Taylor Polynomials

The tangent line at

x = a

has the same value and the same slope as

f

there. A polynomial of higher degree can also match the second derivative, the third, and so on. It usually stays close to

f

over a wider stretch.

Matching derivatives

Write a polynomial in powers of

x - a

\begin{aligned} &c_0 + c_1(x - a) + c_2(x - a)^2 \\[4pt] &\qquad + \cdots + c_n(x - a)^n \end{aligned}

Differentiating

k

times turns

(x - a)^k

into

k!

and sends every lower power to 0. Every higher power still contains a factor of

x - a

. So at

x = a

the

k

th derivative is

k!\,c_k

, and matching

f^{(k)}(a)

forces the coefficient.

A Taylor polynomial about

x = 0

is also called a Maclaurin polynomial. In many cases, raising the degree makes

P_n

approximate

f

well over a longer interval around

a

Interactive: Taylor Polynomials

Below,

f(x) = \sin x

and

a = 0

. The derivatives of

\sin x

0

repeat

0, 1, 0, -1

, so every even-power term is 0. Raising the degree from odd to even leaves the polynomial unchanged.

Example 1

Find the third-degree Taylor polynomial for

f(x) = \ln(x^2 + 3)

about

x = 1

, and use it to approximate

\ln 4.21

Find the derivatives and evaluate them at

x = 1

\begin{aligned} f(x) &= \ln(x^2 + 3) & f(1) &= \ln 4 \\[4pt] f'(x) &= \frac{2x}{x^2 + 3} & f'(1) &= \tfrac{1}{2} \\[4pt] f''(x) &= \frac{2(3 - x^2)}{(x^2 + 3)^2} & f''(1) &= \tfrac{1}{4} \\[4pt] f'''(x) &= \frac{4x(x^2 - 9)}{(x^2 + 3)^3} & f'''(1) &= -\tfrac{1}{2} \end{aligned}

Divide each by the matching factorial.

\begin{aligned} P_3(x) &= \ln 4 + \frac{1}{2}(x - 1) \\[4pt] &\qquad + \frac{1}{8}(x - 1)^2 \\[4pt] &\qquad - \frac{1}{12}(x - 1)^3 \end{aligned}

Since

\ln 4.21 = f(1.1)

, evaluate at

x = 1.1

, where

x - 1 = 0.1

\begin{aligned} \ln 4.21 &\approx P_3(1.1) \\[4pt] &= \ln 4 + 0.05 + 0.00125 \\[4pt] &\qquad - 0.0000833 \\[4pt] &\approx 1.43746 \end{aligned}

y = ln(x² + 3) with its Taylor polynomials of degree 1, 2, and 3 about x = 1. Near x = 1, the degree-3 polynomial stays close to the curve over the longest stretch.

When f is known only through its derivatives

A Taylor polynomial needs only the values of

f

and its derivatives at one point. Those values can come from a table, a graph, or a differential equation, with no formula for

f

at all.

Example 2

The function

y = f(x)

satisfies

\frac{dy}{dx} = y^2 - x

, and

f(0) = 1

. Find the third-degree Taylor polynomial for

f

about

x = 0

, and use it to approximate

f(0.2)

The equation gives

f'(0) = 1^2 - 0 = 1

. Differentiate it implicitly, remembering that

y

is a function of

x

\begin{aligned} y'' &= 2yy' - 1 \\[4pt] f''(0) &= 2(1)(1) - 1 = 1 \\[4pt] y''' &= 2(y')^2 + 2yy'' \\[4pt] f'''(0) &= 2(1)^2 + 2(1)(1) = 4 \end{aligned}

\begin{aligned} P_3(x) &= 1 + x + \frac{1}{2!}x^2 + \frac{4}{3!}x^3 \\[4pt] &= 1 + x + \frac{x^2}{2} + \frac{2x^3}{3} \end{aligned}

Then evaluate at

x = 0.2

\begin{aligned} f(0.2) &\approx P_3(0.2) \\[4pt] &= 1 + 0.2 + 0.02 + 0.00533 \\[4pt] &\approx 1.2253 \end{aligned}

Example 3

The third-degree Taylor polynomial for a function

h

about

x = -1

4 - 3(x + 1) + 5(x + 1)^2 - \frac{1}{2}(x + 1)^3

. Find

h''(-1)

and

h'''(-1)

, and decide whether the graph of

h

is concave up or concave down at

x = -1

Each coefficient is

\frac{h^{(k)}(-1)}{k!}

, so multiply by

k!

to undo the division.

\begin{aligned} h''(-1) &= 2! \cdot 5 = 10 \\[4pt] h'''(-1) &= 3! \cdot \Big(-\frac{1}{2}\Big) = -3 \end{aligned}

Since

h''(-1) = 10 > 0

, the graph of

h

is concave up at

x = -1