Chapter Contents

• Infinite Series Expansion
• 1. Taylor Series
• 2. Maclaurin Series
• 3. How Does a Calculator Work?
• Infinite Series Expansion Problem Solver

• Comments, Questions?

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From the math blog...

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 '(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.

Answer

We plot our answer

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.

Answer

Exercise

Find the Maclaurin Series expansion of cos x.

Answer

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:

The value of this derivative when x = 0 is 1.

Similarly for the subsequent derivatives:

f ''(0) = 0

f '''(0) = -2

f ^iv(0) = 0

f ^v(0) = 24

Now we can substitute into the Maclaurin Series formula:

Considering that (see the 45-45 triangle)

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

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.

(We know now that π = 3.141 592 653 5...)

I wonder whether Leibniz was aware it was not such a good approximation...?