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Note: 1 or 1.5 lecture, §8.1 in [EP], §5.1 in [BD]

Many functions can be written in terms of a power series

If we assume that a solution of a diﬀerential equation is written as a power series, then perhaps we can use a method reminiscent of undetermined coeﬃcients. That is, we will try to solve for the numbers . Before we can carry out this process, let us review some results and concepts about power series.

As we said, a power series is an expression such as

(7.1) |

where and are constants. Let

denote the so-called partial sum. If for some , the limit

exists, then we say that the series (7.1) converges at . Note that for , the series always converges to . When (7.1) converges at any other point , we say that (7.1) is a convergent power series. In this case we write

If the series does not converge for any point , we say that the series is divergent.

is convergent for any . Recall that is the factorial. By convention we deﬁne . In fact, you may recall that this series converges to .

We say that (7.1) converges absolutely at whenever the limit

exists. That is, the series is convergent. If (7.1) converges absolutely at , then it converges at . However, the opposite implication is not true.

converges absolutely for all in the interval . It converges at , as converges (conditionally) by the alternating series test. But the power series does not converge absolutely at , because does not converge. The series diverges at .

If a power series converges absolutely at some , then for all such that (that is, is closer than to ) we have for all . As the numbers sum to some ﬁnite limit, summing smaller positive numbers must also have a ﬁnite limit. Therefore, the series must converge absolutely at . We have the following result.

Theorem 7.1.1. For a power series (7.1), there exists a number (we allow ) called the radius of convergence such that the series converges absolutely on the interval and diverges for and . We write if the series converges for all .

See Figure 7.1. In Example 7.1.1 the radius of convergence is as the series converges everywhere. In Example 7.1.2 the radius of convergence is . We note that is another way of saying that the series is divergent.

A useful test for convergence of a series is the ratio test. Suppose that

is a series such that the limit

exists. Then the series converges absolutely if and diverges if .

Let us apply this test to the series (7.1). That is we let in the test. Compute

Deﬁne by

Then if the series (7.1) converges absolutely. If , then the series always converges. If , then the series converges absolutely if , and diverges if . That is, the radius of convergence is . Let us summarize.

be a power series such that

exists. If , then the radius of convergence of the series is . Otherwise the radius of convergence is .

Example 7.1.3: Suppose we have the series

First we compute,

Therefore the radius of convergence is , and the series converges absolutely on the interval .

The ratio test does not always apply. That is the limit of might not exist. There exist more sophisticated ways of ﬁnding the radius of convergence, but those would be beyond the scope of this chapter.

Functions represented by power series are called analytic functions. Not every function is analytic, although the majority of the functions you have seen in calculus are.

An analytic function is equal to its Taylor series^{1} near a point . That is, for near we have

(7.2) |

where denotes the derivative of at the point .

For example, sine is an analytic function and its Taylor series around is given by

In Figure 7.2 we plot and the truncations of the series up to degree 5 and 9. You can see that the approximation is very good for near 0, but gets worse for further away from 0. This is what happens in general. To get a good approximation far away from you need to take more and more terms of the Taylor series.

One of the main properties of power series that we will use is that we can diﬀerentiate them term by term. That is, suppose that is a convergent power series. Then for in the radius of convergence we have

Notice that the term corresponding to disappeared as it was constant. The radius of convergence of the diﬀerentiated series is the same as that of the original.

Example 7.1.4: Let us show that the exponential solves . First write

Now diﬀerentiate

We reindex the series by simply replacing with . The series does not change, what changes is simply how we write it. After reindexing the series starts at again.

That was precisely the power series for that we started with, so we showed that .

Convergent power series can be added and multiplied together, and multiplied by constants using the following rules. First, we can add series by adding term by term,

We can multiply by constants,

We can also multiply series together,

where . The radius of convergence of the sum or the product is at least the minimum of the radii of convergence of the two series involved.

Polynomials are simply ﬁnite power series. That is, a polynomial is a power series where the are zero for all large enough. We can always expand a polynomial as a power series about any point by writing the polynomial as a polynomial in . For example, let us write as a power series around :

In other words , , , and all other . To do this, we know that for all as the polynomial is of degree 2. We write , we expand, and we solve for , , and . We could have also diﬀerentiated at and used the Taylor series formula (7.2).

Let us look at rational functions, that is, ratios of polynomials. An important fact is that a series for a function only deﬁnes the function on an interval even if the function is deﬁned elsewhere. For example, for we have

This series is called the geometric series. The ratio test tells us that the radius of convergence is . The series diverges for and , even though is deﬁned for all .

We can use the geometric series together with rules for addition and multiplication of power series to expand rational functions around a point, as long as the denominator is not zero at . Note that as for polynomials, we could equivalently use the Taylor series expansion (7.2).

Example 7.1.5: Expand as a power series around the origin () and ﬁnd the radius of convergence.

First, write . Now we compute

where using the formula for the product of series we obtain, , , , etc…. Therefore

The radius of convergence is at least 1. We use the ratio test

So the radius of convergence is actually equal to 1.

When the rational function is more complicated, it is also possible to use method of partial fractions. For example, to ﬁnd the Taylor series for , we write

Exercise 7.1.6: Determine the Taylor series for around the point , and ﬁnd the radius of convergence.

Exercise 7.1.8: Determine the Taylor series and its radius of convergence of around . Hint: You will not be able to use the ratio test.

Exercise 7.1.10: Suppose that the ratio test applies to a series . Show, using the ratio test, that the radius of convergence of the diﬀerentiated series is the same as that of the original series.

Exercise 7.1.102 (challenging): Is the power series convergent? If so, what is the radius of convergence?

Exercise 7.1.105 (challenging): Imagine and are analytic functions such that for all large enough . What can you say about ?

^{1}Named after the English mathematician Sir Brook Taylor (1685–1731).