GILBERT STRANG: OK, I'm going
to explain Fourier series,
and that I can't
do in 10 minutes.
It'll take two,
maybe three, sessions
to see enough examples
to really use the idea.
Let me start with what
we're looking for.
We have a function.
And we want to write it as
a combination of cosines
and sines.
So those our basis functions--
the cosines and the sine.
And a n's and the b n's
are the coefficients
that we have to look for.
That tells us how much of
cosine nx is in the big function
f of x.
Notice that the cosines start at
n equals 0, because cosine of 0
is 1.
So there's an a0 in our sum.
But there isn't a b0, because
n equals zero of the sine
would be zero, and we
don't get anything there.
So we're looking for
the a n's and b n's.
And, really, I want to
show you, at the same time,
the complex form
with coefficient cn.
And now n goes from minus
infinity to infinity.
That's really the
more beautiful form
because that one formula
for cn does the job,
whereas here I will need
a separate formula for a n
and for bn.
OK.
So this is natural when
the function is real,
but in the end, and for the
discrete Fourier transform,
and for the fast
Fourier transform,
the complex case will win.
And, of course, everybody
sees that e to the inx,
by Euler's great formula, is
a combination of cosine nx
and sine nx.
So, I can use those, or I
can use cosine and sine.
OK.
So, how do you
find these numbers?
The key is orthogonality.
So that's the first central
idea here in Fourier series,
is the idea of orthogonality.
Now what does that mean?
That means perpendicular.
And for a vector,
and a second vector,
we have an idea of what
perpendicular means.
The 90 degree
angle between them.
And we check that by the dot
product-- or inner product,
whichever name you like--
between the two vectors
should be 0.
OK.
But here we have functions--
like cosine functions.
So here's one cosine, and
here's a different cosine.
So those are two different basis
functions-- say, cosine of 7x
and cosine of 12 x.
The coefficients a7
and a12 would tell us
how much of cosine 7x
is in the function.
You see, we're separating the
function into frequencies.
We're looking into pure
oscillations, pure harmonics.
And we expect, probably,
that's the lower harmonics
the smoother ones cos
x, cos 2x, cos 3x,
have most of the energy.
And the high harmonics,
cosine 12x, cosine 100x,
probably those are
quickly alternating,
those contain noise,
and high frequency.
Quick changes in the
function will show up
in the high frequencies.
OK.
So what's the answer to
this integral-- cosine
of 7x times cosine of 12x dx,
over the range minus pi to pi?
Orthogonality comes
in, the answer is 0.
That's the crucial fact.
That's what makes it possible
to separate out a7 and a12
and get hold of them.
So let me show you
how to do that.
So I'm going to use this
fact, which is the function
version of 90 degree angle.
So, you see, it's a
little like a dot product.
Well, let me remember, a dot
product would be something
like c1 d1 plus c2 d2 equals
0, if I had a vector c1 c2
and a vector d1 d2.
That would be the dot
product, and it would be 0
if the vectors are orthogonal.
Here, instead of
adding, I'm integrating
because I have functions.
So just that's the
meaning of dot product--
the integral of
one function times
the other function gives 0.
OK.
I'll use that now.
OK, how will I use this?
I will look what I want.
This is my goal.
I'll multiply both sides of
this equation by cosine kx.
And then I'll integrate.
And the beauty is, that when
I multiply by cosine kx,
and I integrate, everything
goes to zero except what I want.
By the way, all the sines
times cosine kx integrate to 0.
All the sines are orthogonal
to all the cosines.
And all the cosines will
be orthogonal to all
the other cosines.
So let me show you what I get.
So I multiply my f
of x by cosine kx,
and I integrate
from minus pi to pi.
OK?
Now, on the right-hand
side, this is my integral
from minus pi to pi, of my
big sum of all these terms,
0 to infinity, a n cos nx,
etcetera-- including the sines
but I'm not even put
them in because they're
going to get killed by this
integration-- times cosine kx
dx.
All I did was take the f
of x equal that formula,
multiplied both sides by
cosine kx, and integrated.
And, now the
orthogonality pays off,
because this times
this, when I integrate
gives 0, with one exception.
When n equals k, then
I do get the integral.
The only term I get is
ak, cosine kx, twice dx.
Only k equal n
survives this process.
And then that integral
of cosine squared
happens to be pi, so
this is just ak times pi.
Look, I've discovered
what ak is.
I've discovered the k
Fourier cosine coefficient.
I just divide by pi.
So can I just divide by pi
to get this formula for ak?
Ak is 1 over pi.
The integral from minus pi
to pi of my function, times
cosine kx dx.
That's the formula.
That tells me the coefficient.
And I could only do
that with orthogonality
to knock out all but one term.
And now, if I wanted
the sine coefficients,
bk, it would be the same formula
except that would be a sine.
And if I wanted the
complex coefficient,
ck, it turns out it'd be
the same formula expect--
well maybe it's 2 pi
there, 1 over 2 pi--
and this becomes an
e to the minus ikx.
In a complex case, the complex
conjugate e to the minus ikx
shows up.
So this is really the dot
product, the inner product,
of the function with the cosine.
OK.
So let me do some examples.
Maybe I should write up the sine
formula that I just mentioned.
So bk is the integral 1 over
pi, the integral of my function,
times sine kx dx.
And there's one exception.
A0 has a little bit
different formula,
the pi changes to 2 pi.
I'm sorry about that.
When k is 0 or it's the integral
of 1, from minus pi to pi,
and I get 2 pi.
So, a0 is 1 over 2 pi-- the
integral of f of x times
when k is zero
cosine-- this is 1 dx.
That has a simple meaning.
That's the average of f of x.
OK.
So the basis function was
just 1 when k was zero.
When k is 0, the function
of my cosine is just one,
and I get the integral
of the function
times 1 divided by 2 pi.
Could we just do an example?
So I want to take a function.
And in this video
why don't I take
an easy, but very important,
function-- the delta function.
So I plan to use these
formulas on the delta function.
Let me draw a little picture
of the delta function.
I'm only going between
minus pi and pi,
and the delta function,
as we know, is 0,
it's infinite, at the
spike, and 0 again.
The reason I wanted to draw it
is, that's an even function.
That's a function which is
symmetric between x and minus
x.
And in that case,
there will be no sines.
Sine functions are odd.
The integral from minus pi to
pi of an odd function gives 0.
The odd means that when
you cross x equals 0
you get minus the result
for x greater than 0.
So my point is, this
is an even function--
delta of x is the same as delta
of minus x, and only cosines.
Good.
The sine coefficients
automatically dropped our 0 so,
of course, the
integral would show it.
But we see it even
before we integrate.
OK I'm ready for
the delta function.
So I'm going to
write delta of x,
and we remember what
the delta function
is-- a combination of cosines.
OK.
That's the delta function
between minus pi and pi.
OK.
And what's our
formula for the a n?
Well, you remember we had a
special formula for a0, which
was 1/2 pi times the
integral, from minus pi to pi,
of our function, which is delta,
times the basis function, which
n equals 0, the basis
function is 1 dx.
OK, we know the answer to that.
We can integrate
the delta function.
The one key thing about the
integral of the delta function
is, it's always 1-- if we cross
x equals 0, which we will.
So that integral is 1
so I'm getting 1/2 pi.
What about the other
for a coefficient?
So that's 1/pi, now.
The integral from minus pi to
pi of all of my function times
cosine kxdx.
You know what I'm doing.
I'm using my formula to
find the coefficients.
My formula says take the
function, whatever it is--
and in this example,
it's the delta function--
multiply by the
cosine, integrate,
and divide by the factor pi.
OK.
Well, of course, we
can do that integral.
Because when you integrate
a delta function,
times some other function, all
the action is at x equals 0.
At x equals 0,
this function is 1.
And I don't care what it
is elsewhere, it's just 1.
So this is the same as
integrating delta of x times 1,
which gives us-- well, the
interval the delta function 1.
So that integral is one,
so I'm getting 1/pi.
Good.
OK.
So now, do you want
me to write out
the series for the
delta function?
It looks kind of unusual.
This is telling us
something quite remarkable.
It's telling us that all these
coefficients are the same.
All the frequencies,
all the harmonics,
are in the delta function
in equal amounts.
Usually, we would see a big
drop off of the coefficients ak,
but for the delta function,
which is so singular,
all a big spike at
one point, there's
no drop off and no decay
in the coefficients,
they just constant.
OK.
So I'm saying that the delta
function is the constant term,
1/2pi, and then 1/pi times
cosine of x, and cosine of 2x,
and so on.
OK.
All frequencies
there are the same.
And I'll stop with
that one example here.
So the key points
were orthogonality,
the formulas for the the
coefficients, and this example.
Thank you.
