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PROFESSOR: Last time we
left off with a question
having to do with
playing with blocks.
And this is supposed to give us
a visceral feel for something
anyway, having to
do with series.
And the question was whether
I could stack these blocks,
build up a stack so that--
I'm going to try here.
I'm already off
balance here, see.
The question is
can I build this so
that the-- let's
draw a picture of it,
so that the first
block is like this.
The next block is like this.
And maybe the next
block is like this.
And notice there
is no visible means
of support for this block.
It's completely to the
left of the first block.
And the question is,
will this fall down?
Or at least, or more precisely,
eventually we'll ask,
you know, how far can we go?
Now before you
answer this question,
the claim is that
this is a kind of
a natural, physical
question, which
involves some important answer.
No matter whether the answer
is you can do it or you can't.
So this is a good
kind of math question
where no matter what the
answer is, when you figure out
the answer, you're going to
get something interesting out
of it.
Because they're
two possibilities.
Either there is a limit to how
far to the left we can go --
in which case that's a
very interesting number --
or else there is no limit.
You can go arbitrarily far.
And that's also
interesting and curious.
And that's the difference
between convergence
and divergence,
the thing that we
were talking about up to
now concerning series.
So my first question
is, do you think
that I can get it so that
this thing doesn't fall down
with-- well you see I have
about eight blocks here or so.
So you can vote now.
How many in favor
that I can succeed
in doing this sort of thing with
maybe more than three blocks.
How many in favor?
All right somebody
is voting twice.
That's good.
I like that.
How about opposed?
So that was really
close to a tie.
All right.
But I think the there was
slightly more opposed.
I don't know.
You guys who are in the
back maybe could tell.
Anyway it was pretty close.
All right.
So now I'm going-- because
this is a real life thing,
I'm going to try to do it.
All right?
All right.
So now I'm going to tell
you what the trick is.
The trick is to do it backwards.
When most people are
playing with blocks,
they decide to build
it from the bottom up.
Right?
But we're going to build
it from the top down,
from the top down.
And that's going
to make it possible
for us to do the optimal
thing at each stage.
So when I build it from the
top down, the best I can do
is well, it'll fall off.
I need to have it you
know, halfway across.
That's the best I can do.
So the top one I'm going
to build like that.
I'm going to take it as
far to the left as I can.
And then I'm going to
put the next one down
as far to the left as I can.
And then the next one as
far to the left as I can.
That was a little too far.
And then I'm going to do the
next one as far to the left
as I can.
And then I'm going
to do the next one --
well let's line it up first --
as far to the left as I can.
OK?
And then the next one as
far to the left as I can.
All right.
Now those of you who are in
this line can see, all right,
I succeeded.
All right, that's over the edge.
All right?
So it can be done.
All right.
All right.
So now we know that we can
get farther than you know,
we can make it overflow.
So the question now
is, how far can I get?
OK.
Do you think I can get to here?
Can I get to the end over here?
So how many people think I
can get this far over to here?
How many people think
I can get this far?
Well you know,
remember, I'm going
to have to use more
than just this one
more block that I've got.
I don't, right?
Obviously I'm
thinking, actually I do
have some more blocks at home.
But, OK.
We're not going to.
But anyway, do you think
I can get over to here?
How many people say yes?
And how many people say no?
More people said no then yes.
All right.
So maybe the stopping place
is some mysterious number
in between here.
All right?
Well OK.
So now we're going
to do the arithmetic.
And we're going to figure out
what happens with this problem.
OK?
So let's do it.
All right, so now
again the idea is,
the idea is we're going to start
with the top, the top block.
We'll call that
block number one.
And then the farthest,
if you like, to the right
that you can put a
block underneath it,
is exactly halfway.
All right, well, that's
the best job I can do.
Now in order to make my
units work out easily,
I'm going to decide to call
the length of the block 2.
All right?
And that means if I
start at location 0,
then the first place where I
am is supposed to be halfway.
And that will be 1.
OK so the first step in the
process is 1 more to the right.
Or if you like, if I
were building up --
which is what you would actually
have to do in real life --
it would be 1 to the left.
OK now the next one.
Now here is the
way that you start
figuring out the arithmetic.
The next one is based
on a physical principle.
Which is that the farthest
I can stick this next block
underneath is what's called the
center of mass of these two,
which is exactly halfway here.
That is, there's a
quarter of this guy,
and a quarter of that
guy balancing each other.
Right?
So that's as far as I can go.
If I go farther than
that, it'll fall over.
So that's the absolute
farthest I can do.
So the next block is
going to be over here.
And a quarter of 2 is 1/2.
So this is 3/2 here.
All right so we went to 1.
We went to 3/2 here.
And then I'm going to keep on
going with this eventually.
All right so we're
going to figure out
what happens with this stack.
Question?
AUDIENCE: How do
you know that this
is the best way to optimize?
PROFESSOR: The
question is how do I
know that this is the
best way to optimize?
I can't answer that question.
But I can tell you
that it's the best
way if I start with
a top like this,
and the next one like this.
Right, because I'm
doing the farthest
possible at each stage.
That actually has a name
in computer science, that's
called the greedy algorithm.
I'm trying to do the best
possible at each stage.
The greedy algorithm
starting from the bottom
is an extremely bad strategy.
Because when you do that,
you stack it this way,
and it almost falls over.
And then the next time
you can't do anything.
So the greedy algorithm is
terrible from the bottom.
This is the greedy algorithm
starting from the top,
and it turns out
to do much better
then the greedy algorithm
starting from the bottom.
But of course I'm not addressing
whether there might not
be some other incredibly clever
strategy where I wiggle around
and make it go up.
I'm not addressing
that question.
All right?
It turns out this is
the best you can do.
But that's not clear.
All right so now, here
we have this thing.
And now I have to figure out
what the arithmetic pattern is,
so that I can figure out what
I was doing with those shapes.
So let's figure out a
thought experiment here.
All right?
Now the thought
experiment I want
to imagine for
you is, you've got
a stack of a bunch of blocks,
and this is the first N blocks.
All right?
And now we're going to
put one underneath it.
And what we're
going to figure out
is the center of mass
of those N blocks,
which I'm going to
call C sub N. OK.
And that's the place
where I'm going
to put this very next block.
I'll put it in a
different color here.
Here's the new--
the next block over.
And the next block over
is the (N+1)st block.
And now I want you to think
about what's going on here.
If the center of mass
of the first N blocks
is this number, this new
one, it's of length 2.
And its center of
mass is 1 further
to the right than the center
of mass that we had before.
So in other words, I've added to
this configuration of N blocks
one more block,
which is shifted.
Whose center mass
is not lined up
with the center of mass
of this, but actually
over farther to the right.
All right so the
new center of mass
of this new block-- And this is
the extra piece of information
that I want to observe,
is that this thing has
a center of mass at C_N + 1.
It's 1 unit over because
this total length is 2.
So right in the middle there is
1 over, according to my units.
All right now this is
going to make it possible
for me to figure out what
the new center of mass is.
So C_(N+1) is the center
of mass of N+1 blocks.
Now this is really only in the
horizontal variable, right?
I'm not keeping track of the
center of mass-- Actually
this thing is hard to build
because the center of mass
is also rising.
It's getting higher and higher.
But I'm only keeping track of
its left-right characteristic.
So this is the
x-coordinate of it.
All right so now
here's the idea.
I'm combining the white
ones, the N blocks,
with the pink one, which
is the one on the bottom.
And there are N
of the white ones.
And there's 1 of the pink one.
And so in order to get the
center of mass of the whole,
I have to take the weighted
average of the two.
That's N*C_N plus 1 times the
center of mass of the pink one,
which is C_N + 1.
And then I have to divide --
if it's the weighted average
of the total of N +
1 blocks -- by N + 1.
This is going to give me
the new center of mass
of my configuration
at the (N+1)st stage.
And now I can just
do the arithmetic
and figure out what this is.
And the two C_Ns combine.
I get (N+1)C_N +
1, divided by N+1.
And if I combine these two
things and do the cancellation,
that gives me this
recurrence formula,
C_(N+1) is equal to C_N
plus-- There's a little extra.
These two cancel.
That gives me the C_N.
But then I also have 1/(N+1).
Well that's how much gain I
can get in the center of mass
by adding one more block.
That's how much I can
shift things over,
depending on how we're
thinking of things,
to the left or the
right, depending on which
direction we're building them.
All right, so now I'm going
to work out the formulas.
First of all C_1, that was
the center of the first block.
I put its left end at 0; the
center of the first block
is at 1.
That means that C_1 is 1.
OK?
C_2 according to this
formula-- And actually I've
worked it out, we'll check
it in a-- C_2 is C_1 + 1/2.
All right, so that's
the case N = 1.
So this is 1 + 1/2.
That's what we already did.
That's the 3/2 number.
Now the next one is C_2 + 1/3.
That's the formula again.
And so that comes out
to be 1 + 1/2 + 1/3.
And now you can see
what the pattern is.
C_N-- If you just
keep on going here,
C_N is going to be 1
+ 1/2 + 1/3 + 1/4...
plus 1/N.
So now I would like
you to vote again.
Do you think I can-- Now
that we have the formula,
do you think I can
get over to here?
How many people think
I can get over to here?
How many people think I
can't get over to here?
There's still a lot
of people who do.
So it's still almost 50/50.
That's amazing.
Well so we'll address
that in a few minutes.
So now let me tell
you what's going on.
This C_N of course, is the same
as what we called last time
S_N.
And remember that we actually
estimated the size of this guy.
This is related to what's
called the harmonic series.
And what we showed
was that log N
is less than S_N, which
is less than S_N + 1.
All right?
Now I'm going to
call your attention
to the red part, which
is the divergence
part of this estimate, which
is this one for the time being,
all right.
Just saying that this
thing is growing.
And what this is saying is
that as N goes to infinity,
log N goes to
infinity, So that means
that S_N goes to infinity,
because of this inequality
here.
It's bigger than log N.
And so if N is big enough,
we can get as far as we like.
All right?
So I can get to here.
And at least half of you,
at least the ones who
voted, that was-- I don't know.
We have a quorum here,
but I'm not sure.
We certainly didn't have
a majority on either side.
Anyway this thing
does go to infinity.
So in principle, if
I had enough blocks,
I could get it over to here.
All right, and
that's the meaning
of divergence in this case.
On the other hand, I want
to discuss with you--
And the reason why
I use this example,
is I want to discuss
with you also what's
going on with this
other inequality here,
and what its significance is.
Which is that it's going to
take us a lot of numbers N,
a lot of blocks, to get
up to a certain level.
In other words, I can't do it
with just eight blocks or nine
blocks.
In order to get
over here, I'd have
to use quite a few of them.
So let's just see
how many it is.
So I worked this out carefully.
And let's see what I got.
So to get across the
lab tables, all right.
This distance here, I
already did this secretly.
But I don't actually even have
enough of these to show you.
But, well 1, 2, 3,
4, 5, 6, and 1/2.
I guess that's enough.
So it's 6 and a half.
So it's two lab tables
is 13 of these blocks.
All right.
So there are 13 blocks,
which is equal to 26 units.
OK, that's how far
to get across I need.
And the first one is already 2.
So it's really 26
minus 2, which is 24.
Which that's what I need.
OK.
So I need log N to be equal
to 24, roughly speaking,
in order to get that far.
So let's just see
how big that is.
All right.
I think I worked this out.
So let's see.
That means that N
is equal to e^24--
and if you realize that these
blocks are 3 centimeters high--
OK let's see how many
that we would need here.
That's kind of a lot.
Let's see, it's 3
centimeters times e^24,
which is about 8*10^8 meters.
OK.
And that is twice the
distance to the moon.
So OK, so I could do it maybe.
But I would need
a lot of blocks.
Right?
So that's not very
plausible here, all right.
So those of you who
voted against this
were actually sort
of half right.
And in fact, if you
wanted to get it
to the wall over there,
which is over 30 feet,
the height would be
about the diameter
of the observable universe.
That's kind of a long way.
There's one other thing
that I wanted to point out
to you about this shape here.
Which is that if you
lean to the left, right,
if you put your
head like this --
of course you have to be on
your side to look at it --
this curve is the shape
of a logarithmic curve.
So in other words, if you think
of the vertical as the x-axis,
and the horizontal that
way is the vertical,
is the up direction,
then this thing
is growing very, very,
very, very slowly.
If you send the x-axis all
the way up to the moon,
the graph still hasn't gotten
across the lab tables here.
It's only partway there.
If you go twice the distance
to the moon up that way,
it's gotten finally to that end.
All right so that's how
slowly the logarithm grows.
It grows very, very slowly.
And if you look at it another
way, if you stand on your head,
you can see an
exponential curve.
So you get some sense as
to the growth properties
of these functions.
And fortunately these
are protecting us
from all kinds of
stuff that would
happen if there
weren't exponentially
small tails in the world.
Like you know, I could
walk through this wall
which I wouldn't like doing.
OK, now so this is
our last example.
And the important
number, unfortunately we
didn't discover another
important number.
There wasn't an amazing number
place where this stopped.
All we discovered again is
some property of infinity.
So infinity is
still a nice number.
And the theme here is just that
infinity isn't just one thing,
it has a character which
is a rate of growth.
And you shouldn't
just think of there
being one order of infinity.
There are lots of
different orders.
And some of them have
different meaning from others.
All right so that's
the theme I wanted
to do, and just have a
visceral example of infinity.
Now, we're going to move
on now to some other kinds
of techniques.
And this is going to
be our last subject.
What we're going
to talk about is
what are known as power series.
And we've already seen
our first power series.
And I'm going to
remind you of that.
Here we are with power series.
Our first series was this one.
And we mentioned last time
that it was equal to 1/(1-x),
for x less than 1.
Well this one is known
as the geometric series.
You didn't use the letter x
last time, I used the letter a.
But this is known as
the geometric series.
Now I'm going to show you
one reason why this is true,
why the formula holds.
And it's just the
kind of manipulation
that was done when these
things were first introduced.
And here's the idea of a proof.
So suppose that this sum
is equal to some number
S, which is the sum of
all of these numbers here.
The first thing
that I'm going to do
is I'm going to multiply by x.
OK, so if I multiply by x.
Let's think about that.
I multiply by x on both the
left and the right-hand side.
Then on the left side, I get x
+ x^2 + x^3 plus, and so forth.
And on the right
side, I get S x.
And now I'm going
to subtract the two
equations, one from the other.
And there's a very, very
substantial cancellation.
This whole tail here
gets canceled off.
And the only thing
that's left is the 1.
So when I subtract, I get
1 on the left-hand side.
And on the right-hand
side, I get S - S x.
All right?
And now that can be
rewritten as S(1-x).
And so I've got my formula here.
This is 1/(1-x) = S. All right.
Now this reasoning has one flaw.
It's not complete.
And this reasoning
is basically correct.
But it's incomplete because
it requires that S exists.
For example, it doesn't make
any sense in the case x = 1.
So for example in
the case x = 1,
we have 1 + 1 +
1 plus et cetera,
equals whatever we call S. And
then when we multiply through
by 1, we get 1 + 1 + 1 plus...
equals S*1.
And now you see that
the subtraction gives us
infinity minus infinity is equal
to infinity minus infinity.
That's what's really going on
in the argument in this context.
So it's just nonsense.
I mean it doesn't give
us anything meaningful.
So this argument, it's great.
And it gives us the right
answer, but not always.
And the times when it gives us
the answer, the correct answer,
is when the series
is convergent.
And that's why we care
about convergence.
Because we want manipulations
like this to be allowed.
So the good case,
this is the red case
that we were
describing last time.
That's the bad case.
But what we want is the good
case, the convergent case.
And that is the case
when x is less than 1.
So this is the convergent case.
Yep.
OK, so they're much
more detailed things
to check exactly
what's going on.
But I'm going to just say
general words about how
you recognize convergence.
And then we're not going
to worry about-- so much
about convergence, because
it works very, very well.
And it's always easy
to diagnose when
there's convergence
with a power series.
All right so here's
the general setup.
The general setup
is that we have
not just the coefficients 1 all
the time, but any numbers here,
dot, dot, dot.
And we abbreviate that with
the summation notation.
This is the sum a_n x^n,
n equals 0 to infinity.
And that's what's known
as a power series.
Fortunately there is
a very simple rule
about how power series converge.
And it's the following.
There's a magic number R which
depends on these numbers here
such that-- And
this thing is known
as a radius of convergence.
In the problem that we had,
it's this number 1 here.
This thing works
for x less than 1.
In our case, it's
maybe x less than R. So
that's some symmetric
interval, right?
That's the same as minus
R less than x less than R,
and so where
there's convergence.
OK, where the series converges.
Converges.
And then there's
the region where
every computation that you
give will give you nonsense.
So x greater than R is
the sum a_n x^n diverges.
And x equals R is very
delicate, borderline,
and will not be used by us.
OK, we're going to stick inside
the radius of convergence.
Now the way you'll be
able to recognize this,
is the following.
What always happens
is that these numbers
tend to 0 exponentially
fast, fast for x in R,
and doesn't even tend to 0
at all for x greater than R.
All right so it'll
be totally obvious.
When you look at
this series here,
what's happening
when x less than R
is that the numbers are getting
smaller and smaller, less
than 1.
When x is bigger
than 1, the numbers
are getting bigger and bigger.
There's no chance that
the series converges.
So that's going to be the
case with all power series.
There's going to be a cutoff.
And it'll be one
particular number.
And below that it'll be obvious
that you have convergence,
and you'll be able
to do computations.
And above that every
formula will be wrong
and won't make sense.
So it's a very clean thing.
There is this very
subtle borderline,
but we're not going to
discuss that in this class.
And it's actually not used in
direct studies of power series.
AUDIENCE: How can you tell
when the numbers are declining
exponentially fast, whereas
just-- In other words
1/x [INAUDIBLE]?
PROFESSOR: OK so,
the question is
why was I able to tell
you this word here?
Why was I able to tell you
not only is it going to 0,
but it's going
exponentially fast?
I'm telling you
extra information.
I'm telling you it always
goes exponentially fast.
You can identify it.
In other words, you'll see it.
And it will happen
every single time.
I'm just promising you
that it works that way.
And it's really
for the same reason
that it works that way
here, that these are powers.
And what's going on
over here is there are,
it's close to powers,
with these a_n's.
All right?
There's a long discussion
of radius of convergence
in many textbooks.
But really it's not necessary,
all right, for this purpose?
Yeah?
AUDIENCE: How do you find R?
PROFESSOR: The question
was how do you find R?
Yes, so I just said,
there's a long discussion
for how you find the radius
of convergence in textbooks.
But we will not be
discussing that here.
And it won't be
necessary for you.
Because it will be obvious in
any given series what the R is.
It will always
either 1 or infinity.
It will always work for
all x, or maybe it'll
stop at some point.
But it'll be very
clear where it stops,
as it is for the
geometric series.
All right?
OK, so now I need to
give you the basic facts,
and give you a few examples.
So why are we looking
at these series?
Well the answer is we're
looking at these series
because the role that
they play is exactly
the reverse of
this equation here.
That is -- and this is a theme
which I have tried to emphasize
throughout this course -- you
can read equalities in two
directions.
Both are interesting, typically.
You can think, I don't know
what the value of this is.
Here's a way of evaluating.
And in other words,
the right side
is a formula for the left side.
Or you can think
of the left side
as being a formula
for the right side.
And the idea of series is
that they're flexible enough
to represent all
of the functions
that we've encountered
in this course.
This is the tool which is very
much like the decimal expansion
which allows you to
represent numbers
like the square root of 2.
Now we're going to be
representing all the numbers,
all the functions that we know:
e^x, arctangent, sine, cosine.
All of those functions
become completely flexible,
and completely available to us,
and computationally available
to us directly.
So that's what
this is a tool for.
And it's just like
decimal expansions
giving you handle
on all real numbers.
So here's how it works.
The rules for
convergent power series
are just like polynomials.
All of the manipulations
that you do for power series
are essentially the
same as for polynomials.
So what kinds of things
do we do with polynomials?
We add them.
We multiply them together.
We do substitutions.
Right?
We take one function
of another function.
We divide them.
OK.
And these are all really not
very surprising operations.
And we will be able to do
them with power series too.
The ones that are interesting,
really interesting
for calculus, are the last two.
We differentiate them,
and we integrate them.
And all of these
operations we'll
be able to do for
power series as well.
So now let's explain
the high points of this.
Which is mainly just
the differentiation
and the integration part.
So if I take a series
like this and so forth,
the formula for its derivative
is just like polynomials.
That's what I just said,
it's just like polynomials.
So the derivative of
the constant is 0.
The derivative of
this term is a_1.
This one is plus 2 a_2 x 2 x.
This one is 3 a_3
x^2, et cetera.
That's the formula.
Similarly if I
integrate, well there's
an unknown constant
which I'm going
to put first rather than last.
Which corresponds sort
of to the a_0 term which
is going to get wiped out.
That a_0 term suddenly
becomes a_0 x.
And the anti-derivative of
this next term is a_1 x^2 / 2.
And the next term is a_2
x^3 / 3, and so forth.
Yeah, question?
AUDIENCE: Is that a
series or a polynomial?
PROFESSOR: Is this a
series or a polynomial?
Good question.
It's a polynomial if it ends.
If it goes on infinitely
far, then it's a series.
They look practically the
same, polynomials and series.
There's this little
dot, dot, dot here.
Is this a series
or a polynomial?
It's the same rule.
If it stops at a
finite stage, this one
stops at a finite stage.
If it goes on forever,
it goes on forever.
AUDIENCE: So I thought that the
series add up finite numbers.
You can add up terms
of x in series?
PROFESSOR: So an
interesting question.
So the question
that was just asked
is I thought that a series
added up finite numbers.
You could add up x?
That was what you said, right?
OK now notice that
I pulled that off
on you by changing the
letter a to the letter
x at the very beginning
of this commentary here.
This is a series.
For each individual value
of x, it's a number.
So in other words, if
I plug in here x = 1/2,
I'm going to add 1
+ 1/2 + 1/4 + 1/8,
and I'll get a
number which is 2.
And I'll plug in a number over
here, and I'll get a number.
On the other hand, I can do
this for each value of x.
So the interpretation of this
is that it's a function of x.
And similarly this
is a function of x.
It works when you plug
in the possible values
x between -1 and 1.
So there's really no
distinction there, it's
just I slipped it passed you.
These are functions of x.
And the notion of a
power series is this idea
that you put
coefficients on a series,
but then you allow
yourself the flexibility
to stick powers here.
And that's exactly
what we're doing.
OK there are other
kinds of series
where you stick other
interesting functions in here
like sines and cosines.
There are lots of other
series that people study.
And these are the simplest ones.
And all those
examples are extremely
helpful for
representing functions.
But we're only going to
do this example here.
All right, so here
are the two rules.
And now there's only one
other complication here
which I have to explain
to you before giving you
a bunch of examples to show you
that this works extremely well.
And the last thing
that I have to do
for you is explain
to you something
called Taylor's formula.
Taylor's formula is the way you
get from the representations
that we're used to of functions,
to a representation in the form
of these coefficients.
When I gave you
the function e^x,
it didn't look
like a polynomial.
And we have to figure
out which of these guys
it is, if it's going to
fall into our category here.
And here's the formula.
I'll explain to you how
it works in a second.
So the formula is
f(x) turns out--
There's a formula in terms
of the derivatives of f.
Namely, you
differentiate n times,
and you evaluate it at 0, and
you divide by n factorial,
and multiply by x^n.
So here's Taylor's formula.
This tells you what
the Taylor series is.
Now about half of our job
for the next few minutes
is going to be to
give examples of this.
But let me just explain
to you why this has to be.
If you pick out this number
here, this is the a_n,
the magic number a_n here.
So let's just illustrate it.
If f(x) happens to be a_0 + a_1
x + a_2 x^2 + a_3 x^3 plus dot,
dot, dot.
And now I differentiate
it, right?
I get a 1 + 2 a_2 x + 3 a_3 x.
If I differentiate it another
time, I get 2 a_2 plus 3--
sorry, 3*2*a_3 x
plus dot, dot, dot.
And now a third time, I
get 3*2 a_3 plus et cetera.
So this next term is really in
disguise, 4*3*2 x a-- sorry,
a_4 x.
That's what really comes down if
I kept track of the fourth term
there.
So now here is my function.
But now you see if
I plug in x = 0,
I can pick off the third term.
f triple prime of 0
is equal to 3*2 a_3.
Right, because all the rest of
those terms, when I plug in 0,
are just 0.
Here's the formula.
And so the pattern here is this.
And what's really going on here
is this is really 3*2*1 a_3.
And in general a_n is
equal to f, nth derivative,
divided by n!.
And of course, n!, I remind
you, is n times n-1 times
n-2, all the way down to 1.
Now there's one more
crazy convention
which is always used.
Which is that there's something
very strange here down at 0,
which is that 0 factorial turns
out, has to be set equal to 1.
All right, so that's
what you do in order
to make this formula work out.
And that's one of the
reasons for this convention.
All right.
So my next goal is to
give you some examples.
And let's do a couple.
So here's, well
you know, I'm going
to have to let you see
a few of them next time.
But let me just tell
you this one, which
is by far the most impressive.
So what happens with e^x -- if
the function is f(x) = e^x --
is that its derivative
is also e^x.
And its second
derivative is also e^x.
And it just keeps
on going that way.
They're all the same.
So that means that these
numbers in Taylor's formula,
in the numerator--
The nth derivative
is very easy to evaluate.
It's just e^x.
And if I evaluate it
at x = 0, I just get 1.
So all of those
numerators are 1.
So the formula here, is the sum
n equals 0 to infinity, of 1/n!
x^n.
In particular, we now have
an honest formula for e
to the first power.
Which is just e.
Which if I plug it in, x = 1, I
get 1, this is the n = 0 term.
Plus 1, This is the n = 1 term.
Plus 1/2!
plus 1/3!
plus 1/4!
Right?
So this is our first
honest formula for e.
And also, this is
how you compute
the exponential function.
Finally if you take a
function like sin x,
what you'll discover is that
we can complete the sort
of strange business that we did
at the beginning of the course
-- or cos x -- where we took
the linear and quadratic
approximations.
Now we're going to get complete
formulas for these functions.
sin x turns out to be
equal to x - x^3 / 3!
+ x^5 / 5!
- x^7 / 7!, et cetera.
And cos x = 1 - x^2 / 2!
-- that's the same as this
2 here -- plus x^4 / 4!
minus x^6 / 6!, plus et cetera.
Now these may feel like they're
hard to memorize because I've
just pulled them out of a hat.
I do expect you to know them.
They're actually extremely
similar formulas.
The exponential here just has
this collection of factorials.
The sine is all the odd
powers with alternating signs.
And the cosine is all the even
powers with alternating signs.
So all three of them form
part of the same family.
So this will actually
make it easier
for you to remember,
rather than harder.
And so with that, I'll
leave the practice
on differentiation
for next time.
And good luck, everybody.
I'll talk to you individually.
