Good way of memorizing Taylor series for common functions?

Question

avuglantsaew · Accepted Answer

Some ways to manipulate series and functions

\begin{matrix} (1) & \frac{1}{1 - x} = \sum_{k = 0}^{\infty} x^{k}, | x | < 1 \end{matrix}

and

\begin{matrix} (2) & e^{x} = \sum_{k = 0}^{\infty} \frac{x^{k}}{k!}, for all x \end{matrix}

Formally if we integrate

(1)

we get

- \log (1 - x) = \sum_{k = 0}^{\infty} \frac{x^{k + 1}}{k + 1} = \sum_{k = 1}^{\infty} \frac{x^{k}}{k}, | x | < 1

and then

\log (1 + x) = - \sum_{k = 1}^{\infty} \frac{(- 1)^{k} x^{k}}{k}, | x | < 1

Computing

\arctan x

is similar since by

(1)

(\arctan x)^{'} = \frac{1}{1 + x^{2}} = \sum_{k \geq 0} (- 1)^{k} x^{2 k}

Also if we use

\sin x = (e^{i x} - e^{- i x}) / 2 i

in conjunction to

(2)

we get

2 i \sin x = \sum_{k = 0}^{\infty} \frac{(i x)^{k}}{k!} - \sum_{k = 0}^{\infty} \frac{(- i x)^{k}}{k!} = \sum_{k = 0}^{\infty} \frac{i^{k} (1 - (- 1)^{k}) x^{k}}{k!}

now

1 - (- 1)^{k} = 0

for even

k

, and

1 - (- 1)^{k} = 2

otherwise. So

2 i \sin x = 2 \sum_{k = 0}^{\infty} \frac{i^{2 k + 1} x^{2 k + 1}}{(2 k + 1)!} = 2 i \sum_{k = 0}^{\infty} \frac{(- 1)^{k} x^{2 k + 1}}{(2 k + 1)!}

where in the last step we used

i^{2 k + 1} = i \cdot i^{2 k} = i \cdot (- 1)^{k}

, and we reach the expansion

\sin x = \sum_{k = 0}^{\infty} \frac{(- 1)^{k} x^{2 k + 1}}{(2 k + 1)!}

szklanovqq · Accepted Answer

Easier way to remember the Taylor for the trig functions about

0

is to use the Taylor expansion for

e^{z}

and hyperbolic identities involving

e

.

s i n h (z) = \frac{1}{2} (e^{z} - e^{- z}) = \frac{1}{2} (\sum_{n = 0}^{\infty} \frac{z^{n}}{n!} - \sum_{n = 0}^{\infty} \frac{(- 1)^{n} z^{n}}{n!})

If you look at the

(- 1)^{n}

you will realise that all the even terms will cancel out leaving you with

\sum_{n = 0}^{\infty} \frac{z^{2 n + 1}}{(2 n + 1)!}

To get the expansion for sine just stick

(- 1)^{n}

in front of the summand
so expansion about

z = 0

will be

s i n (z) = \sum_{n = 0}^{\infty} \frac{(- 1)^{n} z^{2 n + 1}}{(2 n + 1)!}

Answered question