# 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 1: Expanding sin x

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

f(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.

Graph of the approximating Maclaurin Series polynomial, and the original f(x)=sin(x).

We observe that our polynomial (in grey) is a good approximation to f(x) = sin x (in green) near x = 0. In fact, it is quite good between -π ≤ x ≤ π.

### Example 2: Expanding e^x

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

Here's a graph of the answer we just obtained, and the original f(x) = e^x graph:

Graph of the approximating Maclaurin Series polynomial, and the original f(x)=e^x.

We can see the approximation is very close to the original f(x)=e^x in the region -2 < x < 2.5.

## Exercise

Find the Maclaurin Series expansion of cos x.

Here's a graph showing the answer we just obtained, and the original f(x) = cos(x) graph:

Graph of the approximating Maclaurin Series polynomial, and the original f(x)=cos(x).

Once again, we observe that our polynomial (in grey) is a good approximation to f(x) = cos x (in green) between -π ≤ x ≤ π.

## Finding Pi Using Infinite Series

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

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.)

Here's the graph of y = arctan x (in green) compared to the polynomial we just found (in grey).

Graph of the approximating Maclaurin Series polynomial, and the original f(x)=arctan(x).

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