Lec 15 | MIT 18.03 Differential Equations, Spring 2006

Lec 15 | MIT 18.03 Differential Equations, Spring 2006


Well, let’s get started. The topic for today is — Sorry.
Thank you. For today and the next two
lectures, we are going to be studying Fourier series.
Today will be an introduction explaining what they are.
And, I calculate them, but I thought before we do that
I ought to least give a couple minutes oversight of why and
where we’re going with them, and why they’re coming into the
course at this place at all. So, the situation up to now is
that we’ve been trying to solve equations of the form y double
prime plus a y prime, constant coefficient
second-order equations, and the f of t was the input.
So, we are considering inhomogeneous equations.
This is the input. And so far, the response,
then, is the solution equals the corresponding solution,
y of t, maybe with some given initial
conditions to pick out a special one we call the response,
the response to that particular input.
And now, over the last few days, the inputs have been,
however, extremely special. For input, the basic input has
been an exponential, or sines and cosines.
And, the trouble is that we learn how to solve those.
But the point is that those seem extremely special.
Now, the point of Fourier series is to show you that they
are not as special as they look. The reason is that,
let’s put it this way, that any reasonable f of t
which is periodic, it doesn’t have to be even very
reasonable. It can be somewhat
discontinuous, although not terribly
discontinuous, which is periodic with period,
maybe not the minimal period, but some period two pi.
Of course, sine t and cosine t have the
exact period two pi, but if I change the frequency
to an integer frequency like sine 2t or sine 26 t, two pie would still be a
period, although would not be the period.
The period would be shorter. The point is,
such a thing can always be represented as an infinite sum
of sines and cosines. So, it’s going to look like
this. There’s a constant term you
have to put out front. And then, the rest,
instead of writing, it’s rather long to write
unless you use summation notation.
So, I will. So, it’s a sum from n equal one
to infinity integer values of n, in other words,
of a sine and a cosine. It’s customary to put the
cosine first, and with the frequency,
the n indicates the frequency of the thing.
And, the bn is sine nt. Now, why does that solve the problem of general inputs for
periodic functions, at least if the period is two
pi or some fraction of it? Well, you could think of it
this way. I’ll make a little table.
I’ll make a little table. Let’s look at,
let’s put over here the input, and here, I’ll put the
response. Okay, suppose the input is the
function sine nt. Well, in other words,
if you just solve the problem, you put a sine nt
here, you know how to get the answer, find a particular
solution, in other words. In fact, you do it by
converting this to a complex exponential, and then all the
rigmarole we’ve been going through.
So, let’s call the response something.
Let’s call it y. I’d better index it by n
because it, of course, is a response to this
particular periodic function. So, n of t,
and if the input is cosine nt, that also will have
a response, yn. Now, I really can’t call them
both by the same name. So, why don’t we put a little s
up here to indicate that that’s the response to the sine.
And here, I’ll put a little c to indicate what the answer to
the cosine. You’re feeding cosine nt,
what you get out is this function.
Now what? Well, by the way,
notice that if n is zero, it’s going to take care of a
constant term, too.
In other words, the reason there is a constant
term out front is because that corresponds to cosine of zero t,
which is one. Now, suppose I input instead an
cosine nt. All you do is multiply the
answer by an. Same here.
Multiply the input by bn. You multiply the response.
That’s because the equation is a linear equation.
And now, what am I going to do? I’m going to add them up.
If I add them up from the different ends and take a count
also, the n equals zero corresponding to this first
constant term, the sum of all these according
to my Fourier formula is going to be f of t.
What’s the sum of this, the corresponding responses?
Well, that’s going to be summation a n y n c t
plus b n y n, the response to the
sine. That will be the sum from one
to infinity, and there will be some sort of constant term here.
Let’s just call it c1. So, in other words,
if this input produces that response, and these are things
which we can calculate, we’re led by this formula,
Fourier’s formula, to the response to things which
otherwise we would have not been able to calculate,
namely, any periodic function of period two pi will have,
the procedure will be, you’ve got a periodic function
of period two pi. Find its Fourier series,
and I’ll show you how to do that today.
Find its Fourier series, and then the response to that
general f of t will be this infinite series of functions,
where these things are things you already know how to
calculate. They are the responses to sines
and cosines. And, you just formed the sum
with those coefficients. Now, why does that work?
It works by the superposition principle.
So, this is true. The reason I can do the adding
and multiplying by constant, I’m using the superposition
principle. If this input produces that
response, then the sum of a bunch of inputs produces the sum
of the corresponding responses. And, why is that?
Why can I use the superposition principle?
Because the ODE is linear. It’s okay, since the ODE is
linear. That’s what makes all this
work. Now, so what we’re going to do
today is I will show you how to calculate those Fourier series.
I will not be able to use it to actually solve any differential
equation. It will take us pretty much all
the period to show how to calculate a Fourier series.
And, okay, so I’m going to solve differential equations on
Monday. Wrong.
I probably won’t even get to it then because the calculation of
a Fourier series is a sufficient amount of work that you really
want to know all the possible tricks and shortcuts there are.
Unfortunately, they are not very clever
tricks. They are just obvious things.
But, it will take me a period to point out those obvious
things, obvious in my sense if not in yours.
And, finally, the third day,
we’ll solve differential equations.
I will actually carry out the program.
But the main thing we’re going to get out of it is another
approach to resonance because the things that we are going to
be interested in are picking out which of these terms may
possibly produce resonance, and therefore a very crazy
response. Some of the terms in the
response suddenly get a much bigger amplitude than this than
you would normally have thought they had because it’s picking
out resonant terms in the Fourier series of the input.
Okay, well, that’s a big mouthfu.
Let’s get started on calculating.
So, the program today is calculate the Fourier series.
Given f of t periodic, having two pi as a period,
find its Fourier series. How, in other words,
do I calculate those coefficients,
an and bn. Now, the answer is not
immediately apparent, and it’s really quite
remarkable. I think it’s quite remarkable,
anyway. It’s one of the basic things of
higher mathematics. And, what it depends upon are
certain things called the orthogonality relations.
So, this is the place where you’ve got to learn what such
things are. Well, I think it would be a
good idea to have a general definition, rather than
immediately get into the specifics.
So, I’m going to call u of x, u of t, I think I will use, since Fourier analysis is most
often applied when the variable is time, I think I will stick to
independent variable t all period long, if I remember to,
at any rate. So, these are two continuous,
or not very discontinuous functions on minus pi.
Let’s make them periodic. Let’s say two pi is a period.
So, functions, for example like those guys,
sine t, sine nt, sine 22t, and so on, say two pi is a
period. Well, I want them really on the
whole real axis, not there.
Define for all real numbers. Then, I say that they are
orthogonal, perpendicular. But nobody says perpendicular.
Orthogonal is the word, orthogonal on the interval
minus pi to pi if the integral,
so, two are orthogonal. Well, these two functions,
if the integral from minus pi to pi of u of t v of t,
the product is zero, that’s called the orthogonality
condition on minus pi to pi. Now, well, it’s just the definition.
I would love to go into a little song and dance now on
what the definition really means, and what its application,
why the word orthogonal is used, because it really does
have something to do with two vectors being orthogonal in the
sense in which you live it in 18.02.
I’ll have to put that on the ice for the moment,
and whether I get to it or not depends on how fast I talk.
But, you probably prefer I talk slowly.
So, let’s compromise. Anyway, that’s the condition.
And now, what I say is that that Fourier,
that blue Fourier series, —
— what finding the coefficients an and bn depends
upon is this theorem that the collection of functions,
as I look at this collection of functions, sine nt
for any value of the integer, n, of course I can assume n is
a positive integer because sine of minus nt is the same as sine
of nt. And, cosine mt,
let’s give it a different, so I don’t want you to think
they are exactly the same integers.
So, this is a big collection of functions, as n runs from one to
infinity– Here, I could let m be run from zero
to infinity because cosine of zero t means something. It’s a constant,
one– that any two distinct ones, two distinct,
you know, how can two things be not different?
Well, you know, you talk about two coincident
roots. I’m just killing,
doing a little overkill. Any two distinct ones of these,
two distinct members of the set of this collection of,
I don’t know, there’s no way to say that,
any two distinct ones are orthogonal on this interval.
Of course, they all have two pi as a period for all of them.
So, they form into this general category that I’m talking about,
but any two distinct ones are orthogonal on the interval for
minus pi to pi. So, if I integrate from minus pi to pi sine of three t times
cosine of four t dt, answer is zero. If I integrate sine of 3t times the sine of 60t,
answer is zero. The same thing with two
cosines, or a sine and a cosine. The only time you don’t get
zero is if you integrate, if you make the two functions
the same. Now, how do you know that you
could not possibly get the answer is zero if the two
functions are the same? If the two functions are the
same, then I’m integrating a square.
A square is always positive. I’m integrating a square.
A square is always positive, and therefore I cannot get the
answer, zero. But, in the other cases,
I might get the answer zero. And the theorem is you always
do. Okay, so, why is this?
Well, there are three ways to prove this.
It’s like many fundamental facts in mathematics.
There are different ways of going about it.
By the way, along with the theorem, I probably should have
included, so, I’m far away.
But you might as well include, because we’re going to need it.
What happens if you use the same function?
If I take U equal to V, and in that case,
as I’ve indicated, you’re not going to get the
answer, zero. But, what you will get is,
so, in other words, I’m just asking,
what is the sine of n t squared.
That’s a case where two of them are the same.
I use the same function. What’s that?
Well, the answer is, it’s the same as what you will
get if you integrate the cosine, cosine squared n t dt. And, the answer to either one of these is pi.
That’s something you know how to do from 18.01 or the
equivalent thereof. You can integrate sine squared.
It’s one of the things you had to learn for whatever exam you
took on methods of integration. Anyway, so I’m not going to
calculate this out. The answer turns out to be pi.
All right, now, the ways to prove it are you
can use trig identities. And, I’m asking you in one of
the early problems in the problem set, identities,
identities for the product of sine and cosine,
expressing it in a form in which it’s easy to integrate,
and you can prove it that way. Or, you can use,
if you have forgotten the trigonometric identities and
want to get some more exercise with complex– you can use
complex exponentials. So, I’m asking you how to,
in another part of the same problem I’m asking you how to do
it, do one of these, at any rate,
using complex exponentials. And now, I’m going to use a
mysterious third method another way.
I’m going to use the ODE. I’m going to do that because
this is the method. It’s not just sines and cosines
which are orthogonal. There are masses of orthogonal
functions out there. And, the way they are
discovered, and the way you prove they’re orthogonal is not
with trig identities and complex exponentials because those only
work with sines and cosines. It is, instead,
by going back to the differential equation that they
solve. And that’s, therefore,
the method here that I’m going to use here because this is the
method which generalizes to many other differential equations
other than the simple ones satisfied by sines and cosines.
But anyway, that is the source. So, the way the proof of these
orthogonality conditions goes, so I’m not going to do that.
And, I’m going to assume that m is different from n so that I’m
not in either of these two cases.
What it depends on is, what’s the differential
equation that all these functions satisfy?
Well, it’s a different differential equation depending
upon the value of n, —
— but they look at essentially the same.
These satisfy the differential equation, in other words,
what they have in common. The differential equation is,
let’s call it u. It looks better.
It’s going to look better if you let me call it u.
u double prime plus, well, n squared,
so for the function sine n t cosine n t, satisfy u double prime plus n squared times u. In other words,
the frequency is n, and therefore,
this is a square of the frequency is what you put here,
equals zero. In other words,
what these functions have in common is that they satisfy
differential equations that look like that.
And the only thing that’s allowed to vary is the
frequency, which is allowed to change.
The frequency is in this coefficient of u.
Now, the remarkable thing is that’s all you need to know.
The fact that they satisfy the differential equation,
that’s all you need to know to prove the orthogonality
relationship. Okay, let’s try to do it.
Well, I need some notation. So, I’m going to let un and vm
be any two of the functions. In other words,
I’ll assume m is different from n.
For example, this one could be sine nt,
and that could be sine of mt,
or this could be sine nt and that could be
cosine of mt. You get the idea.
Any two of those in the subscript indicates whether what
the n or the m is that are in that.
Any two, and I mean really two, distinct, well,
if I say that m is not n, then they positively have to be
different. So, again, it’s overkill with
my two’s-ness. And, what I’m going to
calculate, well, first of all,
from the equation, I’m going to write the equation
this way. It says that u double prime is
equal to minus n squared u. That’s true for any of these guys.
Of course, here, it would be v double prime is
equal to minus m squared times v.
You have to make those simple adjustments.
And now, what we’re going to calculate is the integral from
minus pi to pi of un double prime times vm dt. Now, just bear with me. Why am I going to do that?
I can’t explain what I’m going to do that.
But you won’t ask me the question in five minutes.
But the point is, this is highly un-symmetric.
The u is differentiated twice. The v isn’t.
So, those two functions– but there is a way of turning them
into an expression which looks extremely symmetric,
where they are the same. And the way to do that is I
want to get rid of one of these primes here and put one on here.
The way to do that is if you want to integrate one of these
guys, and differentiate this one to make them look the same,
that’s called integration by parts, the most important
theoretical method you learned in 18.01 even though you didn’t
know that it was the most important theoretical method.
Okay, we’re going to use it now as a basis for Fourier series.
Okay, so I’m going to integrate by parts.
Now, the first thing you do, of course, when you integrate
by parts is you just do the integration.
You don’t do differentiation. So, the first thing looks like
this. And, that’s to be evaluated
between negative pi and pi. In doing integration by parts
between limits, minus what you get by doing
both. You do both,
the integration and the differentiation.
And, again, evaluate that between limits.
Now, I’m just going to BS my way through this.
This is zero. I don’t care what the un’s,
which un you picked and which vm you picked.
The answer here is always going to be zero.
Instead of wasting six boards trying to write out the
argument, let me wave my hands. Okay, it’s clear,
for example, that a v is a sine, sine mt. Of course it’s zero because the
sine vanishes at both pi and minus pi.
If the un were a cosine, after I differentiate it,
it became a sine. And so, now it’s this side guy
that’s zero at both ends. So, the only case in which we
might have a little doubt is if this is a cosine,
and after differentiation, this is also a cosine.
In other words, it might look like cosine,
after, this cosine nt times cosine mt.
But, I claim that that’s zero, too.
Why? Because the cosines are even
functions, and therefore, they have the same value at
both ends. So, if I subtract the value
evaluated at pi, and subtract the value of minus
pi, again zero because I have the same value at both ends.
So, by this entirely convincing argument, no matter what
combination of sines and cosines I have here, the answer to that
part will always be zero. So, by calculation,
but thought calculation; it’s just a waste of time to
write anything out. You stare at it until you agree
that it’s so. And now, I’ve taken,
by this integration by parts, I’ve taken this highly
un-symmetric expression and turned it into something in
which the u and the v are treated exactly alike.
Well, good, that’s nice, but why?
Why did I go to this trouble? Okay, now we’re going to use
the fact that this satisfies the differential equation,
in other words, that u double prime is equal to
minus n, I’m sorry, I should have
subscripted this. If that’s the solution,
then this is equal to, times.
You have to put in a subscript otherwise.
The n wouldn’t matter. All right, I’m now going to
take that expression, and evaluate it differently.
un double prime vm dt is equal to,
well, un double prime, because it satisfies the
differential equation is equal to that.
So, what is this? This is minus n squared
times the integral from negative pi to pi,
and I’m replacing un double prime by minus n
squared un. I pulled the minus n squared
out. So, it’s un here,
and the other factor is vm dt. Now, that’s the proof.
Huh? What do you mean that’s the
proof? Okay, well, I’ll first state
it, why intuitively that’s the end of the argument.
And then, I’ll spell it out a little more detail,
but the more detail you make for this, the more obscure it
gets instead of, look, I just showed you that
this is symmetric in u and v, after you massage it a little
bit. Here, I’m calculating it a
different way. Is this symmetric in u and v?
Well, the answer is yes or no. Is this symmetric at u and v?
No. Why?
Because of the n. The n favors u.
We have what is called a paradox.
This thing is symmetric in u and v because I can show it is.
And, it’s not symmetric in u and v because I can show it is.
I can show it’s not symmetric because it favors the n.
Now, there’s only one possible resolution of that paradox.
Both would be symmetric if what were true?
Pardon? Negative pi.
All right, let me write it this way.
Okay, never mind. You see, the only way this can
happen is if this expression is zero.
In other words, the only way something can be
both symmetric and not symmetric is if it’s zero all the time.
And, that’s what we’re trying to prove, that this is zero.
But, instead of doing it that way, let me show you.
This is equal to that, and therefore,
two things according to Euclid, two things equal to the same
thing are equal to each other. So, this equals that,
which, in turn, therefore, equals what I would
have gotten. I’m just saying the symmetry of
different way, what I would have gotten if I
had done this calculation. And, that turns out to be minus
m squared times the integral from minus pi to pi
of un vm dt. So, these two are equal because they are both equal to this.
This is equal to that. This equals that.
Therefore, how can this equal that unless the integral is
zero? How’s that?
Remember, m is different from n.
So, what this proves is, therefore, the integral from
negative pi to pi of un vm dt is equal to zero, at least if m is different from
n. Now, there is one case I didn’t
include. Which case didn’t I include?
un times un is not supposed to be zero.
So, in that case, I don’t have to worry about,
but there is a case that I didn’t.
For example, something like the cosine of nt
times the sine of nt. Here, the m is the same as the n.
Nonetheless, I am claiming that this is zero
because these aren’t the same function.
One is a cosine. Why is that zero?
Can you see mentally that that’s zero?
Mentally? Well, this is trying to be in
another life, it’s trying to be one half the
sine of two nt, right? And obviously the integral of sine of two nt is zero between
minus pi and pi because you integrate it, and it turns out to be zero.
You integrate it to a cosine, which has the same value of
both ends. Well, that was a lot of
talking. If this proof is too abstract
for you, I won’t ask you to reproduce it on an exam.
You can go with the proofs using trigonometric identities,
and/or complex exponentials. But, you ought to know at least
one of those, and for the problem set I’m
asking you to fool around a little with at least two of
them. Okay, now, what has this got to
do with the problem we started with originally?
The problem is to explain this blue series.
So, our problem is, how, from this,
am I going to get the terms of this blue series?
So, given f of t, two pi s a period.
Find the an and the bn. Okay, let’s focus on the an.
The bn is the same. Once you know how to do one,
you know how to do the other. So, here’s the idea.
Again, it goes back to the something you learned at the
very beginning of 18.02, but I don’t think it took.
But maybe some of you will recognize it.
So, what I’m going to do is write it.
Here’s the term we’re looking for here, this one.
Okay, and there are others. It’s an infinite series that
goes on forever. And now, to make the argument,
I’ve got to put it one more term here.
So, I’m going to put in ak cosine kt.
I don’t mean to imply that that k could be more than n,
in which case I should have written it here.
I could have also used equally well bk sine kt
here, and I could have put it there.
This is just some other term. This is the an,
and this is the one we want. And, this is some other term.
Okay, all right, now, what you do is,
to get the an, what you do is you multiply
everything through by, you focus on the one you want,
so it’s dot, dot, dot, dot,
dot, and you multiply by cosine nt.
So, it’s ak cosine kt times cosine nt. Of course, that gets
multiplied, too. But, the one we want also gets
multiplied, an. And, it becomes,
when I multiply by cosine nt, cosine squared nt, and now, I hope you can see
what’s going to happen. Now, oops, I didn’t multiply
the f of t, sorry.
It’s the oldest trick in the book.
I now integrate everything from minus, so I don’t endlessly
recopy. I’ll integrate by putting it up
in yellow chalk, and you are left to your own
devices. This is definitely a colored
pen type of course. Okay, so, you want to integrate
from minus pi to pi? Good.
Just integrate everything on the right hand side,
also, from minus pi to pi. Plus, these are the guys just
to indicate that I haven’t, they are out there,
too. And now, what happens?
What’s this? Zero.
Every term is zero because of the orthogonality relations.
They are all of the form, a constant times cosine nt
times something different from cosine nt, sine kt, cosine kt,
or even that constant term. All of the other terms are
zero, and the only one which survives is this one.
And, what’s its value? The integral from minus pi to
pi of cosine squared, I put that up somewhere.
It’s right here, down there?
It is pi. So, this term turns into an pi,
an, dragged along, but this, the integral of the
square of the cosine turns out to be pi.
And so, the end result is that we get a formula for an.
What is an? an is, well,
an times pi, all these terms of zero,
and nothing is left but this left-hand side.
And therefore, an times pi is the integral
from negative pi to pi of f of t times cosine nt dt. But, that’s an times pi. Therefore, if I want just an,
I have to divide it by pi. And, that’s the formula for the
coefficient an. The argument is exactly the
same if you want bn, but I will write it down for
the sake of completeness, as they say,
and to give you a chance to digest what I’ve done,
you know, 30 seconds to digest it. Sine nt dt. And, that’s because the
argument is the same. And, the integral of sine
squared nt is also pi. So, there’s no difference there.
Now, there’s only one little caution.
It have to be a little careful. This is n one,
two, and so on, and this is also n one,
two, and unfortunately, the constant term is a slight
exception. We better look at that
specifically because if you forget it, you can get them to
gross, gross, gross errors.
How about the constant term? Suppose I repeat the argument
for that in miniature. There is a constant term plus
other stuff, a typical other stuff, an cosine,
let’s say. How am I going to get that
constant term? Well, if you think of this as
sort of like a constant times, the reason is the constant is
because it’s being multiplied by cosine zero t.
So, that suggests I should multiply by one.
In other words, what I should do is simply
integrate this from negative pi to pi, f of t dt. What’s the answer?
Well, this integrated from minus pi to pi is how much?
It’s c zero times two pi, right?
And, the other terms all give me zero.
Every other term is zero because if you integrate cosine
nt or sine nt over a complete
period, you always get zero. There is as much area above the
axis or below. Or, you can look at two special
cases. Anyway, you always get zero.
It’s the same thing with sine here.
So, the answer is that c zero is equal to,
is a little special. You don’t just put n equals
zero here because then you would lose a factor of two.
So, c zero should be one over two pi times
this integral. Now, there are two kinds of
people in the world, the ones who learn two separate
formulas, and the ones who just learn two separate notations.
So, what most people do is they say, look, I want this to be
always the formula for a zero. That means, even when n
is zero, I want this to be the formula.
Well, then you are not going to get the right leading term.
Instead of getting c zero, you’re going to get
twice it, and therefore, the formula is,
the Fourier series, therefore, isn’t written this
way. It’s written– If you want an a
zero there, calculate it by this formula.
Then, you’ve got to write not c zero, but a zero over two. I think you will be happiest if
I have to give you advice. I think you’ll be happiest
remembering a single formula for the an’s and bn’s,
in which case you have to remember that the constant
leading term is a zero over two if you insist on
using that formula. Otherwise, you have to learn a
special formula for the leading coefficient, namely one over two
pi instead of one over pi.
Well, am I really going to calculate a Fourier series in
four minutes? Not very likely,
but I’ll give it a brave college try.
Anyway, you will be doing a great deal of it,
and your book has lots and lots of examples, too many,
in fact. It ruined all the good examples
by calculating them for you. But, I will at least outline.
Do you want me to spend three minutes outlining a calculation
just so you have something to work on in the next boring class
you are in? Let’s see, so I’ll just put a
few key things on the board. I would advise you to sit still
for this. Otherwise you’re going to hack
it, and take twice as long as you should, even though I knew
you’ve been up to 3:00 in the morning doing your problem set.
Cheer up. I got up at 6:00 to make up the
new one. So, we’re even.
This should be zero here. So, here’s minus pi.
Here’s pi. Here’s one, negative one.
The function starts out like that, and now to be periodic,
it then has to continue on in the same way.
So, I think that’s enough of its path through life to
indicate how it runs. This is a typical square-away
function, sometimes it’s called. It’s an odd function.
It goes equally above and below the axis.
Now, the integrals, when you calculate them,
the an is going to be, okay, look, the an is going to
turn out to be zero. Let me, instead,
and you will get that with a little hacking.
I’m much more worried about what you’ll do with the bn’s.
Also, next Monday you’ll see intuitively that the an is zero,
in which case you won’t even bother trying to calculate it.
How about the bn, though?
Well, you see, because the function is
discontinuous, so, this is my input.
My f of t is that orange discontinuous function.
The bn is going to be, I have to break it into two
parts. In the first part,
the function is negative one. And there, I will be
integrating from minus pi to pi of the function,
which is minus one times the sine of nt dt. And then, there’s another part, sorry, minus pi to zero.
The other part I integrate from zero to pi of what?
Well, f of t is now plus one. And so, I simply integrate sine nt dt.
Now, each of these is a perfectly simple integral.
The only question is how you combine them.
So, this is, after you calculate it,
it will be (one minus cosine n pi) all over n. And, this part will turn out to
be (one minus cosine n pi) over n also. And therefore, the answer will be two minus
two cosine, two over n times, right, two minus,
two times (one minus cosine n pi) over n. No, okay, now,
what’s this? This is minus one if n is odd.
It’s plus one if n is even. Now, either you can work with
it this way, or you can combine the two of them into a single
expression. Its minus one to the nth power
takes care of both of them.
But, the way the answer is normally expressed,
it would be minus two over n, two over n times,
if n is even, I get zero.
If n is odd, I get two.
So, times two, if n is odd,
and zero if n is even. So, it’s four over n,
or it’s zero, and the final series is a sum
of those coefficients times the appropriate– cosine or sine?
Sine terms because the cosine terms were all coefficients,
all turned out to be zero. I’m sorry I didn’t have the
chance to do that calculation in detail.
But, I think that’s enough sketch for you to be able to do
the rest of it yourself.

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