Taylor and Maclaurin Series — AP® Calculus BC

Part 1 · Hook

Why this matters

A Taylor polynomial stops after a few terms and is only an estimate. Let it run forever and, for

\sin x

,

\cos x

and

e^x

, the infinite sum is not an estimate any more: it is the function, for every

x

. Four of these series are worth knowing by heart; the rest of the unit builds on them.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is the coefficient of $(x - a)^n$ in the Taylor polynomial for $f$ about $x = a$ ?

$\frac{f^{(n)}(a)}{n!}$
$f^{(n)}(a)$
$\frac{f^{(n)}(0)}{n!}$
$\frac{f^{(n)}(a)}{n}$

Show the answer

The $n$ th derivative at the center over $n!$ .

Correct: $\frac{f^{(n)}(a)}{n!}$ :
$f^{(n)}(a)$ :
$\frac{f^{(n)}(0)}{n!}$ :
$\frac{f^{(n)}(a)}{n}$ :

2. What is the interval of convergence of $\sum_{n=0}^{\infty} x^n$ ?

$(-1, 1)$
$[-1, 1)$
$[-1, 1]$
all real numbers

Show the answer

It is geometric with ratio $x$ ; at $x = \pm 1$ the terms do not approach 0.

Correct: $(-1, 1)$ :
$[-1, 1)$ :
$[-1, 1]$ :
all real numbers:

Part 4 · See it

See it first

The graph of y = e to the x from x = -3 to 2.5 with the first three Taylor polynomials about 0: the line 1 + x, the parabola 1 + x + x squared over 2, and the cubic that adds x cubed over 6. Each new term brings the polynomial closer to e to the x over a wider stretch; at x = 2 their values are 3, 5 and about 6.333, approaching e squared, about 7.389. — The partial sums $1 + x$ , $1 + x + \frac{x^2}{2!}$ , $1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}$ close in on $e^x$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Keep adding terms $\frac{f^{(n)}(a)}{n!}(x - a)^n$ without stoppingthe result is a power series, the Taylor series of $f$ about $a$ , and $P_n$ is its partial sum
The Taylor series converges and its partial sums approach $f(x)$ the series equals the function on its interval of convergence
Know four series by heart: $\frac{1}{1 - x}$ , $\sin x$ , $\cos x$ , $e^x$ many sums and limits can be read off, and other series can be built from them

Part 6 · Key ideas

Key ideas

Taylor series of $f$ about $a$ : $\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^n$ . About $a = 0$ it is a Maclaurin series.
$e^x = \sum \frac{x^n}{n!}$ , $\sin x$ $= \sum \frac{(-1)^n x^{2n+1}}{(2n+1)!}$ , $\cos x$ $= \sum \frac{(-1)^n x^{2n}}{(2n)!}$ , for every $x$ .
$\frac{1}{1 - x}$ $= \sum x^n$ $= 1 + x + x^2 + \cdots$ only for $-1 < x < 1$ .
A series is written with " $+ \cdots$ " or with $\sum$ and a general term; a Taylor polynomial is not.

Part 7 · Misconception

A common mistake

The wrong idea: The Taylor series for $e^x$ about $x = 1$ is $\sum \frac{(x - 1)^n}{n!}$ , the same coefficients as about 0.

What actually happens: The coefficients use derivatives at the center: $f^{(n)}(1) = e$ , so the series is $\sum \frac{e\,(x - 1)^n}{n!}$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

$\sin x$ and two of its Taylor polynomials

The graph shows

y = \sin x

(solid) with its Taylor polynomials about

x = 0

of degree 3,

P_3(x)

(dashed, the one that turns down past

x \approx 1.4

), and degree 5,

P_5(x)

(the one that stays close longer), for

-4 \le x \le 4

.

1. What does the graph show about the Taylor polynomials as the degree goes from 3 to 5?

The higher degree matches $\sin x$ exactly for $-4 \le x \le 4$
The higher degree stays close to $\sin x$ farther from 0
Both polynomials are closest to $\sin x$ near $x = \pm 4$
The lower degree is the better estimate near $x = 2$

Show the answer

Both polynomials match $\sin x$ near the center. $P_3$ turns away past about $|x| = 1.5$ , while $P_5$ stays close to about $|x| = 2.5$ : more terms, a closer fit over a wider interval.

The higher degree matches $\sin x$ exactly for $-4 \le x \le 4$ : $P_5$ still pulls away from $\sin x$ near $x = \pm 3$ ; no polynomial equals $\sin x$ on an interval.
Correct: The higher degree stays close to $\sin x$ farther from 0: Right: $P_5$ follows $\sin x$ farther out than $P_3$ .
Both polynomials are closest to $\sin x$ near $x = \pm 4$ : They are closest near the center, $x = 0$ , and pull away farther out.
The lower degree is the better estimate near $x = 2$ : Near $x = 2$ , $P_5$ is much closer to $\sin x$ than $P_3$ is.

2. Write $P_5(x)$ , the fifth-degree Taylor polynomial for $\sin x$ about $x = 0$ .

Type an expression for $P_5(x) =$ .

Show the answer

$\sin x$ $= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$ , so $P_5(x)$ $= x - \frac{x^3}{6} + \frac{x^5}{120}$ .

Answer: $P_5(x) =$ $x - \frac{x^3}{6} + \frac{x^5}{120}$

3. Which is the general term of the Maclaurin series for $\sin x$ ?

$\frac{(-1)^n x^{2n}}{(2n)!}$
$\frac{x^n}{n!}$
$\frac{(-1)^n x^{2n+1}}{2n + 1}$
$\frac{(-1)^n x^{2n+1}}{(2n+1)!}$

Show the answer

$\sin x$ $= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$ $= x - \frac{x^3}{3!}$ $+ \frac{x^5}{5!} - \cdots$ for every real $x$ .

$\frac{(-1)^n x^{2n}}{(2n)!}$ : Even powers give the series for $\cos x$ .
$\frac{x^n}{n!}$ : That is the series for $e^x$ .
$\frac{(-1)^n x^{2n+1}}{2n + 1}$ : The denominators are factorials, $(2n+1)!$ .
Correct: $\frac{(-1)^n x^{2n+1}}{(2n+1)!}$ : Right: odd powers, odd factorials, alternating signs.

4. According to the graph, at $x = 2$ is $P_3(2)$ above or below $\sin 2$ ?

Below: $P_3(2) = \frac{2}{3}$ , while $\sin 2 \approx 0.91$
Above: $P_3(2) = \frac{10}{3}$ , while $\sin 2 \approx 0.91$
Equal: a Taylor polynomial passes through $(2, \sin 2)$
Above: an odd-degree polynomial lies above $\sin x$ for $x > 0$

Show the answer

$P_3(2)$ $= 2 - \frac{2^3}{6}$ $= \frac{2}{3}$ $\approx 0.67$ , and $\sin 2 \approx 0.91$ , so $P_3(2)$ is below $\sin 2$ , as the graph shows.

Correct: Below: $P_3(2) = \frac{2}{3}$ , while $\sin 2 \approx 0.91$ : Right: the dashed curve is under the solid one at $x = 2$ .
Above: $P_3(2) = \frac{10}{3}$ , while $\sin 2 \approx 0.91$ : $P_3(2) = 2 - \frac{8}{6} = \frac{2}{3}$ ; subtract the cubic term.
Equal: a Taylor polynomial passes through $(2, \sin 2)$ : Taylor polynomials about 0 match $\sin x$ at the center, not at $x = 2$ .
Above: an odd-degree polynomial lies above $\sin x$ for $x > 0$ : Read the graph: at $x = 2$ the dashed $P_3$ is below $\sin x$ .

5. Find the exact sum of $\sum_{n=0}^{\infty} \frac{2^n}{n!}$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$e^x$ $= \sum_{n=0}^{\infty} \frac{x^n}{n!}$ for every $x$ , so with $x = 2$ the sum is $e^2$ .

Answer: $e^{2}$ ≈ 7.389

6. In the Taylor series for $e^x$ about $x = 1$ , find the coefficient of $(x - 1)^3$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Every derivative of $e^x$ is $e^x$ , so $f'''(1) = e$ and the coefficient is $\frac{e}{3!} = \frac{e}{6}$ .

Answer: $\frac{e}{6}$ ≈ 0.453

Part 9 · Summary

Summary

The Taylor series of

f

about

a

is

\sum \frac{f^{(n)}(a)}{n!}(x - a)^n

; its partial sums are the Taylor polynomials. The Maclaurin series of

e^x

,

\sin x

and

\cos x

converge to the function for every

x

, and

\frac{1}{1 - x} = \sum x^n

for

|x| < 1

.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

sin⁡x\sin x and two of its Taylor polynomials

Summary

What comes next

Connections

$\sin x$ and two of its Taylor polynomials