MATH3705 A | Test 3: Wed. 2:35pm{3:25pm, Mar 6
Name and Student Number:
Total points: 20. No partial marks for Questions 1-3.
Closed book! Non-programmer calculators are allowed!
1. Let f(x) =
x+ x2 x3; 0 x < 1;2 + x; 1 x < 2. Let fodd(x) be the
4-periodic odd extension of[2]
f(x). Which of the following is the expression of fodd(x) when 1
< x < 0?
(a) 2 + x (b) x x2 + x3 (c) x+ x2 + x3 (d) x x2 x3 (e) x x2 +
x3
Solution: (d)
fodd(x) = f(x) = (x+ x2 + x3) = x x2 x3:
2. Let f(x) =
1 x; 0 x < 1;0:5; 1 x < 2. Determine the value to which
the Fourier sine series[2]
of f(x) converges at x = 12:3 .
(a) 0:3 (b) 0:5 (c) 0:3 (d) 0:7 (e) 0:7
Solution: (d). The period of the Fourier sine series of f(x) is
4.fodd(12:3+)+fodd(12:3)
2= fodd(12:3) = fodd(12 + 0:3) = fodd(0:3) = 1 0:3 = 0:7:
3. Let f(x) be 2-periodic and f(x) =
3; x < 0;0; 0 x < . Find the Fourier sine coe-[2]
cient b3.
(a) 2 (b) 3=2 (c) 3=2 (d) 0:5 (e) 1=2
Solution: (a).
b3 =1
Z
f(x) sin(3x)dx =1
Z 0
3 sin(3x)dx+
Z 0
0 sin(3x)dx
=
Z 0
3 sin(3x)dx = cos(3x)j0 = 2:
4. Let f(x) =
1 x; 0 x < 1;2; 1 x < 2. Let an (n = 0; 1; 2; :::) be the
coecients of the Fourier[4]
cosine series. Find a2.
Solution: Note that L = 2.
a2 =2
L
Z L0
f(x) cos(2x
L) dx =
Z 20
f(x) cos(x) dx
=
Z 10
(1 x) cos(x) dx+Z 21
2 cos(x) dx
=
1
(1 x) sin(x) 1
2cos(x)
10
+
2
sin(x)
21
= 12
cos() +1
2=
2
2:
5. Let f(x) be 2-periodic and f(x) =
sin(x); 1 x < 0;1; 0 x < 1. Find the Fourier cosine[4]
coecient a2.
Solution: Note that L = 1. Thus
a2 =1
1
Z 11
f(x) cos(2x)dx =
Z 01
sin(x) cos(2x)dx+
Z 10
1 cos(2x)dx
=
Z 01
sin(x+ 2x) + sin(x 2x)2
dx+1
2sin(2x)j10
=
Z 01
sin(3x) sin(x)2
dx+ 0 =1
2
13
cos(3x) 1cos(x)
01
=1
2
2
3+
1
3cos(3) 1
cos()
=
2
3:
6. The solution of the heat equation wt = 2wxx; 0 x L; t 0,
subject to the boundary[6]
conditions w(0; t) = w(L; t) = 0 is given by
w(x; t) =1Xn=1
bn sinnx
L
e(
nL )
2t:
Find the solution u(x; t) of uxx =116ut, subject to the boundary
conditions u(0; t) =
2; u(2; t) = 4, and the initial condition u(x; 0) = 5 + x.
Solution: Here = 4, L = 2, A = 2, B = 4. Let u(x; t) = v(x) +
w(x; t) with
v(x) =B A
Lx+ A = x+ 2: (1point)
Then w(x; t) = u(x; t) v(x);) w(x; 0) = u(x; 0) v(x) = 5 + x x 2
= 3:)
wxx =1
16wt; w(0; t) = w(2; t) = 0; w(x; 0) = 3: (2points)
Thus
w(x; t) =1Xn=1
bn sinnx
L
e(
nL )
2t =
1Xn=1
bn sinnx
2
e(2n)
2t;
where bn satises
bn =2
L
Z L0
(3) sinnx
L
dx
=
Z 20
3 sinnx
2
dx
=
6n
cos(nx
2)
20
= 6n
[cos(n) 1] = 6[(1)n+1 + 1]
n(2points)
u(x; t) = (x+ 2) +1Xn=1
6[(1)n+1 + 1]n
sinnx
2
e(2n)
2t: (1point)
