MA 1: PROBLEM SET NO. 7 SOLUTIONS

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1 MA 1: PROBLEM SET NO. 7 SOLUTIONS 1. (1) Obtain the number r = 15 3 as an approimation to the nonzero root of the equation 2 = sin() by using the cubic Taylor polynomial approimation to sin(). Proof. The cubic Taylor polynomial approimation of sin() (around 0) is T 3,0 sin() = 3 3!. Thus, to find a nonzero root of the equation 2 = sin() using this approimation, we need to find solutions to 3 3! + 2 = 1 6 ( ). Using the quadratic formula, the quadratic polynomial has roots = 6 ± = 3 ± 15. In particular, r = 15 3 is such a root. (2) Show that the approimation in the previous part satisfies the inequality sin(r) r 2 < 1 200, given that 15 3 < 0.9. Is the difference (sin(r) r 2 ) positive or negative? Give full details of your reasoning. Proof. Since the fourth derivative of sin() at zero is zero, the third and fourth degree Taylor polynomials of sin() around zero coincide. Therefore, using the error for the Taylor epansion, we have sin(r) T 3,0 sin(r) = sin(r) T 4,0 sin(r) = E 4 (r) = 1

2 sin (5) (c) 5! r 5 for some number c between 0 and r. Hence: sin(r) r 2 = T 3,0 sin(r) + E 4 (r) r 2 = E 4 (r) (since r is chosen to be a root of T 3,0 sin(r) r 2 ) = sin(5) (c) r 5 5! < (0.9)5 < 1 5! 200. We have sin(r) r 2 > 0 sine the error E 4 (r) is a positive number for r < Use proper Taylor polynomial approimations to evaluate the following its: (1) 0 sin(a) sin(b) = a b. Solution. Using the Taylor polynomial approimation at 0, for any constant c, we have and so sin(c) = c (c)3 3! + (c)5 5! sin(a) 0 sin(b) = a(1 (a) 2 /3! + (a) 4 /5! ) 0 b(1 (b) 2 /3! + (b) 4 ) = a b. (2) 0 sin(2) 2 3 = 4 3. Solution. We have sin(2) = 0 (2) 3 /3! + (2) 5 /5! (2) 7 /7! + 3 = 8 6 = 4 3. (3) 0 a = log a log 2. 2

3 Solution. We have a 1 = e log a 1 Thus, = log a + ( log a) 2 /2! + ( log a) 3 /3! + = ( log a) + ( log a) 2 /2! + ( log a) 3 /3! + a = ( log a) + ( log a) 2 /2! + ( log a) 3 /3! + 0 ( log 2) + ( log 2) 2 /2! + (2 log a) 3 /3! + = log a log 2. (4) 0 cosh() cos() 2 = 1. Solution. Recall that and cosh() = ! + 4 4! + Therefore, cos() = 1 2 2! + 4 4! cosh() cos() 0 2 (2 2 )/2! /6! + = 0 2 = /6 + 0 = 1. (5) = e 1. Solution. We have 1/(1 ) = ep( 1 log ). Note that 1 1 log 1 = 1 3

4 (say, by L Hopital s rule). Thus, ( ) log 1 1/(1 ) = ep 1 1 ( 1 1 log ) 2 = 1 + log ! = 1 + ( 1) + ( 1) 2 /2! + ( 1) 3 /3! + = ep( 1). 3. (1) Suppose f and g are two functions. Give a definition of the function h() := g() whenever this makes sense. If f, g are differentiable, compute the derivative of h wherever it eists. (2) Use part (a) to differentiate the following functions: 3 (log ) log Proof. (1) For h() being well defined, must be positive. Then we can consider a function H() = ln(h()) which is equal to ln( g() ) = g()ln(). We can easily see H () = h () = h() g ()ln() + g() f (). Therefore h () = g ()h()ln() + h()g() f () by multiplying h() on both sides. (2) 3 By using the formula in (1), the derivative of this function is 3 ln(3) (log) log The derivative of this function is ln(log)(log) log + (log)log log log 4. Let f : (a, b) R be a continuous and strictly monotone function. Show that f((a, b)) = { : (a, b)} is an open interval (possibly unbounded). (Hint: distinguish between the cases f is bounded and f is not bounded, and use the intermediate value theorem. 4

5 Proof. Denote A = f((a, b)). Now let s first prove the lemme. Lemma. If m, n A and m < n, then (m, n) A. Proof. Since m, n A, there are, y (a, b) satisfying = m, f(y) = n. If f is strictly increasing, < y. For arbitrary c between m and n, by intermediate value theorem, there eists z (, y) satisfying f(z) = c. Therefore c A, hence (m, n) A. (When f is strictly decreasing, there is z (y, ) satisfying f(z) = c from the intermediate value theorem.) Now let s consider supremum of A and infimum of A. There are four possible cases: (1) supa =, infa = t for some t R (2) supa =, infa = (3) supa = s for some s R, infa = t for some t R (4) supa = s for some s R, infa = Now let s focus on the first case. Let s show that A = (t, ) which is an open interval. For arbitrary c (t, ), since t is infimum, there is some d A satisfying t < d < c. Also there is some M A such that c < M. From the lemma, c (d, M) A. Therefore c A for arbitrary c. This means (t, ) A. Conversely, from the definition of supremum and infimum, A [t, ). Moreover if t A, there eits (a, b) satisfying = t. If f is increasing f( a+ ) A but it is smaller than = t. If f 2 is decreasing f( b+ ) A but it is smaller than = t. Therefore 2 A (t, ). Conclusively, A = (t, ). Other three cases can be proved similarly. 5. In the following, let f be a real-valued function defined in some open neighborhood of 0. Let α > 0 be a rational number. Proof. (1) Prove that if f satisfies α locally about 0 for some α > 1, then f is differentiable at 0. (2) Prove that if f satisfies α locally about 0 for some 0 < α < 1 and f(0) = 0, then f is not differentiable at 0. (1) If we put = 0 on the given inequality, f(0) 0, hence f(0) f(0) = 0. Now let s show that 0 = 0 0 = 0. α = α 1 5

6 (2) Now taking 0 of both sides, the right hand side goes to zero, hence 0 = 0. α = α 1 = 1 1 α Since 1 α > 0, as goes to zero, 1 α goes to zero, therefore the right hand side diverges to infinity. Therefore f(0) 0 = 0 0 does not eist. 6