AP Calculus BC

Taylor and Maclaurin Series: How to Build Them

Q: What is the difference between a Taylor series and a Maclaurin series?

A Maclaurin series is a Taylor series centered at x = 0. That is the only difference. The general Taylor series is \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n; setting the center a = 0 turns the (x-a)^n factors into x^n and gives the Maclaurin series. Every Maclaurin series is a Taylor series, but a Taylor series centered at a \ne 0 is not Maclaurin.

Q: Which Taylor series do I need to memorize for AP Calculus BC?

At minimum, memorize the Maclaurin series for e^x, \sin x, \cos x, and \frac{1}{1-x}. The CED points to \sin x, \cos x, and e^x as the foundation for building others (LIM-8.F.2), and \frac{1}{1-x} is the geometric building block (LIM-8.F.1). Knowing \ln(1+x) helps too, but you can recover it by integrating the series for \frac{1}{1+x}.

Q: How do you find a Taylor series centered at a point other than zero?

Use the derivative formula directly: compute f(a), f'(a), f''(a), \dots at the center a, form the coefficients \frac{f^{(n)}(a)}{n!}, and attach each to (x-a)^n. The powers must be (x-a)^n, not x^n. Example 3 above builds \ln x at a = 1 this way. Substituting a known series works only if you can rewrite f in terms of (x-a), such as writing \ln x = \ln(1 + (x-1)) and using the \ln(1+t) series with t = x - 1.

A Taylor series centered at x = a sums f^(n)(a)/n! times (x - a)^n over all n from 0 up. A Maclaurin series is that series centered at 0. In practice you rarely compute every derivative: start from a known series like e^x, sin x, cos x, or 1/(1-x) and substitute, multiply, differentiate, or integrate instead.

The Taylor series formula

A Taylor series is what a Taylor polynomial (Topic 10.11) becomes when you never stop adding terms. You pick a center $x = a$ , compute the derivatives of $f$ there, divide the $n$ -th derivative by $n!$ , and attach the result to $(x-a)^n$ . Summing over every $n$ from $0$ up produces the series.

f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots

Dividing by $n!$ is not arbitrary. It is exactly the factor that forces the polynomial's $n$ -th derivative at $a$ to match $f^{(n)}(a)$ , so each new term corrects the next derivative. That is the design principle behind the whole formula.

The center $a$ is the point the series is built around and where its approximation is sharpest. When $a = 0$ the $(x-a)^n$ factors become plain powers of $x$ , and the series gets its own name: the Maclaurin series. Maclaurin is not a separate idea, just Taylor centered at zero.

f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n = f(0) + f'(0)\,x + \frac{f''(0)}{2!}x^2 + \cdots

Every partial sum of a Taylor series is a Taylor polynomial (LIM-8.E.1), so the series is the limit of its polynomials. This is BC Topic 10.14, resting directly on the coefficient $\frac{f^{(n)}(a)}{n!}$ from Topic 10.11. All of Unit 10 is BC-only.

Two ways to build a series, and how to choose

The derivative formula defines a Taylor series, but on the exam it is usually the slow way in. Differentiating a function four or five times to spot a pattern is error-prone, and for a composite like $e^{-x^2}$ the derivatives grow fast: $f'(x) = -2x\,e^{-x^2}$ , then $f''(x) = (4x^2 - 2)e^{-x^2}$ , and it only gets messier. Finding a clean formula for $f^{(n)}(0)$ that way is brutal.

The fast route is almost always to start from a series you already know and reshape it, so the first decision is recognition. Look at $f(x)$ and ask whether it is a close relative of one of the five standard series. If $f$ is a standard function with $x$ replaced by some expression (like $\cos(2x)$ , $e^{-x^2}$ , or $\frac{1}{1+x}$ ), multiplied by a power of $x$ , or the derivative or integral of a standard function, build it from the known series.

Only when the function is genuinely unlike anything on the list, or when the problem asks for a series centered at a nonzero $a$ , should you fall back on computing derivatives one at a time. Example 3 is exactly that case: $\ln x$ centered at $x = 1$ .

Recognition rule

Is the function a known series in disguise (a substitution, a product with $x^k$ , or a derivative or integral of one)? Reshape the known series. Is it truly new, or centered at $a \ne 0$ ? Use the derivative formula.

This is the whole game in Topics 10.14 and 10.15. The CED names $\sin x$ , $\cos x$ , and $e^x$ as the foundation for constructing other series (LIM-8.F.2) and $\frac{1}{1-x}$ as the geometric building block (LIM-8.F.1). Memorizing a handful of series buys you dozens more for almost no work.

The five series worth memorizing

These five carry the overwhelming majority of series questions on the BC exam. The first three are the CED-designated foundation, the fourth is a geometric series, and the fifth comes from substituting $-x$ into the fourth to get $\frac{1}{1+x}$ , then integrating that term by term.

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

The three that converge for every $x$ ( $e^x$ , $\sin x$ , $\cos x$ ) are the workhorses, because you can substitute any expression into them without worrying about an interval. The trig pair also has a memorable structure: $\sin x$ uses only odd powers, because sine is an odd function, while $\cos x$ uses only even powers, because cosine is even, and both alternate in sign. If you can write the $e^x$ series, you can reconstruct $\sin x$ and $\cos x$ as its odd and even parts with alternating signs.

Building from a known series

Once you have a known series, four operations generate new ones, and all four are fair game (Topic 10.15).

Substitution: replace $x$ with any expression. Putting $-x^2$ into $e^x$ gives the series for $e^{-x^2}$ ; putting $2x$ into $\cos x$ gives the series for $\cos(2x)$ .
Multiplication by a power of $x$ : to get the series for $x^2 e^x$ , multiply every term of the $e^x$ series by $x^2$ .
Term-by-term differentiation: differentiating $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$ gives $\frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n\,x^{n-1}$ .
Term-by-term integration: integrating $\frac{1}{1+x^2}$ term by term recovers the series for $\arctan x$ .

You can chain these freely. To build the series for $x^2 \cos(2x)$ , substitute $2x$ into the cosine series, then multiply every term by $x^2$ . Differentiation and integration do not change the radius of convergence (LIM-8.D.6), though the endpoint behavior can shift, so an interval-of-convergence question still needs both endpoints checked on their own (Topic 10.13).

Common mistakes

Forgetting to divide by $n!$ . The coefficient is $\frac{f^{(n)}(a)}{n!}$ , not $f^{(n)}(a)$ . This is the single most common slip.
Losing the $(x-a)$ shift. For a series centered at $a \ne 0$ , the powers are $(x-a)^n$ , not $x^n$ . Writing $x$ alone silently recenters the series at $0$ .
Recomputing derivatives when a substitution would do. If the function is a known series in disguise, differentiating from scratch wastes time and invites errors.
Mishandling signs after substitution. $(-x^2)^n = (-1)^n x^{2n}$ : the exponent acts on both the sign and the variable.
Assuming the interval carries over unchanged. Substituting $u = g(x)$ into a series valid for $|u| < 1$ makes the new condition $|g(x)| < 1$ , which can be a very different set of $x$ -values.

Worked examples

Worked example

Building $e^x$ from the derivative formula

Find the Maclaurin series for $f(x) = e^x$ directly from the coefficient formula.

Every derivative of $e^x$ is again $e^x$ , so $f'(x) = e^x$ , $f''(x) = e^x$ , and in general $f^{(n)}(x) = e^x$ for every $n$ .
Evaluate at the center $x = 0$ . Since $e^0 = 1$ , every derivative gives the same value: $f^{(n)}(0) = 1$ for all $n$ .
Form the coefficient $\frac{f^{(n)}(0)}{n!} = \frac{1}{n!}$ , so the term of degree $n$ is $\frac{x^n}{n!}$ .
Sum over all $n$ : $e^x = \displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$
A ratio test on $\frac{x^n}{n!}$ gives $L = 0$ for every fixed $x$ , so the series converges for all real $x$ . This is the anchor series: the other four in the table are variations on it or on the geometric series.

$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ , converging for every real $x$ .

Worked example

$e^{-x^2}$ by substitution, not brute force

Find the Maclaurin series for $e^{-x^2}$ .

Differentiating $e^{-x^2}$ repeatedly gets ugly fast, so recognize it instead: it is $e^u$ with $u = -x^2$ . Start from the known series $e^u = \sum_{n=0}^{\infty} \frac{u^n}{n!}$ .
Substitute $u = -x^2$ into that series: $e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!}$ .
Simplify the power: $(-x^2)^n = (-1)^n x^{2n}$ , so $e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}$ .
Write out the first terms: $e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots$
Because the $e^u$ series converges for all $u$ , the substituted series converges for all real $x$ . Building it this way took three lines; the derivative formula would have taken a page.

$e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!} = 1 - x^2 + \frac{x^4}{2!} - \cdots$ , valid for all real $x$ .

Worked example

A Taylor series centered at a nonzero point

Find the Taylor series for $f(x) = \ln x$ centered at $x = 1$ .

There is no simpler known series sitting at $x = 1$ , so use the derivative formula. Set $f(x) = \ln x$ with center $a = 1$ ; then $f(1) = \ln 1 = 0$ , so the constant term vanishes.
Differentiate repeatedly: $f'(x) = x^{-1}$ , $f''(x) = -x^{-2}$ , $f'''(x) = 2x^{-3}$ , $f^{(4)}(x) = -6x^{-4}$ .
Evaluate at $x = 1$ : $f'(1) = 1$ , $f''(1) = -1$ , $f'''(1) = 2$ , $f^{(4)}(1) = -6$ . The pattern is $f^{(n)}(1) = (-1)^{n-1}(n-1)!$ for $n \ge 1$ .
Divide by $n!$ to get the coefficient, using $\frac{(n-1)!}{n!} = \frac{1}{n}$ : $\frac{f^{(n)}(1)}{n!} = \frac{(-1)^{n-1}(n-1)!}{n!} = \frac{(-1)^{n-1}}{n}$ .
Attach each coefficient to $(x-1)^n$ : $\ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}(x-1)^n = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \cdots$
Substituting $t = x - 1$ turns this into $\ln(1+t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \cdots$ , the standard $\ln(1+x)$ series from the table. That confirms the work and shows the two are the same series, just shifted. It converges for $-1 < t \le 1$ , that is $0 < x \le 2$ .

$\ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}(x-1)^n$ , converging for $0 < x \le 2$ .

Frequently asked questions

What is the difference between a Taylor series and a Maclaurin series?

A Maclaurin series is a Taylor series centered at $x = 0$ . That is the only difference. The general Taylor series is $\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$ ; setting the center $a = 0$ turns the $(x-a)^n$ factors into $x^n$ and gives the Maclaurin series. Every Maclaurin series is a Taylor series, but a Taylor series centered at $a \ne 0$ is not Maclaurin.

Which Taylor series do I need to memorize for AP Calculus BC?

At minimum, memorize the Maclaurin series for $e^x$ , $\sin x$ , $\cos x$ , and $\frac{1}{1-x}$ . The CED points to $\sin x$ , $\cos x$ , and $e^x$ as the foundation for building others (LIM-8.F.2), and $\frac{1}{1-x}$ is the geometric building block (LIM-8.F.1). Knowing $\ln(1+x)$ helps too, but you can recover it by integrating the series for $\frac{1}{1+x}$ .

How do you find a Taylor series centered at a point other than zero?

Use the derivative formula directly: compute $f(a), f'(a), f''(a), \dots$ at the center $a$ , form the coefficients $\frac{f^{(n)}(a)}{n!}$ , and attach each to $(x-a)^n$ . The powers must be $(x-a)^n$ , not $x^n$ . Example 3 above builds $\ln x$ at $a = 1$ this way. Substituting a known series works only if you can rewrite $f$ in terms of $(x-a)$ , such as writing $\ln x = \ln(1 + (x-1))$ and using the $\ln(1+t)$ series with $t = x - 1$ .