Taylor Polynomials and the Lagrange Error Bound

Q: What is a Taylor polynomial?

The polynomial of degree n whose value and first n derivatives at x = a match those of f. Its coefficients are f⁽ᵏ⁾(a)/k!.

Q: What is a Maclaurin polynomial?

A Taylor polynomial centered at a = 0.

Q: Why is there a factorial in the coefficients?

Differentiating (x − a)ᵏ k times produces k!, so dividing by k! makes the kth derivative of the polynomial equal f⁽ᵏ⁾(a).

Q: What is the Lagrange error bound?

|f(x) − Pₙ(x)| ≤ M|x − a|ⁿ⁺¹/(n + 1)!, where M is at least the largest value of |f⁽ⁿ⁺¹⁾| between a and x.

Q: How do I get a better approximation?

Raise the degree, or approximate closer to the center. Both shrink |x − a|ⁿ⁺¹/(n + 1)!.

Quick answer

The Taylor polynomial of degree n for f at a is the polynomial that matches f and its first n derivatives at a: Pₙ(x) is the sum of f⁽ᵏ⁾(a)/k! times (x − a)ᵏ for k from 0 to n. The degree-1 polynomial is the tangent line; higher degrees bend to follow the curve farther from a. The Lagrange error bound says |f(x) − Pₙ(x)| ≤ M|x − a|ⁿ⁺¹/(n + 1)!, where M bounds the size of the next derivative between a and x.

What you'll learn

Build the Taylor polynomial of a given degree from derivatives
Use Taylor polynomials to approximate function values
Read derivatives from the coefficients of a Taylor polynomial
Bound the error with the Lagrange error bound

Matching more than the slope

The tangent line matches a function’s value and slope at one point, and it is a good approximation nearby. A parabola can match the curvature too, and a cubic the rate at which curvature changes. The Taylor polynomial of degree $n$ for $f$ at $a$ matches $f$ and its first $n$ derivatives there:

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

Centered at $a = 0$ , it is called a Maclaurin polynomial.

Why the coefficients have factorials

Differentiate $(x - a)^k$ $k$ times and you get $k!$ , a constant; differentiate it fewer times and it still vanishes at $x = a$ . So the $k$ th derivative of $P_n$ at $a$ comes only from the term $c_k(x - a)^k$ , and equals $c_k \cdot k!$ . To make it equal $f^{(k)}(a)$ , choose $c_k = \tfrac{f^{(k)}(a)}{k!}$ . Each coefficient is chosen so that one more derivative matches, and the factorial undoes what differentiation does to the power.

y = sin x
P₁
P₃
P₅

sin x and its Taylor polynomials at 0

The Lagrange error bound

How good is the approximation? The error depends on the next derivative, the first one the polynomial does not match.

Lagrange error bound

If $\lvert f^{(n+1)}(t) \rvert \le M$ for every $t$ between $a$ and $x$ , then

\lvert f(x) - P_n(x) \rvert \le \frac{M}{(n + 1)!}\lvert x - a \rvert^{n+1}

It looks like the next term of the polynomial, with the unknown derivative replaced by its largest possible size. Close to $a$ , the power $\lvert x - a \rvert^{n+1}$ is tiny, and the factorial shrinks it further.

Worked examples

Common mistakes

Practice problems

Find $P_2$ for $\cos x$ at $0$ .

Answer

$1 - \tfrac{x^2}{2}$

Full solution

$\cos 0 = 1$ , $-\sin 0 = 0$ and $-\cos 0 = -1$ , so $P_2 = 1 + 0x - \tfrac{1}{2}x^2$ .
Find $P_3$ for $\tfrac{1}{1 - x}$ at $0$ .

Answer

$1 + x + x^2 + x^3$

Full solution

The $k$ th derivative is $\tfrac{k!}{(1 - x)^{k+1}}$ , which is $k!$ at $0$ , so every coefficient is $\tfrac{k!}{k!} = 1$ .
Find $P_2$ for $\sqrt{x}$ at $4$ , and estimate $\sqrt{4.2}$ .

Answer

$2 + \tfrac{x - 4}{4} - \tfrac{(x - 4)^2}{64}$ ; about $2.049375$

Full solution

$f(4) = 2$ , $f'(4) = \tfrac{1}{2\sqrt{4}} = \tfrac{1}{4}$ , $f''(4) = -\tfrac{1}{4 \cdot 4^{3/2}} = -\tfrac{1}{32}$ , and $\tfrac{-1/32}{2!} = -\tfrac{1}{64}$ . At $4.2$ : $2 + 0.05 - 0.000625$ . The true value is $2.04939\ldots$
The Taylor polynomial of $f$ at $1$ is $2 - 3(x - 1) + 5(x - 1)^2 - (x - 1)^3$ . Find $f''(1)$ and $f'''(1)$ .

Answer

$f''(1) = 10$ and $f'''(1) = -6$

Full solution

The coefficient of $(x - 1)^k$ is $\tfrac{f^{(k)}(1)}{k!}$ . So $f''(1) = 2! \cdot 5$ and $f'''(1) = 3! \cdot (-1)$ .
Estimate $e^{0.2}$ with $P_2$ at $0$ .

Answer

$1.22$

Full solution

$1 + 0.2 + \tfrac{0.04}{2} = 1.22$ . The true value is $1.2214\ldots$
Bound the error in problem 5.

Answer

Less than $0.002$

Full solution

The third derivative is $e^t \le e^{0.2} < 1.3$ on $[0, 0.2]$ . The bound is $\tfrac{1.3}{3!}(0.2)^3 = \tfrac{1.3 \cdot 0.008}{6} \approx 0.0017$ .
What degree $n$ of the Maclaurin polynomial of $e^x$ guarantees $e$ to within $0.001$ , using $e < 3$ ?

Answer

$n = 6$

Full solution

The error at $x = 1$ is at most $\tfrac{3}{(n + 1)!}$ . $\tfrac{3}{6!} = \tfrac{3}{720} \approx 0.004$ is too big, but $\tfrac{3}{7!} = \tfrac{3}{5040} \approx 0.0006$ works, so $n + 1 = 7$ .
Find $P_4$ for $\ln(1 + x)$ at $0$ .

Answer

$x - \tfrac{x^2}{2} + \tfrac{x^3}{3} - \tfrac{x^4}{4}$

Full solution

The derivatives at $0$ are $1$ , $-1$ , $2$ and $-6$ . Dividing by $1!$ , $2!$ , $3!$ and $4!$ gives $1$ , $-\tfrac{1}{2}$ , $\tfrac{1}{3}$ and $-\tfrac{1}{4}$ .
Find $P_3$ for $\sin x$ at $\tfrac{\pi}{2}$ .

Answer

$1 - \tfrac{1}{2}\left(x - \tfrac{\pi}{2}\right)^2$

Full solution

At $\tfrac{\pi}{2}$ : $\sin = 1$ , $\cos = 0$ , $-\sin = -1$ and $-\cos = 0$ . Only the constant and squared terms survive.
A student writes the Maclaurin polynomial of degree 3 for $e^x$ as $1 + x + x^2 + x^3$ . What went wrong?

Hint

What is the second derivative of the student’s polynomial at $0$ ?

Answer

The factorials are missing. The polynomial is $1 + x + \tfrac{x^2}{2} + \tfrac{x^3}{6}$ .

Full solution

The student’s polynomial has second derivative $2$ at $0$ , but $e^x$ has second derivative $1$ there. Dividing each coefficient by $k!$ fixes every derivative: the correct cubic has $f''(0) = 1$ and $f'''(0) = 1$ .

Frequently asked questions

What is a Taylor polynomial?