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I'd like to talk.
Thank you.

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One of the things I'd like to
give a little insight into today

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is the mathematical basis for
hearing.

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For example,
if a musical tone,

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a pure musical tone would
consist of a pure oscillation in

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terms of the vibration of the
air.

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It would be a pure oscillation.
So, [SINGS],

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and if you superimpose upon
that, suppose you sing a triad,

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[SINGS], those are three tones.
Each has its own period of

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oscillation, and then another
one, which is the top one,

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which is even faster.
The higher it is,

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the faster the thing.
Anyway, what you hear,

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then, is the sum of those
things.

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So, C plus E plus G,
let's say, what you hear is the

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wave form.
It's periodics,

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still, but it's a mess.
I don't know,

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I can't draw it.
So, this is periodic,

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but a mess, some sort of mess.
Now, of course,

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if you hear the three tones
together, most people,

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if they are not tone deaf,
anyway, can hear the three

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tones that make up that.
So, in other words,

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if this is the function which
is the sum of those three,

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some sort of messy function,
f of t,

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you're able to do Fourier
analysis on it,

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and break it up.
You're able to take that f of

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t, and somehow mentally express
it as the sum of three pure

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oscillations.
That's Fourier analysis.

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We've been doing it with an
infinite series,

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but it's okay.
It's still Fourier analysis if

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you do it with just three.
So, in other words,

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the f of t is going to
be the sum of,

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let's say, sine,
I don't know,

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it's going to be the sign of
one frequency plus the sine of

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another frequency plus the sine
of a third, maybe with

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coefficients here.
So, somehow,

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since you were born,
you have been able to take the

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f of t,
and express it as the sum of

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the three signs.
And, here, therefore,

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the three tones that make up
the triad.

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Now, the question is,
how did you do that Fourier

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analysis?
In other words,

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does your brain have a little
integrator in it,

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which calculates the
coefficients of that series?

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Of course, the answer is no.
It has to do something else.

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So, one of the things I'd like
to aim at in this lecture is

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just briefly explaining what,
in fact, actually happens to do

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that.
Now, to do that,

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we'll have to make some little
detours, as always.

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So, first I'm going to,
throughout the lecture,

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in fact, I gave you last time a
couple of shortcuts for

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calculating Fourier series based
on evenness and oddness,

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and also some expansion of the
idea of Fourier series where we

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use the different,
but things didn't have to be

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periodic or period two pi,
but it can have an arbitrary

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period, 2L, and we could still
get a Fourier expansion for it.

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Let me, therefore,
begin just as a problem,

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another type of shortcut
exercise, to do a Fourier

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calculation, which we are going
to be later in the period to

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explain the music problem.
So, let's suppose we're

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starting with the function,
f of t,

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which is a real square wave,
and I'll make its period

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different from the one,
not two pi.

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So, suppose we had a function
like this.

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So, this is one,
and this is one.

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So, the height is one,
and this point is one as well.

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And then, it's periodic ever
after that.

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I'll tell you what,
let's do like the electrical

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engineers do and put these
vertical lines there even though

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they don't exist.
Okay, so the height is one and

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it goes over.
The half period is one.

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This really is a square wave.
I mean, it's really a square,

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not what they usually call a
square wave.

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So, my question is,
what's its Fourier series?

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Well, it's neither even nor
odd.

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That's a little dismaying.
It sounds like we're going to

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have to calculate an's and bn's.
So, the shortcuts I gave you

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last time don't seem to be
applicable.

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Now, of course,
nor is the period two pi,

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but that shouldn't be too bad.
In fact, you ought to look for

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an expansion in terms of things
that look like sine of n,

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well, what should it be?

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Since L is equal to one,
the half period is equal to

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one.
Remember, the period is 2L,

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not L.
It's n pi over L,

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but if L is one,
we should be looking for an

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expansion in terms of functions
that look like this.

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Now, since we've already done
the work for the official square

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way, which looks something like
this, what you always try to do

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is reduce these things to
problems that you've already

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solved.
This is a legitimate one,

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since I solved it in lecture
for you.

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So, we can consider it as
something we know.

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So, I observed that since I am
very lazy, that if I lower this

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function by one half,
it will become an odd function.

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Now it's an odd function.
Okay, I just cut the work in

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half.
So, let's call this function,

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let's call this,
I don't know,

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S of t.
The green one is the one we

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wanted to start with.
So, f of t is a green

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function.
But, I can improve things even

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more because the function that
we calculated in the lecture is

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a lot like this salmon function.
That's why I called it S.

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But, the difference is that the
function we calculated with this

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one.
In the first place,

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it went down further.
It went not to negative one

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half, which is where that one
goes.

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But, it went down to negative
one, and then went up here to

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plus one.
And, it went over to pi.

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So, it came down again,
but not, but at the point,

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pi.
And here, negative pi went up

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again.
Okay, let me remind you what

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this one was.
Suppose we call it,

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O doesn't look good,
I don't know,

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how about g of u?
Let's, for a secret reason,

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call the variable u this time,
okay?

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So, the previous knowledge that
I'm relying on was that I

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derived the Fourier series for
you by an orthodox calculation.

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And, it's not too hard to do
because this is an odd function.

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And therefore,
you only have to calculate the

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bn's.
And, half of them turn out to

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be zero, although you don't know
that in advance.

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But anyway, the answer was four
over pi times the sum

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of just the odd ones,
the sine of n u,

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and that you had to 
divide by n.

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So, this is the expansion of g,
this function,

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g of u, the Fourier expansion
of this function.

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Since it's an odd function,
it only involves the signs.

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There's no funny stuff here
because the period is now two

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pi.
And, this came from the first

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lecture on Fourier series,
or from the book,

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wherever you want it,
or solutions to the notes.

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There are lots of sources for
that.

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The solution's in the notes.
Okay, now, that looks so much

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like the salmon function,
---

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-- I ought to be able to
convert one into the other.

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Now, I will do that by
shrinking the axis.

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But, since this can get rather
confusing, what I'll do is

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overlay this.
What I prefer to do is I think

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u, okay, I'm changing,
I'm keeping the thing the same.

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But, I'm going to change the
name of the variable,

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the t, in such a way that on
the t-axis, this becomes the

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point, one.
If I do that,

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then this function will turn
exactly into that one,

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except it will go not from
minus a half to a half,

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but it will go from negative
one to one, since I haven't done

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anything to the vertical axis.
So, how I do that?

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What's the relation between u
and t?

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Well, u is equal to pi times t,
or the other way around.

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You know that it's going to be
approximately this.

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Try one, and then check that it
works.

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When t is equal to one,
u is pi, which is what it's

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supposed to be.
So, this is the relation

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between the two.
And therefore,

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without further ado,
I can say that,

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let's write the relation
between them.

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f of t is what I want.
Well, what's f of t if I

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subtract one half of that?
So, that's going to be equal to

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the salmon function plus one
half, right, or the salmon

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function is f of t lowered by
one half.

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One thing is the same as the
other.

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And, what's the relation
between this salmon function and

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the orange function?
Well, the salmon function is,

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so, let's convert,
so, S of t -- it's more

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convenient, as I wrote the
formula g of u.

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Let's start it from that end.
If I start from g of u,

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what do I have to do to convert
it into S of-- into the salmon

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function?
Well, take one half of it.

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So, if I put them all together,
the conclusion is that f of t

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is equal to one half
plus S of t,

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which is one half of g of u,

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but u is pi t.
So, it's four pi,

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four over pi times the sum of
the sine of n.

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And, for u, I will write pi t

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divided by n.
And, sorry, I forgot to say

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that sum is only over the odd
values of n, not all values of

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n.
So, the sum over n odd of that,

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and, of course,
the two will cancel that.

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So, here we have,
in other words,

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just by this business of
shrinking or just stretching or

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shrinking the axis,
lowering it and squishing it

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that way a little bit.
We get from this Fourier

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series, we get that one just by
this geometric procedure.

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I'd like you to be able to do
that because it saves a lot of

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time.
Okay, so let's put this answer

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up in, I'm going to need it in a
minute, but I don't really want

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to recopy it.
So, let me handle it by

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erasing.
So, let's call that plus two

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over pi,
and there is our formula for

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that green function that we
wrote before.

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So, I'll put that in green.
So, we'll have a color-coded

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lecture again.
Now, what we're going to be

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doing ultimately,
to getting at the music problem

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that I posed at the beginning of
the lecture, is we want to

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solve, and this is what a study
of Fourier series has been

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aiming at, to solve second-order
linear equations with constant

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coefficients were the right-hand
side was a more general function

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than the kind we've been
handling.

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So, now, in order to simplify,
and we don't have a lot of time

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in the course,
I'd have to take another day to

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make more complicated
calculations,

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which I don't want to do since
you will learn a lot from them,

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anyway.
I think you will find you've

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had enough calculation by the
time Friday morning rolls

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around.
So, let's look at the undamped

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case, which is simpler,
or undamped spring,

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or undamped anything because it
doesn't have that extra term,

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which requires extra
calculations.

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So, I'll follow the book now
and some of the notes and the

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visuals, and called the
independent variable-- the

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dependent variable I'm going to
call x now.

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And, the independent variable
is, as usual,

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time.
So, this is going to be,

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in general, f of t,
and I'm going to use it by

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calculating example,
this is the actual f of t I'm

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going to be using.
But, the general problem for a

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general f of t is to solve this,
or at least to find a

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particular solution.
That's what most of the work

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is, because we already know how
from that to get the general

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solution by adding the solution
to the reduced equation,

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the associated homogeneous
equation.

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So, all our work has been,
this past couple of weeks,

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in how you find a particular
solution.

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Now, the case in which we know
what to do is,

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so we can find our particular
solution.

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Let's call that x sub p.

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We could find x sub p if the
right hand side is cosine omega,

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well, in general,
an exponential,

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but since we are not going to
use complex exponentials today,

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all these things are real.
And I'd like to keep them real.

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If it's either cosine omega t
or sine omega t,

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or some multiple of that by

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linearity, it's just as good.
We already know how to find the

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thing, and to find a particular
solution.

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So, the procedure is use
complex exponentials,

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and that magic formula I gave
you.

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But, right now,
just to save a little time,

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since I already did that on the
lecture on resonance,

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I solved it explicitly for
that, and you've had adequate

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practice I think in the problem
sets.

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Let's simply write down the
answer that comes out of that.

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The answer for the particular
solution is cosine omega t

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or sine omega t.

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That's the top.
And, it's over a constant.

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And, the constant is omega
naught squared.

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That's the natural frequency
which comes from the system,

246
00:17:19 --> 00:17:25
minus the imposed frequency,
the driving frequency that the

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system, the spring or whatever
it is, undamped spring,

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is being driven with.
Okay, understand the notation.

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Cosine this over that,
or sine, depending on whether

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you started driving it with
cosine or sine.

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So, this is from the lecture,
if you like,

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from the lecture on resonance,
but again it's,

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I hope by now,
a familiar fact.

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Let me remind you what this had
to do with resonance.

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00:17:58 --> 00:18:04
Then, the observation was that
if omega, the driving frequency

256
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is very close to the natural
frequency, then this is close to

257
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that.
The denominator is almost zero,

258
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and that makes the amplitude of
the response very,

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very large.
And, that was the phenomenon of

260
00:18:20 --> 00:18:26
resonance.
Okay, now what I'd like to do

261
00:18:23 --> 00:18:29
is apply those formulas to
finding out what happens for a

262
00:18:28 --> 00:18:34
general f(t),
or in particular this one.

263
00:18:32 --> 00:18:38
So, in general,
I'll keep using the notation,

264
00:18:36 --> 00:18:42
f of t,
even though I've sorted used it

265
00:18:41 --> 00:18:47
for that.
But in general,

266
00:18:44 --> 00:18:50
what's the situation?
If f of t is a sine series,

267
00:18:49 --> 00:18:55
cosine series,
all right, let's do everything.

268
00:18:53 --> 00:18:59
Suppose it's,
in other words,

269
00:18:56 --> 00:19:02
the procedure is,
take your f of t,

270
00:19:00 --> 00:19:06
expand it in a Fourier series.
Well, doesn't that assume it's

271
00:19:07 --> 00:19:13
periodic?
Yes, sort of.

272
00:19:09 --> 00:19:15
So, suppose it's a Fourier
series.

273
00:19:12 --> 00:19:18
I'll make a very general
Fourier series,

274
00:19:15 --> 00:19:21
write it this way:
cosine (omega)n t,

275
00:19:18 --> 00:19:24
and then the sine terms,

276
00:19:22 --> 00:19:28
sine (omega)n t
from one to infinity where

277
00:19:27 --> 00:19:33
the omegas are,
omega n is short for that.

278
00:19:32 --> 00:19:38
Well, it's going to have the n
in it, of course,

279
00:19:35 --> 00:19:41
but I want, now,
to make the general period to

280
00:19:39 --> 00:19:45
be 2L.
So, it would be n pi over L.

281
00:19:42 --> 00:19:48
Of course, if L is equal to

282
00:19:45 --> 00:19:51
one, then it's n pi.
Or, if L equals pi,

283
00:19:48 --> 00:19:54
those are the two most popular
cases, by far.

284
00:19:52 --> 00:19:58
Then, it's simply n itself,
the driving frequency.

285
00:19:56 --> 00:20:02
But, this would be the general
case, n pi over L

286
00:20:01 --> 00:20:07
if the period is the period of f
of t is 2L.

287
00:20:07 --> 00:20:13
So, that's what the Fourier
series looks like.

288
00:20:10 --> 00:20:16
Okay, then the particular
solution will be what?

289
00:20:14 --> 00:20:20
Well, I got these formulas.
In other words,

290
00:20:18 --> 00:20:24
what I'm using is superposition
principle.

291
00:20:21 --> 00:20:27
If it's just this,
then I know what the answer is

292
00:20:25 --> 00:20:31
for the particular solution,
the response.

293
00:20:30 --> 00:20:36
So, if you make a sum of these
things, a sum of these inputs,

294
00:20:34 --> 00:20:40
you are going to get a sum of
the responses by superposition.

295
00:20:39 --> 00:20:45
So, let's write out the ones we
are absolutely certain of.

296
00:20:43 --> 00:20:49
What's the response to here?
Well, it's (a)n cosine omega n

297
00:20:47 --> 00:20:53
t. The only thing is,

298
00:20:51 --> 00:20:57
now it's divided by omega
naught squared.

299
00:20:55 --> 00:21:01
This constant has changed,
and the same thing here.

300
00:21:00 --> 00:21:06
Of course, by linearity,
if this is multiplied by a,

301
00:21:03 --> 00:21:09
then the answer is multiplied
by, the response is also

302
00:21:06 --> 00:21:12
multiplied by a.
So, the same thing happens

303
00:21:09 --> 00:21:15
here.
Here, it's (b)n and over,

304
00:21:11 --> 00:21:17
again, omega naught squared
minus omega times the sine of

305
00:21:14 --> 00:21:20
omega t.

306
00:21:17 --> 00:21:23
So, in other words,
as soon as you have the Fourier

307
00:21:20 --> 00:21:26
expansion, the Fourier series
for the input,

308
00:21:23 --> 00:21:29
you automatically get this by
just writing it down the Fourier

309
00:21:27 --> 00:21:33
series for the response.
That's the fundamental idea of

310
00:21:32 --> 00:21:38
Fourier series,
at least applied in this

311
00:21:35 --> 00:21:41
context.
They have many other contexts,

312
00:21:38 --> 00:21:44
approximations,
so on and so forth.

313
00:21:41 --> 00:21:47
But, that's the idea here.
All right, what about that

314
00:21:45 --> 00:21:51
constant term?
Well, this formula still works

315
00:21:49 --> 00:21:55
if omega equals zero.
If omega equals zero,

316
00:21:52 --> 00:21:58
then this is the constant,
one.

317
00:21:54 --> 00:22:00
The formula is still correct.
Omega is zero here.

318
00:22:00 --> 00:22:06
The only thing you have to
remember is that the original

319
00:22:03 --> 00:22:09
thing is written in this form.
So, the response will be,

320
00:22:07 --> 00:22:13
what will it be?
Well, it's one divided by omega

321
00:22:10 --> 00:22:16
naught squared,
if I'm in the case omega zero

322
00:22:12 --> 00:22:18
is equal to zero.
So, it's a zero divided by two

323
00:22:16 --> 00:22:22
omega naught squared.
And, as you will see,

324
00:22:19 --> 00:22:25
it looks just like the others.
You're just taking omega,

325
00:22:23 --> 00:22:29
and making it equal to zero for
that particular case.

326
00:22:26 --> 00:22:32
Sorry, this should be omega n's
all the way through here.

327
00:22:31 --> 00:22:37

328
00:22:40 --> 00:22:46
All right, well,
let's apply this to the green

329
00:22:45 --> 00:22:51
function.
So, what have we got?

330
00:22:49 --> 00:22:55
We have its Fourier series.
So, if the green function is,

331
00:22:56 --> 00:23:02
if the input in other words is
this square wave,

332
00:23:02 --> 00:23:08
the green square wave,
so in your notes,

333
00:23:06 --> 00:23:12
this guy, this particular f of
t is the input.

334
00:23:15 --> 00:23:21
And, the equation is x double
prime plus omega naught squared

335
00:23:20 --> 00:23:26
x equals f of t.

336
00:23:23 --> 00:23:29
Then, the response is,
well, I can't draw you a

337
00:23:27 --> 00:23:33
picture of the response because
I don't know what the Fourier

338
00:23:32 --> 00:23:38
series actually looks like.
But, let's at least write down

339
00:23:37 --> 00:23:43
what the Fourier series is.
The Fourier series will be,

340
00:23:43 --> 00:23:49
well, what is it?
It's one half.

341
00:23:45 --> 00:23:51
The constant out front is one
half, except it's one over two

342
00:23:50 --> 00:23:56
omega naught squared.

343
00:23:53 --> 00:23:59
So, this is my function,
f of t.

344
00:23:56 --> 00:24:02
That's the general formula for
how the input is related to the

345
00:24:01 --> 00:24:07
response.
And, I'm applying it to this

346
00:24:06 --> 00:24:12
particular function,
f of t.

347
00:24:10 --> 00:24:16
And, the answer is plus.
Well, my Fourier series

348
00:24:15 --> 00:24:21
involves only odd sums,
only the summation over odd,

349
00:24:21 --> 00:24:27
and only of the sign.
So, it is going to be two over

350
00:24:26 --> 00:24:32
pi,
sorry, so it's going to be two

351
00:24:31 --> 00:24:37
over pi out front.
That constant will carry along

352
00:24:37 --> 00:24:43
by linearity.
And, I'm going to sum over odd,

353
00:24:39 --> 00:24:45
n odd values only.
The basic thing in the upstairs

354
00:24:43 --> 00:24:49
is going to be the sine of omega
n t.

355
00:24:47 --> 00:24:53
But, what is (omega)n?
Well, (omega)n is n pi.

356
00:24:50 --> 00:24:56
So, it's n pi t.
And, how about the bottom?

357
00:24:53 --> 00:24:59
The bottom is going to be omega
naught squared minus omega n

358
00:24:57 --> 00:25:03
squared.

359
00:25:01 --> 00:25:07
And, this is my (omega)n,
minus n pi squared.

360
00:25:05 --> 00:25:11
What's that?

361
00:25:07 --> 00:25:13
Well, I don't know.
All I could do would be to

362
00:25:12 --> 00:25:18
calculate it.
You could put it on MATLAB and

363
00:25:16 --> 00:25:22
ask MATLAB to calculate and plot
for you the first few terms,

364
00:25:22 --> 00:25:28
and get some vague idea of what
it looks like.

365
00:25:26 --> 00:25:32
That's nice,
but it's not what's interesting

366
00:25:31 --> 00:25:37
to do.
What's interesting to do is to

367
00:25:35 --> 00:25:41
look at the size of the
coefficients.

368
00:25:38 --> 00:25:44
And, again, rather than do it
in the abstract,

369
00:25:42 --> 00:25:48
let's take a specific value.
Let's suppose that the natural

370
00:25:46 --> 00:25:52
frequency of the system,
in other words,

371
00:25:50 --> 00:25:56
the frequency at which that
little spring wants to go

372
00:25:54 --> 00:26:00
vibrate back and forth,
whatever you got vibrating.

373
00:25:58 --> 00:26:04
Let's suppose the natural
frequency that's omega naught is

374
00:26:03 --> 00:26:09
ten for the sake of
definiteness,

375
00:26:05 --> 00:26:11
as they say.
Okay, if that's ten,

376
00:26:09 --> 00:26:15
all I want to do is calculate
in the crudest possible way what

377
00:26:15 --> 00:26:21
a few of these terms are.
So, the response is,

378
00:26:19 --> 00:26:25
so let's see,
we've got to give that a name.

379
00:26:23 --> 00:26:29
The response is (x)p of t.

380
00:26:26 --> 00:26:32
What's (x)p of t?
I'm just going to calculate it

381
00:26:31 --> 00:26:37
very approximately.
This means, you know,

382
00:26:35 --> 00:26:41
throwing caution to the winds
because I don't have a

383
00:26:39 --> 00:26:45
calculator with me.
And, I want you to look at this

384
00:26:43 --> 00:26:49
thing without a calculator.
The first term is one over 200.

385
00:26:47 --> 00:26:53
Okay, that's the only term I

386
00:26:50 --> 00:26:56
can get exactly right.
[LAUGHTER] Or,

387
00:26:52 --> 00:26:58
I could if I could calculate.
I suppose it's 0.005,

388
00:26:56 --> 00:27:02
right?
That's the constant term.

389
00:27:00 --> 00:27:06
Okay, so the next term,
let's see, two over pi is two

390
00:27:04 --> 00:27:10
thirds.
I'll keep that in mind,

391
00:27:07 --> 00:27:13
right?
Plus two thirds,

392
00:27:09 --> 00:27:15
0.6, let's say,
that's an indication of the

393
00:27:13 --> 00:27:19
accuracy with which these things
are going to be performed.

394
00:27:19 --> 00:27:25
I think in Texas for a long
while, the legislature declared

395
00:27:24 --> 00:27:30
pi to be three,
anyways.

396
00:27:27 --> 00:27:33
One of those states did it to
save calculation time.

397
00:27:31 --> 00:27:37
I'm not kidding,
by the way.

398
00:27:36 --> 00:27:42
All right, so what's the first
term?

399
00:27:38 --> 00:27:44
If n equals one,
I have the sine of pi t.

400
00:27:42 --> 00:27:48
That's the n equals one term.

401
00:27:46 --> 00:27:52
What's the denominator like?
That's about 100 minus 9

402
00:27:50 --> 00:27:56
squared. Let's say it's 91,

403
00:27:53 --> 00:27:59
sine t over 91.
What's the next term?

404
00:27:56 --> 00:28:02
Sine of three pi t, remember,

405
00:28:00 --> 00:28:06
I am omitting,
I'm only using the odd values

406
00:28:04 --> 00:28:10
of n because those are the only
ones that enter into the Fourier

407
00:28:09 --> 00:28:15
expansion for this function,
which is at the bottom of

408
00:28:14 --> 00:28:20
everything.
All right, what's the sine

409
00:28:19 --> 00:28:25
three pi t?
Well, now, I've got 100 minus

410
00:28:26 --> 00:28:32
three pi, --
-- that's 9 squared is 81.

411
00:28:32 --> 00:28:38
So, no, what am I doing?
So, we have 100 minus three

412
00:28:42 --> 00:28:48
times pi is 9,
squared.

413
00:28:46 --> 00:28:52
Well, let's say a little more.
Let's say 85.

414
00:28:53 --> 00:28:59
So, that's 15.
How bout the next one?

415
00:29:02 --> 00:29:08
Well, it's sine 5 pi t.

416
00:29:05 --> 00:29:11
I think I'll stop here as soon
as we do this one because at

417
00:29:09 --> 00:29:15
this point it's clear what's
happening.

418
00:29:12 --> 00:29:18
This is 100 squared minus,
that's 15 squared is 225,

419
00:29:16 --> 00:29:22
so that's about 125 with a
negative sign.

420
00:29:19 --> 00:29:25
So, minus this divided by 125.
And, after this they are going

421
00:29:24 --> 00:29:30
to get really quite small
because the next one will be

422
00:29:28 --> 00:29:34
seven pi squared.
That's 400, and this is

423
00:29:33 --> 00:29:39
becoming negligible.
So, what's happening?

424
00:29:38 --> 00:29:44
So, it's approximately,
in other words,

425
00:29:42 --> 00:29:48
0.005 plus the next coefficient
is, let's see,

426
00:29:48 --> 00:29:54
6/10, let's say 100,
sine pi t.

427
00:29:51 --> 00:29:57
And, what comes next?
Well, it's now 1/20th.

428
00:29:56 --> 00:30:02
It's about a 20th.
Let's call that 0.005 sine

429
00:30:03 --> 00:30:09
three pi t,
and now so small,

430
00:30:08 --> 00:30:14
minus 0.01, let's say times
this last one,

431
00:30:13 --> 00:30:19
sine 5 pi t.
What you find,

432
00:30:16 --> 00:30:22
in other words,
is that the frequencies which

433
00:30:22 --> 00:30:28
make up the response do not
occur with the same amplitude.

434
00:30:30 --> 00:30:36
What happens is that this
amplitude is roughly five times

435
00:30:35 --> 00:30:41
larger than any of the
neighboring ones.

436
00:30:38 --> 00:30:44
And after that,
it's a lot larger than the ones

437
00:30:43 --> 00:30:49
that come later.
In other words,

438
00:30:46 --> 00:30:52
the main frequency which occurs
in the response is the frequency

439
00:30:52 --> 00:30:58
three pi.
What's happened is,

440
00:30:54 --> 00:31:00
in other words,
near resonance has occurred.

441
00:31:00 --> 00:31:06
So, if omega is ten,
very near resonance,

442
00:31:04 --> 00:31:10
that is, it's not too close,
but it's not too far away

443
00:31:10 --> 00:31:16
either, occurs for the frequency
three pi in the input.

444
00:31:16 --> 00:31:22
Now, where's the frequency
three pi in the input?

445
00:31:22 --> 00:31:28
It isn't there.
It's just that green thing.

446
00:31:27 --> 00:31:33
Where in that is the frequency
three pi?

447
00:31:33 --> 00:31:39
I can't answer that for you,
but that's the function of

448
00:31:37 --> 00:31:43
Fourier series,
to say that you can decompose

449
00:31:41 --> 00:31:47
that green function into a sum
of frequencies,

450
00:31:45 --> 00:31:51
as it were, and the Fourier
coefficients tell you how much

451
00:31:50 --> 00:31:56
frequency goes into each of
those f of t's.

452
00:31:54 --> 00:32:00
Now, so, f of t is decomposed
into the sum of frequencies by

453
00:31:59 --> 00:32:05
the Fourier analysis.
But, the system isn't going to

454
00:32:04 --> 00:32:10
respond equally to all those
frequencies.

455
00:32:07 --> 00:32:13
It's going to pick out and
favor the one which is closest

456
00:32:12 --> 00:32:18
to its natural frequency.
So, what's happened,

457
00:32:15 --> 00:32:21
these frequencies,
the frequencies and their

458
00:32:19 --> 00:32:25
relative importance in f of t
are hidden,

459
00:32:23 --> 00:32:29
as it were.
They're hidden because we can't

460
00:32:26 --> 00:32:32
see them unless you do the
Fourier analysis,

461
00:32:30 --> 00:32:36
and look at the size of the
coefficients.

462
00:32:35 --> 00:32:41
But, the system can pick out.
The system picks out and

463
00:32:44 --> 00:32:50
favors, picks out for resonance,
or resonates with,

464
00:32:54 --> 00:33:00
resonates with the frequencies
closest to its natural

465
00:33:04 --> 00:33:10
frequency.
Well, suppose the system had

466
00:33:09 --> 00:33:15
natural frequency,
not ten.

467
00:33:11 --> 00:33:17
This is a put up job.
Suppose it had natural

468
00:33:14 --> 00:33:20
frequency five.
Well, in that case,

469
00:33:17 --> 00:33:23
none of them are close to the
hidden frequencies in f of t,

470
00:33:21 --> 00:33:27
and there would be no
resonance.

471
00:33:25 --> 00:33:31
But, because of the particular
value I gave here,

472
00:33:29 --> 00:33:35
I gave the value ten,
it's able to pick out n equals

473
00:33:33 --> 00:33:39
three as the most important,
the corresponding three pi as

474
00:33:37 --> 00:33:43
the most important frequency in
the input, and respond to that.

475
00:33:44 --> 00:33:50
Okay, so this is the way we
hear, give or take a few

476
00:33:48 --> 00:33:54
thousand pages.
So, what does the ear do?

477
00:33:51 --> 00:33:57
How does the ear,
so, it's got that thing,

478
00:33:55 --> 00:34:01
messy curve,
which I erased,

479
00:33:57 --> 00:34:03
which has a secret,
which just has three hidden

480
00:34:01 --> 00:34:07
frequencies.
Okay, from now on I hand wave,

481
00:34:05 --> 00:34:11
right, like they do in other
subjects.

482
00:34:07 --> 00:34:13
So, we got our frequency.
So, it's got a [SINGS].

483
00:34:11 --> 00:34:17
That's one frequency.
[SINGS] And,

484
00:34:13 --> 00:34:19
what goes in there is the sum
of those three,

485
00:34:16 --> 00:34:22
and the ear has to do something
to say out of all the

486
00:34:20 --> 00:34:26
frequencies in the world,
I'm going to respond to that

487
00:34:24 --> 00:34:30
one, that one,
and that one,

488
00:34:26 --> 00:34:32
and send a signal to the brain,
which the brain,

489
00:34:29 --> 00:34:35
then, will interpret as a
beautiful triad.

490
00:34:34 --> 00:34:40
Okay, so what happens is that
the ear, I don't talk

491
00:34:37 --> 00:34:43
physiology, and I never will
again.

492
00:34:39 --> 00:34:45
I know nothing about it,
but anyway, the ear,

493
00:34:43 --> 00:34:49
when you get far enough in
there, there are little three

494
00:34:46 --> 00:34:52
bones, bang, bang,
bang; this is the eardrum,

495
00:34:50 --> 00:34:56
and then there's the part which
has wax.

496
00:34:52 --> 00:34:58
Then, there's the eardrum which
vibrates, at least if there is

497
00:34:57 --> 00:35:03
not too much wax in your ear.
And then, the vibrations go

498
00:35:01 --> 00:35:07
through three little bones which
send the vibrations to the inner

499
00:35:05 --> 00:35:11
ear, which nobody ever sees.
And, the inner ear,

500
00:35:10 --> 00:35:16
then, is filled with thick
fluid and a membrane,

501
00:35:13 --> 00:35:19
and the last bone hits up
against the membrane,

502
00:35:16 --> 00:35:22
and the membrane vibrates.
And, that makes the fluid

503
00:35:20 --> 00:35:26
vibrate.
Okay, good.

504
00:35:21 --> 00:35:27
So, it's vibrating according to
the function f of t.

505
00:35:25 --> 00:35:31
Well, what then?
Well, that's the marvelous

506
00:35:28 --> 00:35:34
part.
It's almost impossible to

507
00:35:31 --> 00:35:37
believe, but there is this,
sort of like a snail thing

508
00:35:36 --> 00:35:42
inside.
I've forgotten the name.

509
00:35:38 --> 00:35:44
It's cochlea.
And, it has these hairs.

510
00:35:41 --> 00:35:47
They are not hairs really.
I don't know what else to call

511
00:35:45 --> 00:35:51
them. They're not hairs.

512
00:35:47 --> 00:35:53
But, there are things so long,
you know, they stick up.

513
00:35:52 --> 00:35:58
And, there are 20,000 of them.
And, they are of different

514
00:35:56 --> 00:36:02
lengths.
And, each one is tuned to a

515
00:35:59 --> 00:36:05
certain frequency.
Each one has a certain natural

516
00:36:05 --> 00:36:11
frequency, and they are all
different, and they are all

517
00:36:11 --> 00:36:17
graded, just like a bunch of
organ pipes.

518
00:36:16 --> 00:36:22
And, when that complicated wave
hits, the complicated wave hits,

519
00:36:23 --> 00:36:29
each one resonates to a hidden
frequency in the wave,

520
00:36:29 --> 00:36:35
which is closest to its natural
frequency.

521
00:36:35 --> 00:36:41
Now, most of them won't be
resonating at all.

522
00:36:37 --> 00:36:43
Only the ones close to the
frequency [SINGS],

523
00:36:40 --> 00:36:46
they'll resonate,
and the nearby guys will

524
00:36:43 --> 00:36:49
resonate, too,
because they will be nearby,

525
00:36:45 --> 00:36:51
almost have the same natural
frequency.

526
00:36:48 --> 00:36:54
And, over here,
there will be a few which

527
00:36:50 --> 00:36:56
resonate to [SINGS],
and finally over here a few

528
00:36:53 --> 00:36:59
which go [SINGS],
and each of those little hairs,

529
00:36:56 --> 00:37:02
little groups of hairs will
signal, send that signal to the

530
00:37:00 --> 00:37:06
auditory nerve somehow or other,
which will then carry these

531
00:37:03 --> 00:37:09
three inputs to the brain,
and the brain,

532
00:37:06 --> 00:37:12
then, will interpret that as
you are hearing [SINGS].

533
00:37:11 --> 00:37:17
So, the Fourier analysis is
done by resonance.

534
00:37:15 --> 00:37:21
You here resonance because each
of these things has a certain

535
00:37:21 --> 00:37:27
natural frequency which is able,
then, to pick out a resonant

536
00:37:27 --> 00:37:33
frequency in the input.
I'd like to finish our work on

537
00:37:32 --> 00:37:38
Fourier series.
So, for homework I'm asking you

538
00:37:35 --> 00:37:41
to do something similar.
Taken an input.

539
00:37:38 --> 00:37:44
I gave you a frequency here,
a different omega naught,

540
00:37:42 --> 00:37:48
a different input,
as you by means of this Fourier

541
00:37:46 --> 00:37:52
analysis to find out which it
will resonate,

542
00:37:50 --> 00:37:56
which of the hidden frequencies
in the input the system will

543
00:37:54 --> 00:38:00
resonate to, just so you can
work it out yourself and do it.

544
00:38:00 --> 00:38:06
Now, I'd like to first try to
match up what I just did by this

545
00:38:05 --> 00:38:11
formula with what's in your
book, since your book handles

546
00:38:10 --> 00:38:16
the identical problem but a
little differently,

547
00:38:14 --> 00:38:20
and it's essentially the same.
But I think I'd better say

548
00:38:19 --> 00:38:25
something about it.
So, the book's method,

549
00:38:22 --> 00:38:28
and to the extent which any of
these problems are worked out in

550
00:38:28 --> 00:38:34
the notes, the notes do this,
too.

551
00:38:32 --> 00:38:38
Use substitution.
Base uses differentiation of

552
00:38:36 --> 00:38:42
Fourier series term by term.
The work is almost exactly the

553
00:38:42 --> 00:38:48
same as here.
And, it has a slight advantage,

554
00:38:46 --> 00:38:52
that it allows you,
the book's method has a slight

555
00:38:51 --> 00:38:57
advantage that it allows you to
forget this formula.

556
00:38:56 --> 00:39:02
You don't have to know this
formula.

557
00:39:01 --> 00:39:07
It will come out in the wash.
Now, for some of you,

558
00:39:04 --> 00:39:10
that may be of colossal
importance, in which case,

559
00:39:08 --> 00:39:14
by all means,
use the book's method,

560
00:39:10 --> 00:39:16
term by term.
So, it requires no knowledge of

561
00:39:14 --> 00:39:20
this formula because after all,
I base this solution,

562
00:39:17 --> 00:39:23
I simply wrote down the
solution and I based it on the

563
00:39:21 --> 00:39:27
fact that I was able to write
down immediately the solution to

564
00:39:26 --> 00:39:32
this and put as being that
response.

565
00:39:30 --> 00:39:36
And for that,
I had to remember it,

566
00:39:32 --> 00:39:38
or be willing to use complex
exponentials quickly to remind

567
00:39:36 --> 00:39:42
myself.
There's very,

568
00:39:38 --> 00:39:44
very little difference between
the two.

569
00:39:41 --> 00:39:47
Even if you have to re-derive
that formula,

570
00:39:44 --> 00:39:50
the two take almost about the
same length of time.

571
00:39:48 --> 00:39:54
But anyway, the idea is simply
this.

572
00:39:50 --> 00:39:56
With the book,
you assume.

573
00:39:52 --> 00:39:58
In other words,
you take your function,

574
00:39:55 --> 00:40:01
f of t.
You expand it in a Fourier

575
00:39:58 --> 00:40:04
series.
Of course, which signs and

576
00:40:01 --> 00:40:07
cosines you use will depend upon
what the period is.

577
00:40:07 --> 00:40:13
So, you assume the solution of
the form-- Well,

578
00:40:10 --> 00:40:16
if I, for example,
carried out in this particular

579
00:40:14 --> 00:40:20
case, I don't know if I will do
all the work,

580
00:40:18 --> 00:40:24
but it would be natural to
assume a solution of the form,

581
00:40:22 --> 00:40:28
since the input looks like the
green guy.

582
00:40:26 --> 00:40:32
Assume a solution which looks
the same.

583
00:40:30 --> 00:40:36
In other words,
it will have a constant term

584
00:40:33 --> 00:40:39
because the input does.
But all the rest of the terms

585
00:40:38 --> 00:40:44
will be sines.
So, it will be something like

586
00:40:42 --> 00:40:48
(c)n times the sine of n pi t.

587
00:40:46 --> 00:40:52
The only question is,
what are the (c)n's?

588
00:40:50 --> 00:40:56
Well, I found one method up
there.

589
00:40:53 --> 00:40:59
But, the general method is just
plug-in.

590
00:40:56 --> 00:41:02
Substitute into the ODE.
Substitute into the ODE.

591
00:41:02 --> 00:41:08
You differentiate this twice to
do it.

592
00:41:04 --> 00:41:10
So, I'll do the double
differentiation and I won't stop

593
00:41:08 --> 00:41:14
the lecture there,
but I will stop the calculation

594
00:41:12 --> 00:41:18
there because it has nothing new
to offer.

595
00:41:15 --> 00:41:21
And, this is the way all the
calculations in the books and

596
00:41:19 --> 00:41:25
the solutions and the notes are
carried out.

597
00:41:22 --> 00:41:28
So, I don't think you'll have
any trouble.

598
00:41:25 --> 00:41:31
Well, this term vanishes.
This term becomes what?

599
00:41:30 --> 00:41:36
If I differentiate this twice,
I get summation,

600
00:41:33 --> 00:41:39
so, this is one to infinity
because I don't know which of

601
00:41:37 --> 00:41:43
these are actually going to
appear.

602
00:41:40 --> 00:41:46
Summation one to infinity,
(c)n times, well,

603
00:41:43 --> 00:41:49
if you differentiate the sine
twice, you get negative sine,

604
00:41:48 --> 00:41:54
right?
Do it once: you get cosine.

605
00:41:50 --> 00:41:56
Second time:
you get negative sine.

606
00:41:53 --> 00:41:59
But, each time you will get
this extra factor n pi from the

607
00:41:57 --> 00:42:03
chain rule.
And so, the answer will be

608
00:42:00 --> 00:42:06
negative (c)n times n pi squared
times the sine of n pi t.

609
00:42:05 --> 00:42:11
And so, the procedure is,

610
00:42:10 --> 00:42:16
very simply,
you substitute (x)p double

611
00:42:13 --> 00:42:19
prime into the differential
equation.

612
00:42:16 --> 00:42:22
In other words,
if you do it,

613
00:42:17 --> 00:42:23
we will multiply this by omega
naught squared.

614
00:42:22 --> 00:42:28
And, you add them.
And then, on the left-hand

615
00:42:25 --> 00:42:31
side, you are going to get a sum
of terms, sine n pi t

616
00:42:29 --> 00:42:35
times coefficients
involving the (c)n's.

617
00:42:34 --> 00:42:40
And, on the right,
so, you're going to get a sum

618
00:42:37 --> 00:42:43
involving the (c)n's,
and the sines n pi t,

619
00:42:41 --> 00:42:47
and on the right,
you're going to get the Fourier

620
00:42:44 --> 00:42:50
series for f of t,
which is exactly the same kind

621
00:42:49 --> 00:42:55
of expression.
The only difference is,

622
00:42:52 --> 00:42:58
now the sines have come with
definite coefficients.

623
00:42:56 --> 00:43:02
And then, you simply click the
coefficients on the left and the

624
00:43:01 --> 00:43:07
coefficients on the right,
and figure out what the (c)n's

625
00:43:05 --> 00:43:11
are.
So, by equating coefficients,

626
00:43:10 --> 00:43:16
you get the (c)n's.
Would you like me to carry it

627
00:43:15 --> 00:43:21
out?
Yeah, okay, I was going to do

628
00:43:19 --> 00:43:25
something else,
but I wouldn't have time to do

629
00:43:24 --> 00:43:30
it anyway.
So, why don't I take two

630
00:43:28 --> 00:43:34
minutes to complete the
calculation just so you can see

631
00:43:34 --> 00:43:40
you get the same answer?
All right, what do we get?

632
00:43:40 --> 00:43:46
If you add them up,
you get c naught,

633
00:43:43 --> 00:43:49
out front, plus (c)n is
multiplied by what?

634
00:43:47 --> 00:43:53
Well, from the top it's
multiplied by omega naught

635
00:43:52 --> 00:43:58
squared.
On the bottom,

636
00:43:55 --> 00:44:01
it's multiplied by n 
pi squared.

637
00:44:00 --> 00:44:06
Ah-ha, where have I seen that
combination?

638
00:44:05 --> 00:44:11
The sum is equal to,
sorry, one half plus what is

639
00:44:15 --> 00:44:21
it, sum over n odd of sine n pi
t over n.

640
00:44:29 --> 00:44:35
So, the conclusion is that--
I'm sorry, it should be c naught

641
00:44:34 --> 00:44:40
times omega naught squared.

642
00:44:38 --> 00:44:44
So, what's the conclusion?
If c zero is one over two omega

643
00:44:44 --> 00:44:50
naught squared,

644
00:44:48 --> 00:44:54
and that (c)n,
only for n odd,

645
00:44:51 --> 00:44:57
the others will be even.
The others will be zero.

646
00:44:55 --> 00:45:01
The (c)n is going to be equal
to two over pi here.

647
00:45:01 --> 00:45:07
So, it's going to be two pi,

648
00:45:04 --> 00:45:10
two over pi times one over n
times one over omega naught

649
00:45:10 --> 00:45:16
squared minus n over
pi squared.

650
00:45:15 --> 00:45:21
This is terrible,

651
00:45:20 --> 00:45:26
which is the same answer we got
before, I hope.

652
00:45:26 --> 00:45:32
Did I cover it up?
Same answer.

653
00:45:30 --> 00:45:36
So, that answer at the
left-hand end of the board is

654
00:45:35 --> 00:45:41
the same one.
I've calculated,

655
00:45:38 --> 00:45:44
in other words,
what the c zeros are.

656
00:45:42 --> 00:45:48
And, I got the same answer as
before.