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

932 ratings

The Ohio State University

932 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's Theorem.

[SOUND]

Well, thus far we've been looking for

Taylor's Theorem without actually concerning ourselves with convergence.

So I would like to do is you know, start with some function, write

down it's Taylor's series around' 0' maybe

I'll assume this function as infinite differentiable.

So I can actually write down its Taylor Series around'

0' and I mean it's probably too much to hope

that these are equal but at least you know, I could replace this

infinity with some big number N and maybe I can't hope for equality.

But I could, at least hope that my function's

being approximated by this partial sum of its Taylor series.

Taylor's Theorem is a result that's going to let us pull that off.

Well, here's a statement of Taylor theorem, or, I mean,

there's actually a lot of results that are called Taylor theorem

that are all sort of like this.

This is kind of a weak result, and I'm just doing it around 0.

But anyhow, okay, it's Taylor's theorem. Here's a statement of Taylor's theorem.

Suppose that I've got some function from the real numbers to the real numbers, and

just to make this a little bit easier, I'm going to say it's a smooth function.

So it's infinitely differentiable, and I've picked some number big

N, and then I'm writing out f of x equals.

The big nth partial some of its Taylor series

around the point 0.

And I'm saying that f of x is equal to this plus this

error term, or what's usually called the remainder term, so big R for remainder.

And of course, this is necessarily true.

I mean f of x is equal to anything plus some error, you know,

so the big deal here isn't that I can write f of x like this.

It's that I know something about the error term.

Here's the upshot.

Then, the error term, or the remainder

term, is given by this formula.

The big N plus 1th derivative of f evaluated at

some point z divided by big N plus 1 factorial.

Times x to the big N plus 1, and this

point z is just some point between x and 0.

We'll eventually give a proof of Taylor's theorem, but in

the meantime, let's just see how we can use Taylor's theorem.

So we've already considered

the Taylor series for sine x around the origin.

It's the sum n goes from 0 to infinity of minus 1 to the n.

Over 2 n plus 1 factorial, times x to the 2 n plus 1.

And what I want to show now, right, what the, the big goal is, is to show that

sin x is actually equal to its Taylor series, regardless of what x is.

To get there, we're going to use Taylor's theorem

to control that error term, or that remainder term.

Just to make it a little bit easier to talk about this.

So I have f of x equals sin x and you

just write it out then what Taylor's theorem is telling us.

Taylor's theorem is telling us that sin x minus the big Nth partial sum, so

the sum little n goes from zero to big N. Of the Taylor series, which is the nth

derivative of f at 0 divided by n factorial times x to the n.

I'm just writing out the Taylor series in general.

This is this remainder term, big R sub N of x.

And what Taylor's theorem tells us is that

I've got a formula for this remainder term.

I can control the remainder in terms of the

next, the big N plus 1th derivative of f.

And it's telling me that big R sub N

of x is equal to f, the big

N plus oneth derivative of f at some point z divided by

big N plus one factorial. Times x to the big N plus 1 power

for some point z which is between 0 and x. But I

know something about how big the big N plus 1th derivative of f is.

Well, I don't know anything about big N. But I know that if I were to differentiate

sin any number of times, right, it doesn't matter how many times I differentiate sin.

I'm either getting, you know, plus or minus

sin, or maybe I'm getting plus or minus cos.

But regardless of how many times it differentiates sin, I'm getting one or

the other of these things, maybe with a plus or a minus sin.

And what that means, is that regardless of what big N plus 1 is.

The absolute value of the big N plus 1, derivative of sin

at some point x can be no bigger than 1 in absolute value.

So I can use this knowledge to say something about the remainder.

Because this is true about the big N plus 1

derivative of f, being no bigger than 1, that tells me.

That at least in absolute value, the remainder is no bigger

than well how big could this be in absolute value was 1.

So it's no bigger than 1 over n plus 1 factorial

times x to the n plus 1th power in absolute value.

So I'd like that to be small, or even

better, I'd like to say something about the limit.

So, fix some value of x. I don't

need to consider varying values of x.

I just want to consider a fixed value of x and

I want to show that this is small in the limit.

I want to show that the limit as big N goes to infinity of the absolute

value of x to the big N plus oneth power over big N plus one factorial.

I'll make sure that the limit of this is zero.

Why would thinking about the limit be a good idea?

Well, if I know the limit of this

is zero, then I know that another limit vanishes.

Then I know the limit as big N approaches infinity

of something that's smaller than this, something closer to zero.

Well here's something. Right.

Then I know that the limit of the remainder term is also equal to zero.

Because this is bigger than this.

But, why would I care about knowing the limit of the remainder term is zero?

Well,

remember what this is measuring right.

Big R is measuring the distance between the function and

the big nth partial sum of its tailor series around zero.

So, if I know that in the limit as big N approaches infinity, this goes to 0.

That means in the limit, this goes to this.

But what's the limit of this as big N goes to infinity?

You want me to write that down, right.

What I'm saying is that sin

of x that must be equal to the sum n goes from 0 and I'd write though the big' N'

here but I am taking the limit as big' N'

goes to infinity and that's exactly what this series notation means.

Right so, what I am saying is that if the remainder goes to 0 in the limit then sin

is actually equal to its Taylor series around zero, alright?

That means that sin of x is actually equal to

its Taylor series.

And I know another way of writing down the Taylor series, right?

Then I can say that sin of x is actually equal to the sum and goes from

zero to infinity and instead of writing down this

where I haven't actually described what these coefficients are.

I could write down minus one to the n over two

n plus one factorial times x to the two n plus one.

Because we've already seen that if we differentiate sin a

bunch of times, this is exactly what we get here.

So that is a great goal, right?

So I've got to figure out some way to show that this limit is zero.

Well, how do I deal with that limit?

Well, there's a trick. Claim.

[LAUGH]

The sum big N goes from zero to infinity of the absolute value of x to the big

N plus one over big N plus one factorial. Converges.

If I can show that this series converges then, I

know that the limit of this is equal to zero.

But, how do I know that that series converges?

Well, let's use the ratio test.

So, I should look

at the limit as big N goes to infinity.

It seems like it's going to be worse, but it's going to be better.

Of the N N plus oneth term here where you know N is replaced

by N plus one, divided by this term here, so let me write that out.

It's going to be the limit of x to the big N plus two, divided by

big N plus two factorial and that's divided by x to the big N

plus one over big N plus one factorial. Taking absolute value of this.

And how do I evaluate this limit.

I mean this maybe, seems worse but there is lot of

[UNKNOWN]

cancellation.

So this limit ends up being the limit as big N approaches infinity where I've got

an x to the big N plus two divided by x to the big N plus 1.

So this is just, x, just one copy of x survives.

Here I've got an n plus 1 factorial in the denominator of the denominator.

So something's going to move that up to the numerator,

so we get n plus 1 factorial in the numerator.

And then in the denominator of the numerator, I've got n

plus 2 factorial. Let's put that in the denominator.

Okay, so let's calculate this limit, but now what's this limit?

Well, I've got an n plus 1 factorial in the numerator, and

an n plus 2 factorial in the denominator, so this is the limit.

As N approaches infinity of x over just N plus 2.

And what happens now, x is fixed.

Big N is approaching infinity, so this limit is equal to 0.

0 is less than 1, so by the ratio test, this series converges.

And therefore, by the nth term test if you like because

this series converges that means that this limit is equal to zero.

Let's put it all together.

So really the argument starts when I fix some value for x.

I just pick a single value of x and I'm

going to consider just that value of x from now on.

Now what happens.

I fix that value of x, and for fixed value of

x, the limit as big N approaches infinity of that fixed

value of x, just a constant as far as big N

is concerned, divided by some number that's going to be enormous.

This is 0.

But this is also calculating the limit of

the ratio between subsequent terms in this series.

That means by the ratio test since this limit is 0 and

0 is less than one, which is the cut-off for the ration test.

That means that this series converges.

Now, because that series converges, that means the limit of its nth term must be 0.

But because that limit is 0, I know something now about the remainder.

Right, the remainder is bounded above by this term, so if

this limit is 0, that tells us something about the remainder.

It's telling me that the limit of the remainder

as big N goes to infinity is equal to 0.

But if the remainder term is

going to zero in the limit, right. That tells me that the difference between

sin and it's big Nth partial sum is going to zero as big N goes to infinity.

But that's just to say that sin of x is actually equal to its Taylor series.

Because the error between sin of x and the

big Nth partial sum over here is going to zero.

So when the limit right,

when big N is going to infinity sin of

x must actually be equal to it's Taylor series.

We've already found it's Taylor series, here it is.

And this is true for all real x because the

beginning of this argument I just picked one value of x.

But I didn't use anything about that value of x,

so this must be true regardless of what x is.

That is truly a triumph.

Is a huge success, this famous

[UNKNOWN]

function sin turns out to be given

everywhere by this relatively nice looking power series.

[SOUND]

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