The Uiversity of Sydey School of Mathematics ad Statistics Solutios to Tutorial Week 4 MATH2962: Real ad Complex Aalysis Advaced Semester 1, 2017 Web Page: http://www.maths.usyd.edu.au/u/ug/im/math2962/ Lecturer: Florica C. Cîrstea Questios marked with * are more difficult questios. Material covered 1 Defiitio ad properties of limits, limit iferior ad limit superior; 2 Limits ad the limit laws; Iequalities such as the arithmetic-geometric mea iequality. Outcomes This tutorial helps you to 1 be able to work with iequalities, limits ad limit iferior/superior; 2 have solid foudatios i the more formal aspects of aalysis, icludig a kowledge of precise defiitios, how to apply them ad the ability to write simple proofs; Summary of essetial material Mootoe sequeces i R always have a proper or improper limit. A proper limit meas that the limit exists i R, ad a improper limit meas the sequece diverges to + or. Usig that fact we defie limit iferior ad limit superior of a arbitrary sequece i R as follows. For N set a := if{, +1, +2, +,... } = if x k b := sup{, +1, +2, +,... } = sup x k Because {+1, +1, +2, +,... } {, +1, +2, +,... }, properties of ifimum ad supremum imply that a is icreasig ad b is decreasig. Hece lim if := lim a = lim if x k, := lim b = lim sup x k exist either as a proper or as a improper limit. Facts about limit iferior ad limit superior: lim if ; lim if = if ad oly if lim exists as a proper or improper limits. If that is the case all three are equal. Copyright c 2017 The Uiversity of Sydey 1
We call a sequece k k N or simply k a subsequece of if k is strictly icreasig ad k the latter is automatic if k is strictly icreasig, but we still make that explicit. Subsequeces have the followig properties: The limit iferior of coicides with the smallest accumulatio poit of ; or. The limit superior of coicides with the largest accumulatio poit of, or + ; Every bouded sequece has a coverget subsequece Theorem of Bolzao-Weierstrass. Questios to complete durig the tutorial 1. Cosider the sequece = 1 2 2 + 1, 0. a Sketch the graph of. It helps to look at mootoicity properties of by multiplyig by 2 / 2 ad doig a completio of squares. Multiplyig by 2 / 2 = 2 2 + 1 = 1 1 + 1 2 = 1 1 2 2 + 4 for all 1. Sice 1 1 2 2 is icreasig for 2 it follows that is decreasig for 2. Moreover, x 0 = 0, x 1 =, x 2 = 4 ad x = 27/7. Hece the graph looks as follows: 1 2 4 5 6 7 8 9 10 11 12 1 14 N b Fid a = if x k ad b = sup x k. From the graph i the previous part ad the values computed we have 27/7 for = 0, 1, 4 for = 0, 1, a = +1 for 2 eve, b = for 2 eve, for odd, for odd. c Hece compute lim if ad. Because 1 1 + 1 2 +1 We get lim if = lim x 2+1 = ad = lim x 2 =. 2
2. Compute the limit iferior ad limit superior of the followig sequeces usig directly the defiitio, ad the usig the fact that they are the smallest ad largest poit of accumulatio. The latter method is the oe commoly used. { eve a = 1/ odd for all 0. Hece We clearly have lim if a = if x 1 k = if k = 0. = lim a = lim 0 = 0. Sice the sequece is ot bouded from above =. We illustrate the above usig the graph of : 1 2 4 5 6 7 8 9 10 11 12 We ow idetify the possible limits of coverget subsequeces. From the defiitio of the sequece we see that lim x 2 = ad lim x 2+1 = 0. Hece, ay subsequece of which has a limit will either ted to or to 0 as. Hece, lim if = 0 ad =. That is the same as we obtaied before. b 1 + 1 1. N ad Note that 1 is decreasig. Hece we have a = if x 1 1 if is odd k = 1 1 if is eve + 1 b = sup x k = 1 + 1 + 1 1 + 1 if is odd if is eve
This is best see from the graph of : 1 2 4 5 6 7 8 9 10 11 12 1 14 N We see that a 1 ad also b 1. Both coicide with the limit of. Rather tha compute the limit iferior ad limit superior we could compute the limit, ad coclude the latter are the same.. Compute the limit iferior ad limit superior of the followig by usig the fact that they are the smallest ad largest poit of accumulatio. 1 /2 eve + 1 a = 2 1 odd 2 2 + 1 We idetify the accumulatio poits of by lookig for coverget subsequeces. We ote that x 4 1 x 4+1 1 2 x 4+2 1 x 4+ 1 2 as. Hece ay coverget subsequece has either limit 1, 1 or 1/2. As the smallest ad largest are 1 ad 1, respectively, we have b s = 1 k k=0 We have lim if = 1, = 1. s 0 = 1 s 1 = 1 1 = 0 s 2 = 1 1 + 1 = 1 s = 1 1 + 1 1 = 0 s 4 = 1 1 + 1 1 + 1 = 1. s = 0 s = 1 if is odd if is eve. Oe should really do this by iductio, observig that s 0 = 1 ad s +1 = s + 1 +1. Hece lim if s = 0 ad s = 1. 4
4. Let ad y be bouded sequeces i R. a Let ad y be sequeces i R. Prove that + y Clearly x l + y l sup + y. x k + sup y k for all l. This meas that sup x k + sup y k is a upper boud for x l + y l for all l. By defiitio of a supremum every upper boud is larger or equal tha the supremum sup l x l + y l sup x k + sup y k for all N. By defiitio of the limit superior ad the limit laws + y b Explai why y = lim if y, ad hece particular, y + y. lim if y. For ay set A R we kow that sup A = if A. Hece, i sup x k = if x k for all N. Hece by takig limits, usig the defiitio of limit iferior ad limit superior, we get as claimed. = lim sup x k = lim if x k = lim if c Usig the previous parts, show that + y = if at least oe of the sequeces coverges. + y. 1 Suppose that y coverges just reame the sequeces if coverges. We kow from part a + y + y = lim + y. 2 We wat to prove the reverse iequality. Applyig b we get Hece = + y y + y lim if y. y + lim if Comparig with Combiig 2 we get 1 if + y. y = lim if y. We kow that this is the case if y is coverget. 5
d By givig a couter example, show that strict iequality is possible. It is clear from the previous part that both sequeces have to diverge. We set := 1 ad y = 1 +1 for all N. The = y = 1. O the other had, + y = 1 1 1 = 0 for all N, so + y = 0. Hece, 0 = + y < + y = 1 + 1 = 2. More geeral examples ca be obtaied by lookig at sequeces of the form { { a 1 if is eve, b 1 if is eve, = y = if is odd, if is odd, a 2 by choosig a 1, a 2, b 1 ad b 2 appropriately. We have = max{a 1, a 2 }, y = max{b 1, b 2 } ad +y = max{a 1 +b 1, a 2 +b 2 }. By choosig a 1, a 2, b 1 ad b 2 i R with max{a 1 + b 1, a 2 + b 2 } < max{a 1, a 2 } + max{b 1, b 2 }, we obtai a strict iequality i a. That is exactly what we have doe i the first example: a 1 = 1, a 2 = 1 ad b 1 = 1, b 2 = 1. b 2 5. Let x R ad cosider the sequece give by := 1 + for N. Use the arithmetic-geometric mea iequality to solve the followig problems. a Suppose that q N ad p > 0. Show that for all N with x. + x + pq +q p q 4 + q Note that 1+x/ 0 for x. By the arithmetic-geometric mea iequality with + q factors we have p q = for all x. 1 + 1 + x p q + pq + q b Show that is icreasig for all N with x. for all x. If we choose p = q = 1 i 4, the = 1 1 +q + x + pq +q = + q + x + 1 +1 = 1 + x +1 = x+1 + 1 + 1 6
c Show that is bouded, ad therefore it coverges. We choose p = 1/2 ad q such that + x + q/2 + q, that is, x q/2. Applyig 4 we get + x + q +q 2 q 2 2 q + q +q = 2 q + q + q for all x. *d For discussio: Ca you guess what the limit of is? Ca you compute it for certai x R, ad why ot for others? Formally we ca rewrite i the form [ = 1 + 1 ] x x x Settig m := /x it reads = [ 1 + 1 m m ] x, ad usig the elemetary limit 1 + 1/m m e as m we expect that e x as. Note however, that we eeded to assume that m N for 1 + 1/m m e, ad that is ot the case for all x. We kow that coverges. Hece to determie its limit it is sufficiet to cosider ay subsequece Why?. Assume that x is ratioal ad positive, that is, x = p q for some p, q N, q 0. The /x N if is divisible by p, that is = kp for k N. We the have [ x q kp = 1 + 1 kq ] p e p kq as k by a usig a stadard limit. If we take the p-th root we get the limit e p/q. For irratioal x we caot presetly compute the limit as p/x N for all p N. We could however defie e x to be that limit, ad the prove all the properties of the expoetial fuctio by startig from x ratioal, the approximate. Extra questios for further practice 6. Let ad y be bouded sequeces i R with, y 0 for all N. a Show that y If l, usig that x k, y k 0, we have x l y l sup y. x k sup y k. This meas that sup x k sup y k is a upper boud for xl y l for all l. By defiitio of a supremum every upper boud is larger or equal tha the supremum sup l xl y l sup x k sup y k for all N. By defiitio of the limit superior ad the limit laws y 7 y.
b By givig a couter example, show that strict iequality is possible. We set := 2 + 1 ad y = 2 + 1 +1 for all N. We the have, y > 0 ad = y =. O the other had, y = 2 + 1 2 1 = for all N, so = y < y =. Hece, y = = 9. c If oe of the sequeces coverge, ad the product is ot of the form 0 or 0, show that y = y. Assume that y y. If y 0 ad is bouded, the y 0. Hece equality holds. Assume therefore that y y with y > 0. Applyig a y = y = y y y y y y y = y lim y = y Together with a equality follows. The above argumet works eve if is ot bouded from above. 7. *a Suppose that a is a sequece i R with a 0 for all N. If a +1 /a N is a bouded sequece, prove that 5 lim if a +1 a lim if a a +1 a a. We oly prove the last iequality. The middle oe is obvious from the defiitio of the limit superior ad iferior, ad the first oe ca be obtaied by reversig the sigs. Sice a +1 /a N is bouded, we have that s := a +1 a exists i R. We fix ε > 0. The by defiitio of the limit superior there exists N 1 depedig o ε such that a k+1 sup a k < s + ε for all N. I particular, a k+1 / a k < s + ε for all k N. Give > N we apply the iequality for each k = N, N + 1,..., 1 to obtai a a N = a a 1 a 1 a 2 a N+1 a N s + ε N. 8
Therefore a s + ε 1 N/ a N for all > N. We kow that a N 1 as ad similarly s + ε N/ 1 as. Hece, by the limit laws for the limit superior a s + ε. The above argumet works for every choice of ε > 0, ad hece a s as claimed. b Use part a to compute the limit of =!/ as. the we have Note that If we set a +1 a = a :=!, = a + 1! + 1 +1! = = 1 + 1 1 + 1 e. From part a ad the squeeze law, we coclude that lim = lim a +1 a = lim = 1 a e. 9