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PROFESSOR: Hi, everyone.

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Welcome back.

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So today we'd like to tackle
a problem in Fourier series.

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And specifically, we're just
going to compute the Fourier

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series for a simple function.

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So the function we're interested
in is f of t, which we're told

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is periodic with period 2pi--
f of t is 1 from minus pi to 0,

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and then it's minus
1 from 0 to pi.

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So first off, we're interested
in sketching f of t.

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Secondly, we'd like to compute
the Fourier series for f of t.

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And then thirdly, we'd like to
sketch the first non-zero term

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of the Fourier series.

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And we can specifically
sketch this single term

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on top of f of t.

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So I'll let you think
about this problem for now,

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and I'll be back in a moment.

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Hi, everyone.

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Welcome back.

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So let's take a look
at sketching f of t.

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So for part a, we
have our axes, t.

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And we're told f of t
within some interval.

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So we might as well plot
f of t on that interval.

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So minus pi, pi and
0, we know that f of t

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is 1 from minus pi to 0.

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We're also told that it's
minus 1 from 0 to pi.

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And now to fill in the blanks
or to complete the picture of f,

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we're told that it
has a period of 2pi.

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So note that
they've told us what

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f looks like over
the range of minus pi

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to pi, which is
the length of 2pi.

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So basically what we can do
is we can use this as a stamp

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and just pick up
this entire picture,

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shift it over one
period 2pi, and just

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thinking of this picture in
stamping it in multiple places.

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So just filling
this in it's going

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to look like a square wave,
which jumps between minus 1

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and 1 at every multiple of pi.

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So this concludes part a.

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For part b, which is the
real meat of the problem,

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we're interested in computing
a Fourier series for f of t.

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Now, we can always write
down a Fourier series

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for any periodic function.

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And specifically in
this case, for part b,

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the periodic function we're
interested in has period 2pi.

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So for the class
notes, we've identified

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L with half the period.

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So in this case,
L is 2pi divided

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by 2, which gives us pi.

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And just to recall what a
Fourier series is, what we do

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is we try and take
our function f of t

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and write it down as a
summation of sines and cosines.

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So in this case for function
f of t, which is 2pi periodic,

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it's going to look
something like this.

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It's going to a of 0 plus
sum from n equals 1--

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and there's going to be
infinitely many terms,

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but in this case we have a
of n times cosine of n*t.

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And it's n*t here because
we have period 2pi.

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Plus b_n sine n_t.

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So this is the general form.

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And when asked to compute the
Fourier series of a function,

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the main difficulty is to
compute these coefficients a_n

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and b_n.

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However, that
essentially boils down

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to working out some integrals.

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So let's take a look
at what a of 0 is.

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So the formula for a_0 is 1
over 2L-- so in this case,

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it's 1 over 2pi--
times the integral

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over 1 period of the
function, from minus pi to pi,

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of just f of t.

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So notice how a_0 is just
the average of the function.

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So if we take a look at
the function f of t, f of t

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spends exactly half of its
time at 1 and half of its time

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at minus 1.

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So immediately we could
guess that the average value

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of f of t is going to be 0.

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If you wanted to work
it out specifically,

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we would have 1 over 2pi
minus pi to 0, f of t

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takes on the value of plus 1.

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And then from 0 to pi, f of t
takes on the value of minus 1.

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So we would end up getting
pi minus pi, which is 0.

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For a_n, the formula is
1 over half the period.

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So note how a of 0 is
just a special case.

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We always have the full period
in a_0, but in a_n and b_n,

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the factor that divides the
integral is always going to be

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half the period times minus pi
to pi, f of t cosine n*t dt.

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And I should point out
that, in general, we only

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need to integrate over one
period of the function.

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So in some sense there's nothing
special about minus pi and pi.

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It's just very often
these are the easiest

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bounds of integration
to integrate over.

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But in practice, we
could have used 0

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to 2pi or any other interval,
as long as it's exactly

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one period of the function.

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So in this case, I'd
just like to take a look

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at the symmetry of f of t.

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And we note that
the function f of t

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is actually odd
about the origin.

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So if f of t is odd and cosine
t is an even function, then

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an odd times an even function
is going to be an odd function.

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And when you integrate an odd
function from minus any value

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to the same positive value, so
in this case minus pi to pi,

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we always get 0.

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So this is actually
0, because we're

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integrating an odd function
over a symmetric interval.

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So lastly, we have the values
of b_n, which are 1 over pi,

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minus pi to pi, f
of t of sine n*t dt.

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And if we were to look at just
the symmetry argument again,

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f of t is an odd
function, sine t

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is an odd function, an odd
times an odd function is

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an even function.

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When you integrate
an even function

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over a symmetric bound,
you will essentially

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get twice the value
of the integral

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from 0 to one of the bounds.

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So b of n in this
case doesn't vanish,

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which means we actually
have to do some work.

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So what we do?

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Well, we know the value of
f of t on two intervals,

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so we're just going to have
to work out each interval.

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Minus pi to 0, it takes
on the value of 1.

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So we have sine n*t.

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And then from 0 to pi, f of t
takes on the value of minus 1,

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sine n*t dt.

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And you'll note that these
integrals are actually

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the same.

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So this is negative 2 over
pi, zero to pi, sine n*t dt,

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which if we integrate is
negative 1 over n cosine n*t

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evaluated between 0 and pi.

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So if I work this out,
we get minus and a minus,

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minus 1 over n cosine
n*pi plus 1 over n.

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So note that cosine
of 0 is just 1.

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And now if we take a
look at cosine n*pi,

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we see that cosine n*pi
oscillates between minus 1

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and 1.

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So cosine of pi is negative
1, cosine of 2pi is 1,

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cosine of 3pi is minus 1.

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Dot, dot, dot.

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So this term right here is
actually negative 1 to the n.

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So we have 2 over n*pi 1
minus negative 1 to the n.

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And now if we just plug in
some values of b of 1, b of 2,

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b of 3, b of 4, we can see what
pattern emerges in the b's.

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So b of 1, if I plug in 1,
I get 1 minus negative 1.

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It's going to be 2.

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So I get minus 4 over pi.

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b of 2 is going to be 1 minus
minus 1 squared is just 1.

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So this vanishes.

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b of 3 is going to
be 1 minus minus 1

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cubed, which is negative 1.

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So again, we get
negative 4 over 3pi.

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b of 4 is going to be 0.

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So it's sometimes
useful the write out

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what the Fourier
series looks like.

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So I'll just write
it out right here.

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So we have f of t is going to
be negative 4 over pi times

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sine of t plus 1/3 sine of 3t
plus 1/5 sine of 5t plus dot,

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dot, dot.

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So this concludes part b.

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And now lastly,
for part c, we're

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asked to sketch what does the
first Fourier term look like.

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So in this case, the
first Fourier term

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is going to be negative
4 over pi times sine t.

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So I'm going to go back to
our diagram from part a.

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So let's go back to our
diagram from part a.

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Now what is minus 4 over
pi sine t look like?

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Well, it's a sine wave that
has exactly period 2pi,

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and it's going to line up
exactly with this square wave.

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In addition, minus 4 over pi
is just slightly larger than 1.

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So we're going to
end up with sin,

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which peaks just slightly
above 1 and slightly below 1.

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It's going to go
through 0, and it's

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going to go through
each multiple of pi.

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So it might look
something like this.

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So this is the first
Fourier term in the series.

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And notice how this first
Fourier term is actually

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pretty good approximation to the
square wave, considering it's

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just one term in a series.

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As we add more
terms in the series,

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we're going to get something
which looks closer and closer

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to a square wave function.

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So I'd just like
to quickly recap.

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When computing the Fourier
series for a periodic function,

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the first useful thing
to do is just write down

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the formula for
a Fourier series,

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and then write down the
formulas for the coefficients

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of the Fourier series.

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So write down the formulas
for a_0, a_n, b_n.

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When computing a_0,
you can often just

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look at the average
of the function.

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When computing a_n and
b_n, it's also useful

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look at the symmetry
of your function.

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And if it's either
even or odd symmetric

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then typically, either all the
a_n's or all the b_n's will

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vanish.

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And then when you work
out the integrals,

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you can then reconstruct
the Fourier series.

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So I would like
to conclude here,

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and I'll see you next time.