Solutions to Exercise Session, March 30, 2015

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1 Analysis II Prof. Jan Hesthaven Spring Semester ÉC O L E P O L Y T E C H N I Q U E FÉ DÉR A L E D E L A U S A N N E Solutions to Exercise Session, March 30, Implicit functions I. Show that the equation ln x + e y x = 1 defined in the neighborhood of the point 1 is an implicit function y = g(x such that g(1 = 0. Give the equation of the tangent to the curve y = g(x at 1. We define the function f : U R with U = R + \ {0} R by f(x, y = ln x + e y x 1 Then, the function f is of class C 1 (U (it is even of class C k (U for all k 1 and for all (x, y U: D f(x, y = e y x x. Moreover, f(1, 0 = 0 and D f(1, 0 = 1 0. So, by the theorem of implicit functions, there exists an interval I =]1 ɛ, 1 + ɛ[ and a unique function g : I R of class C 1 (I such that g(1 = 0 and f(x, g(x = 0. The derivative of g is given by g (x = D 1f(x, g(x D f(x, g(x = g(x x e g(x x So g (1 = 1. Hence, the equation of the tangent to the curve y = g(x at x = 1 is y = g(1 + g (1(x 1 = 1 x.. Implicit functions II. Show that the equation cos(x + y + sin(x + y + e x3y = defined in the neighborhood of the point 0 is an implicit function y = g(x such that g(0 = π/. Show that the function g has a local maximum at 0. We define the function f : R R by f(x, y = cos(x + y + sin(x + y + e x3y So, the function f is of class C k, for all k 1, and for all (x, y R : D f(x, y = sin(x + y + cos(x + y + x 3 e x3y. Moreover, f(0, π/ = 0 and D f(0, π/ = 1 0. So, by the theorem of implicit functions, there exists an interval I =] ɛ, ɛ[ and a unique function g : I R of class C 1 (I such that g(0 = π/ and f(x, g(x = 0. The derivative of g is given by g (x = D 1f(x, g(x D f(x, g(x and D 1 f(x, y = x sin(x +y+cos(x+y+3x ye x3y. So g (0 = 0. The second derivative in x = 0 is g (0 = D 11f(0, π/ D f(0, π/ = 3.

2 3. Implicit functions III. Show that the equation x 5 + xyz + y 3 + 3xz 4 = defined in the neighborhood of the point (1, 1 is an implicit function z = g(x, y such that g(1, 1 = 1. Give the equation of the plane tangent to the surfuce z = g(x, y in (1, 1. We define the function f : R 3 R by f(x, y, z = x 5 + xyz + y 3 + 3xz 4 Then, the function f is of class C k, for all k 1, and for all (x, y, z R 3 : D 3 f(x, y, z = xy + 1xz 3. Moreover, f(1, 1, 1 = 0 and D 3 f(1, 1, 1 = Then, by the theorem of implicit functions, there exists a neighborhood B ɛ (1, 1 R and a unique function g : B ɛ (1, 1 R of class C k (B ɛ (1, 1 such that g(1, 1 = 1 and f(x, y, g(x, y = 0. The equation of the tangent plane to the surface z = g(x, y in (1, 1 is given by 0 = f(1, 1, 1, x 1 y + 1 = 0 z 1 i.e. using we find 5x 4 + yz + 3z 4 f(x, y, z = xz + 3y xy + 1xz 3 7x + 4y + 11z = Quadratic forms I. Calculate the symmetric matrices A and the eigenvalues of the following quadratic forms q(x = 1 Ax, x : q : R R, q(x, y = x + 19 y + 56xy Study their stationary points. q : R 3 R, q(x, y = x + y + z(x y Solution - 1. A = Its characteristic polynomial is given by ( p A (λ = det(λid A = (λ 4 (λ The eigenvalues are λ 1 = 45 and λ = 68. The matrix is then invertible. So (x, y = (0, 0 is the unique solution of x 0 q(x, y = A = y 0 The matrix A is indefinite. Hence, (x, y = (0, 0 is a saddle point of q(x, y. Page

3 Solution -. Its characteristic polynomial is given by A = p A (λ = det(λid A = (λ λ 4(λ 4(λ = (λ (λ λ 8. The eigenvalues are λ 1 =, λ = and λ 3 = 4. The matrix is then invertible. So (x, y, z = (0, 0, 0 is the unique solution to x 0 q(x, y, z = A y = 0 z 0 The matrix A is indefinite. Hence, (x, y, z = (0, 0, 0 is a saddle point of q(x, y, z. 5. Quadratic forms II. Let A M n,n (R be a positive-definite, symmetric matrix. Let v R n. Show that the function f(x defined by f(x = 1 Ax, x v, x has a unique stationary point at a = A 1 v. Then show that f(x f(a > 0 for all x a. Note first that A positive-definite implies that A is invertible. The stationary points of f are given by the solutions of the equation f(x = 0 so Ax v = 0. When A is invertible, this equation has as unique solution, the vector a = A 1 v. It is a strict local minimum as Hess (f(a = A > 0. Moreover, f(a = 1 AA 1 v, A 1 v v, A 1 v = 1 v, A 1 v and (note that A 1 is also symmetric hence, writing v, x = 1 Ax, A 1 v + 1 AA 1 v, x we find f(x f(a = 1 A(x A 1 v, (x A 1 v > 0 for all x A 1 v. 6. Study the nature of the stationary points of the function f : R R given by f(x, y = (1 x sin y. The stationary points are given by the solutions of f(x, y = 0 i.e. The function f has four families of stationary points: x sin y = 0 (1 x cos y = 0. P k = (0, π + kπ, Q k = (0, 3π + kπ, S k = ( 1, kπ, T k = (1, kπ for k Z. Proof: If x = 0, then D x f = 0, and D y f = 0 if and only if cos y = 0 hence the P k, Q k. If Page 3

4 sin y = 0, then y = kπ and D y f = 0 if and only if x = 1 or x = +1, hence the S k, T k. To study the stationary points, we calculate the Hessian matrix f: sin y x cos y (Hess f(x, y = x cos y (1 x sin y We get hence some local maxima, hence some local minima (Hess f(s k = hence some saddle points. 7. Let f : R R be defined by (Hess f(p k = (Hess f(q k = ( ±, (Hess f(t ± 0 k = f(x, y = (x y 3 + 4x 3x + 3y. ( 0 ± ± 0 (a Give the stationary points of f and study their nature. Calculate f at these points. (b Let T be the domain given by: T = {(x, y R : y 0, y x 4 y}. Give the minimum and the maximum of f on T. In particular, i. Show that T is bounded. ii. Show that T T and conclude that T is closed. iii. Show that T is a triangle and give its summits. iv. Explain why f has its maximum and minimum on T. v. Give f on the boundary of T, i.e. f T and then study f T. vi. Give the minimum and the maximum of f on T. (a 3(x y f(x, y = + 8x 3 3(x y + 3 ( 6x 6y + 8 6x + 6y Hessf(x, y = 6x + 6y 6x 6y Stationary points: f(x, y = 0 x = 0, (x y = 1 P 1 = (0, 1, P = (0, 1 ( Hessf(0, 1 =, Hessf(0, 1 = P 1 : min. loc. as det and trace > 0, f(0, 1 = P : saddle point as det < 0, f(0, 1 = Page 4

5 (b (i. The definition of T gives us the inequalities 0 y x 4 y 4. So which implies that T is bounded. T [0, 4] [0, 4] (ii. The boundary T is given by the segments S 1 = {(x, y R : y = 0, x [0, 4]} S = {(x, y R : y = 4 x, x [, 4]} S 3 = {(x, y R : y = x, x [0, ]} which are in T (seen by the signs and not < in the definition of T. So T is closed (see exercise 8 of chapter 1. (iii. The boundary of T is given by 3 segments which have 3 intersection points (the summits: A = (0, 0, B = (4, 0, C = (, (iv. f is continuous (polynomial and T bounded and closed hence we conclude that f has its min and max on T. (v. f 1 (x := f S1 = f(x, 0 = x 3 + 4x 3x, x [, 4] f (x := f S = f(x, 4 x = 8x 3 44x + 90x 5, x [, 4] f 3 (x := f S3 = f(x, 4 x = 4x, x [0, ] f 1(x := 3x + 8x 3, x = 1/3, f 1 (0 = 0, f 1 ( 1 3 = 14/7, f 1(4 = 116 f (x := 4x 88x + 90, no stationary points., f ( = 16, f (4 = 116 f 3(x := 8x, x = 0, f 3 (0 = 0, f 3 ( = 16 (vi. min f T = 14/7, max f T = Find the maximum and the minimum of f(x, y = sin x + sin y + cos(x + y. on the set S given by S = [0, π] [0, π]. Uselful formulas - trigonometric functions. sin 0 = 0 sin π 6 = 1 sin π 4 = sin π 3 = 3 sin π = 1 cos 0 = 1 cos π 6 = 3 cos π 4 = cos π 3 = 1 cos π = 0 sin x + cos x = 1, sin( π x = sin(π + x, cos(π x = cos(π + x sin(x + y = sin x cos y + cos x sin y, cos(x + y = cos x cos y sin x sin y Page 5

6 Solutions. We consider a continuous function on a closed and bounded set. So f(x, y has a maximum and a minimum among its stationary points or on the boundary of the set S. To find the stationary points, we have to solve f(x, y = 0, i.e. cos(x sin(x + y = 0 cos(y sin(x + y = 0 So cos(y = cos(x. The function cos is strictly decreasing on [0, π] and hence x = y. To solve the equation cos(x sin(x = 0 we use the relation sin(x = sin x cos x to find cos x(1 sin x = 0, where the roots are x 1 = π 6, x = π or else x 3 = 5π 6 (i.e. x k = (k+1π 6, k = 1,, 3. f has three stationary points P k = (x k, x k. To study the nature of the stationary points, we use the Hessian matrix of f given by sin x cos(x + y cos(x + y (Hess f(x, y = cos(x + y sin y cos(x + y Denote A k the Hessian matrix of f in P k. Then ( 1 1 A 1 = 0 1, A 1 = 1 0, 1 ( 1 1, A 3 = 1 1. The function f has a local maximum in P 1 and P 3 and a saddle point in P. We get f(p 1 = f(p 3 = 3, f(p = 1 On the boundary S the function f is given by: L 1 = [0, π] {0} : f(x, 0 = sin x + cos x and f (x, 0 = cos x sin x hence a stationary point π/4 min f L1 = f(0, π = 1, max f L1 = f(π/4, 0 =. L = {π} [0, π] : f(π, y = sin y + cos(π + y = sin y cos y and min f L = 1, max f L = f(π, 3 π/4 =. L 3 = [0, π] {π} : f(x, π = sin x + cos(π + x and min f L3 = 1, max f L3 =. L 4 = {0} [0, π] : f(0, y = sin y + cos y and min f L4 = 1, max f L4 =. Hence, min f S = 1, max f S = 3. Page 6

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