Selected Solutions for Homework 2
Math 2280
Sept. 22, 2003
1. (Problem 2.3.13)
We have a function f given by the graph below.
1
-1
1
This function is 1 for 0 < x < 1 and 0 for ?1 < x < 0. There are two ways we can compute the Fourier series:
either by direct computation or by using the Fourier series in problem 1. First we will use the Fourier series in
problem 1. Let F be dened by
0<x<1 :
F (x) = ?11 ?
1<x<0
Then F has a Fourier series
F (x) =
1
4X
1
0 2k + 1
sin((2k + 1)x):
Now notice that
1 1
1 1
1
1 4X
1 2X
f (x) = (F (x) + 1) = [
sin((2
k + 1)x) + 1] = +
2
2 0 2k + 1
2 0 2k + 1 sin((2k + 1)x):
For the sake of completeness, we will actually compute the Fourier coecients. First,
a0 =
Next, for k 1 we have
ak =
Finally, we have
bk =
Z1
?1
Z1
?1
1 Z 1 f (x)dx = 1 Z 1 dx = 1 :
2 ?1
2 0
2
f (x) cos(kx)dx =
f (x) sin(kx)dx =
Z1
Z1
0
cos(kx)dx = 0:
1 cos(kx)j1 = 1 (1 ? (?1)k ):
sin(kx)dx = ? k
0 k
0
This last coecient is 2 if k is odd and 0 if k is even. Thus we have
1 1
1 2 X 1 sin(kx) = 1 + 2 X
f (x) = +
2 k
2 0 2k + 1 sin((2k + 1)x):
k odd
2. (Problem 2.3.17)
We start with the Fourier series
for ?p x p.
1 (?1)k
p2 4p2 X
cos(kx=p)=k2
x2 = +
3
1 k2
(a) Evaluating this series at x = 0, we nd
p2
0= 3
Substituting p = =2 we nd
1 (?1)k
4
p2 X
:
+
2
0 = 12 + which we can rearrange to read
12 =
1 (?1)k+1
X
k2
1
1
k2
1 (?1)k
X
1
k2
;
= 1 ? 212 + 312 ? 412 + (b) Now we return to the Fourier series we were given and evaluate at x = p to obtain (using cos(k) = (?1)k )
1 1
p2 4p2 X
:
p2 = +
3
1 k2
We can rearrange this series to read
1 1
X
=
6 1 k2 :
Adding this series to the series we obtained in part (a) we have
1 + 1 ? ) + (1 + 1 + 1 + ) = 2(1 + 1 + 1 + ):
3 =
=
+
=
(1
?
4 12 12 6
22 32
22 32
32 52
Dividing by 2, we have
1 1
8 = 1 + 32 + 52 + 3. (Problem 2.3.25)
We have a 2p-periodic function f , which is piecewise smooth and continuous. We are also suppose to assume
that f 0 is piecewise smooth. Let f and f 0 have the Fourier series:
f (x) = a0 +
X
[ak cos(kx=p) + bk sin(kx=p)]
f 0(x) = a00 + [a0k cos(kx=p) + b0k sin(kx=p)]:
We are to derive relationships between the Fourier coecients for f and those for f 0 .
First let's look at a00 :
Zp
1
1
0
a0 =
f 0(x)dx = f (x)jp?p = 0:
2p
2p
?p
Next we look at a0 :
k
a0k =
1 Z p f 0 (x) cos(kx=p)dx = f (x) cos(kx=p)jp + 1 Z p f (x) k sin(kx=p)dx
?p p
p
p
?p
Z
k 1 p
k
= [f (p) cos(k) ? f (?p) cos(?k)] + p p
f (x) sin(kx=p)dx = bk :
p
?p
Here we have integrated by parts, using cos(kx=p) = u and f 0 dx = dv. We have also used the formula for the
Fourier coecient bk .
Similarly, we have
?p
Z
Z
1 p f 0 (x) sin(kx=p)dx = f (x) sin(kx=p)jp ? k 1 p f (x) cos(kx=p)dx
b0k =
?p p p ?p
p ?p
a:
= ? k
p k
4. (Problem 2.3.32) (Remark: problem 26 of this section is essentially the same as problem 25; I didn't think it
would be good to assign the same problem twice.)
Here we have the sawtooth function, with Fourier series
X1
f (x) =
sin(kx):
k
(a) f is piecewise smooth (piecewise linear, in fact) and it is 2-periodic. Also, f 0 is piecewise constant, so it
is certainly piecewise smooth. However, f is not continuous at x = 2k, for any integer k (the left and
right limits disagree).
(b) Dierentiating the Fourier series term by term, we get
X
cos(kx):
Evaluating this sum at zero, we get 1 + 1 + 1 + , which diverges to 1. In general,
lim cos kx 6= 0;
k!1
P
which means the sum cos kx cannot converge. This is easiest to see if x = p=q, where p and q are
integers (i.e. if x is a rational multiple of ). Then if k = 2nq for any integer n,
cos kx = cos 2np = 1;
so cos kx cannot tend to zero as k ! 1. To complete the argument, use the fact that rational multiples
of are dense in the real numbers.
5. (Problem 2.4.6a)
We want to compute both of the half-range Fourier expansions for
f (x) = cos x
0 < x < :
The even half-range expansion is just cos x. The easiest way to see this is to note that f is already a sum
of cos functions with the right period, so it is already written in Fourier series. But one can actually do the
computation. The function cos x has mean zero between 0 and , so a0 = 0. To see that ak = 0 for k 2, use
the identity
cos(x) + cos(kx) = 21 [cos((k + 1)x) + cos((k ? 1)x)]:
To see that a1 = 1, use the identity
cos2 x = 21 (1 + cos(2x)):
Next we have to compute the odd half-range expansion. This will have the form
Z
X
2
cos(x) sin(kx)dx:
cos x = b sin(kx) where b =
k
This time we will have to use the identity
k
0
cos(x) sin(kx) = 12 [sin((k + 1)x) + sin((k ? 1)x)]:
Then
bk =
1 Z [sin((k + 1)x) + sin((k ? 1)x)]dx = 1 [? 1 cos((k + 1)x)j ? 1 cos((k ? 1)x)j ]
0 k?1
0
k+1
0
k
= 1 [? k +1 1 ((?1)k+1 ? 1) ? k ?1 1 ((?1)k?1 ? 1)] = 1 + (?1) [ k +1 1 + k ?1 1 ]
1)k ) :
= 2k(1(k+2 (??1)
Thus the Fourier coecient is 0 if k is odd and 4k=((k2 ? 2)) if k is even and the Fourier series is
X k
cos x = 4
sin kx:
k2 ? 1
k even
6. (Problem 2.4.11)
The sine series expansion of f (x) = sin(x) for 0 < x < 1 if just sin(x). The reasoning is very similar to the
rst part of problem 2.4.6.
7. (Problem 2.4.17)
We have a function f with the graph shown in the textbook.
(a) First we derive a formula for f . Notice that f is piecewise linear, and the endpoints of the line segments
are (0; 0), (a; h) and (p; 0) (in order). The rst line segment goes through the origin and has slope h=a.
Thus we must have
hx
f (x) =
0 x a:
a
The other line segment has slope ?h=(p ? a) = h=(a ? p), and so
f (x) =
hx
a?p
a x p:
+b
Matching the function values, f (a) = h, we see
b=h?
h(a ? p) ? ah
hp
=
=
?
;
a?p
a?p
a?p
ah
f (x) =
h(x ? p)
a?p
a x p:
(b) Now we want to nd the sine series representation for f . We have that
f (x) =
1
X
1
bk sin(kx=p)
where
Zp
Z a hx
Z p h(x ? p)
2
2
2
sin(kx=p)dx + p
sin(kx=p)dx
f (x) sin(kx=p)dx =
bk =
p 0
p 0 a
a?p
a
We will do these two integrals separately. First,
2 Z a hx sin(kx=p)dx = 2h [? p x cos(kx=p)ja + p Z a cos(kx=p)dx]
p 0 a
0 k 0
pa k
2h [?a cos(ka=p) + p sin(ka=p)] = ? 2h cos(ka=p) + 2hp sin(ka=p):
= ak
k
k
ak2 2
Next we have
2 Z p h(x ? p) sin(kx=p)dx =
2h [? p x cos(kx=p)jp + p Z p cos(kx=p)dx ? p Z p sin(kx=p)dx]
a k
p a a?p
p(a ? p) k
a
a
2
h
p2
pa
p2
= p(a ? p) [? k cos(k) + k cos(ka=p) + k cos(k)
2
2
p
cos(ka=p) + k2p2 sin(kx=p)jpa ]
? k
2
= p(a2?h p) [ p(ak? p) cos(ka=p) ? kp2 2 sin(ka=p)]
2h cos(ka=p) ?
2hp sin(ka=p)
= k
k2 2 (a ? p)
Putting this all together, we get
2hp
bk = sin(ka=p)[ 2 2
ak 2
? (a ?2php)k2 2 ] = a(p ?2hpa)k2 2 sin(ka=p):
Thus the Fourier series of f is
f (x) =
1 1
2hp2 X
sin(ka=p) sin(kx=p):
a(p ? a)2 1 k2
8. (Problem 3.2.6)
In this problem, we want to model the motion of a hanging chain. The top end of the chain is anchored at the
position x = L; u = 0 and the density is constant. See the gure in the text for a picture.
(a) In the equilibrium position, the chain hangs along the vertical line u = 0. The piece of chain located at
position (x; 0) supports a mass equal to x. This is because all the pieces of chain below this particular
piece pull down on this piece. As the only force acting on the chain is gravity, the tension (i.e. force)
acting on the chain at position x is (x) = gx.
(b) Now apply Newton's law: force = mass times acceleration. The tension forces (x) and (x + x) point
tangent to the chain. In the horizontal direction, this becomes
@2u
x 2 = (x + x) sin ? (x) sin :
@t
If we assume (as in the case with a vibrating string) that the chain is close to the equilibrium position
compared to the length of the chain, then
@u
sin tan = @x
(x; t)
Then we have
@u
sin tan = @x
(x + x; t):
@u
@u
@2u
x 2 = (x + x; t) (x + x; t) ? (x; t) (x; t):
@t
@x
@x
Rearranging this last equation, we nd
1
@u
@u
@2u
2 =
@t
x [ (x + x) @x (x + x; t) ? (x) @x (x; t)]:
(c) Now we take the limit as x ! 0. Then the last equation of part (b) becomes
@2u @
@u
2 = [ (x) ]:
@t
@x
@x
Next we substitute = gx to nd
@u
@u
@2u
@2u @
2 = [gx ] = g[ + x 2 ]:
@t
@x
@x
@x
@x
Dividing by , this last equation becomes
@2u
@u
@2u
=
g
[
]:
+
x
@t2
@x
@x2
It is interesting to notice here that the density divides out, and does not appear in the nal equation!
This should not exactly shock you, because gravity tends to act the same on all objects, regardless of their
mass (c.f. the famous experiment Galileo did in dropping rocks from the Tower of Pisa).