9TAqtm-06

9TAqtm-06 - ASSIGNMENT 6 SOLUTIONS MAT 473 SPRING 2012...

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Unformatted text preview: ASSIGNMENT 6 SOLUTIONS MAT 473 SPRING 2012 Exercise 10.3. The equation x3 + y 3 + z 3 - 3xyz = 0 can be solved for z as a function of (x, y) near (-1, -1, 2). If z = g(x, y) denotes such a function, find the degree 2 Taylor polynomial of g for (x, y) near (-1, -1). Solution. Define f : R3 R1 by f (x, y, z) = x3 + y 3 + z 3 - 3xyz. Then [f (x, y, z)] = 3x2 - 3yz 3y 2 - 3xz 3z 2 - 3xy ; in particular, DII f (-1, -1, 2) = 3(22 ) - 3(-1)(-1) = 9 is invertible. Since f is evidently C 1 , it follows from the Implicit Function Theorem that there exist open neighborhoods I of (-1, -1) in R2 and J of 2 in R1 such that for each (x, y) I, there is a unique z = g(x, y) J such that f (x, y, g(x, y)) = 0. The formula provided by the Implicit Function Theorem then gives g (x, y) = -DII f (x, y, g(x, y))-1 DI f (x, y, g(x, y)), whence 2 2 1 x - yz y 2 - xz 2 3x - 3yz 3y - 3xz = [g (x, y)] = - 2 , 3z - 3xy xy - z 2 xy - z 2 which is to say (10.1) D1 g(x, y) = x2 - yz xy - z 2 and D2 g(x, y) = y 2 - xz xy - z 2 for all (x, y) I. Notice that if we just assume the result of Exercise 10.2, we can still arrive at a formula for g (x, y) by implicitly differentiating the equation f (x, y, g(x, y)) = 0. That is, implies (differentiating with respect to x) and (with respect to y) f (x, y, g(x, y)) = x3 + y 3 + g 3 (x, y) - 3xyg(x, y) = 0 3x2 + 3g 2 (x, y)D1 g(x, y) - 3yg(x, y) - 3xyD1 g(x, y) = 0 which is the same as (10.1). (When x and y are close to -1, z will be close to 2, so xy - z 2 will not be zero.) Now, since we know that g(-1, -1) = 2, we have D1 g(-1, -1) = D2 g(-1, -1) = (-1)2 - (-1)(2) = -1. (-1)(-1) - (2)2 3y 2 + 3g 2 (x, y)D2 g(x, y) - 3xg(x, y) - 3xyD2 g(x, y) = 0, Date: February 16, 2012. S. Kaliszewski, School of Mathematical and Statistical Sciences, Arizona State University. 1 In order to compute the second partial derivatives, we use (10.1), remembering that z is a differentiable function of x and y, so the (one-variable) Chain Rule comes into play: (2x - yD1 g)(xy - z 2 ) - (y - 2zD1 g)(x2 - yz) , (xy - z 2 )2 (-z - yD2 g)(xy - z 2 ) - (x - 2zD2 g)(x2 - yz) D12 g(x, y) = D21 g(x, y) = , (xy - z 2 )2 (2y - xD2 g)(xy - z 2 ) - (x - 2zD2 g)(y 2 - xz) D22 g(x, y) = . (xy - z 2 )2 Evaluating at x = y = -1, z = 2, and D1 z = D2 z = -1 gives, sadly, D11 g(x, y) = Thus, the second-degree Taylor polynomial for g(x, y) at (-1, -1) is the same as the firstdegree Taylor polynomial, namely P2 (x, y) = P1 (x, y) = g(-1, -1) + D1 g(-1, -1)(x - (-1))1 + D2 g(-1, -1)(y - (-1))1 Finally, note (by direct calculation) that z = -x-y does indeed solve the original equation x3 + y 3 + z 3 - 3xyz = 0 for z as a function of x and y near (-1, -1, 2) (in fact for all (x, y) R2 ). If we had noticed this earlier, then by the uniqueness clause in the Implicit Function Theorem we could have deduced that g(x, y) = -x - y and computed the Taylor polynomial(s) directly. = 2 - (x + 1) - (y + 1) = -x - y. D11 g(-1, -1) = D12 g(-1, -1) = D21 g(-1, -1) = D22 g(-1, -1) = 0. 2 Exercise 11.2. Let (X, M, ) be a measure space. Prove that if B1 B2 in M and (B1 ) < , then (Bi ) = limi (Bi ). Proof. For each i, let Ai = B1 \ Bi . Then B1 = Ai Bi is a disjoint union, so (B1 ) = (Ai ) + (Bi ). Since (B1 ) < , we have (Ai ) < , and it follows that Similarly, (Bi ) = (B1 ) - (Ai ). B1 = is a disjoint union, so and therefore Ai i=1 Bi i=1 (B1 ) = Finally, note that A1 A2 is an increasing sequence, so putting the above together with part (viii) of Proposition 11.3 gives Bi = (B1 ) - Ai = (B1 ) - lim (Ai ) i i i Bi = (B1 ) - Ai + Bi , Ai . = lim ((B1 ) - (Ai )) = lim (Bi ). Alternative: Write B1 as a disjoint union B1 = Bi (B1 \ B2 ) (B2 \ B3 ) and argue as in the proof of part (viii) of Proposition 11.3. 3 ...
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This document was uploaded on 03/11/2012.

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