Summary of Elementary Calculus


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1 Summry of Elementry Clculus Notes by Wlter Noll (1971) 1 The rel numbers The set of rel numbers is denoted by R. The set R is often visulized geometriclly s numberline nd its elements re often referred to s points. The following specil nottion nd terminology will be used for certin specil subsets of R: The set of nonzero rels is denoted by R := R\{0} = {t R t 0}. The set of positive rels is denoted by P := {t R t 0}. The set of strictly positive rels is denoted by P := P\{0} = {t R t > 0}. Let, b R with b. Then the open intervl ], b, [, the closed intervl [, b], the leftopen intervl ], b ], nd the rightopen intervl [, b [ from to b re defined by ], b [ := {t R < t < b} [, b] := {t R t b} ], b ] := {t R < t b} [, b [ := {t R t < b} The numbers nd b re clled the endpoints of these intervls. If S is ny subset of R we write S := { s s S}. In prticulr, P = {t R t 0} is the set of ll negtive rels nd P = {t R t < 0} the set of ll strictly negtive rels. If S nd T re ny subsets of R, we write 1
2 S + T := {s + t s S, t T }, S T := S + ( T ) = {s t s S, t T }. If R nd T R, we bbrevite + T := {} + T = { + t t T }, T := + ( T ) = { t t T }. The set + P is clled the closed right hlfline nd + P the open right hlfline from R. Similrly, P is clled the closed left hlfline nd P the open left hlfline from. In either cse, we sy tht is the endpoint of the hlfline. Sometimes the nottions [, + [ := + P, ], + [ := + P, ], ] := P, ], [ := P, ], + [ := R re used. The term intervl is pplied not only to intervls from some R to some b R, with b >, s defined before, but lso to hlflines, singleton subsets of R, the whole of R, nd the empty set. One cn prove, then, tht subset I of R is n intervl if nd only if it hs the following property: For ll, b I with < b, [, b] I. The empty set, singletons, nd intervls with endpoints in R re clled bounded intervls. Hlflines nd the whole of R re clled unbounded intervls. We sy tht n intervl is genuine if it is neither empty nor singleton. There re two bsic opertions involving rel numbers: ddition nd multipliction. If, b R, we cn form the sum +b by ddition nd the product b by multipliction. Two specil rel numbers re of centrl importnce: zero 0 nd one 1. Given ny, b R, the eqution?x R, + x = b hs exctly one solution x R, which is denoted by b nd clled the difference between b nd or b minus. The bbrevition := 0 is 2
3 used nd clled minus (insted of zero minus ). Given ny R nd b R the eqution?x R, x = b hs unique solution, which is denoted by b nd clled the quotient of b nd, or b over. One clls 1 the reciprocl of R. If R nd n N, one defines the nth power n in the following wy: 0 := 1, 1 :=, 2 :=, 3 := 2, etc. If R nd m N, one defines m by ( ) m 1 m :=. The set R is totlly ordered (red: less thn ). The corresponding strict order on R is < (red: strictly less thn ). The reverse orders re denoted by ( greter thn ) nd > ( strictly greter thn ). Any nonempty finite subset F or R hs gretest element which is clled the mximum of F nd is denoted by mxf ; it lso hs smllest element, which is clled the minimum of F nd is denoted by minf. The bsolute vlue of P of number R is defined by := mx {, }. 2 Rel Functions Let S be subset of R. A mpping f : S R is clled relvlued function of rel vrible, or simply rel function. The set of ll rel functions on S is denoted by Fun(S) := Mp(S, R) Rel functions cn often be visulized concretely by their grphs. For exmple, the Figure shows the grph of some function f Fun([, b]). 3
4 We cn define two bsic opertions involving rel functions: vluewise ddition nd vluewise multipliction. If f, g Fun(S) we define the vluewise sum f + g Fun(S) by (f + g)(t) := f(t) + g(t) for ll t S, nd we define the vluewise product fg Fun(S) by (fg)(t) := f(t)g(t) for ll t S. We cn lso define vluewise differences of functions by vluewise subtrction: if f, g Fun(S) then f g Fun(S) is defined by (f g)(t) := f(t) g(t) for ll t S. Note tht if f, g Fun(S), then f g is the unique solution of the eqution?x Fun(S), g + x = f. The definition of quotients of functions by vluewise division hs to be hndled with cre. If f, g Fun(S), nd if g(t) 0 for ll t S then f(t) g(t) mkes sense for ll t S nd we cn define f Fun(S) by g ( f g ) (t) := f(t) g(t) for ll t S. In most situtions, however, we hve g(t) = 0 for some vlues of t S but g(t) 0 for other vlues of t S. If this is the cse, we cn still define quotient function f, but this quotient function will hve restricted domin. g To be precise, the set of ll t R for which g(t) 0 is S\g < ({0}) nd we define f g Fun(S\g< ({0})) = Fun(g < (R )) by ( ) f (t) := f(t) g g(t) for ll t S\g< ({0}) = g < (R ). Of prticulr interest is the multipliction of functions by constnts. If f Fun(S) nd R, then f is defined to be the vluewise product of the constnt s R Fun(S) nd the function f. Thus, (f)(t) := (f(t)) for ll t S. We write f for ( 1)f. Then f is the sme s the difference 0 S R f. 4
5 For ny n N, the vluewise nth power f n Fun(S) of f Fun(S) is defined by f n (t) := (f(t)) n for ll t S. For m N the domin of the vluewise mth power f m must be restricted to S\f < ({0}), nd we hve f m := ( ) m 1. f Let f, g Fun(S) nd let h : T S, where S nd T re subsets of R. Then we hve (f + g) h = (f h) + (g h), (fg) h = (f h)(g h), If R, we hve f h (S\g< ({0})) g T \(g h) < ({0}) = f h g h. S R h = T R or simply h = when we do not distinguish, in nottion, between constnts nd their vlues. 3 Elementry Functions Since the identity mpping 1 R of R will be used very often, we use the shorter nottion ι := 1 R. Thus, ι := R R is given by ι(t) := t for ll t R. Using the opertions explined in the previous section, we cn form lrge number of functions from ι nd constnts lone. For exmple, if n N nd 0, 1,..., n R, we cn form wht is clled polynomil function (or simply polynomil) p Fun(R) by p := n ι n + n 1 ι n ι ι
6 The quotient p of two polynomil functions p nd q is clled rtionl q function. The domin of p is q R\q< ({0}), which, in generl, is not the whole of R. Of course, ι is invertible nd it is its own inverse. Most rel functions encountered in elementry mthemtics re not invertible. However, one cn often construct invertible functions from noninvertible ones by suitble djustment of their codomins or domins. For exmple, the squring function ι 2 : R R is not invertible, but the djusted squring function ι 2 P P is. Its inverse, fter extension of the codomin to R, is clled the squre root function nd is denoted by Fun(P), so tht ( ) ι 2 P R P. The vlue of t t P is denoted by t (rther thn the stndrd (t)). We hve ι 2 P =, ι 2 = ι, P where : R R is the bsolutevlue function. If n N is odd, then ι n Fun(R) is invertible. Its inverse is clled the n t h root function nd is denoted by n Fun(R), so tht n := (ι n ). We hve n ι n = ι n n for odd n N. If n N is even, ι n is not invertible, but the djusted function ι n P P is. Its inverse, with codomin extended to R, is clled the nth root function nd is denoted by n Fun(P), so tht n := (ι n ) R. We hve n ι n P =, ι n n = ι P for even n N. The vlue of n t t is written n t (rther thn the stndrd n (t)). Recll tht the composite f g of two functions f nd g mkes sense only if Dom f = Cod g. However, it is often possible to djust the codomin or both domin nd codomin of g in such wy tht f g mkes sense for n pproprite djustment g of g. For exmple, (ι 2 + 1) mkes no sense but ((ι 2 + 1) P ) does. We use then the nottion ι2 + 1 := ((ι 2 + 1) P ). As nother exmple, we see tht ι mkes no sense but ι since ι P P = 1 P, we hve ι P =. P P P does nd 6
7 More generlly, if p is ny polynomil nd if n N we write n p := n p if n is odd, n p := n P p if n is even. p < (P) If n is even, it cn be difficult to determine the domin p < (P) = {t R p(t) 0} of n p. The bsic elementry functions re (i) the constnts, (ii) the identity function ι : R R, (iii) the exponentil function exp : R R, nd (iv) the sine function, sin : R R. Precise definitions of exp nd sin re difficult nd will not be given here. A rel function is clled n elementry function if it cn be obtined from the bsic elementry functions by repeted ppliction of the following opertions: vluewise ddition, subtrction, multipliction, nd division, nd composition nd inversion. When pplying division, composition, nd inversion, one must often suitbly djust the domin or the codomin or both, s we hve seen before. Using poetic license, we often omit the symbols tht indicte these djustments. Also, we do not distinguish, in nottion, between constnt nd its vlue. Exmples of elementry functions re the polynomil nd rtionl functions nd lso the nth root functions mentioned before. The following dditionl elementry functions hve stndrd nmes ttched to them: ( π ) cos := sin 2 ι, tn := sin cos ( ) rcsin := sin [ 1,+1] R [, π s,+ π 2 ] ( ) R rccos := cos, [ 1,+1] [0,π] ( ) R rctn := tn ] π, 2,+ π 2 [ log := (exp P ) sinh := 1 (exp (exp ( ι))), 2 7
8 cosh := 1 (exp +(exp ( ι))), 2 tnh := sinh cosh. For every α R, the α t h power function ι α : P R, with vlues ι α (t) = t α, is defined by ι α := exp (α log). In the cse when α is n integer, it coincides with restriction to P of the vluewise power function of ι ccording to the definition t the end of Sect. 2. If α is rtionl, so tht α = n m with n Z nd m N, we hve ι n m = m ι n P = m P ιn P P. 4 Limits nd continuity Let f be rel function nd let R be number, which my or my not belong to the domin of f. We ssume, however, tht Dom f includes genuine intervl which hs s n endpoint. Then ] δ, +δ[ Dom f\{} is nonempty for ll δ P. Definition: We sy tht f hs the limit λ R t R nd we write lim f = λ or lim t f(t) = λ if for every ε P there exists δ P such tht i.e., f(t) λ < ε for ll t ] δ, + δ[ (Dom f\{}). f(t) λ < ε whenever 0 < t < δ nd t Dom f. This definition gives precise mening to the intuitive ide tht f(t) pproches λ s t pproches. Roughly speking, the definition sys tht f(t) comes rbitrrily close to λ when t moves sufficiently close to. 8
9 The desired closeness of f(t) to λ is described by ε nd the required closeness of t to is described by δ. One cn prove tht f cnnot hve more thn one limit t. It is for this reson tht the nottion lim f = limf(t) t is legitimte. Exmples: 1. The nturl domin of the function ι2 1 is R\{1}, becuse the denomintor vnishes t 1. The domin R\{1} includes, for exmple, the ι 1 intervls 1 + P nd 1 P, which hve 1 s n endpoint. Hence it mkes sense to sk whether ι2 1 hs limit t 1, even though 1 does ι 1 not belong to the domin. Since ( ) ι 2 1 (t) = t2 1 ι 1 t 1 = (t + 1)(t 1) t 1 = t + 1 for ll t R\{1} (but not for t = 1, it is cler tht for every ε P we hve ( ) ι 2 1 (t) 2 ι 1 = t = t 1 < ε whenever t 1 < δ := ε nd t 1. We conclude tht ι 2 1 lim 1 ι 1 = lim t 2 1 t 1 t 1 = The nturl domin of the function 1 is ι R, becuse the denomintor vnishes t 0. As in 1, it mkes sense to sk whether 1 hs limit t ι 0. If 1 hd limit λ t 0, we could find, for every choice of ε ι P, δ P such tht 1 t λ < ε whenever t 0 = t < δ nd t 0. If we choose ε := 1, we then hve 1 t λ < 1 whenever t < δ nd t 0. 9
10 This is impossible when t is smller thn both δ nd 1 λ +27. Hence 1 ι hs no limit t 0. (Note tht 27 is whimsicl choice; ny other rel number greter then 1 would lso do.) Let f : I R be rel function whose domin I is genuine intervl, nd let t I. It is mengful to sk whether or not f hs limit t t. We sy tht f is continuous t t if the following two conditions re stisfied: (i) f hs limit t t. (ii) The limit of f t t is equl to the vlue of f t t. The two conditions together mount to the sttement lim t f = f(t). A function f Fun(I) is sid to be continuous if it is continuous t every t I. If the domin of f is genuine intervl, the grph of f is often connected curve which cn be drwn with pencil without lifting it. If this is the cse, then f is continuous. However, there re weird continuous functions whose grph cnnot esily be represented by pencil drwing. Let f be rel function whose domin is the union of collection of genuine intervls. (For exmple, tn hs this property.) We then sy tht f is continuous if the restriction f I is continuous for every intervl I included in the domin. The nottion is often used. C(I) := {f Fun(I) f is continuous} All elementry functions re continuous, but most continuous functions re not elementry. 10
11 5 Differentibility, derivtives Let f : I R be rel function whose domin I is genuine intervl nd let t I be number. Consider the new function (f (t + ι)) f(t) : ( t + I)\{0} R. ι Here, we hve omitted the symbols tht indicte necessry djustments of domin or codomin. Also, we hve denoted constnts by their vlues. Hd we not done so, this function would hve been written f ( t ( t+i) R + ι ( t+i) ) I f(t)( t+i) R ι ( t+i) Clerly, it mkes sense to sk whether this new function hs limit t 0. Definition: We sy tht f : I R is differentible t t I if the limit f (t + ι) f(t) t f := lim 0 ι exists. If this is the cse, then t f is clled the derivtive of f t t. We sy tht f is differentible if it is differentible t every t I. If this is the cse, the derivtive f : I R of f is defined by Remrk: df, df(t) d, dt dt dy dx f (t) := t f for ll t I. There re vrious other populr nottions for derivtives: f(t), f dt t, or f t for t f nd f, f or Df for f. The nottion when y = f(x), which is very populr, doesn t relly mke ny sense. If f is differentible t t I then f is lso continuous t t. The converse if flse: A function cn be continuous t point without being differentible t tht point. For exmple, the bsolutevlue function (t t ) : R R is continuous but not differentible t 0. One cn even construct function f :]0, 1[ R which is continuous but not differentible t every t ]0, 1[. (The construction is very difficult.) Let f be rel function whose domin is the union of collection of genuine intervls. We then sy tht f is differentible if f I is differentible 11
12 for every intervl I included in the domin. If this is the cse, the derivtive f : Dom f R cn be defined in such wy tht (f I ) = f I for ll intervls I included in the domin. Given S R, we sy tht f : I S is differentible if f R is differentible nd we write f := (f R ). Not ll elementry functions re differentible. For exmple, 3 : R R nd : P R re not differentible t 0. However, these nd ll other elementry functions become differentible fter few exceptionl points re deleted from their domins. Let I be genuine intervl or union of collection of such. We write Diff (I) := {f Fun(I) f is differentible}. We now list, without proof, the bsic fcts of differentil clculus. Here, I nd J denote genuine intervls or unions of collections of such. Theorem 5.1. (linerity rule) Given f, g Diff(I) nd c R, we hve f + g Diff(I), cf Diff(I) nd (f + g) = f + g, (cf) = c(f ). Theorem 5.2. (product rule) IGiven f, g Diff(I) we hve fg Diff(I) nd (fg) = f(g ) + (f )g. Theorem 5.3. (chin rule) Given f Diff(I) with Rngf J nd g Diff(J), we hve g f J Diff(I) nd (g f J) = ( g f J) f. Theorem 5.4. (inversion rule) Assume tht f : I J is invertible nd differentible, nd tht its derivtive f : I R is continuous. Let Then f J I := I\f < ({0} = {t I f (t) 0}, J := f > (I ). is differentible nd (f J ) = 1 f ( f I J ). 12
13 Note tht there re two sttements in ech of these theorems. The first sserts tht certin function is differentible, the second gives formul tht tells how to obtin the derivtive. It is evident, from the definition, tht ll constnts re differentible nd hve the sme derivtive, nmely the constnt zero. The following converse if much hrder to prove: Theorem 5.5. Let genuine intervl I nd f Diff(I) be given. If f = 0, then f is constnt. We sy tht function F Diff(I) is n ntiderivtive of given function f Fun(I) if F = f. If I is genuine intervl nd f hs such n ntiderivtive F, then {F + c c is constnt} is the set of ll ntiderivtives of f. This is consequence of Theorems 1 nd 5. 6 Derivtives nd ntiderivtives of elementry functions It follows immeditely from the definition tht the identity function ι is differentible nd its derivtive is the constnt 1: (ι ) = 1. (1) One cn prove, on the bsis of the definition, tht 1 : ι R R is differentible nd tht its derivtive is given by ( ) 1 = 1 ι ι. (2) 2 The bsic elementry functions sin nd exp re differentible nd their derivtives re given by ( π ) sin = cos := sin 2 ι (3) exp = exp. (4) 13
14 Using only the formuls (1)  (4), one cn determine, in routine mnner, the derivtives of ll elementry functions with the help of Theorems 14 of the previous section, without ever going bck to the definition of derivtive. The following fcts re helpful: If the given function f : I R is differentible, then 1 : I\f < ({0}) R cn be represented in the form f 1 = 1 f R. Hence, by the chin rule (Theorem 3), 1 f ι I\f < f ({0}) nd ( ) 1 = f f f. 2 is differentible Let f, g Diff(I) be given. If we pply the product rule (Theorem 2) to the cse when g is replced by 1 nd I by g I\g< ({0}) we obtin the quotient rule: f : g I\g< ({0}) R is differentible nd ( f g ) = f g fg g 2. Exmple: We wish to determine rctn. Since cos := sin ( π ι) we cn 2 use the chin rule to obtin ( ( π )) ( π ) ( π ) cos = sin 2 ι 2 ι = cos 2 ι. Since ( π ι) ( π ι) = ι, we hve 2 2 ( π ) ( ( π )) ( π ) cos 2 ι = sin 2 ι 2 ι = sin nd hence The quotient rule gives cos = sin. tn = ( ) sin = sin cos sin cos = cos2 + sin 2 = 1 + tn 2. cos cos 2 cos 2 Using the inversion rule, we finlly get 14
15 ] rctn 1 π 2 = 1 + (tn ] π 2,+ π 2 [ ) rctn,+ π 2 [ = ι. 2 For the record, we list the derivtives of the elementry functions mentioned t the end of Sect. 3: cos = sin, tn = 1 + tn 2 = 1 cos 2, rcsin = 1 1 ι 2, rccos = 1 1 ι 2, rctn = 1 1+ι 2, log = 1 ι, sin h = cos h, cos h = sin h, tn h = 1 tn h 2 = 1 cos h 2 (ι α ) = αι α 1 for ll α R. The derivtives of elementry functions re gin elementry functions. But not every elementry function hs ntiderivtives tht re elementry functions. Even if it hs, it is, in generl, very much hrder to find n ntiderivtive thn it is to compute the derivtive. The following tricks work sometimes: (i) Given function f, find functions h nd g nd number α R\{1} such tht f = hg nd such tht n ntiderivtive K of k := h g + αf cn be determined. Then F := 1 (hg K) 1 α 15
16 is n ntiderivtive of f, s is esily proved using the product rule. Exmple: If f := cos 2, we my tke h := cos, g := sin, nd α := 1. Then k = cos sin cos 2 = sin 2 cos 2 = 1, which hs the ntiderivtive K := ι. Hence is n ntiderivtive of cos 2. F := 1 (cos sin +ι) 2 (ii) Given function f, find functions h nd g such tht f = (h g)g nd such tht n ntiderivtive H of h cn be determined. Then F := H g is n ntiderivtive of f, s is evident from the chin rule. Exmple: If f := 1, we my tke g := exp, so tht 1+exp f = 1 ( ) ( ) 1 exp ι exp = 1 + ι exp exp = (1 + ι)ι exp exp Thus we hve f = (h g)g with h := 1 = 1 1, which hs the ntiderivtive H := log log (ι + 1). Hence F := H g = ι log (exp +1) is (1+ι)ι ι ι+1 n ntiderivtive of f. 7 Integrls Let S be subset of R. We sy tht f Fun(S) is positive nd write f 0 if f(t) 0 for ll t S. Let I be n intervl nd let, b I, with b. Given positive continuous function f Fun(I), we hve n intuitive notion of the re 16
17 mesure of the region between the grph of f, the horizontl xis R {0}, nd the verticl lines {} R nd {b} R. (The region is shded in Figure 1.) The integrl f should be defined in such wy tht it gives this remesure. In prticulr, f should be zero. If f Fun(I) is continuous but not necessrily positive, we consider the functions f + nd f defined by f + (t) := mx {f(t), 0}, f (t) := mx { f(t), 0} for ll t I. It is obvious tht f + nd f re positive nd tht f = f + f. Also, it is esy to prove tht f + nd f re continuous. The grphs of f + nd f re indicted in Figure 2 by the dotted nd the dshed curves, respectively). The integrl of f from I to b I, with b, is defined in such wy tht f := f + where the two terms on the right cn be interpreted s remesures. Note tht b is defined by f, f = 0 for ll I. If, b I nd > b, the integrl of f from to f := This eqution then holds for ll, b I. The considertion bove reduces the definition of integrls to tht of remesures. However, rigorous definition for remesure is not ny esier thn rigorous definition for integrls. All we hve gined is some intuitive guide. A rigorous definition of integrls will not be given here. b f. 17
18 On the surfce, derivtives nd remesures seem to be unrelted. It is somewht surprising, therefore, tht there is connection: F (t) := t f for ll t I is n ntiderivtive of f, i.e., F = f. Of course, proof of this theorem cn be given only fter integrls hve been rigorously defined. It follows from Theorem 7.1 tht ll continuous functions hve ntiderivtives;(but noncontinuous functions sometimes hve ntiderivtives, too.) Theorem 7.1 reduces, for continuous functions, the problem of finding ntiderivtives to the problem of finding integrls. The following corollry of Theorem 7.1 does the reverse: it reduces the problem of finding integrls to the problem of finding ntiderivtives. Theorem 7.1. Let I be n intervl, let f : I R be continuos nd let F be n ntiderivtive of f. Then for every, b I, we hve Proof. By Theorem 7.1, t f = F (b) F (). t f is n ntiderivtive of f. Hence, by Theorem 5.5, it cn differ from the given ntiderivtive F only by constnt, sy c, so tht t f F (t) = c for ll t I. Evluting this t t := nd then t t := b nd tking the difference, we obtin the desired result. The following rules re useful for deling with integrls. The first two re vlid when I is n intervl nd, b I, f, g C(I) nd c R. The third requires, in ddition, tht f, g Diff(I) nd f, g C(I). 18
19 (f + g) = f + g, (cf) = c f, f g = (fg)(b) (fg)() (integrtion by prts). The fourth rule is vlid when I is n intervl,, b I, g Diff(I) with g C(I), nd f C(J), where J is n intervl such tht Rng g J : g(b), re given. g(b) g() f = (f g)g (substitution rule). These rules re immedite consequences of Theorem 7.2 nd Theorems 13 of Section 5. fg 19
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