Hi.
Welcome back to recitation.
In class, Professor Jerison
and Professor Miller
have taught you a little
bit about Taylor series
and some of the manipulations
you can do with them,
and have computed a bunch
of examples for you.
So I have three
more examples here
of functions whose Taylor
series are nice to compute.
So the first one is cosh x.
That's the hyperbolic cosine.
So just to remind
you, this can be
written in terms of the
exponential function as e
to the x plus e to
the minus x over 2.
The second one is
the function 2 times
sine of x times cosine of
x, just your regular sine
and cosine here.
And the third one is x times
the logarithm of the quantity 1
minus x cubed.
So why don't you pause
the video, take some time
to work out the Taylor series
for these three functions,
come back, and we can
work them out together.
So here we have three
functions whose Taylor series
we're trying to compute.
Let's start with the first
one and go from there.
So this first one is
the hyperbolic cosine
that's given by the formula e
to the x plus e to the minus
x over 2.
So there are a
couple different ways
you could go about this one.
This is actually,
the hyperbolic cosine
is very susceptible to
the method of just using
the formula that you have.
So if you remember,
the derivative
of the hyperbolic cosine
is the hyperbolic sine.
The derivative of
the hyperbolic sine
is the hyperbolic cosine again.
So this function has very
easy-to-understand derivatives,
which you can see, you know,
just by looking at its formula.
It's easy to understand, because
the exponential function has
very simple derivatives,
and e to the the minus
x also has very
simple derivatives.
So you could do it like that.
The other thing you could
do, is that you already
know the Taylor
series for e to the x.
And I believe you've also
seen the Taylor series for e
to the minus x, and even if you
haven't, you can figure it out
just by substitution.
So if you remember, so e to the
x is given by the sum from n
equals 0 to infinity of x
to the n over n factorial.
I'm going to pull
the 1/2 out in front.
And e to the minus x is
given by the same thing,
if you put in minus x for x.
So it's n equals 0 to infinity,
so that works out to minus 1
to the n x to the
n or n factorial.
Now, when you add these
two series together,
what you see is that when
n is even, over here,
you have x to the n over n
factorial, and over here,
you have x to the
n over n factorial.
So what you get is,
well, you get 2 x
to the n over n factorial, and
then you multiply by a half,
so you just get x to
the n over n factorial.
When n is odd, here you have
x to the n over n factorial,
and here you have minus 1 x
to the n over n factorial.
So you add them and you get 0.
So what happens is that
this series looks just
like the series for e to the
x, except the odd terms have
died off.
So we're left with just 1 plus x
squared over 2 factorial plus x
to the fourth over 4 factorial
plus x to the sixth over 6
factorial, and so on.
And if you wanted to write
this in summation notation,
you could write it
as the sum from n
equals 0 to infinity of x
to the 2n over 2n factorial.
So this is the Taylor series for
the hyperbolic cosine function.
Also, if you wanted, say,
the hyperbolic sine function,
you could do something
very similar,
or you could remember that
the hyperbolic sine is
the derivative of the
hyperbolic cosine,
and just take a derivative
right from this expression.
One other thing that
you should notice
is that this looks very similar
to the expression of the Taylor
series for cosine of x.
So more of our sort
of funny coincidences
between regular trig functions
and hyperbolic trig functions.
All right.
That's the first one.
How about the second one?
So here we have just some
regular trig functions.
We have 2 sine x cosine x.
Let me see where
I've got some space.
I can do it right here.
Let me box off a little
space for myself.
So 2 sine x cosine x-- there
are a couple different ways
you could proceed
with this function.
So one is, you know the Taylor
series for sine x and cosine x
already.
So if all you wanted was a few
terms of this Taylor series,
one natural thing to do would be
to take the series for sine x,
take the series for cosine
x, multiply them together
like you would
multiply polynomials,
and what you would get
is the Taylor series
for this expression,
for this function.
That's one way to proceed.
That works perfectly well.
Another thing you
could do, is you
could try taking derivatives.
You could have a situation
where every time you
take a derivative, you
apply product rule.
It's going to get more
and more complicated.
It still works.
It's a little complicated
to do it that way,
if you wanted more
than just a few terms.
The other thing you could do,
is you could remember your trig
identities.
So if you look at
this expression,
this should be familiar to you,
because it's just sine of 2x.
So once you realize
that this is sine of 2x,
there's a much,
much shorter path
available to you, which is that
you already know the Taylor
series for sine of x,
so what you can do,
is you can just plug in 2x
into that Taylor series.
So sine of x is-- well, so OK.
So sine of x is x--
so in this case,
that's going to be 2x--
then minus, so in sine of x,
we have x cubed
over 3 factorial.
So here we're going to have 2x
cubed over 3 factorial plus--
OK.
So then, you know, and so on.
So here we'll have 2x to
the fifth over 5 factorial
minus-- so on.
If you wanted to write
this in summation notation,
you could write it as the sum
from n equals 0 to infinity.
Well, the denominator has got
to be 2n plus 1 factorial,
because we want it to
go through the odds.
And then we've got minus 1 to
the n times 2 to the 2n plus 1
times x to the 2n plus 1.
So this is 2x.
What we've got here, if you
didn't have the 2's there,
that would just be the
series for the regular sine.
OK.
So this is the series for this
function, 2 sine x cosine x.
And I'll go over here
to do the third one.
So what is the third one?
It's x ln 1 minus x cubed.
Well, what can we
do with this series?
The x out front is just
multiplying this logarithm
part.
That's something we
can save until the end.
If we can figure out what the
Taylor series for the ln of 1
minus x cubed part is, then
we just multiply x into it,
and that'll give us the Taylor
series for this whole thing.
So the x out front
is pretty simple.
So now what about this ln
of 1 minus x cubed stuff?
Well, a thing to remember
is, does it remind you
of anything we've done before?
Well, we have a Taylor series
for a logarithm function,
right?
We've already seen in
lecture, I believe,
we've seen that
ln of 1 plus x is
equal to x minus x squared
over 2 plus x cubed over 3,
minus x to the fourth over four,
and so on, alternating signs.
Notice that the denominators,
when you have a logarithm,
these are not factorials.
These are just the integer 2,
the integer 3, the integer 4,
unlike for exponentials
and trig functions.
So this is what
log of 1 plus x--
this is the Taylor series
for log of 1 plus x.
Well, how does that help us?
Well, log of 1 minus x cubed we
can get from log of 1 plus x,
with the appropriate
substitution.
So in particular,
we just have to put
minus x cubed in for x here.
So what does that give us?
It gives us the ln of 1 minus
x cubed is equal to-- well,
minus x cubed minus, so we
put minus x cubed in here,
we square it, and we
just get x to the sixth.
x to the sixth over 2.
Then, all right.
So minus x cubed quantity
cubed is minus x to the ninth.
So minus x to the ninth over 3,
minus x to the twelfth over 4,
and so on.
And so finally, x ln
of 1 minus x cubed,
we just get by multiplying this
whole expression through by x.
So this is equal to minus
x to the fourth minus x
to the seventh over 2 minus
x-- whoops, not ten-- minus
x to the tenth over 3, minus
x to the 13 over 4, and so on.
And I'll leave it as
an exercise for you
to figure out how to write
this in summation notation,
if you wanted.
So just quickly to summarize,
we had these three power series,
these three functions
that we started out with,
and we used a bunch of
different techniques
that we've learned in order
to compute their power series.
So over here, we took the
function that we'd seen,
and we knew a formula for it
in terms of other functions
that we already knew, and so we
plugged in those power series,
and used our addition
rule for power series.
We could have also done this one
directly from the definition,
if we had wanted to.
For the second one, for
the 2 sine x cosine x,
we recognized that
as something that
is susceptible to a
substitution, although also,
with a little more,
work, we could
have done it by a couple
of different methods.
For example, by multiplying
two power series together.
And finally, for this
third one, for the x ln
of 1 minus x cubed, we first
saw the substitution here
that we could make, and then
we just did a multiplication
by a polynomial, which is
a relatively easy thing
to do for power series.
So that's what we did
in this recitation,
and I'll leave it at that.
