Maclaurin series of eˣ, sin x, cos x and 1/(1 − x)

Warm-up

Three quick questions from earlier lessons.

Discover

Every derivative of $e^x$ at 0 is 1. So its Maclaurin series is $\displaystyle\sum \dfrac{x^n}{n!}$ . What about $\sin x$ , whose derivatives at 0 cycle $0, 1, 0, -1$ ?

$x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dotsb$
$1 + x + \dfrac{x^2}{2!} + \dotsb$

Key idea

Four series to know by heart.

e^x = \displaystyle\sum_{n \ge 0} \dfrac{x^n}{n!}

;

\sin x = \displaystyle\sum_{n \ge 0} \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}

;

\cos x = \displaystyle\sum_{n \ge 0} \dfrac{(-1)^n x^{2n}}{(2n)!}

(all

x

);

\dfrac{1}{1 - x} = \displaystyle\sum_{n \ge 0} x^n

(

|x| < 1

).

Worked example

The first terms of $e^{-x^2}$ .

1 + (-x^2) + \dfrac{(-x^2)^2}{2!} + \dotsb

Substitute u = −x² into eᵘ.

1 - x^2 + \dfrac{x^4}{2} - \dotsb

Simplify.

Your turn, guided

Your turn: the coefficient of $x^2$ in $\cos(3x)$ .

-\tfrac{9}{2}

−(3x)²/2!.

Practice ★

Find the coefficient of $x^3$ in the Maclaurin series of $e^{ 5x}$ .
Find the coefficient of $x^3$ in the Maclaurin series of $e^{ 6x}$ .

$\displaystyle\sum_{n=0}^{\infty} \dfrac{(-1)^n (1x)^{2n+1} }{(2n + 1)!}$ is the Maclaurin series of which function?
1. $\sin(1x)$
2. $\cos(1x)$
3. $e^{ 1x}$

Watch out!

Practice ★★

Find the coefficient of $x^4$ in the Maclaurin series of $\cos(2x)$ .
Find the coefficient of $x^4$ in the Maclaurin series of $\cos(5x)$ .

Exit ticket

Three questions to check you have got it.

Find the coefficient of $x^3$ in the Maclaurin series of $e^{ 5x}$ .
Find the coefficient of $x^3$ in the Maclaurin series of $e^{ 6x}$ .
Find the coefficient of $x^3$ in the Maclaurin series of $e^{ 4x}$ .

Summary card

Known Maclaurin series

$e^x$ , $\sin x$ (odd), $\cos x$ (even), $\dfrac{1}{1 - x}$ .

Watch out: signs alternate in sin and cos.

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