Math 361: Homework 3
Monday, February 17
1. Use the contraction principle to show that the equation x3 + x2 − 6x + 1 = 0 has a unique real solution
over the interval [−1, 1]
Proof. Consider the map f : [−1, 1] → R defined by f (x) = 16 (x3 + x2 + 1). We intend to show f is a
contraction and thus has a unique fixed point.
We first compute the derivative f 0 (x). Since f is a single variable real valued function, we use singlevariable calculus to find f 0 (x) = 21 x2 + 13 x .
We next compute the second derivative of f (x) and obtain f 00 (x) = x + 13 .
We compute the maxima and minima of f 0 (x) by seeing where f 00 (x) = 0 and obtain that f 0 (x) attains
−1
0 −1
0
a critical point at −1
3 , and f ( 3 ) = 18 . We also check the value of f (x) at the edges of our domain,
1
5
0
0
and obtain that f (1) = 6 and f (−1) = 6 . (Note that in order to define the derivative at the endpoints
-1 and 1, we must temporarily consider the function on the domain [−1 − , 1 + ]). Therefore, the
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value of f 0 (x) is between −1
18 and 6 on the domain.
We apply the mean value theorem to arbitrary points x,y in the domain and obtain that |f (x) −
f (y)| ≤ |f 0 (z)||x − y| for some z between x and y. Since |f 0 | is bounded on the domain, we have that
|f (x) − f (y)| ≤ 56 |x − y|. Thus we may write |f (x) − f (y)| < .99|x − y| and therefore f is a contraction
on the domain. Thus ∃ a unique point x in the interval [−1, 1] such that x = 61 (x3 + x2 + 1), i.e. a
unique x such that x3 + x2 − 6x + 1 = 0
2. Use the contraction principle to prove that the equation x = cos x has a unique solution.
Proof. We know cos x maps R to [−1, 1]. Thus any solution of the equation x = cos x must occur on
that interval. We thus restrict ourselves to the interval [−1, 1].
We wish to show cosine is a contraction on [−1, 1]. By MVT,| cos x − cos y| = | sin(c)||x − y| for some
c ∈ (x, y). However, on [−1, 1], sine achieves no minima or maxima except at the boundaries (since its
derivative is cosine, which is 0 only at multiples of π2 ). Thus since | sin(1)| = | sin(−1)| = .84, we have
that | cos x − cos y| < .99|x − y|
Thus cosine is a contraction on [−1, 1] and thus has a unique fixed point, i.e a unique solution to
cos(x) = x. Since there are no solutions outside [-1,1] we thus have that x = cos x has a unique
solution in R
3. Rudin Chapter 9 # 16. Consider f (t) = t+2t2 sin( 1t ). Show that if t 6= 0, and f (0) = 0, then f 0 (0) = 1,
f 0 is bounded in (−1, 1), but f is not one-to-one in any neighborhood of 0.
Proof. We first wish to show that f 0 (0) = 0.
f 0 (0) is defined by limt→0
f (t)−f (0)
t
= limt→0 ( tt +
1
2t2 sin( 1t )
)
t
= limt→0 (1 + 2t sin( 1t ))
We now consider the limit as t approaches 0 of t sin( 1t )). If this exists, then we will be able to split
the limit and show f 0 (0) exists. Since −1 ≤ sin( 1t )) ≤ 1∀t, we can multiply by t and obtain that
−t ≤ t sin( 1t )) ≤ t. Thus by the squeeze theorem the limit as t approaches 0 of t sin( 1t )) is 0.
Thus f 0 (0) = 1.
We now wish to show that f 0 is bounded in (−1, 1). We consider only t non-zero, since we considered
1
1
0
t = 0 above. f 0 (t) = 1+4t sin( 1t )+2t2 cos( 1t )∗ −1
t2 = 1+4t sin( t )−2 cos( t ). Thus |f (t)| ≤ 1+4+2 = 7
Thus f’ is bounded in (-1,1).
We now wish to show f is not injective in any neighborhood of 0. Since f is continuous, if we can show
that in any δ-neighborhood of 0, f’ changes sign, then f is not monotonic and thus not injective.
Therefore, consider any interval (−δ, δ). Recall that f 0 (t) = 1 + 4t sin( 1t ) − 2 cos( 1t ) for t non-zero.
Consider the point x = (2πN )−1 , where N is some integer large enough that x is in(−δ, δ) (this is
allowed by the Archimedian principle). f 0 (x) = 1 + 4x sin(2πN ) − 2 cos(2πN ) = −1.
Thus since f 0 (x) = −1 and f 0 (0) = 1, f 0 changes sign and thus f is not monotonic and thus not
injective.
4. Consider the function f : R2 \ (0, 0) → R2 defined by f (x, y) = (x2 − y 2 , 2xy). Determine its range and
show that the derivative of f is invertible everywhere in the domain but not one-to-one on R2 \ (0, 0)
Proof. We claim the range of f is R2 \ (0, 0).
√√
√√
√ √
√ √√
a2 +b2 +a
2 a2 +b2
a2 +b2 +a− 2a
a2 +b2 +a
√
For any (a, b) in R2 \(0, 0) such that b 6= 0, let x =
and
let
y
=
.
2b
2
Then f(x,y) = (a,b) (this is algebraic manipulation, done by wolfram alpha). Note that this solution
is not necessarily unique.
p
For any (a, b) in R2 \ (0, 0) such that b = 0, we consider two cases. If apis positive, let x = (a) and
let y = 0. Then f (x, y) = (a, 0). If a is negative, let x = 0 and let y = (|a|). Then f (x, y) = (a, 0).
Note that (a, b) = (0, 0) is not attainable since f (x, y) = (0, 0) implies 2xy = 0, i.e. x or y is 0, and
x2 − y 2 = 0, i.e |x| = |y|. Thus x = y = 0, which is not in the domain.
!
The matrix of partial derivatives of f is A =
∂f1
∂x
∂f2
∂x
∂f1
∂y
∂f2
∂y
2x −2y
We compute the determinant and find
2y 2x
det(A) = (4x2 + 4y 2 ) which is strictly nonzero for (x, y) ∈ R2 \ (0, 0).
After computing these partials, we find A =
Thus since f has continuous partials, the derivative of f is the matrix of partial derivatives, which is
invertible in the domain.
However f is clearly not 1-to-1 on R2 \ (0, 0), as f(1,1) = f(-1,-1) = (0,2).
5. Consider the equation f (x, y) = (x2 + y 2 − 2)(x2 − y 2 ) = 0∀(x, y) ∈ R2 . Find the solutions to the
equation ∇f =< fx , fy >= 0 and hence that the assumption of the implicit function theorem do not
hold. Then, by graphing the solution set of the equation f (x, y) = 0, show that the conclusions of the
implicit function theorem do not hold for the set of points previously found.
Proof. fx = 4x(x2 − 1), fy = 4y(1 − y 2 ), and thus ∇f = 0 ⇐⇒ (x, y) = (0,0) or (1,0) or (-1,0) or
(0,1) or (1,1) or (-1,1) or (0,-1) or (1,-1) or (-1,-1).
Thus the implicit function theorem does not apply since the derivative of f is a matrix of zeros at those
points thus has no invertible square submatrices.
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Figure 1: The plot of the solution set of the equation f(x,y) = 0
.
See the plot of the solution set in figure 1. The plot clearly shows that at (1,0), (-1,0), (0,1), and (0,-1),
either the horizontal or vertical line tests do not hold, and that at (1,1), (1,-1), (-1,1), (-1,1), and (0,0),
the implicit function is not unique.
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6. Consider the equation ex + y 2 + z − 4xy 3 − 1 = 0 and think of x as an implicit function in y and z
∂x
determined by the equation. Find ∂x
∂y and ∂z
2
2
x
− 4y 2 ,
Proof. Consider f (x, y, z) = ex + y 2 + z − 4xy 3 − 1. Taking partials, we find ∂f
∂x = 2xe
∂f
∂f
2
∂y = 2y − 12xy , and ∂z = 1. We see that each partial is continuous and thus that the derivative
exists. Thus
adopting the notation used
in class during the proof of the implicit function theorem, Df
2
x
2
2
= A = 2xe − 4y 2y − 12xy 1 .
We now want to think of x as implicit function of y and z, and thus f (x, y, z) is viewed as f (g(y, z), y, z).
From the professor’s email, we assume the square submatrix Ax is invertible.
Therefore, apply
∂x
∂x
0
ing the result of the implicit function theorem, g = Dg = ∂y ∂z is equal to −A−1
x Ay =
1
− 2xex2 −4y3 2y − 12xy 2 1 .
Therefore
∂x
∂y
2
2y−12xy
= − 2xe
and
x2 −4y 3
∂x
∂z
= − 2xex21−4y3
7. Consider the equations x + y + z = 0 and xyz = 1. Regard (y,z) as an implicit function in x. Find
∂z
and ∂x
.
Proof. We apply the same procedure as in the previous problems. We first compute A.
! ∂f1
∂f1
∂f1
1
1
1
∂x
∂y
∂z
A = ∂f2 ∂f2 ∂f2 =
yz xz xy
∂x
∂y
∂z
1
1
1
We consider Ax =
and Ay =
xz xy
yz
xy −1
1
Then, A−1
=
x
xy−xz
−xz 1
∂y xy −1
1
1
0
−1
∂x
Therefore, g (x) = Dg = ∂z = −Ax Ay = − xy−xz
−xz
1
yz
∂x
3
∂y
∂x
Thus
∂y
∂x
=
yz−xy
xy−xz
and
∂z
∂x
=
xz−yz
xy−xz
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