- This problem was suggested to me by Matt Heiner. Earlier there was a problem in which
two surfaces intersected at a point and this implied that in fact, they intersected in a smooth
curve. Now suppose you have two spheres x
^{2}+ y^{2}+ z^{2}= 1 and^{2}+ y^{2}+ z^{2}= 1. These intersect at the single point. Why does the implicit function theorem not imply that these surfaces intersect in a curve? - Maximize 2x + y subject to the condition that +≤ 1. Hint: You need to consider interior points and also the method of Lagrange multipliers for the points on the boundary of this ellipse.
- Maximize x + y subject to the condition that x
^{2}++ z^{2}≤ 1. - Find the points on y
^{2}x = 16 which are closest to. - Let f= x
^{2}−2yx+2z^{2}−4z +2. Identify all the points where Df = 0. Then determine whether they are local minima local maxima or saddle points. - Let f= x
^{4}−2x^{2}+2y^{2}+1. Identify all the points where Df = 0. Then determine whether they are local minima local maxima or saddle points. - Let f= −x
^{4}+2x^{2}−y^{2}−2z^{2}−1. Identify all the points where Df = 0. Then determine whether they are local minima local maxima or saddle points. - Let f : V → ℝ where V is a finite dimensional normed vector space. Suppose f is convex which
means
whenever t ∈

. Suppose also that f is differentiable. Show then that for every x,y ∈ V,Thus convex functions have monotone derivatives.

- Suppose B is an open ball in X and f : B → Y is differentiable. Suppose also there exists
L ∈ℒsuch that
for all x ∈ B. Show that if x

_{1},x_{2}∈ B,Hint: Consider Tx = f

− Lx and argue< k. - Let f : U ⊆ X → Y , Dfexists for all x ∈ U, B⊆ U, and there exists L ∈ℒ, such that L
^{−1}∈ℒ, and for all x ∈ BShow that there exists ε > 0 and an open subset of B

,V , such that f : V → Bis one to one and onto. Also Df^{−1}exists for each y ∈ Band is given by the formulaHint: Let

for

<, consider {T_{y}^{n}}. This is a version of the inverse function theorem for f only differentiable, not C^{1}. - In the last assignment, you showed that if Dfis invertible, then locally the function f was one to one. However, this is a strictly local result! Let f :ℝ
^{2}→ ℝ^{2}be given byThis clearly is not one to one because if you replace y with y + 2π, you get the same value. Now verify that Df

^{−1}exists for all. - Show every polynomial, ∑
_{|α| ≤k}d_{α}x^{α}is C^{k}for every k. - Suppose U ⊆ ℝ
^{2}is an open set and f : U → ℝ^{3}is C^{1}. Suppose Dfhas rank two andShow that for

near, the points fmay be realized in one of the following forms.or

This shows that parametrically defined surfaces can be obtained locally in a particularly simple form.

- Minimize ∑
_{j=1}^{n}x_{j}subject to the constraint ∑_{j=1}^{n}x_{j}^{2}= a^{2}. Your answer should be some function of a which you may assume is a positive number. - A curve is formed from the intersection of the plane, 2x + 3y + z = 3 and the cylinder x
^{2}+ y^{2}= 4. Find the point on this curve which is closest to. - A curve is formed from the intersection of the plane, 2x + 3y + z = 3 and the sphere
x
^{2}+ y^{2}+ z^{2}= 16. Find the point on this curve which is closest to. - Let A = be an n × n matrix which is symmetric. Thus A
_{ij}= A_{ji}and recall_{i}= A_{ij}x_{j}where as usual sum over the repeated index. Show= 2 A_{ij}x_{j}. Show that when you use the method of Lagrange multipliers to maximize the function, A_{ij}x_{j}x_{i}subject to the constraint, ∑_{j=1}^{n}x_{j}^{2}= 1, the value of λ which corresponds to the maximum value of this functions is such that A_{ij}x_{j}= λx_{i}. Thus Ax = λx. Thus λ is an eigenvalue of the matrix, A. - Let x
_{1},,x_{5}be 5 positive numbers. Maximize their product subject to the constraint that - Let f= x
_{1}^{n}x_{2}^{n−1}x_{n}^{1}. Then f achieves a maximum on the set,If x ∈ S is the point where this maximum is achieved, find x

_{1}∕x_{n}. - Maximize ∏
_{i=1}^{n}x_{i}^{2}(≡ x_{1}^{2}×x_{2}^{2}×x_{3}^{2}××x_{n}^{2}) subject to the constraint, ∑_{i=1}^{n}x_{i}^{2}= r^{2}. Show the maximum is^{n}. Now show from this thatand finally, conclude that if each number x

_{i}≥ 0, thenand there exist values of the x

_{i}for which equality holds. This says the “geometric mean” is always smaller than the arithmetic mean. - Show that there exists a smooth solution y = yto the equation
which is valid for x,y both near 0. Find y

^{′}at a pointnear. Then find y^{′′}for such. Can you find an explicit formula for y? - The next few problems involve invariance of domain. Suppose U is a nonempty open set in
ℝ
^{n},f : U → ℝ^{n}is continuous, and suppose that for each x ∈ U, there is a ball B_{x}containing x such that f is one to one on B_{x}. That is, f is locally one to one. Show that fis open. - ↑ In the situation of the above problem, suppose f : ℝ
^{n}→ ℝ^{n}is locally one to one. Also suppose that lim_{|x| →∞}= ∞. Show that it follows that f= ℝ^{n}. That is, f is onto. Show that this would not be true if f is only defined on a proper open set. Also show that this would not be true if the condition lim_{|x| →∞}= ∞ does not hold. Hint: You might show that fis both open and closed and then use connectedness. To get an example in the second case, you might think of e^{x+iy}. It does not include 0 + i0. Why not? - ↑ Show that if f : ℝ
^{n}→ ℝ^{n}is C^{1}and if Dfexists and is invertible for all x ∈ ℝ^{n}, then f is locally one to one. Thus, from the above problem, if lim_{|x| →∞}= ∞, then f is also onto. Now consider f : ℝ^{2}→ ℝ^{2}given byShow that this does not map onto ℝ

^{2}. In fact, it fails to hit, but Dfis invertible for all. Show why it fails to satisfy the limit condition.

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