# 2. Maclaurin Series

By M. Bourne

In the last section, we learned about Taylor Series, where we found an approximating polynomial for a particular function in the region near some value `x = a`.

We now take a particular case of Taylor Series, in the region near `x = 0`. Such a polynomial is called the **Maclaurin Series**.

The infinite series expansion for `f(x)` about `x = 0` becomes:

`f(x)~~f(0)+f^'(0)x+(f^('')(0))/(2!)x^2+(f^(''')(0))/(3!)x^3+(f^("iv")(0))/(4!)x^4+...`

`f^'(0)` is the first derivative evaluated at `x = 0`, `f^('')(0)` is the second derivative evaluated at `x = 0`, and so on.

[**Note: **Some textbooks call the series on this page "Taylor Series" (which they are, too), or "series expansion" or "power series".]

### Example: Expanding sin *x*

Find the Maclaurin Series expansion for `f(x) = sin\ x`.

We plot our answer

`sin\ x=x-1/6x^3` `+1/120x^5` `-1/5040x^7+...`

to see if the polynomial is a good approximation to `f(x) = sin\ x`.

We observe that our polynomial (in red) is a good approximation to `f(x) = sin\ x` (in blue) near `x = 0`. In fact, it is quite good between -3 ≤ *x* ≤ 3.

### Example: Expanding `e^x`

Find the Maclaurin Series expansion of `f(x) = e^x`.

## Exercise

Find the Maclaurin Series expansion of `cos\ x`.

## Finding Pi Using Infinite Series

In the 17th century, Leibniz used the series expansion of arctan *x* to find an approximation of π.

We start with the first derivative:

`d/(dx)arctan\ x=1/(1+x^2)`

The value of this derivative when `x = 0` is `1`. That is, `f^'(0)=1`.

Similarly for the subsequent derivatives:

`d^2/(dx^2)arctan\ x=-(2x)/((1+x^2)^2`

`f^('')(0)=0`

`d^3/(dx^3)arctan\ x=2(3x^2-1)/((1+x^2)^3`

`f^(''')(0)=-2`

`d^4/(dx^4)arctan\ x=-24x(x^2-1)/((1+x^2)^4`

`f^("iv")(0)=0`

`d^5/(dx^5)arctan\ x=24(5x^4-10x^2+1)/((1+x^2)^5`

`f^("v")(0)=24`

Now we can substitute into the Maclaurin Series formula:

`{:(arctan\ x, ~~0+(1)x+0/2x^2-2/(3!)x^3+0/(4!)x^4+24/(5!)x^5),(,=x-x^3/3+x^5/5-x^7/7+...):}`

Considering that (see the 45-45 triangle)

`arctan 1=pi/4`

we can substitute `x = 1` into the above expression and get the following expansion for π

`pi=4(1-1/3+1/5-1/7+...)`

All very well, but it was not a good way to find the value of π because this expansion converges very slowly.

Even after adding 1000 terms, we don't have 3 decimal place accuracy.

`4sum_(n=0)^1000((-1)^n)/(2n+1)=3.142591654...`

(We know now that π = 3.141 592 653 5..., and we know many other more efficient ways to find `pi`.)

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