x^2}\,dx = \arctan x$
Nothing on the list, or a center $a \ne 0$	the derivative formula	compute $\frac{f^{(n)}(a)}{n!}$ and attach to $(x-a)^n$	$\ln x$ centered at $a = 1$

You can chain these operations: for $x^2 \cos(2x)$ , substitute $2x$ into the cosine series, then multiply every term by $x^2$ , giving $x^2 \cos(2x) = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n} x^{2n+2}}{(2n)!}$ . The one case with no shortcut is a series centered at $a \ne 0$ , such as $\ln x$ at $x = 1$ ; there you compute derivatives at $a$ and attach each coefficient to $(x-a)^n$ , not $x^n$ . The Taylor and Maclaurin series guide works that example in full.

Watch-outs when using the table

A handful of slips account for most lost points once you start pulling series from this table.

Factorials belong to $e^x$ , $\sin x$ , and $\cos x$ only. The geometric series and its integrals ( $\frac{1}{1-x}$ , $\ln(1+x)$ , $\arctan x$ ) have plain integer denominators or none at all.
Signs travel through a substitution: $(-x^2)^n = (-1)^n x^{2n}$ , so replacing $x$ with $-x^2$ changes both the sign pattern and the powers.
Intervals do not carry over unchanged. Substituting $u = g(x)$ into a series valid for $|u| < 1$ gives the condition $|g(x)| < 1$ , which you still have to solve for $x$ .
Mind the starting index. The $\ln(1+x)$ series starts at $n = 1$ (its first term is $x$ ), while $e^x$ , $\cos x$ , and $\frac{1}{1-x}$ start at $n = 0$ .