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Math 508 Exam 2 Jerry L. Kazdan
December 8, 2006 12:00 – 1:
Directions This exam has two parts, Part A has 3 shorter problems (8 points each, so 24 points), Part B has 5 traditional problems (15 points each, so 75 points). Closed book, no calculators – but you may use one 3′′^ × 5 ′′^ card with notes.
Part A: Short Problems (3 problems, 8 points each).
A–1. A continuous function f : R → R has the property that ∫ (^) x
0
f (t) dt = cos(x) e−x^ + C,
where C is some constant. Find both f (x) and the constant C. Solution: Letting x = 0, we find that 0 = 1 + C , so C = −1. To compute f , use the fundamental theorem of calculus. Thus take the derivative of both sides
f (x) = d dx
[cos(x) e−x^ + C] = − sin(x) e−x^ − cos(x) e−x.
A–2. A function h : R → R with two continuous derivatives has the property that h(0) = 2, h(1) = 0, and h(3)=1. Prove there is at least one point c in the interval 0 < x < 3 where h′′(c) > 0 by finding some explicit m > 0 (such as m = 3/2) with h′′(c) ≥ m. Solution: By the mean value theorem applied twice, there is some a ∈ (0, 1) and some b ∈ (1, 3) so that
h′(a) = h(1) − h(0) 1 − 0
= − 2 , h′(b) = h(3) − h(1) 3 − 1
Thus, by the mean value theorem again there is some c ∈ (a, b) so that
h′′(c) = h′(b) − h′(a) b − a
1 2 + 2 b − a
A–3. Say a smooth function u(x) satisfies u′′^ − c(x)u = 0 for 0 ≤ x ≤ 1 (here c(x) is some given contunuous function). If c(x) > 0 everywhere, show that there is no point where u(x) is both positive and has a local maximum. If we also knew that u(0) = 0 and u(1) = 0, why can we conclude that u(x) = 0 for all 0 ≤ x ≤ 1? Solution: If u has a positive maximum at some point p, then u′′(p) ≤ 0 and u(p) > 0. Consequently u′′(p) − c(p)u(p) < 0, which contradicts u′′^ − c(x)u = 0. If u(0) = 0 and u(1) = 0 but u is not identically zero, then u must either be positive or negative somewhere. Say u is positive somewhere (otherwise replace u by −u). Then since u is continuous on the compact set [0, 1], it has a positive maximum at some interior point. But we saw just above that this can’t happen.
Part B: Traditional Problems (5 problems, 16 points each)
B–1. Given that two functions f : R → R and g : R → R are differentiable at a point x = c, prove that their product h(x) = f (x)g(x) is also differentiable at x = c. Solution:. h(c + k) − h(c) k
[f (c + k)g(c + k) − f (c)g(c + k)] + [f (c)g(c + k) − f (c)g(c)] k =
f (c + k) − f (c) k g(c + k) + f (c)
g(c + k) − g(c) k Now let k → 0. Since f and g are both assumed to be differentiable at x = c, we see that h is differentiable there and get the usual formula: h′(c) = f ′(c)g(c) + f (c)g′(c).
B–2. Let α(t) and β(s) describe smooth curves in R^3 that do not intersect. Say the points p = α(t 0 ) and q = β(s 0 ) minimize the distance between the curves. Show that the line from p to q is perpendicular to both of these curves. Solution:. To avoid square roots, let Q be the square of the distance from α(t) to the point β(s), so Q(s, t) = ‖α(t) − β(s)‖^2 = 〈α(t) − β(s), α(t) − β(s)〉 Then Q(t, s) has its minimum at (t 0 , s 0 ), Consequently both ∂Q/∂t = 0 and ∂Q/∂s = 0 at (t 0 , s 0 ). But ∂Q ∂t = 2〈α(t) − β(s), α′(t)〉 and
∂Q
∂s = − 2 〈α(t) − β(s), β′(s)〉.
Evaluated at (t 0 , s 0 ) the first gives α(t 0 ) − β(s 0 ) ⊥ α′(t 0 ), whiile the second gives the other othogonality, α(t 0 ) − β(s 0 ) ⊥ β′(s 0 ).
B–3. Compute lim λ→∞
0
|sin(λx)| dx.
Solution: First make the substitution t = λx. and say nπ ≤ λ < (n + 1)π. Since |sin(λt)| is periodic with period π , then the integral becomes
1 λ
∫ (^) λ
0
|sin t| dt =
λ
[∫
π
0
∫ (^2) π
π
∫ (^) nπ
(n−1)π
∫ (^) λ
nπ
|sin t| dt
]
λ
[
n
∫ (^) π
0
sin t dt +
∫ (^) λ
nπ
|sin t| dt
]
2 n λ
λ
∫ (^) λ
nπ
|sin t| dt.
Because nπ ≤ λ < (n + 1)π , then λ/n → π so 2n/λ → 2 /π. Also
1 λ
∫ (^) λ
nπ
|sin t| dt <
λ
∫ (^) (n+1)π
nπ
dt = π λ
Consequently the limit is 2/π.
In a general metric space, the assertion in part a) is false, even if you also assume T bounded.
For example, in ` 2 , let S = { 0 } (the origin) and T = {(1 + (^1) n )en, n = 1, 2 , 3 ,.. .} where e 1 , e 2 , e 3 ,... are the standard basis vectors. Then dist (S, T ) = 1 although there are no points p ∈ S , q ∈ T with ‖p − q‖ = 1. The problem is that although T is closed and bounded, it has no convergent subsequence.
In a general metric space, the assertion in part a) is true if both S and T are compact since a continuous function on a compact set achieves its minimum. One can use this to give a slightly different proof of part a) as follows. As observes above, if m = inf x∈S, y∈T
‖x − y‖, then to find
the minimum distance, we need only use the points y ∈ T that are within distance m + 1 from S , that is, Q := {y ∈ T | | d(x, y) ≤ m + 1 for all x ∈ S}
But since S is compact, it is bounded, so Q is bounded (and closed). Since Q ∈ Rn^ , it is compact. Thus for x ∈ S and y ∈ Q, the function d(x, y) is a continuous function on a compace set so it achieves its minimum at some point of the set. Because the sets are disjoint, this minimum is strictly positive.
