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1 Exercise 8.51 The solution of dx dy = sin φ cos θdr + r cos φ cos θdφ r sin φ sin θdθ sin φ sin θdr + r cos φ sin θdφ r sin φ cos θdθ is dr dφ = dz cos φdr r sin φdφ dθ sin φ cos θdx + sin φ sin θdy + cos φdz r (cos φ cos θdx + cos φ sin θdy sin φdz. (r sin φ 1 ( sin θdx + cos θdy Exercise 8.52( (x, y e (u, v = u + cos v u sin v (x, y e u. The matrix is invertible if and only if det + sin v u cos v (u, v = (e u + cos vu cos v + (e u + sin vu sin v = u(e u (sin v + cos v The condition is the same as u 0 and sin v + cos v e u (u, v. When the condition is satisfied, we have (x, y = ( 1 ( (x, y 1 u cos v u sin v = (u, v u(e u (sin v + cos v + 1 e u sin v e u. + cos v Exercise 8.53 ((1 (x, y 1 1 (s, t =. The matrix is invertible if and only if s t. When the condition 2s 2t ( ( 1 ( ds 1 1 dx is satisfied, we have dz = 3s 2 ds + 3t 2 dt = 3(s 2 t 2 = 3(s dt 2 t 2 = 2s 2t dy 3stdx (s + tdy. Thus z x = 3st = 3 2 (x2 y, z y = 3 2 x. Exercise 8.53 ((2 (x, y e (s, t = s+t e s+t e s t e s t. The matrix is always invertible and we have dz = tds + sdt = ( ( ds e s+t e (t s = (t s s+t 1 ( dx dt e s t e s t = 1 dy 2 [e s t (t + sdx + e s+t (t s]dy. Thus z x = 1 2 e s t (t + s = log x 2x, z y = 1 2 e s+t (t s = log y. y Exercise = (x + yu x (x yu y = u ( x + y = u ( (r, θ x + y ( (x, ( y x y (r, θ (x, y x + y cos θ sin θ r(cos θ + sin θ = (u r u θ r 1 sin θ r 1 = ru cos θ r( cos θ + sin θ r u θ. Exercise 8.55 Let F (X = X 2. Let D be a diagonal matrix with λ 1,..., λ n as diagonal entries. Then F (D(H = DH + HD = ((λ i + λ j h ij, H = (h ij. The linear transform F (D is invertible if and only if λ i + λ j 0 for any i and j.

2 In general, let A be diagonalizable. Then A = P DP 1, where D is diagonal with eigenvalues λ i of A as the diagonal entries. The square map is the composition X at A P 1?P P 1 XP at D? 2 (P 1 XP 2 = P 1 X 2 P at D 2 P?P 1 P (P 1 X 2 P P 1 = X 2 at A 2. The first and the third maps are invertible linear maps. Therefore the square map is locally invertible at A if and only if the second map is locally invertible at D. Since the second map is still the square map, we see that the condition is that λ i + λ j 0 for any i and j. Exercise 8.56 (1 Taking the differential of the equation, we get (3x 2 3aydx+(3y 2 3axdy = 0. If y 2 ax, then y can be locally written as a function of x, with dy dx = ay x2 y 2 ax. Exercise 8.56 (2 Taking the differential of the equation, we get (2x 4dx + (2y + 6dy + (2z 2dz = 0. If z 1, then z can be locally written as a function of (x, y, with z x = x 2 z 1, z y = y + 3 z 1. If x 2, then x can be locally written as a function of (y, z, with x y = y + 3 x 2, x z = z 1 x 2. Exercise 8.56 (3 Taking the differential of the equations, we get 2xdx + 2ydy = 2zdz + 2wdw and dx + dy + dw + dz = 0. If z w, then the system can be solved to get dz = 1 ((x + wdx + (y + wdy and w z dw = 1 ((x + zdx + (y + zdy. Thus (z, w can be locally written as a function of (x, y, z w (z, w with (x, y = 1 ( x + w y + w. z w x z y z If x + w 0, then the system can be solved to get dx = 1 (( y wdy + (z wdz x + w and dw = 1 (( x + ydy + ( x zdz. Thus (x, w can be locally written as a function x + w (x, w of (y, z, with (y, z = 1 ( y w z w. x + w x + y y z Exercise 8.56 (4 Taking the differential of the equation, we get dz = (f 1 + yzf 2 dx + (f 1 + xzf 2 dy + (f 1 + xyf 2 dz. If f 1 + xyf 2 1, then z can be locally written as a function of (x, y, with z x = f 1 + yzf 2, z y = f 1 + xzf 2. If f 1 +yzf 2 0, then x can be locally written as a function 1 f 1 xyf 2 1 f 1 xyf 2 of (y, z, with x y = f 1 + xzf 2, x z = f 1 + xyf 2 1. f 1 + yzf 2 f 1 + yzf 2

3 Exercise 8.57 (1 Taking the differential of the equation, we get f 1 dx + f 2 dy (af 1 + bf 2 dz = 0. f 1 az x + bz y = a + b = 1. af 1 + bf 2 af 1 + bf 2 f 2 Thus Exercise 8.57 (2 Taking the differential of the equation, we get 2xdx + 2ydy + 2zdz = (f zy 1 f dy + f dz. Thus (x 2 y 2 z 2 z x + 2xyz y = (x 2 y 2 z 2 2x f 2z + 2xy 2y f + 2zy 1 f = 2x f 2z f 2z (x2 + y 2 z 2 yf + 2zf = 2x f 2z ( 2z2 + 2zf = 4xz.

4 Exercise 8.59 The graph of a differentiable function f( x defined on an open subset U R n 1 is a hypersurface S = {( x, f( x: x U}. The hypersurface can be considered as a level g( x, y = y f( x = 0. The tangent space of S ( f, 1 is orthogonal to g = ( f, 1. The normal vector of the graph surface is n = f Exercise 8.60 (1 f is orthogonal to (a, b, or parallel to (b, a = (bx + ay. This suggests the levels of f and bx + ay are the same, and we should have f = h(bx + ay. For a rigorous argument, let g(u, v be defined by f(x, y = g(ax by, bx + ay. Then f x = ag u + bg v and f y = bg u + ag v. By af x = bg y, we get g u = 0. Therefore g is independent of its first variable, and we have (x, y = h(bx + ay. Exercise 8.60 (2 f is parallel to ( y, x. Since y ( y, x ( y =, we expect f = h. x ( x 2 x y Let g(u, v be defined by f(x, y = g. x, y Then xf x +yf y = xg u ( y +y ( x 2 y yg v. Therefore the equation means yg v = 0, and we get f = h. x ( 1 g u x + g v = Exercise 8.60 (3 f is parallel to (1, 1, 1 = (x + y + z. So we expect f = h(x + y + z. Let g(u, v, w be defined by f(x, y, z = g(x + y + z, x + y, x. Then f x = g u + g v + g w, f y = g u + g v, f z = g u, and the equality f x = f y = f z becomes g w = g v = 0. Therefore f = h(x + y + z. Exercise 8.60 (4 f is parallel to (yz, zx, xy = xyz. So we expect f = h(xyz. Let g(u, v, w be defined by f(x, y, z = g(xyz, xy, x. Then f x = yzg u + yg v + g w, f y = xzg u + xg v, f z = xyg u, and the equality xf x = yf y = zf z becomes g w = g v = 0. Therefore f = h(x + y + z. Exercise 8.61 Denote the linear transform L( x = ( b 1 x, b 2 x,..., b n 1 x, a x and let h( y = f(l 1 ( y. Then we have f( x = h( b 1 x, b 2 x,..., b n 1 x, a x. Consider any straight line φ(t = x 0 + t a in the direction of a, we have L(φ(t = y 0 + t(0,..., 0, a 2, y 0 = L( x 0, so that a 2 h ( y 0 = dh(l(φ(t y n dt = df(φ(t t=0 dt = D a f( x 0 = 0. t=0 Since y 0 can be any point in R n. This implies that h is independent of its last variable y n. Exercise 8.62

5 φ (t = (1 cos t, sin t. When t 2nπ, we have x (t 0, so that y can be locally written as a function of x, and we have dy dx = y t = sin t x t 1 cos t. Exercise 8.63 The parametrized torus has derivative b sin φ cos θ (a + b cos φ sin θ (x, y, z = b sin φ sin θ (a + b cos φ cos θ. (φ, θ b cos φ 0 ( (x, y b sin φ cos θ (a + b cos φ sin θ If φ nπ, then (φ, θ = is invertible, the surface is b sin φ sin θ (a + b cos φ cos θ regular, and z can be written as a function of (x, y. We have z (x, y = z ( 1 (φ, θ b sin φ cos θ (a + b cos φ sin θ = (b cos φ 0 = cos φ (cos θ sin θ. (φ, θ (x, y b sin φ sin θ (a + b cos φ cos θ sin φ Thus z x = cot φ cos θ, z y = cot φ sin θ. ( (2k + 1π (y, z b sin φ sin θ (a + b cos φ cos θ If φ = nπ, θ, then 2 (φ, θ = is invertible, b cos φ 0 the surface is regular, and x can be written as a function of (y, z. We have x (y, z = x ( (φ, θ b sin φ sin θ (a + b cos φ cos θ = ( b sin φ cos θ (a + b cos φ sin θ (φ, θ (y, z b cos φ 0 1 = (sin θ cos θ sin φ. cos φ cos θ Thus x y = sin θ cos φ, x z = tan φ cos θ. If φ = π and θ nπ, then the surface is also regular, and y can be written as a function of 2 (x, z. Altogether, the parametrized sphere is regular everywhere. Exercise 8.64 We have σ t = (x, y cos θ, y sin θ, σ θ = (0, y sin θ, y cos θ. The two vectors are linearly independent if and only if x 0 and y 0. However, the regularity assumption is that (x, y is never zero. So σ is not necessarily regular everywhere. Exercise 8.65 The condition is that (λ, µ is a regular value of the map ( x 2 F (x, y, z = a + y2 2 b + z2 2 c, x 2 2 A + y2 2 B z2 : R 3 R 2. 2 C 2 We have ( x F y (x, y, z = 2 a 2 b 2 x A 2 y B 2 z c 2 z C 2 1

6 We find that F has rank 2 as long as xz 0 or yz 0. So we only need to verify the rank 2 when xz = yz = 0. The first possibility is x = y = 0, in which case F cannot have rank 2. However, substituting x = y = 0 into the ellipsoid and the hyperboloid, we get z2 = λ and z2 = µ. This implies c 2 C 2 λ 0, µ 0, λc 2 + µc 2 = 0. Therefore if λ < 0, or µ > 0, or λc 2 + µc 2 0, then we cannot have x = y = 0. The second possibility is x 0, y 0 and z = 0. In which case F has rank 2 if and only if det ( x y a 2 b x y 2 A 2 B 2 = xy ( 1 a 2 B 2 1 b 2 A 2 0. Since a, b, c, A, B, C > 0 (the usual assumption for the parameters, this means a b. A B Thus we conclude that the intersection is a curve if and only if a b, and λ < 0, or µ > 0, A B or λc 2 + µc 2 0. Exercise 8.66 Let f a ( x = x a 2 2 be the square of the Euclidean distance for a. Then S( a, r = f 1 a (r2. The intersection of S( a, r and S( b, s is the level (f a, f b 1 (r 2, s 2. By Proposition 8.4.3, the intersection is an (n 2-dimensional submanifold when (r 2, s 2 is a regular value of the map (f a, f b. The derivative of the map is (f a, f b ( u = (f a( u, f b ( u = 2(( x a u, ( x b u: R n R 2. The regularity of the value (r 2, s 2 means that, if x a 2 = r and x b 2 = s, then the linear transformation above is surjective. The surjective is equivalent to that x a and x b are linearly independent. So the condition becomes that, if x lies on the intersection of the two spheres, then x a and x b are not parallel. Note that u and v are parallel if and only if one of the following happens 1 : u v 2 = u 2 + v 2, u v 2 = u 2 v 2, or u v 2 = v 2 u 2. Then the regularity condition above means that none of r + s, r s, s r is a b 2. The intersection of three spheres S( a, r, S( b, s, S( c, t is an (n 3-dimensional submanifold if x a 2 = r, x b 2 = s, x c 2 = t implies that x a, x b, x c are linearly independent. Geometrically, this means the following: 1. Draw a triangle with a b 2, a c 2, b c 2 as three sides. If there is no such triangle (due to the failure of triangle inequality, then the three spheres do not intersect. 2. Draw circles of radii r, s, t centered at the three vertices of the triangle. 3. The regularity condition is equivalent to that the three circles do not intersect at the same point. 1 This is a general fact for strictly convex norms. See Exercise 7.89

7 t ( c r a c 2 b c 2 ( a a b 2 ( b s We remark that the picture describes the case of n = 2, when (n 2-dimensional submanifolds should be empty. Moreover, the intersection of the three spheres is actually non-empty if and only if the intersection of three disks is not empty. The geometric description can be easily generalised. The intersection of k Euclidean spheres in R n is (n k-dimensional submanifold if the similar intersection for the lowest dimensional case of n = k 1 is empty. Exercise 8.67 By Proposition 8.4.3, we only need to show that 1 is a regular value of det function. Since det A = 1 implies that A is invertible, by Exercise 8.11 or Example 8.2.4, we find that det (A(H = (det Atr(A 1 H = tr(a 1 H. Since A 1 H can be any matrix for arbitrary H, tr(a 1 H can be any number. Therefore det (A is onto. This shows that 1 is a regular value of det, and SL(n is a submanifold. The same argument works if 1 is replaced by any nonzero number. Exercise 8.68 Suppose F (A = I. Then A is invertible. By Exercise 8.10, the derivative F (A(H = A T H + H T A = A T H + (A T H T. Since a matrix is symmetric if and only if it is of the form X + X T, we see that F (A: R n2 R n(n+1 2 is onto. The argument shows that I R n(n+1 2 is a regular value of F. Therefore O(n = F 1 (I is a hypersurface in R n2. Exercise 8.69 By the computation in Exercise 8.63, the tangent plane is spanned by σ φ = b( sin φ cos θ, sin φ sin θ, cos φ, σ θ = (a + b cos φ( sin θ, cos θ, 0. This is the same as spanned by By solving ( sin φ cos θ, sin φ sin θ, cos φ, ( sin θ, cos θ, 0. u sin φ cos θ v sin φ sin θ + w cos φ = 0, u sin θ + v cos θ = 0,

8 we get (u, v, w being a multiple of (cos θ, sin θ, tan φ. The tangent plane is then given by x cos θ + y sin θ + z tan φ = 0. Exercise 8.70 (1 f = (2x + y + z + 1, 2y + z + x + 1, 2z + x + y + 1. We have f = 0 if and only if x = y = z = 1 ( 4. Since f 1 4, 1 4, 1 = 8 4 3, any number c 8 is a regular value. The 3 tangent space of f 1 (c at (x 0, y 0, z 0 is (2x 0 + y 0 + z 0 + 1(x x 0 + (2y 0 + z 0 + x 0 + 1(y y 0 + (2z 0 + x 0 + y 0 + 1(z z 0 = 0. Exercise( 8.70 ( F =. F y + z z + x x + y is not onto if and only if y + z = z + x = x + y, which is the same as x = y = z. Since F (x, x, x = (3x, 3x 2, any vector (a, b satisfying a 2 3b is a regular value. The tangent space of f 1 (a, b at (x 0, y 0, z 0 is given by the system (x x 0 + (y y 0 + (z z 0 = 0, (y 0 + z 0 (x x 0 + (z 0 + x 0 (y y 0 + (x 0 + y 0 (z z 0 = 0, which is the same as x + y + z = x 0 + y 0 + z 0 = a, (y 0 + z 0 x + (z 0 + x 0 y + (x 0 + y 0 z = 2(x 0 y 0 + y 0 z 0 + z 0 x 0 = 2b. Exercise 8.71 The condition for the gradient (xyz = (yz, zx, xy to be nonzero is that at least two of x, y, z are nonzero. Thus λ is a regular value if and only if λ 0. At the place where the two surfaces are tangent, the gradient (xyz is parallel to the ( x 2 gradient a + y2 2 b + z2 2 c 2 = 2x a 2 yz = The first two equalities tell us x2 ( 2x a, 2y 2 b, 2z 2 c 2 a 2 2y b 2 xz = = y2 b 2. The point is also on the ellipse. Thus we have 2z c 2 xy, x 2 a 2 + y2 b 2 + z2 c 2 = 1. = z2. Substituting into the second equality, we see c2 x 2 a 2 = y2 b 2 = z2 c 2 = 1 3. Therefore 27(xyz2 = (abc 2, and λ = ±abc 3 3. Exercise 8.72 (x 2 + y 2 + z 2 (x 2 + y 2 b 2 z 2 = (2x, 2y, 2z (2x, 2y, 2b 2 z = 4(x 2 + y 2 b 2 z 2 = 0. The last equality is due to the fact that the point is on the cone. In general, the sphere x x x 2 n = 1 is orthogonal to generalized cone a 1 x a 2 x a n x 2 n = 0 at the intersections.

9 Exercise 8.74 We have B( x 0 + x, y 0 + y = B( x 0, y 0 + B( x, y 0 + B( x 0, y + B( x, y The term B( x 0, y 0 is constant. The terms B( x, y 0 and B( x 0, y are linear in ( x, y. The term B( x, y is quadratic in ( x, y. Therefore the second order derivative is For a triple linear T ( x, y, z, we have Exercise 8.75 We have B ( x 0, y 0 ( u, v = 2B( u, v. T ( x 0, y 0, z 0 ( u, v, w = 2T ( u, v, z 0 + 2T ( u, y 0, w + 2T ( x 0, v, w. (A + H k = A k + i+j=k 1 A i HA j + i 1 +i 2 +i 3 =k 2 A i 1 HA i 2 HA i 3 +, where a finite sum of products of i 3 copies of H and k i copies of A in all possible order. The norms of terms in are A k i H i ɛ H 2. Thus (X k (A(H = 2 i 1 +i 2 +i 3 =k 2 Ai 1 HA i 2 HA i 3. Exercise 8.76 We have (I + H = I H + H 2 H 3 + = I H + H 2 H 3 (I + H 1. As argued in Exercise 8.13, for H < 1, we have (I + H H. Thus H3 (I + H 1 H 3 1 H ɛ H. This shows that the quadratic map H2 of H gives the second order derivative: (X 1 (I(H = 2H 2. Exercise 8.77 Let Then y = L( x P ( x + ( x 2, z = K( y Q( y + ( y 2. z = K(L( x K(P ( x Q(L( x + ( x 2. Therefore (G F ( x 0 = G ( y 0 F ( x 0 + G ( y 0 F ( x 0,

10 or 2 z x = z 2 y 2 y x + 2 z 2 y y 2 x. In terms of the coordinates, the left side is 2 z x 2 ( v = ij ( ( z 2 y y x ( v + 2 z y 2 x 2 x ( v = k = ij 2 z x i x j v i v j, and the right side is ( z 2 y k v i v j + ( ( 2 z y k y l v i v j y k x ij i x j y k y l x kl i i x i j ( z 2 y k + 2 z y k y l v i v j. y k x i x j y k y l x i x j kl k Therefore 2 z x i x j = k z 2 y k + y k x i x j kl 2 z y k y l y k x i y l x j. For f(x(u, v, y(u, v, this means 2 f u v = f x 2 x u v + f y 2 y u v + 2 f x x 2 u x v + 2 f y y 2 u y v + 2 f x y x y u v + 2 f y x x y v u. Exercise 8.78 By f xy = (f x y = 0, we get f x (x, y = a(x depending only on x. Let g(x = x x 0 a(tdt. Then (f(x, y g(x x = f x a(x = 0. Therefore f(x, y g(x = h(y depends only on y. We conclude that f(x, y = g(x + h(y. In general, if f xi x j = 0 for every i j, then f(x 1,..., x n = g 1 (x g n (x n. Exercise 8.79 ( fy (log f xy = = f xyf f x f y = 0. By Exercise 8.78, we have log f = A(y + B(x f x f 2 and f = e A(y e B(x. By the continuity of f, this implies f(x, y = g(xh(y. Exercise 8.80 By restricting to x = t v, we see that f( x 2 is second order differentiable away from 0 if and only if f(t is second order differentiable for t > 0. Now consider the second order differentiability of f( x 2 at 0. From the discussion in Exercise 8.36, the first order differentiability already implies f(0 = lim t 0 + f(t and f +(0 = 0. So the quadratic approximation is f(0 + q( x for a quadratic form q. For any ɛ > 0, there is δ > 0, such that x 2 < δ implies f( x 2 f(0 q( x ɛ x 2 2. By writing x = t v, where v has unit Euclidean length and t > 0, we get 0 < t < δ implies f(t f(0 t 2 q( v ɛt 2. In other words, f(t is right second order differentiable at 0, with f(0 + t 2 q( v as the quadratic approximation. This further implies that q( v is actually independent of v. We conclude that f( x 2 is second order differentiable at 0 if and only if f(t is right second order differentiable at 0, and f +(0. Moreover, the quadratic approximation is given by f(0 + c x 2 2.

11 Exercise 8.81 [Answer incomplete] Note that the second order partial derivatives and second order differentiability do not imply each other. So the discussion needs to be separate. (1 We have f(x, 0 = x 2p sin 1 x 2. Thus f xx(0, 0 exists if and only if p > 1. Conversely, for p > 1, then all second order partial derivatives exist and vanish at (0, 0. By Exercise 8.80, the second order differentiability means x 2p sin 1 is second order right x2 differentiable at 0, and the first order derivative vanishes. This is equivalent to p > 1. (2 By Exercise 8.14, we must have either p > 0 or q > 0. If q > 0, then f(x, 0 = 0, and we always have f x (0, 0 = f xx (0, 0 = 0. If q = 0, then p > 0, and f(x, 0 = x p sin 1 x 2. If f xx(0, 0 exists, then f(x, 0 is second order differentiable at 0, which implies p > 2. Conversely, if p > 2, then it is not difficult to conclude f x (0, 0 = f xx (0, 0 = 0. We conclude that f xx (0, 0 exists if and only if q > 0 or q = 0 and p > 2. By similar reason, f yy (0, 0 exists if and only if p > 0 or p = 0 and q > 2. Now consider f xy (0, 0 = d dy f x (0, y. First assume p > 0. Then f x (0, y exists for y y=0 near 0 if and only if x p is differentiable near 0, which means p > 1. Conversely if p > 1, then f x (0, y = 0, so that f xy (0, 0 = 0. Next assume p = 0. Then f x (0, y = 0 for y 0. If q > 0, then we also have f(x, 0 = 0, so that f x (0, 0 = 0. If q = 0, then f(x, 0 = sin 1 x 2, and f x(0, 0 does not exist. Thus we must have q > 0, in which case we have f x (0, y = 0 for all y. This further implies f xy = 0. We conclude that f xy (0, 0 exists if and only if p > 1 or p = 0 and q > 0. By similar reason, f yx (0, 0 exists if and only if q > 1 or q = 0 and p > 0. Next we turn to the second order differentiability. By Exercise 8.14, the condition for the first order differentiability is p > 2, q = 0, or p = 0, q = 2, or p > 0, q > 0, p + q > 1. When p > 0, we have f(0, y = 0 for all y. The partial derivative f y (0, 0 = 0 exists. When y q sin 1 if y 0 p = 0, we have f(0, y = y 2, and f y (0, y exists (and must be zero if and 0 if y = 0 only if q > 1. Similarly, f x (0, 0 exists if and only if q > 0 or p > 1, q = 0. Suppose p > 0 and q > 0. If the function is differentiable at (0, 0, then by the computation of the partial derivatives, the linear approximation must be 0. Therefore we need x p y q lim (x,y (0,0 max{ x, y } sin 1 = 0. By restricting the limit to the line x = y, we get x 2 + y2 p + q > 1. Now assume p > 0, q > 0 and p + q > 1, then f(x, y (x, y p+q ɛ (x, y 1 when (x, y < ɛ p+q 1. Therefore the function is differentiable at (0, 0 with 0 as the linear approximation. (3 By f(x, 0 = f(0, y = 0, the function has partial derivatives f x (0, 0 = f y (0, 0 = 0. If the function is differentiable at (0, 0, then the linear approximation is 0. This means

12 that lim x,y 0 + When restricted to x m = y n, we have x p y q (x m + y n k (x + y = 0. x p+q m n 2 k+1 x mk x x p y q 1 n min{m,n} (x m + y n k (x + y = x p+q m n 2 k x mk (x + x m n x p+q 2 k x mk x. 1 n min{m,n} Therefore the restriction converges to 0 if and only if p + q m n > mk + 1 min{m, n}. This is the n same as p m + q min{m, n} > k +. n mn When restricted to x = y, we have x p+q 2 k+1 x min{m,n}k x x p y q (x m + y n k (x + y = x p+q 2(x m + x n k x x p+q 2x min{m,n}k x. Therefore the restriction converges to 0 if and only if p + q > min{m, n}k + 1. Conversely, assume the two inequalities hold. Without loss of generality, assume m n. Then the two inequalities mean that Then we consider three regions. For y x m n, we have p m + q n > k + 1, p + q > nk + 1. m x p y q (x m + y n k (x + y xp+q x mk x = m xp+q n mk 1. By p + q m ( p n mk 1 = m m + q n k 1 > 0, the restriction of the limit on the region m converges to 0. For x m n y x, we have x p y q 2 k y nk 2x x p y q (x m + y n k (x + y xp y q y nk x. x p y q Therefore lim m x n y x;x,y 0 + (x m + y n k (x + y = 0 if and only if lim x m x p 1 y q nk = n y x;x,y For the given x, the maximum and the minimum of x p 1 y q nk for the y in the range are x p 1 x q nk = x p+q nk 1 and x p 1 x m n (q nk = x m( p m + q n k m 1. Therefore limx m n y x;x,y 0 + x p 1 y q nk = m n m n

13 0 if and only if lim x 0 + x p+q nk 1 = 0 and lim x 0 + x m( p m + q n k m 1 = 0. This is true, given the two inequalities. For y x, we have x p y q (x m + y n k (x + y yp+q y nk y = yp+q nk 1. Since p + q > nk + 1, the restriction of the limit on the region converges to 0. We conclude that the function is differentiable at (0, 0 if and only if p m + q n min{m, n} > k +, p + q > min{m, n}k + 1. mn (4 By f(x, 0 = x pr mk and f(0, y = y qr nk, the function has partial derivatives if and only if pr mk > 1 and qr nk > 1. Moreover, we have f x (0, 0 = f y (0, 0 = 0 when the condition is satisfied. If the function is differentiable at (0, 0, then pr mk > 1 and qr nk > 1, the linear approximation is 0, and we have Denote lim x,y 0 + (x p + y q r (x m + y n k (x + y = 0. { p λ = min m n}, q, µ = min{m, n}, ν = min{p, q}. As in Exercise 6.23(2, restricting the limit to x m = y n, we have x λmr (x p + y q r 2 r x λmr, (x m + y n k = 2 k x mk, x µ n x + y 2x µ n. x λmr Therefore the restriction has limit 0 if and only if lim x 0 + x mk x µ n λmr > mk + µ, which is the same as n { p r min m n}, q min{m, n} > k +. mn On the other hand, restricting the limit to x = y, we have = 0. This means that x νr (x p + y q r 2 r x νr, x µk (x m + y n k 2 k x µk, x + y = 2x. Therefore the restriction has limit 0 if and only if lim x 0 + µk + 1, which is the same as x νr x µk x r min{p, q} > k min{m, n} + 1. = 0. This means that νr > Conversely, assume the two inequalities hold. Without loss of generality, assume m n. Then the two inequalities mean that { p rλ = r min m n}, q > k + 1, rν = r min{p, q} > kn + 1. m

14 Then we consider three regions. For y x m n, we have (x p + y q r (x m + y n k (x + y (xp + x q 2r x λmr x mk x x mk x. ( Since λmr mk 1 = m rλ k 1 > 0, the restriction of the limit on the region converges m to 0. For y x, we have m n r (x p + y q r (x m + y n k (x + y (yp + y q r y nk y 2r y νr y nk y. Since νr nk 1 > 0, the restriction of the limit on the region converges to 0. For x m n y x, we have Moreover, we have y nk x (x m + y n k (x + y 2 k y nk 2x. 1 2 (ur + v r max{u r, v r } (u + v r (2 max{u, v} r = 2 r (max{u, v} r 2 r (u r + v r. By substituting u = x p and v = y q, we see that the restriction of the limit to the region converges to 0 if and only if the limit of xpr + y qr converges to 0, which is the same as the y nk x x pr limits of y nk x and yqr x pr converge to 0. For fixed x, the maxima and minima of y nk x y nk x and y qr y nk x for the y in the range are xpr nk 1, x pr mk 1, x qr nk 1, x qr m n mk 1. Therefore the limit converges to zero if and only if pr nk 1 > 0, pr mk 1 > 0, qr nk 1 > 0, qr m n mk 1 > 0. Since the two inequalities imply all four inequalities above, we conclude that the two inequalities are necessary and sufficient. In conclusion, the function is differentiable at (0, 0 if and only if { p r min m n}, q > k + min{m, n}, r min{p, q} > k min{m, n} + 1. mn Exercise 8.82 The second order partial derivative f yx (x 0, y 0 is defined by a repeated limit f yx (x 0, y 0 = lim x x0 f y (x, y 0 f y (x 0, y 0 x x 0 = lim x x0 f(x, y f(x, y 0 f(x 0, y + f(x 0, y 0 lim. y y0 (x x 0 (y y 0

15 The other second order partial derivative f xy (x 0, y 0 is defined by switching the two limits in the repeated limit. We will show that the existence of f y, f xy near (x 0, y 0 and the continuity of f xy at (x 0, y 0 actually imply the convergence of the whole limit f(x, y f(x, y 0 f(x 0, y + f(x 0, y 0 f xy (x 0, y 0 = lim. x x 0,y y 0 (x x 0 (y y 0 Since the existence of f x means the convergence of f y (x, y 0 f y (x 0, y 0 x x 0 f(x, y f(x, y 0 f(x 0, y + f(x 0, y 0 = lim, y y0 (x x 0 (y y 0 by Proposition 6.2.3, the repeated limit f yx (x 0, y 0 exists and is equal to f xy (x 0, y 0. For fixed y 0 and y, we apply the mean value theorem to the function g(x = f(x, y f(x, y 0. By the existence of f x near (x 0, y 0, we get f(x, y f(x, y 0 f(x 0, y + f(x 0, y 0 = g(x g(x 0 = g (c(x x 0 = (f x (c, y f x (c, y 0 (x x 0, for some c between x 0 and x. Then we fix c and apply the mean value theorem to the function f x (c, y of y. By the existence of f xy near (x 0, y 0, we get f(x, y f(x 0, y f(x, y 0 + f(x 0, y 0 = f xy (c, d(x x 0 (y y 0, for some d between y 0 and y. Then the continuity of f xy at (x 0, y 0 further tells us f(x, y f(x 0, y f(x, y 0 + f(x 0, y 0 lim x x 0,y y 0 (x x 0 (y y 0 = lim f xy (c, d = lim f xy (c, d = f xy (x 0, y 0. x x 0,y y 0 c x 0,d y 0 Exercise 8.83 Consider We have xy(x 2 y 2, if(x, y (0, 0, f(x, y = x 2 + y 2 0, if(x, y = (0, 0. y(x 4 + 4x 2 y 2 y 4, if(x, y (0, 0, f x = (x 2 + y 2 2 0, if(x, y = (0, 0, x(x 4 4x 2 y 2 y 4, if(x, y (0, 0, f y = (x 2 + y 2 2 0, if(x, y = (0, 0.

16 However, the second order partial derivatives f xy (0, 0 = lim x 0 f y (x, 0 f y (0, 0 x f yx (0, 0 = lim y 0 f x (0, y f x (0, 0 y = lim x 0 x(x 4 x(x 2 2 = 1, = lim y 0 y( y 4 y(y 2 2 = 1, are not equal. We also note that f xx (0, 0 = f yy (0, 0 = 0. If the function were second order differentiable, then by the first order partial derivatives, the linear approximation would be 0. Therefore the quadratic approximation is ax 2 +by 2 +cxy. Restricting to y = 0, we see that ax 2 is the quadratic approximation of f(x, 0 = 0, so that a = 0. Restricting to x = 0, we similarly get b = 0. Restricting to x = y, we see that cx 2 is the quadratic approximation of f(x, x = 0, so that c = 0. Thus we conclude that the quadratic approximation must be 0, and we must have f(x, y lim x,y 0 x 2 + y = lim xy(x 2 y 2 = 0. 2 x,y 0 (x 2 + y 2 2 Since this is not true, we conclude that f is not second order differentiable at (0, 0. Exercise 8.84 The function axy, if y x 2, f(x, y = bxy, if x y 2, 0, otherwise, is second order differentiable, with 0 as the quadratic approximation. The Taylor expansion is 1 (a + bxy, which is the quadratic approximation if and only if a + b = 0. 2 Exercise 8.85 The equalities D 1 = b 1 and D 11 = 2c 11 are obtained by restricting the quadratic approximation of the function to the x 1 -axis. The other two equalities are obtained by restricting to the x 2 -axis. Conversely, suppose the four equalities hold. Then based on Exercise 8.84, the function p(x 1, x 2 + D 12 x 1 x 2, if x 2 x 2 1, f(x 1, x 2 = p(x 1, x 2 + D 21 x 1 x 2, if x 1 x 2 2, p(x 1, x 2, otherwise, has assigned first and second order partial derivatives, is second order differentiable, and has p as the quadratic approximation. For more than two variables, the condition is D i = b i and D ii = 2c ii. Exercise 8.86 (1

17 We have y log x = (4 + y log(1 + x = (4 + y ( x x2 2 + x3 3 + o( x3 = (4 + y ( x x2 + 4 x3 + o( x Therefore x y = e y log x = 1 + (4 + y ( x x2 + 4 x3 2 3 = 1 + (4 + y ( x x2 + 4 x ( 2 2 (4 + y2 x x x 3 + o( x ( y(1 x x x3 + o( x 3 = x + 6 x 2 + x y 4 x x2 y + o( x 3. Exercise 8.86 (2 sin(x 2 + y 2 = (x 2 + y 2 + O((x 2 + y 2 3 = x 2 + y 2 + o( x 6. Exercise 8.86 (3 By (1, we have y log x + z log y = (1 + y ( x x2 + x3 + (1 + z ( y y2 + y o( x 3. Therefore x y y z = e y log x+z log y = 1 + (1 + y ( x x2 + x3 + (1 + z ( y y2 + y ( 2 (1 + y ( x x2 + (1 + z ( y y ( x + y3 + o( x 3 = 1 + x + y + x y + y z x2 y + x y y2 z + x y z + o( x 3. Exercise 8.86 (4 1 0 (1 + x t2y dt = = e t2 y log(1+x dt = ( 1 + t 2 y 1 0 (x x2 2 = yx 1 6 yx2 + o( x 3. e t2 y (x x2 2 +o(x2 dt t4 y 2 x 2 + o( x 3 dt Exercise 8.87

18 The Taylor expansion T k ( x = f( x 0 + f ( x 0 ( x + 1 2! f ( x 0 ( x k! f (k ( x 0 ( x is a k-th order polynomial approximating f at x 0. If x 0 = L( y 0, then the T k L( y = T k ( x 0 + L( y = f( x 0 + f ( x 0 (L( y + 1 2! f ( x 0 (L( y k! f (k ( x 0 (L( y is a k-th order polynomial approximating f L at y 0. Moreover, f (k ( x 0 (L( y is still a k-th order form. Therefore we conclude that (f L (k ( y 0 = f (k (L( y 0 L. Specifically, let us consider the third order partial derivatives of g(x, y = f(ax+by, cx+dy. The equality (f L ( y 0 = f (L( y 0 L becomes g xxx u 3 + 3g xxy u 2 v + 3g xyy uv 2 + g yyy v 3 =f xxx (au + bv 3 + 3f xxy (au + bv 2 (cu + dv + 3f xyy (au + bv(cu + dv 2 + f yyy (cu + dv 3. Comparing the coefficients of u 3, we get (f xxx means f xxx (ax + by, cx + dy, etc 3 (f(ax + by, cx + dy x 3 = a 3 f xxx + 3a 2 cf xxy + 3ac 2 f xyy + c 2 f yyy. Comparing the coefficients of u 2 v, we get 3 (f(ax + by, cx + dy x 2 y = 3a 2 bf xxx + 3(a 2 d + 2abcf xxy + 3(2acd + bc 2 f xyy + 3c 2 df yyy. Exercise 8.88 By f xi + f y y xi = 0, we have y xi = f x i f y. Then y xi x j = (f x i x j + f xi yy xj f y f xi (f xj y + f yy y xj f 2 y = f x i f xj f yy + f xi x j fy 2 (f xi f xj y + f xj f xi yf y. fy 3 Exercise 8.89 The argument in Example can be easily generalized. Let K {1, 2,..., n} be a subset. Let K be the number of numbers in K. Let A K be the square submatrix of A made up of (i, j-entries, with both i, j K. Then det (k (I(H = k! det A K. For the case k > n, we have det (k (I(H = 0. Exercise 8.90 By (A + H k = A k + i 1 +i 2 =k 1,i 1,i 2 0 Ai 1 HA i 2 + i 1 +i 2 +i 3 =k 2,i 1,i 2,i 3 0 Ai 1 HA i 2 HA i 3 + i 1 +i 2 +i 3 +i 4 =k 3,i 1,i 2,i 3,i 4 0 Ai 1 HA i 2 HA i 3 HA i 4 +, we have F (A(H = 6 K =k i 1 +i 2 +i 3 +i 4 =k 3,i 1,i 2,i 3,i 4 0 A i 1 HA i 2 HA i 3 HA i 4.

19 Exercise 8.91 For invertible A and small H, we have (A+H 1 = (I+A 1 H 1 A 1 = A 1 A 1 HA 1 + A 1 HA 1 HA 1 A 1 HA 1 HA 1 HA 1 +. Thus F (k (A(H = ( 1 k k!(a 1 H k A 1. Exercise 8.92 The function is homogeneous of degree 0. In other words, the restriction to any straight line passing through the origin is a constant away form (0, 0. Such functions are not continuous (let alone differentiable at (0, 0, unless they are constantly 0. By induction, we can show that, we can take all the partial derivatives up to (k 1-st order, and these partial derivatives are of the form { f x i y j(x, y = x k i y k j g(x, y, if (x, y (0, 0, 0, if (x, y = (0, 0, where g(x, y is a homogeneous bounded continuous function of degree 0. Then we can show that f has all the partial derivatives of k-th order.