Math 409 Final Exam Solutions May 9, 2005

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1 Math 49 Final Exam Solutions May 9, ) State and prove the monotone convergence theorem. Then use it to show that the recursive sequence defined below converges to some number l. What must l equal? a 1 =1,a n+1 = 2+a n. The monotone convergence theorem says the following: let a n be a monotone sequence of numbers. That is, either a n a n+1 or a n a n+1 for each n. If the sequence is bounded, then it converges. For convenience, let s assume that the sequence is monotone increasing and bounded above. By the completeness axiom of real numbers, the collection of numbers {a n } n=1 has a least upper bound. Call it l. Note, since the numbers a n are bounded above we have l a finite real number. To see that a n converges to l, firstnotethata n lfor each n. For if there is some N such that l<a N,thenlis not an upper bound of the set {a n } n=1. Secondly, for any ɛ>thereisannsuch that l ɛ<a N. For if not, then l ɛ is an upper bound of the set {a n } n=1, contradicting the fact that l is the least upper bound for this set. Now let ɛ>. Let N be such that l ɛ<a N. Then for all n N we have l ɛ<a N a n l<l+ɛ. Thus, for n N we have a n l <ɛ. That is, the sequence a n converges to l. Toseethatthesequencea 1 =1,anda n+1 = 2+a n converges, we ll show that it is monotone increasing and bounded above by 2. To see that 2 is an upper bound for the numbers a n, note that a 1 =1 2. Now suppose that a n 2. Then we have a n+1 = a n =2. So by induction we have a n 2 for all n. To see that a n is increasing note that a 2 = 3 1=a 1. Using induction again suppose that a k a k 1.Then a k+1 = 2+a k 2+a k 1 =a k. Since the sequence is bounded and increasing, the monotone convergence theorem implies that the sequence has a limit. Call it l. Thenlmust satisfy l n+1 an +2= l+2 n n l 2 = l+2. The solutions to this quadratic equation are 2 and 1. Since a n > for all n, wemusthave l=2.

2 2. 25) State and prove Rolle s theorem. Next, suppose a function f has two derivatives on an interval [a, b], and there are three points x i, such that a<x 1 <x 2 <x 3 <bfor which f x 1 )=fx 2 )=fx 3 ). Show such a function must have a point ξ at which f ξ) =, where x 1 <ξ<x 3. Rolle s theorem states that if f is continuous on the closed bounded interval [a, b] and differentiable on the open interval a, b), and f a) =fb), then there is a point ξ a, b) such that f ξ) =. One proof of Rolle s theorem goes as follows. Since f is continuous on the closed bounded interval there are points x m and x M in [a, b], where f attains its infimum and supremum respectively. If one of those points is in the open interval, then f has a local extremum at that point, in which case f s derivative is zero there. On the other hand if neither of the points x m or x M is in the open interval a, b), then both of them are one of the endpoints. However, since f a) = fb) this implies that we have fx m ) fx) fx M )=fx m ). That is, f is constant on the closed interval [a, b]. In this case we have f x) = for all x a, b), and we may pick ξ equal to any interior point. For the function with the properties described above we have f x 1 )=fx 2 )=fx 3 ). Thus, by Rolle s theorem there are points ξ 1 and ξ 2 such that x 1 <ξ 1 <x 2 <ξ 2 <x 3, and f ξ 1 )==f ξ 2 ). A second application of Rolle s theorem to the function f x) says there is a point ξ, whereξ 1 <ξ<ξ 2 and f ξ) =. 3. 3) Define what it means for a bounded function f to be Riemann integrable on a bounded interval [a, b]. Using the definition, show that 5, x<1 f x) = 1, x =1 3, 1 <x 2 is Riemann integrable on [, 2], and calculate f x) dx. A bounded function defined on the interval [a, b] is Riemann integrable on this interval if for all ɛ>, there is a partition P = {x,, x n } of the interval such that U f, P ) L f, P ) <ɛ, where U f, P )andlf,p) are the upper and lower Riemann sums of f with respect to the partition P.Thatis, Uf,P) = Lf, P) = M i f)x i x i 1 ) m i f)x i x i 1 ), 2

3 where M i f) =sup{fx):x i 1 x x i }and m i f) =inf{fx):x i 1 x x i }. To see that the given f is Riemann integrable on [, 1], let ɛ>, and assume that ɛ<14. Let { P =, 1 ɛ 14,1+ ɛ } 14,2. The partition P gives us 3 subintervals of the interval [, 2]. We have M 1 f) = 5, M 2 f)=5,m 3 f)=3 m 1 f) = 5, m 2 f)= 1, m 3 f) =3. Thus, [ U f, P ) L f, P ) = 5 1 ɛ ) [ ɛ 14 ɛ = 6 <ɛ. ɛ +5 ) + 1) ɛ ))] 14 2 ɛ ɛ 17 ))] Nowthatweknowfis Riemann integrable on [, 2], we can use some results of Riemann integration to evaluate the integral f x) dx = 1 ɛ + f x) dx + 1 ɛ 1 f x) dx f x) dx + lim ɛ + 51 ɛ) + lim 31 ɛ) ɛ + ɛ + = 8. 1+ɛ f x) dx 4. 3) Suppose f and g are bounded functions on [a, b] and that both are Riemann integrable on [a, b]. Determine the the truth or falsity of each of the following statements. If a statement is true give a proof; if it is false supply a counterexample. a) If f x) g x) for all x [a, b], then b a f x) dx b a g x) dx. This statement is true. One way to see this is to note that since the functions are integrable, their integrals equal the limit of a sequence of Riemann sums. And for any Riemann sum we have f t i ) x i g t i ) x i. Taking limits we get b a f x) dx b a g x) dx. b) If f x) x [a, b] and b afx)gx) dx =,thengx)= x [a,b]. This is not true. A counter example is f x) =1on[ 1, 1] and g x) =x. 3

4 c) Let F x) = x a ft) dt for x [a, b]. Then F x) =fx) x a,b). This is not true. If we assume that f is continuous then it is true, but if f is not continous at a point x,thenf x ) may not exist. As an example consider the function f in problem 3. For that f, wehave { Fx)= 5x, x 1 3x 1) + 5, 1 <x 2. Note that F 1) does not exist, since the right hand and left hand limits of the difference quotient are not equal F 1 + h) F 1) lim h + h F 1 + h) F 1) lim h h 3h h + h =3 5h h h = ) Suppose that f is continuous on the open interval, 1), and there is a constant K> such that x, 1), f x) K.Foreachx, 1), define F x) = x 1/2ft) dt. a) Show that there is a constant ˆK such that for any x and y in,1)wehave Fx) Fy) ˆK x y. Fx) Fy) = = x 1/2 x y ft) dt f t) dt 1/2 max{x, y} f t) dt f t) dt y max{x, y} min{x, y} min{x, y} Kdt=K x y. Set ˆK equal to K. b) Deduce that F is uniformly continuous on the open interval, 1), and that it has a continuous extension to the closed interval [, 1]. Thus, showing that f is improperly Riemann integrable on the closed interval [, 1]. Clearly this inequality implies that F is uniformly continuous on, 1). To see this implies that F has a continuous extension to the closed interval, we ll show that for any sequence of points x n,1)thatconvergesto1or,thatfx n )mustalsoconverge. Since x n converges it is a Cauchy sequence; the fact that F is uniformly continuous, or use the inequality that F satisfies, implies that F x n ) is a Cauchy sequence, and this implies that F x n ) converges. We need to show that this limiting value of F does not depend upon the sequence x n that converges to one of the interval endpoints. So suppose x n and y n both converge to 1. Consider the sequence x 1, y 1, x 2, y 2,.That is z 2n 1 = x n,andz 2n =y n for any n N. Then z n converges to 1, and F z n )must 4

5 also converge. This means that F x n )andfy n )havetoconvergetothesamevalue, since subsequences of a convergent sequence converge to the same limit as the original sequence. 6. 2) Define what it means to say that the sequence of functions f n converges uniformly to f on the set E. Show that the sequence f n, defined below, converges uniformly to f x) =x, on the set E = [, 1]. f n x) =nsin x n. A sequence of functions f n converges uniformly to f on a set E if for every ɛ>thereisan Nsuch that if n>n,then f n x) fx) <ɛ x E. To see that the given sequence converges uniformly on the interval [, 1], set h x) =sinx x. Then we have f n x) f x) = n sin x n x = n sin x n x x = n h h ) n n) = n h ξ) x = cos ξ 1 x n cos ξ 1. Since <ξ< x n 1 we see that as n the value ξ goes to independent of x. Thus, n we have cos ξ 1 tending to uniformly in x [, 1]. 5