Sample Problems for Third Midterm March 18, 2013

Transcription

1 Mat 30. Treibergs Sampe Probems for Tird Midterm Name: Marc 8, 03 Questions 4 appeared in my Fa 000 and Fa 00 Mat 30 exams (.)Let f : R n R n be differentiabe at a R n. (a.) Let g : R n R be defined by g(x) x f(x), (dot product.) Find te tota derivative (differentia) Dg(a)() were R n. (b.) Witout using te product teorem, prove your answer. (.) For eac part, determine weter te statement is TRUE or FALSE. If te statement is true, give a justification. If te statement is fase, give a counterexampe. You may use teorems. (a.) Statement. Let f : R R be continuous. Suppose tat for a (x, y) R bot f(x +, y) f(x, y) im 0 exist. Ten f is differentiabe at (, ). and f(x, y + ) f(x, y) im 0 (b.) Statement. Suppose f : R R is a C function for wic te tird partia derivatives f xxy (x, y) exist for a (x, y) R suc tat f xxy (x, y) is continuous at (0, 0). Ten f xyx (0, 0) and f yxx (0, 0) exist and are equa: f xxy (0, 0) f xyx (0, 0) f yxx (0, 0). (3.) Suppose f, g : R R. Assume tat g is differentiabe at x 0 R and tat for some α > and M < we ave f(x) g(x) M x x 0 α for a x R. Sow tat f is differentiabe at x 0 and tat Df(x 0 ) Dg(x 0 ). (4.) Let F : R 5 R be given by F (f, f ) were f (v, w, x, y, z) v + w + x + y, f (v, w, x, y, z) vy + wz. Suppose tat tere is an open neigborood U R 3 containing te point (3, 4, 5) and C functions G : U R were G (g, g ) so tat for a (x, y, z) U, g (3, 4, 5), g (3, 4, 5) ; f (g (x, y, z), g (x, y, z), x, y, z), f (g (x, y, z), g (x, y, z), x, y, z) 4. Wat is te differentia DG(3, 4, 5)(,, )? More probems. (E.) Suppose f : R 3 R is given by f(x, y, z) (xy + x z 3, x 4 + y + y 5 z 6 ). Is f differentiabe on R 3? If so, find te differentia df(x, y, z)(,, ). (E.) Find te extrema of φ(x, y, z) x + y + z subject to te constraints x y and y z. (E3.) Suppose tat r and α are positive, E R n is a convex set suc tat Ē B r(0), and tat tere exists a sequence x E suc tat x 0 as. If f : B r (0) R is continuousy differentiabe and f(x) x α for a x E, prove tat tere is an M < suc tat f(x) M x for x E.

2 (E4.) Sow tat f(x, y) as partia derivatives for a (x, y) R but f is not differentiabe at (0, 0), were x 3 + y 3 + 4xy f(x, y) x + y, if (x, y) (0, 0); 0, if (x, y) (0, 0). (E5.) Suppose tat f : R p R is differentiabe at a R p and df(a) 0. Sow tat f(a) points in te direction of fastest increase (tat as te argest directiona derivative.) (E6.) Let S R 4 be a ocay parameterized C -dimensiona surface near b (0,, 0, ), were S {(x, y, z, w) R 4 : x + y + z + w 8, x + y + z w 0}. Sow tat te tangent pane to S at b is {(s, s t, t, + s + t ) : s, t R}. (E7.) Suppose tat f : R p R is C 3 and some point a R p is critica df(a) 0 and d f(a) as bot a positive and a negative eigenvaue. Sow tat te critica point a is a sadde point: for every δ > 0 tere are points x, y B δ (a) so tat f(x) < f(a) < f(y). (E7*.) (Variation of above.) Suppose tat f : R p R is C function suc tat some point a R p is is a critica point: df(a) 0. Assume tat te second partia derivatives exist in a neigborood of a, are continuous at a and tat te Hessian d f(a) as bot a positive and a negative eigenvaue. Sow tat te critica point a is a sadde point: for every δ > 0 tere are points x, y B δ (a) so tat f(x) < f(a) < f(y). (E8.) Let F : R 3 R 3 be given by F (x, y, z) (x + y, xz, y 3 z 3 ). Assume tat tere is an open set U about P 0 (3,, ) and tat V F (U) is an open set about Q 0 F (P 0 ) on wic F as an inverse function F C (V, U). Find D[F ](Q 0 ). (E9.) Suppose G : R 5 R 3 is given by G(p, q, x, y, z) (px + y, q z, py qz + x). Assume tat tere is an open set U around (3, ) and a C function H : U R 3 so tat H(3, ) (, 5, 4) and for a (p, q) U we ave G(p, q, H(p, q)) (8, 6, 8). Find DH(3, ). Soutions. (.)Let f : R n R n be differentiabe at a R n. Let g : R n R be defined by g(x) x f(x), (dot product.) Find te tota derivative (differentia) Dg(a)() were R n. Witout using te product teorem, prove your answer. Te product rue gives Dg(x)() D(x f(x))() g(x) + x Df(x)(). Tis is te differentia because g(x + ) g(x) Dg(x)() im 0 (x + ) f(x + ) x f(x) f(x) x Df(x)() im 0 x (f(x + ) f(x) Df(x)()) + (f(x + ) f(x)) im 0 im 0 { f(x + ) f(x) Df(x)() x x , were we ave used te Scwarz Inequaity. + } f(x + ) f(x) (a.) Statement. Let f : R R be continuous. Suppose tat for a (x, y) R bot f(x +, y) f(x, y) im 0 and f(x, y + ) f(x, y) im 0

3 exist. Ten f is differentiabe at (, ). FALSE! Te two imits are noting more tan f x (x, y) and f y (x, y), te partia derivatives. Tere are functions were te partia derivatives exist at a points, but te function is not differentiabe. (If it were nown tat f x and f y are continuous at some point a R, ten our teorem says tat te function woud be differentiabe at a.) An exampe of suc a function is { (x ) (y ) f(x, y) (x ) +(y ), if (x, y) (, ); 0. if (x, y) (, ). Away from (, ), te denominator avoids zero, so te partia derivatives exist and are continuous, ence f is differentiabe. Aso, observe tat f(, y) f(x, ) 0 for a x, y. Hence f y (, y) f x (x, ) 0 so te partia derivatives exist at (, ). If te function were differentiabe at (, ), ten te vanising of te partia derivatives impies tat te differentia woud ave to be T (, ) 0 a,. But te imit f( +, + ) f(, ) T (, ) f( +, + ) im im (,) (,) (, ) (,) (,) (, ) does not exist. To see tis, consider te first approac (, ) (t, 0) as t 0. Te numerator vanises so aong tis approac te imit woud be zero. Ten consider te second approac (, ) (t, t) for t > 0. Ten f( + t, + t) t/ and (t, t) t. Ten te difference quotient tends to /. Since te two approaces ave different imits, tere is no two dimensiona imit: te function is not differentiabe at (, ). (b.) Statement. Suppose f : R R is a C function for wic te tird partia derivatives f xxy (x, y) exist for a (x, y) R suc tat f xxy (x, y) is continuous at (0, 0). Ten f xyx (0, 0) and f yxx (0, 0) exist and are equa f xxy (0, 0) f xyx (0, 0) f yxx (0, 0). TRUE! Tis is just an appication of te equaity of cross partias teorem to f x and f y wic are C by assumption, since f is C. We are given tat (f x ) xy exists and is continuous at (0, 0). But, tis is sufficient to be abe to assert te existence of te oter mixed partia derivative, and tat it is equa to te first at te point (f x ) xy (0, 0) (f x ) yx (0, 0). But since f C (R ), we aso ave tat f xy f yx for a of R. Hence, a tird derivatives exist and are equa f xxy (0, 0) (f x ) xy (0, 0) (f x ) yx (0, 0) (f xy ) x (0, 0) (f yx ) x (0, 0) f yxx (0, 0). (3.) Suppose f, g : R R. Assume tat g is differentiabe at x 0 R and tat for some α > and M < we ave f(x) g(x) M x x 0 α for a x R. Sow tat f is differentiabe at x 0 and tat Df(x 0 ) Dg(x 0 ). It suffices to sow tat te difference quotient of f imits to zero wit differentia dg(x 0 ). Now, by adding and subtracting, using te triange inequaity, te ypotesis and α >, f(x 0 + ) f(x 0 ) Dg(x 0 )() im 0 f(x 0 + ) g(x 0 + ) f(x 0 ) + g(x 0 ) + g(x 0 + ) g(x 0 ) Dg(x 0 )() im 0 f(x 0 + ) g(x 0 + ) + g(x 0 ) f(x 0 ) + g(x 0 + ) g(x 0 ) Dg(x 0 )() im 0 { α im α + g(x } 0 + ) g(x 0 ) Dg(x 0 )() im α g(x 0 + ) g(x 0 ) Dg(x 0 )() im

4 (4*.) (Sigt generaization.) Let F : R 5 R be given by F (f, f ) were f (v, w, x, y, z) v+w +x+y, f (v, w, x, y, z) vy+wz. Sow tat tere is a neigborood U R 3 containing te point (3, 4, 5) and C functions G : U R were G (g, g ) so tat g (3, 4, 5), g (3, 4, 5) and for a (x, y, z) U, f (g (x, y, z), g (x, y, z), x, y, z), f (g (x, y, z), g (x, y, z), x, y, z) 4. Find te tota derivative DG(3, 4, 5)(, j, ). You were given te first concusion in probem (4.) wic can be answered nowing te cain rue. (4*) is an appication of te Impicit Function Teorem. (Tis teorem says tat if tere is enoug differentiabiity, and if te probem can be soved for te inear approximations given by te differentia, ten, at east in a sma neigborood, te noninear probem can be soved as we.) Te function F : R +3 R is C on R 5 suc tat F (,, 3, 4, 5) (, 4) c. To sove for v and w in terms of (x, y, z) we need to be abe to sove te inearization. If we put u (v, w) and x (x, y, z), we are ooing for G : R 3 R so tat F (G(x), x) c and G(3, 4, 5) (, ). Taing D x gives D u F (G(x), x) DG(x) + D x F (G(x), x) 0 wic says tat we may sove for te differentia DG(x) wenever D u F (G(x), x) is invertibe and ten DG(x) [D u F (G(x), x)] D x F (G(x), x). At te center point (3, 4, 5), te matrix of te transformation is D u F (G(3, 4, 5), (3, 4, 5)) f v f v f w f w w 4 y z 4 5 x(3,4,5) (u,x)(,,3,4,5) wic is invertibe. Hence tere is an open set U R 3 suc tat (3, 4, 5) U and a C function G : U R so tat G(, 3, 4) (, ) and F (G(x), x) c for a x U. Tus, we ave ceced te differentiabiity and te soubiity of te inearized probem is satisfied. Te IFT gives te existence of a G C (U, R ) so tat G(, 3, 4) (, ) and F (G(x), x) c for a x U. In tis probem, ypu were given tis as a ypotesis. Ten tae te tota derivative of F (G(x), x) c wit respect to x and sove for DG(x) at te given point, as above.) By te formua for te differentia DG(3, 4, 5) [D u F (G(3, 4, 5), (3, 4, 5))] D x F (G(3, 4, 5), (3, 4, 5)) f f v w f f f f f v w f f f x(3,4,5) x(3,4,5) w 0 y z 0 v w (u,x)(,,3,4,5) (u,x)(,,3,4,5) 4

5 DG(3, 4, 5) (E.) Suppose f : R 3 R is given by f(x, y, z) (xy + x z 3, x 4 + y + y 5 z 6 ). Is f differentiabe on R 3? If so, find te differentia df(x, y, z)(,, ). YES! Te partia derivatives are f x y z y + xz 3 4x 3, f x y z x + 5y 4 z 6, f x y z 3x z 6y 5 z 5 Since f is a poynomia function, its first partia derivatives exist at a points and are poynomia functions. But by te teorem giving conditions for differentiabiity, since te partia derivatives are continuous at a points, te function is differentiabe at a points. Te differentia is given by te 3 Jacobian matrix df x y z f f f f f f y + xz 3 x 3x z 4x 3 + 5y 4 z 6 6y 5 z 5. (E.) Find te extrema of φ(x, y, z) x + y + z subject to te constraints x y and y z. Let g(x, y, z) x y and (x, y, z) y z. Ten using Lagrange Mutipiers, te extrema occur as soutions (x, y, z, λ, µ) of te system g 0, 0, φ λ g + µ. Hence (x, y, z) λ(,, 0) + µ(0, y, z). Tus x λ, y λ + µy and z µz. From te ast equation, eiter z 0 or µ. If µ ten λ 4y. Since x λ we get x + y 0. Now g 0 impies y /3. However, 0 impies 0 /9 z wic is a contradiction. If z 0 ten 0 impies y ±. Since g 0 we ave x wen y and x 0 wen y. Tus te ony candidates for extrema subject to te constraints g 0 are φ(,, 0) 5 and φ(0,, 0). To see weter tese are maxima or minima, consider te geometric interpretation. Te probem ass to find te cosest or fartest points of te soution set g 0 to te origin. However 0 corresponds to a yperboic cyinder (union of ines parae to te x-axis tat pass troug a yperboa in te y-z pane.) Te constraint g 0 is a pane parae to te z-axis tat crosses te yperbooid. Tus te constraint set consists of two yperboas in te g 0 pane. Te eve curves on φ are circes in tis pane. As tere are ony 5

6 two candidates for extrema, tese occur wen te circes meet (are tangent to) te two arcs of te yperboa. Larger circes cross te arcs. Tis means tat (0,, 0) is a minimum and (,, 0) is a oca but not goba minimum, and tere is no maximum since one can attain arbitrariy arge φ on te constraint set. (E3.) Suppose tat r and α are positive, E R n is a convex set suc tat Ē B r (0), and tat tere exists a sequence x E suc tat x 0 as. If f : B r (0) R is continuousy differentiabe and f(x) x α for a x E, prove tat tere is an M < suc tat f(x) M x for x E. Since te partia derivatives are continuous, te function x df(x) is continuous. As Ē is cosed and bounded, it is compact so te continuous function attains its maximum M sup{ df(x) : x Ē} <. Since E is convex, for any pair of points x, x E te ine segment [x, x] is in E. Since f is differentiabe in a neigborood of [x, x], we may appy te Mean Vaue Teorem. Tere is a c [x, x] so tat f(x) f(x ) + df(c)(x x ) Estimating using triange and Scwarz inequaities, and te ypotesis f(x) f(x ) + df(c) x x x α + M x x. Now since α > 0, by passing to te imit as we obtain te estimate f(x) 0 + M x. (E4.) T eorem. f(x, y) as partia derivatives for a (x, y) R but f is not differentiabe at (0, 0), were x 3 + y 3 + 4xy f(x, y) x + y, if (x, y) (0, 0); 0, if (x, y) (0, 0). Since f is a rationa function wose denominator is nonzero away from (0, 0), te partia derivatives exist and are rationa functions tere. At (0, 0), and f f(, 0) f(0, 0) 3 0 (0, 0) im im 0 0 f f(0, ) f(0, 0) 3 0 (0, 0) im im 0 0 so bot partia derivatives exist at (0, 0) as we. Terefore, supposing tat f were differentiabe at (0, 0), its differentia woud be df(0, 0) (, ). Now ets cec if tis differentia we-approximates f near zero. im (,) (0,0) f(0 +, 0 + ) f(0, 0) df(0, 0) ( ) im + 0 ( + ) (, ) (0, 0) (,) (0,0) + im (,) (0,0) ( + ) +. Aong te pat (, ) (t, 0) te imit is zero. However, taing te pat (, ) (t, t), te imit is /(3 ). Because te imits aong two approac pats disagree, tere is no two dimensiona imit. Te function is not we approximated by te ony possibe affine function, ence it is not differentiabe at (0, 0). 6

7 (E5.) T eorem. Suppose tat f : R p R is differentiabe at a R p and df(a) 0. Ten f(a) points in te direction of fastest increase (tat as te argest directiona derivative.) Let u be a unit vector. Ten te directiona derivative in te u direction is D u f(a) f(a) u. By te Scwarz inequaity D u f(a) f(a) u f(a) wit equaity ony if u ± f(a)/ f(a). Tus D u f(a) f(a) if and ony if u f(a)/ f(a), wic is in te gradient direction. (E6.) T eorem. S R 4 is a ocay parameterized -dimensiona surface near b (0,, 0, ), were S {(x, y, z, w) R 4 : x + y + z + w 8, x + y + z w 0}. Te tangent pane to S at b is {(s, s t, t, + s + t ) : s, t R}. Te surface is te intersection of a 3-pane troug te origin and te 3-spere about te origin of radius wic is a great -spere. Let s write te surface near b as a grap over te (x, z)-pane, for x +z <. Note tat te x and z coordinates of b are in tis open neigborood, and tat te 3-pane is not perpendicuar to te x z-pane. Soving te second equation gives y w x z. Substituting x u and z v into te first and soving for w gives G : U R 4 were we tae te + square root because G(0, 0) (0,, 0, ) and U B ((0, 0)): ( G(u, v) u, u v + 6 3u uv 3v, v, u + v + ) 6 3u uv 3v G C (U, R 4 ) because 6 3(u + v ) uv > 6 3 since uv u + v <. One cecs tat G(u, v) S. Let V {(x, y, z, w) : x + z <, y > 0, w > 0}. One cecs tat b S V G(U). We need tat G : U S V is one-to-one. But if (u i, v i ) U and G((u, v )) G((u, v )) ten x- and z-coordinate functions give u u and v v so G is one-to-one. Finay we cec tat G is two dimensiona. dg((u, v)) is a 4 matrix given by dg((u, v)) 0 3u v 6 3u uv 3v u 3v 0 6 3u uv 3v 3u v 6 3u uv 3v u 3v 6 3u uv 3v 0, dg((0, 0)). 0 dg is ran two because te first and tird rows are independent, ence G(U) is a parametrized -surface. Te tangent space is b + dg((0, 0))(R ). (E7.) T eorem. Suppose tat f : R p R is C 3 and some point a R p is critica df(a) 0 and d f(a) as bot a positive and a negative eigenvaue. Ten te critica point a is a sadde point: for every δ > 0 tere are points x, y B δ (a) so tat f(x) < f(a) < f(y). By continuity of te tird partia { derivatives, a continuous function taes its maximum on a ( compact set, namey M 3 ) f(u) sup i,j, : u a }. It foows tat for any i j R p, and c B (a) tat d 3 f(c)() 3 3 f(c) i,j, i j i j M 3 by appying te Scwarz inequaity to eac sum. Now by assumption tere are v, w R p unit eigenvectors suc tat d f(a)v λ v and d f(a)w λ w wit λ < 0 < λ. Assume tat 0 < δ < is so sma tat Mδ < min{ λ, λ }. Since second derivatives are differentiabe, we can appy Tayor s 7

8 formua to second order. We get for eac 0 < t < δ some c (0, ) suc tat f(a + tv) f(a) + df(a)(tv) + t v d f(a)v + 6 t3 d 3 f(a + ctv)(v) 3 f(a) λ t v v + 6 t3 M v 3 f(a) + ( t λ + ) 3 tm f(a) + (λ t + ) 3 λ Simiary, for eac 0 < t < δ, tere is a c (0, ) so tat < f(a). f(a + tw) f(a) + λ t w w 6 t3 M w 3 f(a) + t ( λ 3 tm ) f(a) + (λ t ) 3 λ > f(a). (E7*.) T eorem. Assume tat f : R p R is C function suc tat some point a R p is is a critica point: df(a) 0. Assume tat te second partia derivatives exist in a neigborood of a, are continuous at a and tat te Hessian d f(a) as bot a positive and a negative eigenvaue at a. Sow tat te critica point a is a sadde point: for every δ > 0 tere are points x, y B δ (a) so tat f(x) < f(a) < f(y). Lemma. Suppose A and B are p p rea symmetric matrices suc tat A as bot positive and negative eigenvaues λ < 0 < λ and suc tat te operator norm A B < min{ λ, λ }. Ten te corresponding unit eigenvectors v i of A satisfy v Bv < λ and v Bv > λ. Proof of Lemma. Let v i be unit eigenvectors of A so tat Av i λ i v i for i,. Ten Simiary v Bv v Av + v (B A)v λ v + B A v λ v + λ v λ v. v Bv v Av + v (B A)v λ v B A v λ v λ v λ v. Proof of Teorem. By assumption tere are v i R p unit eigenvectors suc tat d f(a)v i λ i v i wit λ < 0 < λ. Assume tat 0 < δ < is so sma tat te second partia derivatives of f exist on B δ (a) and by continuity d f(b) d f((a) < min{ λ, λ } wenever b B δ (a). We appy te Lemma to b a + ctv i. Since first derivatives are differentiabe, we can appy Tayor s formua to first order. We get for eac 0 < t < δ some c (0, ) suc tat at b a + ctv, f(a + tv ) f(a) + df(a)(tv ) + t v d f(a + ctv )v f(a) λ t v < f(a). Simiary, for eac 0 < t < δ, tere is a c (0, ) so tat at b a + ctv, f(a + tv ) f(a) + df(a)(tv ) + t v d f(a + ctv )v f(a) λ t v > f(a). 8

9 (E8*.) (Sigt generaization.) T eorem. Let F : R 3 R 3 be given by F (x, y, z) (x + y, xz, y 3 z 3 ). Ten tere is an open set U about P 0 (3,, ) so tat F is invertibe on U and tat F (U) is an open set about Q 0 F (P 0 ) on wic F is C. We find D[F ](Q) were Q F (U). Te function F (x, y, z) poynomia, terefore C. We cec tat te determinant of te Jacobian matrix F (P 0 ) 0 and a of te concusions foow from te Inverse Function Teorem. Te fact tat te inearization was invertibe at te point enabes you to concude te existence of an inverse function. You were, owever, given tis in te probem. Let G C (V, R 3 ) be te inverse function of F. Tus in U we ave te equation F (G(x, y, z)) (x, y, z). Appy te cain rue, and sove for DG at te point. Tus D(F G) DF (P 0 ) DG(Q 0 ) I so DG(Q 0 ) (DF (P 0 )). Te matrix of DF (P 0 ) is te Jacobian matrix DF (P ) F F F 3 F F F 3 F F F 3 x y z 0 x ; DF (P 0 ) y 3z 0 3 Tus F (P 0 ) det(df x (P 0 )) 6 0. Finay, for Q near (0, 6, 7) Q 0 F (P 0 ), so DF (Q) [DF (F (Q))] were (x, y, z) F (Q). 6yz 3 6x y 3xy 3z 3 3zy 4 6yz 6xz 6xy ; DF (Q 0 ) 4 3 xy x yz

10 (E9*.) (Sigt generaization.) T eorem. Suppose G : R 5 R 3 is given by G(p, q, x, y, z) (px+ y, q z, py qz+x). Ten tere is an open set U around T 0 (3, ) and a C function H : U R 3 so tat H(3, ) (, 5, 4) X 0 and for a (p, q) U we ave G(p, q, H(p, q)) (8, 6, 8). We find DH(3, ) and DH(p, q). Te function G is poynomia so C. We ave to cec tat te inearization is soube at (3,,, 5, 4). Let T 0 (3, ) and X 0 (, 5, 4). Tis foows if te D x G part of te Jacobian matrix is invertibe. D x G G G G 3 G G G 3 G G G 3 p y q ; D x G(T 0, X 0 ) p q 3 wic is invertibe since its determinant is 4. Te Impicit Function Teorem appies and yieds te C function H as desired. (Again, you were given tat tere is H C (U) satisfying F (p, q, H(p, q)) (8, 6, 8) for a (p, q) U. Find te differentia of H by differentiating te equation using te cain rue. Tin of H : U U R 3 is given by H(p, q) (p, q, H(p, q)), and ten differentiate G H const using te cain rue. Tus D t H ( I D th) so 0 Dt (G H) D t G + D x G D t H. Here te tota derivative matrix as coumns associated to t (p, q) and coumns associated to x (x, y, z) drivatives, DG (D t G, D x G). To find te tota derivative of H we need te oter part of te Jacobian D t G G p G p G 3 p G q G q G 3 q x qz ; D t G(T 0, X 0 ) 0 6 ; DH(T 0 ) y z since tota derivative of impicit function DH(T ) [D x G(T, H(T ))] D t G(T, H(T )) pq q 0 x 0 xpq yq pq y p 0 qz xyq + y yp yq pq 0 y z xyq were q (p y) and (x, y, z) H(p, q). q 3 z p z pq z yz pq 3 z 0