Solutions to Homework # 5. Section 2.2 # 12a: Prove that every convergent sequence is a Cauchy sequence.

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1 Solutions to Homework # 5 Section 2.2 # 12a: Prove that every convergent sequence is a Cauchy sequence. Proof: Suppose that {x n } is a sequence which converges to a R k. Let ɛ > 0. Choose N so that if n > N, then x n a < ɛ/2. Then, by the triangle inequality, x n x m = x n a + a x m < ɛ if m, n > N. Hence, {x n } is a Cauchy sequence. Section 2.2 #12b: If a subsequence of a Cauchy sequence converges, then the sequence converges. Proof: Suppose that {x n } is a Cauchy sequence and that {x nk } is a subsequence that converges to a R l. Let ɛ > 0. Since the sequence is Cauchy, there exists an N 1 such that if m, n > N 1, then x m x n < ɛ/2. Since the subsequence converges to a, there is a K 1 such that if k > K 1, then x nk a Mɛ/2. Let N be the maximum of N 1 and K 1. Let k = N + 1. Then, by the triangle inequality, if n > K, then x n a x n x nk + x nk a ɛ, since n k k > N. Therefore, {x n } converges to a. Section 2.2 #13: Prove that every Cauchy sequence is contained in some ball centered at the origin. Proof: Suppose {x n } is a Cauchy sequence. Then, choosing ɛ = 1, we know that there exists an N such that if m, n > N, then x m x n < 1. Let R be the largest value among x 0,..., x N+1. And let R = R + 1. I claim that x n B(0, R) for every n N. This is true by design if n {0,..., N + 1}. And if n > N + 1, then x n x N+1 < 1. So, x n < x N R. This proves the claim. Section 2.2 #14a: Suppose that {x n } is a sequence of real numbers in the interval [a, b]. Prove that {x n } has a convergent subsequence. Strategy: In Homework #2 (see the solution to exercise #10 in section 2.2), we proved that if I 0 = [a, b] and for each n N, I n+1 I n and each I n, then there exists a real number c which belongs to the intersection of all I n. I will appeal to this result in order to solve the current exercise.

2 The point is that since {x n } is an infinite sequence (meaning infinitely many terms, though not necessarily infinitely many values for these terms), for any subset S [a, b], either there are infinitely many n such that x n S or there are infinitely many n such that x n / S. By choosing S to be either the left half or the right half of [a, b], we can narrow our focus to the infinite subsequence in either the left or right interval; since this interval is half as long, repeating this process will result in a sequence of nonempty nested intervals. Proof: Let I 0 = [a, b]. Let n 0 = 0. Assuming that I k and n k has been defined, define I k+1 and n k+1 as follows: let m k be the midpoint of I k, l k the left endpoint of I k and r k the right endpoint of I k. If there are infinitely many n > n k such that x n [l k, m k ] (i.e. belonging to the left half of I k ), the let I k+1 = [l k, m k ] and choose some n k+1 > n k such that x nk+1 I k+1. Otherwise, let I k+1 = [m k, r k ] (i.e. the right half of I k ) and choose n k+1 > n k such that x nk+1 I k+1. The above procedure recursively defines the subsequence {x nk }. By design, x nk I k for each k N, and I k+1 I k. By exercise #10 in section 2.2, there exists a real number c k=1 I k. I claim that {x nk } c. Let ɛ > 0. Choose K such that (b a)/2 K < ɛ. Notice that for each k, (b a)/2 k is the length of the interval I k. Therefore, x nk c (b a)/2 k < (b a)/2 K < ɛ for every k > K. Hence, {x nk } c. Section 2.2 #14b: Prove that every Cauchy sequence in R n is convergent. Proof: By exercise 13, a Cauchy sequence in R is contained in [ R, R] (the ball of radius R centered at 0) for some R. By exercise 14a, this Cauchy sequence has a convergent subsequence in [ R, R], and by exercise 12b, the original sequence converges. Section 2.2 #14c: Prove that every Cauchy sequence in R l converges. Proof: By exercise 13, there is an R > 0 such that the Cauchy sequence is contained in B(0, R). Therefore, the sequence is contained in the larger set [ R, R] [ R, R] R l. Each coordinate determines a Cauchy sequence (why is it Cauchy?), and by 14b, each of the coordinate sequences converges to some c i [ R, R] for i = 1,..., l. Finally, since a sequence of

3 vectors converges if and only if each coordinate converges, the original sequence of vectors converges. Section 3.2 #10: Find the derivative of the map ( ) [ x x f = 2 y 2 ] y 2xy at the point a. Show that whenever a 0 that the linear map Df(a) is a scalar multiple of a rotation matrix. Solution: Let f 1 (x, y) = x 2 y 2 and f 2 (x, y) = 2xy be the coordinate functions. Since the derivative is represented by the Jacobian matrix (the matrix of partial derivatives), we have that Df ( ) a = b [ 2a 2b 2b 2a To see this is a scalar multiple of a rotation matrix if a 0, write a as its length times a unit vector in the same direction as a. So, if s = a/ a, t = b/ a, and u = [s, t] T, then a = a u. Let θ [0, π] be the angle that the vector u makes with the positive x-axis, i.e. the angle between the vectors u and e 1. Then cos(θ) = u e 1 = ]. a a Since θ [0, π], the sine of such an angle is non-negative (hence the absolute value around b in the following equation); therefore, sin θ = 1 cos 2 θ = b a Thus, the derivative matrix is a rotation matrix scaled by a factor of 2 a. The rotation matrix (1/2 a )Df(a) represents a rotation by θ in a clockwise sense if b < 0 and in a counterclockwise sense if b > 0. Section 3.2 #12: Let f(x, y) = x 2 y/(x 4 + y 2 ) if (x, y) (0, 0) and f(0, 0) = 0. Show directly that f fails to be C 1 at the origin. Strategy: Compute the partial derivatives and decide that one (or both) are not continuous at the origin.

4 Solution: If (a, b) 0, then and At the origin, and, similarly, f y (0, 0) = 0. f x (a, b) = (x4 + y 2 )(2xy) x 2 y(4x 3 ) (x 4 + y 2 ) 2 f y (a, b) = (x4 + y 2 )x 2 x 2 y(2y) (x 4 + y 2 ) 2. f(t, 0) f(0, 0) f x (0, 0) = lim = 0, t 0 t If f C 1, then both f x and f y are continuous at the origin. However, lim (x,y) (0,0) f y (x, y) does not exist: If y = 2x 2 and x 0, then (x 4 + 4x 4 )x 2 x 2 2x 2 (4x 2 ) x 6 lim x 0 (x 4 + 4x 4 ) 2 = lim x 0 5x 8 = lim 1 x 0 5x 2 does not exist. So, in particular, lim (x,y) (0,0) f y (x, y) is not equal to f y (0, 0). Section 3.2 #15: Let a R n and δ > 0. Suppose that f : B(a, δ) R is differentiable at a. Suppose that f(a) f(x) for a all x B(a, δ). Prove that Df(a) = 0. Proof: It suffices to show that each partial derivative at a is zero. By considering the left-hand and right-hand limits of f(a + te i ) f(a) lim t 0 t we see that the these limits are non-negative and non-positive, respectively since f(a + te i ) f(a) for every t ( δ, δ). Therefore, x i (a) = 0 for each i = 1,..., n. Section 3.3 # 1: Suppose that f : R 3 R is differentiable and Df(a) = [2, 1, 1], where a = [1, 1, 1] T. If compute (f g) (0). g(t) = [cos t + sin t, t + 1, t 2 + 4t 1] T,

5 Solution: Observe that g(0) = [1, 1, 1] T. Direct computation shows that g (0) = [1, 1, 4] T. So, by the chain rule, (f g) (0) = Df(g(0))g (0) = [2, 1, 1][1, 1, 4] T = 1. Section 3.3 #2: Suppose that f(x, y) = [2y sin x, e x+3y, xy + y 3 ] T, and g(x, y, z) = [3x + y z, x + yz + 1] T. Calculate D(f g)(0, 0, 0) and D(g f)(0, 0). Solution: Observe that both f and g are differentiable. (This follows, for instance, from the fact that one can easily check that every partial derivative exists and is continuous, and so f and g belong to the class C 1. Therefore, we can apply the chain rule. Direct computation shows that cos x 2 Df(x, y) = e x+3y 3e x+3y y x + 3y 2 and another computation shows that Dg(x, y, z) = [ 3 1 ] 1 1 z y Now, a direct application of the chain rule shows that D(f g)(0, 0, 0) = Df(0, 1)Dg(0, 0, 0) = 1 2 [ ] e 3 3e = compare with answer in textbook Another computation using the chain rule shows that D(g f)(0, 0) = Dg(0, 1, 0)Df(0, 0) =

6 [ ] = compare with answer in textbook Section 3.3 #3: In this problem, you are asked to compute the derivative of a composite (f g) (t). If the calculation is done correctly, you will find that this derivative is always equal to zero. If you have trouble obtaining this answer, look for a way to apply the double angle formula for the sine function. Since the composite has derivative zero, this means that the composite function is constant. This can be verified directly by computing (f g)(t). (If you did not use the chain rule, then you may have already done this in the first place.) The constant value is three if you work it out. You will want to use the half angle formula for the sine function to see this. Section 3.3 #4: An ant moves along a helical path with trajectory g(t) = [3 cos t, 3 sin t, 5t] T. At what rate is the ant s distance from the origin changing at t = 2π? If the temperature in space is given by f(x, y, z) = xy + z 2, at what rate does the ant detect the temperature to be changing at t = 3π/4? Solution: The first question asks for the computation of (h g) (2π), where h(x, y, z) = x 2 + y 2 + z 2. The second question asks for the computation of (f g) (3π/4). One computes that [ Dh(x, y, z) = x x 2 +y 2 +z 2 y x 2 +y 2 +z 2 ] z x 2 +y 2 +z 2 and that Since g(2π) = [3, 0, 10π] T, Dg(t) = [ 3 sin t, 3 cos t, t] T. (h g) (0) = Dh(3, 0, 10π)Dg(0) = 50π π 2. And one computes that Df(x, y, z) = [y, x, 2z], g(3π/4) = [ 3/ 2, 3/ 2, 15π/4] T, and g (3π/4) = [ 3/ 2, 3/ 2, 5] T. And so, after some computation, one finds that (f g) (3π/4) = 75π 2.

7 Section 3.3 # 5: An airplane is flying near a tower. When the plane is 3 miles West of the tower, the plane has altitude of 4 miles and ground speed of 450 miles per hour and a climb rate of 5 miles per hour. Suppose that (a) the plane is flying due East or that (b) the plane is flying due Northeast. In each case, what is the rate at which the plane is approaching the tower? Solution: We need to choose how to associate compass directions to the rectangular coordinates in R 3. I ll choose North to refer to the positive y-axis, East to refer to the positive x-axis, and positive altitude to refer to the positive z-axis. In case (a), the plane has an initial velocity vector of v(0) = [450, 0, 5] T and has an initial position vector of s(0) = [ 3, 0, 4]. In case (b), the plane has an initial velocity vector of v(0) = [450/ 2, 450/ 2, 5] T and an initial position vector of s(0) = [ 3.0, 4]. (The factors of 2 arise from the fact that the ground speed of 450 miles per hour in a Northeasterly direction decomposes into two components: one in a Northerly direction and one in an Easterly direction.) In both cases, one is being asked to compute (h s) (0), where h(x, y, z) = x 2 + y 2 + z 2 measures the distance from the plane to the tower and s(t) is the position of the plane. Since Ds(0) = v(0), we have sufficient information to answer these questions if we apply the chain rule: (h s) (0) = Dh(s(0))Ds(0) = Dh(s(0))v(0). The derivative of h is computed as in the previous problem. You can check your answer with the answer in the back of the textbook. Exercise: Suppose that f : U R 2 R is expressed in rectangular coordinates by f(x, y). The expression of f in polar coordinates is given by writing x = r cos θ and y = r sin θ, so that f may be viewed as a function of (r, θ). Determine an expression for f x and f y in terms of f r and f θ. And determine an expression for f r and f θ in terms of f x and f y. Solution: In polar coordinates, x = r cos θ and y = r sin θ. By the chain rule, r = x x r + y = cos θ + sin θ y r x y

8 and, θ = x x θ + y y θ = r sin θ + r cos θ x y. The above two equations expres f r and f θ in terms of f x and f y. To express f x and f y in terms of f r and f θ, one solves the above system of two linear (in the symbols f x, f y, f r, and f θ ) equations. Observe that the above two equations can be expressed by a single vector equation: [ ] [ ] [ ] cos θ r sin θ fr f θ = fx f y sin θ r cos θ To express f x and f y in terms of f r and f θ, we need to multiply on the right by the inverse of the 2 2 matrix. It s inverse is [ ] 1 r cos θ r sin θ r sin θ cos θ Therefore, In other words, and, [ ] [ ] [ ] cos θ sin θ fx f y = fr f θ sin θ r = cos θ x r 1 sin θ r θ y cosθ r = sin θ r + 1 cos θ r θ. Exercise: Show that Laplace s equation f xx + f yy = 0 has the following expression in polar coordinates: 2 f r r r f r 2 θ 2 = 0. Solution: Compute the partial derivatives of f r and f θ with respect to r and θ respectively, starting with the results from the exercise above. It may be helpful to interpret the operations of partial differentiation with respect to r and θ as follows: r = cos θ x + sin θ y

9 and So, r θ = r sin θ x + r cos θ y. ( ) = ( cos θ ) + sin θ r r x y = cos 2 θ 2 f x cos θ sin θ 2 f x x + sin2 θ 2 f y 2. Similarly, one computes ( ) = θ θ θ ( r sin θ ) + r cos θ. x y = r cos θ r sin θ x y +r2 sin 2 θ 2 f x 2 2r2 cos θ sin θ 2 f x x +r2 cos 2 θ 2 f y 2. To obtain the above, be certain that you use the product rule when computing the partial derivative with respect to θ on a term such as r sin θf x. The above can be regrouped as follows: 2 ( f θ 2 = r cos θ + sin θ x y )+r 2 ( sin 2 θ 2 f x 2 2 cos θ sin θ 2 f x x + cos2 θ 2 f y 2 So, it is now a matter of putting the pieces together to deduce that Laplace s equation as the form stated in the exercise. ), Exercise: Suppose that f, g : U R R 3. Let h(t) = f(t) g(t), where is the cross product (a.k.a. vector product). Determine a rule which relates the derivative of h(t) to the derivatives of f(t) and g(t). Prove your assertion. Solution: The answer is as follows: D(f g)(t) = Df(t) g(t) + f(t) Dg(t). To see this, one uses the definition of the cross product. Writing f 1, f 2, f 3 and g 1, g 2, g 3 for the component functions, one has, on the one hand, f 2 g 3 g 2 f 3 f g = f 3 g 1 f 1 g 3. f 1 g 2 f 2 g 1

10 So by the product rule, f 2 g 3 + f 2 g 3 (g 2 f 3 + g 2 f 3 ) D(f g) = f 3 g 1 + f 3 g 1 (g 1 f 3 + g 1 f 3 ) f 1 g 2 + f 1 g 2 (g 2 f 1 + g 2 f 1 ) And, on the other hand, one has that f (f 2 g 3 g 2 f 3 f 2 g g) = f 3 g 1 f 1 g 3, (f g 3 g 2 f 3 ) = f 3 g f 1 g 2 f 2 g 1 f 1g 3 1 f 1 g 2 f 2g 1 and it is easy to see that the sum of the two vectors above is D(f g).