Calculus Two: Sequences and Series is an introduction to sequences, infinite series, convergence tests, and Taylor series. The course emphasizes not just getting answers, but asking the question "why is this true?"

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From the course by The Ohio State University

Calculus Two: Sequences and Series

931 ratings

The Ohio State University

931 ratings

Calculus Two: Sequences and Series is an introduction to sequences, infinite series, convergence tests, and Taylor series. The course emphasizes not just getting answers, but asking the question "why is this true?"

From the lesson

Taylor Series

In this last module, we introduce Taylor series. Instead of starting with a power series and finding a nice description of the function it represents, we will start with a function, and try to find a power series for it. There is no guarantee of success! But incredibly, many of our favorite functions will have power series representations. Sometimes dreams come true. Like many dreams, much will be left unsaid. I hope this brief introduction to Taylor series whets your appetite to learn more calculus.

- Jim Fowler, PhDProfessor

Mathematics

Taylor Series.

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Suppose I know that f is represented by a power series.

For example, suppose that I know that the function f is given by a power series.

This sum n goes from zero to infinity of some coefficients a sub n times x to the n

at least say when x is between negative R and R.

In other words, f of x is equal

to a sub 0 plus a sub 1 times x plus a sub 2 times x squared and so on.

In this case, I can figure out what the coefficients, the a

sub n's must be equal to in terms of the function f.

Well, here's what I mean.

Take a look at, say, f of 0.

Well, if f is really given by this power series on this interval, then f of 0 is

a sub 0 plus a sub 1 times 0 plus a sub 2

times 0 squared plus a sub 3 times 0 cubed and so on.

But all of these terms have a zero in them, right, a sub 1 times 0 is

0, a sub 2 times 0 squared is 0, a sub 3 times 0 cubed is 0.

All of these terms is zero except for a sub 0.

And that means that I can

conclude that a sub 0 is just equal to f of 0.

So if f has this nice power series representation, the a sub

0 coefficient must actually be equal to the functions value at zero.

But wait, there's more.

The next coefficient, a sub 1, that coefficient can

be calculated in terms of the derivative of f.

What I can use is this power series again.

Can differentiate this power series term by term, and I'll get a

power series for the derivative of f, at least valid on this interval.

So let me write that down.

So what I've got is that the derivative of f at x is

given by this power series, this sum n goes from 1 to infinity of

a sub n times the derivative of x to the n, which is

n times x to the n minus 1 and this is a least valid

when the absolute value of x is less than R.

Now what am I going to do with this?

Well, let me write out the first few terms of this

power series just to give you an idea what this looks like.

And, of course, it looks exactly like the derivative up here.

But I'm going to just plug in n equals 1 and I get a

sub 1 times 1 times x to the 0th power, which is just 1.

So this is the first term is just a sub 1 times 1.

The n equals 2 term, well, that's a sub 2 times 2 times x to the 1st power.

The n equals 3 term, well that's a sub 3

times 3 times x squared and then it keeps on going.

And you can see this is just the derivative of this, right, the derivative

of a sub zero is just the derivative of a constant which goes away.

The derivative of a sub 1 times x is right here,

a sub 1 times 1.

The derivative of a sub 2 times x squared is right here,

a sub 2 times the derivative of x squared, which is 2x.

It keeps on going. Okay.

Now what happens when I evaluate this power series at x equals 0?

Well, every single term has an x in it

except for the constant term, which is a sub 1.

So the derivative of f at 0 is just

equal to a sub 1, which is telling me that if I've got a power series

representation for my function f, I know what the a sub 1 coefficient has to be.

It has to be the derivative of my function at 0.

And I can just keep on doing this.

The coefficient in front of x squared, a sub 2, that

coefficient can be calculated in terms of the second derivative of f.

Let me differentiate this again

in order to get a power series

representation for the second derivative of f.

So f double prime is the term by term derivative of this power series,

which is the sum n goes not from 1, but from 2 to infinity, of the derivative

of this, which is a sub n times n times the derivative of x to

the n minus 1, which is n minus 1 times x to the n minus 2.

And this is valid at least when the absolute value of x is less than R.

Now we'll look at the first few terms.

So when I plug in n equals 2, I get what, a sub 2 times 2

times 2 minus 1 times x to the 2 minus 2x to the 0, that's just times 1.

Alright, so a sub 2 times 2 times 1 times

1 plus what happens when I plug in n equals 3?

a sub 3 times

3 times 3 minus 1, so times 2 times x to the 3 minus 2, so just times x.

What about when n equals 4?

Well, that's a sub 4 times 4 times 4 minus 1

times 3 times x to the 4 minus 2, times x squared.

And it's going to keep on going.

And here, what happens again is that all of these terms have an x in there,

alright?

So when I plug in x equals 0, this term dies, this term dies, all

the other terms die except for that constant term, which is a sub 2 times 2.

So what is this telling me?

This is telling me that the second derivative of f at

0 is just this constant term, it's 2 times a sub 2.

And then, if I divide both sides by 2, what

I'm finding is that a sub 2 is the second derivative of f at 0 divided by 2.

Let's try to figure out a sub 3.

Well, I've got the original function, its derivative, its second derivative.

Now I'm going to calculate the third derivative.

So f triple prime at x is the sum n goes, not

from 2, but from 3 to infinity of the derivative of this,

which is a sub n times n times n minus 1 times

the derivative of x to the n minus 2, so times n minus

two times x to the n minus third power and this is

valid at least when the absolute value of x is less than R.

Now, what's the constant term?

Well, the constant term is just the n equals 3 term.

So that's x to the 3 minus 3 and that's exactly what the third derivative at 0 is

equal to. So let me write that down.

The third derivative at 0 is just the constant term in this power series, which

is the n equals 3 term, which is a sub 3 times 3 times 3 minus 1 times 3 minus 2.

In other words, what this formula is telling me is that a sub

3 is equal to the third derivative of f at 0 divided by 6.

But it'll be better to think of that six as being three factorial.

Instead of writing down six let me write down three factorial.

An then I can write down the general fact, alright, that a sub n, the

nth coefficient in the power series for my function f, right, I'm

assuming that my function f has this nice power series representation,

and then a sub n is given by computing the nth derivative

of f at the point 0 and dividing that by n factorial.

So yeah, in general, the coefficient on x to the n, that a

sub n, that can be calculated in terms of the nth derivative of f.

But why is this important, right?

Why do we care about this?

Well, the whole point is that now we're in a position to work backwards.

So

I mean that I'll assume that my functions,

some functions that I'm interested in, has a power

series representation and it's given by, say, this power

series on the interval from minus R to R.

Now if I assume that that's the case, then, by differentiating,

I can figure out what these coefficients have to be, right?

And that's exactly what I'm claiming, that the nth coefficient

is given by calculating the nth derivative of f at 0 and dividing by n factorial.

Let me say this a bit differently.

If this function f can be written as a power series on the interval from minus R

to R, then, f of x is equal to this power series, the sum

n goes from 0 to infinity of the nth derivative of f at the point

0 divided by n factorial, that's what the coefficient has to be, times x to the n.

This has a name, or rather, has two names. One of those names is Maclaurin series.

You'll often see this thing called the Maclaurin series for f.

The other name that you'll see you'll see this

thing called the Taylor series for f centered around zero.

Regardless of what you want to call this, I should warn you

about something.

Suppose that you just start with a

function that you don't know very much about.

Maybe all you know about the function f is that

you can differentiate it as many times as you like.

Well then, you could write down the Maclaurin series for f.

You could write down this, the Taylor series for f centered at zero.

All you need to be able to do is just differentiate f as many times

as you like at the point 0 in order to write down this power series.

The issue, though, is that nothing that we've

said so far actually tells you that this power series

is related to the original function f in any way.

This whole story has been predicated on an assumption.

It's been predicated on the assumption that f is equal to a power series.

And if f is equal to a power series, then

this must be the power series that f is equal to.

But if you don't know that, if you're just

starting with some random function, then there's no guarantee

whatsoever that if you write this thing down that it's

going to have any relationship at all to the original function.

In relating f and this, its Maclaurin series or its Taylor series centered

at zero, it's going to be a huge part of what we do the rest

of this week.

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