Notes on Calculus. Dinakar Ramakrishnan Caltech Pasadena, CA Fall 2001

Transcription

1 Notes on Clculus by Dinkr Rmkrishnn Cltech Psden, CA Fll

2 Contents 0 Logicl Bckground Sets Functions Crdinlity EquivlenceReltions Rel Complex Numbers DesiredProperties Nturl Numbers, Well Ordering, Induction Integers RtionlNumbers OrderedFields RelNumbers AbsoluteVlue ComplexNumbers Sequences Series Convergenceofsequences Cuchy scriterion ConstructionofRelNumbersrevisited Infinite series TestsforConvergence Alterntingseries Bsics of Integrtion Open, closed compct sets in R Integrls of bounded functions Integrbility of monotone functions Computtion of b x s dx Exmple of non-integrble, bounded function Propertiesofintegrls The integrl of x m revisited,polynomils Continuous functions, Integrbility LimitsContinuity Sometheoremsoncontinuousfunctions Integrbility of continuous functions Trigonometricfunctions Functionswithdiscontinuities

3 4 Continuous functions, Integrbility The most importnt bounded functions on closed intervl [, b] re continuous functions. Intuitively, continuous function is one whose grph one cn drw without tking the pen (or whtever else one uses) off the pper. Polynomils sin x cos x re the functions which immeditely spring to mind. There re wys to produce lots of others from simple ones by tking sclr multiples, sums, products, quotients composites. In prctice, even when function is not continuous, one often wnts to pproximte it by continuous one. It is nturl to sk if continuous functions re integrble over [, b] mzingly, the nswer is yes. This will be estblished in this chpter. One cn lso sk bout the integrbility of bounded functions f o n[, b] which re continuous out side set S which is smll in some suitble sense. The simplest cse is when S is finite set of points, in this cse f will be integrble. The right generl nswer is tht the set S of discontinuities of f, i.e, he set of points in [, b] wheref is not continuous must be negligible in suitble sense, i.e, hve mesure zero. This will be discussed t the end of the chpter. If this concept becomes forbidding for some reson, one should try to t lest lern tht ll is well when S is finite. 4.1 Limits Continuity We looked t limits of sequences in the Chpter 2. Now we need to go further consider limits of vlues of function. Let A be subset of R. Consider ny function f : A R At every x in A, f(x) will denote the vlue of f t x. Let be rel number such tht f is defined on n open intervl round, except possibly t itself. Let L R. We will sy tht f(x) hs limit L s x pproches, denoted symboliclly by (4.1.1) lim x f(x) =L, iff the following holds: For every open intervl I centered t L, we cn find n open intervl I centered t such tht (i) I {} A, (ii) x I {} = f(x) I. In simple words, we try to pproch from within A, from either direction, see if f(x) tends to L. In prticulr, if {x 1,x 2,...,} is ny sequence in A with limit, then the 51

4 sequence {f(x 1 ),f(x 2 ),...} converges with limit L. Sotoprove tht some f(x) doesnot hve limit s x pproches, it suffices toexhibit sequence x n converging to show tht f(x n ) does not hve limit s n goes to. Note tht to hve lim f(x) =L, it is necessry sufficient tobe ble, for every ε>0, x tofind some δ>0 such tht x <δ = f(x) L <ε. Proposition Let R, letf,g functions defined on n open intervl round, though not necessrily t itself. Suppose f(x) g(x) hve limits, sy L, L respectively, s x pproches. Then we hve, for ll α, β R, lim x (αf + βg)(x) =αl + βl, lim (fg)(x) x =LL, moreover, if g is non-zero in n open intervl round with L 0, lim x f g (x) = L L. The proof is stright-forwrd will be left s n exercise. However, the result is very importnt, we will use it often. Exmple: Let us try toevlute the following limit: 1+x 1 x L = lim x 0 x Since the numertor the denomintor both pproch 0 s x goes to 0, we cnnot directly pply the Proposition bove to express L s the rtio of the limits. How cn one resolve this 0/0 problem? The nswer is ctully simple. One multiplies the numertor the denomintor by 1+x + 1 x, gets L = lim x 0 (1 + x) (1 x) x( 1+x + 1 x) = lim x x + 1 x. Evidently, since the numertor is 1 the denomintor hs the vlue 2 t x =0,weseeby the Proposition bove tht L is 1. The set A where function f is defined is clled the domin of f. When lies in A, f hs vlue t so one cn sk for more, this leds to the notion of continuity. The function f is sid tobe continuous t point in A iff we hve lim f(x) =f() x 52

5 In other words, for f to be continuous t, we need twothings, nmely (1) tht f(x) hs limit s x pproches, (2) tht this limit is none other thn f(). f is sid tobe continuous function on A iff it is continuous t every point in A. Simple exmples of continuous functions re the constnt function f(x) = c (for some c R), the liner function f(x) = mx + c, thesine cosine functions. Another importnt exmple is the power function f(x) =x n, for ny non-negtive integer n. Let us now verify tht it is indeed continuous t ny point in R. We my ssume tht n>0sx 0 = 1 is constnt function, hence continuous. We need to check, for ny fixed n>0, tht given ny ε>0 one cn find δ>0 such tht Now δ <x <δimplies tht δ <x <δ = ε<x n n <ε. ( δ) n n <x n n < ( + δ) n n. Sowe hve torrnge tht In other words, we need We cn chieve this by tking ( + δ) n n ε ( δ) n n ε. δ ( n + ε) 1/n δ ( n ε) 1/n. δ = min{( n + ε) 1/n, ( n ε) 1/n }, which is esily seen tobe positive. Thus f(x) =x n is continuous function for ny integer n 0. If n is negtive integer, then f(x) =x n is defined continuous outside x =0. Let us next consider the squre root function f(x) = x, which is continuous t ny in its domin, i.e., for ny 0. Let us verify it now. Pick ny ε>0. We hve tofind suitble δ>0 such tht ( ) δ<x<+ δ = ε< x< + ε. When we write δ<x, we hve toremember the constrint tht x 0 for its squre-root tomke sense. When = 0, we cn chieve ( ) bytkingδ = ε 2. Sossume >0, choose δ tobe smller thn. Then δ<x<+ δ = δ< x< + δ, 53

6 sowe need + δ + ε δ ε. This is stisfied by tking δ = min{( + ε) 2, ( ε) 2 }. Done. More generlly, by similr rgument one cn prove the following Proposition For ny t R, thepowerfunctionf(x) =x t is continuous in its domin. To cook up more continuous functions from the known ones we need nother structurl result: Proposition Let f,g be continuous functions on subset A of R. Then ny constnt multiple cf of f, given by x cf(x), thesumfunctionf + g, given by x f(x) +g(x), the product function fg, given by x f(x)g(x), re ll continuous on A. Furthermore, if g is non-zero on A, then the rtio f/g, given by x f(x)/g(x), is lso continuous. In prticulr, ny polynomil function or function of the form f(x) = x + 2 x n x n f(x) = 1 x x 3 x +2 is continuous. So is rtionl function f(x)/g(x), with f,g polynomils g 0on A. Onekno wsthtco sx is zero only t the odd integrl multiples of π/2. Hence the tngent function, defined by tnx = sinx is continuous on ny subset A of R voiding cosx {(2k +1)π/2 k Z}. Suppose we hve two functions g : B R f : A R. IfA contins the imge g(b) of g, then we cn define the composite function (4.1.5) h = f g : B R by h(x) =f(g(x)), x B. Proposition If f,g re continuous, then so is their composite h = f g, defined s bove. This llows us to construct lot of continuous functions, for exmple, rtionl functions of trigonometric functions (when defined). 54

7 4.2 Some theorems on continuous functions Intermedite Vlue Theorem Let f be continuous function on closed intervl [, b]. Then for every t between f() f(b) we cn find c in [, b] such tht f(c) =t. Before proving this result, let us use it to show tht there is some rel number x such tht sin x = x 1. Let us put f(x) =x 1 sin x. We hve tofind x such tht f(x) = 0. Note tht f(0) = 0 1 0= 1 < 0, while f(π) =π 1 sin π = π 1 > 0, becuse sin π =0π>1. Since f(x) is continuous on [0,π], in fct on ll of R, wemy pply the intermedite vlue theorem to conclude the existence of some c in [0,π] such tht f(c) =0. Done. Proof. There is nothing to prove if t is f() orf(b). Sowe myssume tht f() f(b). Suppose f() <t<f(b). Put g(x) =f(x) t. Then g() =f() t is negtive g(b) =f(b) t is positive. Put (4.2.1) X = {u [, b] g(x) < 0 if x u}. Since lies in X, this set is non-empty, moreover, b is n upper bound for X s g(b) > 0. Put c = sup(x), which exists by the properties of R. We clim tht g(c) =0. Ifg(c) > 0, then g(x) will be positive in n open intervl centered t c, sothtg(c δ) > 0forsomeδ>0. Since c is the lest upper bound, ny number in (, c), in prticulr c δ will be in X. This gives contrdiction, s g(c δ) 2 is positive, violting the definition of X. Hence g(c) is cnnot be positive. Suppose g(c) is negtive. Then we cn find δ>0such tht g(x) remins negtive in (c δ, c + δ), contrdicting the fct tht c is n upper bound. So the only possibility is for g(c) tobe0. Then f(c) = g(c)+ t will be t, s desired. We hve tostill nlyze the cse when f() >t>f(b). Here we pply wht we hve just proved to f toconclude wht we wnt. Here is nice consequence of this result. Corollry Every polynomil f(x) with rel coefficients of odd degree hs root in R. Proof. Let n be the (odd) degree of f. Write f(x) = x + 2 x n x n, 55

8 with n 0. Put g(x) = 1 n f(x). The coefficients b j = j n re still rel soit suffices toprove tht g(x) hs rel root. Intuitively, the rgument is cler s the x n term should dominte ll the others when x is lrge, so g(x) should be positive for lrge positive x negtive for lrge negtive x. This is wht we will now prove rigorously. Write ( g(x) =x n 1+ b n 1 x b 1 x + b ) 0. n 1 x n By pplying the tringle inequlity repetedly, we get b n 1 x b 0 x n b n 1 x b 0 x n. Now choose x sotht (4.2.3) x mx{1, b n j 1 j n}. Then b n j is bounded bove by 1, x j b n 1 x b 0 x n 1 2. Sowe get with g(x) =x n (1 + h(x)) 1 2 h(x) 1 2 if (4.2.3) holds. Now if x is positive (resp.negtive) stisfies (4.2.3), then g(x) will be positive (resp. negtive), we re done by the intermedite vlue theorem. Theorem Let f : A R be continuous function. Then f tkes ny closed intervl [, b] contined in A tocompctsetinr. Inprticulr,f([, b]) is bounded. Moreover, it ttins its extreml vlues on [, b], i.e., there exist c, d in [, b] such tht (4.2.5) f(x) f(c) x [, b] f(x) f(d) x [, b]. f(c) is clled the minimum of f on[, b], f(d) themximum. Thefcttht continuous function ttins its extrem on ny closed intervl is very importnt oft-used fct. One cn generlize further prove tht the imge of ny compct set C under continuous function f is compct. 56

9 When we combine Theorem with the intermedite vlue theorem, we get the following useful Corollry Let f be continuous, monotone incresing function on [, b]. Put (4.2.7) M =sup{f(t) t [, b]} m =inf{f(t) t [, b]}. Then for every t in [m, M] there exists c [, b] such tht f(c) =t. Proof of Theorem Put B = {x [.b] f([, x]) bounded}. It is bounded, s it is subset of [, b], non-empty becuse is in B. Soby the properties of R, there is lest upper bound t of B. Suppose t<b. Then the continuity of f t t implies in prticulr tht for every ε>0, there is some δ>0 such tht x<t+ δ = f(x) (f(t) ε, f(t)+ε). This implies tht f is bounded in [, t + ε/2], contrdicting the fct tht t is the lest upper bound of B. Hence t cnnot be less thn b, we must hve t = b. So f is bounded on [, b]. Suppose z is boundry point of f([, b]). Then we cn find sequence z 1,z 2,... ofpoints lying in f([, b]) converging to z. For ech n 1 write z n = f(y n )withy n [, b]. The sequence {y n } is bounded (s it lies in [, b]) hence contins subsequence {y nm }which converges, sy to y R. Such y must lie in the closure of [, b], i.e., in [, b] itself. Then f(y) will, by the continuity of f, be the limit of {z nm }, which is subsequence of {z n }.But the sequence {z n } is convergent, so must hve the sme limit s the subsequence {z nm }. This forces z tobe f(y), thus z lies in f([, b]). Thus f([, b]) is closed bounded, i.e., compct. Now define M,m s in (4.2.7), which mkes sense becuse f([, b]) is bounded. But the supremum M must be boundry point, for otherwise we cn choose smller upper bound, contrdicting the fct tht M is the lest upper bound. Since f([, b]) is closed, M must belong to f([, b]). Similrly, the infemum m lsobelongs tof([, b]). Hence there re points c, d in [, b] such tht m = f(c) M = f(d). Now (4.2.5) is evident. 4.3 Integrbility of continuous functions The min result of this section is the following Theorem Every continuous function f on closed intervl [, b] is integrble. The proof rests on bsic property of continuous functions on closed intervls, which we will now expose. 57

10 Tobegin, let us define the spn of f on ny closed intervl [c, d] tobe spn f ([c, d]) = mx f([c, d]) min f([c, d])). The Smll Spn Theorem For every ε>0, there exists prtition P : = t 0 <t 1 <...<t n = b of [, b] such tht spn f ([t j 1,t j ]) <ε, for ech j =1, 2,...,n. Proof. We will prove this by contrdiction. Suppose the theorem is flse. Then there exists some ε>0, cll it ε 0 toindicte tht it is fixed, such tht for every prtition P : = t 0 <t 1 <... < t n = b, spn f ([t j 1,t j ]) ε 0 for some j. Subdivide [, b] intotwo closed intervls [, c], [c, b] withc being the midpoint ( + b)/2. Then the theorem must be flse for one of these subintervls, cll it J 1,forthesme ε 0. Dothis gin gin, we finlly end up with n infinite sequence of nested closed intervls [, b] =R 0 R 1 R 2..., such tht, for every m 0, the spn of f is t lest ε 0 for ny prtition P m = {J j,m } of R m onsomej j,m. Let x m denote the left endpoint of R m, for ech m 0. Then the sequence {x m } m 0 is bounded, so we my find the lest upper bound (sup) γ, sy,of x m. Then γ will be boundry point of [, b], hence must lie in it, s [, b] isclosed. Since f is continuous t γ, we cn find closed subintervl I of[, b] contining γ such tht spn f (S) <ε 0. But by construction R m will hve tolie inside I if m is lrge enough, sy for m M. This gives contrdiction to the spn of f being ε 0 onsomeopensetofevery prtition of R M. Thus the smll spn theorem holds for (f,[, b]). Proof of Theorem Recll tht it suffices toshow tht, given ny ε>0, there is prtition P of[, b] such tht U(f,P) L(f,P) <ε. Put ε 1 = ε/(b ). Applying the smll spn theorem, we cn find prtition P = {J j } such tht spn f (J j ) <ε 1,forllj. Then clerly, U(f,P) L(f,P) <ε 1 (b ) =ε. Done. 4.4 Trigonometric functions We will ssume, s we hve ll long, tht the students tking this clss re fmilir with bsic trigonometric functions, like the sin x, cosx, thentnx, which is defined tobe sin x cos x whenever cos x is non-zero, s well s the functions csc x = 1 sin x, sec x = 1 1, cot x = cos x tn x = cos x sin x. We will ccept the following from Bsic Trigonometry without proof: 58

11 Theorem () sin x cos x re periodic functions with period 2π, i.e., sin(x ± 2π) =sinx cos(x ± 2π) =cosx, with sin 2 x +cos 2 x =1. (b) sin x is n odd function, while cos x is n even function, i.e., sin( x) = sin x cos( x) = co s x. (c) sin x cos x re both non-negtive, monotone functions in [0,π/2], withsin x incresing (from 0 to 1) cos x decresing (from 1 to 0) sx moves from 0 to π/2. (d) (Addition theorems) sin(x ± y) =sinx cos y ± cos x sin y cos(x ± y) =cosx cos y ±sin x sin y. (e) For every x in the open intervl (0,π/4), cos x< sin x x < 1. As consequence of (e), we cn compute some importnt limits. For exmple, we obtin Proposition (i) sin x lim x 0 x =1, (ii) 1 cos x lim x 0 x =0. sin x Proof. For ll smll x, is cught, by prt (e) of Theorem 4.4.1, between the x constnt vlue 1 cos x. Butcosxbecomes 1 s x pproches 0. This gives (i). Toget (ii), multiply the numertor (1 cos x) the denomintor x by 1 + cos x note tht (1 cos x)(1 + cos x) =1 cos 2 x =sin 2 x. So we get, by pplying Proposition 4.1.2, ( )( ) 1 cos x sin x sin x lim = lim lim =1 0=0. x 0 x x 0 x x 0 1+cosx 59

12 Here we hve used (i) s well s the fcts tht sin 0 = cos 0 = 2. The following result is bsic. Theorem Let [, b] be ny closed intervl. Then the functions sin x cos x re integrble on [, b]. Explicitly, b sin xdx =cos cos b b cos xdx =sinb sin. Proof. By periodicity (prt () of Theorem 4.4.1) dditivity of the integrl (Proposition 3.x), we my ssume tht 0 <b 2π. Moreover, for 0 x π, the oddness ofsinximplies (4.4.4) sin(π + x) =sin( π + x) = sin(π x), while the evenness ofcosx implies cos(π + x) =cos( π + x) =cos(π x). Moreover, the trigonometric definition of sin x cos x gives immeditely the following identities: (4.4.5) sin(π x) =sinx cos(π x) = cos x. Thnks to (4.4.4) (4.4.5) it suffices to prove the ssertion of the Theorem when 0 <b π/2. Furthermore, since b f(x)dx = b f(x)dx f(x)dx 0 0 for ny function f(x), it suffices toprove tht for ny in (0,π/2], the functions sin x cos x re integrble on [0,]tht (4.4.6) sin xdx =1 cos cos xdx =sin

13 We will prove the formul for the integrl of sin x over [0,] leve the proof of the corresponding one for cos x tothe reder. For every n 1, define prtition P n of[0,]tobegivenby (4.4.7) 0 = t 0 <t 1 = n <t 2 = 2 n <...<t n =, sotht t j+1 t j = for ll j n 1. n Since sin x is monotone incresing function in [0,π/2], the upper lower sums re given by (4.4.8) U(sin x, P n )= n n 1 j=0 L(sin x, P n )= n sin( n 1 j=0 (j +1) ) n sin( j n ). By the ddition theorem for cos x (prt (d) of Theorem 4.4.1), we hve (4.4.9) 2 sin x sin y =cos(x y) cos(x + y). In prticulr, when y = (j+1) x =,weget n (4.4.10) 2 sin( +1) )sin((j )=cos( n n Applying this in conjunction with (4.4.8), we get U(sin x, P n )= / ( sin( ) (cos( 3 ) cos( which simplifies to (4.4.11) U(sin x, P n )= By Proposition 4.4.2, (2j +1) ) cos( (2j +3) ). 1) )) +...+(cos(( ) cos( ( sin( ) cos( ) +1) ) cos(( ). ) ( +1) )), (4.4.12) lim sin( ) n =1. Also, since cos 0 = 1, (4.4.13) lim cos( +1) ) cos(( )=1 cos. n 61

14 Combining (4.4.11) through (4.4.13), we get (4.4.14) lim n U(sin x, P n )=1 cos. The nlog of (4.4.11) for the n-th lower sum is (4.4.15) L(sin x, P n )= We get the sme limit, nmely sin( ) ( 1) cos( ) ( 1) ). (4.4.16) lim L(sin x, P n )=1 cos. n In view of (4.4.14), (4.4.15) Lemm 3.2.8, the desired identity (4.4.6) follows for the sine integrl. The rgument is entirely nlogous for the cosine integrl. 4.5 Functions with discontinuities One is very often interested in being ble to integrte bounded functions over [, b] which re continuous except on subset which is very smll, for exmple outside finite set. To be precise, we sy tht subset Y of R is negligible, orthtithsmesure zero, ifffor every ε>0, we cn find countble number of closed intervls J 1,J 2,... such tht (4.5.1) (i) Y i=1j i, (ii) i=1 l(j i) <ε. If we cn dothis with finite number of closed intervls {J i } (for ech ε), then we will sy tht Y hs content zero. Exmples: (4.5.2) (1) Any finite set of points in R hs content zero. (Proof is obvious!) (2) Any subset Y of R which contins non-empty open intervl (, b) isnot negligible. Proof. It suffices toprove tht (, b) hs non-zero mesure for <bin R. Suppose (, b) is covered by finite union of countble collection of closed intervls J 1,J 2,... in R. Then clerly, m S := length(j i ) l(, b) =b. i=1 Sowe cn never mke S less thn b. Theorem Let f be bounded function on [, b] which is continuous except on subset Y of mesure zero. Then f is integrble on [, b]. Proof. Let M > 0 be such tht f(x) M, forllx [, b]. Since Y hs content zero, we cn find closed subintervls J 1,J 2,... of[, b] such tht 62

15 (i) Y m i=1j i, (ii) i=1 l(j i) < ε. 4M Extend {J 1,...,J m } to prtition P = {J 1,...,J r }, m<r,of[, b]. Applying the smll spn theorem, we my suppose tht J m+1...,j r re so chosen tht (for ech i m +1) spn f (J i ) < ε. (We cn pply this theorem becuse f is continuous outside the union 2l([,b]) of J 1,...,J m.) Sowe hve U(f,P) L(f,P) 2M m r l(j i )+ spn f (J i )l(j i ), i=1 i=m+1 which is becuse r i=m+1 l(j i) b. ( ) ε ε < (2M) + 2M 2l([, b]) r i=m+1 l(j i ) ε 2 + ε 2 = ε, Remrk We cn use this theorem to define the integrl of continuous function f on ny compct set B in R if the boundry of B is negligible. Indeed, in such cse, we my enclose B in closed intervl [, b] define function f on[, b] bymking it equl f on B 0on[, b] B. Then f will be continuous on ll of [, b] except on the boundry of B, which hs content zero. So f is integrble on [, b]. Since f is 0 outside B, it is resonble to set (4.5.5) B f = b f. 63