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HERBERT GROSS: Hi.
Welcome to our final lecture in part three of calculus.
And I suppose it would be nice to have an appropriate story
to tell.
And the only story I can think of
is, on the transatlantic flight, the stewardess coming in
to the passenger's cabin midway over the Atlantic Ocean
and saying to the passengers, well,
I have some good news and some bad news for you.
The bad news is that we're almost out of gas.
And the good news is that we're making wonderful time.
And the reason I thought of this particular story
is that, in a sense, we are not almost out of gas,
but in terms of this being a fundamentals type of course
in which we're trying to give the ingredients of what
goes into all advanced mathematics courses,
we have sort of come to the end of the road.
We have finished the so-called ABC's of mathematics.
As far as making wonderful time is concerned,
I would say we've done deliberately just the opposite,
that we have essentially taken one or two central themes
and tried to unify all of elementary calculus
in terms of these central themes.
And when you do something like this,
it's very difficult to come to the end of a course.
You sort of don't know where to stop.
So you go back to a little trick that you
learned in about the third grade when you wrote compositions
where you learned that you could write anything
that you wanted as long as the last line was,
we all had a very nice time.
So we now must look for a line along
those particular thoughts.
And if we recall that we began as our major theme
in this course the idea of mathematical structure,
the idea of what new math really meant, not new meaning brand
new, but new meaning the opposite of passe, new meaning
meaningful, I can think of no nicer way
of ending the course than by picking as an application
a topic that existed long before the so-called new math was
born, and which exists in an even
more enlightened way in the spirit of the new mathematics.
And the topic that I have in mind to conclude with--
and it also ties in with our lecture of last time
on the dot product--
is the subject called orthogonal functions,
which has as a special subtopic, Fourier series.
Now the way this comes about is the following.
And you have to know nothing about vector spaces
to appreciate what we're going to do now, that in a sense
we imitate what Fourier actually did in trying to solve
a particular type of problem.
We suppose that we have an infinite sequence of, functions
which I'll denote by the family phi sub
n of x, and a particular closed interval
from a to b such that each member of my family phi sub n
of x is integral on the interval from a to b.
And we'll also assume that the integral from a
to b of the product of any two of these functions, two
different members of the set, that that integral is 0.
In other words, the interval from a to b
phi sub i of x times phi sub j of x dx
is 0, provided i is unequal to j.
So we're assuming this particular property,
remembering of course that the product of two
integral functions is, again, integrable
so that this thing makes sense.
Now traditionally, when a family of functions
had this particular property, that family
was set to be orthogonal on the interval from a to b.
And I will leave it for a little bit
later to show you how this jibes with the meaning of orthogonal
as we defined it in terms of dot products last time.
Let me show you how, for example, Fourier
decided to use this type of information.
Suppose we had a given function f
of x that also happened to be integrable on the interval
from a to b.
Or written more succinctly, we assume
that the integral from a to b f of x dx exists.
Let's also suppose that f of x happens
to be a linear combination of the members
of the orthogonal family phi sub n of x, in other words,
that for the given orthogonal family, f of x
is summation Cn phi n of x as n goes from 1 to infinity.
And the question is, under--
or with the knowledge that this is an orthogonal family,
how can we determine each coefficient C sub
n if this condition here is the whole?
Or to make this more concrete and keep
it as non-symbolic as possible, let's
suppose I wanted to figure out here what C sub 3
happened to be.
The trick is this.
I recognize that, from this basic definition,
if I multiply phi sub 3 of x by any other member of this family
other than phi sub 3 itself and integrate from a
to b, that term will be 0.
So what I do is I come back to this relationship.
And I multiply both sides of this equation by phi sub 3
of x, keeping in mind if I wanted to find C sub k,
I would have multiplied both sides by phi sub k of x.
I'm just picking k equals 3 here.
I multiply both sides by phi sub 3 of x.
And because phi sub 3 of x is a constant subscript, meaning
it doesn't depend on the summation n which only affects
the subscript here, I can bring phi sub up 3
of x inside the summation sign.
Now the trick is I integrate both sides of this equality
from a to b.
In other words, I put an integral
from a to b on both sides here, augment this with a dx.
And I now know that this is the case.
My next gimmick, if you want to call it that--
and if you don't want to call it that,
you should, because it is a gimmick,
as we'll explain in the moment--
I simply interchange the order of summation and integration,
remembering that the integral of the sum
is the sum of the integrals.
I interchange this order.
And I now have the summation n goes from 1
to infinity integral from a to b Cn phi n of x phi 3 of x dx.
And now remembering that the integral
from a to b, phi sub n of x times phi sub 3 of x dx,
by definition of orthogonality is 0 unless n equals 3,
all of these terms drop out except when n equals 3.
Now what happens here?
When n equals 3, this term here is C sub 3.
This is phi sub 3 squared x dx integrated from a to b.
And all the remaining terms are 0, because the integrand is 0.
In other words, the right-hand side simply becomes--
well, since C3 is a constant, I can take it
outside the integral sign.
It's C3 integral from a to b C sub 3 squared of x dx.
And now you see from this relationship here,
I can solve for C sub 3.
Namely, I simply divide both sides of this equation
by this number here.
Now what I said by gimmick is simply this.
When we said that the integral of a sum
is the sum of the integrals, that was only for a finite sum.
You may remember in part one of our course
when we talked about uniform convergence, one of the things
that we pointed out was that for infinite sums,
you weren't always guaranteed that you could interchange
the order of summation and integration.
Or as I usually put it, you could always interchange them,
but you may not get the same answer.
So the first question that comes up
is, how do I know that I am allowed
to interchange this order?
By the way, the second question that comes up is--
and this is a more subtle question,
but going back to what we're talking about before--
how do I even know that there is a linear combination of the phi
sub n's that comes out to be the given f of x?
You see you notice that our demonstration here
presupposes that f of x can be written this way.
You see the two questions that come up
is, given an f of x, first of all,
can it be written this way?
And secondly, even if it can, do we
have the right to interchange the order of integration
and summation?
By the way, assuming for the moment
that we can do this mechanically at least,
notice that what C3 turns out to be
is the integral from a to b f of x phi sub
3 of x dx divided by the integral form a to b phi sub 3
of x squared times dx, and again, replacing the 3 by a k.
This is how every one of these coefficients
could have been determined.
And now you see, as a very, very brief
aside which I will go into much more detail in the study guide
when we talk about the exercises and the various discussion
of the exercises, given a series of functions which
is integrable, a sequence of functions
integrable on a given interval from a to b,
one usually defines the dot product of these two functions
to be the integral from a to b the product of the two
functions times dx.
You see, remember last time we showed
that all you needed to have a dot product
was a rule that showed you how to combine
two members of your set to give you a number?
And certainly, this thing here is a number.
And it also had to have three or four basic properties.
And as one of the exercises, we will
show that those properties for a dot product are obeyed here.
But this is where the word "orthogonal functions"
come from.
Notice that in terms of this definition of a dot product,
two vectors are orthogonal if their dot product is 0.
And now they're saying that this equals
0 is the same as saying that this integral is 0.
And that's precisely what the definition of orthogonal was.
So you see, this is the connection
between the vector space interpretation of orthogonal
and the traditional interpretation of orthogonal.
And as I say, I'll leave further discussion
of that for the notes.
But the important point is this.
If I mechanically solve for these coefficients--
in other words, suppose I'm given f of x.
And I'm given a family of functions
phi sub n of x which is orthogonal on the interval
from a to b.
Suppose that without even knowing whether f of x
can be represented as a linear combination of the phis,
and that without even knowing whether I can interchange
these two operations, the order in which I perform these two
operations, let's suppose that I mechanically
invent as my coefficients the C's in this particular way.
In other words, regardless of whether it
means anything or not, I can always
multiply f of x by phi sub k of x, integrate from a to b.
You see, the product of two integrable functions
is always integrable.
I can always divide that by the square of the function phi sub
k of x from a to b, keeping in mind, by the way, that
as long as phi sub 3 is not identically 0
or if it's not 0 other than on a set of points of measure
0, what have you, notice that the square of a number
can't be negative.
So if phi sub 3 of x is different from 0
in enough places, by squaring it,
I have a positive integrand.
And when I integrate a positive integrand
I can't get 0 for an answer.
So I don't have a zero denominator here.
But the point is this.
If I choose the C's as I did here,
see, with the C's as just defined, we write that f is x--
and then we put a little squiggle here.
See, don't know if we had the right
to assume that this series here is equal to f of x,
but we write this little squiggle mark here.
We call the resulting infinite series capital F of x.
And what we say is that capital F
of x is the Fourier representation of little f of x
with respect to the family phi sub n of x.
You see, we don't say that these two are equal.
But when we choose these coefficients the way I just
mentioned, this is, in any event,
called the Fourier representation of the function
little f of x.
I see there are many questions that come up.
And in the form of an overview, let me just state them.
First of all, how do we know that capital F of x
is even a convergent series?
After all, this is an infinite series.
How do we know it even converges?
Secondly, if it does converge, how do we know
that it converges to f of x?
The answer is we don't know.
We really don't know.
But there is a rather remarkable result.
And the result is not that difficult to prove.
But in the form of an overview, I
prefer to omit the proof here.
It will be an exercise in the study guide,
but I think it's important to see what
the statement of the result is.
And then we can talk about the ramifications of that.
But the proof would actually obscure
the beauty of the result.
And the aside, which I leave as an exercise, is simply this.
Suppose you took a linear combination
of the first n of the terms phi 1 of x up to phi n of x.
If you choose the C sub k's to be the so-called Fourier
coefficients the way I just mentioned,
what is interesting is this, that even if the Fourier
representation doesn't converge to f of x,
it has the property that it gives you the best
least square approximation.
In other words, if I were to compute
the difference between f of x and the sum of the first n
terms in the Fourier representation
where the C's were chosen according to the method I just
mentioned to you before and square
that difference-- you see, this is why it's
called the mean square idea.
You see, large errors and small errors, in other words,
a large negative error can cancel a large positive error
because the algebraic signs tend to cancel one another.
But by squaring large errors, all of these become what?
Squares of large magnitudes are large positive magnitudes.
By squaring, I really magnify the errors.
The positives can't cancel the negatives.
And what this says is that if I use
this linear combination of the phi sub n's,
that the square approximation error is less
using these coefficients than if I had used
any other possible coefficients, that if I picked
any other coefficients in the whole world other than the ones
that I found by the method indicated just a little while
ago, if I summed these terms, subtracted that
from f of x squared and integrated the results from a
to b, I would get a number which was at least as large.
And in most cases in fact, unless the d sub k's
were chosen to equal the C sub k's, I
would get a number which was larger than this.
Now you see what this thing says?
What this says is that there can't be any real interval
where this gets very, very far away
from f of x, because you see, if it did, if it did get very,
very far away, this would be a large magnitude.
Squaring it would give me a positive large magnitude.
Integrating that from a to b would make this thing still
very, very large.
And why should it stay smaller than any other linear
combination that I can get?
And I'll emphasize this more from a concrete point
of view in a few moments.
But for the time being, what the impact of this beautiful result
is-- and notice, it doesn't look like much when
you look at it like this.
It just looks like some abstract remark.
But the beauty of this remark the impact of it
is that, when you write the so-called Fourier
representation, as opposed to power series, which I'll also
mention at the end of the lecture what
the contrast really is, the Fourier representation, if it
does converge at all, converges to the function, what we call
in the large, that over the whole interval
it starts to fit the function very nicely.
Now I'll come back to that in terms
of a more specific example in a few moments.
Let me point out that, Fourier, being basically an engineer
and a scientist, worked with one particular set
of orthogonal functions, a set of orthogonal functions that
were very, very natural to anybody working
with wave motion or things of this type,
as you'll see in a moment.
The special case that I have in mind is this.
Consider the family of functions--
well, I can call this cosine 0x, which
is another way of writing 1.
But I have 1 cosine x, cosine 2x, cosine 3x, cosine 4x,
et cetera, and also included in this family,
sine x, sine 2x, sine 3x, et cetera.
I claim that this family of functions
is orthogonal on the interval from negative pi to pi.
What I mean by that is what?
That if I take any two different members of this family,
multiply them together, integrate from minus pi to pi,
the result will always be 0.
Now the-- I won't go through this whole thing.
That will also be one of the exercises in the study guide
will be to verify this.
But let me give you a head start in doing this.
Let me pick two members of the cosine, two different numbers
of the cosine family here, and multiply them together,
and integrate from minus pi to pi.
In other words, let me take integral from minus pi
to pi cosine mx cosine nx dx where m is unequal to n.
Now again, without belaboring this point,
remember that the formula for cosine a cosine b
comes from looking at the formulas for cosine a plus b
cosine of a minus b and adding these two results.
And the sine a sine b terms drop out.
To make a long story short, or somewhat shorter,
cosine mx times cosine nx is another way
of writing one half the sum of cosine m
plus nx plus cosine m minus n x.
Now you see, to integrate this, very simply,
this just gives me what?
A 1/2 here.
This is sine m plus n x over m plus n.
This is sine m minus x--
m minus n times x over m minus n.
And those, by the way, the fact that m is unequal to n,
guarantees me that my denominator here can't be zero.
And at any rate, noticing that when I evaluate this
from x equals minus pi to pi, that these are integral,
meaning whole-number multiples of pi,
both these terms are zero.
And we've established the fact that this is zero.
So at least the members of the cosine family are orthogonal.
One can also show the members of the sine family are orthogonal.
And one can also show that any member of the cosine family
taken--
multiplied by any member of the sine family integrated
from minus pi to pi also gives zero.
So the point then is, fair enough,
that this family here is an orthogonal family
on the interval from minus pi to pi.
And the beauty of this result--
and let me point out that in textbooks
on advanced mathematics, the multitude of pages
I use to prove the results, but the statements
of the terms that you prove are not
that difficult to keep track of.
Let me just give one major result, a result that
is very, very crucial, and one that I will
emphasize in the study guide.
And that is this.
Suppose that I am given a particular function little
f of x, which is piecewise smooth on the interval
from minus pi to pi.
And remember what I mean by piecewise smooth.
I mean it's a differentiable function except that it
may have a finite number of jump discontinuities,
that if I plot the graph y equals f of x,
I get a smooth curve with the possible exception
that there may be a few jumps in it, by few meaning what?
A not-too-great infinite number, a set of measures
zero type thing.
But we won't belabor that point right now.
But the idea is this.
Suppose that I have this piecewise smooth function
little f of x.
And suppose capital F of x is the Fourier
representation of little f of x relative to my sines
and cosines.
In other words, capital F of x is the linear combinations
of the cosine terms and the sine terms
where the coefficients are chosen by that little recipe
that I showed you at the beginning of the lecture,
and which, as I say, I'll reinforce
as we do the exercises.
At any rate, let's call this capital F of x.
And in this particular case, in other words,
what particular case?
If little f of x is piecewise smooth
and the orthogonal functions are the sines and cosines
on the interval from minus pi the pi, then
the Fourier representation of f of x, capital F of x,
is given by the following.
And this is a remarkable result. It
says that capital F of x at any value of x
in the interval from minus pi to pi
is little f of x plus plus little f of x minus over 2.
Now, what does f of x plus mean?
That means the limit of f of x as you approach the given value
of x from the positive side.
And this f is x minus is the limit
of f of x as you approach the given value of x
from the negative side.
In fact, to keep this easier to understand
so I don't use x in two different ways, what I'm saying
is capital F of x sub 0 is computed as follows.
See what little f of x sub 0 is as x approaches x0
from the left.
See what f of x0 of f of x approaches as x approaches x0
from the right.
Add these two results and divide by 2.
I hope that you can see that what I'm doing here
is taking the average of these two results.
Half the sum is the average.
Also observe that if f happens to be continuous at x0,
f of x0 plus is the same as f of x0 minus.
You see, the only time that these two things can
be different is if there is a jump in the function at x0,
because if there is no jump, as I approach x0 from the left
and approach x0 from the right, the two values I get
must be the same.
In other words, notice that at the points where
f is continuous, this is nothing more than f of x0 itself.
And what this famous result says is
that, except at the points at which little f of x
is discontinuous, capital F, the Fourier representation,
is little f itself, that the Fourier
representation converges to the function itself.
And what it does that's remarkable
is that where there is a jump in the function,
it equals the average of the jump.
Well, let me show you that in terms of an example.
Let me pick a very simple function to begin with.
Let me define the function little f
of x to be negative 1 when x is in the interval from minus pi
to 0.
And let it be, say, positive 1 when x is
in the interval from 0 to pi.
So graphically, you see how this thing is going to look.
It goes like this.
By the way, notice that this is what mathematically we
call an odd function.
F of x is the negative of f of minus x.
Notice that the sine terms are odd.
The cosine terms are even.
You see, cosine of x is the same as cosine of minus x.
But sine of x is minus sine minus x.
So the sine terms are odd.
The cosine terms are even.
So the first hint or the first nice simplification
that takes place by choosing this example
is that because this is an odd function,
the cosine terms will all be absent.
And it turns out that if I use the procedure indicated earlier
in the lecture--
and again that will be one of the exercises
so that you get the drill that's necessary here,
again, I only want to give you the overview--
it turns out that the Fourier series representation
of this particular function, capital F of x,
turns out to be 4 over pi times the summation sine
n x over n for n odd.
And what that means is it's 4 over pi times
sine x plus sine 3x over 3 plus sine 5x over 5, et cetera.
By the way, these are very difficult things to draw.
You're not used to working with these.
I should point out this, that if there is anything that
works nicely in the laboratory, it's sine and cosine terms,
because those are things that you can produce
very nicely on an oscilloscope.
Analytically, these are very nasty terms to handle.
But electronically, it's a very nice way
of getting a feeling for Fourier series.
But I don't have electronic chalk here,
so I just talk about it and draw these things for you.
But for example, to show you what I mean
here, let me just take the first two terms.
Let me plot the curve y equals 4 over pi times
the quantity sine x plus sine 3x over 3.
And if I plot that curve, which I've done here
in the accentuated chalk, look at how nicely
that curve starts already to fit the graph y equals f of x.
You see how it fits over here?
No perfect fit any place, but notice that, by and large, it
fits the curve very nicely without too big an error all
the way along here.
It has roughly the right shape.
And notice also it goes through the origin,
because when x is 0, capital F of x is 0.
And that's exactly the average of the jump over here.
So you jump from minus 1 to 1.
So the origin is the average over here.
By the way, even though I don't draw very well, what
I have tried to do here is show what
happens if I tack on just a couple of more terms.
I've tacked onto this the term sine 5x over 5
and sine 7x over 7.
And then the graph would look something like this.
You see how nicely this starts to fit?
It fits very nicely along here.
But wherever there's a jump, it jumps to-- it averages
out this way.
By the way, one other observation--
obviously, the function is I'm constructing here
is periodic with period 2 pi.
If I replace x by x plus 2 pi, I don't change the sine any.
And so obviously, the function which
I'm calling capital F of x represents more
than this given function.
Rather, it represents the given function reproduced infinitely
in both directions with period 2 pi.
But again, I'll leave that for the exercises.
To summarize what's happened over here,
you see how nicely this thing fits?
If I actually took capital F of x and computed this limit
for each value of x, what the graph would look
like-- now obviously, I can't get this
on the oscilloscope, because no matter how many terms I use,
no matter how large a number, n is still finite.
But if I actually took the infinite sum,
the graph that I would get mathematically
is I would actually get the straight line x equals--
I'm sorry, y equals negative from minus pi to 0.
It would be y equals positive 1 from 0 to pi.
And at 0 itself, the function would be 0.
You see, it splits the gap.
It splits the jump.
And then of course, the function would repeat itself
at a regular interval of--
regular period of 2 pi, that this is what capital F of x
would look like if you graphed it.
By the way, to give you an example of an even function,
if we were to use this same technique for getting
the Fourier series for f of x equals the absolute value of x
on the interval from minus pi to pi,
it turns out that capital F of x would be pi over 2 minus 4
over pi summation over n odd cosine nx over n squared.
By the way, if this looks a little bit
like the previous problem, notice that this
is more than just coincidence.
If you differentiate the square root of x--
I'm sorry, if you differentiate the absolute value
of x, what is the derivative?
It's minus 1 for x negative and positive 1 for x positive.
It's the function that we were just talking about before,
only it doesn't exist at x equals 0, you see?
What I'm driving at over here is that it's not surprising
that the terms in here are the integrals of the terms that
were in our previous summation.
But notice that the absolute value of x is an even function.
Consequently, it's the cosine terms which appear here
rather than the sine terms.
And again, I've run the risk of drawing my own diagrams here.
I have plotted this for n equals 3,
meaning I've plotted y equals pi over 2 minus 4 over pi times
the quantity cosine x plus cosine 3x over 9.
And that's represented by this curve
drawn in the accentuated chalk over here.
And again, notice how even for just a small number of terms,
we have already captured in the large the shape of the curve y
equals the absolute value of x, which by the way
contrasts very sharply with power series.
Remember how power series worked?
First of all, you had to require that the function that you
were talking about possessed derivatives of all orders.
Secondly, when you wanted to fit the function by its power
series, you picked a particular point.
Remember what you got?
You first got a tangent line approximation.
Then you got a quadratic approximation
that fit the curve in a neighborhood at this point
better than the straight line did.
But after a while, the thing would sag off.
Then you added a cubic term.
And that fit even better for a while.
And then again something weird might happen.
And you kept on going like this so that ultimately,
even though you had uniform convergence, what you were
doing was you were getting a splendid fit near one
particular point and slowly but surely, and actually very, very
slowly, and also actually, very surely, but you
may not have noticed it, these approximations were
starting to fit better and better as you went out here.
But the least square approximations
would not be too good.
In other words, out here someplace,
there was starting to become some drastically large errors.
On the other hand, the Fourier representation does what?
It fits the curve that you're talking about very, very
nicely very, very quickly.
And this is one reason why Fourier series
are as powerful as they are.
And by the way, again, to end on a nice theme,
Fourier series probably have as much space
devoted to them in pure mathematics
as they do in applied mathematics.
It's a beautiful topic from both points of view.
But I felt that for our lesson today,
since it is a finishing touch to essentially the equivalent
of three-plus semesters of calculus,
that we should juxtaposition the old and the new
and show the peaceful coexistence versus
this idea of polarization where you either
have to be a traditionalist or a modernist,
where you either have to be an applied man or a purist,
that all of these ideas mingle side by side.
And about the only choice you have as a human
being depending on what your interests are
is what proportion you're going to mix them in.
Well, at any rate, this brings us
to the part of the course that always gives me
a little bit of remorse.
And that is, I hate to let my captive audience go.
We have now been through, as I say,
the equivalent of a couple of years of mathematics together.
It has been a great pleasure for me.
I hope it has been for you.
And once again, I would like to thank everybody who has worked
with me on this, but in particular,
the usual three people who have been with me through thick
and thin on this, John [? Fitchar, ?] project
self-study manager, who has also done a lot of work with me
in helping me write the study guide,
who has given me a lot of advice on how to give these lectures
and how to prepare them for the taping,
and has also been our director for the taping,
to Miss Elise [? Pelletier, ?] who in addition to her usual
secretarial work, has typed much of our study guide and has
worked as a video technician for us, and to our chief--
I'll just call him chief of audio visual effects
at the center here, Charles [? Patton. ?]
I am particularly indebted to these three people,
indebted to a lot of other people.
I thank you all for being with me this long.
And I hope that our paths will cross again.
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