Introduction to Series

So far in this course, we've seen lots of examples of functions: polynomials, exponentials and trigonometric to name a few. Then we could combine functions in ways such as composition to create some quite complicated functions.

But the easiest to work with have always been polynomials like $ p(x) = x^4 + 3x^2 - 10x + 2 $ . They are easy to evaluate (all we need is a four function calculator) and to differentiate and integrate. Wouldn't it be nice if every function were a polynomial? Of course, lots of functions are not polynomials, but we'll see how we can approximate most of the functions we've seen in this course by polynomials. It oftentimes turns out that this approximation gives us all the information we really want anyway.


The Geometric Series: an example

We'll first consider an example we studied at the very beginning of the term: the geometric series. Here we wanted to evaluate a sum like

\[ S_n = 1 + r + r^2 + r^3 + \ldots + r^n \]

One fast way to evaluate this sum is to multiply it by the ratio r . This leads to

\[ \begin{array}{rcl} S_n = & 1+ & r + r^2 + \ldots + r^n \\ rS_n = & & r + r^2 + \ldots + r^n + r^{n+1} \end{array} \]

Now subtract the second line from the first and notice that there are lots of cancelations: $ (1-r)S_n = 1-r^{n+1} $ or

\[ S_n = \frac{1-r^{n+1}}{1-r} \]

Let's consider what happens when $ r = \frac 12 $ . In this case,

\begin{eqnarray*} S_n & = & \frac{1-(\frac 12)^{n+1}}{1-\frac 12} \\ & = & 2( 1 - (\frac 12)^{n+1}) \\ & = & 2 - \frac{1}{2^n} \end{eqnarray*}

The demonstration below will show you this sum for different values of n .

The thing to notice here is what happens when you add in more and more terms: you can see that the sum gets closer and closer to 2. You can see this from our expression above as well:

\[ \lim_{n\to\infty} S_n = \lim_{n\to\infty} 2 - \frac 1{2^n} = 2 \]

In this case, we write

\[ \sum_{n=0}^\infty (\frac 12)^n = 1 + \frac 12 + (\frac 12)^2 + \ldots = 2 \]

Notice that when $ r = 2 $ , we have

\[ S_n = \frac{1-2^{n+1}}{1-2} = 2^{n+1} - 1 \]

Here you can see that as n grows larger, the sum continues to grow indefinitely large. In this case, this sum does not converge.

This is an important point: we are dealing with the sum of an infinite number of terms. What we have seen so far is that sometimes this sum will converge to a finite result. However, other times the sum will not converge and instead exhibit a less friendly behaviour such as growing indefinitely large.

More generally, if $ |r| < 1 $ , we have $ \lim_{n\to\infty} r^{n+1} = 0 $ and so

\[ \sum_{n\to\infty}r^n = 1 + r + r^2 +\ldots = \frac 1{1-r} \]

This is a very interesting expression: to study it further, we let the ratio r vary. Here we will write this ratio as x instead of r and we will write the equality above in the reverse order. This means that

\[ \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots \hspace{.25in} \mbox{ when } |x| < 1 \]

This is a remarkable relationship: it is telling how to write the function on the left as an "infinite polynomial" which we call the Taylor series based at $ c = 0 $ .

Taylor series are a bit unfamiliar at this stage, but polynomials aren't. Notice that the Taylor series provides us with a sequence of polynomials:

\begin{eqnarray*} T_0(x) & = &1 \\ T_1(x) & = & 1+ x \\ T_2(x) & = & 1+ x +x^2\\ T_n(x) & = & 1+ x + x^2 + \ldots + x^n\\ \end{eqnarray*}

We call this last polynomial the Taylor polynomial of degree n.

The following demonstration shows that, for $ |x| < 1 $ , the Taylor polynomials provide a better and better approximation for the function $ \frac{1}{1-x} $ as n gets larger and larger.

In fact, this is one way to think of the expression we wrote above:

\[ \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots \]

The series on the right hand side is the result of considering the polynomials with increasingly large values of n. Of course, we could never actually add all the terms up in the series: there are infinitely many of them so our job would never be done. However, this expression says that, by taking an increasingly large degree in the Taylor polynomial, we obtain a better and better approximation for the original function. This is an idea we will explore in much more detail in the near future.


Convergence

You may wonder about the restriction that $ |x| < 1 $ in the expression:

\[ \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots \]

Let's consider what happens when $ x = 1 $ : in this case, the left hand side is $ \frac{1}{1-1} $ which is undefined. Consider the right hand side: it is telling to perform the sum $ 1 + 1 + 1 + \ldots $ . Clearly this sum is never going to converge.

Also, when $ x = -1 $ , we have the sum: $ 1 + (-1) + 1 + (-1) + \ldots $ . If we imagine performing the addition, we add the first two terms to get zero. Then we add in the third term to obtain 1. Adding in the fourth term gives zero for the sum again. You can see that the sum is going to keep flipping back and forward between 0 and 1. Here again, the sum is never going to converge.

In this example, we need the restriction $ |x| < 1 $ so that the infinite sum will actually converge to a finite result. This example exhibits two ways in which a series can diverge: first, the sum may large indefinitely large. Secondly, the sum could stay bounded but just not settle down near any one number. This is an important point which we will study in more detail.


Summary

So far in this section, we've just considered an example. It shows us that a somewhat complicated function, $ \frac{1}{1-x} $ , can be approximated by a sequence of polynomials. We can think of this sequence of polynomials as being an "infinite polynomial" which we called the Taylor series.

One complication is that the series may not converge for some values of x; that is, the infinite sum given to us by the Taylor series may not be computable. If this is the case, we cannot hope to express the original function in terms of the series like we did above. To understand this phenomenon a bit better, we are going to have to think about what kind of infinite sums we can actually perform.