Lecture 10 Limits (cont d) One-sided its (Relevant section from Stewart, Sevent Edition: Section 2.4, pp. 113.) As you may recall from your earlier course in Calculus, we may define one-sided its, were te independent variable is restricted to approac te point a from eiter te rigt (i.e., > a) or from te left (i.e., < a). Left-sided or Left-and it: Te statement f() = L, means tat for any ǫ > 0, tere eists a δ > 0 suc tat a d < < a implies tat f() L < ǫ. Rigt-sided or Rigt-and it: Te statement f() = L, + means tat for any ǫ > 0, tere eists a δ > 0 suc tat a < < a + δ implies tat f() L < ǫ. And for te it to eist at = a, bot one-sided its must be equal, i.e., f() = f() = f() = L. + If te one-sided its are not equal, ten te it f() does not eist. Eample: Let s return to te Heaviside function, defined in an earlier lecture: 1, 0, H() = 0, < 0. (1) We must consider tree cases: 67
y y = H() 1 0 Grap of Heaviside function H(). 1. a > 0: In tis case, H() = 1 = 1. (2) Te it eists at all a > 0. No matter ow close te point a is to 0, we can always approac it from bot sides. 2. a < 0: In tis case, H() = 0 = 0. (3) Te it eists at all a < 0. Once again, no matter ow close te point a is to 0, we can always approac it from bot sides. 3. a = 0: In tis case, we must consider rigt- and left-sided its: 0 0 0 H() = 1 = 1. (4) + + 0 Since te two one-sided its are not equal, 0 does not eist. H() = 0 = 0. (5) 68
Infinite its (Relevant section from Stewart, Seventy Edition: Section 2.4, pp. 115-116.) You may recall tat we may treat infinity (or ) as a it point: If a positive function f() beaves as follows, ten we may write f() as a + andf() as a, (6) f() =. (7) An eample is te function f() = 1 2. As 0 from eiter side, f(), as te following sketc of its grap sows. y y = 1 2 0 In contrast, te function f() = 1 does not ave a it as 0, since f() as 0 and f() as 0 +. Te question tat remains is, Can we translate te statement in (7) into an ǫ,δ-type statement used for finite its? Te answer is Yes, but wit a little care. Recall tat for finite its, we understand tat te statement, means te following: Given an ǫ > 0, tere eists a δ > 0, suc tat f() = L, (8) 0 < a < δ implies tat f() L < ǫ. (9) Te problem we now face wit infinite its is tat L is. Does it make sense to write te inequality, f() < ǫ? (10) 69
Te answer is No : If we interpret te LHS as te distance between f() and, ten if f() is finite, tis distance is infinite. And no matter ow muc greater we made f(), e.g., multiplying it by 10 243, te distance is still infinite. So ow can we define closeness to infinity in a reasonable way, if at all? Te answer is to consider 1 te reciprocal of f(): If f() is large, ten is small. If we identify 0 as being te reciprocal of f(), ten for positive functions, 1 f() 0 = 1 f() = 1 (11) f() may be viewed as te closeness of f() to. An ǫ, δ definition for (7) could ten be prased as follows: Given any ǫ > 0, tere eists a δ > 0 suc tat 0 < a < δ implies tat 1 < ǫ. (12) f() But tis is peraps a little cumbersome: We can rewrite te inequality on te rigt by taking reciprocals, 1 f() < ǫ implies f() > 1 ǫ. (13) And as ǫ 0 +, 1 ǫ. Instead of working wit te reciprocal of ǫ, we define M = 1 ǫ te relation in (12) as follows: and re-epress f() = means tat given an M > 0, tere eists a δ > 0, suc tat 0 < a < δ implies tat f() > M. (14) Te idea is tat we now let M become arbitrarily large. Tis means tat f() can be made arbitrarily large by taking sufficiently close to a. (Note tat we do not want to write tat f() > M. Tis could allow f() to approac on one side of a and on te oter side, wic is contrary to te idea of f itself being arbitrarily large and positive.) A grapical sketc is given below. 70
y = f() M a δ a a + δ A simple eample of ow δ may be determined in terms of M is given in Eample 5 of Stewart s tet, p. 116. Here, f() = 1 2 and we find tat (not surprisingly) δ = M 1/2. A similar definition for its to may be developed in te same way, replacing M > 0 wit N < 0, wit te idea tat N will become arbitrarily large. Continuity of functions (Relevant section from Stewart, Sevent Edition: Section 2.5) We ave now discussed, in some detail, te concept of te it of a function, i.e., f() = L. (15) Tis statement implies, very rougly, tat as approaces te value a te value f() approaces te value L. (As we ve seen, te ǫδ definition is a precise formulation of tis idea.) y y = f() L a Te function f possesses a it at a: f() = L Of course, tis says absolutely noting about wat f does at = a. Te function f may or may 71
not even be defined at = a, i.e., f(a) may or may not even eist. As far as we re concerned, te above it statement leaves a ole in te grap of f. As you already know, if f is nice and continuous at = a, ten te it L is precisely te value f(a). Formally, ten Definition: A function f is continuous at a if f() = f(a). (16) Note tat tere are tree pieces of information in te above equation, namely, 1. f(a) is defined (i.e., te function f is defined at a), 2. f() eists (i.e., te it L of f eists), 3. Te it L is equal to f(a). As suc, te ole in te above grap is now filled in, as sketced below. You ve probably eard or read tis before: If f is continuous at all in some interval [c,d], ten te grap of f over tis interval can be drawn witout taking your pen/pencil off te paper. y y = f() f(a) a Te function f is continuous at a: f() = f(a) A more formal ǫδ definition of continuity would be as follows: Definition: A function f is continuous at a if, given any ǫ > 0, tere eists a δ > 0, suc tat f() f(a) < ǫ for all suc tat 0 < a < δ. (17) In oter words, we ave replaced L in te ǫδ it definition wit f(a). 72
Anoter way to write Eq. (16) is as follows, f(a + ) = f(a). (18) 0 We ave simply let = a+ ere. Tis formulation will be convenient in our discussion of derivatives. An important consequence from te definition of continuity is te following: If f is continuous at a, ten for close to a, f() is close to f(a). Tis sounds rater simple, but it is a profound concept tat is te basis of many metods in bot teory and applications. 73
Lecture 11 Continuous functions (cont d) One-sided continuity Because tere are one-sided its, we can define one-sided continuity : 1. A function f is left-continuous or continuous from te left at a if f() = f(a). (19) 2. A function f is rigt-continuous or continuous from te rigt at a if f() = f(a). (20) + Of course, for te it in Eq. (16) to eist, i.e., for f to be continuous at a, te two one-sided its must be equal: f is continuous at a if and only if f() = f() = f(a). + Eample: Te Heaviside function discussed earlier, 1, 0, H() = 0, < 0. (21) Once again, we must consider tree cases: y y = H() 1 0 Grap of Heaviside function H(). 74
1. a > 0: In tis case, H() = 1 = 1 = H(a). (22) Terefore H() is continuous at all a > 0. We can also write tat H() is continuous at all > 0. 2. a < 0: In tis case, H() = Terefore H() is continuous at all < 0. 3. a = 0: In tis case, we must consider rigt- and left-sided its: 0 = 0 = H(a). (23) 0 0 0 H() = 1 = 1 = H(0). (24) + + 0 H() = 0 = 0 H(0). (25) Terefore H() is not continuous at = 0. However, it is rigt continuous at = 0. In summary: Te Heaviside function H() is continuous at all 0 and rigt-continuous at = 0. Eac of te results of te following Teorem (Teorem 4 from Stewart, p. 122) follows from te appropriate Limit Law from a previous lecture. Teorem: If f and g are continuous at a, and c R is a constant, ten te following functions are continuous at a: 1. f ± g 2. cf 3. fg 4. f, provided tat g(a) 0. g For eample, to prove te first result, [f() ± g()] = f() ± g() = f(a) ± g(a). (26) 75
We are now in a position to determine te continuity properties of some important functions. For eample, from te it relation, it follows from te Limit Law for Products tat = a, (27) ( )( ) 2 = We can continue tis procedure to obtain te result = a 2. (28) n = a n. (29) In oter words, te functions f() = n are continuous for n = 0,1,2,.... (Te case n = 0 is trivial.) From te above Teorem, it follows tat any linear combination of tese functions is also continuous. In oter words, Any polynomial, P() = c 0 + c 1 + + c n n, (30) were te c i are constants and n 0, is continuous at any R. From te above Teorem, it also follows tat any ratio of polynomials, i.e., R() = P() Q() = c 0 + c 1 + + c n n d 0 + d 1 + + d m m, (31) for m,n 0, is continuous at any for wic Q() 0. Tis is te same as stating tat te rational function R() is continuous at all in its domain of definition. We can continue to build functions in tis way to produce a list of functions tat are continuous on teir domain of definition (see Teorem 7 of Stewart, p. 124). In addition, we would epect tat if a function were continuous and one-to-one, ten its inverse function would also be a continuous function. As a result, te following functions are continuous on teir domains of definition: 1. polynomials 2. rational functions 3. root functions (e.g., ) 76
4. trigonometric functions 5. inverse trigonometric functions 6. eponential functions 7. logaritmic functions Composition of functions We now consider te continuity properties of composite functions, f g() = f(g()). Te following important result may be found in Stewart, Teorem 8, p. 125: Teorem: If g() = b and f is continuous at b, ten f(g()) = f(b). In oter words, ( ) f(g()) = f g(). (32) An ǫδ proof of tis teorem is given in Appendi F of Stewart s tet. We can view tis result grapically as follows: By letting approac b, we can make te value of g() approac b. (But we y y b. g() f(b)... f(t) a b g() t don t even consider = b.) As g() approaces b, f(g()) approaces f(b). And because f is assumed to be continuous, te it of te latter process is f(b). Eample: Let f() = and g() = 1 + sin. We ave tat ( g() = 1 + sin ) = 1 + 1 = 2. (33) 0 0 77
sin (In a few lectures, we ll prove tat = 1.) Note tat g(0) is not defined. Te function 0 f() = is continuous at all > 0, in particular, = 2. From te above teorem, 1 + sin ( 0 = 1 + sin ) = 2. (34) 0 In te following Teorem, we make te additional assumption tat g() = g(a), i.e., g is continuous at a. Teorem: If g is continuous at a and f is continuous at g(a), ten te composite function f g is continuous at a, i.e., (f g)() = (f g)(a). (35) In oter words, a continuous function of a continuous function is a continuous function. Te proof of tis result is quite straigtforward, following from te previous Teorem. ( ) f(g()) = f g() = f(g(a)). (36) 78
Continuity of functions (cont d) Intermediate Value Teorem We start wit a definition: Definition: A function f is said to be continuous over an interval I if it is continuous at all I. In te case tat I = [a,b], it is understood tat f is continuous at all (a,b), rigt-continuous at a and left-continuous at b. Continuous functions on closed intervals [a, b] eibit some special properties, one of wic we now discuss. A possible case is sketced below. Note tat we consider te case tat f(a) f(b). Of course, te grap of f is a continuous curve tat starts at te point (a,f(a)) and ends at te point (b,f(b)). As suc, if we consider te line y = N, were N is any value tat lies between f(a) and f(b), te grap of f will ave to cross tis line at least once. y f(b) y = f() N f(a) a c b We now state tis property more formally as a teorem. Intermediate Value Teorem: Suppose tat f is continuous on [a,b], wit f(a) f(b). Let N be any real number between f(a) and f(b). (In oter words, N f(a) and N f(b).) Ten tere eists a number c (a,b) suc tat f(c) = N. Notes: 1. Te number N is an intermediate value, i.e., it lies between f(a) and f(b). 2. Te prase tere eists a number c could be read as tere eists at least one number c. Tere could be several distinct numbers c 1,c 2,,c n at wic f assumes te value N. 79
3. Te teorem does not tell you were c is located. It is only an eistence teorem. 4. Tis teorem relies on te assumption tat f is continuous on [a, b]. If it were not continuous, it could ave jumps tat would allow te grap to avoid values between f(a) and f(b). 5. In te case tat f(a) = f(b), no suc c value need eist. Can you sketc an eample? You may well encounter a proof of te IVT in a course on real analysis, e.g., AMATH 331. Application of IVT: Estimating zeros of functions Once again, assume tat f is a continuous function over te closed interval [a,b]. Now assume tat f(a) and f(b) are nonzero and ave opposite signs, i.e., f(a)f(b) < 0. Tis means tat N = 0 is an intermediate value. Te IVT guarantees te eistence of at least one number c (a,b) at wic f(c) = 0, i.e., c is a zero of te function f. Tis is sketced below. y f(b) y = f() f(a) 0 a c b Many applications rely on te numerical estimation of zeros of functions. Tere are various powerful metods to accomplis tis estimation, including te Newton-Rapson metod, to be discussed later. For te moment, owever, we can see te use of te IVT in narrowing down te location of te zeros of a function. If we sample a continuous function at various -values, call tem n, and if it canges sign at two consecutive -values, say K and K+1, ten a zero of f is located in te interval ( K, K+1 ). Eample: Consider te function f() = 4 3 6 2 + 3 2. Note tat f(0) = 2, f(1) = 1, f(2) = 12. (37) 80
We may conclude tat a zero of f we ll call it lies in te open interval (1,2), i.e., 1 < < 2. Of course, we may not be content wit tis observation, being interested in a more accurate estimate of. In oter words, we require a more refined searc. One way would be to simply divide te interval [1, 2] into n subintervals of lengt 1/n and ten evaluating f() at te partition points. If we do tis for n = 10, we find tat a sign cange occurs ere: f(1.2) = 0.128, f(1.3) = 0.548. (38) As a result, we now know tat 1.2 < < 1.3. And if we now divide [1.2,1.3] into n = 10 subintervals, we find tat a sign cange occurs ere: f(1.22) = 0.007008, f(1.23) = 0.056068. (39) We now know tat 1.22 < 1.23. And te procedure can be continued until we ave estimated to a desired degree of accuracy. Te above metod is rater inefficient: Eac subdivision will increase te accuracy by possibly one digit but we may ave to searc over many subintervals in order to locate te one over wic tere is a sign cange. A more efficient metod is obtained if we perform a binary searc over subintervals. Tis is te basis of te bisection metod: If we start wit an interval [a,b] tat is known to contain a zero of a function f, we ten split te interval in two and ten ceck eac subinterval. We ten take te one at wic tere is a sign cange and repeat te procedure, splitting it in two, etc.. Tis iterative algoritm is outlined below: Bisection metod for estimating zeros of a continuous function: We start wit an interval [a 1,b 1 ] for wic f(a 1 )f(b 1 ) < 0, implying tat (a 1,c 1 ). Tis algoritm will produce a set of subintervals [a n,b n ] in te following way: At eac step n, 1. Compute te midpoint of te interval [a n,b n ], i.e., c = 1 2 (a n+b n ). Ceck if f(c) = 0 to te desired accuracy, i.e., to m digits of accuracy. If so, ten you can stop. You may also wis to stop if b n a n < ǫ, were ǫ represents some desired accuracy. Oterwise: 2. (a) If f(a n )f(c) < 0, set a n+1 = a n and b n+1 = c. (b) If f(c)f(b n ) < 0, set a n+1 = c and b n+1 = b n. Ten replace n wit n + 1 and go to Step No. 1. 81
We illustrate te bisection metod wit te function f() = 2 2 as a way of estimating 2, one of its zeros. We ll start wit te interval [a 1,b 1 ] = [1,2], noting tat f(1) = 1 and f(2) = 2. Te results of te first 20 iterations are listed in te table below. n a n b n 1 1.0000000 2.0000000 2 1.0000000 1.5000000 3 1.2500000 1.5000000 4 1.3750000 1.5000000 5 1.3750000 1.4375000 6 1.4062500 1.4375000 7 1.4062500 1.4218750 8 1.4140625 1.4218750 9 1.4140625 1.4179688 10 1.4140625 1.4160156 11 1.4140625 1.4150391 12 1.4140625 1.4145508 13 1.4140625 1.4143066 14 1.4141846 1.4143066 15 1.4141846 1.4142456 16 1.4141846 1.4142151 17 1.4141998 1.4142151 18 1.4142075 1.4142151 19 1.4142113 1.4142151 20 1.4142132 1.4142151 21 1.4142132 1.4142141 22 1.4142132 1.4142137 23 1.4142134 1.4142137 Results of bisection metod applied to f() = 2 2 to estimate te zero 2. Te actual value of 2, to 7 digits, is 1.4142135. 82
Lecture 12 Limits at infinity Note: Tis section was accidentally omitted from te Friday, October 5 lecture, but will be included ere. It will be covered at te beginning of Lecture 13 on Wednesday, October 10 (Relevant section from Stewart, Sevent Edition: Section 2.6) We are concerned wit te beaviour of a function f() in two particular cases: (i) and (ii). For te moment, we sall consider te particular case. It is certainly possible tat a function f beaves in te following way: f() = L. (40) In oter words, as > 0 gets larger and larger, te values f() gets closer and closer to L. Tis implies tat te line y = L is a orizontal asymptote of te grap of f. A couple of possibilities are sketced below. y y y = f() y = f() L L f() = L A sligtly more matematical description of tis iting beaviour is tat f() approaces te it L arbitrarily closely for sufficiently large. An ǫδ-type definition is as follows: Given any ǫ > 0, tere eists an M > 0 suc tat f() L < ǫ for all suc tat > M. (41) 83
Note tat our δ is replaced by M, wit te idea tat M. Recall tat te closeness of a number to infinity is measured by ow large it is. Tis idea is illustrated in te figure below. y L ǫ L L + ǫ M Eample: Te function f() = 1. As, it seems clear tat 1 In oter words, we conjecture tat is getting smaller and smaller. 1 = 0. (42) Let us prove tis wit te above ǫ definition. Wit an eye to (41), given an ǫ > 0, can we find an M > 0 suc tat f() L = 1 0 Keeping in mind tat > 0, we ave tat < ǫ for all suc tat > M? (43) 1 < ǫ implies tat > 1 ǫ. (44) Terefore, we can coose M = 1. (As ǫ 0, M.) We ave terefore proved Eq. (42). ǫ In te same way, we can sow tat 1 = 0 for any r > 0. (45) r Question: For a given r, wat is te largest M-value tat corresponds to an ǫ > 0? Rational functions Consider te rational function, f() = 52 + 2 2 2 + 4 + 3. (46) 84
We ask te question, How does f() beave as? Stepping back for a moment, we simply eamine te numerator and denominator: 1. Numerator: For very large, e.g., = 10 20, te term 5 2 will dominate, i.e., te term 2 will be insignificant. 2. Denominator: For very large, e.g., = 10 20, te term 2 2 will dominate, i.e., 10 40 vs. 10 20 and 3. From tese observations, we suspect tat f() 52 2 2 = 5 2 for very large. (47) In oter words, we conjecture tat f() = 5 2. (48) Tis is fine, but can we make tis argument more matematically rigorous, in terms of it laws? Te answer is Yes. Let s divide te numerator and denominator of f() by 2, Ten we can write tat f() = 52 + 2 2 2 + 4 + 3 = 5 + 3 2 2 + 4 + 3 2. (49) f() = = = = 5 2. 5 2 + 2 2 2 + 4 + 3 5 + 3 2 2 + 4 + 3 2 (5 + 3 2 ) (2 + 4 + 3 2 ) (by it law for quotients bot its eist) (50) Infinite its at infinity It is certainly possible tat functions become arbitrarily large in magnitude as gets very large. A simple eample is f() =. Recalling tat we can consider te value or as a it, we can write tat As well, we can write tat =. (51) 2 =. (52) 85
Can we define te statement, f() =, (53) more precisely, in an ǫδ way? In tis statement, f() is getting closer to as gets closer to. Recalling tat closeness of someting to is defined by ow large someting is. We apply tis idea to bot f() and. As a result, we state tat Eq. (53) means te following: Given any N > 0, tere eists an M > 0 suc tat f() > N for all > M. (54) In te particular case, f() =, we ave tat M = N. So wat about te function f() = 2? Wat is Can we do te following operation, invoking it laws? (2 )? (55) Te question is: Wat is? (2 ) = 2 =. (56) Te answer is tat is meaningless. Te above procedure is incorrect. Te use of it laws is incorrect wen we obtain meaningless epressions suc as te above, or 0. Wat we can do is te following: (2 ) = ( 1) =. (57) If is large, ten 1 is also large, implying tat te product ( 1) is large. Here is anoter eample, taken from Stewart s tetbook. Does te following it eist: First of all, we cannot write tat ( 2 + 1 )? (58) ( 2 + 1 ) = 2 + 1 =, (59) to for reasons mentioned earlier: is meaningless. 86
If we do te following, 2 + 1 = ( 2 + 1 ) 2 + 1 + 2 + 1 + = (2 + 1) 2 2 + 1 + = 1 2 + 1 +. (60) As, it looks like te final epression goes to zero. We ll divide numerator and denominator by : We can now write tat 1 1 2 + 1 + =. (61) 1 + 1 + 1 2 ( 2 + 1 ) = = = 1 2 + 1 + 1 1 + 1 2 + 1 1 ( 1 + 1 + 1) 2 (it law for quotients) = 0 2 = 0. (62) Finally, te case can be treated in a similar way but we must be careful about te signs of terms. For eample is negative, 2 is positive, 3 is negative, etc.. Please read Section 2.6 of Stewart s tetbook for more details. 87
Derivatives and rates of cange Note: Because muc of tis material is review, it was delivered by means of te data projector, wic eplains te larger-tan-normal number of pages covered. (Relevant section from Stewart, Sevent Edition: Section 2.7) We ave arrived at anoter important concept in tis course te derivative, wit wic you are already familiar. Only a few minutes were available at te end of tis lecture period, so te discussion was quite brief. We sall return to some of tese ideas in more detail in te net lecture. Te following discussion will refer to te figure below. y = f() f() Q y f(a) P tangent line at = a a In tis discussion, we consider te point a as a reference point. For a given a, consider te points P at (a,f(a)) and Q at (,f()). Te slope of te line segment PQ is given by m PQ = f() f(a) a = y. (63) Te slope of te tangent line to te curve y = f() at = a is given by te following it (provided tat it eists), m = m y PQ = Q P 0 = f() f(a). (64) a Quite often in practice, we let a represent te displacement from a and define = a, ence = a +, (65) so tat Eq. (64) becomes f(a + ) f(a) m =, (66) 0 were it is understood tat can be positive or negative (but not zero). 88
As you know, m represents te instantaneous rate of cange of f wit respect to and is called te derivative of f at a and denoted as Te important quantity, f f(a + ) f(a) (a) =. (67) 0 f(a + ) f(a), (68) is known as te Newton quotient of f at a. (It is also called te difference quotient.) An important application to Pysics: In fact, it was tis problem tat spawned te development of te Calculus. Te following is simply a translation of te previous discussion into one involving pysical quantities. Suppose tat a particle is moving in one dimension, e.g., along te -ais, and tat its position as a function of time is denoted by te function (t), as plotted below. y y = (t) (t 2 ) (t 1 ) tangent line at t = t 1 slope = (t 1 ) t 1 t 2 t t Now consider a time interval [t 1,t 2 ] for t 1 < t 2. Te quotient, (t 2 ) (t 1 ) t 2 t 1 = t, (69) defines te average velocity of te particle over te time interval [t 1,t 2 ]. We now let t 2 approac t 1, in an effort to get a better estimate of te rate of cange of position wit respect to time of te particle at t = t 1. In te it, provided tat it eists, we ave (t 2 ) (t 1 ) = t 2 t 1 t 2 t 1 t 0 t. (70) Once again, we may let t 2 = t 1 +. Te instantaneous rate of cange of te position wit respect to time of te particle at t 1 is given by (t 1 + ) (t 1 ) (t 1 ) =, (71) 0 89
te derivative of (t) wit respect to t at t 1. As you well know, tis is te velocity of te particle at time t 1, denoted as v(t 1 ). Historically, te idea of letting te time t 2 approac t 1 in an effort to describe te motion of te particle over smaller and smaller time represented a monumental jump in tougt wic led to te development of Calculus. Some more comments on tis procedure will be given in tis week s tutorials. We now return to do some matematics. Te derivative as a function Given a function f(), we now define its derivative at a general point (as opposed to an ancor point a): f f( + ) f() () =, (72) 0 provided tat te it eists. Te derivative is now a function of, i.e., we could write, g() = f (), (73) to empasize tat it is a new function. Important note: In response to an ecellent question in class, it is wortwile to mention tat for f () to eist, te function f must be defined at, i.e., f() must eist! Oterwise, te Newton quotient of f at, i.e., would not be defined, since f() is not defined. f( + ) f() Let us now apply te Newton quotient metod to some simple functions. Eample 1: Te world s simplest function f() = C, were C is a constant. (You know te answer to tis, but let s do it anyway.) Te Newton quotient of f at an arbitrary is f( + ) f() = 1 1 = 0. (74) 90
As a result, f f( + ) f() () = = 0 = 0. (75) 0 0 Eample 2: Te world s net-to-simplest function f() =. (Yes, once again, you know te answer.) Te Newton quotient of f at an arbitrary is f( + ) f() = ( + ) = = 1. (76) As a result, f f( + ) f() () = = 1 = 1. (77) 0 0 Eample 3: Te world s net-to-net-to-simplest function f() = 2. Te Newton quotient of f at an arbitrary is f( + ) f() = ( + )2 2 = 2 + 2 = 2 +. (78) As a result, f () = f( + ) f() 0 = 0 (2 + ) = 0 2 + 0 = 2. (79) Eample 4: Te general function f() = n for n = 1,2,. Te Newton quotient of f at an arbitrary is f( + ) f() = ( + )n n = n + n n 1 + + n n 1 + n n = n n 1 + A 1 + A 2 2 + A n 1 n 1 (were te A k are constants). (80) 91
As a result, f () = f( + ) f() 0 = 0 (n n 1 + A 1 + A 2 2 + A n 1 n 1 ) = n n 1. (81) Eample 5: Te absolute value function f() =. We must go back to its fundamental definition,, 0 = (82), < 0. Because of te special nature of te point = 0, we ll ave to consider tree cases: 1. Case 1: > 0. In tis case, te Newton quotient of f is (for sufficiently small, i.e., < ), f( + ) f() = ( + ) = = 1. (83) As a result, f f( + ) f() () = = 1 = 1, (84) 0 0 as epected. (We could ave simply used our result for f() = from before.) 2. Case 2: < 0. In tis case, te Newton quotient of f is (for sufficiently small, i.e., < ), f( + ) f() = ( + ) + = = 1. (85) As a result, f f( + ) f() () = = 1 = 1. (86) 0 0 3. Case 3: = 0. In tis case, te Newton quotient of f at 0 is f() f(0) =. (87) We ave eamined tis function before, but let s go troug te details again. We ave to consider two cases, i.e., te rigt- and left-sided its as 0: (a) > 0: 0 + = 0 + = 1 = 1. 0 + (b) < 0: 0 = 0 = 1 = 1. 0 92
Since te two its are not equal, it follows tat f (0) does not eist. In summary, te derivative of te absolute value function is given by 1, > 0 f () = 1, < 0 undefined, = 0. (88) We may represent tis result grapically as follows, y y = f () Eample 6: Te Heaviside function H() eamined earlier: 1, 0 H() = 0, < 0. (89) Once again, we ll ave to consider tree cases: 1. > 0: Here H() = 1. We can simply state te result tat H () = 0 from Eample 1, were C = 1. (Or we could ave used te Newton quotient it is always zero, implying tat H () = 0.) 2. < 0: Here H () = 0. Again, from Eample 1, it follows tat H () = 0. 3. = 0: Before going on to eamine te Newton quotient, let s step back and consider if te following procedure is valid: We ave found tat H () = 0 for > 0 and for < 0. Can we conclude tat H (0) = 0 H () = 0 + H () = 0? (90) 93
Te answer is NO!!!!! Just because te two one-sided its eist and are equal to eac oter, implying te eistence of te it, we may not conclude tat H (0) = 0. Tat would imply tat te function H () is continuous at = 0, and we ave no guarantee tat tis is true. To determine weter or not H (0) eists, we must eamine te Newton quotient at = 0, i.e., Case 1: (rigt-sided it) For > 0, H(0 + ) H(0) = H() 1. (91) Case 2: (left-sided it) For < 0, H(0 + ) H(0) H() 1 0 = = = 0. (92) 0 + 0 + 0 + H(0 + ) H(0) H() 1 1 = = = (since < 0). (93) 0 0 0 Before we go on, let s try to understand tese results grapically: For > 0, te slope of te line segment PQ is always zero, implying tat te rigt-sided it of te Newton quotient is zero. But for < 0, te slope of te line segment PQ is 1 wic goes to as 0. P y Q y = H() Q Because te two one-sided its are not equal, we conclude tat H (0) does not eist. In summary, we ave A plot of H () is given below. H 0, 0 () = undefined, = 0. (94) Eample 7: At tis point it is instructive to eamine te function f() = for R. Because of te presence of te absolute value function, we may suspect tat te point = 0 is a bad point. 94
y y = H () From te definition of te absolute value function, we ave 2, > 0 f() = 2, < 0 0, = 0. (95) Te grap of f() is sketced below. y y = f () For 0, we may use te piecewise-defined formulas of f() to compute f (): f () = 2 wen > 0 and f () = 2 wen < 0. It remains to see weter f (0) eists. Once again, we must eamine te Newton quotient at = 0, f(0 + ) f(0) = f(). (96) Case 1: (rigt-sided it) For > 0, Case 2: (left-sided it) For < 0, f(0 + ) f(0) f() = 0 + 0 + = 2 0 + = = 0. (97) 0 + f(0 + ) f(0) f() = 0 0 = 2 0 = = 0. (98) 0 ( ) 95
Since bot its are equal to zero, it follows tat f (0) = 0. We may terefore summarize our results as follows, f () = 2, > 0 2, < 0 0, = 0. (99) Te grap of f () is sketced below. y 4 y = f () 2-1 0 1 You ll notice tat te grap of f () looks someting like tat of te absolute value function. In fact, from te piecewise definition of f (), it is not difficult to see tat it may be rewritten in terms of te absolute value function as follows, f () = 2. (100) Te moral of tis story is tat just because an absolute value term appears in a function, it doesn t necessarily mean tat te function is nondifferentiable. Te reader may wis to investigate te more general family of functions f() = n for n = 1,2,. As n increases, te function f() will ave even iger order derivatives, e.g., f (). Differentiability implies continuity We ave now arrived at an important concept involving differentiability of a function. Teorem: If f is differentiable at a, ten it is continuous at a. Let s rewrite tis teorem sligtly, empasizing te difference between ypotesis and conclusion : 96
Teorem: If Hypotesis: f is differentiable, ten Conclusion: f is continuous at a. Proof: From te ypotesis, i.e., tat f is differentiable at a, we ave tat We want to sow tat f f() f(a) (a) =. (101) a f() = f(a). (102) Tis implies tat f() approaces f(a), so we ll eamine te term f() f(a). We ll rewrite it as follows, f() f(a) = We now take its, using te Product Law: [f() f(a)] = = = f (a) 0 f() f(a) a [ f() f(a) a [ ] f() f(a) a ( a) for a. (103) ] ( a) ( a) (since bot its eist) = 0. (104) We re essentially done. It follows tat f() = [f(a) + f() f(a)] = f(a) + [f() f(a)] = f(a) + 0 = f(a). (105) Tis implies tat f is continuous at a and te teorem is proved. Recall tat te contrapositive to te statement A B is not B not A. (106) Te contrapositive to te above teorem is not C not H, i.e., 97
If f is not continuous at a, ten f is not differentiable at a. Tis eplains te fact tat te Heaviside function is not differentiable at = 0 it s not even continuous tere. Wat about te reverse of te teorem: If f is continuous at a, is it differentiable at a? Te answer is NO te absolute value function f() = illustrates tat tis statement is not true. Differentiability of a function is a stronger requirement tan continuity Te fact tat differentiability implies continuity, but not te oter way around, indicates tat differentiability is a stronger requirement. A differentiable function is generally smooter tan a continuous function. Just compare te graps of f() = 2 and g() = at te point = 0: f is differentiable tere but g is not. Some differentiation rules (Relevant section from Stewart, Sevent Edition: Sections 3.1-3.3) In te previous lecture, we derived te well known Power rule of differentiation: d d n = n n 1, n = 0,1,2,. (107) In a little wile, we ll sow tat tis rule etends to all real values of n. From te it definition of te derivative (i.e., via Newton quotient) and te Limit Laws, te following familiar rules for differentiation may be derived (for more details, see Stewart, Section 3.1): Constant multiple rule: Sum/difference rule: d d [cf()] = cf () for any constant c R d d [f() ± g()] = f () ± g () Tere are two oter well-known differentiation rules wic require a little more effort to prove: 98
Product rule: Quotient rule: d d [f()g()] = f ()g() + f()g () d d [ ] f() = f ()g() f()g () g() g() 2 Here we sall prove te Product Rule - you can find a proof of te Sum Rule in Stewart s tetbook. Proof of Product Rule: In wat follows, we assume tat f and g are differentiable at. First define p() = f()g(). (108) We ll ave to consider te numerator of te Newton quotient of p(), namely, It s convenient to define te following, so tat From Eq. (109), p( + ) p() = f( + )g( + ) f()g(). (109) f = f( + ) f(), g = g( + ) g(), (110) f( + ) = f() + f, g( + ) = g() + g. (111) p( + ) p() = f( + )g( + ) f()g() = [f() + f][g() + g] f()g() = f()g() + f() g + g() f + f g f()g() = f() g + g() f + f g. (112) Te Newton quotient for p ten becomes [ ] [ p( + ) p() g( + ) g() f( + ) f() = f() + g() [ f( + ) f() ] + ] [g( + ) g()]. (113) We now take te it of bot sides as 0. Since f() and g() are constants, we may write [ ] [ ] p( + ) p() g( + ) g() f( + ) f() = f() + g() + 0 0 0 [ ] f( + ) f() [g( + ) g()]. (114) 0 0 99
In te last term, te it of te product is te product of te its, since te its eist. Tese its become p () = f()g () + g()f () + f () 0. (115) Te zero in te last term follows from te fact tat g is continuous at, wic follows from te assumption tat g is differentiable at. Eq. (115) is identical to te product rule, i.e., [f()g()] = f ()g() + f()g (). (116) Eponential functions We now wis to find te derivatives of te eponential functions, defined as follows, f a () = a, a > 0. (117) We must consider te Newton quotient definition for te derivative of f a (), f a () = f a ( + ) f a () 0 a + a = 0 = a a 1. (118) 0 If we consider te special value = 0 in te above equation, ten a 0 = 1 and we ave As a result, we can rewrite Eq. (118) as follows, or f a(0) a 1 =. (119) 0 f a () = f a (0)a, (120) d d a = f a (0)a. (121) In oter words, te derivative of te function a is proportional to a. Te coefficient of proportionality, f a(0) is given by Eq. (119). A couple of values of tis proportionality constant are given in te table below. 100
a 1 a 0 2 0.693147 3 1.098612 It is a fact tat tis factor f a(0) increases wit a. From a look at te table, tere is a number a between 2 and 3 suc tat te proportionality constant is 1. Tis special value of a is called e. Definition: e is te number suc tat f e(0) e 1 = = 1. (122) 0 Consequently, from Eq. (121), d d e = e. (123) In oter words, e is its own derivative. Te slope of te grap of e is 1 at = 0. As increases, te slope of te grap increases. 101