Math 242: Principles of Analysis Fall 2016 Homework 6 Part B Solutions. x 2 +2x = 15.

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1 Math 242: Principles of Analysis Fall 2016 Homework 6 Part B Solutions 1. Use the definition of a it to prove that x 2 +2x = 15. Solution. First write x 2 +2x 15 = x 3 x+5. Next let δ 1 = 1. If 0 < x 3 < 1, then we have 2 < x < 4, so 7 < x + 5 < 9 and thus x + 5 < 9. Hence we have x 3 x+5 9 x 3 < ǫ when x 3 < ǫ. So let δ 9 2 = ǫ and define δ = min{δ 9 1,δ 2 }. Then, when 0 < x 3 < δ we have x 3 < 1 and x 3 < ǫ, so x+5 < 9 and 9 thus x 2 +2x 15 = x+5 x 3 9 x 3 < 9 ǫ 9 = ǫ. The figure below shows the unit circle. Recall that when an angle is measured in radians it is given by the length of the corresponding arc on the unit circle. In the figure, this is represented by the arc from C to B. Also recall that the coordinates of the point B on the unit circle are defined to be (cos,sin). B D 1 O A C 2. Prove that sin() = 0, as follows: (a) Use the figure to explain why 0 < sin() < for any angle 0 < < π/2. Solution. For 0 < < π/2, sin() = AB (the length of the line segment AB), and clearly this is positive. Also, since triangle ABC is a right triangle with hypotenuse BC, we have AB < BC. Now the shortest path from B to C is the line segment BC. Thus its length, BC is less than the length of the arc from B to C, which has length. BC <. Hence sin(theta) <. (b) Show that +sin() = 0. Solution. Since +0 + = 0, and 0 < sin() < for 0 < < π/2, the Squeeze Theorem implies +sin() = 0.

2 (c) Use the fact that sin( ) = sin() for any, together with the result of part (b), to show that sin() = 0. Solution. For < 0, define x =. Then x > 0. Hence sin() +sin( x) + sin(x) = + sin(x) = 0 = Use the result of Exercise 2 to show that cos() = 1. Solution. First, sincecos() > 0for ( π/2,π/2),wehavecos() = + 1 sin 2 () for any ( π/2,π/2). Since 0 1 sin 2 () 1 it follows that 1 sin 2 () 1 sin 2 () for all. Thus 1 sin 2 () cos() 1 for all ( π/2,π/2). Therefore, since 1 sin 2 () = 1 0 = 1, the Squeeze Theorem implies that cos() = Show that sin(x) = sin(c) for every c R. Solution. Since setting = x c gives sin(x) = sin(x c+c) = sin(x c)cos(c)+cos(x c)sin(c), sin(x) sin(x c)cos(c)+cos(x c)sin(c) sin()cos(c)+cos()sin(c) = 0 cos(c)+1 sin(c) = sin(c). 5. Show that cos(x) = cos(c) for every c R. Solution. Since setting = x c gives 6. Prove sin cos(x) = cos(x c+c) = cos(x c)cos(c) sin(x c)sin(c), cos(x) cos(x c)cos(c) sin(x c)sin(c) cos()cos(c) sin()sin(c) = 1, as follows: = 1 cos(c) 0 sin(c) = cos(c).

3 (a) Find the areas of the triangle OCD and the sector BOC (not the triangle, but the portion of the circle), and use this to explain why < tan() when 0 < < π. 2 Solution. Since OCD and OAB are similar triangles, it follows that CD = AB = sin() = tan() and the area of OCD is 1 OA cos() 2 tan(). The area of the sector BOC is 1 2. Since the sector is clearly contained in the triangle, we have 1 2 < 1 2 tan(). (b) Combine this with the inequality in Exercise 2 to show that cos() < sin() < 1 for 0 < < π 2. Solution. We know from Exercise 2 that sin() <, and thus sin() < 1 for 0 < < π sin(). From the previous part, we know < tan(), and thus cos() < 2 for 0 < < π. 2 sin() (c) Show that = 1. + sin() Solution. +cos() +1 = 1, and cos() < < 1 for 0 < < π/2, the sin Squeeze Theorem implies = 1. + sin() (d) Use symmetry and the result of part (c) to show that = 1. Solution. For < 0, define x =. Then x > 0, so sin + sin( x) x + sin(x) x + sin(x) x 1 cos() 7. Use the result of Exercise 6 to prove = 0. Solution. Using the identity sin 2 () = 1 cos 2 () = (1 cos())(1+cos()) gives 1 cos() sin 2 () (1+cos() sin() = 1. sin() 1+cos() = = Let f(x) = x. (This denotes the greatest integer that is less than or equal to x. So for example 3.1 = 3, 3 = 3, and 2.9 = 2.) Evaluate +f(x), f(x), and f(x). Solution. Since f(x) = 3 for 3 < x < 4, +f(x) +3 = 3. Since f(x) = 2 for 2 < x < 3, f(x) 2 = 2. Since +f(x) f(x), f(x) does not exist. 9. Show that x = for any c > 0. Solution. Since c > 0, we have > 0 so for any x > 0 we have x = x c x+ x c.

4 Let ǫ > 0. To ensure that x > 0 when x c < δ, we need δ c. To enure that x c < ǫ when x c < δ, we need x c < ǫ. Thus we will choose δ = min{c,ǫ }. Then for any x such that 0 < x c < δ, we have x c < c so x > 0 and thus x x c < ǫ = ǫ. 10. Suppose f(x) = 0 and there exists M > 0 such that g(x) M for all x. Show that f(x)g(x) = 0. Solution. Let ǫ > 0. Since f(x) = 0 there exists δ > 0 such that f(x) 0 < ǫ M for all x such that 0 < x c < δ. Then f(x)g(x) 0 = f(x)g(x) M f(x) < M for all x such that 0 < x c < δ. Thus f(x)g(x) = Prove that sin(1/x) does not exist. ǫ M = ǫ Solution. Let x n = 1 and y 1 πn n =. Then x π/2+2πn n 0 and y n 0 for all n and x n y n = 0. Since sin(1/x n ) sin(πn) 0 = 0 and sin(1/y n ) = sin(π/2+2πn) 1 = 1, it follows that cos(1/x) does not exist. { 3x x Q 12. Let f(x) = x 2 x Q c (a) Prove that f(x) does not exist if c 0,3. Solution. Fix c 0,3. Let x n be any sequence in Q such that x n c for all n and x n = c, and let y n be any sequence in Q c such that y n c for all n and y n = c. Then f(x n ) 3x n = 3c and f(y n ) yn 2 = c2. Since c 0 and c 3, c 2 3c and thus f(x n ) f(y n ). Thus f(x) does not exist. (b) Evaluate f(x). Prove your assertion. Solution. For 0 < x < 3, x 2 f(x) 3x. Thus, since x 2 3x = 0, the Squeeze Theorem implies +f(x) = 0. For x < 0, 3x f(x) x2, so the f(x) = 0. Squeeze Theorem also implies f(x) = 0. Hence (c) Evaluate f(x). Prove your assertion. Solution. For 0 < x < 3, x 2 f(x) 3x. Thus, since x 2 3x = 9, the Squeeze Theorem implies f(x) = 0. For x > 3, 3x f(x) x2, so the f(x) = 9. Squeeze Theorem also implies +f(x) = 9. Hence

5 13. Show that the equation x 2 = 3+sin(x) has at least two solutions. Solution. The equation is equivalent to f(x) = 0 where f(x) = x 2 sin(x) 3. Since f(0) = 3 < 0 and f(3) = 6 sin(3) > 5 > 0 and f is continuous on [0,3], the Intermediate Value Theorem implies that there exists c 1 (0,3) such that f(c 1 ) = 0. Since f( 3) = 6 sin( 3) > 5 > 0 and f is continuous on [ 3,0], the Intermediate Value Theorem implies that there exists c 2 ( 3,0) such that f(c 2 ) = 0. Since c 1 > 0 and c 2 < 0, these roots are distinct. 14. Suppose f and g are continuous functions on some interval [a,b], f(a) > g(a) and f(b) < g(b). Prove that there exists some c [a,b] such that f(c) = g(c). Solution. Define h(x) = f(x) g(x). Then h is a continuous function on [a,b], and h(a) = f(a) g(a) > 0 and h(b) = f(b) g(b) < 0. Thus the Intermediate Value Theorem implies that there is some c [a,b] such that h(c) = 0. Thus f(c) g(c) = 0, so f(c) = g(c).