# 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`.

Answer

Recall the natural logarithm, `ln x`. Recall also the graph of *y* = ln *x*:

Graph of `f(x)=ln(x)`.

Our aim is to find a good polynomal approximation to the curve in the region near *x* = 10.

We need to use the Taylor Series with *a* = 10.

The first term in the Taylor Series is *f*(*a*). In this example,

`f(a)` ` = f(10) = ln 10 = 2. 302\ 585\ 093.`

Now for the derivatives.

Recall the derivative of ln *x*, which is `1/x`. So

`f'(x)=1/x`

We need `f'(10)`, which is `1/10 = 0.1`.

Now for the second derivative:

`f''(x)=-1/x^2`

At `x = 10`, this has value `-0.01`.

`f'''(x)=2/x^3`

The third derivative at `x = 10` has value `0.002`.

`f^("iv")(x)=-6/x^4`

At `x = 10`, this has value `-0.0006`.

You can see that we could continue forever. This function is infinitely differentiable.

Now to substitute these values into the Taylor Series:

`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+...`

We have:

`f(x)~~f(10)` `+f’(10)(x-10)` `+(f’’(10))/(2!)(x-10)^2` `+(f’’’(10))/(3!)(x-10)^3` `+(f^("iv")(10))/(4!)(x-10)^4+...`

`~~2.302585093+0.1(x-10)` `+(-0.01)/(2!)(x-10)^2` `+(2xx0.001)/(3!)(x-10)^3` `+(-6xx0.0001)/(4!)(x-10)^4+...`

Expanding this all out and collecting like terms, we obtain the polynomial which approximates `ln x`:

`ln x ~~ 0.21925 ` `+ 0.4x` ` − 0.03x^2 ` `+ 0.00133x^3` ` − 0.000025x^4+ ...`

This is the approximating polynomial that we were looking for.

We see from the graph that our polynomial (in grey) is a good approximation for the graph of the natural logarithm function (in green) in the region near `x = 10`.

Graph of the approximating polynomial, and `f(x)=ln(x)`.

Notice that the graph is not so good as we get further away from `x = 10`. The regions near `x = 0` and `x = 20` are showing some divergence.

Let's zoom out some more and observe what happens with the approximation:

Graph of the approximating polynomial, and `f(x)=ln(x)`, zoomed further out.

Clearly, it is no longer a good approximation for values of *x* less than 3 or greater than 20. How do we get a better approximation? We would need to take more terms of the polynomial.

Easy to understand math videos:

MathTutorDVD.com

Don't miss the Taylor and Maclaurin Series interactive applet where you can explore this concept further.

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

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