Taylor Polynomials As Approximations and the Error Formula:
We have seen that Taylor polynomials give a way of approximating a function for values of near some specific value , and that in some cases, higher and higher degree Taylor polynomials give better and better approximations to the function . For example, here are some Taylor polynomials approximating near :
As the degree of the Taylor polynomial increases, the approximation to sine becomes better and better, and the polynomial is a reasonable approximation to the function over a larger and larger interval.
In contrast, here are some Taylor polynomials approximating near :
As the degree of the Taylor polynomial increases, the approximation to becomes better and better, but only for values of between and ; outside of this interval, we don't seem to be getting a good approximation to the function.
We have seen how to use the error formula for Taylor polynomial approximations to determine whether the Taylor polynomials really do give better and better approximations to as gets large. For the function , we can use the error formula to show that for every value of ,
The error formula is rather unwieldy and difficult to apply. Today we will learn a new method of finding the values of for which
Taylor Series: Infinite Taylor Polynomials:
First, some terminology. Recall that we can write
If, for some particular value of , the values of approach a limit as ,
We saw from the error formula that for , this Taylor series actually gives the values of the function,
For most of the functions that we know, the Taylor series either gives us the function , as the Taylor series for around does when , or else diverges (doesn't give us a number at all), as the Taylor series for around does when . In the next section we will see a method that can give us an interval of 's for which the Taylor series is guaranteed to converge (to give us an answer.)
Notice that this method will guarantee that the Taylor series converges, NOT that the thing it converges to is the function . Almost always, that is, for virtually all the functions we meet in this course, if the Taylor series converges to anything then it converges to . To be absolutely sure of that for any one particlar function, we'd have to go back to the error formula (or to use some facts we'll learn later about how you can build new Taylor series from old ones.)
Infinite Sums and the Ratio Test:
You may have been staring at the equation
This last example is a particular case of a geometric series, a series in which each term is gotten from the preceding one by multiplying by a fixed number , in this case . Putting this another way, if we write the series as
It turns out that for the same reason this series converges (equals a finite number) when , it converges whenever . And it also turns out, amazingly enough, that we don't need to know that
The Ratio Test: To find out whether an infinite series
Let's see how we can apply the ratio test to Taylor series. First we'll look at the Taylor series for around ,
(In general, if we want to tell whether a Taylor series converges for the values of for which , we need to know some other convergence tests for infinite series besides the ratio test. We will not cover that topic in this course. Chapter 9 of the textbook develops some of the general theory of infinite series.)
Now let's look at the Taylor series for around ,
Of course, in both these cases, the ratio test does not tell us what value the Taylor series converges to. We need to go back to the error formula for Taylor polynomials to see that it does converge to the function itself.
Radius of Convergence:
For , the Taylor series around converged for all ; for , it converged for . In both cases, the Taylor series converged for in some interval centered at . In the case of , it was an infinite interval, .
>From the ratio test, we can see that something like this always happens. Let's look at a Taylor series,
So we have , and the ratio test tells us that if , or , then the series converges, and if , or , then the series diverges. (In the case where , we have convergence for all , and in the case where , we have convergence only for .) In other words, there is a radius of convergence , which may be zero or infinity or anything in between, such that if then the Taylor series converges and if then the Taylor series diverges. The point , being the center of the interval of 's for which the series converges, is sometimes called the center of convergence.
For (about ) the radius of convergence was infinity, and for (about ) the radius of convergence was 1.
Let's do one more example. Let's find the radius of convergence of the Taylor series for around . It turns out that there is a slick way to find the Taylor series: Just replace by in the Taylor series for . So we get
A Secret Trick Using Complex Numbers:
It isn't surprising that the radius of convergence for the Taylor series for about has radius of convergence 1. The function goes off to infinity at , so we might expect the Taylor series to be stymied at that point. But what about ? That is a perfectly good function on the whole real number line. Why doesn't its Taylor series work for every ?
The answer is that Taylor series are good not only for real numbers, but also for complex numbers. And the radius of convergence works the same way, except now we're talking modulus instead of absolute value. The Taylor series for about converges for all complex values of for which , and diverges when . In the complex plane, we have this picture:
The circle around the point consists of all complex numbers with modulus equal to 1. The Taylor series converges inside this circle, and diverges outside it.
And now we can see why this circle can't be pushed outward any farther: The function goes off to infinity at , and that's what stops the Taylor series from converging inside any larger circle.
This always happens for functions like this. Say is a quotient of polynomials. To find the radius of convergence for the Taylor series for about , find all the points in the complex plane where the denominator is zero, so goes off to infinity.1 The radius of convergence will be the radius of the largest circle with center that does not surround any of those points at which the function goes off to infinity.
This fact is beyond the purview of this course and you're not ``responsible'' for it. If you take a course in complex analysis, you'll learn about this, and other fun things as well.