# Taylor Series

A Taylor Series is an expansion of some function into an **infinite sum of terms**, where each term has a larger exponent like x, x^{2}, x^{3}, etc.

### Example: The Taylor Series for e^{x}

*e*^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + \frac{x^{5}}{5!} + ...

^{x}

^{2}/2! + x

^{3}/3! + ... etc

(Note: ! is the Factorial Function.)

Does it really work? Let's try it:

### Example: e^{x} for x=2

Well, we already know the answer is **e ^{2}** = 2.71828... × 2.71828... =

**7.389056...**

But let's try more and more terms of our infinte series:

Terms | Result | |
---|---|---|

1+2 | 3 | |

1+2+\frac{2^{2}}{2!} | 5 | |

1+2+\frac{2^{2}}{2!}+\frac{2^{3}}{3!} | 6.3333... | |

1+2+\frac{2^{2}}{2!}+\frac{2^{3}}{3!}+\frac{2^{4}}{4!} | 7 | |

1+2+\frac{2^{2}}{2!}+\frac{2^{3}}{3!}+\frac{2^{4}}{4!}+\frac{2^{5}}{5!} | 7.2666... | |

1+2+\frac{2^{2}}{2!}+\frac{2^{3}}{3!}+\frac{2^{4}}{4!}+\frac{2^{5}}{5!}+\frac{2^{6}}{6!} | 7.3555... | |

1+2+\frac{2^{2}}{2!}+\frac{2^{3}}{3!}+\frac{2^{4}}{4!}+\frac{2^{5}}{5!}+\frac{2^{6}}{6!}+\frac{2^{7}}{7!} | 7.3809... | |

1+2+\frac{2^{2}}{2!}+\frac{2^{3}}{3!}+\frac{2^{4}}{4!}+\frac{2^{5}}{5!}+\frac{2^{6}}{6!}+\frac{2^{7}}{7!}+\frac{2^{8}}{8!} | 7.3873... |

It starts out really badly, but it then gets better and better!

Try using "**2^n/fact(n)**" and **n=0** to 20 in the Sigma Calculator and see what you get.

Here are some common Taylor Series:

Taylor Series expansion | As Sigma Notation | |

| ||

| ||

| ||

(There are many more.)

## Approximations

We can use the first few terms of a Taylor Series to get an approximate value for a function.

Here we show better and better approximations for **cos(x)**. The red line is **cos(x)**, the blue is the approximation (try plotting it yourself) :

1 − x^{2}/2! | |

1 − x^{2}/2! + x^{4}/4! | |

1 − x^{2}/2! + x^{4}/4! − x^{6}/6! | |

1 − x^{2}/2! + x^{4}/4! − x^{6}/6! + x^{8}/8! |

You can also see the Taylor Series in action at Euler's Formula for Complex Numbers.

## What is this Magic?

How can we turn a function into a series of power terms like this?

Well, it isn't really magic. First we say we **want** to have this expansion:

f(x) = c_{0} + c_{1}(x-a) + c_{2}(x-a)^{2} + c_{3}(x-a)^{3} + ...

Then we choose a value "a", and work out the values c_{0} , c_{1} , c_{2} , ... etc

And it is done using **derivatives** (so we must know the derivative of our function)

### Quick review: a derivative gives us the slope of a function at any point.

These basic derivative rules can help us:

- The derivative of a constant is
**0** - The derivative of
**ax**is**a**(example: the derivative of**2x**is**2**) - The derivative of
**x**is^{n}**nx**(example: the derivative of^{n-1}**x**is^{3}**3x**)^{2}

We will use the little mark ’ to mean "derivative of".

OK, let's start:

To get c_{0}, choose x=a so all the (x-a) terms become zero, leaving us with:

f(a) = c_{0}

So **c _{0} = f(a)**

To get c_{1}, take the derivative of f(x):

f’(x) = c_{1} + 2c_{2}(x-a) + 3c_{3}(x-a)^{2} + ...

With x=a all the (x-a) terms become zero:

f’(a) = c_{1}

So **c _{1} = f’(a)**

To get c_{2}, do the derivative again:

f’’(x) = 2c_{2} + 3×2×c_{3}(x-a) + ...

With x=a all the (x-a) terms become zero:

f’’(a) = 2c_{2}

So **c _{2} = f’’(a)/2**

In fact, a pattern is emerging. Each term is

- the next higher derivative ...
- ... divided by all the exponents so far multiplied together (for which we can use factorial notation, for example 3! = 3×2×1)

And we get:

f(x) = f(a)_{} + \frac{f'(a)}{1!}(x-a) _{} + \frac{f''(a)}{2!}(x-a)^{2} + \frac{f'''(a)}{3!}(x-a)^{3} + ...

Now we have a way of finding our own Taylor Series:

For each term: take the next derivative, divide by n!, multiply by (x-a)^{n}

### Example: Taylor Series for cos(x)

Start with:

f(x) = f(a)_{} + \frac{f'(a)}{1!}(x-a) _{} + \frac{f''(a)}{2!}(x-a)^{2} + \frac{f'''(a)}{3!}(x-a)^{3} + ...

The derivative of **cos** is **−sin**, and the derivative of **sin** is **cos**, so:

- f(x) = cos(x)
- f'(x) = −sin(x)
- f''(x) = −cos(x)
- f'''(x) = sin(x)
- etc...

And we get:

cos(x) = cos(a) _{}− \frac{sin(a)}{1!}(x-a) _{} − \frac{cos(a)}{2!}(x-a)^{2} + \frac{sin(a)}{3!}(x-a)^{3} + ...

Now put **a=0**, which is nice because **cos(0)=1** and **sin(0)=0**:

cos(x) = 1 _{}− \frac{0}{1!}(x-0) _{} − \frac{1}{2!}(x-0)^{2} + \frac{0}{3!}(x-0)^{3} + \frac{1}{4!}(x-0)^{4} + ...

Simplify:

cos(x) = 1 − x^{2}/2! + x^{4}/4! − ...

Try that for sin(x) yourself, it will help you to learn.

Or try it on another function of your choice.

**The key thing is to know the derivatives of your function f(x).**

Note: A **Maclaurin Series** is a Taylor Series where **a=0**, so all the examples we have been using so far can **also** be called Maclaurin Series.