Math 509: Advanced Analysis

Homework 5

Lawrence Tyler Rush

<me@tylerlogic.com>

February 17, 2015

http://coursework.tylerlogic.com/courses/upenn/math509/homework05

1 Problem 12 from Slides

Define f : R⁴ → R³ by

f(x,y,z,u ) = (3x + y- z + u2,x - y+ 2z + u,2x + 2y- 3z + 2u)

With this we can define four functions on R × R³:

f₁(x,(y,z,u))	= f(x,y,z,u)	(1.1)
f₂(y,(x,z,u))	= f(x,y,z,u)	(1.2)
f₃(z,(x,y,u))	= f(x,y,z,u)	(1.3)
f₄(u,(x,y,z))	= f(x,y,z,u)	(1.4)

Thus we have f_i(0,(0,0,0)) = f(0,0,0,0) = (0,0,0,0) for all i = 1,2,3,4. Denote the second argument of each of f_i by v, then we have the following derivatives

(∂f₁∕∂v)_(0,(0,0,0))	=
(∂f₂∕∂v)_(0,(0,0,0))	=
(∂f₃∕∂v)_(0,(0,0,0))	=
(∂f₄∕∂v)_(0,(0,0,0))	=

with the corresponding determinants

det	= 1(4 - (-3)) - (-1)(-2 - 2) + (0)(3 - 4) = 11
det	= 3(4 - (-3)) - (-1)(2 - 2) + (0)(-3 - 4) = 21
det	= 3(-1 - 2) - (1)(2 - 2) + (0)(2 - (-2)) = -9
det	= 3(3 - 4) - (1)(-3 - 4) + (-1)(2 - (-2)) = 0

Therefore because (∂f₁∕∂v)_(0,(0,0,0)), (∂f₂∕∂v)_(0,(0,0,0)), and (∂f₃∕∂v)_(0,(0,0,0)) all have nonzero determinant, then the implicit function theorem tells us there exist functions g₁ : R → R³, g₂ : R → R³, and g₃ : R → R³ such that

f₁(0,g₁(0))	= f₁(0,(0,0,0)) = (0,0,0)
f₂(0,g₂(0))	= f₂(0,(0,0,0)) = (0,0,0)
f₃(0,g₃(0))	= f₃(0,(0,0,0)) = (0,0,0)

In light of the defintion of f and equations 1.1 through 1.3, this implies that the equations

3x + y - z + u²	= 0	(1.5)
x - y + 2z + u	= 0	(1.6)
2x + 2y - 3z + 2u	= 0	(1.7)

have solutions for

y,z,u in terms of x, due to g₁
x,z,u in terms of y, due to g₂
x,y,u in terms of z, due to g₃

On the other hand, because the determinant of (∂f₄∕∂v)_(0,(0,0,0)) is zero then there is no function g₄ : R → R³ with f₄(0,g₄(0)), implying that there is no solution to equations 1.5 through 1.7 for x,y,z in terms of u.

2 Problem 13 from Slides

Define f : R² × R → R by

f(x,y,z) = z2x +ez + y

Then because

( ∂f) det ∂z- = det(2xz + ez)(1,-1,0) = det(1) = 1 (1,-1,0)

the implicit function theorem tells us that there exists a g : R² → R such that g(1,-1) = 0 and f(x,y,g(x,y)) = 0 for all (x,y) in some neighborhood of (1,-1). Furthermore, since

∂f- ( ∂f-)( ∂g) ∂x = ∂g ∂x

and we know that

∂f-= z2 ∂x

and

(∂f ) g -∂g = 2gx+ e

then we can solve for ∂∂gx to get

∂g-= z2(2gx + eg)-1 ∂x

thus

∂g 2 g(1,-1)- 1 2 ∂x-(1,- 1) = z (2g(1,- 1) + e ) = z

Similarly we have

( )- 1 ( ) ∂g-= ∂f- ∂f- = 2g(x,y)x + eg(x,y) ∂y ∂x ∂g

so that

∂g --(1,- 1) = 1 ∂y

3 Problem 14 from Slides

Define a function f : R × R³ → R³ by

2 3 3 3 2 2 2 f(t,(x,y,z)) = (t + x + y + z ,t+ x + y +z ,t+ x + y+ z)

Let u denote the second component in R³ of f(t,u). Then the partial derivative with respect to u at the point (t,u) = (0,(-1,1,0)) is

( ) ( ) [ ] 3x2 3y2 3z2 3 3 0 ∂f- = ( 2x 2y 2z ) = ( - 2 2 0 ) ∂u (0,(-1,1,0)) 1 1 1 (0,(-1,1,0)) 1 1 1

which has determinant of 3(2 - 0) - 3(-2 - 0) + 0 = 12. The above matrix is therefore nonsingular, and hence the Implicit Function Theorem yields the existence of a function g : R → R³ defined on a neighborhood of 0 such that g(0) = (-1,1,0) and f(t,g(t)) = f(0,(-1,1,0)) = (0,2,0) for all points in that neighborhood. In turn, g implicitly defines three real-valued functions on R, x,y,z, where g(t) = (x(t),y(t),z(t)). With this notation, the previously mentioned result of the the Implicit Function Theorem can be restated as: there exists a neighborhood around (0,-1,1,0) such that f(t,(x(t),y(t),z(t))) = (0,2,0) for all t. In other words, given the definition of f, the following equations have solutions around (t,x,y,z) = (0,-1,1,0)

t² + (x(t))³ + (y(t))³ + (z(t))³	= 0
t + (x(t))² + (y(t))² + (z(t))²	= 2
t + x(t) + y(t) + z(t)	= 0

4 Problem 15 from Slides

Define three functions f′,g′,h′ : R² × R → R by

f′((x,y),z)	= xz + sin(xy) + cos(xz)
g′((y,z),x)	= xz + sin(xy) + cos(xz)
h′((x,z),y)	= xz + sin(xy) + cos(xz)

then we have the following derivatives

f′_z	= z + y cos(xy) - z sin(xz)
g′_x	= xcos(xy)
h′_y	= x + xcos(xz)

so that

f′_z(0,1,1)	= 2
g′_x(0,1,1)	= 0
h′_y(0,1,1)	= 0

Thus the Implicit Function Theorem tells us that there is a differentiable function f : R² → R such that z = f(x,y) but that there are no such functions g,h : R² → R with x = g(y,z) or y = h(x,z)

5 Problem 16 from Slides

Let F(x,y) be a C² function such that F(x,f(x)) = 0 and (∂F∕∂y)(x,f(x))≠0 for all x ∈ R. Then we have that

∂F ∂F ∂F ∂x-(x,f(x )) = ∂x-(x,y)+ ∂y f ′

(5.8)

By assumption ∂F--
∂y ≠0 so that we may solve for f′ as

( ∂F) -1( ∂F ∂F ) f′ = --- ---(x,f (x))- ---(x,y) ∂y ∂x ∂x

(5.9)

Again taking the derivative with respect to x in equation 5.8 yields

∂2F ∂2F ∂2F ∂F -2--(x,f (x)) = -2-(x,y)+ --2-f′ +---f′′ ∂ x ∂ x ∂ y ∂y

substituting equation 5.9 into this equation and then solving for f′′ leaves us with

( )- 1( 2 2 2 ( )- 1( ) ) f′′ = ∂F- ∂-F2-(x,f(x))- ∂2F-(x,y)- ∂-F2- ∂F- ∂F-(x,f(x))- ∂F-(x,y) ∂y ∂ x ∂ x ∂ y ∂y ∂x ∂x

6 Problem 17 from Slides

Define F : R³ → R by

F (x,y,z) = x2 + 4y2 - 2yz - z2

(a) Verify the hypotheses of the Implicit Function Theorem

(b) Find the largest neighborhood U of (2, 1,-4) on which ∂F∕∂z≠0

We have that

∂F-= - 2y - 2z = - 2(y + z) ∂z

which means that ∂F-
∂z ≠0 whenever y≠-z. Hence any point of the form (x,y,-y) will have = 0 which implies that the largest neighborhood of (2,1,-4) will be the open ball with radius r where r is the distance from (2,1,-4) to the closest point of the form (x,y,-y). Since y≠ - z at the point (2,1,-4), we know that such an r will exist. We just need to find it.

To do so, we can find the minimum of the distance between (2,1,-4) and any point (x,y,-y), or equivalently the minimum of the square of the distance between (2,1,-4) and any point (x,y,-y). So define f : R² → R by

f(x,y) = (x- 2)2+ (y- 1)2+ (- y- (- 4))2 = x2 - 4x + 4+ y2- 2y+ 1 + y2+ 8y+ 16 = x2 - 4x+ 2y2+ 6y+ 21

With this definition we have

f_x	= 2x - 4
f_y	= 2y + 6

indicating that there is a critical point at (x,y) = (2,-3), which we simply need to confirm that this is indeed a minimum. We compute the Hessian matrix of this function