# Taylor Series

See how to write e^{x}, sin x, cos x in taylor series.

3 examples and their solutions.

## Taylor Series

### Formula

f(x) = f(a)0!(x - a)

= ∑n = 0∞f

A Taylor series is an approximation of y = f(x) near x = a. ^{0}+ f'(a)1!(x - a)^{1}+ f''(a)2!(x - a)^{2}+ f'''(a)3!(x - a)^{3}+ ... + f^{(n)}(a)n!(x - a)^{n}+ ...= ∑n = 0∞f

^{(n)}(a)n!(x - a)^{n}It is useful because

you can approximate the value of a non-polynomial function

(trigonometric, exponential, logarithmic functions, etc)

by changing it to a polynomial function.

(Taylor series: Polynomial)

This is how a calculator finds the value

of a non-polynomial function.

f

^{(n)}: nth derivative

### Formula: Maclaurin Series

f(x) = f(0)0!x

= ∑n = 0∞f

A Maclaurin Series is the special case of a Taylor series^{0}+ f'(0)1!x^{1}+ f''(0)2!x^{2}+ f'''(0)3!x^{3}+ ... + f^{(n)}(0)n!x^{n}+ ...= ∑n = 0∞f

^{(n)}(0)n!x^{n}near x = 0.

### Example

Taylor series of e

Solution ^{x}near x = 0? f(x) = e

f(0) = e

= 1

f'(x) = e

f'(0) = e

= 1

f''(x) = e

f''(0) = e

= 1

f'''(x) = e

f'''(0) = e

= 1

...

f

f(x) = 10!x

= ∑n = 0∞1n!x

^{x}f(0) = e

^{0}= 1

f'(x) = e

^{x}- [1]f'(0) = e

^{0}= 1

f''(x) = e

^{x}f''(0) = e

^{0}= 1

f'''(x) = e

^{x}f'''(0) = e

^{0}= 1

...

f

^{(n)}(0) = 1f(x) = 10!x

^{0}+ 11!x^{1}+ 12!x^{2}+ 13!x^{3}+ ... + 1n!x^{n}+ ...= ∑n = 0∞1n!x

^{n}- [2][2]

Close

### Example

Taylor series of sin x near x = 0?

Solution f(x) = sin x

f(0) = 0 - [1]

f'(x) = cos x - [2]

f'(0) = 1

f''(x) = -sin x

f'(0) = -0

= 0

f'''(x) = -cos x

f'(0) = -1

f''''(x) = -(-sin x)

= sin x = f(x)

f''''(0) = 0

...

f

f(x) = 00!x

+04!x

= 11!x

= ∑n = 0∞(-1)

f(0) = 0 - [1]

f'(x) = cos x - [2]

f'(0) = 1

f''(x) = -sin x

f'(0) = -0

= 0

f'''(x) = -cos x

f'(0) = -1

f''''(x) = -(-sin x)

= sin x = f(x)

f''''(0) = 0

...

f

^{(n)}(0): 0, 1, 0, -1, ... - [3]f(x) = 00!x

^{0}+ 11!x^{1}+ 02!x^{2}+ (-1)3!x^{3}+04!x

^{4}+ 15!x^{5}+ 06!x^{6}+ (-1)7!x^{7}+ ...= 11!x

^{1}+ (-1)3!x^{3}+ 15!x^{5}+ (-1)7!x^{7}+ ... + (-1)^{n}(2n + 1)!x^{2n + 1}= ∑n = 0∞(-1)

^{n}(2n + 1)!x^{2n + 1}- [4][3]

Pattern of f

^{(n)}(0): 0, 1, 0, -1[4]

Find the nth term of f(x).

a

→ Write the Taylor series by using a

a

_{n}= [(-1)^{n}/(2n + 1)!]x^{2n + 1}→ Write the Taylor series by using a

_{n}.Close

### Example

Taylor series of cos x near x = 0?

Solution f(x) = cos x

f(0) = 1

f'(x) = -sin x

f'(0) = -0

= 0

f''(x) = -cos x

f'(0) = -1

f'''(x) = -(-sin x)

= sin x

f'''(0) = 0

f''''(x) = cos x = f(x)

f''''(0) = 1

...

f

f(x) = 10!x

+14!x

= 10!x

= ∑n = 0∞(-1)

f(0) = 1

f'(x) = -sin x

f'(0) = -0

= 0

f''(x) = -cos x

f'(0) = -1

f'''(x) = -(-sin x)

= sin x

f'''(0) = 0

f''''(x) = cos x = f(x)

f''''(0) = 1

...

f

^{(n)}(0): 1, 0, -1, 0 ...f(x) = 10!x

^{0}+ 01!x^{1}+ (-1)2!x^{2}+ 03!x^{3}+14!x

^{4}+ 05!x^{5}+ (-1)6!x^{6}+ 07!x^{7}+ ...= 10!x

^{0}+ (-1)2!x^{2}+ 14!x^{4}+ (-1)6!x^{6}+ ... + (-1)^{n}2n!x^{2n}= ∑n = 0∞(-1)

^{n}2n!x^{2n}- [1][1]

Find the nth term of f(x).

a

→ Write the Taylor series by using a

a

_{n}= [(-1)^{n}/2n!]x^{2n}→ Write the Taylor series by using a

_{n}.Close