UCSC
AMS 211
FALL 2014
Review Questions for Final Exam – Solutions
1. (a) From first principles, find an integrating factor µ(x) for the general first order linear differential equation
dy
+ p(x)y = q(x).
dx
Solution.
We want to find µ(x) such that
µ(x)(y 0 + p(x)y) =
d
(µ(x)y).
dx
dµ
= p(x)µ(x), which in turn implies that
dx
This holds if and only if
1 dµ(x)
= p(x)
µ(x) dx
and integrating both sides leads to
Z
ln(µ(x)) =
p(x) dx,
R
so µ(x) = exp( p(x) dx).
(b) Solve the initial value problems
dy
i. sin x − 2 cos x y = sin3 x; y(π/4) = 0.
dx
Solution. First divide through by sin x:
dy
cos x
−2
y = sin2 x.
dx
sin x
Now, find the integrating factor:
Z
cos x
µ(x) = exp −2
dx = exp(−2 ln(sin x)) = sin−2 x.
sin x
Next, multiply equation (1) through by sin−2 x:
cos x
d
dy
−2
y sin−2 x = 1.
sin x
−2
y =
dx
sin x
dx
Integrate and solve for y:
y sin−2 x = x + C =⇒ y = (x + C) sin2 x.
Finally, use the boundary condition to solve for C:
0 = y(π/4) = (π/4 + C) sin2 (π/4) =
π
π
+ C =⇒ C = − ,
4
4
so
y = (x − π/4) sin2 x.
1
(1)
dy
+ 2xy = (1 − x2 )3/2 ; y(0) = 0.
dx
Solution. Divide through by (1 − x2 ):
ii. (1 − x2 )
dy
2x
+
y = (1 − x2 )1/2
dx 1 − x2
(2)
Integrating factor:
Z
µ(x) = exp
2x
dx
1 − x2
= exp(− ln(1 − x2 )) = (1 − x2 )−1 .
Multiply equation (2) through by µ(x):
2x
dy
d
2 −1
+
y · (1 − x2 )−1 = (1 − x2 )−1/2
(1 − x )
y =
2
dx 1 − x
dx
Integrate and solve for y:
Z
y
dx
√
=
= arcsin x + C =⇒ y = (1 − x2 )(arcsin x + C).
2
2
1−x
1−x
Solve for C:
0 = y(0) = arcsin(0) + C = C,
so
y = (1 − x2 ) arcsin x.
iii.
dy
+ 2xy = 3xy 3 ; y(0) = 1. (This one needs a substitution to make it linear.)
dx
Solution. As alluded to, this is a Bernoulli equation, and the substitution u = y −2 , which
dy
du
implies that
= − 12 y 3 , transforms the original equation to
dx
dx
du
− 4xu = −6x,
dx
(3)
after multiplication by −2y −3 . The integrating factor for (3) is
Z
2
µ(x) = exp −4 x dx = e−2x ,
and multiplying (3) through by this factor gives
du
d −2x2 2
−2x2
e
− 4xu =
ue
= −6xe−2x .
dx
dx
Integrating both sides and solving for u, we have
Z
3
6
3
2
2
2
−2x2
ue
=
−4xe−2x dx = e−2x + C =⇒ u = + Ce2x .
4
2
2
The boundary condition y(0) = 1 implies that u(0) = 1−2 = 1, and using this to solve for C,
we have
3
1
1 = + C =⇒ C = − ,
2
2
2
2
so u = 12 (3 − e2x ), and
r
y=u
−1/2
=
2
.
3 − e2x2
Comment: The last equation can be rewritten as y 0 = x(2 − 3y 3 ) and the function on the
right is continuously differentiable with respect to both variables in the entire xy-plane (which
is an infinite rectangle
q at (0, 1)). Nonetheless, the solution we found is only defined
q centered
ln 3
,
2
in the interval −
ln 3
2
≈ (−0.741, 0.741) around 0.
2. (a) Use the definition to find the Laplace transforms of the following functions.
f (x) = sin(2x),
g(x) = e−2x cos x and h(x) = H(x − 2) − H(x − 4),
where H(x) is the Heaviside function
1 : x≥0
0 : x<0
H(x) =
Note: for the first two, integration by parts is your friend.
Solution.
∞
Z
1
2 ∞
−xs sin(2x)e dx = − sin(2x)e +
L(sin(2x)) =
cos(2x)e−xs dx
s
s 0
0
0
∞
Z ∞
2
4
2
4
sin(2x)e−xs dx = 2 − 2 L(sin(2x)),
= − 2 cos(2x)e−xs − 2
s
s 0
s
s
0
Z
so
L(e
−2x
∞
−xs
4
1+ 2
s
Z
L(sin(2x)) =
∞
so
−2x
Z
L(h(x)) =
e
−2x
2
2
=⇒ L(sin(2x)) = 2
.
2
s
s +4
Z
−xs
∞
dx =
cos x e−(s+2)s dx
0
∞ 0
Z ∞
1
1
−(s+2)x =−
cos x e
sin x e−(s+2)x dx
− s+2
s+2
0
∞0
Z ∞
1
1
1
−(s+2)x =
+
sin x e
−
cos x e−(s+2)x dx
2
2
s+2
(s + 2)
(s + 2) 0
0
1
1
=
−
L e−2x cos x
2
s + 2 (s + 2)
cos x) =
L(e
cos x e
1
cos x) 1 +
(s + 2)2
1
s+2
=⇒ L(e−2x cos x) =
.
s+2
(s + 2)2 + 1
=
∞
−xs
(H(x − 2) − H(x − 4))e
0
Z
dx =
e
2
3
4
−xs
4
1 −xs e−2s − e−4s
dx = − e =
.
s
s
2
(b) Use the Laplace transform method to solve the initial value problems
y 00 − 3y 0 − 4y = sin(x);
y(0) = 0, y 0 (0) = 1
and
y 00 + 2y 0 − 3y = H(x) − H(x − 1);
y(0) = 1, y 0 (0) = 0.
Solutions. The idea is to apply the Laplace transform to both sides of the differential
equation. This incorporates the boundary conditions because:
L(y 0 ) = sY (s) − y(0) and L(y 00 ) = s2 Y (s) − y(0)s − y 0 (0),
where Y (s) = L(y). We then solve the resulting algebraic equation for Y (s) and use the
inverse Laplace transform to find y.†
First equation:
L(y 00 − 3y 0 − 4y) = L(sin x) =⇒ s2 Y (s) − 1 − 3sY (s) − 4Y (s) =
=⇒ Y (s)(s2 − 3s − 4) =
s2
s2
1
+1
1
+1
+1
Therefore
1
s2 + 2
=
(s2 + 1)(s2 − 3s − 4) s2 − 3s − 4
(s2 + 1)(s − 4)(s + 1)
3
3
5
s− 5
s
18/85
3/10
18/85
3/10
= 34 2 34 +
−
= 234 − 2 34 +
−
,
s +1
s−4
s+1
s +1 s +1
s−4
s+1
1
Y (s) =
+
after partial fraction decomposition. Looking up the inverse transforms (and making use of
the linearity of L), we have
3
5
s
18/85
3/10
5
18
3
3
−1
34
34
− 2
+
−
cos x −
sin x + e4x − e−x
=
y=L
2
s +1 s +1
s−4
s+1
34
34
85
10
Second equation:
1 e−s
L(y + 2y − 3y) = L(H(x) − H(x − 1)) =⇒ s Y (s) − s + 2sY (s) − 2 − 3Y (s) = −
s
s
−s
1
−
e
=⇒ Y (s)(s2 + 2s − 3) =
+s+2
s
00
0
2
Therefore,
s2 + 2s + 1
1
− e−s
s(s − 1)(s + 3)
s(s − 1)(s + 3)
1/3
1
1/3
1/3e−s 1/4e−s 1/12e−s
=−
+
+
−
+
+
s
s−1 s+3
s
s−1
s+3
Y (s) =
From section 13.2.2 in the book, we learn that
0 : 0<x≤b
−1
−bs
e F (s) =
= f (x − b)H(x − b)
L
f (x − b) : x > b
†
In the most common cases, we can ‘read’ the inverse transform from a table.
4
where f (x) = L−1 (F (s)). This means that
1/3
1
1/3
1/3e−s 1/4e−s 1/12e−s
−1
−
+
+
−
+
+
y=L
s
s−1 s+3
s
s−1
s+3
1
1
1
1
1
= − + ex + e−3x − H(x − 1) + ex−1 H(x − 1) + e−3(x−1) H(x − 1)
3
3
3
4
12
12.5
10
7.5
5
2.5
-1
-0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Figure 1: Graph of the solution of y 00 + 2y 0 − 3y = H(x) − H(x − 1); y(0) = 1, y 0 (0) = 0.
3. Solve the initial value problem
dy
d2 y
+
2
+ 5y = 3 cos(2t);
dt2
dt
y(0) = y 0 (0) = 0.
(4)
Solution. Two methods...
(i) Undetermined coefficients: First find the general solution of the complementary (homogeneous) equation
d2 y
dy
+ 2 + 5y = 0
(5)
2
dt
dt
√
−2 ± 4 − 20
2
Characteristic equation — r + 2r + 5 = 0 =⇒ r =
=⇒ r = −1 ± 2i, so a basis
2
for the space of solutions of equation (5) is given by
u1 (t) = e(−1+2i)t and u2 (t) = e(−1−2i)t ,
and a basis of real-valued solutions is given by
1
1
y1 = (u1 + u2 ) = e−t cos 2t and y2 = (u1 − u2 ) = e−t sin 2t.
2
2i
Thus, the general solution of equation (5) is
yh = c1 y1 + c2 y2 = e−t (c1 cos 2t + c2 sin 2t).
5
Next, find a particular solution of the differential equation (4) using the method of undetermined coefficients, i.e., look for a solution of the form yp = A cos 2t + B sin 2t. We have
yp0 = −2A sin 2t + 2B cos 2t and yp00 = −4A cos 2t − 4B sin 2t, and substituting these into the
original equation gives
yp00 + 2yp0 + 5yp = cos 2t(−4A + 4B + 5A) + sin 2t(−4B − 4A + 5B) = 3 cos 2t,
and leads to the pair of linear equations
A + 4B = 3
−4A + B = 0
with solution
3 4 0 1 3
=
A = 17
1 4 −4 1 and
1 3 −4 0 12
=
B = 17
1 4 −4 1 It follows that the general solution of (4) is
y = yp + yh =
3
12
cos 2t +
sin 2t + e−t (c1 cos 2t + c2 sin 2t).
17
17
Finally, we use the initial conditions, y(0) = y 0 (0) = 0 to determine c1 and c2 . First, we have
0 = yp (0) =
6
Second, yp0 = − 17
sin 2t +
24
17
3
3
+ c1 =⇒ c1 = − .
17
17
cos 2t + e−t ((2c2 − c1 ) cos 2t − (2c1 + c2 ) sin 2t) , so
0 = yp0 (0) =
27
27
24
+ 2c2 − c1 = 2c2 +
=⇒ c2 = − ,
17
17
34
and the solution to the initial value problem is
y=
3
12
1
cos 2t +
sin 2t − e−t (6 cos 2t + 27 sin 2t).
17
17
34
(ii) Laplace transform: First take Laplace transforms of both sides of equation (4), using the
given boundary conditions
2
dy
dy
3s
2
+
2
+
5y
=
L(3
cos(2t))
=⇒
s
Y
(s)
+
2sY
(s)
+
5Y
(s)
=
,
L
dt2
dt
s2 + 4
where Y (s) is the Laplace transform of the (as-of-yet unknown) solution y. Next, solve the
equation above for Y (s):
Y (s)(s2 + 2s + 5) =
3s
3s
=⇒ Y (s) = 2
.
+4
(s + 4)(s2 + 2s + 5)
s2
Now use a partial fraction decomposition of the rational function on the right in order to more
easily identify the inverse Laplace transform. Note that both quadratic factors in the denominator
6
cannot be factored over the real numbers, and this leads to a partial fraction decomposition of
the form
As + B
Cs + D
3s
=
+
(s2 + 4)(s2 + 2s + 5)
s2 + 4
s2 + 2s + 5
(As + B)(s2 + 2s + 5) + (Cs + D)(s2 + 4)
=
(s2 + 4)(s2 + 2s + 5)
(A + C)s3 + (2A + B + D)s2 + (5A + 2B + 4C)s + (5B + 4D)
.
=
(s2 + 4)(s2 + 2s + 5)
Comparing the coefficients of s3 , s2 , s and the constant coefficient on both sides of the equation
above, yields a system of four linear equations in the unknown coefficients A, B, C and D:
A+C
2A + B + D
5A + 2B + 4C
5B + 4D
=0
=0
=3
=0
The first and fourth equation show that C = −A and D = − 45 B, and substituting these expressions in the second and third equation gives the pair of equations
1
2A − B = 0
4
A + 2B = 3
with solution
A = 0 −1/4 3
2 3/4
3
=
=
17/4
17
2 −1/4 1
2 and
B = 2 0 1 3 2 −1/4
1
2
6
24
=
=
.
17/4
17
3
It follows that C = − 17
and D = − 30
, and therefore
17
3s
+ 4)(s2 + 2s + 5)
3
3
s + 24
s + 30
17
= 17 2 17 − 217
s +4
s + 2s + 5
3
s
12
2
3
s+1
27
2
=
· 2
+
· 2
−
·
−
·
.
2
17 s + 4 17 s + 4 17 (s + 1) + 4 34 (s + 1)2 + 4
Y (s) =
(s2
Note that the coefficients of the rational functions in the second row have been distributed so that
the rational functions in the third row all have the form
s+a
(s + a)2 + b2
or
b
,
(s + a)2 + b2
and
e−at sin(bt)
whose inverse Laplace transforms are
e−at cos(bt)
7
respectively. It follows that the solution is
3
s
12
2
3
s+1
27
2
−1
−1
y(t) = L (Y (s)) = L
·
+
·
−
·
−
·
17 s2 + 4 17 s2 + 4 17 (s + 1)2 + 4 34 (s + 1)2 + 4
which is (not surprisingly) the same solution we found the first time around.
Conclusion: The algebra is unavoidable, so suck it up.
4. Use Stokes’ theorem
Z
I
(∇ × a) · dS =
a · dr,
S
C
where S is an open surface in R3 bounded by the closed curve C and a is a vector field, to derive
Green’s theorem in the plane.
Solution. Suppose that a = P (x, y)i + Q(x, y)j is a vector field confined to the x, y-plane and
that R is an open surface in the plane bounded by the closed curve C. Then, on the one hand,
i
∂ ∂j k
∂Q ∂P
∂
−
k and dS = k dx dy,
∇ × a = ∂x ∂y ∂z =
∂x
∂y
P Q 0 so the left-hand side of Stokes’ theorem (above) is
Z Z Z
∂Q ∂P
∂Q ∂P
(∇ × a) · dS =
−
k · k dx dy =
−
dx dy.
∂x
∂y
∂x
∂y
R
R
R
On the other hand,
a · dr = (P i + Qj) · (dxi + dyj) = P dx + Q dy,
so the right-hand side of Stokes’ theorem gives
I
I
P dx + Q dy,
a · dr =
C
C
and equating the right and left hand sides yields Green’s theorem,
Z I
∂Q ∂P
−
dx dy =
P dx + Q dy.
∂x
∂y
R
C
5. Use Green’s theorem to evaluate the integral
I
x2 y dx + 2xy 2 dy,
C
where C is the triangle in R2 with corners (0, 0), (0, 2) and (1, 4).
Solution. Setting P = x2 y and Q = 2xy 2 , we have Py = x2 and Qx = 2y 2 . Invoking Green’s
theorem, we have
I
Z
2
2
x y dx + 2xy dy =
2y 2 − x2 dx dy,
C
T
8
where T is the triangular region bounded by C:
T = {(x, y) : 0 ≤ x ≤ 1 and 4x ≤ y ≤ 2x + 2}.
It follows that
I
2
Z
2
1
Z
2x+2
2y 2 − x2 dy dx
x y dx + 2xy dy =
Z0 1
C
=
4x
2
(2x + 2)3 − (4x)3 − x2 (2x + 2 − 4x) dx
3
0
Z
2 1
=
−53x3 + 21x2 + 24x + 8 dx
3 0
2
53
55
=
− + 7 + 12 + 8 =
3
4
6
Z
6. Compute the surface integral
x2 + 2y 2 dS over the surface
S
S = {(x, y, z) : x2 + y 2 = z 2 and 0 ≤ z ≤ 1}.
(Better) Suggestion: Use the parametrization
r = ρ cos θi + ρ sin θj + ρk
for the cone, with 0 ≤ ρ ≤ 1 and 0 ≤ θ < 2π.
Solution: With the suggested parametrization we have
rρ = cos θi + sin θj + k and rθ = −ρ sin θi + ρ cos θj,
from which it follows that
dS = krρ × rθ k dA = k − ρ cos θi − ρ sin θj + ρkk dA =
√
2ρ dρ dθ.
Therefore,
Z
2
2
√
ρ2 cos2 θ + 2ρ2 sin2 θ 2ρ dρ dθ
0
0
Z 1
√ Z 2π
2
= 2
(1 + sin θ)
ρ3 dρ dθ
0
√ Z0 2π
√
2
3 2π
2
1 + sin θ dθ =
=
4 0
4
Z
2π
Z
1
x + 2y dS =
S
7. If ϕ(x, y, z) is a scalar field and v(x, y, z) is a vector field, show that
∇ × (∇ϕ) = 0 and ∇ · (∇ × v) = 0.
9
Solution: For the scalar field ϕ, we have
∇·ϕ=
∂ϕ
∂ϕ
∂ϕ
i+
j+
k,
∂x
∂y
∂z
so, assuming that all the second order derivatives are continuous, so that ϕxy = ϕyx , ϕxz = ϕzx
and ϕyz = ϕzy , it follows that
i
j
k
∇ × (∇ · ϕ) = ∂/∂x ∂/∂y ∂/∂z = (ϕzy − ϕyz )i + (ϕxz − ϕzx )j + (ϕyx − ϕxy )k = 0.
ϕx
ϕy
ϕz For the vector field v = v1 (x, y, z)i + v2 (x, y, z)j + v3 (x, y, z)k, we have
∂v3 ∂v2
∂v1 ∂v3
∂v2 ∂v1
∇×v =
−
i+
−
j+
−
k.
∂y
∂z
∂z
∂x
∂x
∂y
Therefore (once again assuming that all second order partial derivatives are continuous)
∇ · (∇ × v) =
∂ 2 v3
∂ 2 v2
∂ 2 v1
∂ 2 v3
∂ 2 v2
∂ 2 v1
−
+
−
+
−
= 0.
∂x∂y ∂x∂z ∂y∂z ∂y∂x ∂z∂x ∂z∂y
8. Suppose that u = (φ(x, y), ψ(x, y), 0) is a continuous vector field confined to R2 that is both
solenoidal and irrotational. Show that the functions φ and ψ are both harmonic (potential)
functions (i.e., they are each solutions of Laplace’s equation).
Solution. If u is solenoidal, then its divergence is 0, i.e.,
∇·u=
∂φ ∂ψ
+
= 0,
∂x ∂y
so ψy = −φx . If u is irrotational, then its curl is 0, i.e.,
i
j
k
∂ ∂ ∂ ∂ψ
∂φ
∇ × u = ∂x ∂y ∂z =
−
k = 0,
∂x
∂y
φ ψ 0 so ψx = φy . Assuming that both of these conditions hold, we have
∂ ∂ψ
∂ ∂φ
∂ ∂φ
∂
∂ψ
∂ 2ψ
∂ 2ψ
=
=
=
=
−
=
−
,
∂x2
∂x ∂x
∂x ∂y
∂y ∂x
∂y
∂y
∂y 2
so ψxx + ψyy = 0. Likewise,
∂ 2φ
∂ ∂φ
∂ ∂ψ
∂ ∂ψ
∂
=
=
=
=
2
∂y
∂y ∂y
∂y ∂x
∂x ∂y
∂x
∂φ
−
∂x
∂ 2φ
= − 2,
∂x
so φxx + φyy = 0.
(Note that in this case, the function f (x + iy) = ψ(x, y) + iφ(x, y) is holomorphic in C).
10
9. A (left) stochastic matrix A is an n × n matrix,


a11 · · · a1n

..  ,
..
A =  ...
.
. 
an1 · · · ann
with nonnegative coefficients whose column-sums are all 1, i.e., for which
n
X
aij = 1 for each j.
i=1
(a) Show that a stochastic matrix always has an eigenvalue equal to 1.
Proof 1. If I is the n × n identity matrix, then the column-sums of A − I are all 0, because
the sum of the entries in the j th column of A − I is
a1j + a2j + · · · + (ajj − 1) + · · · + anj = (a1j + · · · + anj ) − 1 = 0.
It follows that the sum of the rows of A − I is the zero (row) vector, i.e., the rows of A − I
are linearly dependent, so rank(A − I) ≤ n − 1. This implies that det(A − I) = 0, so λ = 1
is an eigenvalue of A.
Proof 2: If x = [1 1 · · · 1]T , then

  
  
a11 a21 · · · an1
1
a11 + a21 + · · · + an1
1
 a12 a22 · · · an2   1   a12 + a22 + · · · + an2   1 

  
  
AT x =  ..
 =  ..  ,
.. . .
..   ..  = 
..
 .




  . 
.
.
.
.
.
a1n a2n · · · ann
a1n + a2n + · · · + ann
1
1
i.e., AT x = x, which means that x is an eigenvector of AT with eigenvalue λ = 1. But the
eigenvalues of A and AT are the same, so λ = 1 is an eigenvalue of A too.
(b) Find the eigenvalues and corresponding eigenvectors of the stochastic matrix
0.7 0.8
A=
.
0.3 0.2
Solution. Eigenvalues:
0.7 − λ
0.8
det (A − λI) = 0.3
0.2 − λ
= λ2 − 0.9λ − 0.1 = (λ − 1)(λ + 0.1),
so the eigenvalues of A are λ1 = 1 and λ2 = −0.1.
Eigenvectors: we solve the equations (A − I)x = 0 and (A + 0.1I)x = 0.
3
−0.3
0.8
x
0
λ1 = 1 :
=
=⇒ −0.3x + 0.8y = 0 =⇒ y = x,
0.3 −0.8
y
0
8
8
so x1 =
is an eigenvector with eigenvalue λ1 = 1.
3
0.8 0.8
x
0
λ2 = −0.1 :
=
=⇒ x + y = 0 =⇒ y = −x,
0.3 0.3
y
0
1
so x2 =
is an eigenvector with eigenvalue λ2 = −0.1.
−1
11
1
(c) Let u0 =
, and for n = 1, 2, 3, . . ., let un+1 = Aun . Find a vector w ∈ R2 such that
1
−→
un n→∞
w (and prove the convergence).
Solution. Note that the eigenvalues of A are distinct, which implies that the corresponding
eigenvectors are linearly independent (which we can also verify by inspection), and hence the
vectors x1 and x2 that we found above form a basis for R2 . Furthermore, we observe that
since Ax1 = x1 , it follows that An x1 = x1 , and likewise, since Ax2 = −0.1x2 , it follows
that An x2 = (−0.1)n x2 .
Therefore, expressing u0 as a linear combination of x1 and x2 , u0 = c1 x1 + c2 x2 , we see that
un = An u0 = An (c1 x1 + c2 x2 ) = c1 An x1 + c2 An x2 = c1 x1 + c2 · (−0.1)n x2 −→ c1 x1 ,
since (−0.1)n → 0 (rapidly). I.e., w = c1 x1 , and it remains to find c1 , which we do by
(partially) solving the system
8
1
c1
1
=
.
3 −1
c2
1
For this we can use Cramer’s rule
c1 = 1
1 1 −1 2
= ,
11
8
1 3 −1
8/11
hence w = 2
.
3/11
a
(d) Show that if u =
, then there is a vector w that depends only on a + b such that
b
An u → w.
Solution. Repeating the arguments of the previous part, if
a
8
1
u=
= c1
+ c2
,
b
3
−1
8
n
then it follows that A u −→ c1
, where
3
a
1 b −1 a+b
=
,
c1 = 11
1 8
3 −1 8/11
n
i.e., A u −→ (a + b)
.
3/11
10. Consider the function f (x) = ex defined on the interval [0, 1].
(a) Sketch the graph of its periodic extension to R with period 1, as well as its even and odd
periodic extensions to R with period 2.
12
Solution: The periodic extension of period 1 is the function f1 (x) = ex for 0 < x ≤ 1, then
continued periodically so that for every integer n, if n < x ≤ n + 1, then f1 (x) = ex−n , see
Figure 7 for its graph.
The even periodic extension of period 2 is the function f2e (x), defined by f2e (x) = ex for
0 < x ≤ 1, f2e (x) = e−x for −1 < x ≤ 0, and then continued periodically, so that for every
integer n, if 2n − 1 < x ≤ 2n, then
x−2n
e
: 2n < x ≤ 2n + 1
f2e (x) =
e2n−x : 2n − 1 < x ≤ 2n
Its graph is displayed in Figure 3.
The odd periodic extension of period 2 is the function f2o (x), defined by f2o (x) = ex for
0 < x ≤ 1, f2o (x) = −e−x for −1 < x ≤ 0, and then continued periodically, so that for
every integer n, if 2n − 1 < x ≤ 2n, then
ex−2n : 2n < x ≤ 2n + 1
f2o (x) =
−e2n−x : 2n − 1 < x ≤ 2n
Its graph is displayed in Figure 4.
3
2
1
-3
-2
-1
0
1
2
3
Figure 2: Periodic extension with period 1: y = f1 (x)
3
2
1
-3
-2
-1
0
1
2
3
Figure 3: Even periodic extension with period 2: y = f2e (x)
13
2.5
-3
-2
-1
0
1
2
3
-2.5
Figure 4: Odd periodic extension with period 2: y = f2o (x)
(b) Which of these periodic extensions will yield the best Fourier series expansion? Why?
Solution: The even periodic extension should produce the most quickly converging Fourier
series expansion because it is the only one that yields a continuous function.
(c) Compute the Fourier coefficients for all three periodic extensions. Were you right?
Solution:
(i) Period 1 extension: Denoting the coefficients of cos(2πnx) and sin(2πnx) by an and bn
respectively, we have
Z 1
ex cos(2πnx) dx
an = 2
0
Z 1
1
x
ex sin(2πnx) dx
= 2e cos(2πnx) + 4πn
0
0
Z 1
1
2 2
x
ex cos(2πnx) dx
= 2(e − 1) + 4πne sin(2πnx) − 8π n
0
0
2 2
= 2(e − 1) − 4π n an
so
(1 + 4π 2 n2 )an = 2(e − 1) =⇒ an =
2(e − 1)
.
1 + 4π 2 n2
Likewise,
1
Z
ex sin(2πnx) dx
bn = 2
0
Z
1
= 2e sin(2πnx) − 4πn
x
0
1
ex cos(2πnx) dx
0
= −2πnan ,
so
bn = −
4πn(e − 1)
.
1 + 4π 2 n2
14
Thus the Fourier series for f1e is
f1 (x) = (e − 1) 1 +
∞
X
2 cos(2πnx)
4πn sin(2πnx)
−
2
2
4π n + 1
4π 2 n2 + 1
n=1
!
In the figure below, the graph of f1 is displayed together with the truncation
!
7
X
2 cos(2πnx) 4πn sin(2πnx)
(e − 1) 1 +
−
4π 2 n2 + 1
4π 2 n2 + 1
n=1
of its Fourier series (dashed red line).
3
2
1
-3
-2
-1
0
1
2
3
Figure 5: Approximation of y = f1 (x), truncating its Fourier series at n = 7.
(ii) Even extension (period 2): Since f2e is an even function, its Fourier series will consist
only of cosine terms. Denoting the coefficients by an again, we have
Z 1
Z 1
ex cos(πnx) dx
(because f2e (x) cos(πnx) is even).
an =
f2e (x) cos(πnx) dx = 2
−1
0
Z 1
1
x
= 2e cos(πnx) + 2πn
ex sin(πnx) dx
0
0
Z 1
1
x
2 2
= 2(e − 1) + 2πne sin(πnx) − 2π n
ex cos(πnx) dx
0
n
0
2 2
= 2((−1) e − 1) − π n an .
Hence
an =
2((−1)n e − 1)
,
π 2 n2 + 1
and therefore
f2e (x) = (e − 1) + 2
∞
X
((−1)n e − 1) cos(πnx)
n=1
15
π 2 n2 + 1
.
In the figure below, the graph of f2e is displayed together with the truncation
(e − 1) + 2
7
X
((−1)n e − 1) cos(πnx)
π 2 n2 + 1
n=1
of its Fourier series (dashed red line).
3
2
1
-3
-2
-1
0
1
2
3
Figure 6: Approximation of y = f2e (x), truncating its Fourier series at n = 7.
(iii) Odd extension (period 2): Since f2o is an even function, its Fourier series will consist
only of sine terms. Denoting the sine coefficients by bn again, and using the fact that the product
of odd functions is even, we have
Z 1
Z 1
bn =
f2o (x) sin(πnx) dx = 2
ex sin(πnx) dx
(because f2o (x) sin(πnx) is even).
−1
0
Z 1
1
x
= 2e sin(πnx) − 2πn
ex cos(πnx) dx
0
0
Z 1
1
x
2 2
= − 2πne cos(πnx) − 2π n
ex sin(πnx) dx
0
= 2πn((−1)
n+1
0
e + 1) − π 2 n2 bn .
Hence
bn =
and therefore
f2e (x) = 2π
2πn((−1)n+1 e + 1)
,
π 2 n2 + 1
∞
X
n((−1)n+1 e + 1) sin(πnx)
π 2 n2 + 1
n=1
.
In the figure below, the graph of f2o is displayed together with the truncation of its Fourier series
at n = 10 (dashed red line).
16
3
2
1
-3
-2
-1
0
1
2
3
-1
-2
-3
Figure 7: Approximation of y = f2o (x), truncating its Fourier series at n = 10.
The coefficients of the even periodic extension are all on the order of 1/n2 , while the other two
periodic extensions have coefficients on the order of 1/n, so, yes, I was right. This is also visible
in the graphs — the truncated Fourier series of the even periodic extension yields a much better
approximation of f2e than the other two truncated series do for their respective functions.
11. Let p(z) = an z n + · · · + a1 z + a0 be a non constant polynomial with complex coefficients (i.e.,
n > 0 and an 6= 0).
(a) By applying Liouville’s Theorem (see section 24.10 in the text) to the function 1/p(z), show
that p(z) has at least one root in C.
(b) Using long division of polynomials and mathematical induction, conclude that every polynomial of degree n and (real or) complex coefficients may be factored into linear factors.
Solution.
(a) Assume to the contrary that p(z) 6= 0 for all z ∈ C, then it follows that 1/p(z) is holomorphic
(analytic) in the entire complex plane, since
1
p0 (z)
d
=−
dz p(z)
p(z)
is defined in all of C. If we can show that 1/p(z) is bounded in all of C, then it will follow from
Liouville’s theorem that 1/p(z) is constant, and hence so is p(z), contradicting the assumption
that it is nonconstant.
17
To show that 1/p(z) is constant, we first observe that 1/p(z) and therefore the (real-valued)
function |1/p(z)| are both continuous in all of C. Hence, for any r > 0, there is a bound Br > 0
such that
1 p(z) ≤ Br
in the compact (closed and bounded) set Dr = {z ∈ C : |z| ≤ r}. It remains to find an r > 0
such that |1/p(z)| is also bounded for |z| > r.
To that end, note that |p(z)| = |z|n · |an + an−1 z −1 + · · · + aj z j−n + · · · + a0 z −n |, and let
M = max |aj |.
0≤j≤n
If |z| > max(1, 2nM/|an |), then for 0 ≤ j ≤ n − 1, we have j − n ≤ −1 and |z| > 1, so
|z|j−n < |z|−1 < (2nM/|an |)−1 from which it follows that
|aj z j−n | < |aj |(2nM/|an |)−1 =
|an |
|an | |aj |
·
≤
,
2n M
2n
since |aj /M | ≤ 1. Hence, if |z| > max(1, 2nM/|an |), then
an−1 z −1 + · · · + aj z j−n + · · · + a0 z −n < n |an | = |an |
2n
2
and therefore for such z,
an + an−1 z −1 + · · · + aj z j−n + · · · + a0 z −n > an − an = |an | .
2
2
It follows that if |z| > r0 = max(1, 2nM/|an |), then
|an ||z|n
|an ||z|
|p(z)| = |z|n · an + an−1 z −1 + · · · + aj z j−n + · · · + a0 z −n >
≥
> nM,
2
2
and hence |1/p(z)| < 1/nM . Finally, if B = max(Br0 , 1/nM ), then |1/p(z)| < B for all z ∈ C,
and we’re done.
(b) If n = 1 and a1 6= 0, then p(z) = a1 z + a0 is a linear polynomial, and the statement is true in
this case. Now assume that n > 1 and that if q(z) is any polynomial of degree n − 1, then q(z)
may be factored into (n − 1) linear factors,‡ .
q(z) = (a1 z + b1 )(a2 z + b2 ) · · · (an−1 z + bn−1 ).
Now let p(z) have degree n, then by part (a), there is a ζ ∈ C such that p(ζ) = 0 and it follows
that p(z) = (z − ζ)q(z), where q(z) has degree n − 1. From the induction hypothesis, it follows
that
q(z) = (a1 z + b1 )(a2 z + b2 ) · · · (an−1 z + bn−1 ),
and hence
p(z) = (z − ζ)q(z) = (z − ζ)(a1 z + b1 )(a2 z + b2 ) · · · (an−1 z + bn−1 ),
which show that p(z) may also be factored into linear factors. It follows by the principal of
mathematical induction that every non constant polynomial (with real or complex coefficients)
may be factored into linear factors (over C).
‡
This is the induction hypothesis.
18
12. Chapter 24, problems 1, 3, 10; Chapter 25, problem 9.
Solutions. 24.1, 24.3 and 25.9 — see the hints/solutions in the book.
d
24.10: The function exp(iaz 2 ) is holomorphic in all of C (since dx
exp(iaz 2 ) = 2iaz exp(iaz 2 )
exists in all of C). This means that if γ is any closed curve in C, then by Cauchy’s theorem,
I
exp(iaz 2 ) dz = 0.
γ
2
If z = x is real, then exp(iax
) = cos(ax2 ) + i sin(ax2 ), thus the integral
R∞
real part of the integral 0 exp(iax2 ) dx.
R∞
0
cos(ax2 ) dx is the
The idea then, is to find a closed contour γ that includes the positive real axis such that the
integral of exp(iaz 2 ) dz along the remainder of γ can be computed directly.
Observation 1: If we write z = r(cos θ + i sin θ), then
2
2
exp(iaz 2 ) = exp iar2 (cos 2θ + i sin 2θ) = e−ar sin 2θ eiar cos 2θ .
Now, if 0 < θ < π/2, then sin 2θ > 0 and | exp(iaz 2 )| = e−ar
for fixed θ . This would appear to indicate that
Z
exp(iaz 2 ) dz = 0,
lim
r→∞
2
sin 2θ
which goes to 0 as r → ∞,
γ1 (r)
where γ1 (r) is the eighth of a circle in the first quadrant of radius r with the counterclockwise
orientation, depicted below. More on this later.
𝛄2(r)
𝛄1(r)
π/4
r
Figure 8: The curves γ1 (r) and γ2 (r).
2
Observation 2: If z = teiπ/4 , then z 2 = t2 eiπ/2 = it2 so exp(iaz 2 ) = e−at in this case. It
follows that if γ2 (r) is the ray from 0 to reiπ/4 (oriented towards 0 as depicted above), then
Z
Z
Z 0
1 + i r −at2
2
−at2 iπ/4
exp(iaz ) dz =
e
e
dt = − √
e
dt,
2 0
γ2 (r)
r
19
R
(since along γ2 (r), dz = eiπ/4 dt). Hence as r → ∞, the real part of γ2 (r) exp(iaz 2 ) dz approaches
the limit
r
r
Z ∞
1
1
π
π
−at2
e
dt = − √
=−
,
−√
8a
2 0
2 4a
because
r
√
Z ∞
Z ∞
1
1
π
π
−at2
−x2
e
dt = √
e
dx = √ ·
=
.
4a
a 0
a 2
0
Conclusion:§ Let γr be the closed curve comprised of the segment [0, r] on the real line, γ1 (r)
and γ2 (r) (with the standard counterclockwise orientation), then
Z
Z
I
Z r
2
iaz 2
iax2
iaz 2
e
dz +
eiaz dz = 0
e
dx +
e
dz =
γ1 (r)
0
γr
γ2 (r)
from which it follows, upon comparing real parts, that
Z r
Z
Z r
2
iax2
cos(ax ) dx = <
e
dx = −<
0
0
e
iaz 2
Z
iaz 2
dz +
e
γ1 (r)
dz .
γ2 (r)
Now, if it is indeed true that
Z
2
eiaz dz = 0,
lim
r→∞
γ1 (r)
then it follows that
Z
Z r
Z ∞
2
2
cos(ax ) dx = lim −<
cos(ax ) dx = lim
0
r→∞
r→∞
0
γ1 (r)
e
iaz 2
Z
iaz 2
e
dz +
r
dz
=0+
γ2 (r)
π
,
8a
as claimed.
R
2
The fun part: We want to show that γ1 (r) eiaz dz → 0 as r → ∞, which we do using the
inequality
Z
f (z) dz ≤ max |f (z)| · `(Γ),
z∈Γ
Γ
2
2
where `(Γ) is the length of the curve Γ. In this case, `(γ1 (r)) = πr/4 < r and |eiaz | = e−ar sin 2θ ,
2
and since for a constant c > 0, xe−cx → 0 (rapidly) as x → ∞, we would like to say that
2
re−ar sin 2θ → 0 (which implies that the integral goes to 0). But it’s not quite that simple, because
if for example θ = 1/r2 with r large (so θ is very close to 0) then sin 2θ ≈ 2θ = 2/r2 , in which
case
2
2
e−ar sin 2θ ≈ e−ar 2θ = e−2a ,
which is bounded away from 0. On the other hand, this phenomenon only occurs on a very short
portion of γ1 (r), which suggests the following fix.
For a (very small) angle θr > 0,¶ we divide γ1 (r) into two parts, Γ1 (r) and Γ2 (r), where Γ1 is the
(very short) portion of γ1 (r) where 0 ≤ θ ≤ θr and Γ2 is the portion of γ1 (r) where θr < θ ≤ π/4.
The lengths of these two parts are
π
`(Γ1 (r)) = rθr and `(Γ2 (r)) = r
− θr < r.
4
§
¶
Except for the fun part.
Which, as indicated, we choose to depend on r.
20
2
Next, for any θ between 0 and π/4, e−ar sin 2θ ≤ 1 (since sin 2θ ≥ 0) and therefore
Z
2
iaz
≤ `(Γ1 (r)) · 1 = rθr .
e
dz
Γ1 (r)
On the other hand, if π/4 ≥ θ > θr > 0, then sin 2θ > sin 2θr > 0 so
e−ar
2
sin 2θ
< e−ar
2
sin 2θr
.
Furthermore, if 0 < θr < π/6, then sin 2θr > θr ,k and it follows that if θr < π/6, then
Z
2
2
2
iaz 2
e
dz ≤ `(Γ2 (r))e−ar sin 2θr < re−ar sin 2θr < re−ar θr .
Γ2 (r)
Combining these two results, we have
Z
Z
Z
Z
2
2
2
iaz
iaz
iaz
=
≤
e
dz
e
dz
+
e
dz
γ1 (r)
Γ1 (r)
Γ2 (r)
e
iaz 2
Γ1 (r)
Z
dz +
Γ2 (r)
iaz 2
e
2
dz < rθr +re−ar θr ,
Finally, given ε > 0, if θr = ε/2r, then
rθr = ε/2
re−ar
and
2θ
r
= re−(aε/2)r .
Since re−cr → 0 as r → ∞ for any c > 0, it follows that (for fixed ε) there is an Rε such that
if r > Rε , then
ε
re−(aε/2)r < .
2
Hence, for r > Rε
Z
ε ε
iaz 2
< + = ε,
e
dz
2 2
γ1 (r)
and since ε was arbitrary, it follows that
Z
2
eiaz dz = 0,
lim
r→∞
γ1 (r)
as claimed.
k
Because: (i) sin 0 = 0 and (ii) (sin 2x − x)0 > 0 for 0 < x < π/6. Fill in the details.
21