PROFESSOR: Hi, everyone.
Welcome back.
So today we'd like to tackle a problem in Fourier series.
And specifically, we're just going to compute the Fourier
series for a simple function.
So the function we're interested in is f of t, which we're told
is periodic with period 2pi-- f of t is 1 from minus pi to 0,
and then it's minus 1 from 0 to pi.
So first off, we're interested in sketching f of t.
Secondly, we'd like to compute the Fourier series for f of t.
And then thirdly, we'd like to sketch the first non-zero term
of the Fourier series.
And we can specifically sketch this single term
on top of f of t.
So I'll let you think about this problem for now,
and I'll be back in a moment.
Hi, everyone.
Welcome back.
So let's take a look at sketching f of t.
So for part a, we have our axes, t.
And we're told f of t within some interval.
So we might as well plot f of t on that interval.
So minus pi, pi and 0, we know that f of t
is 1 from minus pi to 0.
We're also told that it's minus 1 from 0 to pi.
And now to fill in the blanks or to complete the picture of f,
we're told that it has a period of 2pi.
So note that they've told us what
f looks like over the range of minus pi
to pi, which is the length of 2pi.
So basically what we can do is we can use this as a stamp
and just pick up this entire picture,
shift it over one period 2pi, and just
thinking of this picture in stamping it in multiple places.
So just filling this in it's going
to look like a square wave, which jumps between minus 1
and 1 at every multiple of pi.
So this concludes part a.
For part b, which is the real meat of the problem,
we're interested in computing a Fourier series for f of t.
Now, we can always write down a Fourier series
for any periodic function.
And specifically in this case, for part b,
the periodic function we're interested in has period 2pi.
So for the class notes, we've identified
L with half the period.
So in this case, L is 2pi divided
by 2, which gives us pi.
And just to recall what a Fourier series is, what we do
is we try and take our function f of t
and write it down as a summation of sines and cosines.
So in this case for function f of t, which is 2pi periodic,
it's going to look something like this.
It's going to a of 0 plus sum from n equals 1--
and there's going to be infinitely many terms,
but in this case we have a of n times cosine of n*t.
And it's n*t here because we have period 2pi.
Plus b_n sine n_t.
So this is the general form.
And when asked to compute the Fourier series of a function,
the main difficulty is to compute these coefficients a_n
and b_n.
However, that essentially boils down
to working out some integrals.
So let's take a look at what a of 0 is.
So the formula for a_0 is 1 over 2L-- so in this case,
it's 1 over 2pi-- times the integral
over 1 period of the function, from minus pi to pi,
of just f of t.
So notice how a_0 is just the average of the function.
So if we take a look at the function f of t, f of t
spends exactly half of its time at 1 and half of its time
at minus 1.
So immediately we could guess that the average value
of f of t is going to be 0.
If you wanted to work it out specifically,
we would have 1 over 2pi minus pi to 0, f of t
takes on the value of plus 1.
And then from 0 to pi, f of t takes on the value of minus 1.
So we would end up getting pi minus pi, which is 0.
For a_n, the formula is 1 over half the period.
So note how a of 0 is just a special case.
We always have the full period in a_0, but in a_n and b_n,
the factor that divides the integral is always going to be
half the period times minus pi to pi, f of t cosine n*t dt.
And I should point out that, in general, we only
need to integrate over one period of the function.
So in some sense there's nothing special about minus pi and pi.
It's just very often these are the easiest
bounds of integration to integrate over.
But in practice, we could have used 0
to 2pi or any other interval, as long as it's exactly
one period of the function.
So in this case, I'd just like to take a look
at the symmetry of f of t.
And we note that the function f of t
is actually odd about the origin.
So if f of t is odd and cosine t is an even function, then
an odd times an even function is going to be an odd function.
And when you integrate an odd function from minus any value
to the same positive value, so in this case minus pi to pi,
we always get 0.
So this is actually 0, because we're
integrating an odd function over a symmetric interval.
So lastly, we have the values of b_n, which are 1 over pi,
minus pi to pi, f of t of sine n*t dt.
And if we were to look at just the symmetry argument again,
f of t is an odd function, sine t
is an odd function, an odd times an odd function is
an even function.
When you integrate an even function
over a symmetric bound, you will essentially
get twice the value of the integral
from 0 to one of the bounds.
So b of n in this case doesn't vanish,
which means we actually have to do some work.
So what we do?
Well, we know the value of f of t on two intervals,
so we're just going to have to work out each interval.
Minus pi to 0, it takes on the value of 1.
So we have sine n*t.
And then from 0 to pi, f of t takes on the value of minus 1,
sine n*t dt.
And you'll note that these integrals are actually
the same.
So this is negative 2 over pi, zero to pi, sine n*t dt,
which if we integrate is negative 1 over n cosine n*t
evaluated between 0 and pi.
So if I work this out, we get minus and a minus,
minus 1 over n cosine n*pi plus 1 over n.
So note that cosine of 0 is just 1.
And now if we take a look at cosine n*pi,
we see that cosine n*pi oscillates between minus 1
and 1.
So cosine of pi is negative 1, cosine of 2pi is 1,
cosine of 3pi is minus 1.
Dot, dot, dot.
So this term right here is actually negative 1 to the n.
So we have 2 over n*pi 1 minus negative 1 to the n.
And now if we just plug in some values of b of 1, b of 2,
b of 3, b of 4, we can see what pattern emerges in the b's.
So b of 1, if I plug in 1, I get 1 minus negative 1.
It's going to be 2.
So I get minus 4 over pi.
b of 2 is going to be 1 minus minus 1 squared is just 1.
So this vanishes.
b of 3 is going to be 1 minus minus 1
cubed, which is negative 1.
So again, we get negative 4 over 3pi.
b of 4 is going to be 0.
So it's sometimes useful the write out
what the Fourier series looks like.
So I'll just write it out right here.
So we have f of t is going to be negative 4 over pi times
sine of t plus 1/3 sine of 3t plus 1/5 sine of 5t plus dot,
dot, dot.
So this concludes part b.
And now lastly, for part c, we're
asked to sketch what does the first Fourier term look like.
So in this case, the first Fourier term
is going to be negative 4 over pi times sine t.
So I'm going to go back to our diagram from part a.
So let's go back to our diagram from part a.
Now what is minus 4 over pi sine t look like?
Well, it's a sine wave that has exactly period 2pi,
and it's going to line up exactly with this square wave.
In addition, minus 4 over pi is just slightly larger than 1.
So we're going to end up with sin,
which peaks just slightly above 1 and slightly below 1.
It's going to go through 0, and it's
going to go through each multiple of pi.
So it might look something like this.
So this is the first Fourier term in the series.
And notice how this first Fourier term is actually
pretty good approximation to the square wave, considering it's
just one term in a series.
As we add more terms in the series,
we're going to get something which looks closer and closer
to a square wave function.
So I'd just like to quickly recap.
When computing the Fourier series for a periodic function,
the first useful thing to do is just write down
the formula for a Fourier series,
and then write down the formulas for the coefficients
of the Fourier series.
So write down the formulas for a_0, a_n, b_n.
When computing a_0, you can often just
look at the average of the function.
When computing a_n and b_n, it's also useful
look at the symmetry of your function.
And if it's either even or odd symmetric
then typically, either all the a_n's or all the b_n's will
vanish.
And then when you work out the integrals,
you can then reconstruct the Fourier series.
So I would like to conclude here,
and I'll see you next time.