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Taylor Series.

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[SOUND]

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

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

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

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

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In other words, f of x is equal

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

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

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

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Well, here's what I mean.

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Take a look at, say, f of 0.

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

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

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

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

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

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

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And that means that I can

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

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

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

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But wait, there's more.

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The next coefficient, a sub 1, that
coefficient can

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be calculated in terms of the derivative
of f.

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What I can use is this power series again.

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

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

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So let me write that down.

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

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

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

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

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when the absolute value of x is less than
R.

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Now what am I going to do with this?

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

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

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

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

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

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

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

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

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

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

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

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

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a sub 1 times 1.

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

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

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It keeps on going.
Okay.

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

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Well, every single term has an x in it

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

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So the derivative of f at 0 is just

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

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

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

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And I can just keep on doing this.

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

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

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Let me differentiate this again

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in order to get a power series

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representation for the second derivative
of f.

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

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

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

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

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

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Now we'll look at the first few terms.

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

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

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Alright, so a sub 2 times 2 times 1 times

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

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a sub 3 times

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

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What about when n equals 4?

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

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

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And it's going to keep on going.

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

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alright?

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

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

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So what is this telling me?

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This is telling me that the second
derivative of f at

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

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

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

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Let's try to figure out a sub 3.

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

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Now I'm going to calculate the third
derivative.

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

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

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

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

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

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

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Now, what's the constant term?

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

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

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equal to.
So let me write that down.

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

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

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

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

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

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

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

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

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

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

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

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

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

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But why is this important, right?

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Why do we care about this?

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

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So

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I mean that I'll assume that my functions,

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

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

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series on the interval from minus R to R.

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

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

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

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

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Let me say this a bit differently.

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

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

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

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

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

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

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

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thing called the Taylor series for f
centered around zero.

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

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about something.

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Suppose that you just start with a

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function that you don't know very much
about.

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Maybe all you know about the function f is
that

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you can differentiate it as many times as
you like.

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

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

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

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

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

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said so far actually tells you that this
power series

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is related to the original function f in
any way.

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This whole story has been predicated on an
assumption.

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

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And if f is equal to a power series, then

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

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

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

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

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

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

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

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of this week.

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[SOUND]