Now we'll look at something called Taylor
Series Expansions.
So before I assumed that my function f was
differentiatiable
n times, so if I wanted to make order and
Taylor polynomial so for a linear
approximation I just needed
to assume that it was differentialable at
least one time.
most of the functions were going to
encounter are differentiable
infinitely many times, and this means
there so, the key thing to notice
here is the function zero has a derivative
that's also zero.
So any polynomial is derivative,
defferentiable an infinite number of
times.
So if it's say x squared, I take the
derivative, I get 2x.
I take the derivative, I get two.
I take the derivative, I get zero.
And then I can just continually take
derivatives and I always have zero after
that.
so it's, it's difficult.
I actually was trying to think up a
function that I could put on a quiz.
Or something that wouldn't, that would
look like a normal
function, and would not be differentiable
after a certain point.
And I couldn't really come up with
anything that
was, was pretty easy so there They do
exist and
they do show up when you start making,
kind
of, more complicated types of functions,
for instance, these things
that are defined as limits of other
functions.
But, for something you can write down,
generally, they're going
to have as many derivatives as you as you
want.
So I'm going to make something called a
Taylor series expansion.
So I'm going to use a notation t of x to
mean a Taylor serious expansion, and
that's just going to
be the limit as n goes to infinity of the
Taylor polynomial evaluated at x as n goes
to infinity.
And so what I mean by that is just, the
limit as
this, limit of this sum as the number of
terms gets, infinite.
And it turns out that a Taylor series is
a special case of something called a Power
series, so.
For a Taylor series, we know exactly what
the, the coefficients are.
They're determined by the function f that
I'm trying to approximate.
So, a Power series is just a coefficient a
sub k
[SOUND]
times the quantity x minus a to the kth
power.
And if I look at my definition of the,
of the Taylor polynomial, it just tells me
that
a sub k has to equal the kth derivative of
f evaluated at a, divided by k factorial.
And then another little bit of notation
that
you're going to start seeing pretty often
is.
The limit as n goes to infinity of this
finite sum.
So from k equals 0 up to n is going
to be written the sum from k equals 0 to
infinity.
So it's probably what you would have
guessed it would be anyways, but
the real definition of what's going on
here is this means the limit.
As n goes to infinity of the finite sum
from k equals 1 to n of this series.
So the Power series coefficients, a sub k,
when I'm talking about a Taylor series
expansion, are
just the kth derivative of the function
I'm trying
to approximate, evaluated at a, divided by
k factorial.
And the reason I want to show that the
Taylor series of expansion is a special
case
of Power series, is there a more general
results for convergence of power series,
so there's
convergence results for power series in
general, and
those, because of Taylor series, there's a
special
case of a power series, we can also
use those convergence results for a Taylor
series.
So the important one, the most important
one,
I suppose, is something called the radius
of convergence.
And this is going to be a number capital R
greater than 0, and
I want to look at my Power series, and if
this sums up to
a finite number so when I write less than,
strictly less than infinity,
what I mean is that whatever is on the
less than side, so whatever
is over here is a finite real number.
And then I'm going to define S of x just
to be this
infinite sum and so if this is, if x, so x
here is in this
interval a minus R to a plus R, where R is
the radius of
the convergence, so it's giving me and
interval.
If I take any point
in that interval then when I make this
infinite sum I'm going to get a finite
number.
So clearly, If we don't get a finite
number then
we really can't, you know it's not going
to be useful
for approximation because that's telling
me as I add more terms,
my approximation's just getting bigger,
and bigger, and bigger, and bigger.
So unless I know that f of x is equal to
infinity I, I don't
want to be trying to approximate with
something that's diverging as I add more
terms.
So, the, the whole idea is I want to be
able to add more terms
to my Taylor polynomial to get a
better approximation, not to get a worse
approximation.
Let's see so, it also turns out that this
function
S of x, we can take derivatives of these
too.
So, I can take the derivative of this when
it's finite.
And then S of x we're going to say is not
defined.
So, if I'm outside of the radius of
convergence, then I don't know what
happens to S.
It, it goes to a infinity.
So, I'm not really going to make, be
able to make any meaningful statement
about its property.
It's like I certainly couldn't define a
derivative For something that's equal to
infinity everywhere.
And so this is the convergence property
that comes from Power series
that I want to also be able to use for
Taylor series.
So it turns out, I need, it's, it's going
to depend on These coefficients a sub k.
But if the limit as k goes to infinity.
So I just need to look at these
coefficients a sub k absolute
value, to the one over kth power.
So this is the same thing as taking the
kth root Of the absolute value of a sub k.
So if it was one half, that would be the
square root.
If it was one third, that would be the
cube root.
If it was one over four, that would be the
fourth root.
And so on.
So if you can make sense of that limit, if
that's a finite value, then I
can define the radius of convergence to be
one divided by that limit.
And now for our Taylor series expansion,
I know what these a sub k's have to be.
And you can use a property so the sort of
annoying thing is that a sub k has this
k factorial in it and sometimes that's
going to be
a pain to try and deal with in the limit.
So luckily somebody proved this additional
theorem where
instead of k factorial, I just have k.
And so if this limit so it's k divided by.
This is the kth derivative of f evaluated
at a
and then I take the absolute value of that
to the
one over kth power so it looks kind of
like
this, except I don't have this k factorial
in there anymore.
So, somehow the k factorial.
To the 1 over k has become this k, up in
the, just this k to with no factorial up
in the numerator.
So if this limit exists, then the radius
of
convergence, so this is for a Taylor
series now,
because I'm defining these coefficients in
terms of the,
the derivatives that I'm using to build my
Taylor polynomials.
The radius of convergence is 1 divided by
e times the value of that limit.
And so far, what we've been able to show
then is that T of x is a finite number so
that, that's a good first step but it's
still not doing what we want to do.
What we really want to be able to do is to
use T of x, or, or a polynomial
that we would build on the way to T of x
to approximate our function f.
So if T of x is at least a finite
number, you know, we're doing a lot better
than if
it was infinite, there's still the
possibility that it would work.
But what we'd really like to know is if or
where T of x is equal to
f of x, because if T of x, so if this
infinite sum is converging
To f of x at a point x, then the idea
would be that I can use some
of those terms to make an approximation of
f at that point x.
So we have a, another theorem.
And this is, this one's a little bit
trickier.
So we need to know what the radius of
convergence is
and we can get that from the formula on
the previous slide.
So you can use either the formula for the
Power series in general.
Or you can use the special case for the
Taylor's expansion.
And you, sometimes it's actually easier to
use the general
formula even when you're working with the
Taylor expansion just to, it
depends on whether when you want this k
factorial in there or not.
[COUGH]
Then once you know what the radius of
convergence is, so you
know for what values of x is T of x going
to at least be finite?
We can look at this limit.
So I need to take R to the k, so this is
some number that's
going to be inside the, it's going to be
smaller than the radius of convergence.
So I take R to the kth power, divided by k
factorial.
And then I have to look at the kth
derivative of f.
Evaluated at a point z, that's in between
a minus r and a plus r.
So I'm making an interval around a of
width lower case r.
And I need to find the maximum value of
the kth derivative on that interval.
So, once you're done doing this part, that
should just be a number.
So it's the maximum value of this
derivative on
this interval so, but it might involve
this number r.
And then I'm going to look at that, this
limit as k goes to infinity.
If that limit is equal to 0, then T of x
is equal to f of x.
And so what that's going to tell me is
that every x.
That's within r, of my point a, has the
property that the Taylor series expansion.
So if I just make this infinitely long
Taylor polynomial.
That's going to be exactly equal to the
function that I'm trying to approximate.
So let me work through a quick example of
this.
So the Taylor series expansion for the
function f of
x equals log of 1 plus x around the point
a equals 0 is going to be, so I can
either define this just with k equals 0 to
infinity.
Or I can think about making Taylor
polynomials of order n and then taking
the limit as n goes to infinity, either
one of these they mean the
same thing, they're just two different
ways of communicating
this same idea.
But the one on the right hand side is sort
of the more precise of the two.
So I'm just going to take x minus a.
But here, I've said I want to make the
expansion around the point 0.
So that's going to be x minus 0 to the kth
power, divided by k factorial, times f k,
so the kth derivative of my function f
evaluated at the point 0.
So it turns out, if I, if I have log of 1
plus x, the first
derivative of that is just going to be 1
divided by 1 plus
x, so I can write that as 1 plus x to the
negative 1.
And if I were to take another derivative
of that, I'd have negative
1 times 1 plus x to the negative 2.
If I took another derivative of that I
would
have negative three times whatever
coefficient I have out here.
And because every time I take a derivative
I'm getting a negative sign.
the sign of each one of these derivatives
is going to be alternating.
So the way I'm going to capture that is
just put a minus one To a power.
And then I just choose this power so that
it's even when I want a positive term and
odd when I want a negative term.
So the first derivative, when I have one
prime here,
those are the ones that I want to be
positive.
So when I have an odd derivative, so first
derivative, third derivative,
fifth derivative and so on, I want to have
a positive expression.
So I'm just going to put k plus 1 as my
power on the negative 1,
and that's going to take care of the flip
flopping sign as the, as the sum goes
along.
And then, every time I take a derivative,
I'm going to get 1
and then a two and then a three and then a
four.
But, because the first derivative I got
from
log I'm actually one behind in my count.
So I end up with k minus 1 factorial as my
coeefiecient here.
And then I have 1 plus x to the minus k.
So when I take the first derivative, I
have a minus 1.
The second derivitive would have a minus 2
and so on.
And so this is what I want to plug in
to this term right here, and I'm going to
evaluate.
So this is the kth derivitive at x I
want to evaluate the kth derivative
at the point 0, so when I put in 0 in this
part of the
expression here, this is the only part
that actually has an x in it,
I'm going to get 1 plus 0 to the minus kth
power, and that's just going to
be 1 to the minus kth power.
And 1 to any power is equal to 1.
So when I evaluate this at 0, this final
part of the expression is just going to
drop away.
So I end up with just this alternating
sign.
And then a k minus 1 factorial, for my fk
of 0 derivatives,
and now, when I plug this back into my
Taylor expansion, what
I end up getting is x minus 0 to the k, so
that just going to
be an x to the k in the numerator I have a
k factorial and then
this expression fk of 0 that we just
worked out over here and
this has a k minus one factorial in it.
So the k minus one
factorial time, divided by k factorial is
just going to give me.
Why can't I select you?
Okay I'm not going to try that anymore.
It's just going to give me this factor of
k in the denominator.
So that's k minus one factorial divided by
k factorial.
Everything cancels itself out except for
the k.
So I divide by k In the numerator, I have
x to
the k which came from this x minus 0 to
the kth power.
And then I have this flip flopping
sine term.
So it turns out that my Taylor series
expansion is going to
be x minus x squared over two plus x cubed
over three.
Minus x to the 4th over 4 and so on.
And so now I want to find the radius of
convergence of that.
And so first I'll try, because this is a
Taylor series expansion,
I'll try using the formula we had for the
Taylor series expansion.
So I had the.
Fk evaluated at the point a.
So my point a was 0, so I have an
expression for that.
But now I have this k minus 1 factorial to
the 1 over
kth power in the denominator, and I have a
k in the numerator.
And so I don't really know how to make
sense out of that limit.
So, I said, hmh.
But I have another possibility.
Because this is a special case of a Power
series, I can
go back and try the formula for the power
series as well.
And so that said that if the limit of
these coefficients
existed, then the radius of convergence
would be one over this limit.
So let's
see what happens if I just plug this in.
So I have minus one to the k plus
one divided by k as my sequence of
coefficients.
And when I take the absolute value of
that, the whole numerator's just going to
go away.
So what I end up with is the limit As k
goes to infinity of 1 over k to the one
over kth power, and so notice that
the power is exactly equal to the thing
that I am taking the power.
So I'm going to say this is equal to the
limit as u of u to the
uth power, and if one over k if k is going
to infinity,
then 1 over k is going to 0.
So I can just say, this
is the limit as u goes down to 0 of u to
the uth power.
And if you go back to our lectures on
limits, this
was one of the things that we said was
equal to 1.
So if this limit is equal to 1, the radius
of convergence is 1 divided by that.
So that tells me that the radius of
convergence is equal to 1.
And then the last thing we have to do is
see if Or where t of x is equal to the log
of one plus x.
So this is my function f of x.
So I know that the radius of convergence
is equal to one.
So I'm looking for some number r that's
between 0.
So some number little r that's in between
0 and one
and so again, I'm just going to use the
theorem that I had
on a few slides ago, and that said that if
I
looked at this derivative, but now instead
of being able to evaluate
z it at the point 0, I have to evaluate it
at the point, and
find out where it's maximum on the
interval minus r to r.
And what I end up having here is minus 1
to the n plus 1.
So my absolute value is going to take care
of this flip-flopping sign for me.
I have an n minus 1 factorial in the
numerator and then a 1 plus z to the nth
power in the denominator and I'm trying to
make this as big as I possibly
can.
So I want to make the denominator
as small as I possibly can.
So the smallest I can make this is if I
take one minus r
And so that gives me is that
right?
I must have
had a missing one
line in here.
For what I should do.
but essentially
because r, I can't actually make this
denominator 0,
because r has to be strictly less than 1.
And so what I end up with is, I have
something
to the n down here, and this r to the n
here.
And that actually gives, I can think of
that as then
just r times 1, or, r divided by 1 minus
r.
To the nth power but I'm taking the limit
as n goes to infinity,
and so r divided by 1 minus r, that's just
a constant.
And I have, so I have some constant to the
nth power divided by n
factorial, and that limit is again one of
the results we had from the first week.
That any constant to a power divided by n
factorial, if I take
the limit of that as n goes to infinity,
I'm going to get 0.
And so that tells me, I can make r as
close as I want
to capital R but I can not actually make
it equal to capital R.
So it's just turned this sort of open
interval, this radius
of convergence of being 1 it goes, it's a
closed interval its
now an open interval so our little r can
be any
number if it's smaller than 1 but it can't
actually be 1.
And so that tells me that T of x is equal
to f
of x as long as the absolute value of x is
less than 1.