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Unformatted text preview: 24 SPRING 2012 10. Proof of the Implicit Function Theorem. Proof. The formula for g (x) will follow from the rest of the theorem, as in Exercise 9.1, by defining : Rn Rn Rm by (x) = (x, g(x)), so that (f )(x) = f (x, g(x)) = 0 for all x I, and then using the Chain Rule. Define F : U Rn Rm by F (x, y) = (x, f (x, y)). Then F is C 1 on U , with Id 0 (10.1) F (x, y) = ; DI f (x, y) DII f (x, y) this also shows that F (a, b) is invertible, since DII f (a, b) is. Thus the Inverse Function Theorem provides open sets V and W such that (a, b) V U , F (a, b) W Rn Rm , and F : V W is onetoone and onto with C 1 inverse. By shrinking V and W if necessary we can assume that V = K J for some open sets K Rn and J Rm . Let u and v denote the C 1 component functions of F 1 , so F 1 (s, t) = (u(s, t), v(s, t)) for all (s, t) W . Then which shows that (10.2) for all (s, t) W . Let Then a I since (a, 0) = F (a, b) W . Moreover, I is open, since for every (x, 0) W there are open sets A Rn and B Rm such that (x, 0) A B W ; in particular, A {0} W , so x A I. Now for x I, f (x, v(x, 0)) = 0 by (10.2), and v(x, 0) J because (x, v(x, 0)) = F 1 (x, 0) V = K J. Thus g(x) = v(x, 0) defines a function g : I J such that f (x, g(x)) = 0 for each x I. Moreover, g is C 1 , since it's the composition of the C 1 function v with the C 1 function x (x, 0). For uniqueness, fix x I and suppose y J satisfies f (x, y) = 0. Then f (x, y) = f (x, g(x)), so F (x, y) = F (x, g(x)), so (x, y) = (x, g(x)) since F is onetoone on V , so y = g(x). Exercise 10.1. With notation as in the above proof, use (10.1) to find a formula for (F (a, b))1 in terms of DI f and DII f . Example 10.1. Consider f : R2 R defined by f (x, y) = x2 + y 2  1. We have [f (x, y)] = DI f (x, y) DII f (x, y) = 2x 2y
Date: February 7, 2012 S. Kaliszewski, Department of Mathematics and Statistics, Arizona State University. (s, t) = F (F 1 (s, t)) = F (u(s, t), v(s, t)) = (u(s, t), f (u(s, t), v(s, t))), u(s, t) = s and f (s, v(s, t)) = t I = {x Rn  (x, 0) W } . MAT 473 25 for any (x, y), so the equation x2 + y 2 = 1 can be solved (continuously differentiably) for y as a function of x near any point (a, b) which satisfies a2 + b2 = 1 and b = 0. In this case we'll have 1 x y (x) =  2x =  . 2y y Notice that the formula for y involves y itself. The Implicit Function Theorem never tells us (and it can't tell us) what y is, only y . In this example, once we know that x2 + y 2 = 1 defines y as a differentiable function of x, we can compute y by "implicitly differentiating" with respect to x. This just means that we recognize that g(x) = x2 + y 2 and h(x) = 1 are the same differentiable function of x, so have the same derivative; and we can compute g (x) using the Chain Rule: g (x) = 2x + 2yy . Equating with h (x) = 0 and solving gives y (x) = x/y. (This is exactly the same calculation, albeit in fewer variables, that gives the formula for g (x) in the proof of the Implicit Function Theorem.) Running through the rest of the proof in this special case, with (a, b) = (0, 1): we define F : R2 R2 by F (x, y) = (x, x2 + y 2  1), which is evidently continuously differentiable, with 1 0 [F (x, y)] = , 2x 2y so F (0, 1) is invertible. The Inverse Function Theorem now provides an open neighborhood V of (0, 1) in R2 , which we can take to be of the form V = K J for an open interval K containing 0 and an open interval J containing 1, and an open set W = F (V ) in R2 , such that F is onetoone, with C 1 inverse, on V . Now the points we're interested in in V are those for which x2 + y 2 = 1; these are exactly those sent to K {0} in W . More precisely, any point (x, y) in V with x2 + y 2 = 1 gets sent to (x, 0). Thus, since F is onetoone on V , given x K such that (x, 0) is in W , we can recover the unique y J such that x2 + y 2 = 1 by applying F 1 to (x, 0) and looking at the second coordinate of the result. (This is what v(x, 0) accomplishes in the proof.) Geometrically, the effect of F is to (invertibly, differentiably) distort a block of R2 con taining what we know is the graph of y = 1  x2 near x = 0, in such a way that the curve is straightened out and sent to a piece of the xaxis. Example 10.2. Suppose f : R3 R2 is defined by Then f is C 1 , with [f (x, y, z)] = and f (1, 1, 1) = (0, 0). Since f (x, y, z) = (x ey + y ez , x ez + z ey ). e y x ey + e z y ez , ez z ey x ez + e y 0 e e 0 DII f (1, 1, 1) = Date: February 7, 2012 S. Kaliszewski, Department of Mathematics and Statistics, Arizona State University. 26 SPRING 2012 is invertible, the Implicit Function Theorem says that the system x ey + y ez = 0 x ez + z e y = 0 can be solved for y and z as continuously differentiable functions of x for x near 1 and (y, z) near (1, 1). Further, if we write (y, z) = g(x) for x near 1, we can say that y 1 y x e + ez y ez e g (x) =  z . y z y ze xe + e e Exercise 10.2. Show that the equation x3 + y 3 + z 3  3xyz = 0 can be solved for z as a function of (x, y) near (1, 1, 2). Exercise 10.3. If z = g(x, y) denotes the solution for z as a function of (x, y) from Exercise 10.2, find the degree 2 Taylor polynomial of g for (x, y) near (1, 1). Date: February 7, 2012 S. Kaliszewski, Department of Mathematics and Statistics, Arizona State University. ...
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