Welcome to Calculus. 
I'm Professor Ghrist. We're about to begin Lecture 5 on 
Convergence. Every epic adventure involves facing some 
sort of adversity. It's time that we confronted a challenge. 
Every series involves adding together an infinite number of terms. 
This is dangerous. In this lesson, we'll confront some of 
technicalities associated with adding up an infinite number of terms. 
We'll find that along the way we encounter series that are not only novel, 
but extremely useful. There is clearly a problem with the 
principle that every function has a nice Taylor series about 0. 
It's simply not true. If we look at a function such as log of 
x, then of course, this function is not even defined at x equals 0. 
There's no way that this can be true. One of the reasons behind this is that 
polynomials are simply too nice or simple to capture all of the complexities of 
functions that we see. Another problem is it is dangerous to add 
together an infinite number of terms. One must be careful. 
Let's look at an example where adding together an infinite number of terms can 
lead to something good. Consider the sum 1 plus 1 half plus 1 
half squared plus 1 half cubed, et cetera. 
The sum k goes from 0 to infinity of 1 half to the k. 
As you may be able to guess, from geometry if nothing else, this sum 
converges and converges quite simply to the number 2. 
This is an example of something called a geometric series. 
You may have seen it before. The more general version of the geometric 
series Is 1 plus x, plus x squared, plus x cubed et cetera. 
The sum k goes from 0 to infinity of x to the k. 
The geometric series says that this equals 1 over 1 minus x, so that for 
example, if x equals 1 half, this sum converges to 2. 
How would you how would you prove something like this even for values for x 
not equal to 1 half? Well, let's see. 
Let us take this infinite sum and call it y. 
If we multiply this by x, then we obtain x plus x squared plus x cubed, et cetera. 
If we were to subtract that from our original series y, then we see that 
everything cancels except for the first term. 
So that y minus x times y equals 1. Factoring and solving for y gives that y 
equals 1 over 1 minus x. That is clear as far as an algebraic 
manipulation. But does this actually work? 
Does it make sense? What happens if we plug in x equals 1? 
Then we get 1 plus 1 plus 1 plus 1 plus 1. 
Clearly that is not a finite number. And 1 over 1 minus 1 doesn't even make 
any sense. That is undefined. 
Well, this derivation only holds for values of x less than 1 in absolute 
value, otherwise, this infinite series does not converge. 
Let's take a look at what happens at the other extreme. 
What if x equals minus 1? Is there really a problem here? 
Well, we would be asking for the sum 1 minus 1, plus 1, minus 1, et cetera. 
Maybe this does have a well-defined answer. 
If we paired up the terms, 1 minus 1 plus 1 minus 1, then we would seem to get 0, 
what's wrong with that? Well, if you could do that, then why not 
pair the terms a bit differently? And take 1 minus 0 minus 0 minus 0, so 
that the sum would be equal to 1. If you're not sure which of these is 
correct, perhaps, the best thing to do is to take the average, split the 
difference, and declare it to be 1 half. This is, after all, what one would obtain 
by plugging in a value of negative 1 into the expression 1 over 1 minus x. 
Well, as you might guess, none of these is correct. 
This series does not converge. The value of x equals minus 1 or for any 
value of x that is not less than 1 in absolute value. 
The moral of the story is that this and other Taylor series possess a convergence 
domain, on which the series is well-defined and well-behaved. 
Now, for many of the Taylor series that we've encountered thus far, the domain of 
convergence is the entire real line for the exponential function, sine, cosine, 
hyperbolic sine, hyperbolic cosine. There are no convergence issues. 
It is for series like the geometric series that we have to be more careful. 
Now, within the domain of convergence, you can manipulate the series at will. 
You can rearrange the terms. Differentiate term by term. 
Integrate. Recombine. 
Whatever you wish. There are no problems, as long as you are 
within the interior of that domain of convergence. 
You must be very careful about what happens near the borders. 
We'll talk more about that in Chapter 5. For the moment, I want to focus on what 
we can do. Let's take a look at another example. 
Compute the Taylor series of not log of x, but log of quantity 1 plus x. 
Maybe this will be not so bad, since for small values of x, this function is well 
defined. We could start taking derivatives of this 
function. But manipulating a known series is going 
to be a bit easier. One of the things that you may recall is 
that the log of 1 plus x is equal to the integral of 1 over 1 plus x, dx. 
If you remember that, great. If you don't, don't worry about it. 
Accept it as a given and let's move on from there. 
What could we do with that integrand, 1 over 1 plus x? 
That is really the geometric series evaluated at negative x, and so, we could 
integrate the sum k goes from 0 to infinity of negative x to the k with 
respect to x. By rearranging the terms, a thing that we 
are allowed to do as long as we are within the convergence domain, and 
pulling out the negative 1 to the k, then we get the sum, as k goes from 0 to 
infinity, of negative 1 to the k, times the integral of x to the k. 
That is x to the k plus 1 or k plus 1 if we add an arbitrary constant out in 
front, well, that's actually not going to be so bad. 
Why? When x equals 0 we obtain the log of 1 
which is 0. So forget that constant. 
And now, we have an answer with a bit of re-indexing. 
As the sum k goes from one to infinity, a negative 1 to the k plus 1 times x to the 
k over k. That is perhaps more easily remember as x 
minus x squared over 2 plus X cubed over 3, et cetera. 
Notice there are no factorials here. This is x to the k over k. 
Now are there any problems with this? Well, since we used the geometric series 
in this derivation, we must be careful to remain within the convergence domain for 
this. x must be less than 1 in absolute value. 
This fits well with the principle that Taylor polynomials approximate well only 
on the convergence domain. If we look at the terms of this series as 
an approximation to log of 1 plus x and we get increasingly something that seems 
to fit what the curve is doing. But only on the domain of convergence as 
x goes to negative 1 to 1, that is, we're looking at values of log from 0 to 2. 
Outside of this region these polynomial curves do not become better and better 
approximations. They in fact become worse and worse. 
Let us consider another example, this time the arctan function, one of the 
things that you may recall is that arctan has as its derivative 1 over 1 plus x 
squared. So, we can integrate this to obtain 
arctan. You will notice also that 1 over 1 plus x 
squared can be represented as the geometric series of negative x squared. 
So, that we can integrate the sum of negative x squared to the k. 
k goes from 0 to infinity. Again, by rearranging terms doing a 
little bit of work here. We see that we get the sum, k goes from 0 
to infinity, negative 1 to the k times the integral of x of 2k. 
That is x to the 2k plus 1 over 2k plus 1. 
We have to deal with the constant of integration. 
But again, since arctan of 0 equals 0 it is irrelevant. 
We therefore obtain the Taylor series, x minus x cubed over 3. 
Plus x to the 5th over 5, minus x to the 7th over 7, et cetera. 
Once again, no factorials in those denominators. 
And once again, it is important that x remain less than 1 in absolute value in 
order for this series to converge, because we used the geometric series as 
its basis. There is one last series that will be 
very helpful to us. This is the binomial series. 
It is the Taylor series for 1 plus x to the alpha, where alpha is a constant. 
The series is 1 plus alpha x plus 1 over 2 factorial times alpha times alpha minus 
1 times x squared. Well, there are lot of other terms here. 
Let me say simply that the kth order term is alpha choose k times x to the k, where 
by alpha choose k, I mean a binomial notation. 
This is the number alpha, times alpha minus 1, times alpha minus 2, all the way 
down to alpha minus k plus 1. Take that product and divide by k 
factorial. For example, if alpha equals 2, then what 
does the binomial series give you? 1 plus alpha times x plus 1 half alpha 
times alpha minus 1 times x squared and that simplifies to x squared. 
All other terms involve an alpha minus 2 hence the coefficients are 0. 
And we obtain, of course, what the Taylor series for this polynomial must be. 
When alpha equals negative 1, we're really looking at a variation of the 
geometric series for 1 over 1 plus x. This however, holds for all values of 
alpha. For example, if we take alpha equal to 1 
half. So that we're looking at the square root 
of 1 plus x. Then one can evaluate these binomial 
coefficients and obtain a Taylor series for that. 
In this case, in particular this is only going to remain valid for x less than 1 
in absolute value. We now have a full collection of Taylor 
series. Many of these have domains of convergence 
that are infinite, e to the x, cosine and sine, and cosh and sinh. 
These are the nicest Taylor series that you're going to run into in this course. 
However, from this lesson, we have another collection of Taylor series that 
are guaranteed to convert only when x is less than 1 in absolute value. 
This is the geometric series. The series for log of 1 plus x. 
The series for arctan and the binomial series. 
These one has to be a bit careful with. But as long as you are within the domain 
of convergence, these wonderful series. The moral of this lesson is that, so long 
as you are within the domain of convergence, you can do wonderful things. 
Check out the bonus material for this lecture for an additional example. 
In our next lesson, we'll consider what happens when you wish to focus your 
attention on some specific location, an expansion point.