Welcome to Calculus. I'm Professor Ghrist. We're about to begin Lecture
55 on Taylor Series Redux. Like Odysseus after his long wanderings, we finally come back home to where
we began with Taylor series. Having seen a lot of strange
sights along the way, we come back wiser,
with more experience, all the better for handling whatever challenges
we might face here at the end. In our last lesson,
we explored power series, that is series of the form sum
over n of a sub n x to the n. And we consider this as an operation for turning a sequence into a function, f(x). Now, the question arises,
can one invert this procedure and go from the function to the sequence? Well, of course you already
know the answer to this. It is, yes,
we've been doing this all semester. The answer is, of course, Taylor series. Taylor expansion is really
a means of turning a function into a sequence of coefficients which, when reconstituted into a power series,
gives you the original function back. In Taylor expansion, about, say, x = 0, we know that these
coefficients are the nth derivates of f at 0
divided by n factorial. But now, we can return to the issue
of convergence of Taylor series. Recall that at the beginning
of this course I said, don't worry about convergence too much. Just learn how to use Taylor series, and
we'll get to issues of convergence later. Well, it is now later. And it's now time to worry about
convergence of Taylor series. But now we can say something definite. The following theorem is crucial. If we have a power series of
the form a sub n x to the n, and we call that f(x), perhaps this
is a Taylor expansion about zero. Then, f converges absolutely for values of x that are within R of zero, where R is the radius of convergence, the limit as n goes to infinity of a sub
n over a sub n+1 in absolute value. Now if this were a Taylor
expansion about zero, we see that what really matters is how quickly the
derivatives at zero grow and get large. The ratio of subsequent derivatives
controls the convergence radius. Now more importantly, when you are on this
interval of convergence from -R to R, then not only does your function converge,
but it's differentiable. And the derivative is what you obtain
differentiating the series term by term. Moreover, f is integrable. And the integral within this domain of
convergence is exactly what you would get if you integrate the series term by term, oh, as long as you don't
forget the constant. Now what's so
important about a result like this? Well it tells us exactly what
the convergence domain is and allows us to manipulate the series. Let's see how that's useful. We have claimed in the past what the
Taylor series for arctangent looks like. Knowing the derivative of arctangent, we can express it as the integral
of dx over 1 + x squared. Using the geometric series, we can see that within that domain of
absolute convergence, what do we have? Arctangent is really the integral
of the sum over n of quantity (-x squared) to the n. Now, if we expand that out and
then integrate term by term,
then we get the arctangent as the sum. n goes from- to infinity
of -1 to the n over 2n + 1, times x to the 2n+1. There's a constant of integration, but that's equal to zero,
since arctangent of zero is zero. Therefore, we have our familiar series for
arctangent(x), x- x cubed over 3 + x to the 5th over 5,
etc. That works within the domain
of convergence from -1 to 1, based on the geometric series. Now, what happens at the end points? Well, when x = -1,
then it would seem to suggest that you get pi over 4
as this alternating sum, 1- a third + one-fifth- one-seventh. Does that actually converge? Well yes it does by
the alternating series test. However, this convergence is
conditional right at the boundary. There's really no end to what
one can do with this result. Let's begin with the geometric series and
differentiate it term by term. When we do so,
we obtain the sum over n of n times x to the n-1 = 1 over
(1-x) quantity squared. Now, let's multiply through
by x on both sides. That takes us back to a simpler
looking power series, that of the sum n x to the n. We see that that equals x
over (1- x) quantity squared. If we evaluate this at a particular x,
let's say x = one-tenth, what do we get? We get that this sum, one-tenth + two one-hundredths
+ plus three one-thousandths, etc., is really equal to the right-hand
side evaluated at one-tenth. That is, 10 over 81. This has an interesting
sort of decimal expansion. It's somewhat surprising that
this discrete integral has such a simple answer, but
of course we could keep going. We could differentiate once more and
then multiply through again by x, obtaining the result,
n squared, x to the n, summed over n, yields x times 1+ x,
over 1- x cubed. If we again evaluate this at one-tenth, then we get the not very
intuitive answer that one-tenth + four one-hundredths +
nine one-thousandths, etc., = 110 divided by 729. Such results are very easy to obtain. You will undoubtedly recall that Taylor
series can be useful in defining new functions. Consider the Fresnel Integrals, C(x) = the integral of cosine t
squared as t goes from 0 to x. And a complementary Fresnel integral, S(x) that is the same but
with sine instead of cosine. These functions are very useful
in optics and diffraction. But, it's not so
easy to come up with an antiderivative for sine or cosine of t squared. That's why these functions
have special names. Now you could try to
graph these functions. You would notice that they're oscillatory,
as is expected from their form. But how do you really
get your hands on them? Well, one way to do so
would be to use a Taylor series. Expand out cosine(t squared) or sine(t
squared) using the familiar formulae. And then, integrating term by term,
substituting in the values from 0 to x,
gives us power series formulae for these functions that are extremely useful,
especially for smaller values of x. We know enough now to answer the question
of when a Taylor series converges, but there's one question
we have not answered. How do we know that what
the Taylor series converges to is actually the function you began with? Let's say that we've expanded
f(x) about x equals a. Then how do we know that that
Taylor expansion converges to f? Well, it certainly must be
the case that f is smooth, that is,
all of the derivatives at a exist. Secondly, x must be within the domain
of convergence, that is, within R of a. Very good, but even with both of those
conditions satisfied, it's not enough. In fact, we make the definition
that f is a real analytic function at a if for all values of x close to a, the Taylor series exists and
converges to f(x). Now, why would we need such a definition? Some functions are not real-analytic. Consider the following example. Let f(x) be defined as
e to the -1 over x for x strictly positive and 0 otherwise. Now, your first clue that
something is funny is, if we write out the series definition for e to the -1 over x,
we don't get a power series. We get a power series in 1 over x. That's gonna give us problems
when x is near zero. So there's no way that we're going to be
able to compute the derivatives of this function by looking at
the coefficients of the series. So, let's do it the old fashioned way. Let's compute the derivative of
f at 0 by taking the limit as h goes to 0 of f(h)- f(0) over h. If we do this limit from the right,
then we can evaluate f(h) as e to the -1 over h. Now, f(0) is, of course, 0. And so we're left with the limit
as h goes to 0 from the right of e to the -1 over h divided by h. If we change variables and let t be
1 over h, then this is the same thing as the limit as t goes to to
infinity of t times e to the -t. Now that we know because
exponential beats polynomial. This is 0. So I claim that with a little bit
more work using a similar approach, you can show that all of the derivatives
of f at 0 are exactly 0. That means that if you take the Taylor
expansion of this function, it exists, and it is precisely 0. Every single coefficient
in the Taylor series is 0. However, the function
itself is positive for values of x that
are strictly bigger than 0. This is a smooth function,
but it not is real-analytic. This leads us to an image of
the universe of functions, beginning with the simplest functions. At the very core of this
universe lie the polynomials. These are themselves divided or
graded into different realms, beginning with the constants and
then the first order polynomials, the quadratics, the cubics, etc. Each filling out,
degree by degree, larger and larger subspaces of simple functions. But this is not all there is, since polynomials have power
series that eventually terminate. Beyond the space of polynomials
lie those functions whose Taylor series exist and
converge to the functions. That is the real-analytic functions,
like e to the x, sine of x, cosine of x, all those beautiful functions
we've been working with all term. However, beyond these still lie other functions,
more mysterious functions for which Taylor expansion is not
sufficient to describe them. One normally doesn't run into such things. That's why we haven't focused
too much attention on them. But you should know that they lie
out beyond the real-analytics. This leads us to our last image of
what Taylor expansion really is. Taylor expansion can be
thought of as projection to the space of polynomials,
where computing a Taylor polynomial is a projection to
one of these finite subspaces. Throwing away the higher ordered
terms really is a form of projection. My hope is that you now have a deeper
understanding of what Taylor series are, where they converge, and how they're
situated in the space of all functions. In our next lesson, we're going to
descend from these ethereal heights to tackle a more earthly problem,
that of approximation and error.