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

A Taylor polynomial T_n(x) stops at degree n and is a finite sum, so it is an approximation with a measurable error. A Taylor series continues the pattern infinitely; when it converges to f, it equals the function exactly. The Lagrange bound measures how far a finite T_n is from f.

How do I choose the value of M in the Lagrange error bound?

Pick any number that is at least as large as |f^(n+1)(t)| for every t between the center c and the point x. You want the maximum of the (n+1)th derivative on that interval. If the derivative is something like t or t, M=1 works; often the problem simply tells you a bound to use.

Why is the exponent n+1 and not n in the error formula?

A degree-n Taylor polynomial matches derivatives through order n, so the first unmatched behavior comes from the (n+1)th derivative. Taylor's theorem shows the remainder is exactly f^(n+1)(z)(n+1)!(x-c)^n+1, which is why both the exponent and factorial are n+1.

When should I use the alternating series error bound instead of Lagrange?

Use the alternating series bound only when the series alternates in sign and its terms decrease in magnitude to zero; then the error is at most the first omitted term. Use Lagrange for any Taylor polynomial, especially when the problem provides a bound on the next derivative or the series does not alternate.

AP-CALCBC-10.11-10.12

U10.11 Taylor Polynomials and Lagrange Error Bound

Master AP Calculus BC Taylor polynomials: build Tₙ(x) about x=c from derivatives and bound the error with the Lagrange remainder formula.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U10.11 Taylor Polynomials and Lagrange Error Bound, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

When a function is too messy to evaluate by hand, we replace it near a point with a polynomial that matches its value and derivatives — a Taylor polynomial. The more derivatives we match, the tighter the fit near the center. But an approximation is only trustworthy if we can say how wrong it might be. That is where the Lagrange error bound comes in.

In this lesson you will learn to construct

T_n(x)

about

x=c

, recognize the Maclaurin case (

c=0

), and rigorously bound the remainder

R_n(x)=f(x)-T_n(x)

. These skills appear on both multiple-choice and free-response, often as a two-part problem: build the polynomial, then estimate or bound the error.

Building the Taylor Polynomial

A Taylor polynomial of degree

n

centered at

x=c

is the unique polynomial whose value and first

n

derivatives at

c

match those of

f

. Its formula is

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

Each term corrects the previous approximation using one more derivative. The degree-0 term

f(c)

is just a horizontal line; adding

f'(c)(x-c)

gives the tangent line

T_1

; adding the quadratic term captures concavity, and so on.

When

c=0

the polynomial is called a Maclaurin polynomial, and

(x-c)^k

simplifies to

x^k

. The exam frequently gives you a table of derivative values at

c

rather than a formula, so you must be comfortable plugging numbers directly into the definition.

A common misconception is forgetting the factorial in each denominator, or confusing

f^{(k)}(c)

(a constant) with

f^{(k)}(x)

(a function). Remember: every coefficient is a number evaluated at the center, so

T_n(x)

is a genuine polynomial in

(x-c)

The Remainder and Lagrange Error Bound

The error in using

T_n

is the remainder

R_n(x)=f(x)-T_n(x)

. Taylor's theorem states that for some

z

between

c

and

x

R_n(x)=\frac{f^{(n+1)}(z)}{(n+1)!}(x-c)^{n+1}.

We cannot find

z

exactly, but we can bound it. If

M

is any number with

|f^{(n+1)}(t)|\le M

for all

t

between

c

and

x

, then the Lagrange error bound is

|R_n(x)|\le \frac{M\,|x-c|^{n+1}}{(n+1)!}.

Notice the structure: it looks exactly like the next term of the Taylor series, except

f^{(n+1)}(c)

is replaced by its maximum

M

. This is the single most tested idea in the topic. To apply it you must (1) identify

n

, (2) find or estimate

M

as a bound on the

(n+1)

th derivative, and (3) plug in

|x-c|

.

Choosing

M

trips students up. You want a safe upper bound on

|f^{(n+1)}|

over the interval, not its exact value. For example, if

f^{(n+1)}(t)=\cos(t)

, then

M=1

works because

|\cos t|\le 1

everywhere.

When to Use Lagrange vs. Alternating Series Bound

AP problems offer two error tools, and picking the right one matters. The alternating series bound (from U10.7) applies only when the series alternates and its terms decrease in magnitude; then the error is at most the first omitted term. The Lagrange bound works for any function with the required derivatives, regardless of sign pattern.

Feature	Lagrange error bound	Alternating series bound
Requires alternating terms	No	Yes
Needs a bound $M$ on next derivative	Yes	No
Error estimate	$\frac{M\,\|x-c\|^{n+1}}{(n+1)!}$	first omitted term
Typical prompt	"show error is less than..."	"approximate with error < ..."

If a free-response question hands you a bound like "

|f^{(n+1)}(x)|\le 5

on the interval," it is signaling Lagrange. If it explicitly says the series is alternating and asks for the next-term estimate, use the alternating bound. Both can appear on the same exam, so read carefully. A frequent error is applying the alternating bound to a series that is not actually alternating, or forgetting to verify the terms decrease.

How the Exam Tests This

Expect a scaffolded free-response: part (a) asks for

T_2

T_3

from a derivative table; part (b) uses the polynomial to approximate

f

at a nearby point; part (c) asks you to bound the error with Lagrange, sometimes giving you the maximum of the next derivative.

The key phrase to watch for is "

|f^{(n+1)}(x)|\le M

for all

x

in the interval." That

M

goes straight into the formula. Graders want to see the substituted expression

\frac{M|x-c|^{n+1}}{(n+1)!}

and a final numerical bound, not just the general formula.

Multiple-choice items often ask which term or coefficient appears in

T_n

, or which expression correctly represents the error bound. Watch the off-by-one issue: a degree-

n

polynomial has error controlled by the

(n+1)

th derivative, and the exponent and factorial are both

n+1

. Also be careful that

|x-c|

is raised to the power, not

x

alone. Practicing the mechanical substitution until it is automatic frees you to focus on choosing

M

correctly under time pressure.

Key terms

Taylor polynomial $T_n(x)$ .: The degree- $n$ polynomial centered at $c$ whose first $n$ derivatives match $f$ : $\sum_{k=0}^n \frac{f^{(k)}(c)}{k!}(x-c)^k$ .
Maclaurin polynomial.: A Taylor polynomial centered at $c=0$ , so each term uses $x^k$ instead of $(x-c)^k$ .
Center $c$ .: The point about which the polynomial is built; the approximation is most accurate near $c$ .
Remainder $R_n(x)$ .: The exact error $f(x)-T_n(x)$ when approximating $f$ by its degree- $n$ Taylor polynomial.
Lagrange error bound.: The inequality $|R_n(x)|\le \frac{M|x-c|^{n+1}}{(n+1)!}$ where $M$ bounds $|f^{(n+1)}|$ on the interval.
$M$ (bound on the derivative).: An upper bound for $|f^{(n+1)}(t)|$ for all $t$ between $c$ and $x$ ; used to make the error estimate rigorous.
Taylor's theorem with remainder.: States $R_n(x)=\frac{f^{(n+1)}(z)}{(n+1)!}(x-c)^{n+1}$ for some $z$ between $c$ and $x$ .

Worked example

Let

f(x)=\cos x

. (a) Find the degree-3 Maclaurin polynomial

T_3(x)

. (b) Use it to approximate

\cos(0.5)

. (c) Bound the error using the Lagrange error bound.

For part (a), compute derivatives at

c=0

. We have

f(0)=\cos 0=1

f'(x)=-\sin x

f'(0)=0

f''(x)=-\cos x

f''(0)=-1

, and

f'''(x)=\sin x

f'''(0)=0

.

Plug into the definition:

T_3(x)=1+0\cdot x+\frac{-1}{2!}x^2+\frac{0}{3!}x^3 = 1-\frac{x^2}{2}.

For part (b), substitute

x=0.5

T_3(0.5)=1-\frac{(0.5)^2}{2}=1-\frac{0.25}{2}=1-0.125=0.875.

For part (c), the error for a degree-3 polynomial is controlled by the 4th derivative. Here

f^{(4)}(x)=\cos x

, and

|\cos t|\le 1

for all

t

, so we take

M=1

. With

n=3

c=0

, and

|x-c|=0.5

|R_3(0.5)|\le \frac{M\,|x-c|^{4}}{4!}=\frac{1\cdot(0.5)^4}{24}=\frac{0.0625}{24}\approx 0.0026.

\cos(0.5)\approx 0.875

with error at most about

0.0026

. The true value is

0.8776

, and indeed

|0.8776-0.875|=0.0026

, consistent with the bound.

Practice questions

A function

f

has

f(2)=3

f'(2)=-1

f''(2)=4

, and

f'''(2)=6

. What is the coefficient of

(x-2)^2

in the degree-3 Taylor polynomial of

f

about

x=2

$4$
$2$
$-1$
$1$

Answer: $2$

The coefficient of

(x-2)^2

\frac{f''(2)}{2!}=\frac{4}{2}=2

. A common mistake is to use

f''(2)=4

directly without dividing by the factorial

2!

. Note

f'''(2)

is irrelevant to this particular coefficient.

Suppose

T_4(x)

is the degree-4 Taylor polynomial of

f

about

x=1

, and it is known that

|f^{(5)}(x)|\le 10

for all

x

[1,1.2]

. Find an upper bound for the error

|f(1.2)-T_4(1.2)|

Answer: $\frac{10\,(0.2)^5}{5!}=\frac{10(0.00032)}{120}\approx 2.67\times 10^{-5}$

For a degree-4 polynomial, use

n=4

, so the error involves the 5th derivative and

(n+1)!=5!=120

. Here

M=10

and

|x-c|=|1.2-1|=0.2

. Then

|R_4|\le \frac{10(0.2)^5}{120}=\frac{10\cdot 0.00032}{120}\approx 0.0000267

. The key is matching the exponent and factorial to

n+1=5

The Maclaurin polynomial of degree 2 for

f(x)=e^x

T_2(x)=1+x+\frac{x^2}{2}

. Which expression gives a valid Lagrange bound for the error at

x=0.1

, given that

e^t\le e^{0.1}<1.2

[0,0.1]

$\frac{1.2\,(0.1)^3}{3!}$
$\frac{1.2\,(0.1)^2}{2!}$
$\frac{1.2\,(0.1)^3}{2!}$
$\frac{1.2\,(0.1)^4}{4!}$

Answer: $\frac{1.2\,(0.1)^3}{3!}$

For degree

n=2

, the error uses the third derivative, so the exponent is

n+1=3

and the factorial is

(n+1)!=3!

. Since

f^{(3)}(x)=e^x\le 1.2

on the interval,

M=1.2

. This gives

\frac{1.2(0.1)^3}{3!}

. The other choices use wrong exponents or factorials.

FAQ

What is the difference between a Taylor polynomial and a Taylor series?: A Taylor polynomial $T_n(x)$ stops at degree $n$ and is a finite sum, so it is an approximation with a measurable error. A Taylor series continues the pattern infinitely; when it converges to $f$ , it equals the function exactly. The Lagrange bound measures how far a finite $T_n$ is from $f$ .
How do I choose the value of M in the Lagrange error bound?: Pick any number that is at least as large as $|f^{(n+1)}(t)|$ for every $t$ between the center $c$ and the point $x$ . You want the maximum of the $(n+1)$ th derivative on that interval. If the derivative is something like $\cos t$ or $\sin t$ , $M=1$ works; often the problem simply tells you a bound to use.
Why is the exponent n+1 and not n in the error formula?: A degree- $n$ Taylor polynomial matches derivatives through order $n$ , so the first unmatched behavior comes from the $(n+1)$ th derivative. Taylor's theorem shows the remainder is exactly $\frac{f^{(n+1)}(z)}{(n+1)!}(x-c)^{n+1}$ , which is why both the exponent and factorial are $n+1$ .
When should I use the alternating series error bound instead of Lagrange?: Use the alternating series bound only when the series alternates in sign and its terms decrease in magnitude to zero; then the error is at most the first omitted term. Use Lagrange for any Taylor polynomial, especially when the problem provides a bound on the next derivative or the series does not alternate.

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