If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:6:08

- [Voiceover] So we've
been spending some time now thinking about the idea
of a Fourier series, taking a periodic function
and representing it as the sum of weighted cosines and sines, and some of you might say,
well, how is this constant a weighted cosine or sine? Well, you can view a
sub zero as a sub zero times cosine of zero t, and of course cosine of zero t is just 1, so this, it'll just end up being a sub zero, but that could be a weighted
cosine, if you want to view it that way, and in the last video, we started leveraging
some of the integrals, definite integrals of sines and cosines to establish a formula for a sub zero which intuitively ended up
being the average value of our function over, over 1 of its period, or over zero to 2 pi which is the interval
that we are caring about. What I want to do now is
find a general expression for a sub n, where n is not equal to zero. So for n is greater than zero, for n is an integer greater than zero and I'm going to use a very similar, a very similar technique. To figure out a sub zero, I just multiply, I just took the integral of both sides. What I'm going to do now is
I'm going to multiply both sides of this equation by cosine of nt. So let me do that, so on the left-hand side,
I'm going to multiply by cosine of nt, and on the right-hand side, multiplying it by cosine of nt, if I distribute that cosine
of nt, I'm going to multiply every term by cosine nt. So it's going to be cosine of nt, cosine of nt, you might see where this is going, especially 'cause we took
all that trouble to take, to figure out the properties of, of definite integrals. Cosine of nt, cosine of nt, cosine of nt, or the properties of definite integrals of products of cosines and sines. Cosine of nt. And this is going to be
true for all of the terms. There's an infinite number of terms here. That's what that dot,
dot, dot is representing. We're just going to keep
on going on and on and on. And now, let's take the
definite integral of both sides from zero to 2 pi. So the definite integral
from zero to 2 pi, dt, well that's going to
be the integral of this, from zero to 2 pi, and
actually, let me just take that coefficient out of the integral, so zero to 2 pi, dt, so zero to 2 pi, dt, a little bit monotonous but it's worth it, zero to 2 pi, dt, zero to 2 pi, dt, we might as well have fun while we do it, zero to 2 pi, dt, all right. Zero, my hand is hurting, zero to 2 pi, dt, zero to 2 pi, dt, once again, just integration property. I'm taking the integral from zero, the definite integral from zero to 2 pi of both sides, and I'm just
saying, hey, the integral of this infinite sum is
equal to the infinite sum of the different integrals. Now, what is this going to be equal to? Well, we know, we know from before that if I just take the definite integral from
zero to 2 pi of cosine nt, where n is some non-zero integer, this is just going to be zero, so this whole thing is going to be zero. We also know, and we've
established it before, if we take, so I just used this property, we also know if we take
cosine nt times sine of mt for any integers mt over the
interval from zero to 2 pi, that's going to be zero, and we also know, we also know that if we take
the integral of cosine mt, cosine nt, where m does not equal n, where m does not equal n, then that is going to be zero. And so, we know, well, here, we have, we're assuming that, we're assuming here that
n is not equal to 1, the coefficient here, so that
is going to be equal to zero. This is sines and cosines with
the different coefficients, and frankly, if they had
the same coefficient, that's going to be zero. We established that in
the last few videos. This is going to be zero, same argument. This is going to be zero, this is going to be zero. Everything's going to be
zero except for this thing right over here, but what is this thing? This is the definite, what we're taking the integral of, this is the same thing
as cosine-squared of nt, dt, the definite integral
from zero to 2 pi of cosine-squared nt, dt. Well, what is that? Well, we established that for any m, non-zero m, this is
going to be equal to pi. So all our work is paying off. So, this whole integral, this whole integral right
over here is going to evaluate to pi. We did that in another video and so now we can start
solving for a sub n. We know that a sub n times pi, a sub n times pi, times pi is equal to, is equal to this 'cause everything
else ended up being zero which is very nice. It's equal to the definite
integral from zero to 2 pi of f of t, of f of t times cosine nt, cosine of nt, dt. Cosine of nt, dt, and so we
can now solve for a sub n. We just divide both sides by, we just divide both sides by pi, and we get a sub n is equal to, I think we deserve a
little bit of a drum roll, a sub n is equal to 1 over pi, times all of this business, times the definite
integral from zero to 2 pi, f of t, f of t, cosine nt, cosine nt, dt. Dt. So if you want to find the
nth coefficient for one of our cosines, a sub n, well, you take your
function, multiply it times cosine of nt, and then
take that definite integral from zero to 2 pi, and then divide by pi, so pretty intuitive, and what's cool is that the math kind of works out this way, that you can actually do this, so hopefully you enjoyed
that as much as I did.