Welcome to calculus. 
I'm Professor Greist. We're about to begin Lecture 6 on 
Expression Points. We've seen that the Taylor series 
provides a good approximation to a function near zero. 
What happens if we wish to focus our attention at some other point. 
In this lesson, we'll consider changing the expansion point. 
This will lead us to a broader definition and interpretation of Taylor series. 
As we have seen, Taylor expansion gives an excellent way to approximate a 
function for inputs near zero. However, in many applications, zero is 
not the most interesting input that you can think of. 
Many examples of this, besides finance and economics, where we care about 
nonzero values. What would be nice is a way to 
Taylor-expand a function that works well for inputs that are not necessarily close 
to zero. There certainly is. 
This is an important definition. The Taylor series of f at x equals a, is 
the sum, k goes from 0 to infinity, the kth derivative of f, evaluated at a, 
divided by k factorial times quantity x minus a to the k. 
This is not a polynomial in x, but rather a polynomial in the quantity x minus a, 
where the coefficient in front of each monomial term is the kth derivative of f 
evaluated at a, and then divided by k factorial. 
A little bit of terminology is in order. The constant term is called the zeroth 
order term. Next, comes the first order term, the 
second order term, the third order term, and so on. 
These correlate to the degree of the quantity x minus a. 
If we were to change things, perform the change of variables defining h to be the 
quantity x minus a. Then the above series would become a 
polynomial series in h. However, it would not be giving you f of 
h, but rather f of x, which is a plus h. This series in h is telling you that if 
you want to know values of f that are close to a, then you can substitute in a 
small value for h. And if you're looking for an 
approximation, you can ignore some of the higher order terms. 
Let's look at an example. Compute the Taylor series of log of x. 
Now, we know that we cannot do that about x equals 0. 
So let us do it about a more natural value. 
Let's say about x equals 1. To compute this Taylor expansion, we're 
going to need to start taking some derivatives. 
If we looked at the function log of x, the zeroth order term is obtained by 
evaluating at x equals 1, and that gives us, simply 0. 
The derivative of log of x, you may recall is one over x. 
Evaluating that of x equals 1 gives us a coefficient 1. 
Now, taking the derivative of one over x, gives us minus x to the negative 2. 
We evaluate that and continue differentiating. 
As we go, evaluating each term at x equals 1. 
After computing sufficiently many derivatives, you start to see a pattern. 
It requires a little bit of thinking, and a particularly formal way of thinking 
called induction. But, with a bit of effort, one can 
conclude that the kth derivative of log of x is negative 1 to the k plus 1 times 
k minus 1 factorial times x to the negative k. 
Evaluating that at x equals 1 gives us simply the coefficient, negative one to 
the k plus 1 times k minus 1 factorial. Now, to get the full Taylor Series, we 
need to divide this coefficient by k factorial. 
In so doing, we see a not too unfamiliar series. 
For log of x as x minus 1, minus x minus 1 squared over 2, plus x minus 1 cubed 
over 3, et cetera. The coefficient in front of the degree k 
term is negative 1 to the k plus 1, over k. 
And our summation goes from 1 to infinity. 
Indeed, if we let h be x minus 1, then we obtain the very familiar series, log of 1 
plus h equals sum k goes from 1 to infinity, negative 1 to the k plus 1, h 
to the k over k. We've seen that before with an x instead 
of an h, but it works the same. Do keep in mind that Taylor Series are 
not guaranteed to converge everywhere. Indeed, if we look at the terms for log 
of x and take a finite Taylor polynomial, then the higher and higher degree terms 
only provide a reasonable approximation to log of x within the domain of 
convergence. We know from our last lesson that that 
domain is going to be for values of x between 0 and 2. 
Outside of that, these terms are providing worse and worse approximations 
of log of x. What do we do if we want to approximate 
log of x outside of this domain? Well, you need to do a Taylor expansion 
about a different point, someplace close to where you want to approximate. 
It's a bit easy to get confused with all of the different notation associated with 
Taylor series. So, let's review. 
One way to think about the Taylor expansion about x equals a is to write f 
of x as a series in the quantity x minus a. 
Another way to do it is to write it as a series in the quantity h, where h is 
equal to x minus a. In which case, you're coming up with an 
approximation for f of a plus h. One way to think about this is that the 
more derivatives of f you know at a point a, the better an approximation you get at 
x or at a plus h. This is a perspective that you're 
going to want to keep with you for the remainder of this course. 
The principle is that successive polynomial truncations of the Taylor 
series approximate increasingly well. We've seen that before. 
In this lesson, the point is that where you do the Taylor expansion matters. 
If you are expanding way over here and trying to get information about way over 
there, then you're going to need a lot of derivatives to do that. 
On the other hand, if you approximate about the correct expansion point, you 
might not need so many derivatives in order to get the job done. 
Let's look at an explicit example. What would it take to estimate the square 
root of 10? Well, we would have to look at the Taylor 
series of the function square root of x. If we expand that about a point, x equals 
a, then I'll leave it to you that the first few derivatives work out to a 
Taylor series of root a plus one over 2 root a times quantity x minus a minus 1 
over 8, times the square root of a cubed, times quantity x minus a squared plus 
some higher-order terms in x minus a. Well, to compute the square root of 10, 
what are we going to do? Let's say I expand about 1 because 1 is a 
simple value. I know the square root of 1, that is 
simply 1. That makes the coefficients of the Taylor 
expansion easy to compute. On the other hand, when I'm actually 
trying to estimate the square root of 10, based on this, then I get 1 plus one half 
times 9 minus one eighth times 9 squared or 81, plus higher order terms. 
How good of an approximation is that? Well, that gives me a value of negative 
4.125. Now, I know this is not the square root 
we're looking for. That is a bad approximation. 
So, what if we were to compute an expansion about x equals 9? 
Then, this is something for which I also know the square root. 
The square root of 9 is 3. The coefficients are easy to work with. 
And if I plug in a value of x equals 10 into this Taylor series, I get an 
approximation of 3 plus one sixth minus 1 over 216. 
This gives an approximation of 3.1620 some other stuff. 
The true answer agrees with this up to the first four digits. 
The expansion point, in this case, certainly matters. 
Now, one thing to be cautious of, is that if you're computing a Taylor series of a 
composition, you must expand about the correct values. 
If you have a function, f composed with g, and you want to expand it about some 
input x. Then, you must expand g about x. 
But you must expand f, not about x, but about g of x. 
And it is that term in particular that causes problems. 
In an explicit example, we'll be able to see how this works. 
Compute the Taylor series of e to the cosine of x about x equals zero. 
Well, even the cosine x is a composition. What do you what do you do first? 
First, you take the cosine of x, then you exponentiate it. 
If we are to compute the Taylor series about x equals 0, then we must expand 
cosine about 0, but not e to the X. Cosine of X about 0 is very simple. 
This one, we know. However, for the second term, the 
exponential, we must expand that about an input of 1 because that is what gets fed 
into it. Cosine of 0 is 1. 
Well, the Taylor series of e to the u about u equals 1 is easy to compute. 
There's nothing difficult there. However, what we must then do is 
substitute into it this series for cosine of x. 
That is u is 1 minus x squared over 2 plus x to the 4th over 4 factorial plus 
higher ordered terms. That is [COUGH] a little bit too much 
algebra to fit on this slide, and so I'll leave you to do that. 
Though I might point out a little bit of help here, that this can be rewritten as 
e times e to the cosine of x minus 1. You'll find that to be a little bit 
easier to compute the Taylor series of. This lesson marks the end of our 
introduction to Taylor series. But by no means does it mark the end of 
our use of Taylor series. In our next lesson, we will consider one 
of the primal objects of calculus, the notion of a limit. 
And we'll consider it using the understanding gained from Taylor series.