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Well, let's get started.

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The topic for today is --

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Sorry.
Thank you.

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For today and the next two
lectures, we are going to be

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studying Fourier series.
Today will be an introduction

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explaining what they are.
And, I calculate them,

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but I thought before we do that
I ought to least give a couple

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minutes oversight of why and
where we're going with them,

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and why they're coming into the
course at this place at all.

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So, the situation up to now is
that we've been trying to solve

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equations of the form y double
prime plus a y prime,

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constant coefficient
second-order equations,

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and the f of t was the input.
So, we are considering

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inhomogeneous equations.
This is the input.

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And so far, the response,
then, is the solution equals

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the corresponding solution,
y of t,

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maybe with some given initial
conditions to pick out a special

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one we call the response,
the response to that particular

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input.
And now, over the last few

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days, the inputs have been,
however, extremely special.

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For input, the basic input has
been an exponential,

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or sines and cosines.
And, the trouble is that we

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learn how to solve those.
But the point is that those

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seem extremely special.
Now, the point of Fourier

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series is to show you that they
are not as special as they look.

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The reason is that,
let's put it this way,

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that any reasonable f of t
which is periodic,

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it doesn't have to be even very
reasonable.

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It can be somewhat
discontinuous,

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although not terribly
discontinuous,

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which is periodic with period,
maybe not the minimal period,

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but some period two pi.
Of course, sine t

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and cosine t have the
exact period two pi,

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but if I change the frequency
to an integer frequency like

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sine 2t or sine 26 t,

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two pie would still be a
period, although would not be

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the period.
The period would be shorter.

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The point is,
such a thing can always be

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represented as an infinite sum
of sines and cosines.

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So, it's going to look like
this.

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There's a constant term you
have to put out front.

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And then, the rest,
instead of writing,

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it's rather long to write
unless you use summation

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notation.
So, I will.

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So, it's a sum from n equal one
to infinity integer values of n,

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in other words,
of a sine and a cosine.

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It's customary to put the
cosine first,

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and with the frequency,
the n indicates the frequency

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of the thing.
And, the bn is sine nt.

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Now, why does that solve the

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problem of general inputs for
periodic functions,

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at least if the period is two
pi or some fraction of it?

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Well, you could think of it
this way.

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I'll make a little table.
I'll make a little table.

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Let's look at,
let's put over here the input,

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and here, I'll put the
response.

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Okay, suppose the input is the
function sine nt.

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Well, in other words,
if you just solve the problem,

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you put a sine nt
here, you know how to get the

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answer, find a particular
solution, in other words.

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In fact, you do it by
converting this to a complex

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exponential, and then all the
rigmarole we've been going

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through.
So, let's call the response

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something.
Let's call it y.

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I'd better index it by n
because it, of course,

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is a response to this
particular periodic function.

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So, n of t,
and if the input is cosine nt,

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that also will have
a response, yn.

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Now, I really can't call them
both by the same name.

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So, why don't we put a little s
up here to indicate that that's

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the response to the sine.
And here, I'll put a little c

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to indicate what the answer to
the cosine.

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You're feeding cosine nt,
what you get out is

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this function.
Now what?

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Well, by the way,
notice that if n is zero,

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it's going to take care of a
constant term,

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too.
In other words,

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the reason there is a constant
term out front is because that

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corresponds to cosine of zero t,
which is one.

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Now, suppose I input instead an
cosine nt.

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All you do is multiply the
answer by an.

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Same here.
Multiply the input by bn.

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You multiply the response.
That's because the equation is

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a linear equation.
And now, what am I going to do?

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I'm going to add them up.
If I add them up from the

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different ends and take a count
also, the n equals zero

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corresponding to this first
constant term,

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the sum of all these according
to my Fourier formula is going

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to be f of t.
What's the sum of this,

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the corresponding responses?
Well, that's going to be

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summation a n y n c t 
plus b n y n,

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the response to the
sine.

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That will be the sum from one
to infinity, and there will be

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some sort of constant term here.
Let's just call it c1.

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So, in other words,
if this input produces that

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response, and these are things
which we can calculate,

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we're led by this formula,
Fourier's formula,

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to the response to things which
otherwise we would have not been

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able to calculate,
namely, any periodic function

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of period two pi will have,
the procedure will be,

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you've got a periodic function
of period two pi.

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Find its Fourier series,
and I'll show you how to do

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that today.
Find its Fourier series,

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and then the response to that
general f of t will be this

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infinite series of functions,
where these things are things

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you already know how to
calculate.

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They are the responses to sines
and cosines.

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And, you just formed the sum
with those coefficients.

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Now, why does that work?
It works by the superposition

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principle.
So, this is true.

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The reason I can do the adding
and multiplying by constant,

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I'm using the superposition
principle.

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If this input produces that
response, then the sum of a

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bunch of inputs produces the sum
of the corresponding responses.

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And, why is that?
Why can I use the superposition

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principle?
Because the ODE is linear.

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It's okay, since the ODE is
linear.

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That's what makes all this
work.

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Now, so what we're going to do
today is I will show you how to

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calculate those Fourier series.
I will not be able to use it to

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actually solve any differential
equation.

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It will take us pretty much all
the period to show how to

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calculate a Fourier series.
And, okay, so I'm going to

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solve differential equations on
Monday.

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Wrong.
I probably won't even get to it

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then because the calculation of
a Fourier series is a sufficient

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amount of work that you really
want to know all the possible

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tricks and shortcuts there are.
Unfortunately,

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they are not very clever
tricks.

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They are just obvious things.
But, it will take me a period

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to point out those obvious
things, obvious in my sense if

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not in yours.
And, finally,

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the third day,
we'll solve differential

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equations.
I will actually carry out the

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program.
But the main thing we're going

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to get out of it is another
approach to resonance because

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the things that we are going to
be interested in are picking out

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which of these terms may
possibly produce resonance,

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and therefore a very crazy
response.

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Some of the terms in the
response suddenly get a much

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bigger amplitude than this than
you would normally have thought

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they had because it's picking
out resonant terms in the

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Fourier series of the input.
Okay, well, that's a big

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mouthfu.
Let's get started on

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calculating.
So, the program today is

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calculate the Fourier series.
Given f of t periodic,

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having two pi as a period,
find its Fourier series.

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How, in other words,
do I calculate those

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coefficients,
an and bn.

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Now, the answer is not
immediately apparent,

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and it's really quite
remarkable.

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I think it's quite remarkable,
anyway.

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It's one of the basic things of
higher mathematics.

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And, what it depends upon are
certain things called the

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orthogonality relations.
So, this is the place where

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you've got to learn what such
things are.

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Well, I think it would be a
good idea to have a general

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definition, rather than
immediately get into the

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specifics.
So, I'm going to call u of x,

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u of t, I think I will use,

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since Fourier analysis is most
often applied when the variable

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is time, I think I will stick to
independent variable t all

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period long, if I remember to,
at any rate.

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So, these are two continuous,
or not very discontinuous

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functions on minus pi.
Let's make them periodic.

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Let's say two pi is a period.
So, functions,

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for example like those guys,
sine t, sine nt,

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sine 22t,

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and so on, say two pi is a
period.

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Well, I want them really on the
whole real axis,

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not there.
Define for all real numbers.

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Then, I say that they are
orthogonal, perpendicular.

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But nobody says perpendicular.
Orthogonal is the word,

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orthogonal on the interval
minus pi to pi

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if the integral,
so, two are orthogonal.

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Well, these two functions,
if the integral from minus pi

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to pi of u of t v of t,
the product is zero,

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that's called the orthogonality
condition on minus pi to pi.

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Now, well, it's just the

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definition.
I would love to go into a

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little song and dance now on
what the definition really

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means, and what its application,
why the word orthogonal is

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used, because it really does
have something to do with two

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vectors being orthogonal in the
sense in which you live it in

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18.02.
I'll have to put that on the

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ice for the moment,
and whether I get to it or not

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depends on how fast I talk.
But, you probably prefer I talk

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slowly.
So, let's compromise.

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Anyway, that's the condition.
And now, what I say is that

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that Fourier,
that blue Fourier series,

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--
-- what finding the

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coefficients an and bn depends
upon is this theorem that the

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collection of functions,
as I look at this collection of

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functions, sine nt
for any value of the integer,

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n, of course I can assume n is
a positive integer because sine

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of minus nt is the same as sine
of nt.

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And, cosine mt,
let's give it a different,

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so I don't want you to think
they are exactly the same

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integers.
So, this is a big collection of

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functions, as n runs from one to
infinity-- Here,

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I could let m be run from zero
to infinity because cosine of

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zero t means something.

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It's a constant,
one-- that any two distinct

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ones, two distinct,
you know, how can two things be

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not different?
Well, you know,

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you talk about two coincident
roots.

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I'm just killing,
doing a little overkill.

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Any two distinct ones of these,
two distinct members of the set

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of this collection of,
I don't know,

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there's no way to say that,
any two distinct ones are

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orthogonal on this interval.
Of course, they all have two pi

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as a period for all of them.
So, they form into this general

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category that I'm talking about,
but any two distinct ones are

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orthogonal on the interval for
minus pi to pi.

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So, if I integrate from minus

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pi to pi sine of three t times
cosine of four t dt,

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answer is zero.

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If I integrate sine of 3t times

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the sine of 60t,
answer is zero.

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The same thing with two
cosines, or a sine and a cosine.

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The only time you don't get
zero is if you integrate,

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if you make the two functions
the same.

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Now, how do you know that you
could not possibly get the

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answer is zero if the two
functions are the same?

227
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If the two functions are the
same, then I'm integrating a

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square.
A square is always positive.

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I'm integrating a square.
A square is always positive,

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and therefore I cannot get the
answer, zero.

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But, in the other cases,
I might get the answer zero.

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And the theorem is you always
do.

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Okay, so, why is this?
Well, there are three ways to

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prove this.
It's like many fundamental

235
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facts in mathematics.
There are different ways of

236
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going about it.
By the way, along with the

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theorem, I probably should have
included, so,

238
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I'm far away.
But you might as well include,

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because we're going to need it.
What happens if you use the

240
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same function?
If I take U equal to V,

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and in that case,
as I've indicated,

242
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you're not going to get the
answer, zero.

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But, what you will get is,
so, in other words,

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I'm just asking,
what is the sine of 

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n t squared.
That's a case where two of them

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are the same.
I use the same function.

247
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What's that?
Well, the answer is,

248
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it's the same as what you will
get if you integrate the cosine,

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cosine squared n t dt.

250
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And, the answer to either one

251
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of these is pi.
That's something you know how

252
00:19:06 --> 00:19:12
to do from 18.01 or the
equivalent thereof.

253
00:19:09 --> 00:19:15
You can integrate sine squared.
It's one of the things you had

254
00:19:14 --> 00:19:20
to learn for whatever exam you
took on methods of integration.

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Anyway, so I'm not going to
calculate this out.

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The answer turns out to be pi.
All right, now,

257
00:19:27 --> 00:19:33
the ways to prove it are you
can use trig identities.

258
00:19:33 --> 00:19:39
And, I'm asking you in one of
the early problems in the

259
00:19:37 --> 00:19:43
problem set, identities,
identities for the product of

260
00:19:41 --> 00:19:47
sine and cosine,
expressing it in a form in

261
00:19:44 --> 00:19:50
which it's easy to integrate,
and you can prove it that way.

262
00:19:48 --> 00:19:54
Or, you can use,
if you have forgotten the

263
00:19:51 --> 00:19:57
trigonometric identities and
want to get some more exercise

264
00:19:56 --> 00:20:02
with complex-- you can use
complex exponentials.

265
00:20:01 --> 00:20:07
So, I'm asking you how to,
in another part of the same

266
00:20:05 --> 00:20:11
problem I'm asking you how to do
it, do one of these,

267
00:20:09 --> 00:20:15
at any rate,
using complex exponentials.

268
00:20:13 --> 00:20:19
And now, I'm going to use a
mysterious third method another

269
00:20:18 --> 00:20:24
way.
I'm going to use the ODE.

270
00:20:20 --> 00:20:26
I'm going to do that because
this is the method.

271
00:20:24 --> 00:20:30
It's not just sines and cosines
which are orthogonal.

272
00:20:30 --> 00:20:36
There are masses of orthogonal
functions out there.

273
00:20:33 --> 00:20:39
And, the way they are
discovered, and the way you

274
00:20:36 --> 00:20:42
prove they're orthogonal is not
with trig identities and complex

275
00:20:40 --> 00:20:46
exponentials because those only
work with sines and cosines.

276
00:20:44 --> 00:20:50
It is, instead,
by going back to the

277
00:20:46 --> 00:20:52
differential equation that they
solve.

278
00:20:48 --> 00:20:54
And that's, therefore,
the method here that I'm going

279
00:20:52 --> 00:20:58
to use here because this is the
method which generalizes to many

280
00:20:56 --> 00:21:02
other differential equations
other than the simple ones

281
00:20:59 --> 00:21:05
satisfied by sines and cosines.
But anyway, that is the source.

282
00:21:05 --> 00:21:11
So, the way the proof of these
orthogonality conditions goes,

283
00:21:09 --> 00:21:15
so I'm not going to do that.
And, I'm going to assume that m

284
00:21:14 --> 00:21:20
is different from n so that I'm
not in either of these two

285
00:21:18 --> 00:21:24
cases.
What it depends on is,

286
00:21:20 --> 00:21:26
what's the differential
equation that all these

287
00:21:23 --> 00:21:29
functions satisfy?
Well, it's a different

288
00:21:26 --> 00:21:32
differential equation depending
upon the value of n,

289
00:21:30 --> 00:21:36
--
-- but they look at essentially

290
00:21:35 --> 00:21:41
the same.
These satisfy the differential

291
00:21:38 --> 00:21:44
equation, in other words,
what they have in common.

292
00:21:43 --> 00:21:49
The differential equation is,
let's call it u.

293
00:21:48 --> 00:21:54
It looks better.
It's going to look better if

294
00:21:52 --> 00:21:58
you let me call it u.
u double prime plus,

295
00:21:56 --> 00:22:02
well, n squared,
so for the function sine n t

296
00:22:00 --> 00:22:06
cosine n t, satisfy u double

297
00:22:05 --> 00:22:11
prime plus n squared times u.

298
00:22:11 --> 00:22:17
In other words,
the frequency is n,

299
00:22:13 --> 00:22:19
and therefore,
this is a square of the

300
00:22:16 --> 00:22:22
frequency is what you put here,
equals zero.

301
00:22:19 --> 00:22:25
In other words,
what these functions have in

302
00:22:22 --> 00:22:28
common is that they satisfy
differential equations that look

303
00:22:26 --> 00:22:32
like that.
And the only thing that's

304
00:22:28 --> 00:22:34
allowed to vary is the
frequency, which is allowed to

305
00:22:32 --> 00:22:38
change.
The frequency is in this

306
00:22:36 --> 00:22:42
coefficient of u.
Now, the remarkable thing is

307
00:22:42 --> 00:22:48
that's all you need to know.
The fact that they satisfy the

308
00:22:49 --> 00:22:55
differential equation,
that's all you need to know to

309
00:22:55 --> 00:23:01
prove the orthogonality
relationship.

310
00:22:59 --> 00:23:05
Okay, let's try to do it.
Well, I need some notation.

311
00:23:06 --> 00:23:12
So, I'm going to let un and vm
be any two of the functions.

312
00:23:11 --> 00:23:17
In other words,
I'll assume m is different from

313
00:23:16 --> 00:23:22
n.
For example,

314
00:23:17 --> 00:23:23
this one could be sine nt,
and that could be

315
00:23:22 --> 00:23:28
sine of mt,
or this could be sine nt

316
00:23:26 --> 00:23:32
and that could be
cosine of mt.

317
00:23:33 --> 00:23:39
You get the idea.
Any two of those in the

318
00:23:35 --> 00:23:41
subscript indicates whether what
the n or the m is that are in

319
00:23:40 --> 00:23:46
that.
Any two, and I mean really two,

320
00:23:42 --> 00:23:48
distinct, well,
if I say that m is not n,

321
00:23:45 --> 00:23:51
then they positively have to be
different.

322
00:23:48 --> 00:23:54
So, again, it's overkill with
my two's-ness.

323
00:23:51 --> 00:23:57
And, what I'm going to
calculate, well,

324
00:23:53 --> 00:23:59
first of all,
from the equation,

325
00:23:56 --> 00:24:02
I'm going to write the equation
this way.

326
00:24:00 --> 00:24:06
It says that u double prime is
equal to minus n squared u.

327
00:24:07 --> 00:24:13
That's true for any of these

328
00:24:11 --> 00:24:17
guys.
Of course, here,

329
00:24:13 --> 00:24:19
it would be v double prime is
equal to minus m squared 

330
00:24:20 --> 00:24:26
times v.
You have to make those simple

331
00:24:26 --> 00:24:32
adjustments.
And now, what we're going to

332
00:24:30 --> 00:24:36
calculate is the integral from
minus pi to pi of un double

333
00:24:37 --> 00:24:43
prime times vm dt.

334
00:24:43 --> 00:24:49
Now, just bear with me.

335
00:24:48 --> 00:24:54
Why am I going to do that?
I can't explain what I'm going

336
00:24:53 --> 00:24:59
to do that.
But you won't ask me the

337
00:24:56 --> 00:25:02
question in five minutes.
But the point is,

338
00:24:59 --> 00:25:05
this is highly un-symmetric.
The u is differentiated twice.

339
00:25:05 --> 00:25:11
The v isn't.
So, those two functions-- but

340
00:25:08 --> 00:25:14
there is a way of turning them
into an expression which looks

341
00:25:12 --> 00:25:18
extremely symmetric,
where they are the same.

342
00:25:16 --> 00:25:22
And the way to do that is I
want to get rid of one of these

343
00:25:20 --> 00:25:26
primes here and put one on here.
The way to do that is if you

344
00:25:25 --> 00:25:31
want to integrate one of these
guys, and differentiate this one

345
00:25:29 --> 00:25:35
to make them look the same,
that's called integration by

346
00:25:33 --> 00:25:39
parts, the most important
theoretical method you learned

347
00:25:38 --> 00:25:44
in 18.01 even though you didn't
know that it was the most

348
00:25:42 --> 00:25:48
important theoretical method.
Okay, we're going to use it now

349
00:25:47 --> 00:25:53
as a basis for Fourier series.
Okay, so I'm going to integrate

350
00:25:51 --> 00:25:57
by parts.
Now, the first thing you do,

351
00:25:53 --> 00:25:59
of course, when you integrate
by parts is you just do the

352
00:25:56 --> 00:26:02
integration.
You don't do differentiation.

353
00:25:59 --> 00:26:05
So, the first thing looks like
this.

354
00:26:02 --> 00:26:08
And, that's to be evaluated
between negative pi and pi.

355
00:26:08 --> 00:26:14
In doing integration by parts
between limits,

356
00:26:12 --> 00:26:18
minus what you get by doing
both.

357
00:26:16 --> 00:26:22
You do both,
the integration and the

358
00:26:20 --> 00:26:26
differentiation.
And, again, evaluate that

359
00:26:24 --> 00:26:30
between limits.
Now, I'm just going to BS my

360
00:26:29 --> 00:26:35
way through this.
This is zero.

361
00:26:34 --> 00:26:40
I don't care what the un's,
which un you picked and which

362
00:26:39 --> 00:26:45
vm you picked.
The answer here is always going

363
00:26:43 --> 00:26:49
to be zero.
Instead of wasting six boards

364
00:26:46 --> 00:26:52
trying to write out the
argument, let me wave my hands.

365
00:26:51 --> 00:26:57
Okay, it's clear,
for example,

366
00:26:54 --> 00:27:00
that a v is a sine, sine mt.

367
00:26:57 --> 00:27:03
Of course it's zero because the
sine vanishes at both pi and

368
00:27:02 --> 00:27:08
minus pi.
If the un were a cosine,

369
00:27:06 --> 00:27:12
after I differentiate it,
it became a sine.

370
00:27:09 --> 00:27:15
And so, now it's this side guy
that's zero at both ends.

371
00:27:14 --> 00:27:20
So, the only case in which we
might have a little doubt is if

372
00:27:18 --> 00:27:24
this is a cosine,
and after differentiation,

373
00:27:21 --> 00:27:27
this is also a cosine.
In other words,

374
00:27:24 --> 00:27:30
it might look like cosine,
after, this cosine nt times

375
00:27:28 --> 00:27:34
cosine mt.
But, I claim that that's zero,

376
00:27:34 --> 00:27:40
too.
Why?

377
00:27:35 --> 00:27:41
Because the cosines are even
functions, and therefore,

378
00:27:39 --> 00:27:45
they have the same value at
both ends.

379
00:27:42 --> 00:27:48
So, if I subtract the value
evaluated at pi,

380
00:27:46 --> 00:27:52
and subtract the value of minus
pi, again zero because I have

381
00:27:51 --> 00:27:57
the same value at both ends.
So, by this entirely convincing

382
00:27:56 --> 00:28:02
argument, no matter what
combination of sines and cosines

383
00:28:00 --> 00:28:06
I have here, the answer to that
part will always be zero.

384
00:28:07 --> 00:28:13
So, by calculation,
but thought calculation;

385
00:28:11 --> 00:28:17
it's just a waste of time to
write anything out.

386
00:28:16 --> 00:28:22
You stare at it until you agree
that it's so.

387
00:28:20 --> 00:28:26
And now, I've taken,
by this integration by parts,

388
00:28:25 --> 00:28:31
I've taken this highly
un-symmetric expression and

389
00:28:30 --> 00:28:36
turned it into something in
which the u and the v are

390
00:28:35 --> 00:28:41
treated exactly alike.
Well, good, that's nice,

391
00:28:40 --> 00:28:46
but why?
Why did I go to this trouble?

392
00:28:43 --> 00:28:49
Okay, now we're going to use
the fact that this satisfies the

393
00:28:47 --> 00:28:53
differential equation,
in other words,

394
00:28:50 --> 00:28:56
that u double prime is equal to
minus n,

395
00:28:53 --> 00:28:59
I'm sorry, I should have
subscripted this.

396
00:28:56 --> 00:29:02
If that's the solution,
then this is equal to,

397
00:29:00 --> 00:29:06
times.
You have to put in a subscript

398
00:29:02 --> 00:29:08
otherwise.
The n wouldn't matter.

399
00:29:06 --> 00:29:12
All right, I'm now going to
take that expression,

400
00:29:10 --> 00:29:16
and evaluate it differently.
un double prime vm dt

401
00:29:15 --> 00:29:21
is equal to,
well, un double prime,

402
00:29:18 --> 00:29:24
because it satisfies the
differential equation is equal

403
00:29:22 --> 00:29:28
to that.
So, what is this?

404
00:29:25 --> 00:29:31
This is minus n squared
times the integral from

405
00:29:29 --> 00:29:35
negative pi to pi,
and I'm replacing un double

406
00:29:33 --> 00:29:39
prime by minus n
squared un.

407
00:29:39 --> 00:29:45
I pulled the minus n squared
out.

408
00:29:43 --> 00:29:49
So, it's un here,
and the other factor is vm dt.

409
00:29:47 --> 00:29:53
Now, that's the proof.
Huh?

410
00:29:50 --> 00:29:56
What do you mean that's the
proof?

411
00:29:54 --> 00:30:00
Okay, well, I'll first state
it, why intuitively that's the

412
00:29:59 --> 00:30:05
end of the argument.
And then, I'll spell it out a

413
00:30:06 --> 00:30:12
little more detail,
but the more detail you make

414
00:30:11 --> 00:30:17
for this, the more obscure it
gets instead of,

415
00:30:16 --> 00:30:22
look, I just showed you that
this is symmetric in u and v,

416
00:30:22 --> 00:30:28
after you massage it a little
bit.

417
00:30:26 --> 00:30:32
Here, I'm calculating it a
different way.

418
00:30:30 --> 00:30:36
Is this symmetric in u and v?
Well, the answer is yes or no.

419
00:30:37 --> 00:30:43
Is this symmetric at u and v?
No.

420
00:30:40 --> 00:30:46
Why?
Because of the n.

421
00:30:42 --> 00:30:48
The n favors u.
We have what is called a

422
00:30:46 --> 00:30:52
paradox.
This thing is symmetric in u

423
00:30:50 --> 00:30:56
and v because I can show it is.
And, it's not symmetric in u

424
00:30:55 --> 00:31:01
and v because I can show it is.
I can show it's not symmetric

425
00:31:01 --> 00:31:07
because it favors the n.
Now, there's only one possible

426
00:31:09 --> 00:31:15
resolution of that paradox.
Both would be symmetric if what

427
00:31:19 --> 00:31:25
were true?
Pardon?

428
00:31:22 --> 00:31:28
Negative pi.
All right, let me write it this

429
00:31:29 --> 00:31:35
way.
Okay, never mind.

430
00:31:32 --> 00:31:38
You see, the only way this can
happen is if this expression is

431
00:31:37 --> 00:31:43
zero.
In other words,

432
00:31:39 --> 00:31:45
the only way something can be
both symmetric and not symmetric

433
00:31:44 --> 00:31:50
is if it's zero all the time.
And, that's what we're trying

434
00:31:48 --> 00:31:54
to prove, that this is zero.
But, instead of doing it that

435
00:31:53 --> 00:31:59
way, let me show you.
This is equal to that,

436
00:31:57 --> 00:32:03
and therefore,
two things according to Euclid,

437
00:32:00 --> 00:32:06
two things equal to the same
thing are equal to each other.

438
00:32:07 --> 00:32:13
So, this equals that,
which, in turn,

439
00:32:09 --> 00:32:15
therefore, equals what I would
have gotten.

440
00:32:12 --> 00:32:18
I'm just saying the symmetry of
different way,

441
00:32:15 --> 00:32:21
what I would have gotten if I
had done this calculation.

442
00:32:19 --> 00:32:25
And, that turns out to be minus
m squared times the integral

443
00:32:23 --> 00:32:29
from minus pi to pi
of un vm dt.

444
00:32:28 --> 00:32:34
So, these two are equal because

445
00:32:33 --> 00:32:39
they are both equal to this.
This is equal to that.

446
00:32:38 --> 00:32:44
This equals that.
Therefore, how can this equal

447
00:32:42 --> 00:32:48
that unless the integral is
zero?

448
00:32:46 --> 00:32:52
How's that?
Remember, m is different from

449
00:32:50 --> 00:32:56
n.
So, what this proves is,

450
00:32:52 --> 00:32:58
therefore, the integral from
negative pi to pi of un vm dt is

451
00:32:59 --> 00:33:05
equal to zero,

452
00:33:05 --> 00:33:11
at least if m is different from
n.

453
00:33:10 --> 00:33:16
Now, there is one case I didn't
include.

454
00:33:12 --> 00:33:18
Which case didn't I include?
un times un is not supposed to

455
00:33:16 --> 00:33:22
be zero.
So, in that case,

456
00:33:18 --> 00:33:24
I don't have to worry about,
but there is a case that I

457
00:33:22 --> 00:33:28
didn't.
For example,

458
00:33:24 --> 00:33:30
something like the cosine of nt
times the sine of nt.

459
00:33:28 --> 00:33:34
Here, the m is the same as the

460
00:33:32 --> 00:33:38
n.
Nonetheless,

461
00:33:34 --> 00:33:40
I am claiming that this is zero
because these aren't the same

462
00:33:39 --> 00:33:45
function.
One is a cosine.

463
00:33:42 --> 00:33:48
Why is that zero?
Can you see mentally that

464
00:33:46 --> 00:33:52
that's zero?
Mentally?

465
00:33:48 --> 00:33:54
Well, this is trying to be in
another life,

466
00:33:52 --> 00:33:58
it's trying to be one half the
sine of two nt, right?

467
00:33:57 --> 00:34:03
And obviously the integral of

468
00:34:02 --> 00:34:08
sine of two nt is zero between
minus pi and pi

469
00:34:06 --> 00:34:12
because you integrate it,

470
00:34:09 --> 00:34:15
and it turns out to be zero.
You integrate it to a cosine,

471
00:34:13 --> 00:34:19
which has the same value of
both ends.

472
00:34:16 --> 00:34:22
Well, that was a lot of
talking.

473
00:34:18 --> 00:34:24
If this proof is too abstract
for you, I won't ask you to

474
00:34:22 --> 00:34:28
reproduce it on an exam.
You can go with the proofs

475
00:34:25 --> 00:34:31
using trigonometric identities,
and/or complex exponentials.

476
00:34:31 --> 00:34:37
But, you ought to know at least
one of those,

477
00:34:34 --> 00:34:40
and for the problem set I'm
asking you to fool around a

478
00:34:39 --> 00:34:45
little with at least two of
them.

479
00:34:41 --> 00:34:47
Okay, now, what has this got to
do with the problem we started

480
00:34:47 --> 00:34:53
with originally?
The problem is to explain this

481
00:34:50 --> 00:34:56
blue series.
So, our problem is,

482
00:34:53 --> 00:34:59
how, from this,
am I going to get the terms of

483
00:34:57 --> 00:35:03
this blue series?
So, given f of t,

484
00:35:02 --> 00:35:08
two pi s a period.
Find the an and the bn.

485
00:35:06 --> 00:35:12
Okay, let's focus on the an.
The bn is the same.

486
00:35:11 --> 00:35:17
Once you know how to do one,
you know how to do the other.

487
00:35:16 --> 00:35:22
So, here's the idea.
Again, it goes back to the

488
00:35:21 --> 00:35:27
something you learned at the
very beginning of 18.02,

489
00:35:26 --> 00:35:32
but I don't think it took.
But maybe some of you will

490
00:35:32 --> 00:35:38
recognize it.
So, what I'm going to do is

491
00:35:36 --> 00:35:42
write it.
Here's the term we're looking

492
00:35:40 --> 00:35:46
for here, this one.
Okay, and there are others.

493
00:35:45 --> 00:35:51
It's an infinite series that
goes on forever.

494
00:35:50 --> 00:35:56
And now, to make the argument,
I've got to put it one more

495
00:35:56 --> 00:36:02
term here.
So, I'm going to put in ak

496
00:36:00 --> 00:36:06
cosine kt.
I don't mean to imply that that

497
00:36:07 --> 00:36:13
k could be more than n,
in which case I should have

498
00:36:11 --> 00:36:17
written it here.
I could have also used equally

499
00:36:16 --> 00:36:22
well bk sine kt 
here, and I could have put it

500
00:36:22 --> 00:36:28
there.
This is just some other term.

501
00:36:25 --> 00:36:31
This is the an,
and this is the one we want.

502
00:36:30 --> 00:36:36
And, this is some other term.
Okay, all right,

503
00:36:35 --> 00:36:41
now, what you do is,
to get the an,

504
00:36:38 --> 00:36:44
what you do is you multiply
everything through by,

505
00:36:42 --> 00:36:48
you focus on the one you want,
so it's dot,

506
00:36:46 --> 00:36:52
dot, dot, dot,
dot, and you multiply by cosine

507
00:36:50 --> 00:36:56
nt.
So, it's ak cosine kt times

508
00:36:54 --> 00:37:00
cosine nt.

509
00:36:57 --> 00:37:03
Of course, that gets
multiplied, too.

510
00:37:02 --> 00:37:08
But, the one we want also gets
multiplied, an.

511
00:37:06 --> 00:37:12
And, it becomes,
when I multiply by cosine nt,

512
00:37:11 --> 00:37:17
cosine squared nt,

513
00:37:16 --> 00:37:22
and now, I hope you can see
what's going to happen.

514
00:37:21 --> 00:37:27
Now, oops, I didn't multiply
the f of t,

515
00:37:26 --> 00:37:32
sorry.
It's the oldest trick in the

516
00:37:30 --> 00:37:36
book.
I now integrate everything from

517
00:37:35 --> 00:37:41
minus, so I don't endlessly
recopy.

518
00:37:38 --> 00:37:44
I'll integrate by putting it up
in yellow chalk,

519
00:37:42 --> 00:37:48
and you are left to your own
devices.

520
00:37:46 --> 00:37:52
This is definitely a colored
pen type of course.

521
00:37:50 --> 00:37:56
Okay, so, you want to integrate
from minus pi to pi?

522
00:37:55 --> 00:38:01
Good.
Just integrate everything on

523
00:37:59 --> 00:38:05
the right hand side,
also, from minus pi to pi.

524
00:38:05 --> 00:38:11
Plus, these are the guys just
to indicate that I haven't,

525
00:38:10 --> 00:38:16
they are out there,
too.

526
00:38:13 --> 00:38:19
And now, what happens?
What's this?

527
00:38:16 --> 00:38:22
Zero.
Every term is zero because of

528
00:38:20 --> 00:38:26
the orthogonality relations.
They are all of the form,

529
00:38:25 --> 00:38:31
a constant times cosine nt
times something different from

530
00:38:31 --> 00:38:37
cosine nt, sine kt,

531
00:38:35 --> 00:38:41
cosine kt,
or even that constant term.

532
00:38:42 --> 00:38:48
All of the other terms are
zero, and the only one which

533
00:38:46 --> 00:38:52
survives is this one.
And, what's its value?

534
00:38:50 --> 00:38:56
The integral from minus pi to
pi of cosine squared,

535
00:38:54 --> 00:39:00
I put that up somewhere.
It's right here,

536
00:38:57 --> 00:39:03
down there?
It is pi.

537
00:39:00 --> 00:39:06
So, this term turns into an pi,
an, dragged along,

538
00:39:04 --> 00:39:10
but this, the integral of the
square of the cosine turns out

539
00:39:10 --> 00:39:16
to be pi.
And so, the end result is that

540
00:39:14 --> 00:39:20
we get a formula for an.
What is an?

541
00:39:18 --> 00:39:24
an is, well,
an times pi,

542
00:39:20 --> 00:39:26
all these terms of zero,
and nothing is left but this

543
00:39:25 --> 00:39:31
left-hand side.
And therefore,

544
00:39:28 --> 00:39:34
an times pi is the integral
from negative pi to pi of f of t

545
00:39:34 --> 00:39:40
times cosine nt dt.

546
00:39:40 --> 00:39:46
But, that's an times pi.

547
00:39:45 --> 00:39:51
Therefore, if I want just an,
I have to divide it by pi.

548
00:39:50 --> 00:39:56
And, that's the formula for the
coefficient an.

549
00:39:54 --> 00:40:00
The argument is exactly the
same if you want bn,

550
00:39:57 --> 00:40:03
but I will write it down for
the sake of completeness,

551
00:40:02 --> 00:40:08
as they say,
and to give you a chance to

552
00:40:05 --> 00:40:11
digest what I've done,
you know, 30 seconds to digest

553
00:40:09 --> 00:40:15
it. Sine nt dt.

554
00:40:12 --> 00:40:18
And, that's because the
argument is the same.

555
00:40:16 --> 00:40:22
And, the integral of sine
squared nt is also

556
00:40:20 --> 00:40:26
pi. So, there's no difference

557
00:40:22 --> 00:40:28
there.
Now, there's only one little

558
00:40:24 --> 00:40:30
caution.
It have to be a little careful.

559
00:40:27 --> 00:40:33
This is n one,
two, and so on,

560
00:40:29 --> 00:40:35
and this is also n one,
two, and unfortunately,

561
00:40:33 --> 00:40:39
the constant term is a slight
exception.

562
00:40:35 --> 00:40:41
We better look at that
specifically because if you

563
00:40:39 --> 00:40:45
forget it, you can get them to
gross, gross,

564
00:40:42 --> 00:40:48
gross errors.
How about the constant term?

565
00:40:48 --> 00:40:54
Suppose I repeat the argument
for that in miniature.

566
00:40:54 --> 00:41:00
There is a constant term plus
other stuff, a typical other

567
00:41:01 --> 00:41:07
stuff, an cosine,
let's say.

568
00:41:06 --> 00:41:12
How am I going to get that
constant term?

569
00:41:10 --> 00:41:16
Well, if you think of this as
sort of like a constant times,

570
00:41:16 --> 00:41:22
the reason is the constant is
because it's being multiplied by

571
00:41:22 --> 00:41:28
cosine zero t.
So, that suggests I should

572
00:41:27 --> 00:41:33
multiply by one.
In other words,

573
00:41:31 --> 00:41:37
what I should do is simply
integrate this from negative pi

574
00:41:36 --> 00:41:42
to pi, f of t dt.

575
00:41:40 --> 00:41:46
What's the answer?
Well, this integrated from

576
00:41:44 --> 00:41:50
minus pi to pi is how much?
It's c zero times two pi,

577
00:41:49 --> 00:41:55
right?
And, the other terms all give

578
00:41:52 --> 00:41:58
me zero.
Every other term is zero

579
00:41:55 --> 00:42:01
because if you integrate cosine
nt or sine nt

580
00:42:00 --> 00:42:06
over a complete
period, you always get zero.

581
00:42:06 --> 00:42:12
There is as much area above the
axis or below.

582
00:42:10 --> 00:42:16
Or, you can look at two special
cases.

583
00:42:13 --> 00:42:19
Anyway, you always get zero.
It's the same thing with sine

584
00:42:18 --> 00:42:24
here.
So, the answer is that c zero

585
00:42:21 --> 00:42:27
is equal to,
is a little special.

586
00:42:24 --> 00:42:30
You don't just put n equals
zero here because then

587
00:42:30 --> 00:42:36
you would lose a factor of two.
So, c zero should be one

588
00:42:36 --> 00:42:42
over two pi times
this integral.

589
00:42:40 --> 00:42:46
Now, there are two kinds of
people in the world,

590
00:42:44 --> 00:42:50
the ones who learn two separate
formulas, and the ones who just

591
00:42:50 --> 00:42:56
learn two separate notations.
So, what most people do is they

592
00:42:55 --> 00:43:01
say, look, I want this to be
always the formula for a zero.

593
00:43:02 --> 00:43:08
That means, even when n 
is zero, I want this to be the

594
00:43:07 --> 00:43:13
formula.
Well, then you are not going to

595
00:43:10 --> 00:43:16
get the right leading term.
Instead of getting c zero,

596
00:43:14 --> 00:43:20
you're going to get
twice it, and therefore,

597
00:43:18 --> 00:43:24
the formula is,
the Fourier series,

598
00:43:21 --> 00:43:27
therefore, isn't written this
way.

599
00:43:24 --> 00:43:30
It's written-- If you want an a
zero there,

600
00:43:28 --> 00:43:34
calculate it by this formula.
Then, you've got to write not c

601
00:43:34 --> 00:43:40
zero, but a zero over two.

602
00:43:37 --> 00:43:43
I think you will be happiest if
I have to give you advice.

603
00:43:41 --> 00:43:47
I think you'll be happiest
remembering a single formula for

604
00:43:45 --> 00:43:51
the an's and bn's,
in which case you have to

605
00:43:48 --> 00:43:54
remember that the constant
leading term is a zero over two

606
00:43:52 --> 00:43:58
if you insist on
using that formula.

607
00:43:55 --> 00:44:01
Otherwise, you have to learn a
special formula for the leading

608
00:43:59 --> 00:44:05
coefficient, namely one over two
pi instead of one

609
00:44:03 --> 00:44:09
over pi.
Well, am I really going to

610
00:44:08 --> 00:44:14
calculate a Fourier series in
four minutes?

611
00:44:11 --> 00:44:17
Not very likely,
but I'll give it a brave

612
00:44:14 --> 00:44:20
college try.
Anyway, you will be doing a

613
00:44:17 --> 00:44:23
great deal of it,
and your book has lots and lots

614
00:44:21 --> 00:44:27
of examples, too many,
in fact.

615
00:44:23 --> 00:44:29
It ruined all the good examples
by calculating them for you.

616
00:44:28 --> 00:44:34
But, I will at least outline.
Do you want me to spend three

617
00:44:34 --> 00:44:40
minutes outlining a calculation
just so you have something to

618
00:44:38 --> 00:44:44
work on in the next boring class
you are in?

619
00:44:42 --> 00:44:48
Let's see, so I'll just put a
few key things on the board.

620
00:44:46 --> 00:44:52
I would advise you to sit still
for this.

621
00:44:49 --> 00:44:55
Otherwise you're going to hack
it, and take twice as long as

622
00:44:54 --> 00:45:00
you should, even though I knew
you've been up to 3:00 in the

623
00:44:58 --> 00:45:04
morning doing your problem set.
Cheer up.

624
00:45:03 --> 00:45:09
I got up at 6:00 to make up the
new one.

625
00:45:08 --> 00:45:14
So, we're even.
This should be zero here.

626
00:45:13 --> 00:45:19
So, here's minus pi.
Here's pi.

627
00:45:17 --> 00:45:23
Here's one, negative one.
The function starts out like

628
00:45:24 --> 00:45:30
that, and now to be periodic,
it then has to continue on in

629
00:45:31 --> 00:45:37
the same way.
So, I think that's enough of

630
00:45:37 --> 00:45:43
its path through life to
indicate how it runs.

631
00:45:42 --> 00:45:48
This is a typical square-away
function, sometimes it's called.

632
00:45:48 --> 00:45:54
It's an odd function.
It goes equally above and below

633
00:45:53 --> 00:45:59
the axis.
Now, the integrals,

634
00:45:56 --> 00:46:02
when you calculate them,
the an is going to be,

635
00:46:00 --> 00:46:06
okay, look, the an is going to
turn out to be zero.

636
00:46:08 --> 00:46:14
Let me, instead,
and you will get that with a

637
00:46:11 --> 00:46:17
little hacking.
I'm much more worried about

638
00:46:14 --> 00:46:20
what you'll do with the bn's.
Also, next Monday you'll see

639
00:46:17 --> 00:46:23
intuitively that the an is zero,
in which case you won't even

640
00:46:22 --> 00:46:28
bother trying to calculate it.
How about the bn,

641
00:46:25 --> 00:46:31
though?
Well, you see,

642
00:46:26 --> 00:46:32
because the function is
discontinuous,

643
00:46:29 --> 00:46:35
so, this is my input.
My f of t is that

644
00:46:32 --> 00:46:38
orange discontinuous function.
The bn is going to be,

645
00:46:37 --> 00:46:43
I have to break it into two
parts.

646
00:46:40 --> 00:46:46
In the first part,
the function is negative one.

647
00:46:43 --> 00:46:49
And there, I will be
integrating from minus pi to pi

648
00:46:47 --> 00:46:53
of the function,
which is minus one times the

649
00:46:50 --> 00:46:56
sine of nt dt.

650
00:46:54 --> 00:47:00
And then, there's another part,

651
00:46:57 --> 00:47:03
sorry, minus pi to zero.
The other part I integrate from

652
00:47:02 --> 00:47:08
zero to pi of what?
Well, f of t is now plus one.

653
00:47:06 --> 00:47:12
And so, I simply integrate sine

654
00:47:10 --> 00:47:16
nt dt.
Now, each of these is a

655
00:47:14 --> 00:47:20
perfectly simple integral.
The only question is how you

656
00:47:19 --> 00:47:25
combine them.
So, this is,

657
00:47:21 --> 00:47:27
after you calculate it,
it will be (one minus cosine n

658
00:47:26 --> 00:47:32
pi) all over n.

659
00:47:29 --> 00:47:35
And, this part will turn out to
be (one minus cosine n pi) over

660
00:47:34 --> 00:47:40
n also. And therefore,

661
00:47:40 --> 00:47:46
the answer will be two minus
two cosine, two over n times,

662
00:47:48 --> 00:47:54
right, two minus,
two times (one minus cosine n

663
00:47:55 --> 00:48:01
pi) over n.

664
00:48:01 --> 00:48:07
No, okay, now,
what's this?

665
00:48:03 --> 00:48:09
This is minus one if n is odd.
It's plus one if n is even.

666
00:48:09 --> 00:48:15
Now, either you can work with
it this way, or you can combine

667
00:48:15 --> 00:48:21
the two of them into a single
expression.

668
00:48:19 --> 00:48:25
Its minus one to the nth power
takes care of both of

669
00:48:26 --> 00:48:32
them.
But, the way the answer is

670
00:48:29 --> 00:48:35
normally expressed,
it would be minus two over n,

671
00:48:34 --> 00:48:40
two over n times,
if n is even,

672
00:48:37 --> 00:48:43
I get zero.
If n is odd,

673
00:48:41 --> 00:48:47
I get two.
So, times two,

674
00:48:43 --> 00:48:49
if n is odd,
and zero if n is even.

675
00:48:46 --> 00:48:52
So, it's four over n,
or it's zero,

676
00:48:50 --> 00:48:56
and the final series is a sum
of those coefficients times the

677
00:48:55 --> 00:49:01
appropriate-- cosine or sine?
Sine terms because the cosine

678
00:49:01 --> 00:49:07
terms were all coefficients,
all turned out to be zero.

679
00:49:08 --> 00:49:14
I'm sorry I didn't have the
chance to do that calculation in

680
00:49:13 --> 00:49:19
detail.
But, I think that's enough

681
00:49:16 --> 00:49:22
sketch for you to be able to do
the rest of it yourself.