volq = U(f, P 2 ). This finally gives

Transcription

1 MAT 257Y s to Practice Final (1) Let A R n be a rectangle and let f : A R be bounded. Let P 1, P 2 be two partitions of A. Prove that L(f, P 1 ) (f, P 2 ). The statement is obvious if P 1 = P 2. In general, let P be a common refinement of P 1 and P 2. Then L(f, P 1 ) L(f, P ). Indeed, for any rectangle Q of P contained in a rectangle Q of P 1 we have that m(f, Q) m(f, Q ). therefore L(f, P ) = Q P m(f, Q )volq = Q P Q P Q Q Q Q m(f, Q)volQ = Q P m(f, Q) Q Q m(f, Q)volQ = L(f, P 1 ) Q P m(f, Q )volq volq = Thus L(f, P 1 ) L(f, P ) and similarly (f, P ) (f, P 2 ). This finally gives L(f, P 1 ) L(f, P ) (f, P ) (f, P 2 ) (2) let M = {(x, y) R 2 such that x 2 + y 2 = 1. let f : M R be given by f(x, y) = x 2 + y. Find the minimum and the maximum of f on M. Let g(x, y) = x 2 + y 2. By the Lagrange multiplier method extremum points of f on M can only occur when f = λ g. We have g(x, y) = (2x, 2y) and 1

2 2 f(x, y) = (2x, 1). We need to solve 2x = λ2x 2y = λ x 2 + y 2 = 1 If x 0 the first equation give λ = 1. hence y = 1/2 and x = ± 3 2. If x = 0 then y = ±1. Thus we have four possible points we need to test (0, 1), (0, 1), ( 3 2, 1/2) and ( 3 2, 1/2). Computing f at these points we get f(0, 1) = 1, f(0, 1) = 1, f( 3 2, 1/2) = f( 3 2, 1/2) = 3/4 + 1/2 = 5/4. Thus the maximum of f on M os 5/4 and the minimum is 1. (3) Let T : R 2n = R n R n R be a 2-tensor on R n. Show that T is differentiable at (0, 0) and compute dt (0, 0). Let x = (x 1,... x n ) be the coordinates on the first copy of R n and y = (y 1,..., y n ) on the second. Then by multilinearity we have that T (x, y) = ij T ijx i y j. This function is a polynomial and hence is differentiable. It is also obvious to check that its partial derivatives at (0, 0) are all zero. therefore dt (0, 0) = 0. (4) Let ω = xdy dz+ydz dx+zdx dy (x 2 +y 2 +z 2 ) 3/2 be a 2-form on R 3 \(0, 0, 0). Verify that ω is closed. Hint: One way to simplify the computation is to write ω = f ω where f = 1 (x 2 +y 2 +z 2 ) 3/2 and ω = xdy dz + ydz dx + zdx.

3 We have dω = df ω + ( 1) 0 fd ω df ω = (x 2 + y 2 + z 2 ) 5/2(2xdx+2ydy+2zdz) (xdy dz+ydz dx+zdx) = 3(x2 + y 2 + z 2 )dx dy dz 3dx dy dz = (x 2 + y 2 + z 2 ) 5/2 (x 2 + y 2 + z 2 ) 3/2 1 3dx dy dz fd ω = 3dx dy dz = (x 2 + y 2 + z 2 ) 3/2 (x 2 + y 2 + z 2 ) 3/2 Therefore dω = df ω + ( 1) 0 fd ω = 0. (5) Let f : R 2 R 2 be given by f(x, y) = (e 2y, 2x + y) and let ω = x 2 ydx + ydy. Compute f (dω) and d(f (ω)) and verify that they are equal. dω = (2xydx + x 2 dy) dx + dy dy = x 2 dx dy. f (dω) = (e 2y ) 2 de 2y (2dx + dy) = e 6y dy (2dx + dy) = 2e 6y dx dy. On the other hand, f (ω) = (e 2y ) 2 (2x + y)de 2y + (2x+y)(2dx+dy) = e 6y (2x+y)dy +(2x+y) (2dx+ dy). Finally, d(f (ω)) = d(e 6y (2x+y)) dy+(2dx+dy) (2dx+dy) = (6e 6y (2x+y)dy+e 6y (2dx+dy) dy+0 = 2e 6y dx dy. (6) Determine if ext 0<x 2 +y 2 <1 ln(x2 +y 2 ) exists and if it does compute it. Let = 0 < x 2 + y 2 < 1\{(0, 1) 0}. then ext 0<x 2 +y 2 <1 ln(x2 + y 2 ) exists iff ext ln(x2 + y 2 ) exists and if they both exist they are equal. For the second integral make a polar change of variables x = r cos θ, y = r sin θ where 0 < r < 1, 0 < θ < 2π. Then ext ln(x2 + y 2 ) = 2π 0 ( 1 0 r ln r2 dr)dθ = 3

4 4 4π 1 0 r ln rdr = 4π 1 0 ln rd(r2 /2) = 4π( r2 ln r r d ln r) = 4π(0 1 0 r/2dr) = 4πr3 /6 1 0 = 2π/3. Here we used the fact that lim r 0+ r 2 ln r = 0. Thus ext 0<x 2 +y 2 <1 ln(x2 + y 2 ) = 2π/3 (7) Let, V be open in R n. Let f : R n R be a continuous nonnegative function such that ext f and ext V f exist. Prove that ext V f exists. Hint: use compact exhaustions of and V to construct a compact exhaustion of V. let K i be a compact exhaustion by measurable sets of and C i be a compact exhaustion by measurable sets of V. then we have that K i f is increasing and lim i ext V f. K i f = ext f. Similarly, lim i C i f = Then it s easy to see that K i C i is a compact exhaustion by measurable sets of V. Since f 0 we have that C i K i f = C i f + K i f C i K i f C i f + K i f ext f + ext V f. Therefore lim i K i C i f exists and hence so does ext V f. (8) Let F (x) = x 2 e f(tx)dt where f : R R is C 1. x Show that F (x) is C 1 and find the formula for F (x). Let G(x, a, b) = b a f(tx)dt. Then G is C1 by a theorem from class and. G x (x, a, b) = b d a dx f(tx)dt = b a tf (tx)dt. Also, G G b (x, a, b) = f(bx) and a (x, a, b) = f(ax).

5 Then F (x) = G(x, e x, x 2 ) is C 1 by the chain rule and F (x) = G x (x, ex, x 2 )+ G a (x, ex, x 2 ) (e x ) + G b (x, ex, x 2 ) (x 2 ) = x 2 e tf (tx)dt f(e x x)e x + f(x 3 )2x. x (9) Prove that a compact set is closed. We will show that if C R n is compact then = R n \C is open. let p. let i = {x R n such that d(x, p) > 1/i}. then V i is open and i V i = R n \{p} covers C. By compactness C is covered by finitely many V i s and hence, C V j for some j. This means that B(p, 1/j). since p is arbitrary, this means that is open and C is closed. (10) Let x(t 1, t 2 ) = t 1 cos t 2, y(t 1, t 2 ) = t e t 1t 2. Let f(x, y) be a differentiable function f : R 2 R. Let g(t 1, t 2 ) = f(x(t 1, t 2 ), y(t 1, t 2 )). Express g (1, 0) and g t 2 (1, 0) in terms of partial derivatives of f. By the chain rule g (t 1, t 2 ) = f x x (x(t), y(t)) + f y y (x(t), y(t)) = f x (t 1 cos t 2, t 2 1+e t 1t 2 ) cos t 2 + f y (t 1 cos t 2, t 2 1+e t 1t 2 )(2t 1 + t 2 e t 1t 2 ). Similarly, g t 2 (t 1, t 2 ) = f x (t 1 cos t 2, t 2 1+e t 1t 2 )( t 1 sin t 2 )+ f y (t 1 cos t 2, t e t 1t 2 )(t 1 e t 1t 2 ). (11) Mark true or false. Justify your answer. Let A, B be any subsets of R n. (a) br(a) Lim(A) (b) Lim(A) A (c) br(a B) br(a) br(b). (a) False. Example A = {p}. Then bra = {p} and Lim(A) =. 5

6 6 (b) False. Example A = (0, 1) R. Then Lim(A) = [0, 1] is not contained in A. (c) False. Example A = [0, 2], B = [1, 3]. Then A B = [1, 2] and br(a B) = {1, 2}. On the other hand, br(a) br(b) =. (12) Let M 3 be a compact 3-manifold with boundary in R 3 and let n be the outward unit normal on. Let F = (F 1, F 2, F 3 ) be a vector field on R 3. Prove that divf = F, n M Hint: Convert the integral over to an integral of a form in R 3 and use Stokes formula. recall that divf = F 1 x + F 2 y + F 3 z. Let n = (n 1, n 2, n 3 ). Let ω = F 1 dy dz + F 2 dz dx + F 3 dx dy. then F, n da = ω. Indeed, F, n = F 1n 1 + F 2 n 2 + F 3 n 3. Recall that n 1 da = dy dz, n 2 da = dz dz and n 3 da = dx dy. Therefore, F, n da = (F 1 n 1 + F 2 n 2 + F 3 n 3 )da = F 1 dy dz + F 2 dz dx + F 3 dx dy = ω By Stokes formula this is equal to M dω = M ( F 1 F 2 y + F 3 z )dx dy dz = M divf. (13) let M 2 R 3 be the torus of revolution obtained by rotating the circle (x 2) 2 + z 2 = 1 in the xz plane around the yz axis. Consider the orientation on M induced by the outward normal field N where N(3, 0, 0) = (1, 0, 0). Find M xdy dz x +

7 Let V be the solid obtained by rotating the disk = (x 2) 2 + z 2 1 in the xz plane around the yz axis. then M = V and by Stokes formula M xdy dz = V d(xdy dz) = V dx dy dz = volv. Recall that by a homework problem this is equal to 2π x. sing polar coordinates change of variables x = 2 + r cos θ, y = r sin θ we compute x = 2π (2+r cos θ)rdrdθ = 2π (2r+r2 cos θ)drdθ = 2π. Therefore M xdy dz = 4π2. (14) Let M R n be an oriented manifold. Prove that vol(m) = M da is positive. Let i be a covering of M by orientation preserving coordinate patches f i : W i M and let φ i be the partition of unity subordinate to this covering. Note that 0 φ i 1. Then M da = i M φ ida = i W i fi (φ ida) = i W i (φ i f i )fi (da). Note that φ i f i (x) 0 for any x W i and is positive at some point of W i. We also have that fi (da) = u(x)dx1... dx k where u(x) = da(f i e 1,..., f i e k ) > 0 since f i is orientation preserving. Altogether the above means that W i (φ i f i )fi (da) = W i g i (x) where g i is a continuos nonnegative function with compact support which is positive somewhere. therefore W i g i (x) > 0 and hence M da = i M φ ida > 0. 7