The third case is all x, y, z are nonzero. Then the condition is log(x 2 + y 2 + z 2 ) = x 2 + y 2 + z 2 = 2z 2

Transcription

1 Exercise 894 (1) f x = y 3 x(a 3x y), f y = ( x y (3a 3x 4y) The places where f x = f y = 0 are (x, 0) a ad (0, y) for ay x ad y, ad 3, a ) Exercise 894 () f x = cos x si y, f y = si x cos y If f x = f y = 0, the ta x = ta y, so that (cos y, si y) = ±(cos x, si x), ad the equatio becomes 1 ± 4 cos x si x = 1 ± si x = 0 Thus x = π ± π 6, x = π ± π 1 Ad y = mπ + x = mπ + π ± π 1 Exercise 894 (3) We have f x = yz log(x + y + z ) + ( ) x yz x + y + z = yz log(x + y + z x ) + x + y + z ad similar formulae for f y ad f z There are three cases for the three partial derivatives to vaish The first case is at least two of x, y, z are zero We get (t, 0, 0), (0, t, 0), (0, 0, t) for all t 0 The secod case is oly oe of x, y, z is zero Assume x = 0 ad yz 0 The the coditio is log(y + z ) = 1 We get (0, cos t, si t) for all t By symmetry, we also get (cos t, 0, si t) ad (cos t, si t, 0) The third case is all x, y, z are ozero The the coditio is log(x + y + z ) = x x + y + z = y x + y + z = z x + y + z e We get eight poits (±1, ±1, ±1) 3 Exercise 894 (4) f x1 = x x a 1, ad the formulae for the other partial derivatives are similar If all x 1 the partial derivatives vaish, the a 1 = a = = a = x 1 x x We get x i = λa i, with x 1 x x λ = (a 1 a a ) 1 +1 i case is eve ad λ = ±(a1 a a ) 1 +1 i case is odd Exercise 894 (5) ( ) ( ) x 1 1 The equatio f x1 = 0 is the same as = 0 By f = 0 = 0, (a + x 1 )(x 1 + x ) x 1 f we fid the coditio is x 1 = ax By similar argumet for the other partial derivatives, we fid the coditio for all the partial derivatives to vaish is x 1 = ax, x = x 1 x 3, x 3 = x x 4,, x 1 = x x, x = x 1 b Sice we do ot allow both x i ad x i+1 to vaish at the same time, we have x i 0 ad x 1 a = x = x 3 = = x = b Deote the quotiet by λ The x 1 x x 1 x x 1 = λa, x = λx 1 = λ a,, x = λ a, b = λ +1 a This give us x i, with λ = ( ) 1 b +1 case is eve ad λ = ± i case is odd a Exercise 895 (1) 1 3 ( b a ) 1 +1 i

2 (x )dx + ydy + (z + 6)dz = 0 Thus the coditio is x = 0, y = 0, x + y + z x + 6z = 6 The solutios are (1, 0, 7) ad (1, 0, 1) Exercise 895 () 4(x + y + z )xdx + 4(x + y + z )ydy + 4(x + y + z )zdz = a (xdx + ydy zdz) Thus the coditio is 4(x + y + z )x = a x, 4(x + y + z )y = a y, (x + y + z ) = a (x + y z ) If x = y = 0, we get the solutio (0, 0, 0) If x = 0, y 0, we get y + z = a, (y + z ) = a (y z a ) ad the solutios (0, ± 3, ±1) If x 0, y = 0, the we similarly a get (± 3, 0, ±1) Fially, if x 0, y 0, the the solutio is x + y = 3 8 a, z = ± a Combiig all the cases, we get (0, 0, 0) ad two circles x + y = 3 8 a, z = ± a Exercise 895 (3) (3x ayz)dx + (3y axz)dy + (3z axy)dz = 0 So cosider the system 3x = ayz, 3y = axz, x 3 + y 3 + z 3 axyz = 1 From the first two, we get x 3 = y 3, or x = y If x = y = 0, 1 we get (0, 0, 1) If x = y 0, the we get 3 (a, a, 3) 33 a3 Exercise 896 (1) Sice f 0, the possible local extremes are o the boudary O the boudary part z = x + y 1, we have f = x + y + x + y, ad f x = f y = 0 implies x = y = 1, z = 1 ( The value f 1, 1, 1 ) = 1 O the boudary part x + y 1, z = 1, we have f(x, y) = x + y + 1, ad the maximum is Thus the maximum is ad the miimum is 1 Exercise 896 () f x = cos x cos(x + y) = 0, f y = cos y cos(x + y) implies x = y = π 3 or x = y = 4π The values are f = ad f = The f(x, 0) = si x, with maximum 1 ad miimum 1 Similar situatio happes for f(0, y) Fially, f(x, π x) = si π = 0 Therefore the maximum is 1 ad the miimum is 1 Exercise 897 Let φ(t) = x 0 + t u ad ψ(t) = y 0 + s v be parameterisatios of straight lies L ad K The problem is to fid whe f(s, t) = φ(t) ψ(s) = φ(t) ψ(s), φ(t) ψ(s) is miimized We have f t = t (φ(t) ψ(s)), φ(t) ψ(s) = φ (t), φ(t) ψ(s), f s = s (φ(t) ψ(s)), φ(t) ψ(s) = ψ (s), φ(t) ψ(s)

3 Whe the distace is miimized, we have f t = f s = 0 This meas u = φ (t) φ(t) ψ(s) ad v = φ (s) φ(t) ψ(s) I other words, the distace is miimized whe φ(t) ψ(s) is orthogoal to the directios u ad vof the straight lies Exercise 898 For v close to 0, apply the Mea Value Theorem to the sigle variable fuctio g(t) = f(t v) for t [0, 1] (the assumptio implies that g is cotiuous o [0, 1] ad differetiable o (0, 1)), we get f( v) f( 0) = g(1) g(0) = g (c) = f (c v)( v) = f(c v) v = 1 f(c v) (c v) c I the third equality, we use the differetiability of f at c v 0 ad the chai rule By the assumptio, the right side 0 Therefore 0 is a local miimum Exercise 899 [Exercise 894] (1) f xx = y 3 (a 6x y), f yy = x y(6a 6x 1y), f xy = xy ( (6a 9x 8y) We 1 have h f,(x,0) (u, v) = 0, h f,(0,y) (u, v) = y 3 (a y), h f,( a 3, a ) (u, v) = a4 8 u v + 1 ) 6 uv = ( ( a u + 4 u + 1 ) ) ( a 3 v Therefore 3, a is a local maximum ad the quadratic approximatio is ot strog eough to coclude about (x, 0) ad (0, y) ) By direct ispectio, we fid (x, 0) are ot local extremes, ad (0, y) is a local maximum if y(a y) < 0 ad is a local miimum if y(a y) > 0 () f xx = f yy = 4 si x si y, f xy = 4 cos x cos y At the possible local extremes, we have ta x = ta y = ± ta π 1 or ± cot π 1, ad ( ( h f (u, v) = 4 cos x cos y ta 1) π ) ± (u + v ) + uv Sice 0 < ta π 1 < 1, the quadratic form ( ta π 1) ± (u + v ) + uv is egative defiite or idefiite whe the expoet ( is or The by cosiderig the sig of cos x cos y, we fid π the local extrema are ± π 1, mπ + π 1) ± π with odd Moreover, the local extrema are local maxima whe m is eve ad local miima whe m is odd (3) We have f(t, 0, 0) = f(0, t, 0) = f(0, 0, t) = 0 Moreover, it is easy to see that f has o defiite sig ear (t, 0, 0) ad the similar poits Therefore these poits are ot local extrema We have f(0, si t, cos t) = 0, Agai f has o defiite sig ear (0, si t, cos t), so that the poit ad the similar poits are ot local extrema At e (±1, ±1, ±1), we have 1 log(x + y + z ) = x x + y + z = y x + y + z = z x + y + z = 1 3,

4 ad f xx = xyz(x y z ) = e 1 3 (x + y + z ) 9 3 (±1) x(±1) y (±1) z, ( ) f xy = z log(x + y + z ) + (x4 + y 4 + x z + y z ) (x + y + z ) ad similar formulae for the other secod order partial derivatives The = e (±1) z h f (u, v, w) = e (±1) x(±1) y (±1) z [(±1) x u + (±1) y v + (±1) z w] Thus the hessia is ot sufficiet for us to coclude the local extreme (4) At the possible local extremes, we have a 1 = a = = a = x 1 x x = λ ad x 1 x x f x1 x 1 = a 1 x 3 1 Thus the hessia h f (v 1, v,, v ) = 1 λ 3 ( x1 to coclude local extremes = 1, f λ 3 a x1 x = x 3 x = 1 1 λ 3 a 1 a + x + + x ) This is ot sufficiet for us a 1 a [Exercise 895] (1) From (x 1) + (z + 3)z x = 0, y + (z + 3)z y = 0, we get 1 + zx + (z + 3)z xx = 0, 1 + zy + (z + 3)z yy = 0, z y z x + (z + 3)z xy = 0 At the possible local extremes (1, 0, 7) ad (1, 0, 1), we have z x = z y = 0 Thus we get z xx = z yy = 1 z + 3, z xy = 0 at the two poits At the first poit, z + 3 < 0 ad h z is positive defiite At the secod poit, z + 3 > 0 ad h z is egative defiite Thus the first poit is local miimum ad the secod is local maximum () By (x +y +z )(x+zz x ) = a (x zz x ), we get 4(x+zz x ) +(x +y +z )(1+zx+zz xx ) = a (1 zx zz xx ) ad 4(y + zz y )(x + zz x ) + (x + y + z )(z x z y + zz xy ) = a ( z x z y zz xy ) We also have the similar equality for z yy At the possible local extremes, z xx, z yy, z xy ca be solved by makig use of z x = z y = 0 (0, 0, 0) is ot a local extreme because if z is a solutio, the z is also a solutio Thus z has o defiite sig ear 0 I fact the implicit fuctio theorem fails at (0, 0, 0) Whe x + y = 3 8 a, z = ± a, we have 4x + a (1 + zz xx) = a (1 zz xx ), 4y + a (1 + zz yy ) = a (1 zz yy ), 4xy + a zz xy = a zz xy Thus By h(u, v) = 1 3a z [(a 8x )u + (a 8y )v 16xyuv] (a 8x )(a 8y ) (8xy) = a 4 8(x + y )a = a 4 < 0, a

5 the hessia is ot defiite Therefore the poits are ot local extremes (3) By (3x ayz)+(3z axy)z x = 0, we get (6x ayz x )+(6zz x ay)z x +(3z axy)z xx = 0 ad ( az az y ) + (6zz y ax)z x + (3z axy)z xy = 0 We also have the similar equality for z yy At the possible local extremes, z xx, z yy, z xy ca be solved by makig use of z x = z y = 0 We get 1 h(u, v) = 3z axy ( 6xu 6yv azuv) At (0, 0, 1), the hessia is idefiite, so that the poit is ot a local extreme At we have h(u, v) = 6a 3 33 a 3 (u + v + ) a uv (a, a, 3), 33 a3 This is defiite if ad oly if a 0 ad <, which meas a > or 33 a a < The coclusio is that the poit is a local miimum if a > or 0 < a < , ad is a local maximum if a < 0 Exercise 8100 The hessia at (x 0, y 0 ) is h f (u, v) = f xx (x 0, y 0 )u + f xy (x 0, y 0 )uv + f yy (x 0, y 0 )v If f xx 0, the ) v h f = f xx ( u + f xy = f xx ( u + f xy f xx v uv + f xy f xx fxx ) + f xxf yy fxy v f xx f xy f xx v + f yy (x 0, y 0 )v Therefore f xx > 0 ad f xx f yy f xy > 0 imply that the coefficiets f xx > 0 ad f xxf yy f xy f xx > 0 The hessia is positive defiite ad (x 0, y 0 ) is a local miimum The discussio for the other cases is similar Exercise 8101 We have x 3x y 3 (x y ) if(x, y) (0, 0) f x = (x + y ) 4 0 if(x, y) = (0, 0) The partial derivative f x (0, 0) = 0 follows from f(x, 0) = x The by f x (x, 0) = x ad f x (0, y) = 0, we get f xx (0, 0) = ad f yx (0, 0) = 0 Similarly, we have f y (0, 0) = 0 ad f yy (0, 0) = ad f xy (0, 0) = 0 Therefore f(0, 0) = 0 ad h f (u, v) = u + v is positive defiite

6 O the other had, we have f(x, x) = x ad f(x, x) = x 1 Therefore for small 8 x 0, we have f(x, x) > f(0, 0) > f(x, x) I particular, (0, 0) is ot a local extreme The c 3 fact lim y=cx f(x, cx) = also tells us that f is ot cotiuous at (0, 0) (1 + c ) 3 For a example differetiable at (0, 0), we may try x + y x 3 y if(x, y) (0, 0) f(x, y) = (x + y ) 0 if(x, y) = (0, 0) The differetiability at (0, 0) follows from f(x, y) x + y + 5(x + y ) 1 (usig xy 3 x + y ), which shows that 0 is the liear approximatio of f at (0, 0), so that f(0, 0) = 0 We have x + y 0 x y 3 (4x 3y ) if(x, y) (0, 0) f x = (x + y ) 0 if(x, y) = (0, 0) The by f x (x, 0) = x ad f x (0, y) = 0, we get f xx (0, 0) = ad f yx (0, 0) = 0 Similarly, we have f y (0, 0) = 0 ad f yy (0, 0) = ad f xy (0, 0) = 0 Therefore f(0, 0) = 0 ad h f (u, v) = u + v is positive defiite By f(x, x) = ( (0, 0) is ot a local extreme ) x > f(0, 0) ad f(x, x) = ( 0 ) x < f(0, 0), we see that 8 Exercise 810 The cotiuity assumptio implies that the fuctio is third order differetiable The vaishig of the first ad secod order partial derivatives shows that the third order approximatio is of the form f( x 0 ) + m( x), where m( x) is a third order form Thus for ay ɛ > 0, there is δ > 0, such that x < δ = f( x) f( x 0 ) m( x) < ɛ x 3 If some third order derivative is ozero, the m is ot completely zero Therefore there is some vector v of uit legth, such that m( v) 0 Assume m( v) > 0 ad take ɛ = 1 m( v) The there is δ > 0, such that (ote that m(t v) = t 3 m( v)) t < δ = t v = t < δ = f( x 0 + t v) f( x 0 ) t 3 m( v) < 1 m( v)t3 Thus for small t > 0, we get f( x 0 + t v) f( x 0 ) > t 3 m( v) 1 t3 m( v) = 1 t3 m( v) > 0, ad f( x 0 t v) f( x 0 ) < ( t) 3 m( v) + 1 ( t)3 m( v) = 1 t3 m( v) < 0

7 Thus we coclude that x 0 is ot a local extreme The argumet for the case m( v) < 0 is similar (or use m( v) = m( v) to reduce to the positive case) Exercise 8103 Let ξ = x k ad η = y k The the equatios become ξ z + si(η + w) = 0, e ξ w + η z = 1 The liear approximatios of the equatios at (ξ, η, z, w) = (0 k, 0 k, 0, 0) = (0, 0, 0, 0) are dξ dz + dη + dw = 0, dξ dw + dη dz = 0 This has solutio dz = dξ + dη, dw = 0 By implicit fuctio theorem, ear (ξ, η, z, w) = (0, 0, 0, 0), z ad w are cotiuously differetiable fuctios of ξ ad η Moreover, ear (ξ, η) = (0, 0), we have approximatios z = ξ + η + o( ξ, η), w = o( ξ, η) Substitutig ξ = x k ad η = y k, we see that z ad w are cotiuously differetiable fuctios of x ad y ear (0, 0), ad (ote that ξ = ξ = x k = x k ad similarly for η) z = x k + y k + o( x k, y k ), w = o( x k, y k ) The approximatio of z shows that if k is odd, the (0, 0) is ot a local extreme of z(x, y), ad if k is eve, the (0, 0) is a local miimum of z(x, y) Exercise 8104 (1) x p y q z r = x p y q z r ( p x, q y, r z the gradiets to be parallel is p x = q y = r z ), (x + y + z) = (1, 1, 1) At places xyz 0, the coditio for If p x = q y = r a, the combied with x + y + z = a, we get (x, y, z) = z (p, q, r) p + q + r Exercise 8104 () si x si y si z = (cos x si y si z, si x cos y si z, si x si y cos z), (x + y + z) = (1, 1, 1) The coditio for the gradiets to be parallel is cos x si y si z = si x cos y si z = si x si y cos z Suppose cos x si y si z = si x cos y si z = si x si y cos z = 0 The either two of si x, si y, si z are zero, or cos x = cos y = cos z = 0 I case si x = si y = 0, by x + y + z = π, we get x = mπ, y = π, z = 1 π (m + )π for some itegers m ad The other similar cases give us x = mπ, y = 1 π (m + )π, z = π or x = 1 π (m + )π, y = mπ, z = π I case cos x = cos y = cos z = 0, by x + y + z = π, we get x = 1 π + mπ, y = 1 π + π, z = 1 π (m + )π for some itegers m ad Put together, we get (x, y, z) = (m,, m )π + v,

8 where v is oe of the followig ( 0, 0, 1 ) π, (0, 1, 0 ) π, ( ) 1, 0, 0 π, ( 1, 1 ), 1 π Suppose cos x si y si z = si x cos y si z = si x si y cos z 0 The we get ta x = ta y = ta z, which tells us y = x + mπ, z = x + π for some itegers m ad Combied with x + y + z = π, we get Exercise 8104 (3) (x, y, z) = 1 ((1 m )π, (1 + 4m )π, (1 m + 4)π) 6 (x 1 + x + + x ) = (1, 1,, 1), x 1 x x = x 1 x x ( 1 x 1, 1 x,, 1 x ) The two gradiets are parallel if x 1 = x = = x or two of x 1, x,, x are zero Combied with x 1 x x = a, we get x = a(1, 1,, 1) i the first case I the secod case, we have a = 0, which is ot a regular value of x 1 x x The theory of lagrage multiplier does ot apply Exercise 8104 (4) x 1 x x = x 1 x x ( 1 x 1, 1 x,, 1 x ), (x 1 + x + + x ) = (1, 1,, 1) The two gradiets are parallel if x 1 = x = = x or two of x 1, x,, x are zero Combied with x 1 + x + + x = a, we get x = a (1, 1,, 1) i the first case I the secod case, we get x = (x 1, x,, x 3, a x 1 x x 3, 0, 0) ad the similar vectors obtaied by permutig the coordiates Exercise 8104 (5) ( 1 x 1 x x = x 1 x x, 1,, 1 ), (x 1 + x + + x ) = (1, 1,, 1), (x 1 + x 1 x x x + + x x ) = (x 1, x,, x ) Sice the matrix 1 x x is ivertible if ad oly if x 1 x x 3 two of x 1, x, x 3 are equal We see that the first gradiet is a liear combiatio of the other two if ad oly if all except oe of x 1, x,, x are the same Without loss of geerality, assume x = x 3 = = x The the two costraits become x 1 +( 1)x = a ad x 1 +( 1)x = b, we coclude x 1 = 4a 4a ( )a 1 1 (a b), x = x 3 = = x = a ± 1 1 (a b)

9 Exercise 8104 (6) (x p 1+x p + +x p ) = p(x p 1 1, x p 1,, x p 1 ), (a 1 x 1 +a x + +a x ) = (a 1, a,, a ) The (x p 1 + x p + + x p ) = pλ (a 1 x 1 + a x + + a x ) meas x i = (λa i ) 1 p 1 ( Substitutig p p p ) p 1 p 1 p 1 this ito the costrait, we get a1 + a + + a λ 1 p 1 = b This gives x i = p ba 1 p 1 i p p 1 p 1 a1 + a + + a p p 1 Exercise 8105 (1) Exercise 8104(3) tells us that the oly cadidate for the local extreme of f = x 1 +x + +x subject to the costrait g = x 1 x x = a is v = a(1, 1,, 1) Fix a big umber R The costrait surface S = { x: x 1 x x = a, x i 0} is divided ito two parts S 1 = { x: R 1 < x i < R, x 1 x x = a}, S = { x: some x i R or some 0 < x i R 1, x 1 x x = a} The subset K = S 1 = { x: R 1 x i R, x 1 x x = a} is oempty ad compact Therefore f reaches miimum mi K f o K For big eough R, x S will imply f( x) > f( v) = a Therefore mi K f = mi S1 f is the miimum of f o the whole S Applyig the lagrage multiplier method to f o S 1, we get oly oe cadidate v Therefore f( x) f( v) = a for ay x S This proves that x 1 + x + + x a for ay x satisfyig x 1 x x = a Exercise 8105 () Exercise 8104(6) tells us that the oly cadidate for the local extreme of f = x p 1+x p + +x p subject to the costrait g = x 1 + x + + x = b, x i > 0, is v = b (1, 1,, 1) At the poit we have f( v) = bp p 1 The subset S = { x: x 1 + x + + x = b, x i 0} ca be divided ito pieces S k = { x: x 1 + x + + x k = b, x 1 > 0,, x k > 0, x k+1 = 0,, x = 0} ad the similar pieces obtaied by permutig the idices The lagrage multiplier method may be applied to S k by cosiderig the fuctios of x 1, x,, x k oly The we fid f = bp k p 1 at the possible local extreme o S k (ad the similar permutatio pieces) b p Therefore we coclude that all the possible local extreme values of f i S are k, 1 p 1 b p k By p > 1, amog these values, the miimum is Therefore we coclude that p 1 x 1 + x + + x = b, x i 0, p > 1 = x p 1 + x p + + x p bp p 1 This is the same as xp 1 + x p + + x p ( ) p x1 + x + + x

10 For 0 < p < 1, we have 1 1 p x1 + x 1 p + + x 1 ( ) 1 p x1 + x + + x p > 1 The we have p Replacig x i by x p i ad takig the p-th power, we get xp 1 + x p ( ) + + x p p x1 + x + + x Aother way to derive the iequality is to observe that amog the possible local extreme values b p b p, 1 k, the value is the maximum whe 0 < p < 1 kp 1 p 1 Exercise 8106 Let 1 p + 1 q = 1 ad a i > 0 The p = q Exercise 635(6) tells us that the oly q 1 cadidate for the local extreme of f = x q 1 + x q + + x q subject to the costrait g = a 1 x 1 + a x + + a x = 1, x i > 0, is v = (a 1 q 1 1, a 1 q 1,, a 1 q 1 ) a p 1 + a p + + a p At the poit we have f( v) = ap 1 + a p + + a p (a p 1 + a p + + a p ) = 1 q (a p 1 + a p + + a p ) q p Similar to the discussio of Exercise 8105(), the costrait g = a 1 x 1 +a x + +a x = 1, x i 0 ca be divided ito several pieces accordig to which of x i are > 0 ad which are = 0 O each piece, we may apply lagrage multiplier method to the fuctios of the coordiates 1 that are ozero The result is the potetial local extreme Amog (a p i 1 + a p i + + a p i k ) q p 1 all the possible local extremes over all the pieces, the miimum value is (a p 1 + a p + + a p ) q p Therefore we coclude that a 1 x 1 + a x + + a x = 1, x i 0 = x q 1 + x q + + x q By takig x i = which is Hölder iequality b i a 1 b 1 + a b + + a b, this is the same as the iequality b q 1 + b q + + b q (a 1b 1 + a b + + a b ) q, (a p 1 + a p + + a p ) q p 1 (a p 1 + a p + + a p ) q p Exercise 8107 The problem is to fid local extremes of q( x) = A x x subject to the costrait g( x) = x = x x = 1 By q( v) u = d dt A( v + t u) ( v + t u) = A v u + v A u = A v u, t=0 we have q( v) = A v Similarly, we have g( v) = v Thus by lagrage multiplier method, we have A v = λ v at the costraied local extreme v The coditio meas exactly that v is a eigevector of A with eigevalue λ

11 Exercise 8108 Suppose the covex polyhedro is a triagle with side legth a, b, c Let x, y, z be the legth of a poit to the three sides (the sigs of x, y, z idicate the locatio of the poit with respect to the triagle) The g(x, y, z) = 1 (ax+by+cz) = A is the area of the triagle Let the height be h The the total area of the side faces is f(x, y, z) = 1 (a x + h + b y + h + c z + h ) The problem is to fid the miimum of f uder the costrait g = A By f = λ g, we get ax a + x = λa, by b + y = λb, cz c + z = λc From this we get x a = y = z λ b c = µ = Note that x, y, z should have the same 1 λ sig (the same as λ), ad all of them beig egative is geometrically impossible Therefore we have x, y, z 0 ad x = µa, y = µb, z = µc Substitutig ito the costrait g = A, we get µ (a + b + c ) = A We get the oly potetial local extreme (x 0, y 0, z 0 ) = A (a, b, c) a + b + c Sice f is very big whe oe of x, y, z are very big, we coclude that the local extreme is a global miimum Exercise 8109 The problem is to fid the absolute extremes of f(x, y, z) = x + y + z subject to the costraits g 1 = x+y z = 0 ad g = x +y +z xy yz zx = 1 The lagrage multiplier coditio f = λ 1 g 1 + λ g meas x = λ 1 + λ (x y z), y = λ 1 + λ (y z x), z = λ 1 + λ (z x y) By g 1 = 0, we get z = x + y Substitutig ito above, we get x = λ 1 + λ (x y), y = λ 1 + λ (y x), (x + y) = λ 1 + λ (x + y) Addig all three together, we get λ 1 = 4(x+y) Therefore the third equatio becomes 6(x+y) = λ (x + y) If λ = 6, the substitutig ito the first two equatios ad elimiatig λ 1, we get (x y) = 6(3x 3y) Therefore x = y, z = x Substitutig ito g = 1, we get x = 1 Thus we have two potetial local extremes (1, 1, ) ad ( 1, 1, ), where f = 6 If λ 6, the x = y ad z = 0 Substitutig ito g = 1, we get 3x = 1 Thus we have 1 two potetial local extremes (1, 1, 0) ad 1 ( 1, 1, 0), where f = We coclude the log axis is 6 ad the short axis is 3

12 Exercise 8110 (1) At the potetial local extreme (x 0, y 0, z 0 ) = y + z), with λ = p + q + r x p a 0y0z q 0 r The at (x 0, y 0, z 0 ), we cosider h(u, v, w) = h x p y q z r(u, v, w) λh x+y+z(u, v, w) = = p(p 1) u q(q 1) + v + x 0 y 0 ( p x 0 u + q y 0 v + r z 0 w ) p a p + q + r (p, q, r), we have xp y q z r = λ (x + r(r 1) w + pq uv + pr uw + qr vw z0 x 0 y 0 x 0 z 0 y 0 z 0 u q v r w x 0 y0 z0 By p x 0 = q y 0 = r z 0, the restrictio h to the costrait plae g (x 0, y 0, z 0 )(u, v, w) = u+v +w = 0 is h(u, v, w) = p u q v r (u + v) x 0 y0 z0 Sice this is egative defiite, we coclude that (x 0, y 0, z 0 ) is a local maximum Exercise 8110 ( () At v = m,, m + 1 ) π, we have ( π ) f = ( 1) m si x( 1) si y( 1) m si + z = ( x + o( x))( y + o( y))(1 + o( z )) = x y + o( x ) Therefore h f (u, v, w) = uv Sice h f λh g = h ( f = uv is idefiite for (u, v, w) satisfyig g (u, v, w) = u + v + w = 0, we coclude that m,, m + 1 ) π is ot local extreme Similarly, (m, + 1 ), m π ad (m + 1 ),, m π are ot local extremes ( At v = m + 1, + 1, m 1 ) π, we have ( π ) f = ( 1) m si + x = (1 x + o( x ) ( π ) ( 1) si + y ) (1 y = 1 + x + y + z + o( x ) ( 1) m si ( π ) + z ) ) + o( y ) (1 z + o( z ) Therefore h f (u, v, w) = u + v + w, ad ( h f λh g = h f = u + v + w is positive defiite eve for ucostraied (u, v, w) Therefore m + 1, + 1, m 1 ) is a local miimum

13 At v = π (1 m, 1 + 4m, 1 m + 4), we have 6 f = si(x 0 + x)( 1) m si(x 0 + y)( 1) si(x 0 + z) = ( 1) m+ 1 6 si3 x 0 ( x + y + z ) + ( 1) m+ 1 4 si x 0 cos x 0 ( x y + y z + z x) + o( x ) Sice h g = 0, uder the costrait u + v + w = 0, we have h f λh g = ( 1) ( m+ 1 3 si3 x 0 (u + v + w ) + 1 ) si x 0 cos x 0 (uv + vw + wu) = ( 1) ( m+ 1 3 si3 x 0 (u + v + w ) 1 ) 4 si x 0 cos x 0 (u + v + w ) = ( 1) ( m+ 1 3 si3 x 0 (u + v + w ) 1 ) 4 si x 0 cos x 0 (u + v + w ) = ( 1) m si x 0(3 + si x 0 )(u + v + w ) Therefore uder the costrait u + v + w = 0, the sig of h f λh g is completely determied by the sig of ( 1) m++1 si x 0 The sig is positive whe m + =, 5 mod 6, whe v is a local miimum, ad is egative whe m + = 0, 1, 3, 4 mod 6, whe v is a local maximum Exercise 8110 (3) At the potetial local extreme x 0 = a(1, 1,, 1), we have λ = a 1 1, ad the liear costrait is v 1 + v + + v = 0 By x 1 x x = ( a + x 1 )( a + x ) ( a + x ) = + a 1 x i x j + o( x ), we have h x1 x x = a 1 i<j v iv j Therefore uder the costrait, h = h x1 +x + +x λh x1 x x = λa 1 v i v j = λa 1 (v 1 + v + + v) Sice this is positive defiite, we coclude that x 0 is a local miimum i<j i<j

14 Exercise 8111 A chage of variable is orthogoal if the colums of the Jacobia matrix x are orthogoal y The Jacobia matrix of the iverse trasform is y ( ) 1 x x = y Deote r x y = ( ) ( ) 0 r 0 r 1 u 1 r u r u = UD, U = u1 u u, D = 0 0 r where r i = x yi ad u 1 = x y i The u 1, u,, u form a orthoormal basis, ad U is x yi a orthogoal matrix Therefore y x = (UD) 1 = D 1 U 1 = D 1 U T = r r r 1 The (i, j)-etry of the matrix is y i = r x j i = 1 x j x j y i x yi y i u T 1 u T = u T r1 1 u T 1 r 1 u T r 1 u T r r 1 x T y 1 = x T y r x T y Exercise 811 From the formula for y i Exercise 8111, the rows of the matrix are orthogoal, ad the x colums are ot ecessarily orthogoal Therefore the iverse may ot be a orthogoal chage of variable Exercise 8113 Fix y,, y ad cosider F ( y) as a curve parametrized by y 1 The f is the rate of y 1 chage of f alog the curve Sice x y1 is the taget vector of the curve, we have f = f x y1 y 1 By the similar reaso, we get f = f x yi By orthogoality, we also have y i ( f x y1 y 1 x y1 + f y x y x y + + f y x y x y ) x yi = f y i x yi x yi x yi = f y i = f x yi Sice the vectors x yi form a basis of the Euclidea space, we coclude that x y1 f y 1 x y1 + f y x y x y + + f y x y x y = f Exercise 8114

15 Icludig the coefficiets i the otatio, we deote the polyomial by p( σ, x) Cosider the map P ( σ, x) = (p( σ, x 1 ), p( σ, x ),, p( σ, x )): R R The the Vieta s formulae is the solutio to P ( σ, x) = 0 The differetial of the map P is dp = P σ d σ+p x d x The matrix of P σ is the Vadermot matrix x 1 1 x 1 1 p x 1 x 1 ( ) (x 1 ) 0 0 p( σ, The matrix of P xi ) 0 p (x ) 0 x is = x j, x 1 x p (x ) where p (x) = x 1 + ( 1)σ 1 x + ( )σ x σ The dp = 0 gives us the equatio x 1 1 x 1 1 p (x 1 ) 0 0 x 1 x 1 σ x + 0 p (x ) 0 = O x p (x ) x 1 for the matrix σ x Suppose the roots x i are distict The p (x i ) 0 ad the Vadermot matrix is ivertible Therefore the matrix σ is ivertible By the iverse fuctio theorem, the roots x is locally x a cotiuously differetiable fuctio of the coefficiets σ Exercise 8115 Usig the otatio i Exercise 8114, the problem is whether p( σ, x) = 0 implicitly defied x as a fuctio of σ ear x 0 By implicit fuctio theorem, this is the case whe p x ( σ 0, x 0 ) = p 0(x 0 ) 0 the coditio meas that x 0 is ot a multiple root Exercise 8116 We have u 0,k = x k j = s k Moreover, we have σ l s k l = i 1 <i < <i l x i1 x i x il j x k l j = j=i p, for some p + j i p, for all p where The j=i p, for some p = u l 1,k l+1, j i p, for all p = { u l,k l if k l (l + 1)σ l+1 if k l = 1 s k σ 1 s k 1 + σ s k + ( 1) k 1 σ k 1 s 1 =u 0,k (u 0,k + u 1,k 1 ) + (u 1,k 1 + u,k ) + + ( 1) k (u k 3,3 + u k, ) + ( 1) k 1 (u k, + kσ k ) =( 1) k 1 kσ k Exercise 8117

16 By Newto s idetities, we have s 1 = σ 1, s = σ 1 s 1 σ = σ1 σ, s 3 = σ 1 s σ s 1 + 3σ 3 = σ 1 (σ1 σ ) σ σ 1 + 3σ 3 = σ1 3 3σ 1 σ + 3σ 3, s 4 = σ 1 s 3 σ s + σ 3 s 1 4σ 4 = σ1 4 4σ1σ + σ + 4σ 1 σ 3 4σ 4, We also have σ 1 = s 1, σ = 1 (s σ 1 s 1 ) = 1 ( s + s 1), σ 3 = 1 3 (s 3 σ 1 s + σ s 1 ) = 1 3 (s 3 s 1 s + 1 ( s + s 1)s 1 ) = 1 6 (s 3 3s 1 s + s 3 1), σ 4 = 1 4 (s 4 σ 1 s 3 + σ s σ 3 s 1 ) = 1 4 ( 6s 4 + 8s 1 s 3 + 3s 6s 1s + s 4 1), I geeral it is easy to use iductio to prove that σ k is a polyomial of s 1, s,, s k, ad s k is a polyomial of σ 1, σ,, σ k I particular, σ ad s are related by ivertible maps Therefore x σ is locally ivertible if ad oly if x s is locally ivertible The Jacobia matrix of the later map is x 1 x x s x = x 1 1 x 1 x 1 x 1, x,, x are distict The coditio for the matrix to be ivertible is that Exercise 8118 Suppose f(x) = h(g(x)) The g(x) = g(x ) implies f(x) = f(x ) Coversely, suppose g(x) = g(x ) implies f(x) = f(x ) The we defie h as follows The domai of h is the rage Y of g If y Y lies i the image g(x) Y, the we fid x X satisfyig g(x) = y ad the defie h(y) = f(x) If there is aother x X satisfyig g(x ) = y, the we have g(x) = g(x ) Sice this implies f(x) = f(x ), we fid that h(y) is well defied, at leat o the image of g If y g(x), the we may assig ay value to h(y) This completes the costructio of h The way h is costructed shows that f(x) = h(g(x)) We coclude that the if ad oly if coditio is g(x) = g(x ) implyig f(x) = f(x ) Exercise 8119 Suppose f 1 ( x) = g(f ( x),, f ( x)) The f 1 = g y1 f + g y f g y 1 f Similarly, if f i is a fuctio of f 1, f,, f i 1, f i+1,, f, the f i is a liear combiatio

17 of f 1, f,, f i 1, f i+1,, f Therefore the fuctioal depedecy implies that oe gradiet is a liear combiatio of the other gradiets This is the same as all gradiets to be liearly depedet Exercise 810 If h( y 0 ) 0, the some partial derivative of h is ozero Suppose h x 1 ( x 0 ) 0 By the implicit fuctio theorem, there is a cotiuously differetiable fuctio g, such that h( y) = 0 for y ear y 0 is equivalet to y 1 = g(y,, y ) The h(f 1 ( x), f ( x),, f ( x)) = 0 implies that f 1 ( x) = g(f ( x),, f ( x)) = 0 for x ear x 0 Therefore f 1, f,, f are fuctioally depedet The same argumet works if other partial derivatives of h are ozero Exercise 811 Suppose g ( x 0 ) 0 The the Jacobia of the map (x 1, x,, x ) (g, x,, x ) is ivertible Therefore the map is ivertible ear x 0, ad we have f(x 1, x,, x ) = H(g( x), x,, x ) x 1 for a fuctio H = H( y) We have Therefore f = H g, x 1 y 1 x 1 f = H g + H, i x i y 1 x i y i f = H y 1 g + (0, H y, H y3,, H y ) O the other had, the gradiets are assumed to be depedet, so that f = c g for some c Sice the first coordiate g x 1 of g is assumed to be ozero ear x 0, the compariso of the two equalities tells us c = H y 1 ad (0, H y, H y3,, H y ) = 0 Therefore H depeds o y 1 oly: H(y 1, y,, y ) = h(y 1 ), ad f( x) = h(g( x)) ear x 0 Exercise 81 The exercise exteds Exercise 810 from oe g to k fuctios The similar proof ca be applied Sice the gradiets of g 1, g,, g k are liearly idepedet at x 0, we may assume (by rearragig orders of x i if ecessary) that (g 1, g,, g k ) (x 1, x,, x k ) ( x 0) is ivertible The by applyig the iverse fuctio theorem, the map x y = (g 1 ( x), g ( x),, g k ( x), x k+1,, x ) is ivertible, ad we have f(x 1, x,, x ) = H(g 1 ( x), g ( x),, g k ( x), x k+1,, x ) for a fuctio H = H( y) We have f = H y 1 g 1 + H y g + + H y k g k + (0,, 0, H yk+1,, H y )

18 O the other had, the depedece assumptio o the gradiets tells us f = c 1 g 1 + c g + + c k g k Sice the first k coordiates of g 1, g,, g k form a ivertible matrix (g 1, g,, g k ) (x 1, x,, x k ) ear x 0, the compariso of the two equalities tells us c i = H y i ad (0,, 0, H yk+1,, H y ) = 0 Therefore H depeds o y 1, y,, y k oly: H(y 1, y,, y ) = h(y 1, y,, y k ), ad f( x) = h(g 1 ( x), g ( x),, g k ( x)) ear x 0 Exercise 813 Sice the rak of the gradiet vectors f 1, f,, f m is always k ear x 0, we ca fid k fuctios, say f 1, f,, f k, such that their gradiets are liearly idepedet ear x 0 Moreover, sice the rak of f 1, f,, f m is also k, we see that f k+1,, f m are liear combiatios of f 1, f,, f k For each k < i m, take f i, f 1, f,, f k to be f, g 1, g,, g k i Exercise 66 The we coclude f i is a fuctio of f 1, f,, f k ear x 0 Thus the m k fuctios f k+1,, f m are fuctioally depedet o the k fuctios f 1, f,, f k Exercise 814 (f, g, h) Deote the three fuctios by f, g, h We have (x, y, z) = ( f g h ) T (f, g, h) (1) (x, y, z) = x y z has rak < 3 (ie, ot ivertible) if ad oly if two of 3x 3y 3z x, y, z are equal Therefore the matrix caot have rak < 3 o ay ball The fuctios are fuctioally idepedet (f, g, h) () (x, y, z) = always has rak = Therefore the fuctios are x y z z y fuctioally idepedet I fact, we have f + g = h (3) (f, g, h) (x, y, z) = 1 (x + y + z ) rak = 3 (the determiat is x + y + z xy xz xy x y + z yz always has xz yz x + y z 1 ) Therefore the fuctios are fuctioally ide- (x + y + z ) 4 pedet (4) The fuctios are depedet I fact, f + g + h = 1 Exercise 815 The iequality ca be proved by usig exactly the argumet for the sigle variable case Exercise 816 The defiitio of covexity cocers oly about the restrictio of the fuctio the straight lie coectig x ad y Therefore the covexity depeds oly o the covexity o straight lies

19 Exercise 817 Suppose f(x) is covex ad f( x) L( x), f( y) L( y) The by the covexity assumptio, we have f((1 λ) x + λ y) (1 λ)f( x) + λf( y) (1 λ)l( x) + λl( y) L((1 λ) x + λ y) Therefore the combiatio (1 λ) x + λ y is still i the set This meas the set is covex Coversely, for ay x ad y, costruct ay liear fuctio L, such that f( x) = L( x), f( y) = L( y), the x ad y are iside the set By the covexity of the set, for ay 0 < λ < 1, (1 λ) x + λ y is also i the set Therefore f((1 λ) x + λ y) L((1 λ) x + λ y) = (1 λ)l( x) + λf( y) = (1 λ)f( x) + λf( y) The covexity coditio is verified Exercise 818 Sice the restrictio of the covex fuctio o ay straight lie is covex, the oe-side directioal derivative D( v) = d dt t=0 + f( z + t v) = lim t 0 + f( z + t v) f( z) t exists at z By the proof i the sigle variable case, if we ca fid a liear fuctioal k satisfyig k( v) D( v) for ay v, the the liear fuctio K( x) = f( z) + k( x z) satisfies K( z) = f( z) ad K( x) f( x) for ay x The derivative satisfies D(c v) = cd( v) for c > 0 Moreover, by f( z + t((1 λ) u + λ v)) (1 λ)f( z + t u) + λf( z + t v), we get D((1 λ) u + λ v) (1 λ)d( u) + λd( v) λd( v) = D(λ v), the iequality is the same as Sice (1 λ)d( u) = D((1 λ) u) ad D( u + v)) D( u) + D( v) It remais to prove that if a fuctio D satisfies D(c v) = cd( v) for c > 0 ad D( u + v)) D( u) + D( v), the there is a liear fuctioal k satisfyig k( v) D( v) for ay v Assume the a liear fuctioal k has bee foud for the subspace R m = R m 0 R Im other words, k is a liear fuctioal o R m satisfyig k( v) D( v) for ay v R m The for ay u, v R m ad the (m + 1)-st stad basis vector e = e i+1, we have Therefore k( u) + k( v) = k( u + v) D( u + v) D( u + e) + D( v e) D( u + e) k( u) (D( v e) k( v)) Sice u, v are arbitrary, we ca fid a umber a satisfyig D( u + e) k( u) a (D( v e) k( v))

20 for ay u, v R m The by D(c v) = cd( v) for c > 0 ad the liearity of k, we see that D( u + c e) k( u) + ac, D( v c e) k( v) ac for ay u, v R m Therefore k ( u + c e) = k( u) + ac is a liear fuctioal o R m+1 extedig k ad satisfyig D( w) k( w) for ay w R m+1 Exercise 819 We prove the cotiuity of a covex fuctio f at x 0 Let v 1, v,, v k be fiitely may vectors such that the covex hull B = {λ 1 v 1 + λ v + + λ k v k : 0 λ i 1, λ 1 + λ + + λ k = 1} cotais the origi as a iterior poit By the cotiuity of oe-variable covex fuctios, for ay ɛ > 0, there is δ > 0, such that f( x 0 ± δ v i ) f( x 0 ) < ɛ Now a vector x x 0 + δb is of the form x = x 0 + δ(λ 1 v 1 + λ v + + λ k v k ) = λ 1 ( x 0 + δ v 1 ) + λ ( x 0 + δ v ) + + λ k ( x 0 + δ v k ) By Jese s iequality, we get f( x) λ 1 f( x 0 + δ v 1 ) + λ f( x 0 + δ v ) + + λ k f( x 0 + δ v k ) λ 1 (f( x 0 ) + ɛ) + λ (f( x 0 ) + ɛ) + + λ k (f( x 0 ) + ɛ) = f( x 0 ) + ɛ Applyig Jese s iequality agai to y = λ 1 ( x 0 δ v 1 ) + λ ( x 0 δ v ) + + λ k ( x 0 δ v k ) x 0 δb, we also get f( y) f( x 0 ) + ɛ By x 0 = Therefore x + y, we further have f( x 0 ) f( x) + f( y) f( x) f( x 0 ) f( y) f( x 0 ) (f( x 0 ) + ɛ) = f( x 0 ) ɛ Thus we proves that f( x) f( x 0 ) < ɛ for x x 0 + δb Exercise 8130 If f has cotiuous secod order partial derivatives, the we may use the chai rule to compute the secod order derivative of the restrictio of the fuctio o straight lie: d dt f( x + t v) = h f, x ( v) t=0 By the criterio for the covexity of sigle variable fuctios, the restrictio of the fuctio to straight lies i directio v is covex if ad oly if h f ( v) 0 By Exercise 816, the fuctio is covex if ad oly if h f ( v) 0 for ay v

21 Exercise 8131 The equality (f + g) = f + g follows from (f + g) x i = f x i The equality (fg) = g f +f g + f g follows from (fg) x i g x i Exercise 813 By r = x 1 + x + + x, we get rr xi = x i ad r xi = x i f x i r u = u x i r x i r + u 1 r u x i r x i r = u x i u r r 3 = u + ( 1)r 1 u r + 1 u r x u i Thus f = u r3 r + 1 u r Exercise 8133 By f r = f x cos θ + f y si θ ad f θ = rf x si θ + rf y cos θ, we have The f rr = (f xx cos θ + f xy si θ) cos θ + (f xy cos θ + f yy si θ) si θ = f xx cos θ + f xy cos θ si θ + f yy si θ, + g x i = f g +f g + f + x i x i x i r Thus f x i = u r xi = u x i r, x i + u 1 x r r i u = r 3 f θθ = r( rf xx si θ + rf xy cos θ) si θ + r( rf xy si θ + rf yy cos θ) cos θ = r f xx si θ r f xy si θ cos θ + r f yy cos θ rf x cos θ rf y si θ f rr + r 1 f r + r f θθ = f xx (cos θ + si θ) + f xy (cos θ si θ si θ cos θ) + f yy (si θ + cos θ) + r 1 (f x cos θ + f y si θ) + r ( rf x cos θ rf y si θ) = f xx + f yy Exercise 8135 By takig derivative of f(c x) = c p f( x) for c > 0 with respect to c, we fid homogeeity implies f(c x) x = pc p 1 f( x) = p c cp f( x) = p f(c x) By multiplyig c ad replacig c x with c x, this is the same as the Euler equatio f( x) x = pf( x) Coversely, if the Euler equatio holds, the f(c x) = c p f( x) + g( x) for c > 0 ad some fuctio g( x) idepedet of c By takig c = 1, we get g( x) = 1 Thus f(c x) = c p f( x) for c > 0 Exercise 8136 f satisfies the Euler equatio pf = x i f xi Takig the partial derivatives of the equatio, we get pf xj = f xj + x i f xi x j Multiplyig x j ad addig together, we get p x j f xj = xj f xj + x i x j f xi x j Usig the Euler equatio agai, we get p f = pf + x i x j f xi x j, or p(p 1)f = x i x j f xi x j I geeral, we have p(p 1) (p k + 1)f = x i1 x i x ik f xi1 x i x ik Exercise 8137

22 Note that v i f xi = v f = f ( x)( v), where the derivative f ( x): R R at x is a liear fuctioal ad f ( x)( v) is the value of the liear fuctioal at v The preservatio of the expressio x i f xi the meas (f H) ( z)( z) = f ( x)( x) for x = H( z) Sice (f H) ( z) = f ( x) H ( z), we have f ( x)(h ( z)( z)) = f ( x)( x) = f ( x)(h( z)) for ay fuctio f, or ay liear fuctioal f ( x) Thus we fid the coditio for the equality is H ( z)( z) = H( z) By lookig at idividual coordiates, we fid H ( z)( z) = H( z) meas that each coordiate of H satisfies the Euler equatio for homogeeous fuctios of degree α = 1 i Exercise 6344 Alteratively, here is a direct proof For fixed z, let φ(c) = H(c z) The φ (c) = H (c z)( z) = 1 c H(c z) = 1 ( ) φ(c) φ(c) Thus = φ (c) φ(c) = 0 This implies φ(c) is idepedet of c c c c c c Thus H(c z) = cg( z) By takig c = 1, we get G( z) = H( z) Therefore we coclude H(c z) = ch( z) Exercise 8138 If x is a root of the polyomial with coefficiets (σ 1, σ,, σ ), the Multiplyig t, we have x σ 1 x 1 + σ x + ( 1) 1 σ 1 x + ( 1) σ = 0 (tx) tσ 1 (tx) 1 + t σ (tx) + ( 1) 1 t 1 σ 1 (tx) + ( 1) t σ = 0 Therefore tx is a root of the polyomial with coefficiets (tσ 1, t σ,, t σ ) The weighted homogeeous property shows that, if x 0 is a a root of the polyomial with coefficiets σ 0 = (σ 10, σ 0,, σ 0 ), the alog the path σ(t) = (tσ 10, t σ 0,, t σ 0 ), the root at σ(t) is tx 0 Therefore the restrictio of x( σ) to the curve does ot have local extreme at σ 0 This implies that, without restrictio, the root does ot have local extreme at σ 0 Exercise 8139 The discussio i Exercise 8138 shows that, if we wish the root to have local extreme for σ U, the the local extreme ca oly happe at the boudary poits of U I particular, whe we cosider the uit ball B = { σ : σ 1}, the root will have local extreme oly at the boudary B = { σ : σ = 1} So we study the local extreme of f(x, σ) = x uder the costraits p(x, σ) = x σ 1 x 1 + σ x + ( 1) 1 σ 1 x + ( 1) σ = 0, g(x, σ) = σ 1 + σ + + σ = 1 At local extremes, either the gradiets p = (p (x), x 1, x,, ( 1) 1 x, ( 1) ), g = (0, σ 1, σ,, σ 1, σ ), are parallel, or the gradiets are ot parallel ad f = µ p + λ g

23 I the first case, sice σ 0, we get p = λ g This meas p (x) = 0, λσ 1 = x 1, λσ = x,, λσ 1 = ( 1) 1 x, λσ = ( 1) Substitutig ito p(x, σ) = 0 ad usig g(x, σ) = 1, we get The λ = x ad 0 = x σ 1 x 1 + σ x + ( 1) 1 σ 1 x + ( 1) σ = x + λ(σ 1 + σ + + σ ) = x + λ σ 1 = x 1, σ = x,, σ 1 = ( 1) x +1, σ = ( 1) +1 x Substitutig ito p(x, σ) = 0 ad p (x) = 0, we get x x x 4 x = 0, x 1 ( 1)x 3 ( )x 5 x +1 = 0 Let q(y) = y y 1 y y 1 The two equatios mea that q(x ) = q (x ) = 0 I other words, q(y) has x as multiple root Sice q(y) = y y 1 y 1 = y+1 y + 1 y 1 This implies that the polyomial y +1 y + 1 has multiple root, which meas that the two equatios y +1 y + 1 = 0, (y +1 y + 1) = ( + 1)y y 1 = 0 have commo solutio The first equatio implies y 0, so that the secod equatio implies y = Substitutig ito the first equatio, we get + 1 ( ) +1 ( ) 0 = + 1 = ( ) The equality implies that has to be odd Let = k 1 The we get 1 ( ) k 1 k 1 = 1 k k Uless k = 1, the itegers k 1 ad k have o otrivial commo factors Therefore we get cotradictio uless = 1 The secod case meas µp (x) = 1, µx 1 +λσ 1 = 0, µx +λσ = 0,, ( 1) 1 µx+λσ 1 = 0, ( 1) 1 µ+λσ = 0 Substitutig ito p(x, σ) = 0 ad usig g(x, σ) = 1, we get 0 = x σ 1 x 1 + σ x + ( 1) 1 σ 1 x + ( 1) σ = x λ µ (σ 1 + σ + + σ ) = x λ µ

24 The λ µ = x ad σ 1 = µ λ x 1 = x 1, σ = x,, σ 1 = ( 1) x +1, σ = ( 1) +1 x Substitutig ito p(x, σ) = 0, we get Let a be the uique positive solutio of x x x 4 x = 0 1 = y 1 + y + + y The we get two solutios x = ± a, which are the oly cadidates for the local extrema The largest root is the a

25 Exercise 91 (1) Let A = m i=1 a i, b i ad B = j=1 c j, d j be fiite disjoit uios The A B = m i=1 j=1 a i, b i c j, d j is also a fiite disjoit uio, ad the itersectio a i, b i c j, d j = max{a i, c j }, mi{b i, d j } is still a iterval () Let A = m i=1 a i, b i be a fiite disjoit uio The by rearragig the order of itervals, we may assume a 1 b 1 a b a b This implies R A = (, a 1 b 1, a b, a 3 b, + ) (3) If A, B Σ, the R A, R B Σ by the secod part By the first part, we get A B = A (R B) Σ By the first ad secod parts, we get A B = R (R A) (R B) Σ Exercise 9 Let A = m i=1c i ad B = j=1d j be fiite disjoit uios, with C i, D j C By the assumptio, we have C i D j = k ij k=1 E k,ij for some E k,ij C The is also a fiite disjoit uio of subsets i C Let A = m i=1c i, with C i C The A B = m i=1 j=1 C i D j = m i=1 j=1 k ij k=1 E k,ij X A = m i=1(x C i ) By the assumptio, we already kow each X C i Σ The by what we just proved for itersectio, the fiite itersectio X A Σ Let A, B Σ The by what we just proved, the complemets X A, X B Σ, ad the itersectio (X A) (X B) Σ, ad further the complemet X (X A) (X B) = A B Σ Fially, let A, B Σ The the complemet X B Σ This implies that the itersectio A (X B) = A B Σ Exercise 93 By Propositio 91, we have λ(a B C) = λ(a B) + λ(c) λ((a C) (B C)) = λ(a) + λ(b) λ(a B) + λ(c) [λ(a C) + (B C) λ((a C) (B C)) = λ(a) + λ(b) + λ(c) λ(a B) λ(b C) λ(c A) + λ(a B C)

26 Exercise 94 By Propositio 91, we have λ(a B) λ(a) + λ(b) By iductio, this implies λ(a 1 A ) λ(a 1 A 1 ) + λ(a ) λ(a 1 ) + + λ(a 1 ) + λ(a ) I case A B =, Propositio 91 tells us λ(a B) + λ(a) + λ(b) The i case A i are pairwise disjoit, By iductio, we have λ(a 1 A ) = λ(a 1 A 1 ) + λ(a ) = λ(a 1 ) + + λ(a 1 ) + λ(a ) Exercise 95 Let K = i=1[c i, d i ] Without loss of geerality, we may assume c 1 d 1 c d c d Pick a < c 1 d < b The K U = (a, b), ad U K = (a, c 1 ) (d 1, c ) (d 1, c ) (d, b) Therefore λ(k) = λ(u) λ(u K) = (b a) [(c 1 a)+(c d 1 )+ +(c d 1 )+(b d)] = (d i c i ) Exercise 96 (1) Let U be a ope subset cotaiig L The U also cotais K The cotaimet K L implies U K U L By Propositio 914, we have λ(u K) λ(u L) Therefore λ(k) = λ(u) λ(u K) λ(u) λ(u L) = λ(l) () Let U be a ope subset satisfyig K L U The U K L = (U K) (U L) Sice K ad L are disjoit, we also have U = (U K) (U L) Comparig λ(k) + λ(l) =λ(u) λ(u K) + λ(u) λ(u L), λ(k L) =λ(u) λ(u K L), the proof is reduced to λ(u K L) = λ(u K) + λ(u L) λ(u) The equality is proved as follows: λ(u) =λ((u K) (U L)) = λ(u K) + λ(u L) λ((u K) (U L)) =λ(u K) + λ(u L) λ(u K L) The secod equality makes use of the secod part of Propositio 914 Exercise 97 (1) Let K V, V ope The K U is compact, K U V, ad the iequality becomes λ(v ) λ(v K) λ(v ) λ(v (K U)) + λ(u) This is the same as λ(v (K U)) λ(v K) + λ(u)

27 This is a cosequece of V (K U) (V K) U ad Propositio 914 Moreover, if U K, the V (K U) = (V K) U, ad the iequality becomes equality () Let U K V, V ope The V (V K) U, ad (V K) U = U K By Propositio 914, we have This meas λ(v ) λ((v K) U) = λ(v K) + λ(u) λ(u K) λ(k) = λ(v ) λ(v K) λ(u) λ(u K) Moreover, the equality happes whe V = (V K) U This meas K U Exercise 98 By the defiitio of λ(k), we have λ(k) = λ(v ) λ(v K) λ(v ) By Exercise 97, we have λ(u) = λ(k) λ(k U) λ(k)

28 Exercise 99 Let A = {a :, N} be a coutable subset of R For ay ɛ > 0, costruct U = N (a ɛ, a + ɛ) The λ(u) = µ (U) µ (a ɛ, a + ɛ) = +1 ɛ = ɛ Sice A U, this implies that µ (A) ɛ for ay ɛ > 0 This implies that µ (A) = 0 Therefore A is Lebesgue measurable ad has measure 0 Exercise 910 If K = I i, where I i are itervals, the K = i I i Sice I i is two disjoit itervals obtaied by deletig the middle third of I i, we get λ(i i ) = 3 λ(i i) Therefore λ(k +1 ) = i λ(ii ) = λ(i i ) = 3 3 λ(k ) i By K = K K ad Exercise 96, we get λ(k) λ(k ) = 3 λ(k 1) = ( ) λ(k ) = = 3 ( ) λ(k 0 ) = 3 ( ) 3 Sice this is true for all, µ(k) 0, ad lim ( 3) = 0, we coclude that µ(k) = 0 Exercise 911 Suppose A ad B are measurable The for ay ɛ > 0, there are ope U, V ad closed K, L, such that K A U, λ(u K) = λ(u) λ(k) < ɛ, L B V, λ(v L) = λ(v ) λ(l) < ɛ The K V A B U L, K L A B U V, K L A B U V, where U L, U V, U V are ope, ad K V, K L, K L are closed Moreover, we have (U L) (K V ) [(U L) (K L)] [(K L) (K V )] (U K) (V L), U V K L (U K L) (V K L) (U K) (V L), U V K L (U V K) (U V L) (U K) (V L) By λ((u K) (V L)) λ(u K) + λ(v L) < ɛ, we see that A B, A B, A B are measurable The estimatios also tell us 0 λ(u V ) µ(a B) < ɛ, 0 λ(u V ) µ(a B) < ɛ

29 Combied with 0 λ(u) λ(a) < ɛ, 0 λ(v ) λ(b) < ɛ, λ(u V ) = λ(u) + λ(v ) λ(u V ), we get µ(a B) µ(a) µ(b) + µ(a B) < 6ɛ Sice ɛ is arbitrary, we get µ(a B) = µ(a) + µ(b) µ(a B) Exercise 91 For a closed subset K, take a ope subset U K The U K is still ope By Example 933, U ad U K are measurable By Exercise 911, K = U (U K) is also measurable, ad µ(u) = µ(k) + µ(u K) µ( ) = µ(k) + µ(u K) By λ = µ for ope subsets U ad U K, we have µ(k) = µ(u) µ(u K) = λ(u) λ(u K) = λ(k) The last equality is the defiitio of λ(k) Exercise 913 For ay fixed ad ɛ > 0, there are closed subsets K 1 A 1, K A,, K A, such that λ(k i ) > µ (A i ) ɛ The K 1 K K is a closed subset cotaied i A i Sice A i are disjoit, K i are also disjoit Therefore by Exercises 911 ad 91, µ ( A i ) λ(k 1 K K ) = λ(k 1 ) + λ(k ) + + λ(k ) > µ (A 1 ) + µ (A ) + + µ (A ) ɛ Sice the ɛ ad are arbitrary, this implies that µ ( A i ) µ (A i ) If A i are measurable, the by what just proved ad the subadditivity of µ, µ ( A i ) µ (A i ) = µ(a i ) = µ (A i ) µ ( A i ) This implies A i is measurable ad µ( A i ) = µ(a i ) Exercise 914 Suppose A i, i N, are measurable (ad o loger ecessarily disjoit), the A i = (A i A 1 A A i 1 ) By Exercise 91, A i A 1 A A i 1 is measurable The by Exercise 913, the coutable disjoit uio A i is also measurable If all A i are cotaied i a iterval I, the I A i is measurable by Exercise 91 By what is just proved, the uio I A i = (I A i ) is also measurable By Exercise 91 agai, the coutable itersectio A i = I (I A i ) is measurable Exercise 915

30 By subadditivity, µ (B) µ (A B) µ (A)+µ (B) = µ (B) The first iequality follows from B A B The secod iequality follows from the subadditivity The whole computatio implies µ (A B) = µ (B) To prove µ (A B) = µ (B), we ote that for ay ɛ > 0, there is ope U A ad compact K A B, such that µ (A) λ(u) < ɛ, µ (A B) λ(k) + ɛ By Example 933 ad Exercise 91, U, K ad K U (which is compact) are measurable, with µ(u) = λ(u), µ(k) = λ(k) ad µ(k U) = λ(k U) By Exercise 911 ad K U beig a compact subset of B, we have µ (A B) ɛ λ(k) = µ(k) = µ(k U) + µ(k U) Sice ɛ is arbitrary, we get µ (A B) µ (B) µ(k U) + µ(u) = λ(k U) + λ(u) µ (B) + ɛ Exercise 916 By Exercise 915, we have µ (A B) = µ ((A B) B) = µ (B) ad µ (A B) = µ ((A B) B) = µ (B) By the similar reaso, we have µ (A B) = µ (A) ad µ (A B) = µ (A) Therefore µ (A) = µ (B) ad µ (A) = µ (B) This implies µ (A) = µ (A) if ad oly if µ (B) = µ (B) I other words, A is measurable if ad oly if B is measurable Moreover, the equality µ (A) = µ (B) implies that µ(a) = µ(b) i case both are measurable Exercise 917 (1) Suppose A is measurable For ay ɛ > 0, there are K A U, K closed ad U ope, such that λ(u) λ(k) < ɛ The µ (U A) µ (U K) = λ(u K) = λ(u) λ(k) < ɛ Coversely, suppose for ay ɛ > 0, there is ope U A, such that µ (U A) < ɛ The we have a sequece of ope subsets U A, such that µ (U A) < 1 By Exercise 914, the coutable itersectio W = U is measurable Moreover, we have µ (W A) µ (U A) < 1 for ay, so that µ (W A) = 0 Sice A W, we have µ ((W A) (A W )) = µ (W A) = 0 By Exercise 916, the measurability of W implies the measurability of A () Suppose A is measurable For ay ɛ > 0, there are K A U, K closed ad U ope, such that λ(u) λ(k) < ɛ The µ (A K) µ (U K) = λ(u K) = λ(u) λ(k) < ɛ Coversely, suppose for ay ɛ > 0, there is closed K A, such that µ (A K) < ɛ The we have a sequece of ope subsets K A, such that µ (A K ) < 1 By Exercise 914, the coutable uio W = K is measurable Moreover, we have µ (A W ) µ (A K ) < 1 for

31 ay, so that µ (A W ) = 0 Sice A W, we have µ ((W A) (A W )) = µ (A W ) = 0 By Exercise 916, the measurability of W implies the measurability of A (3) Suppose A is measurable By the first part, for ay ɛ > 0, we have µ (U A) < ɛ for some ope U A Let U = i=1(a i, b i ) Sice (b i a i ) coverges, there is, such that i> (b i a i ) < ɛ For the uio V = (a 1, b 1 ) (a, b ) (a, b ) of fiitely may ope itervals, we have µ (V A) µ (U A) ɛ, ad µ (A V ) µ (U V ) = µ ( i>+1 (a i, b i )) = i>(b i a i ) < ɛ This implies µ ((V A) (A V )) µ (V A) + µ (A V ) < ɛ Coversely, if µ ((U A) (A U)) < ɛ for some ope U, the µ (A U) < ɛ Thus there is ope V A U satisfyig µ (V ) = λ(v ) < ɛ The U V is a ope subset cotaiig A, ad µ (U V A) µ ((U A) V ) µ (U A) + µ (V ) < ɛ By the first part, this implies that A is measurable (4) By the mootoe property ad subadditivity, we have ad µ (U A) < ɛ, µ (A U) < ɛ = µ ((U A) (A U)) < ɛ µ ((U A) (A U)) < ɛ = µ (U A) < ɛ, µ (A U) < ɛ The coditio for measurability the follows from the third part Exercise 918 For ay ɛ > 0, there is a ope U A, such that λ(u) µ (A) + ɛ Let U = (a i, b i ) The Lipschitz coditio implies that f(x ɛ, x + ɛ) (f(x) Lɛ, f(x) + Lɛ) Therefore each f(a i, b i ) is cotaied i some (c i, d i ) satisfyig d i c i L(b i a i ) The Sice f(a) V ad V is ope, we get f(u) f(a i, b i ) V = (c i, d i ) µ (f(a)) µ (V ) (d i c i ) L (b i a i ) = Lλ(U) Lµ (A) + Lɛ Sice this holds for ay ɛ, we coclude that µ (f(a)) Lµ (A) Now we use the secod part of Exercise 917 If A is measurable, the for ay ɛ > 0, there is a closed K A, such that µ (A K) < ɛ By f(a) f(k) f(a K), we get µ (f(a) f(k)) Lµ (A K) < Lɛ Sice Lipshitz fuctios are also cotiuous, f(k) is a closed subset of f(a) The by usig the secod part of Exercise 917 agai, we see that f(a) is measurable