# 1. Taylor Series

By M. Bourne

Our aim is to find a polynomial that gives us a good approximation to some function. (See why we want to do this in the Introduction.)

We find the desired polynomial approximation using the **Taylor Series**.

If we want a good approximation to the function in the region near `x = a`, we need to find the first, second, third (and so on) derivatives of the function and substitute the value of* a.* Then we need to multiply those values by corresponding powers of `(x − a)`, giving us the **Taylor Series expansion** of the function `f(x)` about `x = a`:

`f(x)` `~~f(a)+f’(a)(x-a)+(f’’(a))/(2!)(x-a)^2``+(f’’’(a))/(3!)(x-a)^3``+(f^("iv")(a))/(4!)(x-a)^4``+...`

### Getting Lost?

See some background to why this sum converges to a polynomial in Infinite Geometric Series from an earlier chapter.

We can write this more conveniently using summation notation as:

`f(x)~~sum_(n=0)^oo(f^((n))(a))/(n!)(x-a)^n`

### Conditions

In order to find such a series, some conditions have to be in place:

- The function `f(x)` has to be
**infinitely differentiable**(that is, we can find each of the first derivative, second derivative, third derivative, and so on forever). - The function `f(x)` has to be defined in a region near the value `x = a`.

Let's see what a Taylor Series is all about with an example.

### Example - Expansion of ln *x*

Find the Taylor Expansion of `f(x) = ln x` near `x = 10`.

Next, we move on to the Maclaurin Series, which is a special case of the Taylor Series (and easier :-).

### Search IntMath, blog and Forum

### Online Algebra Solver

This algebra solver can solve a wide range of math problems.

Go to: Online algebra solver

### Calculus Lessons on DVD

Easy to understand calculus lessons on DVD. See samples before you commit.

More info: Calculus videos

### The IntMath Newsletter

Sign up for the free **IntMath Newsletter**. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!