Math 140B - Notes. Neil Donaldson. September 2, 2009

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1 Mth 40B - Notes Neil Doldso September 2, 2009 Itroductio This clss cotiues from 40A. The mi purpose of the clss is to mke bsic clculus rigorous.. Nottio We will observe the followig ottio throughout this course. A fuctio f will, for us, lwys be from subset U R of the rel umbers (usully U is itervl), to the rel umbers, d we write f : U R. U is the domi of the fuctio f (ofte writte dom( f ) = U). The imge of the elemet (umber) x U uder the fuctio f is the elemet f (x) R. The imge of the fuctio f is the uio of the imges of the elemets of U: we will vriously write this s Im f = f (U) = { f (x) : x U}. Whe fuctio is defied without referece to U, the obvious choice of U s the lrgest possible set o which the fuctio could be defied is implied. For exmple f (x) = x 2 hs implied U = R, wheres f (x) = /x hs implied U = R \ {0}. Here re severl other cocepts tht my be required d tht you should be fmilir with. Supremum of set of rel umbers: the smllest rel umber greter th or equl to every elemet of the set. E.g. sup{, 2, 3, 4} = sup{x R : < x 4} = sup{x R : x < 4} = 4. sup X = mx X if X hs mximum vlue, otherwise sup X = mi{y R : y > x, x X}. Ifium of set of rel umbers: the gretest rel umber less th or equl to every elemet of the set. A fuctio f : U R is, or ijective, if f (x) = f (y) x = y. Oto A fuctio f : U V R is oto, or surjective, if for ll v V there exists some u U such tht f (u) = v. Iverses f : U V is bijective, or ivertible, if it is both d oto. It the hs iverse fuctio f : V U (lso bijective) stisfyig f ( f (u)) = u d f ( f (v)) = v for ll u U, v V. 2 Cotiuity Before we c del with differetible fuctios, we must first uderstd cotiuous fuctios (recll tht ll differetible fuctios re cotiuous).

2 2. Cotiuous Fuctios Defiitio 2.. A fuctio f : U R is cotiuous t x U if either of the followig hold:. For ll sequeces (x ) i U covergig to x, the sequece f (x ) i R coverges to f (x). 2. For ll ɛ > 0, there exists δ > 0 such tht y x < δ f (y) f (x) < ɛ. f is sid to be cotiuous o U if it is cotiuous t every poit x U. Recll the defiitio of x x: for ll δ > 0, there exists N N such tht > N x x < δ. The first defiitio my be rewritte most succictly s x x f (x ) f (x). Propositio 2.2. The bove defiitios re equivlet. Proof. Suppose tht f is cotiuous t x ccordig to the first defiitio, but ot ccordig to the secod. The there exists some ɛ > 0 such tht, for ll δ > 0, there exists y U stisfyig y x < δ d f (y) f (x) ɛ. I prticulr, the previous setece holds for δ = /. Whe δ = /, let y be the y s give bove. We thus hve sequece (y ) for which y x < / for ll. Hece y x d so, by the first defiitio, f (y ) f (x). But this is cotrdictio becuse f (y ) f (x) ɛ > 0. Now suppose tht the secod defiitio holds, d let x x. Let ɛ > 0 be give. The there exists δ > 0 such tht x x < δ f (x ) f (x) < ɛ. But x x there exists N N such tht > N x x < δ. Thus > N x x < δ f (x ) f (x) < ɛ d we hve prt. Here re two exmples where we use both methods. Exmples.. Let f (x) = x 5. () Let x x. The, by elemetry lysis, 2 ( ) 5 lim f (x ) = lim x x x x x5 = lim x = x 5 = f (x). x x (b) Let ɛ > 0 be give, d let ( ) ɛ δ = mi, 5 mx( x + 4, x 4. ) The y x < δ y 5 x 5 = y x y 4 + xy 3 + x 2 y 2 + x 3 y + x 4 y x 4 i=0 y i x 4 i (by the -iequlity) < δ 5 mx( x + 4, x 4 ) ɛ. This lst is sice both x, y (x, x + ), so tht y i x 4 i < mx( x + 4, x 4 ). y depeds o δ here. 2 lim(x k ) = (lim x ) k should hve bee proved i previous clss. 2

3 Usig both the limit d the ɛ-δ defiitio, we hve show tht f is cotiuous t ll vlues x R. { x si(/x) x = 0, 2. Let f (x) =. We show tht f is cotiuous t x = 0. 0 x = 0. () Let x 0. By the squeeze theorem, lim(x si(/x ) lim x = 0. (b) Give ɛ > 0, we let δ = ɛ. The 0 < y 0 < δ f (y) f (0) = f (y) = y si(/y) y < δ = ɛ. Usig either method we see tht f is cotiuous t x = 0. I the bove exmples it is cler tht the limit defiitio is esier to use. However, it hs some disdvtges, d s such you will be expected to use the ɛ-δ defiitio s well. You my well come up with differet choices of δ for your give ɛ compred to these exmples. This is to be expected, d your swer my lso be correct. The ɛ-δ defiitio is like plyig gme; give ɛ by the gme show host, you must provide δ tht stisfies the defiitio. Choosig smller δ th oe which lredy works still gives you workig δ. There re therefore ifiite umber of choices of δ which will work for you! Rther th use these defiitios every time, it is esier to prove few theorems tht will llow us to del with huge swthe of fuctios t oce. Theorem 2.3. If f is cotiuous t x, the k f d f re cotiuous t x, where k is y costt. Proof. Let x x. The lim((k f )(x )) = lim(k f (x )) = k lim f (x ) = k f (x). For the secod prt, f (x ) f (x) f (x ) f (x) 0. This iequlity is obvious if you observe tht is becomes equlity whe f (x ) d f (x) hve the sme sig. Theorem 2.4. Let f, g : U R be cotiuous t x U. The. f + g : U R defied by ( f + g)(y) := f (y) + g(y) is cotiuous t x, 2. f g : U R defied by ( f g)(y) := f (y)g(y), is cotiuous t x, 3. f /g : U R defied by ( f /g)(y) := f (y)/g(y), is cotiuous t x if g(x) = 0. Proof. Suppose tht x x.. lim(( f + g)(x )) = lim( f (x ) + g(x )) = lim f (x ) + lim g(x ) = f (x) + g(x). 2. lim(( f g)(x )) = lim( f (x )g(x )) = lim f (x ) lim g(x ) = f (x)g(x). ( ) 3. Suppose tht g(x ) = 0 for ll x. The lim g(x = ) lim g(x ) =. Now pply prt 2. g(x) Theorem 2.5. Suppose tht f, g re fuctios such tht g is cotiuous t x, d f is cotiuous t g(x). The the compositio f g is cotiuous t x. 3

4 Proof. Let x x. The g is cotiuous t x so tht g(x ) g(x). However (g(x )) is sequece covergig to g(x), poit t which f is cotiuous. Hece f (g(x )) f (g(x)). Theorem 2.6. The followig fuctios re cotiuous t ll poits t which they re defied: exp, si, cos, t, y polyomil. Proof. For e x, we use ɛ-δ. Let > ɛ > 0 be give, d let δ = l( + ɛ). Note tht δ < l( ɛ) = l ɛ = l( + ɛ + ɛ2 + ). The y < δ l( ɛ) < y < l( + ɛ) ɛ < e y < + ɛ e y < ɛ, hece the expoetil fuctio is cotiuous t x = 0. Now let x = 0 d let x x. The e x e x = e x e x x 0, sice exp is cotiuous t x = 0. exp is thus cotiuous everywhere. For the remider of the proof we c ppel to the previous theorems: use the defiitios of the trigoometric fuctios i terms of exp, for exmple cos(x) = eix +e ix 2 to see tht these re cotiuous wherever they re defied. 3 Tht x 2 d x 3 re cotiuous is similr rgumet to the erlier exmple o x 5. These fuctios re eough by compositio d dditio to crete ll polyomils. There is somethig little uderhded i this proof. We use the fct tht the expoetil fuctio is icresig, but where does this fct come from? Is it from clculus? If so the we lredy kow tht the expoetil is differetible d thus cotiuous. The issue is of which defiitio we re usig d how. Perhps the oly wy i which this proof would work would be to use the defiitio e x = lim ( + x ). But the you hve the questio of showig tht this first defies fuctio ot lest icresig oe, d tht its iverse fuctio is the logrithm (which is defied how?). You would ot therefore fid such proof i textbook o cotiuity. The precedig theorems tell us tht fuctios such s f (x) = x si(/x), f (x) = exp(si 2 (x 3 t x))), etc., re cotiuous t ll poits t which they re defied. Note tht x si(/x) is ot defied t x = 0; whe we did this exmple erlier we merely showed tht lim x 0 x si(/x) = 0, so tht extedig the domi of the fuctio by ddig i f (0) = 0 cretes cotiuous fuctio. 2.2 Properties of Cotiuous Fuctios I this sectio we cosider how beig cotiuous ffects other properties fuctio might hve. Defiitio 2.7. A fuctio f : U R is bouded if there exists some M R such tht f (x) M for ll x U. 3 We hve ot defied cotiuity for complex-vlued fuctios, though it is exctly the sme defiitio. 4

5 Recll tht sequece (x ) is bouded if there exists some M R such tht x M for ll N. We wish to show tht cotiuous fuctios o closed itervls re bouded d tti their bouds. Before provig this, we recll two importt fcts: Lemm 2.8. Suppose tht (x ) is coverget sequece whose vlues live i the closed itervl [, b]. The the limit of (x ) lso lies i [, b]. Theorem 2.9 (Bolzo Weierstrss). Every bouded sequece i R hs coverget subsequece. Proof of Lemm 2.8. Suppose x x > b. The let ɛ = x b 2 i the defiitio of covergece of sequece. There must exist N N such tht > N x x < ɛ. But the x > x ɛ = x x b = x + b > b, for ll > N, 2 2 cotrdictio. The rgumet to see tht x is similr. Proof of Bolzo Weierstrss. Let (x ) be bouded so tht x < M for ll. The, sice the sequece is ifiite, t lest oe of [ M, 0] or [0, M] cotis ifiite umber of terms of the sequece. Choose such d cll the itervl I, the choose some x I. Agi, t lest oe of the hlves of the itervl I cotis ifiite umber of terms of the sequece; choose such d itervl, cll it I 2, d choose some x 2 I 2 such tht 2 >. Cotiuig i this vei we obti subsequece (x k ). Observe tht I hs legth 2 M, d tht the itervls re stcked: I I I 2 I 3, so tht x k x j < 2 k M whe j > k. But s k, this pproches zero so tht (x k ) is Cuchy sequece. By the completeess of R ll Cuchy sequeces coverge d so this subsequece is coverget. Theorem 2.0. Suppose tht f : [, b] R is cotiuous fuctio. The f is bouded. Proof. Suppose tht f is ot bouded. The for ll N, there exists elemet x [, b] such tht f (x ). Now (x ) is bouded sequece, hece, by the Bolzo Weierstrss theorem, it hs coverget subsequece (x k ). Moreover, sice (x k ) is bouded i the itervl [, b], it follows tht its limit x lso lies i [, b]. But f is cotiuous, d so f (x k ) f (x). However, f (x k ) by ssumptio. Cotrdictio. Theorem 2.. Suppose tht f : [, b] R is cotiuous fuctio. The f ttis its bouds. I.e. there exist x 0, y 0 [, b] such tht f (x 0 ) f (x) f (y 0 ) for ll x [, b]. Proof. Let M = sup{ f (x) : x [, b]}. This is fiite sice f is bouded fuctio (by previous theorem). Now let N. There exists y [, b] such tht M < f (y ) M. Repetig for every, we hve sequece f (y ) whose limit is M. The sequece (y ) is bouded i the closed itervl [, b], d so by Bolzo Weierstrss hs coverget subsequece (y k ) with limit y 0 [, b]. By cotiuity of f we hve f (y 0 ) = lim k f (y k ) = M (sice ( f (y k )) hs the sme limit s ( f (y ))). We thus hve f (y 0 ) = M. The proof for ttimet of the lower boud is the ideticl rgumet pplied to the fuctio f. 5

6 Theorems 2.0 d 2. must be used crefully: they oly pply to cotiuous fuctios f whose domi is closed itervl [, b]. You should be ble to give exmples of the theorems filig whe either of these coditios re dropped. The followig theorem is ituitively obvious from the ïve defiitio of cotiuity (tht you my grph such fuctio without tkig your pe from the pge). Theorem 2.2 (Itermedite vlue theorem). Let f be cotiuous rel-vlued fuctio o itervl I d let, b I with < b d f () = f (b). If ξ lies betwee f () d f (b), the there exists c (, b) such tht f (c) = ξ. Roughly spekig, if cotiuous fuctio tkes two vlues t differet poits, the its vlues must pss through everythig i betwee. Proof. Suppose first tht f () < ξ < f (b) d let S = {x [, b] : f (x) < ξ}. Clerly S so tht S =. Moreover S is bouded, hece c := sup S is well-defied d c [, b]. Clerly f (c) ξ (there exists sequece s S with limit c, f is cotiuous d so lim f (s ) = f (c) ξ). Now defie sequece (t ) by t = mi{b, c + }. Clerly c t c + = lim t = c. Moreover, t [, b] \ S f (t ) ξ for ll. Hece f (c) = lim f (t ) ξ. We thus hve f (c) = ξ. The rgumet for f (b) < ξ < f () is lmost ideticl. Note tht the c chose i the proof is the lrgest elemet of the set {x : f (x) = ξ}. We could hve rrged the proof differetly i order to choose some other vlue c (e.g. the miimum of this set). Try grphig fuctios for rdom vlues of c. Corollry 2.3. If f is cotiuous d rel-vlued o itervl I, the the imge f (I) of f is lso itervl (or poit). Proof. A itervl I is chrcterized by the property tht if x, x 2 I d x < x < x 2, the x I. Cosider f (I) ow d suppose tht it cotis more th oe poit. Ideed suppose tht f (x ) < ξ < f (x 2 ) for some ξ. By the itermedite vlue theorem, ξ = f (c) for some c I (ideed c lies betwee x, x 2 ). f (I) is thus itervl. The itermedite vlue theorem hs gret pplictio i demostrtig the existece of solutios to problems. It c lso be used to help rrow i o their loctio. For exmple, suppose you wt to solve g(x) = 0 for g cotiuous o itervl. By tril d error you fid tht g(0) > 0 d g() < 0. The theorem tells us tht there must exist t lest oe solutio to g(x) = 0 i the itervl (0, ). Sy you ow compute g(.5) > 0; solutio must exist betwee.5 d. Note tht we ve ot ruled out the possibility of solutio betwee 0 d.5, or ywhere else, we simply kow tht there must be t lest oe betwee.5 d. We c keep hlvig the size of the itervl to home i o solutio. For exmple, the sie fuctio is cotiuous o the ifiite itervl R, we kow tht si(2π) = 0 d tht si((2 +.5)π) = for Z. There must therefore exist ifiite umber of solutios to the equtio si(x) =.392, t lest oe lyig i the itervl (2π, (2 +.5)π) for ech Z. Exmple. Prove tht x2 x = for some x (0, ). For this we defie g(x) = x2 x d ote tht g is cotiuous o the itervl [0, ]. Moreover g(0) = < 0 d g() = > 0. By the itermedite vlue theorem, g(x) = 0 hs solutio x (0, ). 6

7 2.3 Uiform Cotiuity Recll the secod (ɛ δ) versio of the defiitio 2. of cotiuity. I this we estblish cotiuity of fuctio f idepedetly t ech poit of its domi. Ideed, we first fix poit y dom f, the, for ech ɛ > 0 we must fid δ > 0 such tht x y < δ = f (x) f (y) < ɛ. The importt poit to ote i this sectio is tht i successful proof of the cotiuity of f over its domi, our δ is fuctio both of ɛ d the loctio: δ = δ(ɛ, y). I some cses it my be possible to fid δ which is idepedet of y. Defiitio 2.4. Let f be rel-vlued fuctio defied o set S R. We sy tht f is uiformly cotiuous o S if: For ll ɛ > 0, there exists δ > 0 such tht for ll x, y S with x y < δ we hve f (x) f (y) < ɛ. We sy tht f is uiformly cotiuous if it is uiformly cotiuous o its domi. Exmples.. f (x) = x is uiformly cotiuous o the set [, ). For this let ɛ > 0 be give d let δ = ɛ. Now suppose tht x y < δ. The x y = y x xy < δ xy = ɛ xy. However x, y xy, so this RHS is < ɛ d we hve uiform cotiuity. 2. The bove exmple c be exteded to show uiform cotiuity o y itervl [, ), where > 0. Here x, y xy, hece we my choose δ = 2 ɛ. Note tht s gets smller, our 2 choice of δ is forced to be smller: it is cler tht the rgumet fils whe preseted with the ope itervl (0, ). 3. f (x) = x is ot uiformly cotiuous o y itervl I of the form (0, ) (for > 0). For this ssume the coverse: tht give ɛ we hve δ for which ll x, y I stisfy x y < δ x y < ɛ. Let x I such tht y = x + δ/2 I. We my lwys choose δ smll eough to ccomplish this if δ stisfies the defiitio of uiform cotiuity, the δ < δ lso stisfies, sice x y < δ x y < δ. But ow x y = y x xy = δ 2xy = δ x(2x + δ). Lettig x 0 we see tht this RHS c be mde lrger th whtever ɛ we were give. Cotrdictio. It follows tht f (x) = x is ot uiformly cotiuous o y itervl of the form (0, ] or (0, ). Rther th provide hudreds of exmples, we c see tht uiform cotiuity is utomtic i lot of cses. Theorem 2.5. If f is cotiuous o closed bouded itervl [, b], the it is uiformly cotiuous o tht itervl. 7

8 Proof. Suppose tht f is ot uiformly cotiuous. The there exists ɛ > 0 such tht for ll δ > 0 there exists x, y [, b] with x y < δ d f (x) f (y) ɛ. For ech N we my choose δ = d see tht there exist sequeces (x ), (y ) i [, b] for which x y < d f (x ) f (y ) ɛ. (x ) is bouded d so hs coverget subsequece (x k ) by the Bolzo Weierstrss theorem. Let its limit be x 0. The we clerly hve (y k ) covergig to x 0 lso. By the cotiuity of f we hve tht f (x 0 ) is the limit of both of the sequeces ( f (x k )) d ( f (y k )). However this implies tht f (x k ) f (y k ) 0 s k, which cotrdicts the ssumptio tht f (x ) f (y ) ɛ. f is therefore uiformly cotiuous. Be creful with this theorem. R is closed itervl but ot bouded oe, d so y cotiuous fuctio o the whole of R eed ot be uiformly so. I prticulr, f (x) = x 2 is cotiuous fuctio o R. Suppose f is uiformly cotiuous. The for fixed ɛ > 0 we must be ble to fid fixed δ > 0 for which x y < δ x 2 y 2 < ɛ. However, lettig x = y + δ/2 we hve x y = δ/2 < δ d x 2 y 2 = δ(y + δ 2 /4). Sice y c be rbitrrily lrge, it is cler tht this right hd side eed ot be less th ɛ. f (x) = x 2 is ot therefore uiformly cotiuous o R. With this theorem we c see how fr uiform cotiuity exteds. For exmple, suppose tht f is cotiuous o ope itervl (, b). The f is uiformly cotiuous o y closed itervl iside this: i.e. if < < b < b, the f is uiformly cotiuous o [, b ]. Note tht there is o mximum closed subitervl of (, b). Similrly, if f is cotiuous o [0, ), the f is uiformly cotiuous o y closed itervl [0, ] for fiite, but ot ecessrily o the full itervl [0, ). Cotiuous extesios Defiitio 2.6. A fuctio f is extesio of f if dom f dom f d f (x) = f (x) for ll x dom f. If f is cotiuous we sy it is cotiuous extesio of f (ote tht f must be cotiuous for cotiuous extesio to exist). Exmples.. f : x x 2 with domi R is extesio of the fuctio f : x x 2 with domi [0, ). 2. Let f (x) = x si(/x) o the itervl (0, ). This is clerly cotiuous fuctio (formed by composig d multiplyig cotiuous fuctios). Cosider the extesio f defied by { x si(/x), x = 0, f (x) = 0, x = 0. The domi of f is ow the whole of R. Is f cotiuous extesio of f? For this we simply eed tht the limit of f s x 0 is f (0) = 0. This is trivil by the squeeze theorem, d so f is cotiuous. 3. Now cosider g defied o the itervl [0, ] by g(x) = f (x) from exmple 2. g is clerly cotiuous, d is uiformly cotiuous by theorem 2.5. If g(x) = x si(/x) with domi (0, ), the g is uiformly cotiuous extesio of g. The figures below show first f, the the restrictio of f to [ /2, /2]. The grph oscilltes ifiite umber of times before rechig zero. 8

9 Coversely to the bove, the fuctio f (x) = si(/x) defied o (0, ) hs o cotiuous extesio, sice it hs o limit s x 0. Tht f (x) here is ot uiformly cotiuous my be see from the followig theorem which we stte without proof. Theorem 2.7. Suppose tht f : (, b) R. The f is uiformly cotiuous iff it hs cotiuous extesio f : [, b] R. This is esier to work with th to prove. Give cotiuous fuctio o closed itervl we lredy kow tht it is uiformly cotiuous. Otherwise, if f is cotiuous o ope itervl (, b) the it is cler tht there c oly be t most oe cotiuous extesio f : [, b] R defied by f (x), x (, b), f (x) = lim y f (y), x =, lim y b f (y), x = b. f, beig cotiuous o closed itervl, is mifestly uiformly so. The issue is simply whether the limits exist. For fil criterio o uiform cotiuity we recll the Me Vlue Theorem (we will prove this towrd the ed of the course). Theorem 2.8 (Me Vlue Theorem). Let f : [, b] R be cotiuous d moreover differetible o (, b). The there exists c (, b) such tht f (c) = f (b) f (). b Theorem 2.9. Let f be cotiuous o itervl I. Let J I be the itervl obtied from I by removig y (o-ifiite) edpoits. 4 If f is differetible d f is bouded o J, the f is uiformly cotiuous o I. Proof. Suppose tht f (x) M o J. Let ɛ > 0 be give d set δ = ɛ M. We show tht this ɛ δ pir stisfy the defiitio of uiform cotiuity. Let x, y I such tht x y < δ = ɛ M. By the Me Vlue Theorem, there exists c (x, y) such tht f f (x) f (y) (c) = x y. But the f (x) f (y) = f (c) x y < Mδ = ɛ. 4 E.g. I = [0, ) J = (0, ) d I = [0, ) J = (0, ). 9

10 This theorem quickly shows us, for exmple, tht o-lier polyomils re ot uiformly cotiuous o R. Ideed if f (x) = x, the f (x) = x is bouded o (0, ) iff. Similrly f is bouded o [0, ) (for fiite ) iff. Thus f (x) = x x 3/2 is uiformly cotiuous wy from 0, while g(x) = x + x 3/2 is uiformly cotiuous o fiite itervls. I prticulr, the oly fuctios of the form f (x) = r R α r x r where ech α r R to be uiformly cotiuous o the whole of R re the lier polyomils f (x) = + bx. 2.4 Limits of Fuctios We hve lredy worked i severl courses with the cocept of limit. I this sectio we gther together some results tht should mostly be fmilir to you. I prticulr we eed to be sure wht we me by left- d right-hded limits d ifiite limits, d how the limit of fuctio my be expressed i terms of ɛ-δ ottio. You should be comfortble workig with the defiitio of the limit of sequece. Defiitio 2.20 (Oe-sided limits). Let R d suppose tht f is defied o itervl (b, ). If lim x f (x) = L exists, the we cll L the left-hd limit of f t. The right-hd limit of f t is defied similrly s L = lim x + f (x) for f defied o some itervl (, b). Defiitio 2.2 (Two-sided limits). Suppose tht f is defied o set S = (b, ) (, c) d tht the left- d right-hded limits of f t both exist. If these oe-sided limits re ideticl (sy = L) the we sy tht L = lim x f (x) is the limit of f t. I ll of these defiitios we hve the followig otio of limit: if for ll sequeces (x ) i dom f covergig to, the limits L = lim x f (x ) re ideticl, the 5 lim x (±) f (x) = L. Defiitio 2.22 (Ifiite limits). Suppose tht f is defied o set (, ). If for every sequece (x ) (, ) covergig to we hve L = lim x f (x ) ideticl, the we write L = lim x f (x). lim x f (x) is defied similrly. Note tht the fuctio f eed ot be defied t. Eve if it is, its limit might ot exist, or be etirely differet. { Exmples.. Let f (x) = x x = 0, 0 x = 0.. Here lim f (x) =, lim x 0 + f (x) = x 0 Note tht lim x 0 f (x) does ot exist, d tht either the left or right limits of f t 0 equl the vlue f (0). { x = 0, 2. Let f (x) = x 2 0 x = 0.. Here lim x 0 f (x) = = lim f (x) = lim f (x). + x 0 x 0 Agi the limit of f t 0 does ot equl the vlue f (0). 5 We write x (±) to cover ll limits; left, right d two-sided. 0

11 3. Let f (x) = x4 x. Here f is ot defied t x =, however, for x = we hve f (x) = x 3 + x 2 + x + x 4 = lim x f (x) = 4. The domi of f is (, ) (, ), wheres the domi of the fuctio g(x) = x 3 + x 2 + x + is the whole of R. As such, g is extesio of f. Theorem Suppose tht f, g re fuctios d R {± } is such tht ll possible limits L = lim x (±) f (x), M = lim x (±) g(x) exist. The, provided the limits mke sese,. lim x (±)( f + g)(x) = L + M. 2. lim x (±)( f g)(x) = LM. 3. lim x (±)( f /g)(x) = L/M (provided M = 0). Proof. All these my be proved by choosig sequeces x, where (x ) lies i the required rge (i.e. (x ) (b, ) for some b whe we re cosiderig x ). The, by defiitio, we hve lim x f (x) = L d lim x g(x) = M. The, e.g., The remider re similr. lim x (±) ( f + g)(x) = lim x ( f + g)(x ) = lim x ( f (x ) + g(x )) = L + M. We hve to be creful with the bove results whe limits re ifiite: for exmple if L = d M =, we cot use the theorem to sy wht lim x ( f + g)(x) is, sice L + M does t me ythig. To evlute such limit we would hve to tke differet pproch. We hve lredy see tht we c use the cocept of limits to exted fuctios cotiuously. Suppose f is cotiuous o itervl I except t fiite umber of poits. We cll these poits sigulrities (either f is ot defied t these poits, or f is defied, but is ot cotiuous there). There re two types of sigulrity: removble d o-removble. Suppose tht f is cotiuous o (, b) except for sigulrity t c (, b). c is removble sigulrity if the limit L = lim x c f (x) exists d is fiite. I this cse { f (x) x = c, f (x) := L x = c, is cotiuous extesio of f (o the domi (, c) (c, b)). I.e. f is cotiuous o (, b) d grees with f wy from c. Note tht we do t cre whether f (c) is defied. If ot, there is o issue, d f is cotiuous extesio of f ; if so, the the vlue f (c) is exctly tht whose removl mkes the sigulrity so. I either cse, we re chgig f i s smll wy s possible to s to mke it cotiuous o the whole itervl (, b). A o-removble sigulrity c is oe for which lim x c f (x) either does ot exist, or is ifiite. I such cse we cot simply chge the vlue (or eve isert vlue) of f t c i order to mke f cotiuous there.

12 si x x < 0, x x (0, ), 2 x =, Exmples.. Let f (x) =. Here f is cotiuous o R except for t its sigulrities t x = 0, d 2. The first two of these re removble sigulrities, while the third is 2 x x (, 2), 2 x = 2, x 3 x > 2. ot. Ideed removig the first two sigulrities leves us with si x x < 0, x x [0, ], f (x) = 2 x x (, 2), 2 x = 2, x 3 x > 2, which is cotiuous everywhere except x = 2. 3 Sequeces d Series of Fuctios You should recll the discussio of power series from Mth2J. I tht course we ecoutered the rdius of covergece of series. I this chpter we will discuss the cotiuity of power series. 3. Power Series First we recll much of the importt mteril from 2J. Defiitio 3.. A ifiite series ) is the sum of ifiite sequece: =0. A series is coverget if the sequece (S k ) = ( k =0 of prtil sums is coverget. A series is bsolutely coverget if =0 coverges. A coverget series tht is ot bsolutely coverget is coditiolly coverget. Otherwise series is diverget. Exmples.. =0 2 is coverget (bsolutely sice ll terms re positive lredy). 2. =0 is diverget. 3. =0 ( ) is coditiolly coverget. Defiitio 3.2. A power series is ifiite sum x, =0 where ( ) is sequece of rel umbers, d x is vrible. A power series evluted t x is simply ifiite series d thus coverges bsolutely, coditiolly or diverges. A power series defies fuctio of x provided it coverges for some vlues of x. 2

13 Theorem 3.3. Give power series =0 x, exctly oe of the followig is true:. The power series coverges bsolutely for ll x R. 2. The power series oly coverges whe x = The power series coverges oly o bouded itervl cetered t 0: i.e. itervl of the form [ R, R], ( R, R), [ R, R) or ( R, R]. I this cse the covergece is bsolute o ( R, R). Defiitio 3.4. The rdius of covergece of power series is the R defied i the bove theorem. I the first cse we hve R = while i the secod R = 0. The rdius of covergece my be clculted i severl wys, the esiest of which is the rtio test. Theorem 3.5. Give power series =0 x, suppose tht the sequece hs limit L. The the rdius of covergece is give by R = L. + The followig theorem mkes the criticl defiitio from which ll the previous results follow: Theorem 3.6. Give power series =0 x defie L = lim sup / d R = L. The R is the rdius of covergece of the power series. The previous theorem is ot ofte the best wy to clculte rdii of covergece sice workig with lim sup c be messy. It is sometimes ecessry however, sice it s esy to cococt exmples for which the rtio test fils. Recll the defiitio: Defiitio 3.7. Let (x ) be sequece. The lim sup x := lim N sup{x : > N}, lim if x := lim N if{x : > N}. Oe of the resos for itroducig these cocepts is tht they work for sequeces tht hve o limit. For exmple lim sup( ) =. We stte the root test for ifiite series without proof. Lemm 3.8 (Root test for ifiite series). Let b be ifiite series d defie α = lim sup b /. The:. If α < the series coverges bsolutely. 2. If α > the series diverges. 3. If α = we cot coclude ythig. 3

14 Proof of theorem 3.6. Fix x R d defie b = x. The α = lim sup b / = x lim sup / = x L. We hve bsolute covergece of b for those x such tht α < ; i.e. wheever x < /L. Coversely we hve divergece wheever α > ; i.e. wheever x > /L. This is precisely the defiitio of R = /L beig the rdius of covergece of power series. Proof of the Rtio test. Recll the lemm tht for y sequece ( ) we hve lim if + lim if / lim sup / lim sup +. If exists the its limit is equl to the left- d right-most terms i the bove iequlity. Hece ll + four re equl, d, by the root test, the limit is ecessrily /R where R is the rdius of covergece. Exmples.. = x. Here = d + = + R =. Moreover, whe x = we hve =0 =0, hece the rdius of covergece is which diverges, d whe x = we hve ( ) which coverges. Hece the power series coverges oly o the itervl [, ). It c be see tht this power series coverges to the fuctio l( x) o the itervl of covergece. Note tht eve though the fuctio l( x) is perfectly well defied for x >, the power series does t coverge there, let loe coverge to this fuctio. 2. =0! x hs + =! (+)! = + 0, hece R =. The power series coverges for ll x R. It c be see tht this power series coverges to the fuctio exp(x) everywhere o R. Ideed this is oe of the more commo defiitios of the expoetil fuctio. 3. =0!x hs + oly coverges whe x = 0. = (+)!! = +, hece the rdius of covergece is zero: the series 4. =0 2 x coverges o the itervl [, ]. 5. =0 x coverges o the itervl (, ). Together with exmples d 4 we see tht if 0 < R <, the itervl of covergece my be either ope, closed, or hlf-ope (just s climed i Theorem 3.3). You lwys eed to check the edpoits idepedetly. 6. I this exmple we see how the rtio test c fil. Let = ( 2 ( 3) if is eve, d 3 2) if is odd. The {( + 2 ) 2+ = 3 eve, ) 2+ odd. ( 3 2 This sequece hs o limit (ideed it hs oe subsequece covergig to zero d other divergig to ). However if we use the root test we hve L = lim sup / = 3 2, so tht the rdius of covergece of the power series =0 x is R = 2 3. By strightforwrd clcultio we c see tht the itervl of covergece is ( 2/3, 2/3). 4

15 7. Filly, other tht c be doe by either the rtio test (suitbly modified), or the full root test. Cosider = 3 x4. To use the rtio test we hve to replce y = x 4 to obti power series i y for which successive coefficiets re o-zero. The we re iterested i the covergece of = 3 y which, by the rtio test hs rdius of covergece 3 R = + ( + ) lim 3 = 3. The origil series thus hs rdius of covergece 4 3. By the root test we re obliged to fid 6 lim sup(3 ) /4 = lim sup(3 / ) /4 = 3 /4, yieldig the sme rdius of covergece R = 4 3. This series diverges t both edpoits of the itervl of covergece: = 3 (±3/4 ) 4 = =. = I the lst exmple the sequece ( m ) is (0, 0, 0, /3, 0, 0, 0, /8,...), with oly 4m beig ozero. We either hd to chge the vrible of the series i order to ivoke the rtio test, or simply use the root test, beig creful to observe tht here lim sup m /m = lim sup 4m /4m. Wht will we do with power series? The gol of this clss is ot to fid sums of power series, but rther to sk questios bout the cotiuity d differetibility of fuctios defied s power series. For exmple, suppose f (x) = k=0 kx k for some sequece ( ) d for ech N. The sequece of fuctios ( f ) coverges poitwise to fuctio f (x) = k=0 kx k wherever the ifiite series coverges. Ech f is simply polyomil, d thus cotiuous. C we clim tht the poitwise limit f is cotiuous? The swer is ot lwys. f will lter be see to be cotiuous o closed subitervls [ R, R ] [ R, R] (where R < R), but eed ot be o the whole itervl of covergece. Similrly, give geerl sequece of cotiuous fuctios ( f ) so tht the poitwise limit is defied (for ech x defie f (x) := lim f (x)), do we hve y fith tht f is itself cotiuous? We cosider this cocept i the ext sectio. 3.2 Uiform Covergece Defiitio 3.9. Let ( f ) be sequece of fuctios f : U R. We sy tht ( f ) coverges poitwise to f : U R if, for ech x U, f (x) f (x). Otherwise sid: f f poitwise if for ech ɛ > 0 d ech x U there exists N N such tht 6 We use /. > N = f (x) f (x) < ɛ. 5

16 I the bove defiitio N depeds o both ɛ d the positio x. I the discussio of uiform cotiuity (Sectio 2.3) we sw tht cocept of cotiuity where δ is idepedet of positio is useful. The correspodig cocept for covergece of fuctios is defied ext. Defiitio 3.0. Let ( f ) be sequece of fuctios f : U R. We sy tht ( f ) coverges uiformly to f : U R if: for ll ɛ > 0 there exist N N such tht > N = f (x) f (x) < ɛ, for ll x U. The ide is simply tht the sme uiform choice of N works for ll x. It is cler tht uiform covergece implies poitwise covergece. Exmples.. Let f (x) = x o [ /2, /2]. Clerly f (x) 0 poitwise. Now let > ɛ > 0 be give d defie N = l ɛ l 2. The > N implies Hece the covergece is uiform. x = x 2 < 2 N = 2 l ɛ l 2 = e l ɛ = ɛ. 2. Now cosider f (x) = x o [0, ]. Here the poitwise limit is f (x) = 0 for x = d f () =. Suppose tht the covergece is uiform d let ɛ = /2. The there exists N N such tht > N implies tht x < /2 for ll x [0, ) d x < /2 for x =. However, for > x > 2 /(N+) we hve x N+ > /2; cotrdictio. The covergece f f is ot uiform. 3. Let f (x) = +2 cos2 (x) for x R. The f coverges poitwise to the zero fuctio: f (x) 3 0. Let ɛ > 0 be give d defie N = 9. The > N implies ɛ 2 Hece f 0 uiformly. + 2 cos 2 (x) 3 < 3 = ɛ. N 4. Let f (x) = ( x ) 2 o [, ]. Clerly f (x) f (x) = x 2 poitwise s. Now let ɛ > 0 be chose d defie N = 3 ɛ. The > N implies ( x ) 2 x 2 = 2 2x ( ) + 2 x 3 < 3 N = ɛ. The covergece is uiform. Theorem 3.. Suppose tht ( f ) is sequece of cotiuous fuctios which coverge uiformly to f. The f is cotiuous. Proof. Let x 0 be fixed - we prove cotiuity of f t x 0. Let ɛ > 0 be give. Sice f f uiformly, there exists N N such tht, for ll x, > N f (x) f (x) < ɛ/3. 6

17 Choose y > N. Sice f is cotiuous t x 0 we hve tht there exists some δ > 0 such tht x x 0 < δ f (x) f (x 0 ) < ɛ/3. Puttig ll this together we hve tht if x x 0 < δ the f is therefore cotiuous t x 0. f (x) f (x 0 ) f (x) f (x) + f (x) f (x 0 ) + f (x 0 ) f (x 0 ) < ɛ 3 + ɛ 3 + ɛ 3 = ɛ. Exmples.. Cosider exmples, 3 d 4 bove. I ll three cses the fuctios f re cotiuous, the covergece is uiform, d the limit f is cotiuous. 2. The theorem does ot go both wys: f my be cotiuous without the covergece beig uiform. Cosider f (x) = x o [0, ). The poitwise limit is clerly f (x) = 0 which is cotiuous. However essetilly the sme rgumet s give i exmple 2 shows tht the covergece is ot uiform. 3. Let f (x) = cos (x) o [ π/2, π/2]. The f (x) f (x) poitwise, where f (x) = 0 if x = 0, d f (0) =. f is ot cotiuous t x = 0 d so the covergece is ot uiform. The followig corollry gives good wy of thikig bout uiform covergece i terms of cotrollig the mximum of the differece betwee f (x) d f (x). Better still it gives criterio by which lck of uiform covergece c be detected. Corollry 3.2. f f uiformly iff sup f (x) f (x) 0. Proof. If f f uiformly, the give ɛ > 0 there exists N N such tht > N f (x) f (x) < ɛ/2. Clerly we the hve sup f (x) f (x) < ɛ d the climed covergece. Coversely, if sup f (x) f (x) 0 the give ɛ > 0 there exists N N such tht > N sup f (x) f (x) < ɛ. However f (x) f (x) sup f (x) f (x) < ɛ, so the covergece is uiform. Exmples.. I exmple 2 bove we clerly hve sup x = Cosider f (x) = ( x ) 2, but this time o R. As we ve see, this sequece coverges poitwise to f (x) = x 2 d the covergece is uiform o the itervl [, ]. O R however, we hve sup f (x) f (x) = sup x R 2 2x =, sice x is ubouded. Covergece is ot uiform o R. 7

18 3. Cosider the followig sequece of fuctios: f (x) = x ( x) o the itervl [0, ]. Sice for x < we hve f + f = + x x <, it follows tht f coverges poitwise to zero. However, we c lso clculte the mximum of ech f : f (x) = x ( ( + )x) = 0 x = 0, +. The mximum of the fuctio is clerly locted t x = +. The loctio is sequece pprochig. Moreover the mximum vlue of ech f is see to be ( ) ( ) ( f = ) ( ) + ( = = ) + e The covergece is therefore ot uiform. 3.3 More o Uiform Covergece I this sectio we outlie severl useful ides d results to do with uiform covergece. Theorem 3.3. Let f f be uiform o [, b] where ll fuctios f d f re itegrble. The lim f (x) dx = f (x) dx. Proof. If f, f re itegrble, the so re the fuctios f f. Let ɛ > 0 be give. The, sice f f uiformly, there exists N N such tht > N implies f (x) f (x) < ɛ b for ll x. Thus if > N we hve f (x) dx f (x) dx = f (x) f (x) dx f (x) f (x) dx ɛ dx = ɛ. b I prticulr the theorem holds wheever the sequece ( f ) cosists of cotiuous fuctios (i which cse f is lso cotiuous). Exmples.. Let f (x) = x. This pproches f (x) = 0 poitwise for ll x R. Differetite +x 2 to fid the mximum: f (x) = x2 + x 2 = 0 x = ±. Sice lim x ± f (x) = 0 we coclude tht f hs its mximum t x = d its miimum t x =. Evlutig we hve Therefore f ( ± ) = ± /2. sup f (x) f (x) = /2 0. The covergece is therefore uiform. Moreover we compute f (x) dx = 2 l( + x2 ) b = 2 l + 2 b = f (x) dx. 8

19 2. We ve lredy see tht the poitwise covergece f 0 where f (x) = x ( x) is ot uiform. However f (x) dx = /( + ) /( + 2) 0, 0 so the sequece of itegrls my coverge eve if the covergece is ot uiform. Next we cosider the cocept of sequece beig uiformly Cuchy. This is logous to the cocept of Cuchy sequece: sice R is complete we kow tht every Cuchy sequece coverges to somethig. The sme ide pplies here: A uiformly Cuchy sequece of fuctios c be see to coverge without kowig wht the limit is. Defiitio 3.4. A sequece of fuctios f : U R is uiformly Cuchy o U if: for ll ɛ > 0 there exists N N such tht m, > N f (x) f m (x) < ɛ for ll x U. Theorem 3.5. A sequece of fuctios f : U R is uiformly Cuchy o U iff it coverges uiformly to some f : U R. Proof. Let ( f ) be uiformly Cuchy o U. Let x 0 U d let ɛ > 0 be give. The there exists N N such tht m, > N f (x 0 ) f m (x 0 ) < ɛ. Hece ( f (x 0 )) is Cuchy sequece. By the completeess of R this sequece coverges: defie f (x 0 ) to be this limit. Repet for ll x U so tht we hve defied fuctio f : U R. We clim tht f f uiformly. Let ɛ > 0 be give. The there exists N N such tht m, > N f (x) f m (x) < ɛ/2. Hece for > N we hve tht f m (x) ɛ/2 < f (x) < f m (x) + ɛ/2. It follows tht the limit f (x) lies i the itervl [ f m (x) ɛ/2, f m (x) + ɛ/2]: i.e. Thus f f uiformly. For the coverse see the homework questios. f m (x) f (x) ɛ/2 < ɛ. The use of this theorem comes whe we do t kow the limit of sequece of fuctios, i prticulr whe cosiderig power series. For exmple cosider the power series =0 x. Cosiderig the sequece of prtil sums f (x) := k=0 kx k we hve tht this sequece is uiformly Cuchy o U iff for ll ɛ > 0 there exists N N such tht > m > N k=m+ k x k < ɛ for ll x U. If we c demostrte this for give power series, the we kow tht the sequece of prtil sums ( f ) coverges uiformly o U to the power series: we sy tht the series coverges uiformly o U. I prticulr, sice ech term i the sequece ( f ) is cotiuous, the series itself is cotiuous o U. x +x is se- There re my other series of fuctios tht we my cosider. For exmple =0 ries: we my sk if the sequece of prtil sums coverges uiformly to some limit by skig if the sequece is uiformly Cuchy. Cosiderig the bove discussios, we hve proved the followig: Theorem 3.6. Let k=0 g k(x) be series of fuctios o U. If ech g k is cotiuous d the series coverges uiformly o U the the series is cotiuous o U. 9

20 Sometimes it c be difficult to tell directly tht series of fuctios coverges uiformly, or is ideed uiformly Cuchy. I this cse the followig result c be helpful. Theorem 3.7 (Weierstrss M-test). Let (M k ) be sequece of o-egtive rel umbers such tht M k <. If g k (x) M k for ll x U the g k (x) coverges uiformly o U. Proof. For y > m we clerly hve k=m+ g k (x) M k. k=m+ However the sequece of prtil sums of the series M k is Cuchy d so for y ɛ > 0 we my fid N N such tht > m > N implies tht the RHS of the bove is less th ɛ. The sequece of prtil sums of the power series is thus uiformly Cuchy. Exmple. Cosider the series =0 +cos 2 (x) 2 + cos 2 (x) 2 si(x). We clerly hve si(x) 2 2. is bouded d so the series coverges uiformly o R. Ideed the series is cotiu- However 2 2 ous o R. 3.4 Differetitio d Itegrtio of Power Series Theorem 3.8. Let =0 x be power series with rdius of covergece R (0, ]. If 0 < R < R the the power series coverges uiformly o [ R, R ]. Moreover =0 x is cotiuous fuctio o the full ope itervl ( R, R). Proof. Sice 0 < R < R we hve bsolute covergece t x = R: i.e. =0 R <. Moreover it is cler tht x R for ll d for ll x [ R, R]. Defiig M = R d pplyig the Weierstrss M-test we see tht the series =0 x coverges uiformly o [ R, R]. The limit is thus cotiuous o [ R, R ] for ll 0 < R < R. Sice every poit x 0 ( R, R) is i such closed itervl for some R we coclude tht the series is cotiuous o the whole of ( R, R). It is importt to ote tht power series might ot coverge uiformly o the whole of ( R, R). For exmple the series =0 x coverges to the fuctio f (x) = x o the itervl of covergece (, ). However, tkig the sequece of prtil sums f (x) = k=0 = x+ x we see tht f (x) f (x) = x x+ x = x x. It is cler tht the sequece of supremums of the bove does ot pproch zero o y set (, ) or (b, ) d so the covergece is ot uiform o (, ). Lemm 3.9. Suppose tht =0 x hs rdius of covergece R. The lso hve rdius of covergece R. x d = 20 x + =0 +

21 Proof. By pullig out fctor of x from the first series d x from the secod we see tht we eed oly show tht x = x d x x =0 + hve rdius of covergece R. However R comes from the root test, so clculte. Firstly sice /. Similrly lim sup / = lim sup /, lim sup This lst follows by pplyig the squeeze theorem to / = lim sup /. + ( ) + / ( < ( + ) / = / = + ) / ( / < + ) /. The rdii of covergece re uchged. I the ext two theorems we see tht for x < R we my term-by-term itegrte or differetite power series. Theorem If f (x) = =0 x hs rdius of covergece R, the x 0 f (y) dy = x +, for x < R. =0 + Proof. Let x < R. O the itervl [ x, x ] the power series coverges uiformly to f. Hece x 0 x f (y) dy = lim 0 k=0 = lim k y k dy = lim k=0 k=0 x k+ k k + = x + =0 +. x k y k dy 0 Theorem 3.2. If f (x) = =0 x hs rdius of covergece R > 0 the f (x) = x for x < R. = Proof. We ve see tht the series g(x) := = x hs rdius of covergece R. Let s itegrte it: x g(y) dy = x = f (x) 0. 0 = Hece f (x) = g(x) by the Fudmetl Theorem of Clculus. Exmple. The series f (x) = =0 obti f (x) = ( ) x 2 (2)! hs rdius of covergece. We my thus differetite to ( ) x 2 (2 )! = = =0 ( ) + x 2+. (2 + )! 2

22 This series is lso vlid for ll x R. We c differetite gi to see tht f (x) = ( ) + x 2 =0 (2)! = =0 ( ) x 2 (2)! = f (x). Sice f (0) = d f (0) = 0 the differetil equtio f (x) = f (x) hs solutio f (x) = cos x. Clerly si x = f (x) = =0 ( ) x 2+ (2+)!. We lredy kow tht power series with rdius of covergece R is cotiuous o the itervl ( R, R). But wht if the series coverges for x = ±R? Is the series cotiuous there? Theorem 3.22 (Abel s theorem). Power series re cotiuous o their etire itervl of covergece. Exmples.. Strtig with =0 x = x, substitute x2 for x to obti =0 ( ) x 2 = + x 2 = d rct x for x <. dx Itegrtig term-by-term, d recllig tht rct 0 = 0 yields rct x = ( ) =0 2 + x2+ for x <. I fct we hve tht the series coverges t x = ± to ± =0 2+. By Abel s theorem we coclude tht the series is cotiuous o [, ] d so its vlues t ± must coicide with those of rct. We thus recover the followig idetity for π: 2. Itegrtig =0 x = x we obti l( x) = ( ) =0 2 + = rct = π 4. =0 x + + = x = ( ) for x <. The derived series coverges t x = d is thus cotiuous there. Hece l 2 = = ( ) + = The derived series does ot coverge t x = so Abel s theorem does ot pply. Filly, for referece, we cosider the itegrtio of sequece d series of fuctios: i prticulr we c see tht = whe series coverges uiformly. Theorem If k=0 g k coverges uiformly to g o [, b] such tht ll g k re itegrble, the g is itegrble d we hve g(x) dx = g k (x) dx = g k (x) dx. k=0 k=0 22

23 Proof. (We omit the proof of the itegrbility of g). Let f (x) = k=0 g k(x). The f uiform. Thus g k (x) dx = lim f (x) dx = lim k=0 k=0 g k (x) = g k (x). k=0 g k is The cse for derivtives is less esy, d we stte it without proof. Theorem Suppose tht f f uiformly such tht ech f is differetible. Moreover ssume tht ( f ) coverges uiformly to some fuctio g. The f is differetible with derivtive g. Exmples.. The sequece f (x) = e x o [0, ) coverges uiformly to zero (sice f (x) 0). However the derivtive stisfies f (x) = e x which oly coverges to zero for x > 0, d oly uiformly so o itervls [, ) where > The sequece f (x) = cos(x) coverges uiformly to the zero fuctio (sice f (x) 0). The derivtive of the zero fuctio is, of course, the zero fuctio. However the derived sequece is f (x) = si(x), which does ot coverge (uiformly or otherwise) to ythig. 3. From the previous theorems, we lredy kow wht hppes for power series. x coverges uiformly o [ R, R ] wheever R < R d so the derived series coverges uiformly o the sme itervl to the derivtive. I prticulr, by cotiuity, the derivtive of power series my be foud ywhere i ( R, R) by term-by-term differetitio. 4 Differetitio 4. Bsic Properties of the Derivtive Defiitio 4.. Let f : (, b) R d let c (, b). The f is differetible t c if the limit f (x) f () lim x c x exists d is fiite. I such cse we defie the derivtive of f t c to be the vlue of this limit, d deote it f (c). If f is differetible t every poit x (, b) the we sy tht f is differetible o (, b) d tht the fuctio f : (, b) R is its derivtive. There re umber of equivlet ottios for the derivtive fuctio: f (x) = d f dx = d dx f (x). Similrly the vlue of the derivtive t x = c my be equivletly writte f (c) = d f = f (x). x=c x=c I geerl fuctio my ot be differetible t ll poits i its domi: thus dom f dom f. If fuctio f is defied o itervl [, b) the the right-derivtive of f t is defied s the limit lim x + f (x) f () x if it exists d is fiite. The left-derivtive of fuctio t poit is defied similrly. If dom f = [, b) the we sy tht f is differetible if it is differetible o (, b) d right-differetible t x =. dx d dx 23

24 It is cler tht if c (, b) the f is differetible t c iff f is both right- d left- differetible t c d tht the left d right derivtives re equl. For exmple cosider the fuctio f (x) = x. This is differetible o the set R \ {0} d is left- d right- differetible t x = 0. However the left derivtive t x = 0 is while the right is : these re ot equl d so f is ot differetible t x = 0. Exmples.. Let f (x) = x for N. The Hece f (x) = x. 2. Let f (x) = x. The x lim x x Hece f (x) = x 2 for x = 0. = lim x (x + x 2 2 x + ) =. x lim x x 3. Let f (x) = x defied o [0, ). The, t (0, ), x = lim x x(x ) = 2. lim x f (x) f () x = x x = x + = 2 /2. Thus f (x) = 2 x /2 o (0, ). f is ot right-differetible t x = 0 d so dom f = (0, ) is ot the etire domi of f. 4. Let f (x) = x 3/2 defied o [0, ). The, t (0, ), lim x f (x) f () x = x3/2 3/2 x = x2 + x + 2 = 32 x 3/2 + 3/2 2 = 3. 3/2 2 Thus f (x) = 3 2 x o (0, ). Ideed, strig t the clcultio we see tht f is right-differetible t x = 0 d so dom f = [0, ). 5. Let f (x) = si x. The lim x f (x) f () x si x si = lim x x si(h + ) si = lim h 0 h si h cos + cos h si si = lim h 0 h (let h = x ) h 2 si 2 = cos h 2 ( cos + 2 si 2 h 2 lim h 0 h 2 si h 2 = lim cos h 2 cos 2 si2 h 2 si h 0 h ) si si 24

25 ( ) h si 2 = cos h 2 + lim. h 0 h/2 si x I the homework we sw tht lim x 0 x = by cosiderig sector of gle x i circle of rdius. It follows tht f (x) = cos x for ll x. Theorem 4.2. If f is differetible t the f is cotiuous t. A similr sttemet c be mde regrdig left- (resp. right-) differetibility implyig left- (resp. right-) cotiuity. Proof. Observe tht f (x) = f (x) f () (x ) + f (). x Tkig limits of both sides s x gives the result (the first term pproches f () lim x (x ) = 0 sice f () is defied d fiite). Theorem 4.3. Let f, g be differetible t. The the fuctios k f (k costt), f + g, f g re differetible t. Moreover, t ll poits where the relevt fuctios re differetible we hve:. (k f ) = k f, 2. ( f + g) = f + g, 3. (Product rule) ( f g) = f g + f g. Proof.. Just clculte from the defiitio: 2. More defiitio-pushig: (k f )(x) (k f )() f (x) f () lim = lim k = k f (). x x x x ( f + g)(x) ( f + g)() f (x) f () g(x) g() lim = lim + = f () + g (). x x x x x 3. Simply dd d subtrct dummy f ()g(x) terms ito the limit s follows: lim x f (x)g(x) f ()g() x f (x) f () = lim x x g(x) g() g(x) + f () = f ()g() + f ()g (). x Theorem 4.4 (Chi Rule). If g is differetible t d f is differetible t g() the f g is differetible t d hs derivtive ( f g) () = g () f (g()). 25

26 Proof. Observe tht sice g is differetible t x = we hve g( + h) g() = hg () + hα, where lim h 0 α = 0. Similrly, f beig differetible t g() is equivlet to where lim H 0 β = 0. Now compute: f (g() + H) f (g()) = H f (g()) + Hβ, ( f g)( + h) ( f g)() = f (g( + h)) f (g()) = f (g() + hg () + hα) f (g()) = f (g() + H) f (g()) (where H = h(g () + α)) = H f (g()) + Hβ = hg () f (g()) + hα f (g()) + Hβ. Therefore ( f g)( + h) ( f g)() = g () f (g()) + f (g())α + (g () + α)β. h Clerly lim h 0 H = 0 d so lim h 0 β = 0. Lettig h 0 i the bove thus yields the result. Corollry 4.5 (Quotiet rule). If f, g re differetible t d g() = 0 the f /g is differetible t d ( f /g) = f g f g g 2. x = x 2 we my pply the Chi rule to g(x) to obti d dx Proof. Sice d dx product rule pplied to f g yields ( f /g) = f /g + f ) ( g g 2 = f g f g g 2. g(x) = g (x). But the the g(x) 2 Exmples.. Sice cos x = si(x + π/2) we hve d dx cos x = d dx si(x + π/2) = cos(x π/2) = si x. 2. Let m N. The d dx x m = d dx hve d dx x = x. ( x m ) = mx m x 2m = mx m. I prticulr, if Z we ow 3. Let f (g(x)) = x. The differetitig both sides we obti g (x) f (g(x)) = g (x) = f (g(x)). We hve thus differetited iverse fuctios: if f is the iverse fuctio of f, the d dx f (x) = f ( f (x)). 26

27 4. Applyig the previous exmple to rtiol powers of x we hve tht if N the d dx x/ = (x / ) = x. I prticulr if m Z we ow hve ( ) d d ( dx xm/ = dx x/ m x /) m = x mx m = m x m. We therefore hve d dx xq = qx q for ll q Q. 5. To fiish off the story of x, recll tht the rel umbers my be defied s the limits of ll Cuchy sequeces of rtiol umbers. Thus let r R be the limit of the sequece (q ) of rtiols d defie f (x) = x q. If f (x) = x r we the clerly hve f f uiformly o y itervl [, b] where 0 < < b <. Moreover we hve f g uiformly o the sme itervl where g(x) = rx r. It follows from Theorem 3.24 tht f is differetible d g(x) = f (x). We hve thus proved tht d dx xr = rx r o (0, ) for ll r R. 6. We c similrly pply the derivtives of iverse fuctios rgumet to fuctios like t x. By d the quotiet rule, dx t x = cos2 x+si 2 x = sec 2 x = + t 2 x. Therefore cos 2 x 4.2 The Me Vlue Theorem d dx rct x = + t 2 (rct x) = + x 2. Theorem 4.6. Suppose tht f is defied o (, b) d tht it ttis its mximum or miimum t poit x 0 (, b). If f is differetible t x 0 the f (x 0 ) = 0. Proof. Suppose f ttis its mximum t x 0. Also suppose tht f (x 0 ) = lim x x0 f (x) f (x 0 ) x x 0 > 0. By the defiitio of limit with ɛ = f (x 0 ), there exists δ > 0 such tht 0 < x x 0 < δ f (x) f (x 0 ) f (x 0 ) x x 0 < f (x 0 ) f (x) f (x 0) > 0. x x 0 I prticulr, if x (x 0, x 0 + δ), the f (x) > f (x 0 ) givig cotrdictio. Supposig tht f (x 0 ) < 0 yields cotrdictio similrly (choose ɛ = f (x 0 ) ). The rgumet t lower boud is simply the bove pplied to f. Theorem 4.7 (Rolle). Let f be cotiuous o [, b], differetible o (, b) d stisfy f () = f (b). The there exists ξ (, b) such tht f (ξ) = 0. Proof. By theorems 2.0 d 2. the fuctio f is bouded d ttis its bouds. Hece there exist x 0, x [, b] such tht f (x 0 ) f (x) f (x ) for ll x [, b]. If {x 0, x } = {, b} the f (x 0 ) = f (x ) implies tht f is costt d so f (x) = 0 for ll x (, b). If ot, the f hs mximum (x ) or miimum (x 0 ) somewhere i (, b) t which, by theorem 4.6, f vishes. 27