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.