‘0 ., 5.2: r - Wilbi. 9 0a. .- ...}... : .y:31.....z.....:.;...r:2. .. ...233533343. 831... I '2315.23::“33:...222?...» .2 ...., W. 1 31. 5.3.15 3': ..v... ; ... E... ’\ ,.....é.......%§ .... .. .31....” . MSU LIBRARIES _—c—. RETURNING MATERIALS: PIace in book drop to remove this checkout from your record. FINES wiII be charged if book is returned after the date stamped below. ELEMENTARY CALCULUS FROM THE STANDPOINT OF ADVANCED CALCULUS by Jilliam Paul Fuller A THESIS Submitted to the Graduate School of Michigan State College of Agriculture and Applied Science in partial fulfilment of the requirements for the degree of MASTER OF SCIENCE Department of Mathematics 1959 TEN Face ACKNOWLEDGMENT To J. E. Powell Whose help and suggestions are responsible for this thesis . 1.232849 TABLE OF CONTENTS INTRODUCTION CHAPTER I. LIMITS AND CONTINUITY 1.1 Limits 1.2 Theorems on Limits 1.3 Definition of Continuous Function 1.4 Continuity of the Elementary Functions CHAPTER II. DIFFERENTIAL CALCULUS 2.1 Formulas for Derivatives 2.2 Rolle's Theorem 2.3 Theorem of the mean 2 .4 Indeterminate Forms CHAPTER III. INTEGRAL CALCULUS 3.1 Theorems on Continuity 3.2 Existence Theorem for Definite Integrals 3.3 Duhamel's Theorem 3.4 Osgood's Theorem #0103?” 11 12 33 33 56 58 59 F3 75 78 79 3.5 Fundamental Theorem of Integral Calculus 80 CHAPTER IV. 4.1 4.2 4.3 '4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 CHAPTER V. 5.1 5.2 5.3 APPLICATIONS OF DEFINITE INTEGRAL Introduction Area Under a Curve Area in Polar Coordinates Volume by the Disc Method Volume by the Shell Method Length of a Curve Surface of Revolution Pressure on Horizontal Submerged Area Work Moment of Inertia Criticism SERIES Positive Term Series Finding the nth Term of a Series Power Series BIBLIOGRAPHY 82 82 85 84 85 87 89 90 91 ‘93 94 95 98 98 100 101 105 INTRODUCTION The writers of most elementary calculus books find that they need many theorems whose proofs, they feel, should be omitted since the average student would not appreciate or understand such proofs. Some writers, in their effort to give simple proofs of these theorems, make statements that are false. Occasionally they fail to acknowledge the necessity of certain assumptions. The writer of the advanced calculus book feels that these omitted proofs are either too elementary or else not a true part of his work and therefore he also omits them. In this way we find omitted from elementary and advanced calculus the proofs of many necessary theorems. For example, we find that the subject of continuity is mentioned in elementary calculus and discussed at.some length in advanced calculus, but in neither case do we find any proof of the continuity of the elementary functions. Again, the derivations of the formulas for .derivatives are based on the assumption that the derivative exists. The purpose of this paper shall be to state and prove those theorems that are important to the elementary calculus and are not proven in most elementary or advanced calculus texts. In deing this we shall assume those theorems of real variable that are ordinarily assumed in advanced calculus. An effort will be made to point out some of the errors that are made in some of the elementary calculus books. Also, numerous examples will be given to illustrate other errors commonly made. No effort will be made to include proofs or definitions generally given correctly in the elementary or advanced calculus book. CHAPTER I LIMITS AND CONT INUITY 1.1 Limits. Since the idea of the limit is basic to the study of calculus it is important that its present- ation be made with care. We define the limit of f(x) at x=a as a number A such that if 6 is any positive number we can find a number I7 such that if Ix n al<flz and x *a then it follows that lf(x)-A’<6 . We write this as (1) Lim f(x) : A x=a The definition of the limit of a function of n variables is very similar to that for one variable. We define the limit of f(x1......xn) at the point x1 : al , x2: a2 , .... 1:11 I an , as a number A such that if 6 is any positive number, we can find a number r7 such that if lxl- a1'07 then it follows that lf(x) - A|< 6 . We write this as (2) Lim f(x) 3A x=oo We define the limit of f(xl....xn) at x,=°o,x,=°°,---, 7‘4"” : “Kw: “I ) 35mm: ‘22, """ I 797: “At-K 1 as a number A such that if e is any positive number we can find a number n? such that if Ix,]>nz , '14)”? , "”4111 - an-k' <7}?- then it follows that ‘f(xl"'xn)' Al<6, In this definition 1: may have any value from I to n. In these definitions we have spoken of the limit at x a a or at x a co and used the notation Lim f(x) line. or Lim f(x) instead of the usual notation Lim f(x) xsoo xea or Lim f(x) . The latter suggests the question of x—no whether x ever reaches a and if so when. The introduction of the element of time in the definition of a limit is, it seems, objectionable and unnecessary. From these definitions it does not follow that a number A, and therefore the limit, will always exist. For example if we let f(x) : sin-:1?— when xz’I; O a 0 when x a 0 then Lim f(x) does not exist. x=O The limit of a sequence, {an} , is included in the definition of Lim f(x) . To see this let f(n) : an . 1:100 Then Lim f(x) would give the usual limit of the runes sequence. We have noticed that most authors of elementary calculus books start with a definition of a limit of a variable. It is difficult to say just what they mean, if anything. Some of their examples indicate that they are concerned with the limit of a sequence. Again it seems that they are talking about the limit point of a set of points. On page 6 of Love's Calculus we find the definition: " When the successive values of a variable x approach nearer and nearer a fixed number (a), in such a way that the difference a - x becomes and remains numerically less than any preassigned positive number however small, the constant (a) is called the limit of x." 6 If the time element is removed from this definition there is nothing left. As an example he gives the sequence .9, .99, .999, ..... as having the limit 1. Further he states that no term of this sequence will ever equal 1. This statement is true. However we can not say the limit is never reached no matter how long the process continues. If this sequence represents distance covered, and if it takes one minute to go .9 of the distance, 1% minutes to go .99 of the distance, 1% minutes to go .999 of the distance, ......., then it is apparent that the limit is reached after 2 minutes. However, if it takes 1 minute to go .9, 2 minutes to go .99, 3 minutes to go .999,..... then obviously the limit is never reached. On page 9 in Slobin and Solt's book in calculus we find the definition: " A constant L is said to be the limit of a variable x, if the variable changes in suCh a way that the difference L98, in absolute value, becomes and remains less than any preassigned positive quantity,however small ". Again the time element is the only part of the definition that gives it any meaning. In the calculus book by Granville, Smith, and Longley we find a very similar definition on page 11 and a similar criticism can be made. Likewise in Dalaker and Hartig's book on page 4. These authors would have done much better if they had used instead a definition such as Neelley and Tracy gives on page '78 of their book in elementary calculus. 1.2 Theorems 2g Limits. Having defined the limit of a function we next prove some well known theorems on limits. These theorems are usually stated in elementary i calculus books without proof. Most advanced calculus texts fail also to prove them. The first theorem is Theorem 1.1 If Lim f(x) -A and Lim F(x) :8, then we have Lim [f(x)+F(x)] = A+B 4e. Since Lim! f(x) - A, we have by definition (1) that for 6, >0 it follows that there exists a117, such that if '1: - aka?" then lf(x) - Al<€, . Likewise for 63>0 it follows that there exists m%such that if‘x - ”(’72 then 'F(x) - Al<63. For anyewsuppose we choose 3568:;- and let 7 be the smaller of the 7, and I7; . Then we have for 'x - al< ’7 that 15°an ~ffl+BJl5 IF(X>"7'+IF®‘BI<€W€2=€ . a Theorems 1.1, 1.2, 1.3, 1.4 are true for a=oo if lilt.-.fl'.. S i’ we make one major change, namely, in selecting an»! we will, for ease , pick the largest of the 7,“?qu instead of the smallest. Stated as a limit this statement becomes :3: [f(x) + Hm]: A+B . This proves the theorem. Theorem 1.2 _2_[_f_ Lim f(x) = A and Lim F(x)= B, then =a x: we have Lim [f(x) - F(x§= A - B . x=a The proof is similar to the proof of theorem 1.1 except that we have Wu) - F(x] - (A - B)lé ‘f(x) - A‘ + ’. F(x) 4.. Bl0 it follows that there exist an ’7, such that if Ix - skull, then lf(x) - AlO it follows that there exists an 073 such that if 'x - a'0 suppose we choose 6'=2% , and 62:?an and let/7 be the smallest of the 07,,nzz,p73 , Then we have for 'x - al<¢7 that [f(xbflxi‘ - AB a f(x)oF(x) - F(x)-A + F(x)-A - AB '- == affix) - N + F(x) [ f(x) - A] , and therefore [f(x) «F(x) - ABléLAHic) - Bl+lF(x)‘ [f(x) - AI . Now by substituting as indicated above, °F -AB'£A¢6+O-§-=-§-+—€—:€' 1mm (x) HEW/C 20 Z Z Stated as a limit we have Lim [f(x)oF(x)] IAB . x3 a This proves our theorem. . _' ... wp. n.13'uqfhfl ...1 «Hialu‘ 10 The theorem for a quotient of functions is Theorem 1.4 _1_[_f_ lim f(x) - A and lim F(x) I B , x-a x-a then we have Lim fl. 3 .5. if B 39 O . _ x-a FIX) 3 Again since lim f(x) = A we have that for e,>0 x-fi it follows that there exists an r7, such that if 'x-alo it follows that there exists an n7, such that forlx-a'<0zz then it follows that 'F(x) - Bl<€2 . Also, since Lim F(x) I B -‘\= 0 it follows that there exists a :ugber O C forlx - “(’73 . This statement can be verified by referring to the corresponding statement of Theorem 1.3. This number C can be found by considering that since 8:? 0 then IBI>0 and so there exists a C such that 00 suppose we choose 65% and €8=i£lfi and let I? be the smallest of the three quantities 7,,»7‘f7, . Then we have for Ix - alO it follows that there exists an»? such that if [x - a'<0[ then it follows that |g(x) - f(x)|(6 . By (3), then, [h(x) - f(x)|<6 also, which is Sufficient to show that lim [h(x) - f(x)]: 0, Since lim f(x) I A then by lemma 1.1 we have Lim h(x) = A . Lemma 1.5 _I_f_ f(x) > O and _i_f_ the lim f(x) = l and _:_I_i_‘_ 3 is any rational number, then lim [f(x)] q_ 1. For let q I3;- where r and s are positive integers. If f(x) < 1 then, 1 f(x) <[r1 then, 1 1 <[f(x)]8 < f(x) . In either case [f(x)] 3' lies between 1 and f(x). 14 By lemma 1.4 we have, mud Lim [f(x)} = l . a .1. I. Since [f(x)]? = fi(x)] 3 then by theorem 1.3 of section 2 we have, Lim [f(x)] 3 = Lim [f(x)] q I l . Lemma 1.6 .1: the lim f(x) = A >'O then the lim [f(x)] q = Aq . Let us write [f(x)] q as Aq[£%.§1]q . Now by theorem 1.4 of section 2 we have, f x) = Lim '—K—' 1 . Therefore, by lemma 1.5, Lim Aq [£%]q = Lim Aq = Aq . Lemma 1):] _I_§ f(x) and g(x) are continuous at x = a, thenfxi x fxv x f(x) if x Oare continuous 33; x I a. Since f(x) and g(x) are continuous then by the definition of continuous functions we have the lim f(x)= f(a) 1:8. and the lim g(x) = g(a) . xIa 15 Now by theorems 1.1 and 1.2 of section 2 we have, Lim [f(x) 1'. g(x)] = f(a)ig(a) . x=a V In a similar manner f(x) - g(x) and flat where g(x)=’pO 900 H can be shown to be continuous at x = lemma 1.8 y; the 11m F(x) - A, lim f[F(x)] :- B, xIa F(x)IA and F(x) 4: A _i_5_1 the neighborhood _o_f_'_ (a), then lim f[F(x)] =B. x=a Suppose a, A, B are finite. Since lim ‘ fEthfl I13 then we have that for 6, >0 it follows thgtxthgre exists an r7,>0 such that if lF(x) - Al0 it follows 3.35:1; there exists an ’72’0 such that if |x - alo it follows $121))..there exists an (7,70 such that if lF(x) - A‘ < ’7, then it follows that lf[F(x)] - eke, . Also, since litiflx) = A, then we have that for 02, >0 it follows Jthat there exists an r73)0 such that if lx‘ >72 then it follows that lF(x) - A] < 0?, . Now by combining these 1 two statements we have that for 6, >0 it follows that ! there exists an "73>0 such that for 'xl >072 it ‘ follows that f[F(x)] - B‘ < 6/ . Written as a limit this becomes Lim f[F(x)]='-B . x-oo Lemma 1.9 If 1.11 3 F1(xl.....Xm), 112 : F2(Xl...xm) , ..... um I Fm(xl...xm), lim u:L 3 b1 , lim uz '-"- b2 , xIa xIa ..... lim um I bm , y I f(ul.......um), lim y = A, xIa uIb and _:_|._f_ uiz’abi _i__n the neighborhood _o_l_‘.’_ (a), then lim y = A. xIa If the interpretation suggested on page three be applied to the proof of lemma 1.8 then we have the proof for this lemma. 1'7 Lemma 1.10 . y; u.=F,(x, ....xj, , %=Fz(x, ....nfl), .... um=1‘,‘.,(x,....xm), _§._I_‘__6_ continuous at x'=a| , xz=52"""xm=%' 229.111.: 1.1.: 1),, uz=bz.,."’ um=bm, _a_t_ xi: 8L and. y =f(u, ...um) _i_s_ continuous _a_1_:_ u,=b. , “:2 b29"°'9um= bm, than y considered as a function of the x's is continuous _a_t_ x,=-. a, , xz=az, 0..., x'"=am O For um continuous at x = a means that k}? FAX, 0.00%) = "1(3, cocoam) and y continuous at u =.- b means that 52% f(uI ...um):f(b, ...bm). Then by lemma 1.9 we have £13. f[F, (x, ...xm)...sm(x,....xm]=r[p, (a, ....m) ...Fm(a, ....am] Therefore y is continuous when considered as a function of x at the point x=a . This lemma states that a continuous function of a continuous function is continuous . Lemma 1.11 . ELISW F(x)=x is W continuous . Since 1:7(x’ then 8=F(a) and therefore for any €>O there will always exist anqsuch that if Ix - a‘<’7 then it follows that [F(x) - F(a)‘(e. An7less than or equal to s will satisfy the above. condition . 18 Theorem 1.5 The rational integral functions are everywhere continuous. For let f(x):= xn where n is any positive integer. Then by lemmas 1.6 and 1.11, f(x) is continuous since 0‘ as f(x): [F(x)] = x . Now by lemma 1.7 q,x¢%-a,xmfl' + ......+a,,Hx + am, is everywhere continuous. This completes the proof for one variable. For the case of several variables let y==2EAX?x:K..x:?' where the s;,i=L'“Wgare positive integers or zero. Now let any term of y be designated by t = Aflx?.....x:’:’. We wish to show that t is continuous at an arbitrary point x, ...xm . Let u,= xf’ , oz:- x? , .... um: xi” ; then t:= u,uz... unis a continuous function of the u,.... umby lemma 1.7. Each uL is, however, a continuous function of the x,, .... x”.. Therefore, t, considered as a function of the x,, .... xm,is continuous by lemma 1.10. Therefore, y, which is the sum of terms of type t,is also continuous by lemma 1.7. Theorem 1.6 The rational functions are continuous wherever they are defined. ZAx/z’ o o o o oX/‘M E' ’ ,I......x4"m — Q is defined everywhere except Where the denominator is The rational function, y: zero 0 19 These undefined points are called the zeros of the denominator, or the poles of y. Since, by theorem 1.6, F and Q are everywhere continuous, then, by lemma 1.7, y is everywhere continuous except at the poles of y. Lemma 1.12 g Lim f(x)-=1, _a__ng_ x = a+bu where bato,__t_h_e_n_ Lim mafia . u=0 For as x ranges over a region D on the x-axis, u ranges over some region A on the u-axis since the equation x = a+bu causes the two axes to be in one to one correspondence. To the point xza corresponds the point u=O. Now let f(x) = f(a+bu)=F(u). Then, since x and u are corresponding points,f has the same value at x as F has at u. Since Lim f(x)==Ag then for any €>O it follows that therfi—gxist an 0? such that if Ix - alo it follows. that there exist an 7, such that if I ul 0 let 7:: angle A'OC and then for any 6 >0 it follows that there exist an 7 such that if [x'l < 42 then it follows that [sin x"< e . Therefore Lim sin x = O x==O O ! I 1 21 In figure 2 the line OD represents cos x. i la 5 Fig. 2 For any €>O, such as line BC, let I? =-.. angle AOC. Then for €>O it follows that there exist an I7 such that if [xl 403 then it follows that [1 - cos xI<6 . Therefore Lim cos x=l . xIO Next we wish to show that the function f(x): sin x 3.3 everywhere continuous. For let x=a-Pu; then sin x = sin (a+u)= sin a-cos u + cos a-sin 11. Now since Lim sin x=- x=a Lim sin (a-l- u) by lemma 1.12 and since Lim sin u=O u=0 u=O and Lim cos u==l then we have Lim sin x==Lim sin a=sin a. u=O x=a In a similar way we see that f(x)=cos x is everywhere continuous by writing cos x=cos (a+u)=cos a-cos u-sin aasin u. Since sin x and cos x are everywhere continuous then it follows that tan 3:, cot x, sec x, csc x are also continuous by lemma 1.7 . 22 Lemma 1.13 if y f(x) _i_§_ _a_r_1 increasirg function in the linear neighborhood gf x=a and _i_f_ULim y==b, —- x=a then ULimxza . 41' "“" y=b For suppose the function y is increasing so that when x 0 be arbitrarily small, and a - € O be such that b -r7>y'. b ”Z I # w Fig. 3 m Under these conditions while y remains in the I? -vicinity of b then x must remain in thee-vicinity of a. This means that U Lim x=a. y=b * The [I will, in general, mean either the right or left hand limit. Throughout this proof [I will represent the left-hand limit. 25 Theorem 1.8 The one-valued functions sin" y, cos" y, tanIIy, cot-’y are continuous _a_t_ every point whege they are definedg By theorem 1.7 Lim sin x==sin a. Therefore, by . x=a lemma 1.13, Lim x=a since sin x is an increasing sin x=sin a ‘ function if single valued. Now let y: sin x and b== sin a; —/ - then we have x=sin y and a: sin [b and therefore we have Lim sin-4y = sin-4b. In a similar manner the 37““ -/ -/ -/ continuity of the functions cos y, tan y, cot y can be shown. Lemma 1.14 Cauchy's Condition. The neeessagy and sufficient condition for a W 93 W9{aq}, there exists a (m) such that \a‘"- apl 4 e _f_g_r_-_ all n,p > m. Since {84,} has a limit then,by definition,for any 6 >0 it follows that there exists an m such that for n>m then it follows that {A - am‘<%. Also for the same €>O and m it follows that for p>m then, ’A - Bp‘<-%— . Adding these inequalities we have that for 6>O it follows that 'a,,- ap‘ m. We next prove that the condition is sufficient. 24 We wish to show that if for any’e>o it follows that there exists an m such that for n,p)m then it follows that 'a”- apl7, then it follows that ‘aw- ap‘<6,. Then there is an interval d;=2€, in length, which contains all ap for p >0z, . Double I, so as to 'obtain an interval 247 in length, 05‘ on either side of am, . Now for azIZé—there exists an m,>oz, such that p )4: then it follows that '8"?- ap < 62. Then there is an interval J2:- Zez in length which contains all ap for p >0»: . Now 4 is not necessarily contained in J,’ but 24} is always contained in 2d; . In general, for saz—Z-IZ- , set up a sequence of 2d; by picking 43:26,; and each d; will contain all ap for p >014 . Then 24‘ contains 2d} contains 2J3 contains..... 2d‘7_,w contains.2d2 ....... . Since the length of 2d}; is 21,12- we have just one point, A, inside all these intervals {2&1} .w * C. Caratheodary, Vorlesungen uber Reelle Funktionen,p.54 25 We will show that A is the limit of the sequence {an}. For any €>0 it follows that Iap -A‘ 5. 'ap- an," +Ia...,;-A‘<€ if i is chosen such that 65-1-5;- and p >014; . That is, if we take m="&£, then for p >m we have 1ap - A‘4 e , Therefore Lim 9m: A n:@ Lemma 1.15 _I_f_ {an} is a sequence 2;; positive rational numbers whose limit 3.3 zero and _i_f_ b>O, then Lim be“: 1. Let b>1; then b“) 1 . Since an > O and lim a”: 0 then for or. sufficiently large a”, is as small as we please. We can take m so large that 21;)? forn >m however large the positive integer g be chosen. By the binomial expansion (1+e)g> l+ge . For sufficiently large g we have 1+-ge'>'b and then (1+e)3>b. 26 Also a'; (1+6) >(1+6)g . To show this let/11,0? be two rational numbers such that ’7 fl oz)” . We wish to show that (1+6) >(1+6) ._ Let 02:741— and ,u=-§- where r,s,t are integers and r>s and t>0. Then .4. (4) (1+é)t>(l+€)% . I --... Raising both sides of equation (4) to thet—thpower we have A. 5 (1+6) >(1+6). which is true since r and s are integers and r > 3. Therefore, (1 + 6 #3) b for n > m. Then we have b“"< 1 +6 . That is for any 6 >0 we can find an m such that if a... n>m then b 4 1+6 . Thus we have proved that Lim be": 1 for b > l. . n=00 For the case b< 1 let bz-é- . Then 0 >1 and, 8Lpplying the preceding case,we have ca"< 1 +3 . ‘r‘ Hence and Therefore 2‘7 as C“ [+5 1-b“"< e . a»: Lim b = l for b0 , then the seguence b" ,b"2,b"’.... has 2 limit. a a - For let dmz b‘“- b P= Mn?" (3 1). Then for ”(>0 it follows that there exists an e such that if Iam- 3P’<€ 4.. then it follows that ‘b - l -a P <0? if we apply lemmas 1.14 and 1.15. Also, since {am} has a limit,it follows that ‘there exists two numbers Q and R such that Ql, we take 7:753- ,and,if b<1,we take 17:? |d«]< b3»? if b>l . 'd..'< be? if bO it follows that there exists an m such that for n,p >m then it follows that [.044 - bapl<€. Now by applying lemma 1.14 our theorem is proved. Lema- 1017 E a, , az ’ 0000 and C, , C: 00000 Eg‘two sequences of rational numbers having the same 62». c4. T limit and}; b>0, then the Lim b = Lim b . ' By lemma 1.16 both limits exist. Let " i d .- do: = ba'f- ban-1b ”(l - be" 47. We must show that Lim d4: 0 . However, Lim (a,,I - 3,, )=0 and therefore nsoo ' bk" nzoo the Lim (l - ("i=0 . Hence 11:00 a Limiszim [b( (l-bc' mil-:20 Lim (bah-be“). n=°° n8” n=ao c Therefore the Lim .04.: Lim b”by applying theorem 1.2. 11:” n=ao We are now ready to define what is meant by an irrational exponent. If 0 is any number and r, ,r, .... is a sequence of rational numbers such that Lim nn=:c, IFRO an then ac is defined to be Lim a . nzao 29 Lemma 1.18 _I_f_ c, ,0? ,.... is _a_ sequence whose limit _i_§ c and _:_l_f_ a>O, then Lim ac“: ac. Il-‘JO For let r, ,r, ...., and s, ,3: ,..... be two sequences of rational numbers whose limits are c and such that ra< c,,,l then it follows that a’“"< aC‘“< as" . However, by lemma 1.16 we have Lim am“: Lim 35,: ac. Therefore, by applying lemma 1.4, nan n=co we have c Lim ac“: a o The proof for aél is similar. Theorem 1.9- The exponential functions are everywhere continuous. we wish to prove that £433. as": ex," or that for e>o it follows that there exists an 7 such that if [x - x,[<’7 then it follows that 'ax - azal< 6 . Let us write ax’ - at” az"(az'x‘- l). as Let x, ,xz ,....x,,,bea decreasing sequence whose limit. is x, , and x', , x}, ....x},, be an increasing sequence whose limit is also x0 . Let ’7 be less than the smaller of ‘xk - x0! and [xp - no, where p and q are picked so that for €>O then lé’.)&11<_d§15 ’- and ‘a” 1"- l' 2% . Consider first the case where a>1. 50 For x>xo within an? distance of XO we have ‘J'k- 1‘ x?— x, and,then ll - axx"<[l — axfz'léfi}. Therefore we have 2 1. toe _ ‘8 - a ‘4 821; - 6 0 Now consider the case when ax,, then we have, since x - x,< x,- x, , that l>aza°> Jr". Then ‘a1‘10_ 1|< - 1kg; and we write ’81 - ahi xfi- 11,, that al< 2’ az’ x: Therefore we have lax - ale The case when a=l causes no difficulty. Lemma 1.19 if {an} _i_s_ a sequence whose limit _i_s_ 1, then Lim log aq=0. For let b, the base of our logarithms, be greater than one. Then for any 6>O we have b6 >1. Let 7:16] >0. Since Lim a4=l we have for any ”(>0 n=¢o that there exists an m such that —»lm . Hence we have (5) l-nz< amm . Then lemma 1.4 gives Lim log tam-1:0. n=oo 51 Lemma 1.20 g; the Lim Eng-=8. >0 and am>0, then the Lim log an, =log a. 52 d 4 Since a”: a-é'l- , then log a“ = log a + log-51"- . However, the lira—245:1 by theorem 1.4 of section 2. Therefore Lim lOg-Z—zo by lemma 1.19 and hence Lim log am .1: log a . Lemma 1.2;; The Lim log x = log a _i_f_ a>0. x=a This is lemma 1.20 stated in a different form. Theorem 1.10 The logarithmic functions are everywhere continuous where they are defined. For if we let f(x)=logb x then we must show that Lim logbx =1ogb a. We will write, using lemma 2.3, x=a f(x) :7. SILTg-gfi—E- and then f(x) will be continuous if e logex is continuous. However, logex is continuous for every x>0 by lemma 1.21. Therefore f(x) is continuous which can be stated as Lim logbx =log6a . x=a CHAPTER II DIFFERENTIAL CALCULUS 1.1 Formulas for Derivatives. The derivation of the formulas for the derivatives of the elementary functions will next be considered. Many times the existence of the derivative is assumed in the derivation. If y::f(x) is defined over an interval (a,b) and‘ x==c is a point of the interval then the quotient Aly._ f(x) - f(cl Ale— x - c where x is in the interval,is called the difference quotient at x=c. If xzcs-h then ‘é;y _ f(cdbh) - f(c) 43x" h Let the 3.933% :0? and if 7 exists then ,7 is called the derivative of f(x) at x==c and is written f'(c). Let D be the points of the interval for which? exists. The values of ’7 define a function of x called the first derivative of f(x) and is written f'(x) or y'. Theorem 2.1 If the derivative f'(c) exists, then f(x) _i_s_ continuous at x=c. 33 54 Since by definition the Lim f<°+h)h‘ f(CL);-:_f'(c) and since the limit does exisE-gnd equals f'(c), then for any é>0 it follows that there exists an ’7 such that if lhl (4? then it follows that - f(c+h) - f(c) __ h f'(c)+6’ where I€II<6 . Then f(c+h)=f(c)+ h[f'©)+ 6’] and by taking the limit of both sides we have Lim f(c+h)=f(c) . 1130 This shows that f(x) is continuous at x=c. Theorem 2.2 if y z % , where V360, and_i_f u' and v' 91:13th D, then y': vu' -zuv' . v Since y z 1.1 then y +Ay _-.= __u +1411 v v +Av and Ay_u+Au _ 2. __ vAu- DAV- l _ u 1 -V+AV v "" Viv-rm} __ 1+Av v v+Av Therefore Az—_}___é‘_1.-11. 1 4.! Ax-v-i-AVAK v v+Zv Ax ' 55 But by theorem 2.1 we have zIalflirano(v-|»Av).-:.v and by h othesis the Lim 4.3.1 : u' and Lim 4.1 = v' . yp AX’O Ax 41:0 AX Passing to the limit we have y,____u_1__uv'__vu' -uvj_ _v — V2 V3 It is permissible to divide by \H-Av since v-t—AvSF 0 if 43x is sufficiently small and since v is continuous and not equal to zero. This proof is sometimes given by writing vy'==u and then taking the derivative of a product and solving the resulting equation for y'. Such a proof assumes the existence of the derivative y'. Theorem 2.5 _I_§ y: f(x), x = g(t),and f'(x,) and _ d d dx 81(t,)_e_x_:!._§_1_: and x,.. g(t,), then ‘%=dxzd_€ . For some values of Alt we may have 43x==0. Let V be the set of values of t,+43t for which 43x==0 and v' the set for which Axaeo. For trrzit in V' we have 23 ._ 23 23x 1%“ZYE'EE ' 56 Since the derivative g'(t,) and f'(x,) exist then 8 Lim AlzLim ALLim AE ( ) At=o At At=0 Axmsa 0A 1+3 which ,by lemma 1.8,becomes All- 1““ At- Atzo Hfi' L-i.a 9'5: f'Hx/ '8'(tl) . dn" ALt‘O This limit is taken for At's for which t,+At is in V'. If in every interval containing t, we have values of t,+-At for which Ax =0 then equation (8) does not prove the theorem and we must consider t,+At in V. \ AX ... Ne have that t .. O and therefore LtimA 02.5 = This limit is taken for just those At's for which t,+At is in V. Since g'(t,) exists it must then be zero. For +43 43 :: t, t in V we have also that fl 0 since the Ax is zero. Then for t,+At in V we have (9) - f'(x,)vg'(t,)==0 . 1312 Therefore, by equation (8) and (9), for 6>O it follows that there exists a d such that if [Atk d ‘fi-f'(x,).g'(t,) < e . then 57 This means that .fl’cfl: ms.) -g'(t,) . The usual proof of this theorem in elementary calculus books fails to mention the possibility olex being equal to zero. If y ::f(x) and x equals a constant we would have a case of A3x==0 for alltzlt's. A better example is y=f(x) and x = tzsin% for t:\=0 and x=0 for t==0. Here, no matter how close we get to the origin,we have At's for which Ax =0. Before the formulas for the derivatives of the exponential and logarithmic functions can be derived the constant 6 must be established and a few properties of sequences proved. Lemma 2.1 if a>b20 and n is a positive integer -1 .. greater than one, then n(a - b)6zb,then we have a’"- b'">(a - b)(b”.‘,+- bad-I- ... n terms) , which shows that a“- b">n(a - b)b”‘" . 58 If we replace b by a when a>b then a“-- bm<(a - b)(a""-41-a”"{i- ... n terms) and a“- b’"<(a - b) ham" . We note that we could have stated this lemma as bfi>a’"" [a - n(a - b):’ by transposing and factoring. Lemma 2.2 ‘5 bounded monotone sequence has a limit. For let A =-a,, at, as..... be an increasing monotone sequence and a¢0 it follows that there exists an m such that if n>m then it follows that 0G . Now adding the first p inequalities of equation (11) we have 84"; - aMQZPe ’ and substituting this value of pa. in equation (12) we have am>e which contradicts the hypothesis. Then there must exist but a finite number of indices m; such that equation (12) holds. Therefore an m can be taken so large that a..- amm. m. Lemma 2.5 The sequence %=(1 +7-1- ) , where n=l,2,3..., has a limit. First, we wish to show that {aa} is increasing or that a£>awq . By lemma 2.1 we have .1 b”>a”' [a - n(a - b2] and,if we let a = 1+;t—{n- and b: 1+7{f ,then we have fit «4 an (“7”?) ””717 [ko‘t'ér‘nmz’n-l "ea-Esta -l 40 This shows that {an} is increasing. Second, we wish to show that the sequence {an} is bounded. Let a=l+-—-al, b=l,and n=m+l and, applying lemma 2.1, we have I on 1>—zL’(]-+2;) for m=1,2,5,oooo ' Now by squaring we have 4>(1+-an- )z" which shows that 32m< 4. But by the first part of this lemma we have 62M_l< azmand since all positive integers are of the form 2m or 2m-l then we have a... < 4 for n=l,2,5,..... We have now shown the sequence {an} to be increasing and bounded and,by lemma 2.2,the limit then exists. This limit is the constant called e . F7 ‘L = . Lemma 2.4 ihe 71.:me (1+; f e . For consider x such that n .5 x s n+1. _L.> ...!— ......L... NOW 1+m/l+x>1+nfl and (1.3) (l+-,';—)”"'> (“7%)") (Ha-£17)“ . 41 However (14) (1+—-)""' =(1+7,L€.)(1+.J_)”‘ and (15) (1+ 31—“) = (14.0”, )m’. 7+ L571, , Now equations (14) and (15) give,in (13), J... J.“ ‘LZ __(_.. 411" / (l6) (l+,,,)(1+ a) >(1+ x)>(1"’oz+/) .7+a&7 _‘ By lemma 2.3 we have / Lim (1 4.7%)”: Lim (1+ 6 nu- +0t+/) and also Lim -lim (l+-i‘) >*55 - 1 Therefore, Lim (l+—L) =c , by lemma 1.4. x=a-ea 1 Lemma 2.5 The %}£(1+-§f=e . For let x: - 11. Then I I- ,‘u‘ c“ (1+7) _ (l we) — (nil-:7) __ / I V — (1+7)(1+V’) where u-l=v . 41 However (14) (1+;L—f‘“: (1+7le>(1+—,£,— Y" and (15) (1+R'ITTP=(1+E£T)MH' 7 + ./_._L. 11 ”Ink! Now equations (14) and (15) give,in (13), z I I 0‘ I I 41'” (16) (in, )(1+—,,-i-) >(1+ ;) > (11-01-71) .7+ 1 . 41+] By lemma 2.5 we have Lim (1+0!) .— Lim (l‘l‘m .- C and also Lim (l+-,é,—) = Lim ...—L7— =/ . «=00 mace /+-;,-;-;; So equation (16) bea>mes,upon taking limits, I Z 8 > 11m (1+7) > 6 - r .1— z- ' ' Therefore, Lim (1+ ) - C , by lemma 1.4. z=+ca 1 Lemma 2.5 The £3133, (1+ff‘=e . For let x=- 11. Then 1 ad (1+3?) = (l 71—) =(1+-—’-—) . / V :: (1+7)(1+‘é‘) where u-l=v . 41 However (14) (l+-’-)"" =(i+7£,-)(i+—L-)”‘ and I 011', I #_. (15) (1+ RT!) =(1+;—,—-—+,) - 7+ 0”,, . Now equations (14) and (15) give,in (15), x J_ .1.“ __,_ I 41+! / (16) (1+“)(14' 01.) >(1+ I) > (11.02.74) 07+ I ”rd-l By lemma 2.5 we have ..L—“a- n+1 Lim (1+“) — Lim (1+ +4317) 5 and also Lim (1+-[-)- = Lim “—LT =/ . A" “3° /+ 021-! So equation (16) becomes,upon taking limits, I Z e>1im (1+-7) >6 - 1 Therefore, Lim (1+-f) =6 , by lemma 1.4. J‘s-m Lemma 2,59 The L32. (1+-’é'ft=e . For let x: - 11. Then (1+-5H: (1 -717)”: (ii-5:) / V 3 (1+7)(1+T5") where u-1=v . 42 'Nhen xz-co then u=+aa and v=+°o . Since Lim (1+-{,— ):1 and Lim (1+-L)V=8 V=+O Val-to V then,if we take the limit in equation (17),we have V Lim (1+—L) = Lim (1+—{7) = e . gas-Q Vzm 3’.- Lemma 2.6 The Lim (l + x) = 3 - x=0 Since Lgim (1+-7 ’ _L)x= 3 let 3:21" Then the right hand limit of (l + u)“:.€ . I I A180 L111} (1*? ) = e . x=a-co Again let x=a-L and the left-hand lfimgt (l + u)£= e . Since both the left and right hand limits equal 6 then ..L Lim (1 + x)z= e . x=a Lemma 2,7 The Litton log(l+-x) : 1 . ..L. The Lim 103 (1+“) :: Lim log (1+x)‘ . x=0 x=0 If L321. f(x) =7 , then,by lemma 1.15,we have Lim 10g f(x) = log/7 =: log Lim f(x) . x=a x=a Therefore ..L Lim log (1+x).£—= log Lim (1+-x)” x=0 x=0 =- loge =1 . 45 Lemma 2.8 Suppose that y:f(x) is; _a_ monotone increasirg gr decreasirg and continuous function. Let x = g(y) _‘ge the inverse function and let x and y _be corresponding points. if: f'(x) exists and _i_§_ different _fgom zero, then g'(y) 1 I ... exists and g (y)- m) . Since f(x) is monotone increasing or decreasing then Ay and £33.76. are not equal to zero. Hence _a—y-z-ZL— has no division by zero. Since y is continuous, x then Lim Ay =0. Taking the limit we have Alt-*0 .. " ' A Air—0 43’ $132? if we apply lemma 1.8. Therefore 3"Y’=r7%xr . Theorem 2.4 _I_f_‘_ y=log x then y'=-zl— . For All A él— 1°a 1. I From y=sec" u we have u=sec y. Therefore du = a: a sec y tan y dx and (24) Y': 1 2L1 , sec y tan y dx Since tan y: IVsec 3?, - l and u = sec y, equation (24) becomes y it 1 du . uVuz-l E 54 Here the plus Sign applies to the first and third quadrants and the minus sign to the second and fourth quadrants. "’ / dd [11' > 10 Writing y=csc-lu as u=csc y, then '31:: - csc y cot y '3‘; and 25 L-—- 1 5.1.1.1 . ( ) y""“"csc y cot y dx Substituting cot y::.+'. Vcsczy - l and u=csc y in equation (25),we have 1 du uI/uz-lfi . y'= 4'7 The plus sign holds for the second and fourth quadrants and the minus sign for the first and third. Theorem 2.15 __If y=u“ , then y'znua-IEE Writing y=u”" as y = oak,“ and letting v.-:n log 11 s we have yzev o 55 v dx we get v n i_eu du._nu du ot-Idu In theorem 2.15 the derivation must not assume the existence of the derivative. If such an assumption is made the formula simply states that if there existsa derivative it can be found in this way. This is a common error in elementary calculus texts. Smith, Salkover, and Justice on page 147 of their book and Neelley and Tracy on page 108 assume the existence of the derivative in their proof of theorem.2.15. On page 47 of the text by Slobin and Solt and on page 91 of Granville, Smith, and Longley we find the existence of the derivative assumed in the proof of theorem 2.7. Some rather surprising results may be obtained if you start with false assumptions. For example, suppose we assume that x3 A Tx - 2)(x+1)= 37-7 W where A and B are constants. If A and B are determined by 56 .3 writing x = A(x+~1)+B(x - 2) and substituting = -1 and x==2 we obtain the obviously false result that 33 s 1 (x - animal: "(:5 "“22: - )"‘3T““Tx 1 ° On page 56 of Slobin and Solt, after having proven theorem 2.15 for the rational numbers, the authors make this statement, "This theorem is also true for n irrational. This is evident since an irrational number may be expressed as the limit of a sequence of rational numbers and, since the theorem is true for every rational number of the sequence, it is true in the limit." This statement implies that if a thing is true before the limit is taken it is true after the limit is taken. This statement is obviously false. 2.2 Rolle's Theorem. Theorem 2.16 'If f(x) 1e continuous and single-valued _ill 2 region aéxéb, f(a)= f(b)=0, and f'(x) exists for a0 be the maximum of f(x). Since f(x) is continuous there is some value a0 then f(C‘Ph) - f(c) £a0 , and f(c -h) -f(c)50 . Then (25) f(c+h) - f(c_) 4 o h —' ’ and (27) f(c - h)h- f(C) z 0 o By equation (26) f'(c) £ 0 and by (27) f'(c) 2.0 . Therefore f'(c) = O . If the maximum is zero then there is a minimum that is negative and a similar proof holds. '3' F1136, p. 24 58 Demonstrations resting upon the fact that the function ‘ in passing from a to b must increase and then decrease or decrease and then increase will be valid as long as we have only a finite number of oscillations of f(x) between a and b. If an infinite number of oscillations exist then this need not be true. For example, Slobin and Solt start their proof with the statement, "If f(x) is not identically zero, then it must increase from f(a)=r0, then decrease to f(b) = 0, or it must decrease from f(a)-=0, then increase to f(b)== 0." Their proof holds then only when the number of oscillations are finite. 2.5 Law of the Mean. Theorem 2.17 .If f(x)'ie continuous in (a,b) and if f'(x) exists fqg a_f the expressions f!(a) f!!(a) fiit(a) ’00.. which _i_s_ not indeterminate, provided this expression 9X1St80 f(x) .. Consider now the case Lim -—(-—)- where Lim f(x)- x=a: x and Lim g(x)=0 . Let x=-E and consider Lim xii). . 3“" 19:0 g(t) Then,by equation (55) 3 / I r(--> szv(-£') f'(x) Lim ——fz_—___ —--Lim = Lim t=0 g( t) t=o it swit)‘ xzoo s'(x5 This result is given by Theorem 2.19 If the derivatives _o_f_ f(x) and g(x), that are involved, exist for all values 9_f_ x greater than 393112 W N and _i_f the fraction g(i) assumes the indeterminate fo_r_m ‘73-‘33}. x =00 , then 62 will b equal _9_ the first e_f_ the expressions ngrwx) 313133 f' '(x) £33 f"'(X) ’ .9 £$ggl(x) £313 g1!(x) %}£g|l1(x) , O... which _i_s_ not indeterminate, prpvided the expression exists. f(X) :: =60 If STE is such that Lim f(x) no and Iii-m g(X) a then if it is desirable to find the Lim f(X) it can mom be shown that theorem 2.19 applies.* By equation (50) we have - g(xf- g(c) g'(§) where c 0 we can x=a find an ’7 such that if Ix - a|-é- . In a similar way Lim f(x)-'00 means . 1:“, that forany €>0 we can find an ’7) 0 such that if x >-'—- then it follows that f(x) 7-i- . ”L 6 62 f A x) q Lim x=a: will be equal _1_:_q the first 9;; the expressions Limof'u) , 230m f"(X) ’ £33; f"'(x) giggqx) £33 g"(x) 71:93; gll|(x) , 0000 which _i_s_ not indeterminate, pgovided the expression exists. f(x) ..- .. If EGU is such that Lim f(x) 0'0 and 1555;!“ g(x) a then if it is desirable to find the Lim 2‘:) it can =00 be shown that theorem 2.19 applies.* By equation (30) we have where c (f < x and c is large - but finite. -)(- By 2:132 f(x)=-o we mean that for any 5 >0 we can find an '7 such that if Ix - al <7 then it follows that f(x) >é— . In a similar way £13.31; f(x)=-°O means that forany €>0 we can find an ,7) 0 such that if x >-’—- then it follows that f(x) 7-i- . "L 6 65 Now by algebra ‘ (a) (35) f(x)_ f'(f) 1 - . g(x)- g'(§?) l - 3 f'(f) .. Assumin that has a limit at -—°0 and callin a __T—‘Ts' f f 13(0) 8 that limit A, we can take 0 so large that'ETFET , and therefore also $.81; , differs from A by less than 5, . By this method c is now fixed and f(c) and g(c) are still finite. Since x'may still vary, we can take x so large c that i_‘_%_ will differ from 1 by 63 . From equation (55), then, EGCT — (A +7, )(l+%) where (mks, and Theorem 2.20 .1; the derivatives-e; f(x)_ehg g(x), that are involved, exist for__l_.1_1ee§ e: x.gneate2.1han _§__ome n__11_____mber N 2.11.611: the f___<11:.i_<1n é—E—Eg— assumes the indeterminate from {33- at x =00 , than f(x) Lim x=a g(x) 64 will be equal to the first 23 the expressions aim f'(x) 53m f"(x) fifHWx) ,... £4353 8'(X) ’ Iiggg 8"(x) ’ £1;an g"'(x) which is not indeterminate, provided this expression exists. The case where Iii-fl f(x) =°O and ,Iz‘y}; g(x)-=- 00 is easily obtained from theorem 2.20. For, letting x==(a +3%), we obtain ru)_fia+’)_m-) ”’7’ m-an‘ir-Efir - Then f x) _. F( )._ F' ) ‘58) %=‘2§m*§’3£“oaT%-§é$e' y by theorem 2.20. Since dx -/ F'( ):= f'(x) ::——-f'(X) 3’ Ex? 33 ’ and G'(yl'==g'(x)-%§=ré%ég'(x) , equation (58) becomes 65 This proves Theorem 2.21 _I_f_‘. the derivatives _03 f(x) and g(x), that are involved, exist _i_._r_1_ the neighborhood _o_i_‘_ x=a and are continuous _a_t x=a, and _i_f_ the fraction .388. assumes the indeterminate form {3- when x=a, then f(x) Lim x=a g(x) will be equal to the first of the expressions fl(a) , ft1(a) f!!!(a) g'la} g"(a5 ’gH‘Ia} , to. Which _i_§ not indeterminate,provided this expression exists. Most of the other indeterminate forms such as 0-69 , «9..., , 00 , 000, l” , can be transfermed into the form—g- org- . The last three forms may be treated by the use of logarithms. A function,sec x - tan x,which becomes co~oo at x=%r- may be written l-sinx sec x - tan x .-.-. cosx 0 becomes 00-00 , then we can write which becomes ~9- when xz-g- . In general, if f(x) - g(x) 1 __ 1 f(x) - g(x): EL! )1 f(x) TTXLgfi) Which is of the form ...Q... 0 66 Also, if f(x).g(x)=0'00,we set f(x)-g(x): f(IX) ' This reduces to the form 7-0,— . f2!) An error that is often made is the assumption that £2.13 5.88. does not exist if £1.13 £4163- does not exist. The falseness of such an assumption can be shown by an example. Consider f(x) = xz sin—£- , g(x):::x , and a=0. 2 I Then Lim f(:) = Lim W=L1m(x sin—XL)=O . x=0 g x=O x x=0 At the same time 1 I £112 53%;) = L m (2x sin? cos‘f’) = 8i( 5 x=0 wnich does not exist at x=O. It should be noticed that in order to say that Lim f'{x) _ f'(a) g'lx} "" g'fai x=a we need the continuity of the derivatives at x=a. Some other mistakes that often occur in the treatment of indeterminate forms are the following. If h(x): f(x) am and f(x) and g(x) vanish at x=a, then the value of h(x) at x=a is undefined. It is incorrect to speak of the true value at x=a because there exists no such value. 67 No transformation or limiting process will bring out a true value when none exists. Many times it is desirable to define h(x) at x::a and h(x) can be defined to be anything you please. For the sake of continuity h(x) is f(x) sometimes defined at x==a to be Lim if this limit is finite. Slobin and Solt are qfiiie emphatic in pointing out that we are discussing undetermined forms and not indeterminate forms. That is, they say that-f;- has a value which is found by means of a limit. Obviously they are giving a definition. To find the value of Lim h(x) some might write =a fikrkf-ffiv .. NEH-k) _ Jc _ (59) h(a+k) -—m - :4*!2" ZdZ K ‘ _f'(a) and conclude that 2:2‘h(x)'”§TTET . This is true if this limit exists and g'(a)3EO. If both f'ta) and g'(a) vanish then it is impossible by this method to conclude that Lim f(X) .. f”(a) x=a g(x) ” 8"(85 ’ . f'(a) since e ation 59 onl s s that Li h x =: and qu ( ) Y :Y( ) 11:1: ( ) m not that Lim h(x) ..-—. Lim ' X x=a x=a g'Ix) 68 It is possible to find Iii-1:13 h(x) by writing f(x) and g(x) in power series. However power series representing such functions as see x, csc x, tan x, cot x, 69w; , are seldom developed satisfactorily in elementary books. Unless they are developed an author would have no right to use them when they occur in finding Lim h(x). x=a In evaluating the form g:- one might write 1 f "(x5 11"" = 3%"; = b l ’ ' f(x) Using theorem 2.20 we would obtain g'(x) Lim h(x) = Lim x : Lim hZUE) gz—E—{S o _ - '(x x-a x—a x=a lexi Then dividing by Lim h(x), we get x=a é'm x=a x or that f'(a) Lim h(x) = my" . xzn Here the error comes in assuming the existence of Lim h(x) which as far as we know does not exist until x=a its existence has been shown. 69 . sin x 1 Consider tne %;8 '—7E—' . Since bObh sin x and x vanish at x==0 let us apply theorem 2.19. Then Lim Sinszim “(’5le . x=0 x x=0 This would be an easy way to dispose of this troublesome limit of section 2.1 were it not for the fact that we used the Lim 3.39% = 1 to develop a formula for the derivativg—gf sin x. Now we turn around and use the derivative to evaluate the limit, thus forming the customary vicious circle. hany of the elementary calculus books have this example as one of their problems in indeterminate forms. In discussing Lim -§%§% when it reduces to f; we x=a find that both Love and McKelvey, in their books, use a method that is incomplete in that it cannot be extended to higher derivatives. This is true since they write f(x) -£(a)___ f'(x,) ETX) - 8(a)'— Sitlzj In the calculus book by Granville, Smith and Longley as well as in the calculus book by meelley and Tracy no mention is made of the fact that the existence and continuity of f'tx) and g'(x) is assimed. Dalaker and Hartig's book is the only one of these books that assumed existence and continuity of f'(x) and g'tx). 70 Smith, Salkover, and Justice as well as Slobin and a . f‘(x) . Solt assume that 1f Lim do not exi t tre “a m es 8 .1 n Lim f(x) does not exist. The former, on page 358 of x=a g x their text, state, "If the fraction -§%%% assumes the 10 indeterminate form Tor age-- when x = a then f(x) LS: am will be equal to the first of the expressions ft(a) fvlga) flit(a) ’00. which is not indeterminate provided this expression exists; and if it fails to exist, the limit sought does likewise." The last part is obviously not true. CHAPTER III THE DEFINITE INTEGRAL 3.1 Theorems 3n Continuity . In order to prove the existence theorem for the definite integral we need two theorems concerning continuity. Lemma 5.1 I; f(x) _i_s_ continuous _i_n the closed interval (a,b) and e is _a_r_1_ arbitrary positive number, then (a,b) can 39 divided into partial intervals such that the difference between the values of f(x) at any 'Ewg points in the same partial interval is less than 6 .in‘absolute.galg§. Divide the interval (a0, be) by a point 0:32.21'39... . Unless the lemma is true, either (a,,c) or (c,b.) has two points that do not satisfy the lemma. Suppose (a, ,c) has two such points. Call a, = a, and cab, and divide the interval (a, ,b,) as (a, ,b, ) was divided. Continue this process. Thus we have two unlimited sequences of points, a,, a,, a,,.... am,...... and b,, b,, b,,...b“,... For every n, a4”, 5 am_. 6 . Now f(x) is continuous at x=A since A is in the interval (a,b) or at one of the end points. Therefore there is an h>0 such that f(x) - f(A)‘«<-§§- for all x's within (a,b) sucn that A - h0 we can, by lemma 3.1, subdivide (a,b) into partial intervals such that f(x') - f(x") <15: if x' and xH are in a partial interval. how if x is in the first partial interval (a,x,) we have {f(x) - f(a)|< 6 , and F(x)! 4 lf(a)}+e . If x is in the seand partial interval (x,,x3) then /;(X) dx exists if f(x) -——'MDu3 a. ______.__. 'ia continuous. Let f(x) be a continuous function in (a,b). Let Mi. = maximum f(x) inAx,’ and m; = minimum f(x) in Axg. '76 Let (42) ”s", = in; Ax; l and (43) -§-0 = 2m; Ax; . I Then Let Dz be composed of D, plus some additional points. Then S94. SD, and iota 50’ since Ax; will be sub-divided and the M; olex; will be replaced in some of the sub- divisions by M}: S M; and therefore ED, .4. Sb] . Likewise i0: 2 £0, . Let D and D be any two divisions. Then SDHS éygo ’-§pfi5 2L§£|"§Lfi3 =55¥7:vifiyfi5;ZiL5 a» .§-D+D- 6" 30*5 3 then Let 44 '1': "" dI==§ . () go an __ 0&0 '77 The greatest lower bound and the least upper bound exist since SD 2:. m(b - a) and SoéMb - a) where m and M are the greatest lower and least upper bound respectively Of f(X) in (a,b). 3-2.1. . From equation (45) we have Since f(x) is a continuous function we have by lemma 3.1 that Then and But Since each term on the right is %3-30 (80 " I) t" E“ C. I m H ' " m" )Ax‘é 6 THE-1) +(1-a0) . positive or zero then 78 Then Li S =l= =L . ADE; 5’ 1; ~33; jLO Therefore Lim s =Lim ‘3' =1 ngo C) ND=O £3 1 which proves the theorem. Corollary 5.1 _I_f_ f(x) la a function with _a_t_ most _a_ finite number _c_>_f_ finite discontinuities, then the definite integral [g(x) dx exists. Let the discontifiuities be K in number and let the sum of the A5x¢ in.which the discontinuities lie be denoted by L. Then for any 6J>0 we can find a d such that if ND<0 _i_._t follows that there engage _a_n mamatllaifl n>m 2.1192 l6£l<01 M for i=l,2,... n, then ”LIigébcr—A. 79 For since ’egl0 _i_t follows that there exists g m such that _i_f n>m then '€;]_;|f(x)l . Therefore 4‘ .....z Lim Emma” f( imam) = o ND=0 I and,by the existence theorem of the definite integral, as b V: kgglozl:(2fl'fxf( fk)AX/()= 277 x f(X) dx 4 89 4.6 Length g£.a Curve. Let the curve whose length is to be found be y==f(x) where f'(x) exists and is continuous at all points. The definition of the length / €E J Awiw ‘9 b of a curve is L==-Lim choc! f3. a:E(lengths of chords) . Therefore L = iaggaA-X—z-l-Ey. 9 ==Lim. Vr1.+(%§§: Azlk' . ”0:0 There exists, by the mean value theorem, a point g; in 13:“ at which the slope of the curve equals the slepe of the chord; that is 9.25: f'( fir)- £31K 90 Therefore ”haZVl + 5"" 5* 83AM ' Applying the existence theorem for the definite integral, Lzfll‘PEfWJSJ? dx . we have 4.7 Surface 22 Revolution. Surface of revolution, 3, is defined to be Lim.‘2E(surface of frustrums of cones). chords-=0 We shall require that f(x) have a continuous derivative. Using figure 9, s .1533sz W ”33354-5: since the surface of the frustrum of a cone equals the average base times the slant height. Then s = £39,102an V1 + [m sigma . mazt §x>+9~AvJW+£rw a as, and since the value of x which satisfies the mean value theorem may not be the value of x so that f(§}) will represent the radius of the average base. 91 Now s = £33022rrfi gs) V1+Tf'( fngxfi/Iiggizmoyle +[f' (mfisxx . Call the first sum 8,, and the second sum Se. Since f(x) is continuous in (a,b) we can for any 6:>0 find a d such that if N DAy/c - I This follows since 5*} is 5K<§K< f}, and,if {K is picked ’ so that Z( QUAyK is the area, the same 5‘ may not be the proper depth to use to get the pressure on this area. 95 As a result, we have, M M p =ZAP = 2A was. + 94AM )Ayx . I I =Z[!( fx) gang + 2% {as 62.55:] . W s, + s, . Now ISZPEEE]W.ZK§L)€%z3yK :éfix wLE]ZS§§[ £EX.W'ND2EV3y4 éXwND (b—a) 9 J 6 i. where X.is the maximum of.£ky0 . Therefore Lim S2==0 . sto Then by use of the existence theorem for definite integrals we have that b P—L3%S,=a/Z(Y)Wydy . 4.9 ‘figgk; The work, W, is defined to be the force, F, times the distance, d, when the force is constant. Let f(x) be a continuous function of x representing a variable force acting on a particle, P, in the direction 0X. Let the particle, P, be moved from x=a to x=b. ... 94 Divide (a,b) into intervals,AxK. Let émd abe the values of x inASxK for which the force, f(x) is a maximum and minimum respectively. Now the work,13w, done in passing overzlxk is greater than f(szdxg and less than f(gxklxk. Then there exists a f“, $IxK . Then w=ZAw =5“ aux. . I By applying the existence theorem for the definite integral,we get a . w =Lim jimmy: fies) dx . ND=0 1 a 4.l0 Moment 22 Inertia. Moment of inertia about a line 1? is defined to be mrz for a point mass, m, at a distance r from.4g. Let a mass, m, be distributed along the x-axis from x = a )0 to x=b>a such that the linear density , f(x), is a continuous function. We wish to find the moment of inertia about the y-axis. Diaide (a,b) into sub-intervalsAxK = Xx - xk-, . Let éand 3; be the values of x where fix) has a maximum and minimum respectively in Axx . 95 Then the moment of inertia ole “Vie less than x:f(la)flxg and greater than fizqf(2;)zlxg'. Therefore there exists a f} such that the moment of inertia of Ax,‘ is 2 (ix +QAXK) f( {[AAXK . This gives the total moment of inertia as a ZUx'f-anx)’: f( 5“,.)de a K=I ...: = ngm seams“ 325.9. + #42:.) f( mm. . Denoting these sums by S, and S: respectively we have that [32‘s (2b +1113) M°ND(b - a) where M is the maximum of f(x) in (a,b). This gives Lim S ==0 ND=0 z ’ and therefore b Itoment of inertia = Lim S, = xzf(x) dx. . .M0=0 a p 4.11 Criticism. It might well be asked why we are so careful in showing that such quantities as area, volume, and pressure can be found by certain definite integrals. 96 It probably seems evident that the approximations usually made are sufficient to allow us to actually find these quantities upon taking the limit. It is not, however, always evident that these usual approximations are sufficient to give us the desired results. To illustrate this fact consider the following example. Suppose we wish to find the surface of revolution, S, formed by revolving the semi-circle, BCA, of figure 11 about the x-axis. 5% at .5 4, Fig. 11 Since gimo(chord)=0 we obtain an approximation to the 1‘ surface by summing the lateral areas, 2nyzlxk, of discs. 97 Then we might assume that n I i=‘//;;y'dxu==2//fl:‘ - x2 dx . -a a Letting x=a sine, dx =a cosede , we get IE 2 3:4,"; 14-00329 d9=4fia2£+sia££1 , ‘0 2 2 2 0 :4naz_§ 277:4? . Obviously this is not the surface of a sphere of radius a. It is now apparent that our approximation was not good enough. At the beginning of the discussion, however, it is not apparent, at least to calculus students, that our approximations are inadequate. This can be shown by trying the example on a class. CHAPTER V SERIES 5.1 Positive Term Series. ‘fle will not attempt, in this chapter, to give a development of series. The definitions and tests are usually given correctly in elementary calculus books and the advanced calculus gives quite a thorough treatment of the theoretical aspects. We shall limit ourselves to a few remarks on topics not usually discussed. The usual comparison test is not always easy for the student to apply. The following variation of it can be used in most cases. Let U4 + uz + u,+- ...fu,+... be a positive term series whose convergence is in question. Let a,-+ az-+ a3+- ...+am f... be a positive term series known to converge (or diverge). If 313510 11:; =K>O, an then Zum converges (diverges); if Lim 11.21... = 0 and 4‘3” 34. Zaa converges, then Eu,” does also; and if u»: _, [2:12. 3:... and 2a,, diverges, then Zu," does also. For if Lim 11.5.. = x then there is an N such that m“ a,” u’” éK+l forn>N, an 98 99 and Since :8", is convergent then Z(K 1" l) a,,, is also, and Zn," is convergent by the comparison test. Also there exists an N such that Ez—zK-d>o , where O0 . forn>N 9 fl and 100 Therefore since is... diverges then E u... diverges. 5.2 Findiag the nth Term.g§ a Series. Lany books make the statement that by inspection of the first few terms of a series we can find the nth term of the series. This is not always true. For example suppose .. ...I __l. 11,—1, 1.12-‘2', 11"": ’00.... and we wish to find the n th term. We will show that a polynomial in jét', such as J (47) f(-,;’fi = b,+ b860,?) + b3(-ng-)’+ b,,(-;,L,) . will serve as an nth term if the b's are properly determined. In fact there are infinitely many such polynomials. ‘He must have b,+b,+b,+b,,=l for n=1, (48) b, +-%‘+-?f+-%L=2’- for n=2 , b, +-%‘+-$"+‘2é7£=‘é' for n=5 . Solving equations (48) for b,,bz,b5 in terms of by we get 101 Substituting these values in equation (48) we have I _ I I z a (49) f(-;,;):: —§L+(l+b4)(7;) - ___ilgb (-,,-.-) + bat-5;) . If in equation (49) we let b,,="-0 9 then f(‘A-g) =7];- 3 which is the nth term that we would expect by inspection of the first few terms. dowever,‘bq may take any value, and therefore we have an infinite number of polynomials any one of which would be satisfactory as an nth term of this series. It is easily seen that if any finite number of terms were given we could carry through a similar discussion. 5.5 Power Series. We can make the statement that every power series defines a function in its region of convergence. To some of these functions we have given names. On the other hand, we cannot say that every function can be expanded in a power series. Consider a function, f(x), which exists and has its first (n-tl) derivatives in the neighborhood of x==0. Then we may write * f(x) = f(O) + f'(O)x + filigié‘+...+f”5.g.}3‘:+a.., O * F. S. Woods, Advanced Calculus, p. 10 . 102 X." I Emil! J i : Now if f(x) possesses all its derivatives the above formula for f(x) may be extended indefinitely. If at the same time aim, IR”! =0 then we have a convergent infinite series representing f(x). For example, consider Kr x2 I 3 )k .... x + .. + where m»! {”18 RZK*I -— ( '1) m 003 f 0 Then tray RZMH":L§ I (2k+57! ’ and, whatever the value of [x] , we have the Lam szfl=0 . Hence we have found an infinite series which represents sin x for all values of x. In order for the above statement to be true it is necessary that Ingram [R‘ =0 . If Lim IR) 4: 0 for x==a then the series may converge but will not represent f(x) for x==a . Consider I f(x) = e79 ., for x=i=0 , :0 , forx=oo 105 This function is continuous and has derivatives of all orders for all values of x. This is evident except at x=0. For x20 we have __I f'(0):Lim 2373—1: Lim £17.: 0 , h=0 h=o 2e” _ l -2971} _- f"(0)= Lim hi ___ Lim -2e , h=o h h=o h = Lim H :: Lim ___..2 q -.-: O , 11:0 3%, q=¢ eq where q: i? . For f”(0) we have a finite number of terms of the type L im 32— o q=ca 6‘1 Therefore f“(0) = 0 for all values of n . The series a at f(O) + mania-.... +L§21£+ .00... obviously converges for all values of x but does not represent the function except at x=0 . In fact, if _ I x:\:0, R = e? for all values of n. 4!. 104 I Likewise let us consider f(x) ='- sin x + e-fi. . Then 1’: - L ——x— ‘0‘. 8; x R 0 If fiflfid is left off and k allowed to become infinite then we get a series of the type f(0) + f'(0)x+ f"(0)xz+ ......+f”(0)x“+.... , 2! n! a 5 x zx+l : -X x --.... -1 x .0000 x 5T +6? +( )(2k+1)l + ’ which converges for all values of x but represents I f(x) = sin x +- e. I. at no value of x,except at x.==0. Therefore we see, that to expand f(x) in a MacLaurin's or Taylor's expansion, it is necessary to do more than just show that f(a) + f'(a)(x - a)+ ....-0-f‘”(9‘)1f1"C " a)“ .... converges. BIBLIOGRAPHY I Elementary Calculus Books Dalaker, H.H. and Hartig, H.E., The Calculus, New York, McGraw-Hill Book Company, 195 6. a Granville, U.A., Smith, P.F., and Longley, W.R., Elements of the Differential and Integgal Calculus, New York, Ginn and Company, I954 . Love, G.E., Differential and Integral Calculus, New York, The Macmillan Company, I957 . McKelvey, J.V., Calculus, ‘New Yerk,'fhe MacMillan Company, 1957 . Neelley, J.H. and.Tracy, J.I., Differential and Into ral Calculus, New Ybrk, The MacMillanfiCompany, I952 . Smith, E.S., Salkover, M., and Justice, H;K., Calculus, New Yerk, John Wiley and Sons, Inc., 1958 . Slobin, H.L., and Solt, M.R., A First Course in Calculus, New Yerk, Farrar and Rinehar E Ihc., 1935* . II Advanced Calculus Books and Books on Analysis Caratheodory, C., Vorlesungen uber Reelle Funktionen, Leipzig und Berlin, VerIag und Druék von B.G.Teubner, 1927 . Fine, H.B., Calculus, New York; The MacMillan Company, 1927 o ' Fits, W.B., Advanced Calculus, New Yerkb The MacMillan Company, I938 . Osgood, W.F., Functions of Real Variable, New Ybrk, G.E. Stechert and Company, 1 . 105 106 Pierpont, J., The Theor of Functions of Real Variables, Vol. I, New org, £55 and Companyj'IGUS . Pierpont, J., Functions of'g Complex Variable, New Yerk, Ginn and company, 1914 . Townsend, E.J., Functions gg Real Variables, New Yerk, Henry Holt and’Company, 192 . Wilson, E.B., Advanced Calculus, New York, Ginn and Company, 1912 . Woods, F.S., Advanced Calculus, New Ybrk, Ginn and Company, 1954 . cl v fish .o ' o "I WI'. ‘ ,\§\N zfiafixcgquhw MICHIGAN STATE UNIV RSITY LIB E RARIES 293 3056 7584 1 3