Hi.
Welcome back to recitation.
We've been talking about Taylor series for a number of
functions and rules by which you can compute Taylor series.
I have here an example that I don't think we did in lecture.
So this is the function f of x equals secant of x.
Now, unlike some of the other ones you've seen, there's not
a really simple formula for the whole Taylor
series of secant x.
So what I'd like you to do is not to find, you know, a
formula for the general term, but rather, just to use some
of the tools that we've learned to compute the first
few terms of the Taylor series for f of x equals secant x.
Say, up through the x to the fourth term, if you wanted, or
even a little further if you were feeling ambitious.
So why don't you pause the video, have a go at that, come
back, and we can do it together.
So welcome back.
I asked you to compute the first few
terms of a Taylor series.
One thing you can always do in this case, is you can go and
you can apply the general formula that we have for
Taylor series, and use it to compute the series that way.
So in order to do that, you just need to compute a few
derivatives of your function.
So remember that the general formula for a Taylor series is
the Taylor series for f of x is equal to the sum from n
equals 0 to infinity of the nth derivative of f taken at
zero divided by n factorial times x to the n.
And since we're only interested in the first few
terms, this is f of 0 plus f prime of 0 x plus f double
prime of 0 over 2x squared, plus dot dot dot.
I'm not going to write out the next few terms.
And if you want to apply this formula to secant of x, what
you would have to do, is you would have to compute these
derivatives.
And so we could try doing that.
So in our case, f of x is equal to secant of x.
So f of 0 is secant of 0, which is just 1.
So then we need to know f prime of x, so that's the
derivative of secant of x.
So this is one you should remember.
This is equal to secant x times 10x.
And when you plug in x equal to 0, well, the tan x is 0.
So f prime of 0 is equal to 0.
And then you could keep going.
So you could compute f double prime of x.
So we would have to compute the derivative of
secant x tan x.
And you would do that by the product rule.
You know, you take derivative of secant of x, and that gives
you secant x tan x times 10x plus--
and then you, OK.
So you leave secant of x, and you multiply it by the
derivative of tangent, which is secant squared x.
And now when you put in 0 here, you get f
double prime of 0.
Well, OK.
So this has a tan x in it, so that part's
going to give you 0.
And here we end up with secant of 0 times secant squared of
0, so that's 1 times 1 times 1.
So that's just 1.
So f double prime of 0 is 1, and you could keep doing this.
Now one thing you'll notice is that this is getting more and
more complicated.
I mean, we can simplify this expression a little bit.
We could write it as secant x tan squared x plus secant
cubed x, and there are, you know, all sorts of trig
manipulations you could do, if you wanted to try and rewrite
that in some simpler form.
But fundamentally, it's more complicated than the first
derivative was, and that's more
complicated in the function.
And it will keep getting more complicated as you compute
more derivatives.
So we can do this.
So so far this shows us, by the way, that this is equal to
1 plus 0x plus x squared over 2 plus dot dot dot.
And if you wanted to compute, you know, up through the
fourth degree term or something like this, that's
something that's manageable.
But I want to suggest that there are maybe some nicer
ways to do it.
So one thing to notice is that secant of x is closely related
to the function cosine of x.
And you know the Taylor series for cosine of x.
So one thing you could think to do, is to leverage the
information that you have about cosine of x in order to
use it to get some information about secant of x.
So one simple way to do that, is that you know that cosine
of x is even.
It's an even function.
Cosine of minus x is equal to cosine of x.
So that means secant of x is also even.
Secant of x is an even function.
And you've seen that even functions, their Taylor
series, all the odd powers have coefficient 0.
So what that means is without ever computing the third
derivative, we can know already that the next term in
this Taylor series is going to be 0x cubed over 6.
OK?
So that's nice.
So the odd terms of the Taylor series for secant of x are 0.
OK.
So that's one thing you can get right away.
So that gives you, if you like, that gives you half of
the terms of the Taylor series.
It's a little bit of a joke, but.
OK, so then you only need to figure out the even terms.
That's one way we can leverage the relationship between
secant and cosine.
The other way is that we can remember that Taylor series
multiply just like polynomials do.
So if secant is 1 over cosine, that means secant times cosine
is equal to 1.
OK?
So idea, secant of x times cosine of x is equal to 1.
Now, that means that the Taylor series for secant of x
times the Taylor series for cosine of x has to
simplify just to 1.
So we can write down that product as a product of two
infinite polynomials, and we can start
multiplying term by term.
And that'll allow us to solve for a bunch of terms, just by
solving some simple linear equations.
So let me show you what I mean.
So we know--
let me write it as cosine times secant.
So we know that cosine of x is 1 minus x squared over 2 plus
x to the fourth over 24.
That's 4 factorial.
Minus x to the sixth over 720, which is 6 factorial
plus dot dot dot.
And so we know that if we multiply this by the series
for secant of x--
well, what is the series for secant of x?
Well, we've already computed a few terms. We know that it's 1
plus x squared over 2, we computed that already.
And we know that the third degree term is 0.
So there's some fourth degree coefficient a4 x to the
fourth, or 4 factorial.
And there's some--
well, we know, we said it's even, so we know the fifth
degree coefficient is 0.
So then the sixth degree coefficient we don't know yet.
So this is plus a sub six times x to the sixth over six
factorial, plus dot dot dot.
So we know that when we multiply these two things
together, it has to give us just 1.
All the higher order terms have to cancel, because over
here we have a 1.
So what you can do, is you can actually try multiplying these
things out.
So it's easy to see, for example, that the constant
term of this product is just 1 times 1, which is 1.
Which is good.
So that's a check on what we've done so far.
And there is no x-term, because there are no x's.
The x squared term here is 1 times x squared over 2 minus x
squared over 2 times 1.
Well, that gives us 0, so that's good.
So this product is equal to-- well, it's 1, plus we saw 0x,
plus 0x squared.
How about the x cubed term?
Well, there are no odd terms in this product, so
there's no x cubed.
How about the x to the fourth term?
Well, OK.
So how do we get an x to the fourth?
We could have an x to the fourth here times a constant.
So that's x to the fourth over 24.
Or, we could have an x squared times an x squared.
So in this case, that gives us minus x squared over 2 times x
squared over 2, which is minus x to the fourth over 4.
Or we could have a constant times an x to the fourth.
So this is plus a4 x to the fourth over 24.
And then we'll have a sixth degree term, and so on.
Notice that there's never any involvement from the higher
order terms in the fourth power here.
Right?
If you ever took an x to the sixth here, well then
everything you multiply it by has at least an x
to the sixth power.
So we don't have to worry about that showing up in the x
to the fourth term.
Well, OK.
We know this is actually equal to 1, so we know that this
thing has to be 0.
This is x to the fourth term.
It has to be 0.
So that means that 1/24 minus 1/4 for plus a4
over 24 equals 0.
So OK.
So now you can multiply everything through by 24, and
rearrange to figure out that a4 is equal to 5.
So I've done that correctly.
And then if you wanted, it would be fairly easy to go
back up and then you look at the x to the sixth term.
And from there, you could figure out that a6 was equal
to, say, 61 or something like that.
I think 61.
And you could keep doing this.
So this is one way to compute more terms of the series for
secant of x.
Another thing you could do-- which I'll just mention very
briefly, I'm not going to show you how to do it-- is that you
can do long division on power series.
So it actually works out very, it works out just like this.
It's mathematically equivalent.
The way you actually do it looks different.
It looks like long division.
When you do long division with polynomials, you start with
the highest order term.
Of course, power series don't have highest order terms. What
you actually do with a power series, is you start with the
lowest order term.
So to divide this into 1, you'd say, oh, you mean the
factor of 1, and then you'll have a, you know, you subtract
off 1 times this, and that gives you an x squared plus 2.
And so, OK, so you say, I need a plus an x squared
plus 2, and so on.
So that was too brief for you to understand it properly, but
you can go and look up somewhere the method of long
division on power series.
So just to recap, we talked about three methods for
computing the coefficients of a power series.
There's the formula that you were given, which works, and
which you could use.
In this case, it's a little complicated.
Then there's the method of using a relationship between
your power series and other known power series.
In this case, we can use the relationship that we know our
power series satisfies a certain product.
We know that our power series times cosine of x equals to 1,
and we can use that to solve for some of the coefficients.
Or you can also, similarly, when you have that situation,
you can also use long division to compute the coefficients.
I'll end there.