Solutions to Homework # 5. Section 2.2 # 12a: Prove that every convergent
sequence is a Cauchy sequence. Proof: Suppose that {xn} is a sequence which ...
Solutions to Homework # 5 Section 2.2 # 12a: Prove that every convergent sequence is a Cauchy sequence. Proof: Suppose that {xn } is a sequence which converges to a ∈ Rk . Let > 0. Choose N so that if n > N , then kxn − ak < /2. Then, by the triangle inequality, kxn − xm k = kxn − ak + ka − xm k < if m, n > N . Hence, {xn } is a Cauchy sequence. Section 2.2 #12b: If a subsequence of a Cauchy sequence converges, then the sequence converges. Proof: Suppose that {xn } is a Cauchy sequence and that {xnk } is a subsequence that converges to a ∈ Rl . Let > 0. Since the sequence is Cauchy, there exists an N1 such that if m, n > N1 , then kxm − xn k < /2. Since the subsequence converges to a, there is a K1 such that if k > K1 , then kxnk − akM /2. Let N be the maximum of N1 and K1 . Let k = N + 1. Then, by the triangle inequality, if n > K, then kxn − ak ≤ kxn − xnk k + kxnk − ak ≤ , since nk ≥ k > N . Therefore, {xn } converges to a. Section 2.2 #13: Prove that every Cauchy sequence is contained in some ball centered at the origin. Proof: Suppose {xn } is a Cauchy sequence. Then, choosing = 1, we know that there exists an N such that if m, n > N , then kxm − xn k < 1. Let R0 be the largest value among kx0 k, . . . , kxN +1 k. And let R = R0 + 1. I claim that xn ∈ B(0, R) for every n ∈ N. This is true by design if n ∈ {0, . . . , N + 1}. And if n > N + 1, then kxn − xN +1 k < 1. So, kxn k < kxN +1 k + 1 ≤ R. This proves the claim. Section 2.2 #14a: Suppose that {xn } is a sequence of real numbers in the interval [a, b]. Prove that {xn } has a convergent subsequence. Strategy: In Homework #2 (see the solution to exercise #10 in section 2.2), we proved that if I0 = [a, b] and for each n ∈ N, In+1 ⊂ In and each In 6= ∅, then there exists a real number c which belongs to the intersection of all In . I will appeal to this result in order to solve the current exercise.
The point is that since {xn } 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 xn ∈ S or there are infinitely many n such that xn ∈ / 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 I0 = [a, b]. Let n0 = 0. Assuming that Ik and nk has been defined, define Ik+1 and nk+1 as follows: let mk be the midpoint of Ik , lk the left endpoint of Ik and rk the right endpoint of Ik . If there are infinitely many n > nk such that xn ∈ [lk , mk ] (i.e. belonging to the left half of Ik ), the let Ik+1 = [lk , mk ] and choose some nk+1 > nk such that xnk+1 ∈ Ik+1 . Otherwise, let Ik+1 = [mk , rk ] (i.e. the right half of Ik ) and choose nk+1 > nk such that xnk+1 ∈ Ik+1 . The above procedure recursively defines the subsequence {xnk }. By design, xnk ∈ Ik for each k ∈ N, and Ik+1 ⊂ Ik . By exercise #10 in section 2.2, there exists a real number c ∈ ∩∞ k=1 Ik . I claim that {xnk } → c. Let > 0. Choose K such that (b − a)/2K < . Notice that for each k, (b − a)/2k is the length of the interval Ik . Therefore, kxnk − ck ≤ (b − a)/2k < (b − a)/2K < for every k > K. Hence, {xnk } → c. Section 2.2 #14b: Prove that every Cauchy sequence in Rn 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 Rl 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] ⊂ Rl . Each coordinate determines a Cauchy sequence (why is it Cauchy?), and by 14b, each of the coordinate sequences converges to some ci ∈ [−R, R] for i = 1, . . . , l. Finally, since a sequence of
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 2 x x − y2 f = y 2xy at the point a. Show that whenever a 6= 0 that the linear map Df (a) is a scalar multiple of a rotation matrix. Solution: Let f1 (x, y) = x2 − y 2 and f2 (x, y) = 2xy be the coordinate functions. Since the derivative is represented by the Jacobian matrix (the matrix of partial derivatives), we have that 2a −2b a . = Df 2b 2a b To see this is a scalar multiple of a rotation matrix if a 6= 0, write a as its length times a unit vector in the same direction as a. So, if s = a/kak, t = b/kak, and u = [s, t]T , then a = kaku. 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 e1 . Then cos(θ) = u · e1 =
a kak
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 =
p |b| 1 − cos2 θ = kak
Thus, the derivative matrix is a rotation matrix scaled by a factor of 2kak. The rotation matrix (1/2kak)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) = x2 y/(x4 + y 2 ) if (x, y) 6= (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.
Solution: If (a, b) 6= 0, then fx (a, b) =
(x4 + y 2 )(2xy) − x2 y(4x3 ) (x4 + y 2 )2
and fy (a, b) =
(x4 + y 2 )x2 − x2 y(2y) . (x4 + y 2 )2
At the origin, f (t, 0) − f (0, 0) = 0, t→0 t
fx (0, 0) = lim and, similarly, fy (0, 0) = 0.
If f ∈ C 1 , then both fx and fy are continuous at the origin. However, lim(x,y)→(0,0) fy (x, y) does not exist: If y = 2x2 and x → 0, then −x6 −1 (x4 + 4x4 )x2 − x2 2x2 (4x2 ) = lim = lim 2 4 4 2 8 x→0 5x x→0 5x x→0 (x + 4x ) lim
does not exist. So, in particular, lim(x,y)→(0,0) fy (x, y) is not equal to fy (0, 0). Section 3.2 #15: Let a ∈ Rn 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 lim
t→0
f (a + tei ) − f (a) t
we see that the these limits are non-negative and non-positive, respectively ∂f (a) = 0 for each since f (a + tei ) ≤ f (a) for every t ∈ (−δ, δ). Therefore, ∂x i i = 1, . . . , n. Section 3.3 # 1: Suppose that f : R3 → R is differentiable and Df (a) = [2, 1, −1], where a = [1, 1, −1]T . If g(t) = [cos t + sin t, t + 1, t2 + 4t − 1]T , compute (f ◦ g)0 (0).
Solution: Observe that g(0) = [1, 1, −1]T . Direct computation shows that g 0 (0) = [1, 1, 4]T . So, by the chain rule, (f ◦ g)0 (0) = Df (g(0))g 0 (0) = [2, 1, −1][1, 1, 4]T = −1. Section 3.3 #2: Suppose that f (x, y) = [2y − sin x, ex+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 3ex+3y Df (x, y) = ex+3y y x + 3y 2 and another computation shows that 3 1 −1 Dg(x, y, z) = 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 e3 3e3 3 1 −1 = compare with answer in textbook 1 0 0 1 3 Another computation using the chain rule shows that D(g ◦ f )(0, 0) = Dg(0, 1, 0)Df (0, 0) =
−1 2 3 1 −1 1 3 = compare with answer in textbook 1 0 1 0 0
Section 3.3 #3: In this problem, you are asked to compute the derivative of a composite (f ◦ g)0 (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? 0 Solution: The p first question asks for the computation of (h ◦ g) (2π), where h(x, y, z) = x2 + y 2 + z 2 . The second question asks for the computation of (f ◦ g)0 (3π/4).
One computes that h x Dh(x, y, z) = √x2 +y2 +z 2
√
y
√
x2 +y 2 +z 2
z
i
x2 +y 2 +z 2
and that Dg(t) = [−3 sin t, 3 cos t, t]T . Since g(2π) = [3, 0, 10π]T , (h ◦ g)0 (0) = Dh(3, 0, 10π)Dg(0) = √
50π . 9 + 100π 2
And one computes √ that √ Df (x, y, z) = [y, x, 2z], √ √ g(3π/4) = [−3/ 2, 3/ 2, 15π/4]T , and g 0 (3π/4) = [−3/ 2, −3/ 2, 5]T . And so, after some computation, one finds that (f ◦ g)0 (3π/4) =
75π . 2
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 R3 . 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, (b), the plane √ 0, 4]. √In case T has an initial velocity vector of v(0) = [450/ 2, 450/ √ 2, 5] 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.) 0 In both cases, p one is being asked to compute (h ◦ s) (0), where h(x, y, z) = x2 + 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 (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 ⊂ R2 → 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 fx and fy in terms of fr and fθ . And determine an expression for fr and fθ in terms of fx and fy . Solution: In polar coordinates, x = r cos θ and y = r sin θ. By the chain rule, ∂f ∂x ∂f ∂y ∂f ∂f ∂f = + = cos θ + sin θ ∂r ∂x ∂r ∂y ∂r ∂x ∂y
and, ∂f ∂f ∂x ∂f ∂y ∂f ∂f = + = −r sin θ + r cos θ . ∂θ ∂x ∂θ ∂y ∂θ ∂x ∂y The above two equations expres fr and fθ in terms of fx and fy . To express fx and fy in terms of fr and fθ , one solves the above system of two linear (in the symbols fx , fy , fr , and fθ ) equations. Observe that the above two equations can be expressed by a single vector equation: cos θ −r sin θ fr fθ = fx fy sin θ r cos θ To express fx and fy in terms of fr and fθ , 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, cos θ fx fy = fr fθ − sinr θ
sin θ
cosθ r
In other words, ∂f ∂f 1 ∂f = cos θ − sin θ ∂x ∂r r ∂θ and, ∂f ∂f 1 ∂f = sin θ + cos θ . ∂y ∂r r ∂θ
Exercise: Show that Laplace’s equation fxx + fyy = 0 has the following expression in polar coordinates: ∂2f 1 ∂f 1 ∂2f + + = 0. ∂r2 r ∂r r2 ∂θ2 Solution: Compute the partial derivatives of fr and fθ 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: ∂ ∂ ∂ = cos θ + sin θ ∂r ∂x ∂y
and
∂ ∂ ∂ = −r sin θ + r cos θ . ∂θ ∂x ∂y
So, ∂ ∂r
= cos2 θ
∂f ∂r
∂ = ∂r
∂f ∂f cos θ + sin θ ∂x ∂y
2 ∂2f ∂2f 2 ∂ f + 2 cos θ sin θ . + sin θ ∂x2 ∂x∂x ∂y 2
Similarly, one computes ∂ ∂f ∂ ∂f ∂f = −r sin θ + r cos θ . ∂θ ∂θ ∂θ ∂x ∂y ∂f ∂2f ∂2f ∂2f ∂f − r sin θ + r2 sin2 θ 2 − 2r2 cos θ sin θ + r2 cos2 θ 2 . ∂x ∂y ∂x ∂x∂x ∂y 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 θfx . = −r cos θ
The above can be regrouped as follows: 2 2 ∂2f ∂f ∂f ∂2f 2 2 ∂ f 2 ∂ f = −r cos θ + sin θ +r sin θ 2 − 2 cos θ sin θ + cos θ 2 , ∂θ2 ∂x ∂y ∂x ∂x∂x ∂y 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 → R3 . 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 f1 , f2 , f3 and g1 , g2 , g3 for the component functions, one has, on the one hand, f2 g3 − g2 f3 f × g = f3 g1 − f1 g3 . f1 g2 − f2 g1
So by the product rule, 0 f2 g3 + f2 g30 − (g20 f3 + g2 f30 ) D(f × g) = f30 g1 + f3 g10 − (g10 f3 + g1 f30 ) f10 g2 + f1 g20 − (g20 f1 + g2 f10 ) And, on the other hand, one has that 0 f2 g3 − g2 f30 (f 0 × g) = f30 g1 − f10 g3 , f10 g2 − f20 g1
0 f2 g3 − g20 f3 (f × g 0 ) = f3 g10 − f1 g30 f1 g20 − f2 g10
and it is easy to see that the sum of the two vectors above is D(f × g).
