MATH2111 Higher Several Variable Calculus Week 2

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1 MATH2111 Higher Several Variable Calculus Week 2 Dr Denis Potapov School of Mathematics and Statistics University of New South Wales Semester 1, 2016 [updated: March 8, 2016] D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32 Open, closed subets, boundary and interior JKress-open-closed-boundary-and-interior.pdf interior-and-boundary-direct-argument.canvas JKress-union-intersection-of-open-closed-subsets.pdf D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32

2 Continuity Let f : D R n R m and let D be open. Denition f is continuous at a D if and only if lim f(x) = f(a). x a f is continuous on D if and only if f is continuous at every a D. D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32 Continuity by components Let and let f : D R n R m f = (f 1, f 2,..., f m ), f k : D R n R. The mapping f is continuous if and only if the every function f k is continuous, k = 1, 2,..., m. D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32

3 Continuity by components continuity-by-components-theorem.canvas (to create): the proof is similar to continuity-per-component.canvas D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32 Continuity elementary functions Let and let f : D R n R x = (x 1, x 2,..., x n ) R n. Denition A function f is called elementary if f is either a a constant; or one of x k cos x k, sin x k, exp x k ; or an inverse of elementary function; or an algebraic combination of elementary functions; or superposition of elementary functions. D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32

4 Continuity elementary functions If f : D R n R is elementary function, then f is continuous everywhere where f is well-dened. D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32 Continuity and open/closed subsets Let f : D R n R m and let preimage is dened by f 1 (U) := {x D : f(x) U}, U R m. 1 The mapping f is continuous on D R n if and only if 1 f (U) D is open for every open U R m ; and 2 the mapping f is continuous on D R n if and only if 1 f (U) D is closed for every U R m closed. D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32

5 Continuity and open/closed subsets Example Show that the set is open. Since V = { (x, y) R 2 : x 2 y 2 < 1 } f (x, y) = x 2 y 2 is elementary, it is continuous everywhere on R 2. Since V = f 1 (U), where U = (, 1) and since U is open, we conclude that is also open. f 1 (U) D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32 Continuity and open/closed subsets continuity-via-preimage.canvas D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32

6 Bounded subsets JKress-bounded-subsets.pdf D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32 Bolzano-Weierstrass theorem For Ω R n, the following are equivalent. (i) Ω is closed and bounded. (ii) Every sequence in Ω has a subsequence that converges to an element of Ω. (iii) Whenever the union of a collection of open sets contains Ω there is always a nite sub-collection thats union also contains Ω [Heine-Borel theorem]. D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32

7 Bolzano-Weierstrass theorem, II JKress-bolzano-weierstrass-comments.pdf D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32 Compact subsets Denition A set Ω is compact if it satises either property from the Bolzano-Weierstrass theorem. Examples: compact R not compact (0, 1) not compact [0, 1] compact [0, 1] [3, 4] compact [0, 1] [0, 1] compact S 2 (the 2-sphere) compact Cantor set compact D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32

8 Path-connected subsets JKress-path-connected.pdf D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32 Image of compact/path-connected subset Let f : Ω R n R m be continuous. Then (i) K Ω and K is compact f(k) is compact. (ii) B Ω and B is path connected f(b) is path connected. D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32

9 Image of compact/path-connected subset, II Consider S 1 = {(x, y) : x 2 + y 2 1} S 2 = {(x, y) : x 2 + y 2 < 1} Is there a continuous function f : R 2 R 2 such that 1 f(s 1 ) = S 2? 2 f(s 2 ) = S 1? 3 f(r 2 ) = S 2? 4 f(r 2 ) = S 1? 5 f(s 2 ) = R 2? 6 f(s 1 ) = R 2? continuous-maps-between-subsets.canvas D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32 Image of compact/path-connected subset, III Consider S 1 = {(x, y) : x 2 + y 2 1} and S 2 = {(x, y) : x 2 + y 2 < 1}. S 1, S 2 and R 2 are path connected but only S 1 is compact. 1. S 1 is compact and S 2 is not. So there can not be a continuous function f1 with f1(s 1 ) = S Consider the function f2 : R 2 R 2 described in terms of polar coordinates by { (2r, θ) for r < 1 (r, θ) 2 (1, θ) for r 1 2. This is continuous and f2(s 2 ) = S Consider the function f3 : R 2 R 2 described in polar coordinates by ( ) 2 (r, θ) π tan 1 r, θ. This is continuous and f3(r 2 ) = S f2 f3 is a continuous function that maps R 2 to S f5 = 1 f3 is a continuous function with f (S 2 ) = R S 1 is compact and R 2 is not. So there can not be a continuous function f6 with f6(r 2 ) = S 1. D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32

10 Image of compact/path-connected subset, IV JKress-compact-path-connected-image-proof.pdf D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32 Min/Max If f : K R n R is a continuous function on a compact subset K, then both maximum and minimum values are attained, i.e., there are xmin, xmax K such that f (xmin) = min x K f (x) and f (xmax) = max f (x). x K D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32

11 Intermediate Value If f : K R n R is a continuous function on a path-connected and compact subset K, then every intermediate value is attained, i.e., for every b R, such that min x K f (x) b max f (x), x K there is such that x K f (x) = b. D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32 Min/Max and Intermediate Value s, II JKress-MM-and-IV-theorems.pdf MM-theorem-proof.canvas IV-theorem-proof.canvas (the proof is explained by the example in JKress-MM-and-IV-theorems.pdf) D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32

12 Dierentiability of f : R R Denition A function f : R R is dierentiable at x = a if and only if the following limit exist: f (x) f (a) L := lim. ( ) x a x a The value L is called derivative of f at x = a and is denoted by f (a). The requirement (*) is equivalent to [ ] f (x) f (a) f (x) f (a) L (x a) ( ) lim L = 0 lim = 0. x a x a x a x a D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32 Dierentiability of f : R n R m Denition A function is dierentiable at f : R n R m x = a R n if and only if there is m n-matrix L such that f(x) f(a) L (x a) lim x a x a = 0. The matrix L is denoted by D a f. D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32

13 Dierentiability examples Suppose f : R n R m is a linear transformation given by f(x) = Ax, where A is a m n-matrix. Is it dierentiable and if so, what is it's derivative? f(x) f(a) A (x a) lim x a x a Hence f is dierentiable and D a f = A. dierentiability-direct-example.canvas = lim x a 0 x a = 0. D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32 Dierentiability examples For f : R 2 R 2 with ( x f(x, y) = 2 ) + 2xy x + y 2, L = ( ) and a = ( ) 1, 1 show that f is dierentiable at a and that the matrix of its derivative is D a f = L. ( x f(x) f(a) L(x a) = 2 ) ( ) ( ) ( ) + 2xy x 1 x + y y 1 ( x = 2 ) + 2xy 4x 2y + 3 y y So for f to be dierentiable at a with derivative L, we need lim (x,y) (1,1) This is true, but takes a bit of work. (x 2 + 2xy 4x 2y + 3) 2 + (y y) 2 (x 1) 2 + (y 1) 2 = 0. D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32

14 Dierentiable continuous Suppose Ω R n is open and f : Ω R n R m is dierentiable on Ω. Then f continuous on Ω. is D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32 Dierentiable continuous, II JKress-dierentiable-so-continuous.pdf D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32

15 Partial derivatives of f : R n R JKress-partial-derivatives.pdf D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32 Jacobian matrix Denition If all partial derivatives of f : Ω R n R m exist at a Ω, then the Jacobian matrix of f at a is f1 (a) f1 (a) f1 x1 x2 x n (a) f2 J a f = (a) f2 (a) f2 x1 x2 x n (a) f m (a) f m (a) f m x1 x2 x n (a) For f : Ω R n R m and an interior point a Ω. If f is dierentiable at a then all partial derivatives f j x i of the components of f exist at a and D a f = J a f. That is, where f is dierentiable, its derivative is given by its Jacobian matrix. D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32

16 Jacobian matrix, Example The Jacobian matrix may exist even when the function is not dierentiable. Example: f : R 2 R with f (x, y) = { 0 for x = 0 or y = 0, 1 otherwise. Clearly f f (0, 0) = (0, 0) = 0. However, the ane function x y T (x, y) = f (0, 0) + J (0,0) f ( ) x = 0 + (0 0) y is not a good approximation to f (x, y) near (0, 0). ( ) x = 0. y Notice that in this example, f is not continuous. Should a dierentiable function be continuous? D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32 Dierentiability, Examples dierentiability-example-1.canvas dierentiability-example-2.canvas dierentiability-example-3.canvas D Potapov (UNSW Maths & Stats) MATH2111 Semester 1, / 32