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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 (onevariable) 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 seconddegree 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 = xy 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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 Spring '09

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