# Taylor Series

How to find the Taylor Series of a function: formula, 3 examples, and their solutions.

## Formula

### Taylor Series

The Taylor series is an approximation of y = f(x) near x = a.

The Taylor series 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.

(The Taylor series is a polynomial.)

This is how a calculator finds the value of a non-polynomial function.

### When x = 0: Maclaurin Series

For x = 0,

the Taylor series becomes quite simple.

This series is called the Maclaurin series.

The Maclaurin series is simpler than the Taylor series.

So the most of the problems you'll see

will say to find the Maclaurin series.

(= the Taylor series near x = 0.)

To make a summation,

find the pattern of the derivative parts:

f(0), f'(0), f''(0), f'''(0), ... .

This is the most important part.

## Example 1

### Example

### Solution

To find the pattern of the derivative parts,

start from finding f(0).

f(0) = 1

Find f'(0).

f(x) = e^{x}

So f'(x) = e^{x}.

Derivative of e^{x}

So f'(0) = 1.

Find f''(0)

f'(x) = e^{x}

So f''(x) = e^{x}.

So f''(0) = 1.

Find f'''(0)

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

So f'''(x) = e^{x}.

So f'''(0) = 1.

Find the pattern of the derivative parts.

f(0) = 1

f'(0) = 1

f''(0) = 1

f'''(0) = 1

The derivative parts are all 1.

Then the Taylor series of f(x) near x = 0 is

[1/0!]⋅x^{0} + [1/1!]⋅x^{1} + [1/2!]⋅x^{2} + ... + [1/n!]⋅x^{n} + ... .

Find the nth term of the polynomial

and make a summation.

The nth term is [1/n!]⋅x^{n}.

n starts from 0.

So the Taylor series is

the sum of [1/n!]⋅x^{n}

as n goes from 0 to infinity.

So this series is the answer.

This means

near x = 0,

e^{x} = [the sum of [1/n!]⋅x^{n} as n goes from 0 to infinity].

## Example 2

### Example

### Solution

To find the pattern of the derivative parts,

start from finding f(0).

f(0) = 0

Sine Values of Commonly Used Angles

Find f'(0).

f(x) = sin x

So f'(x) = cos x.

Derivative of sin x

So f'(0) = 1.

Cosine Values of Commonly Used Angles

Find f''(0).

f'(x) = cos x

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

Derivative of cos x

So f''(0) = 0.

Find f'''(0).

f''(x) = -sin x

So f'''(x) = -cos x.

So f'''(0) = -1.

Find f''''(0).

f'''(x) = -cos x

So f''''(x) = sin x.

So f''''(0) = 0.

f''''(x) and f(x) are both sin x.

So the derivative parts will repeat 0, 1, 0, -1.

f(0) = 0

f'(0) = 1

f''(0) = 0

f'''(0) = -1

f''''(0) = 0

The derivative parts will repeat 0, 1, 0, -1.

Then the Taylor series of f(x) near x = 0 is

[0/0!]⋅x^{0} + [1/1!]⋅x^{1} + [0/2!]⋅x^{2} + [-1/3!]⋅x^{3}

+[0/4!]⋅x^{4} + [1/5!]⋅x^{5} + [0/6!]⋅x^{6} + [-1/7!]⋅x^{7} + ... .

Remove the zero terms.

Then the series is

[1/1!]⋅x^{1} + [-1/3!]⋅x^{3} + [1/5!]⋅x^{5} + [-1/7!]⋅x^{7} + ... .

Find the nth term of the polynomial

and make a summation.

The derivative parts show 1, -1, 1, -1, ... .

So the derivative part of the nth term is (-1)^{n}.

The denominator parts show 1!, 3!, 5!, 7!, ... .

So the denominator of the nth term is (2n + 1)!.

The powers of x show x^{1}, x^{3}, x^{5}, x^{7}, ... .

So the power of x of the nth term is x^{2n + 1}.

So the nth term of the polynomial is

[(-1)^{n}/(2n + 1)!]⋅x^{2n + 1}.

So the Taylor series is

the sum of [(-1)^{n}/(2n + 1)!]⋅x^{2n + 1}

as n goes from 0 to infinity.

So this series is the answer.

This means

near x = 0,

sin x = [(-1)^{n}/(2n + 1)!]⋅x^{2n + 1} as n goes from 0 to infinity].

## Example 3

### Example

### Solution

To find the pattern of the derivative parts,

start from finding f(0).

f(0) = 1

Find f'(0).

f(x) = cos x

So f'(x) = -sin x.

So f'(0) = 0.

Find f''(0).

f'(x) = -sin x

So f''(x) = -cos x.

So f''(0) = -1.

Find f'''(0).

f''(x) = -cos x

So f'''(x) = sin x.

So f'''(0) = 0.

Find f''''(0).

f'''(x) = sin x

So f''''(x) = cos x.

So f''''(0) = 1.

f''''(x) and f(x) are both cos x.

So the derivative parts will repeat 1, 0, -1, 0.

f(0) = 1

f'(0) = 0

f''(0) = -1

f'''(0) = 0

f''''(0) = 1

The derivative parts will repeat 1, 0, -1, 0.

Then the Taylor series of f(x) near x = 0 is

[1/0!]⋅x^{0} + [0/1!]⋅x^{1} + [-1/2!]⋅x^{2} + [0/3!]⋅x^{3}

+[1/4!]⋅x^{4} + [0/5!]⋅x^{5} + [-1/6!]⋅x^{6} + [0/7!]⋅x^{7} + ... .

Remove the zero terms.

Then the series is

[1/0!]⋅x^{0} + [-1/2!]⋅x^{2} + [1/4!]⋅x^{4} + [-1/6!]⋅x^{6} + ... .

Find the nth term of the polynomial

and make a summation.

The derivative parts show 1, -1, 1, -1, ... .

So the derivative part of the nth term is (-1)^{n}.

The denominator parts show 0!, 2!, 4!, 6!, ... .

So the denominator of the nth term is 2n!.

The powers of x show x^{0}, x^{2}, x^{4}, x^{6}, ... .

So the power of x of the nth term is x^{2n}.

So the nth term of the polynomial is

[(-1)^{n}/2n!]⋅x^{2n}.

So the Taylor series is

the sum of [(-1)^{n}/2n!]⋅x^{2n}

as n goes from 0 to infinity.

So this series is the answer.

This means

near x = 0,

cos x = [(-1)^{n}/2n!]⋅x^{2n} as n goes from 0 to infinity].