MATH 147 Homework 4 1. Defie the sequece {a } by a =. a) Prove that a +1 a = 0. b) Prove that {a } is ot a Cauchy sequece. Solutio: a) We have: ad so we re doe. a +1 a = + 1 = + 1 + ) + 1 ) + 1 + 1 = + 1 + 1 2 = 0 b) To prove that {a } is ot a Cauchy sequece, we eed to prove the followig statemet: There is some ɛ > 0 such that for all atural umbers N, there are two atural umbers ad m with > N ad m > N such that a a m ɛ. This is tricky. If you do t immediately see why the setece above is what we eed to prove, go back to the defiitio of Cauchy sequece ad work it out carefully. It s worth it. We eed to show that there is some ɛ with a certai property. So, let ɛ = 1 we ll show that this particular choice of ɛ will work. I fact, every value of ɛ > 0 is OK, but we ll choose this oe.) We do t get to pick N, though that s foisted upo us. But with our give N, we eed to fid ad m bigger tha N such that a a m ɛ = 1. That is to say, we eed to fid ad m bigger tha N such that m 1. Not too hard, really, but let s do it carefully ad rigorously. Let s pick = N + 1) 2 ad m = N + 2) 2. Sice N is a atural umber, we kow N > 0, so i particular = N + 1) 2 = N + 1 > N ad m = N + 2) 2 = N + 2 > N. Better yet, we have a a m 1 1, so we have succeeded! The sequece {a } is ot Cauchy.
2. a) For ay real umber a > 1, prove that the fuctio f) = a is a strictly icreasig fuctio of that is, if > y the a > a y. b) For ay real umbers a > 1 ad b > 0, show that there is a uique real umber such that a = b. Call this real umber. c) For ay real umber a > 1, show that the fuctio f) = log a is cotiuous o the iterval 0, ). Solutio: a) If ad y are itegers, the this is easy: if m >, the sice a > 1, we get a m > a. If we broade our horizos ad let = m/ ad y = t/u be ratioal umbers, the m/ > t/u meas mu > t, so a mu > a t. It s temptig to raise each side to the power 1/u, but we eed to make sure that that wo t chage the directio of the iequality. But this is easy: for ay positive iteger N ad real umbers z > 1 ad w > 1, if z N > w N, the z > w because z w implies z N w N. So we ve verified the statemet if ad y are ratioal. I geeral, for ay real umbers ad y with > y, choose ratioal umbers p ad q so that > p > q > y. The a a p by elemetary properties of its choose a sequece { } of ratioal umbers covergig to with > p for all, ad ote that a a p ). Similarly, a q a y. Sice we already kow that a p > a q, we re doe. b) Sice a is a strictly icreasig fuctio, there ca be at most oe real umber with a = b. Sice the sequece a is ubouded ad icreasig, ad the sequece a is decreasig ad coverges to 0, the fuctio a achieves values both larger ad smaller tha b. By the Itermediate Value Theorem, it therefore achieves the value b. c) First, ote that log a + log a y = log a y, because if = a X ad y = a Y, the y = a X a Y = a X+Y by elemetary properties of sequeces. Similarly, log a log a y = log a /y. Choose ay ɛ > 0, ad ay b 0, ), ad ay sequece {b } covergig to b. We wat to fid N such that if > N, the < ɛ. Choose N large eough so that b b < mi{ b a ɛ b, a ɛ b b } for all > N. The a bit of maipulatio shows that b /b is betwee a ɛ ad a ɛ, so: = /b) < ɛ ad we re doe.
3. Let a R be a real umber with a > 1. From the defiitios we used, prove the log base chage formula for ay base b > 1: log b = log a [Note: If you use the fact that a ) y = a y for irratioal or y, you must prove it.] Solutio: We will use the fact metioed i the Note, so we ll prove that first, to get it out of the way. Let ad y be real umbers, ad let a > 0 be a real umber. Choose ay sequeces of ratioal umbers { } ad {y } with = ad y = y. The we compute: ) y a ) y = a ) ym = m a ) = m a ) ym ) = m aym = m aym ) = a y as desired. Notice that m a ym ) = m a ym ) because the sequece { y m } for varyig is a sequece of ratioal umbers covergig to y m. Now we ca verify the log base chage rule. To check it, otice that log b is the uique real umber such that b log b =. So to prove the rule, we just eed to prove the followig: b log a = By defiitio of logarithms, we kow that b = a
Substitutig this ito our alleged log base chage formula yields: b log a = a ) log a = a log a = a log a = just like we wated. 4. Let a < b be real umbers, ad let f : [a, b] R be a cotiuous fuctio. Prove that the image f[a, b]) = {r R r = f)for some [a, b]} is also a closed iterval. Solutio: By the Etreme Value Theorem, there are elemets M ad m of [a, b] such that fm) = sup{f) [a, b]} ad fm) = if{f) [a, b]}. Therefore, we have f[a, b]) [fm), fm)]. Coversely, if C [fm), fm)] is ay elemet, the sice fm) < C < fm), ad because f is cotiuous, the Itermediate Value Theorem says that there is some c [a, b] such that fc) = [a, b]. Therefore f[a, b]) is the closed iterval [fm), fm)]. 5. Compute a +b for ay real umbers a ad b with a > 1. Solutio: The it is zero, regardless of what a ad b are. If you did t kow this already, you ca guess it by pluggig very big values of ito this formula.) To prove this, otice that the ratio of cosecutive terms is as follows: We compute: + 1 a +1 + b + 1 a + b = + 1 a + b a +1 + b = 1 + 1 ) = 1
ad a + b a +1 + b = 1 a a +1 + ba a +1 + b = 1 a 1 + ba b a +1 + b ) = 1 a So both those its eist, ad we ca deduce: + 1 = + 1 = 1 a a + b a +1 + b a + b a +1 + b The relevace of this calculatio is that we ca ow compare our sequece to a geometric sequece, whose it we ca easily calculate. To wit: there is some atural umber N such that if > N, we have: + 1 a + b a +1 + b < α where 1 < α < 1. Use ɛ = α 1, ad remember that everythig is positive.) a a Therefore, sice the quatity o the left is the ratio betwee cosecutive terms of the sequece > N, we have: a +b, we deduce by a simple iductio that for all 0 < a + b < ) N α N a N + b Sice the sequece o the right coverges to zero, the Squeeze Theorem tells us that the sequece i the middle also teds to zero. But that s what we wated to prove. 6. Let f : R R be the fuctio f) = log a. Compute f) for ay a > 1. Solutio: The it is zero, regardless of what a is.
To prove this, choose ay ɛ > 0. We eed to fid a real umber N such that if > N, the log a < ɛ. The first, easiest thig to ote is that we do t eed the absolute value bars: for a > 1 ad > 1, both ad log a are positive. Net, ote that log a log a + 1) sice log a ad are both icreasig fuctios of. The otatio meas the largest iteger that is ot bigger tha. I other words, it s, rouded dow to the earest iteger.) If we ow set k = log a + 1), the we get: log a k a k 1 If we write l = k, the we further deduce: log a l + 1 a l 1 We d like to use the previous problem to compute the it of the iteger thig o the right side to be zero, ad the deduce that the left side must also go to zero. We compute: l l + 1 a l 1 = l l a l 1 + l 1 a l 1 = 0 But this is ot quite eough! Maybe l does t go to ifiity as does. Of course, this does ot happe: l = log a + 1), so if is ubouded above, the is also ubouded above, ad so l is ubouded above. We therefore kow that for all real umbers N, there is some B such that if > B, the l > N. If we choose N large eough so that l+1 < ɛ, we a l 1 therefore deduce that if > B, the log a < ɛ, as desired.