AP Calculus BC

Taylor Series Builder: Coefficients from Derivatives

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

A Maclaurin series is a Taylor series centered at a = 0, nothing more. Every Maclaurin series is a Taylor series; the name just signals that the center is 0, so the terms are powers of x and you evaluate the derivatives at 0. When a problem centers the expansion anywhere else, a = 1, a = \frac{\pi}{2}, and so on, it is called a Taylor series, and you plug that center into \frac{f^{(n)}(a)}{n!} and use (x-a)^n.

Q: Why do you divide by n! in a Taylor series?

The factorial cancels the one that appears when you differentiate the term c_n(x-a)^n exactly n times: that term becomes n!\,c_n at the center while every other term vanishes. To make the polynomial's nth derivative match f^{(n)}(a), you need n!\,c_n = f^{(n)}(a), which forces c_n = \frac{f^{(n)}(a)}{n!}. Without the factorial the higher derivatives would not line up.

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

A Taylor polynomial stops at a chosen degree n; a Taylor series keeps going forever. Formally, a Taylor polynomial is a partial sum of the Taylor series (LIM-8.E.1). The builder shows a polynomial because you pick a finite degree, and as you raise that degree you are switching on more terms of the full series, tightening the fit around the center.

Q: Do I have to compute every derivative to find a Taylor series?

Often no, and recognizing when to skip it is the point. If f is e^x, \sin x, \cos x, or \frac{1}{1-x} after a substitution, constant multiple, derivative, or integral, start from that known Maclaurin series instead of differentiating repeatedly (Topic 10.15). Reserve the from-derivatives method for a nonzero center or a function that does not reduce to one of the standard series, like \ln x centered at 1.

A Taylor series is an infinite polynomial (a power series) reverse-engineered so its derivatives match f at a center a: the coefficient of the (x-a)^n term is the nth derivative of f at a divided by n factorial. A Maclaurin series is just this centered at 0. Grind the derivatives, or reuse a known series.

BCMaclaurin series (Taylor series centered at a = 0)

f(x) = e^xP_{3}(x)

Degreen = 3

Maclaurin polynomial, degree 3

P_{3}(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}

Converges to f(x) = e^x on (-\infty, \infty).

f(0.5)

1.6487

P_3(0.5)

1.6458

error

0.0029

Every coefficient is c_k = \dfrac{f^{(k)}(0)}{k!}. That is why raising the degree adds one more term and pulls P_{3} closer to the curve.

k	f^{(k)}(0)	k!	c_k = f^{(k)}(0)/k!
0	1	1	1
1	1	1	1
2	1	2	\frac{1}{2}
3	1	6	\frac{1}{6}
4	1	24	\frac{1}{24}

This tool builds a Taylor polynomial one derivative at a time. Pick a function $f$ , a center $a$ , and a degree $n$ ; the builder fills a table of derivatives $f, f', f'', \dots$ , evaluates each one at the center, divides by a factorial, and drops the results in as coefficients. Then it graphs the polynomial on top of $f$ so you can watch it hug the curve near $x = a$ . Finding these approximations is CED Topic 10.11, and the one fact doing all the work is that the coefficient of the degree- $n$ term is $\frac{f^{(n)}(a)}{n!}$ (LIM-8.A.1).

The reason that formula works is worth seeing, because it is the whole idea of the tool. A Taylor polynomial is reverse-engineered to agree with $f$ at the center as closely as a polynomial of its degree can: it matches the value $f(a)$ , the slope $f'(a)$ , the concavity $f''(a)$ , and every higher derivative up to order $n$ , all at $x = a$ . Each term you add pins down exactly one more derivative. That is why more terms mean a tighter fit near the center, and why the fit is sharpest right at $x = a$ and loosens as you move away.

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

Where do the factorial and the power $(x-a)^k$ come from? Differentiate a single term $c_k(x-a)^k$ a total of $k$ times: every term below it has already vanished, every term above it still carries a factor of $(x-a)$ that dies at $x = a$ , and the term itself becomes $k!\,c_k$ . Setting that equal to $f^{(k)}(a)$ forces $c_k = \frac{f^{(k)}(a)}{k!}$ . The tool's derivative table is exactly this calculation laid out column by column: one row per derivative order, its value at the center, and the factorial you divide by to get the coefficient.

A Maclaurin series is not a different object; it is a Taylor series with the center fixed at $a = 0$ . Set the center to $0$ in the builder and two things simplify: every $(x-a)^k$ collapses to a clean power $x^k$ , and you evaluate each derivative at $0$ , usually the easiest input to plug in. That is why the three series worth memorizing, $e^x$ , $\sin x$ , and $\cos x$ , are always quoted as Maclaurin series (LIM-8.F.2). Centering anywhere else, say $a = 1$ for $\ln x$ , gives a genuine Taylor series, and that is where the from-derivatives method earns its keep.

Two ways to get a series: pick the shorter one

Grinding out derivatives is the definition, not always the fast route, and knowing when to skip it is the real skill (Topic 10.15). Before you build a table, check whether $f$ is a known Maclaurin series in disguise. If $f$ is $e^x$ , $\sin x$ , $\cos x$ , or $\frac{1}{1-x}$ after a substitution, a constant multiple, a derivative, or an integral, start from the known series instead: to get the series for $e^{-x^2}$ , substitute $-x^2$ into the series for $e^x$ rather than differentiating $e^{-x^2}$ four times. Build from derivatives when the center is not $0$ , or when $f$ does not reduce to one of the four standard series.

Function	Maclaurin series	Reach for it when
$e^x$	$\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \cdots$	any exponential, or a substitution like $e^{-x^2}$
$\sin x$	$\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \cdots$	odd-powered, sign-alternating targets
$\cos x$	$\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \cdots$	even-powered, sign-alternating targets
$\frac{1}{1-x}$	$\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + \cdots$ for $\|x\| < 1$	rational functions, via algebra or substitution

Worked example

Degree-3 Taylor polynomial for ln x centered at a = 1

Build the third-degree Taylor polynomial $P_3(x)$ for $f(x) = \ln x$ centered at $a = 1$ , taking each coefficient straight from a derivative.

Set up the coefficient formula at this center. With $a = 1$ , the polynomial is $P_3(x) = \sum_{k=0}^{3} \frac{f^{(k)}(1)}{k!}(x-1)^k$ , so you need $f$ and its first three derivatives, each evaluated at $1$ .
Differentiate three times, rewriting as powers so the pattern is visible: $f(x) = \ln x$ , then $f'(x) = x^{-1}$ , $f''(x) = -x^{-2}$ , and $f'''(x) = 2x^{-3}$ .
Evaluate every one at the center $x = 1$ : $f(1) = \ln 1 = 0$ , $f'(1) = 1$ , $f''(1) = -1$ , and $f'''(1) = 2$ . These are the raw derivative values, before any factorials.
Divide each by its factorial to get the coefficients: $c_0 = \frac{0}{0!} = 0$ , $c_1 = \frac{1}{1!} = 1$ , $c_2 = \frac{-1}{2!} = -\frac{1}{2}$ , and $c_3 = \frac{2}{3!} = \frac{2}{6} = \frac{1}{3}$ .
Attach each coefficient to its matching $(x-1)^k$ and add: $P_3(x) = 0 + 1(x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3$ .
Drop the zero term to read off the answer: $P_3(x) = (x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3$ . Sanity-check at the center: $P_3(1) = 0 = \ln 1$ and $P_3'(1) = 1 = f'(1)$ , so the polynomial matches $f$ in value and slope at $x = 1$ , exactly as designed.

$P_3(x) = (x-1) - \frac{1}{2}(x-1)^2 + \frac{1}{3}(x-1)^3$

Frequently asked questions

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

A Maclaurin series is a Taylor series centered at $a = 0$ , nothing more. Every Maclaurin series is a Taylor series; the name just signals that the center is $0$ , so the terms are powers of $x$ and you evaluate the derivatives at $0$ . When a problem centers the expansion anywhere else, $a = 1$ , $a = \frac{\pi}{2}$ , and so on, it is called a Taylor series, and you plug that center into $\frac{f^{(n)}(a)}{n!}$ and use $(x-a)^n$ .

Why do you divide by $n!$ in a Taylor series?

The factorial cancels the one that appears when you differentiate the term $c_n(x-a)^n$ exactly $n$ times: that term becomes $n!\,c_n$ at the center while every other term vanishes. To make the polynomial's $n$ th derivative match $f^{(n)}(a)$ , you need $n!\,c_n = f^{(n)}(a)$ , which forces $c_n = \frac{f^{(n)}(a)}{n!}$ . Without the factorial the higher derivatives would not line up.

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

A Taylor polynomial stops at a chosen degree $n$ ; a Taylor series keeps going forever. Formally, a Taylor polynomial is a partial sum of the Taylor series (LIM-8.E.1). The builder shows a polynomial because you pick a finite degree, and as you raise that degree you are switching on more terms of the full series, tightening the fit around the center.

Do I have to compute every derivative to find a Taylor series?

Often no, and recognizing when to skip it is the point. If $f$ is $e^x$ , $\sin x$ , $\cos x$ , or $\frac{1}{1-x}$ after a substitution, constant multiple, derivative, or integral, start from that known Maclaurin series instead of differentiating repeatedly (Topic 10.15). Reserve the from-derivatives method for a nonzero center or a function that does not reduce to one of the standard series, like $\ln x$ centered at $1$ .