Math 49 Final Exam Solutions May 9, 25 1. 25) 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 2+2=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. 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
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 ) [ 14 5 1 ɛ 14 ɛ = 6 <ɛ. ɛ +5 ) + 1) +3 2 1+ ɛ ))] 14 2 ɛ +3 1+ 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
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 =5. 5. 2) 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
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