18.03 Lecture #21 Oct. 26, 2009: notes
Topic for today is more details and explanations about Fourier series.
Last time I said that a Fourier series starts with a 2
π
periodic function
f
(
t
) (meaning
f
(
t
+2
π
) =
f
(
t
) for all
t
) and tries to write it as a sum of “pure sine waves”:
f
(
t
)
?
=
a
0
2
+
∞
summationdisplay
n
=1
(
a
n
cos(
nt
) +
b
n
sin(
nt
))
.
(Fourier series of
f
)
(I also promised to explain why it was convenient to write the first term as
a
0
/
2.) You should not
worry a great deal about the question mark over the equality. It’s possible to make examples in
which Fourier series do not converge, and it’s
easy
to make examples in which they converge very
slowly. I’ll talk more about this on Friday, but for now it’s enough to know that “nice” periodic
functions have a convergent Fourier series.
[The part about complex Fourier series that follows was not in the lecture.] You know that
sin
t
and cos
t
are very closely related to
e
int
: there are simple formulas for going back and forth
between them, like
e
int
= cos(
t
) +
i
sin(
t
)
,
cos(
t
) =
1
2
(
e
it
+
e
−
it
)
and so on.
Because of these formulas, it is almost the same thing to write
f
using complex
exponentials:
f
(
t
)
?
=
c
0
+
∞
summationdisplay
n
=1
(
c
n
e
int
+
c
−
n
e
−
int
)
.
(Complex Fourier series of
f
)
Complex Fourier series are often easier to work with algebraically than ordinary Fourier series;
we’ve seen that already in trying to solve differential equations using sin and cos.
I said that in order to figure out what coefficient to use in a Taylor series
g
(
x
)
?
=
C
0
+
C
1
x
+
C
2
x
2
2!
+
C
3
x
3
3!
+
· · ·
,
you should first evaluate both sides at
x
= 0. All the terms on the right except the first give 0, so
we get
g
(0) =
C
0
. Next, differentiate the Taylor series and evaluate at 0. The first step gives
g
′
(
x
) = 0 +
C
1
+
C
2
x
+
C
3
x
2
2!
+
· · ·
.
After evaluation at 0, all the terms on the right disappear except the second: so
g
′
(0) =
C
1
. Et
cetera.
We’ll do something parallel with Fourier series. But Fourier series are about not
values
of
f
at
one point, but rather things like
average values
. Suppose we take the average value of both sides of
(Fourier series of
f
). On the left we get
1
2
π
integraltext
π
−
π
f
(
t
)
dt
. On left, we get first of all
a
0
/
2. For the rest,
the integral of a
sin
or
cos
function over a multiple of the period is always zero
; so all the other
terms in the Fourier series have average value zero. (This is written as Facts (A) and (B) below.)
Adding up, we get
1
2
π
integraldisplay
π
−
π
f
(
t
)
dt
=
a
0
/
2
(Formula for zeroth Fourier coefficient of
f
)
1
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To continue, we’ll need some fancier facts about integrals of trig functions. It’s convenient to
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 Fall '09
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 Fourier Series, Periodic function, break up, 2k, Sine wave, 2 M

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