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AMATH 231 ASSIGNMENT # 10 Fourier Series Fall 2014
Due Monday, December 1, 2014 at 2pm in box 7, slot 11 (A-M) and
12 (N-Z), located acrossfrom MC4066. Late assignments or
assignments submitted to the incorrect dropbox willreceive a grade
of zero. Write your solutions clearly and concisely. Marks will be
deductedfor poor presentation and incorrect notation.
1. Show that the Fourier series of f(x) = x2 on pi < x <
pi is
pi2
3 4
n=1
(1)n1 cos(nx)n2
.
[3 marks]
Solution:First, note that x2 is even and therefore we know bn =
0. The coefficients for thisFourier cosine series are computed
directly. First for n 1,
an =2
pi
pi0
x2 cos(nx) dx,
=2
npi
[x2 sin(nx)
]pi0 4npi
pi0
x sin(nx) dx,
= 4n2pi
( [x cos(nx)]pi0 +
pi0
cos(nx) dx
),
= 4n2pi
(pi cos(npi) + 1
n[sin(nx)]pi0
),
= 4n2
(1)n+1.
Then for n = 0,
a0 =2
pi
pi0
x2 dx,
=2
3pi
[x3]pi0,
=2pi2
3.
Therefore the Fourier series is as suggested. Note that the
constant term is 12a0.
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2. Consider the fullwave rectification function g defined by
g(t) = | sin(t)| for t (pi, pi).[5 marks]
(i) Sketch the graph of the function to which the series
converges pointwise in R.Justify your answer.
By the pointwise convergence theorem (Theorem 5.6 in the
lectures notes), theFourier series converges to the periodic
extension of g, denoted gp, since g ispiecewise C1. Note that g(t)
has no points of discontinuity. See figure 1 below.
Figure 1: Sketch to which Fourier series converges pointwise on
R.
(ii) Find the complex Fourier series of g(t). Hint: To evaluate
the integral, it may behelpful to make an appropriate shift of the
interval of integration and also rewritesin(t) using Eulers
formula.
We can see from the sketch in (i) that g(t) has period = pi and
hence 0 =2pi/ = 2.
The complex Fourier coefficient is
cn =1
pi
pi2
pi2
| sin(t)|einotdt.
Since | sin(t)| is periodic with period pi, then we can rewrite
cn as
cn =1
pi
pi0
sin(t)einotdt
and using Eulers formula, replace sin(t) as follows
cn =1
2ipi
pi0
(eit eit)ein2tdt = 12ipi
pi0
(ei(2n1)t ei(2n+1)t)dt.
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Solving the integral leads to
cn =1
2ipi
[ei(2n1)t(2n 1)i +
ei(2n+1)t
(2n+ 1)i
]pi0
=1
2pi
[ei(2n1)pi
(2n 1) ei(2n+1)pi
(2n+ 1) 1
2n 1 +1
2n+ 1
].
From Eulers formula, we have that eipi = cos(pi) + i sin(pi) =
cos(pi) = 1.In a similar manner, e2inpi = 1. It follows that
cn =1
2pi
[ 12n 1 +
1
2n+ 1 1
2n 1 +1
2n+ 1
]=
1
2pi
[ 22n 1 +
2
2n+ 1
]=
2pi(4n2 1) .
Therefore, the complex form of the Fourier series of f(t) = |
sin(t)| is
n=
2e2intpi(4n2 1) .
(iii) Recover the real form of the Fourier series from the
complex Fourier series.
For n Z, sincecn =
an ibn2
,
then 2pi(4n2 1) =
an ibn2
which implies bn = 0 and
an =4
pi(4n2 1) .
Therefore, the real form of the Fourier series is
2
pi 4pi
n=1
1
4n2 1 cos(2nt). (1)
As an exercise, determine the real form of the Fourier series
directly. Your answershould be the same as that in (1).
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3. Apply Parsevals Theorem tox2
4=pi212
+n=1
(1)n1n2
cos(nx) for pi < x < pi, to
determine the value ofn=1
1
n4. [2 marks]
Sincex2
4=pi212
+n=1
(1)n1n2
cos(nx), pi < x < pi
we have
f(x) =x2
4, a0 =
pi26, an =
(1)n1n2
, bn = 0, = 2pi.
Applying Parsevals Theorem leads to
1
2pi
pipi
(x24
)2dx =
n=
|cn|2 = |c0|2 +n=1
|cn|2 +n=1
|cn|2
where
cn =1
2(an ibn) = (1)
n1
2n2 |cn|2 = |cn|2 = 1
4n4for n 6= 0
c0 =a02
for n = 0 |c0|2 =a0
2
2 = 14
pi4
36
It follows that
1
pi
pi0
x4
16dx
pi4
5 16
=1
4
pi4
36+ 2
n=1
1
4n4=
1
4
pi4
36+n=1
1
2n4
and rearranging leads ton=1
1
2n4=
pi4
5 16 pi4
4 36 .
Therefore,n=1
1
n4=pi4
90
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4. Let f be a C1 function on [pi, pi]. Prove that the Fourier
coefficients of f satisfy
|an| Kn, |bn| L
n, n = 1, 2, . . .
for some constants K and L. [0 marks]
The Fourier coefficients of f is
an =1
pi
pipif(x) cos(nx)dx
and applying integration by parts gives
an =1
pif(x)
sin(nx)
n
pipi 1pi
pipif (x)
sin(nx)
ndx = 1
pi
pipif (x)
sin(nx)
ndx
Taking the absolute value leads to
|an| 1npi
pipi|f (x) sin(nx)| dx
=1
npi
pipi|f (x)|| sin(nx)| dx
1npi
pipi|f (x)| dx since | sin(x)| 1.
Given that f is a C1 function on [pi, pi], this means f is
continuous on [pi, pi] andhence bounded on [pi, pi]. It follows
that there exists a constant M 0 such that|f (x)| M for [pi, pi].
Therefore,
|an| 1npi
pipiMdx =
2M
n=K
n
where K = 2M .
Similarly,
|bn| 1pi
pipi f(x) sin(nx)dx
=1
pi
f(x)cos(nx)npipi
+
pipif (x)
cos(nx)
ndx
1pi
f(pi) f(pi)n+ 1pi
pipi f (x)cos(nx)n dx (triangle inequality)
1npi|f(pi) f(pi)|+ 1
npi
pipi|f (x)|| cos(nx)|dx
1npi|f(pi) f(pi)|+ 1
npi
pipi|f (x)|dx since | cos(x)| 1
1npi|f(pi) f(pi)|+ 2M
n
1npi|f(pi)|+ 1
npi|f(pi)|+ 2M
n(triangle inequality).
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Since f is C1, then f is continuous and hence there exists m
such that |f(x)| m forall x [pi, pi]. Therefore,
|bn| 1n
(2m
pi+ 2M
)=L
n
where L = 2(mpi
+M).
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5. Define f on the interval 0 < x < pi to be
f(x) =
{12, |x pi/2| <
x, otherwise,
where 0 < < pi/2. Find the Fourier series of f , assuming
an odd extension to theinterval pi < x < pi.Solution:Assuming
an odd extension we have that an = 0 for all n. Next we find the
coefficientsin the Fourier sine series,
bn =2
pi
pi0
f(t) sin(nx) dx,
=2
pi
[ pi/20
x sin(nx) dx+
pi/2+pi/2
1
2sin(nx) dx+
pipi/2+
x sin(nx) dx
].
But the antiderivatives of the first and third terms arex
sin(nx) dx = x
ncos(nx) +
1
n
cos(nx) dx = x
ncos(nx) +
1
n2sin(nx).
The first integral above becomes, pi/20
x sin(nx) dx =
[xn
cos(nx) +1
n2sin(nx)
]pi/20
,
=
[(pi/2 )
ncos(n(pi/2 )) + 1
n2sin(n(pi/2 ))
].
The third integral is, pipi/2+
x sin(nx) dx =
[xn
cos(nx) +1
n2sin(nx)
]pipi/2+
,
= pin
cos(npi) +(pi/2 )
ncos(n(pi/2 )) 1
n2sin(n(pi/2 )).
The second integral is, pi/2+pi/2
1
2sin(nx) dx = 1
2n[cos(nx)]
pi/2+pi/2 ,
=1
2n[cos(n(pi/2 )) cos(n(pi/2 + ))] .
Therefore, the coefficients are,
bn =2
pi
[(pi/2 )
ncos(n(pi/2 )) + 1
n2sin(n(pi/2 ))
],
+2
pi
1
2n[cos(n(pi/2 )) cos(n(pi/2 + ))]
+2
pi
[pin
cos(npi) +(pi/2 )
ncos(n(pi/2 )) 1
n2sin(n(pi/2 ))
].
One could simplify this but it is messy and this answer is
sufficient.
