0:41

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?

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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.

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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.

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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.