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### Practice Exam 1 with solution

Course: ECONOMICS 2301, Spring 2012
School: UT Dallas
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Word Count: 1649

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of University Texas at Dallas Practice Exam #1 Last Name: ............................. First Name and Initial: ...................................... Course Name: Mathematical Analysis 2 Number: MATH 4302 Due Date: March 1, 2012 Student's Signature: ...................................... Instructor: Wieslaw Krawcewicz E-mail Address: ............................. Question 1. Weight 10 Your Score...

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of University Texas at Dallas Practice Exam #1 Last Name: ............................. First Name and Initial: ...................................... Course Name: Mathematical Analysis 2 Number: MATH 4302 Due Date: March 1, 2012 Student's Signature: ...................................... Instructor: Wieslaw Krawcewicz E-mail Address: ............................. Question 1. Weight 10 Your Score Comments 2. 10 3. 10 4. 10 5. 10 6. 10 7. 10 8. 10 9. 10 10. 10 Total: 100 2 Problem 1. Consider the function f : [-, ] R, defined by f (x) = x2 sin(x). (a): Compute the n-th derivative of the function f . (b): For a given n N, write the Taylor polynomial Tn (x) of f (centered at xo = 0). (c): For a given > 0, find n N such that sup{|f (x) - Tn (x)| : x [-a, a]} < . SOLUTION: Notice that we have f (x) = 2x sin(x) + x2 cos(x), f (x) = 2 sin(x) + 4x cos(x) - x2 sin(x), and for n 2 we have (by Leibnitz formula) ( ) ( ) ( ( ) n 2 n ) f (n) (x) = x sin x + n + 2x sin x + (n - 1) 0 2 1 2 ( ) ) ( n + 2 sin x + (n - 2) 2 2 ( ( ) ) 2 = x sin x + n + 2nx sin x + (n - 1) 2 2 ( ) n! sin x + (n - 2) + (n - 2)! 2 Then we have and for n 3 f thus f (2k+1) (0) = (-1)k+1 Therefore, we have for n 3 2 2k+1 k+1 x . Tn (x) = (-1) (2k - 1)! n f (0) = f (0) = f (0) = 0 (n) ( ) n! (0) = sin (n - 2) (n - 2)! 2 (2k + 1)! . (2k - 1)! k=1 3 By applying the Lagrange formula, we obtain that for some c (-, ) rn (x) = f (n+1) (c) n+1 x = (n + 1)! c2 sin(x + (n + 1)) + 2(n + 1)c sin(c + n) + (n + 1)n sin(c + (n - 1) n+1 2 2 2 x (n + 1)! Therefore, by applying the inequality (n + 1)! |rn (x)| (n+2)n+1 en+1 , we obtain c2 + 2c(n + 1) + n(n + 1) n+1 (n + 1)! (n + 1 + )2 n+1 (n + 1)! (n + 1 + )2 (e)n+1 (n + 2)n+1 < 1, thus (since n+5 n+2 Therefore, if n 8 then e n+2 2) |rn (x)| Therefore, for a given > 0 if (n + 5)2 4(e)3 (e)3 (n + 2)3 n+2 { } 4(e)3 n max 8, -2 then for x [-, ] 4(e)3 (n + 5)2 3 (e) < . |f (x) - Tn (x)| = |rn (x)| (n + 2)3 n+2 4 only if the epigraph of f is a convex set of Rn R. SOLUTION: The epigraph of f is defined by Problem 2. Let f : Rn R be a function. Show that f is convex if and epi(f ) := {(x, ) Rn R : f (x) }. Assume that f is convex, i.e. x,yRn t[0,1] f (tx + (1 - t)y) tf (x) + (1 - t)f (y). Then, for (i.e. f (x) 1 and f (y) 2 ) we have for t [0, 1] t(x, 1 ) + (1 - t)(y, 2 ) = (tx + (1 - t)y, t1 + (1 - t)2 ). Since f ((tx + (1 - t)y) tf (x) + (1 - t)f (y) t1 + (1 - t)2 it follows that (tx + (1 - t)y, t1 + (1 - t)2 ) epi(f ). Conversely, assume that epi(f ) is a convex set, i.e. for all (x, 1 ), (y, 2 ) epi(f ) and t [0, 1] w have t(x, 1 ) + (1 - t)(y, 2 ) = (tx + (1 - t)y, t1 + (1 - t)2 ) epi(f ). Since for all x, y Rn , (x, f (x)), (y, f (y)) epi(f ), it follows that for t [0, 1] we have t(x, f (x)) + (1 - t)(y, f (y)) = (tx + (1 - t)y, tf (x) + (1 - t)f (y)) epi(f ) = f (tx + (1 - t)y) tf (x) + (1 - t)f (y), which implies that f (x) is convex. 5 convex non-decreasing function. Show that the function g f : Rn R, (g f )(x) = g(f (x)), x Rn , is convex. Give an example of two convex functions f , g : R R such that g f is not convex. SOLUTION: Assume that x, y Rn and t [0, 1]. Since f is convex, we have f (tx + (1 - t)y) tf (x) + (1 - t)f (y), and since g is non-decreasing, (1) implies ( ) ( ) g f (tx + (1 - t)y) g tf (x) + (1 - t)f (y) . On the other hand g is also convex, thus ( ) g tf (x) + (1 - t)f (y) tg(f (x)) + (1 - t)g(f (x)). By combining (2) and (??) we get ( ) g f (tx + (1 - t)y) tg(f (x)) + (1 - t)g(f (x)). which implies that g f is convex. As a counter example that the compositions of convex functions may not be a complex function in general, consider f, g : R R defined by f (x) = x2 , g(x) = -x, x R. (1) Problem 3. Let f : Rn R be a convex function and g : R R a (2) (3) Clearly f (x), g (x) 0, thus these two functions are convex. However the function (g f )(x) = -x2 is not convex. 6 Problem 4. Show that every continuous function f : [a, b] R is Riemann integrable on [a, b]. Solution: Check out the textbook. 7 arbitrary partition of [a, b]. Suppose that Q := P {c}, where c (xj-1 , xj ) for some j {1, 2, 3, . . . , n}. Show that s(f, P ) s(f, Q) S(f, Q) S(f, P ). SOLUTION: Put for k = 1, 2, . . . , n mk := min{f (x) : x [xk-1 , xk ]}, and m1 := min{f (x) : x [xj-1 , c]}, m2 := min{f (x) : x [c, xj ]}, j j Mj1 := max{f : (x) x [xj-1 , c]}, Mj2 := max{f (x) : x [c, xj ]}. Also define for k = 1, 2, . . . , n, n + 1 mk if k < j 1 m if k = j j mk := , m2 if k = j + 1 j m if k > j + 1, k-1 Since Q = {tk }n+1 , where k=0 xk tk = c xk-1 Since Mj (xj - xj-1 ) Mj1 (c - xj-1 ) + Mj2 (xj - c) and mj (xj - xj-1 ) m1 (c - xj-1 ) + m2 (xj - c) j j 8 Problem 5. Let f : [a, b] R be a bounded function and P = {xk }n an k=0 Mk := max{f (x) : x [xk-1 , xk ]}. Mk := Mk 1 M j Mj2 if k < j if k = j if k = j + 1 if k > j + 1. M k-1 if k < j if k = j if k > j. we have S(f, P ) - S(f, Q) = = - + n k=1 j-1 k=1 j-1 Mk (xk - xk-1 ) - n+1 k=1 Mk (tk - tk-1 ) n k=j+1 Mk (xk - xk-1 ) + Mj (xj - xj-1 ) + Mk (xk - xk-1 ) Mk (xk - xk-1 ) - Mj1 (c - xj-1 ) - Mj2 (xj - c) Mk (xk - xk-1 ) k=1 n k=j+1 = Mj (xj - xj-1 )) - Mj1 (c - xj-1 ) - Mj2 (xj - c) 0. Similarly, s(f, P ) - s(f, Q) = = - + n k=1 j-1 k=1 j-1 sk (xk - xk-1 ) - n+1 k=1 mk (tk - tk-1 ) n k=j+1 mk (xk - xk-1 ) + mj (xj - xj-1 ) + mk (xk - xk-1 ) mk (xk - xk-1 ) - m1 (c - xj-1 ) - m2 (xj - c) j j mk (xk - xk-1 ) k=1 n k=j+1 = mj (xj - xj-1 )) - m1 (c - xj-1 ) - m2 (xj - c) j j 0. 9 Problem 6. Compute Var(f ) for the following functions f : [a, b] R [a,b] 1 (a) f (t) = 2 3 2 for t = a; for a < t < b; fort t = b, (b) f (t) = et , where a < 0 < b. SOLUTION: (a): Since the function f is non-decreasing Var(f ) = f (b) - f (a) = 3 - 1 = 2. [a,b] (b): Since f (t) = 2tet , f (t) < 0 for t < 0 and f (t) > 0 for t > 0, the function f is decreasing on [a, 0] and increasing on [0, b]. Therefore, we have 2 Var(f ) = Var(f ) + Var(f ) = f (a) - f (0) + f (b) - f (0) = ea + eb - 2. [a,b] [a,0] [0,b] 2 2 10 Problem 7. Given f (x) = if 0 x < 1 if 1 x < 2 if 2 x 3. Find Var(f ). [0,3] SOLUTION: Notice that we have Var(f ) = Var(f ) + Var(f ) + Var(f ) [0,3] [0,1] [1,2] [2,3] = | - | + | - | + | - | = | - | + | - |. 11 Problem 8. Compute Var(f ) for f (x) = sin2(nx), where n N. [0,] SOLUTION: Notice that 1 f (x) = sin2 (nx) = (1 - cos(2nx)). 2 Since cos(2nx) = 1 for x = l n, l = 0, 1, 2, . . . , n, and Var (cos(2nx)) = 2, ] [ l (l+1) n, n therefore 1 Var(f ) = (2n) = n. 2 [0,] 12 Problem 9. Let f : [a, b] R be function such that Var(f ) = 0. Show that [a,b] the function f is constant. SOLUTION: Since Var(f ) = sup{v(f, P ) : where P is a partition of [a, b]}. [a,b] therefore, or every partition P = {xk }n we have k=0 0 v(f, P ) = n k=1 |f (xk ) - f (xk-1 )| Var(f ) = 0, [a,b] which means that for every partition P = {xk }n we have k=0 n k=1 |f (xk ) - f (xk-1 )| = 0 We will show that for every x [a, b] we have f (x) = f (a). Indeed, consider the partition P = {a, x, b}, where a < x b. Then we have 0 = v(f, P ) = |f (x)-f (a)|+|f (b)-f (x)| |f (x)-f (a)| 0 |f (x)-f (a)| = 0. 13 Problem 10. Compute: t2 (a) Var(f ) for f (t) = 4 [0,4] 4-t [0, ] for 0 t 1; for 1 < t 2; for 2 < t 4. (b) Var (f ) for f (t) = sin t2 . SOLUTION: (a): Notice that the function f is non-decreasing on the interval [0, 2] and it is decreasing on the interval [2, 4]. Therefore, we have Var(f ) = Var(f ) + Var(f ) [0,4] [0,2] [2,4] = (f (2) - f (0)) + (f (2) - f (4)) = (4 - 0) + (4 - 0) = 8. (b): Notice that f (t) = 2t cos(t2 ). Therefore, f[ is increasing on the interval (t) [ ] ] 0, 2 , and f (t) is decreasing on the interval , . Therefore, we have 2 Var (f ) [0, ] = Var (f ) + (f ) Var [0, ] [ , ] 2 2 [ ( ) ] ] [ ( ) = f - f (0) + f - f ( ) 2 2 = (1 - 0) + (1 - 0) = 2 14 Problem 11. Let a < c < b be given numbers, f : [a, b] R a continuous function, and g : [a, b] R be defined by { for t c , g(t) = for t > c, where = . Compute the Riemann-Stieltjes integral b f (t)dg(t). a SOLUTION: Notice that g (t) = 0 for t = c, g(c+ ) = and g(c- ) = . Therefore, we have b b f (t)dg(t) = f (t)g(t)dt + f (c)(g(c+ ) - g(c- )) = f (c)( - ). a a 15 Problem 12. Consider the function f : [0, 2] R, f (t) = et and the function g : [0, 2] R be defined by { t2 for 0 t 1 g(t) = . 2 for 1 < t 2, Compute the Riemann-Stieltjes integral 2 f (t)dg(t). 0 2 SOLUTION: Notice that { 2t for 0 t < 1 g (t) = 0 for 1 < t 1, g(1+ ) = 2, g(1- ) = 1. Therefore, we have 2 2 f (t)dg(t) = f (t)g (t)dt + f (1)(g(1+ ) - g(1- )) 0 0 1 1 t2 t2 e 2tdt + e(2 - 1) = e = + e = 2e - 1. 0 0 16
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