Math 2280-001
Wed Apr 22
9.1-9.3 Fourier Series. (On Friday we'll revisit forced oscillations, section 9.4, and explain the results we
obtained playing "the resonance game" with convolution integrals last Friday.)
, Finish Monday's notes. The key points to recall from Monday are that Fourier series are a way of
expressing piecewise continuous functions f on the interval Kp, p (or equivalently, a 2 pKperiodic
functions f), as infinite sum of trigonometric functions. If f has Fourier series
N
N
a0
fw
C
an cos n t C
bn sin n t
2
n=1
n=1
Then the partial sums
>
fN t =
a0
2
>
N
C
N
> a cos nt C > b sin nt
n=1
n
n=1
n
with
1
a0 d
p
p
bn d f, sin n t
VN = span
2
Kp
an d f, cos n t
are the projections of f onto
a0
f t dt
1
=
p
=
1
p
=! f,
1
2
O
1
2
p
f t cos n t dt, n 2 ;
Kp
p
f t sin n t dt, n 2 ;
Kp
1
, cos t , cos 2 t , ..., cos N t , sin t , sin 2 t , ... sin N t .
2
These partial sums converge to f as N/N in the ways described in Monday's notes.
,
After finishing with Monday's notes, continue with today's...
Exercise 1 If a function ("vector") is already in a subspace, then projection onto that subspace leaves the
function fixed. Use that fact to very quickly compute the 2 pKperiodic Fourier series for
a) f t = sin 5 t K 8 cos 10 t
b) g t = cos2 3 t
Fourier series for 2 LKperiodic functions:
Theorem: Consider the vector space of piecewise continuous, 2 LKperiodic functions. Then the inner
product
1
g, h d
L
L
g u h u du
KL
makes
p
2p
kp
p
2p
kp
u , cos
u ,... cos
u ,... sin
u , sin
u , ...sin
u ....
L
L
L
L
L
L
2
into an orthonormal collection of functions.
p
L
proof: The substitution u = t, equivalently u =
t converts between 2 LKperiodic functions
L
p
L
L
g u , h u and 2 pKperiodic functions g
t ,h
t . Also (verify this!!!)
p
p
1
, cos
1
L
L
KL
p
1
g u h u du =
p
g
Kp
L
t h
p
L
t dt .
p
A
Thus the Fourier series for a 2 LKperiodic function f is defined by
N
N
a0
p
p
fw
C
an cos n
u C
bn sin n u
2
L
L
n=1
n=1
with
>
>
1
a0 =
L
an d
bn d
f, cos n
f, sin n
p
u
L
=
p
u
L
=
1
L
1
p
L
g u du
KL
L
f u cos n
KL
p
f u sin
Kp
p
u du, n 2 ;
L
p
u du, n 2 ;
L
(and then we usually use the dummy variable t rather than u). As a result, Fourier series for 2 LKperiodic
functions along with convergence theorems, are "equivalent" to ones for 2 pKperiodic ones, via this
isometry of the two vector spaces.
Exercise 2)
a) Use the Fourier series for tent t :
3
2
1
K3 p
K2 p
0
Kp
p
2p
3p
t
p
4
1
K
cos n t
2
p n odd n2
to deduce the Fourier series for the related function f u with period 2 that has graph
>
tent w
1
0.6
0.2
K3
K2
K1
0
1
2
3
t
solution: f w
1
4
K 2
2
p
>
n odd
1
n2
cos n p u
We talked about differentiating Fourier series term by term in Monday's notes. There is also:
Theorem If f is piece-wise continuous, 2 LKperiodic, with Fourier series
N
N
a0
p
p
f t w
C
an cos n
t C
bn sin n t
2
L
L
n=1
n=1
then the antiderivative may be computed by term by term antidifferentiation, and the corresponding series
will converge for each t:
>
t
f s ds =
0
a0
2
N
tC
>a
n=1
n
L
np
>
N
p
sin n
t C
bn
L
n=1
L
np
>
Kcos n
p
t C1 .
L
Exercise 3) (This is the first part of your homework exercise 9.3.19)
Start with
N
t = saw t w 2
K1
>
nC1
sin n t
n
and integrate to get the 2 pKperiodic function that on Kp, p is given by
n=1
N
2
t2
p
K1 n C 1
=
C2
cos n t .
2
6
n=1
n2
a0
Hint: The value of the constant term is easiest to compute as
. If you compare to the definite integral
2
formula in the Theorem you will reproduce one of the "magic" series.
>
4
1
K3 p
K2 p
Kp
0
p
t
2p
3p
In your homework you will antidifferentiate twice more to get a formula for the periodic extension of
t4
g t =
.
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