# 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.

We need to find the first, second, third, etc derivatives and evaluate them at x = 0. Starting with:

f(x) = sin x

f(0) = 0

First dervative:

f '(x) = cos x

f '(0) = cos 0 = 1

Second dervative:

f ''(x) = −sin x

f ''(0) = −sin 0 = 0

Third dervative:

f '''(x) = −cos x

f '''(0) = −cos 0 = −1

Fourth dervative:

f iv(x) = sin x

f iv(0) = sin 0 = 0

We observe that this pattern will continue forever.

Now to substitute the values of these derivatives into the Maclaurin Series:

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

We have:

sin\ x=0+(1)(x) +0 +(-1)/(3!)x^3 +0 +1/(5!)x^5 +0 +(-1)/(7!)x^7+...

This gives us:

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

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.

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

### Example 2: Expanding e^x

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

Recall that the derivative of the exponential function is f^'(x) = e^x. In fact, all the derivatives are e^x.

f '(0) = e0 = 1

f ''(0) = e0 = 1

f '''(0) = e0 = 1

We see that all the derivatives, when evaluated at x = 0, give us the value 1.

Also, f(0) = 1, so we can conclude the Maclaurin Series expansion will be simply:

e^x~~1+x+1/2x^2 +1/6x^3 +1/24x^4 +1/120x^5+...

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.

f(x) = cos x

f(0) = 1

First dervative:

f '(x) = −sin x

f '(0) = −sin 0 = 0

Second dervative:

f ''(x) = −cos x

f ''(0) = −cos 0 = -1

Third dervative:

f '''(x) = sin x

f '''(0) = sin 0 = 0

Fourth dervative:

f iv(x) = cos x

f iv(0) = cos 0 = 1

When substituting these values into the expansion formula, we obtain:

cos x=1-1/2x^2 +1/24x^4 -1/720x^6 +1/40320x^8-...

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

### Example 3

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)=3.142\ 591\ 654...

(We noe know 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).

### Example 4

Let's now find the value of pi using the same expansion, but this time we substitute x=1/sqrt(3). We'll use the fact tan (pi/6) = 1/sqrt(3).

Now arctan\ x ~~x-x^3/3+x^5/5-x^7/7+...

 ~~1/sqrt(3)-(1/sqrt(3))^3/3+(1/sqrt(3))^5/5-(1/sqrt(3))^7/7 +...

The value of pi will be 6 times the above value.

Using this expansion, we only need to use 18 steps to get 10 decimal place accuracy for pi. That is:

6sum_(n=0)^18((-1)^n)(1//sqrt(3))^(2n+1)/(2n+1) =6 xx 0.523\ 598\ 775\ 61 =3.141\ 592\ 653\ 6...

#### Why is it so much more accurate here?

The approximating curve in this case has what's called a radius of convergence (limited domain) where it "works" (that is, converges). Outside that domain, the approximation diverges.

We'll see another example of this situation in the Taylor and Maclaurin Series interactive applet.

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