CHAPTER IV DIFFERENTIATION, LOCAL BEHAVIOR

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1 CHAPTER IV DIFFERENTIATION, LOCAL BEHAVIOR e iπ = 1. I tis capter we will fially see wy e iπ is 1. Alog te way, we will give careful proofs of all te stadard teorems of Differetial Calculus, ad i te process we will discover all te familiar facts about te trigoometric ad expoetial fuctios. At tis poit, we oly ow teir defiitios as power series fuctios. Te fact tat si 2 + cos 2 = 1 or tat e x+y = e x e y are ot at all obvious. I fact, we ave t eve yet defied wat is meat by e x for a arbitrary umber x. Te mai teorems of tis capter iclude: (1) Te Cai Rule (Teorem 4.7), (2) Te Mea Value Teorem (Teorem 4.9), (3) Te Iverse Fuctio Teorem (Teorem 4.10), (4) Te Laws of Expoets (Corollary to Teorem 4.11 ad Exercise 4.20), ad (5) Taylor s Remaider Teorem (Teorem 4.19). THE LIMIT OF A FUNCTION Te cocept of te derivative of a fuctio is wat most people ti of as te begiig of calculus. However, before we ca eve defie te derivative we must itroduce a id of geeralizatio of te otio of cotiuity. Tat is, we must begi wit te defiitio of te limit of a fuctio. DEFINITION. Let f : S C be a fuctio, were S C, ad let c be a limit poit of S tat is ot ecessarily a elemet of S. We say tat f as a limit L as z approaces c, ad we write lim f(z) = L, z c if for every ɛ > 0 tere exists a δ > 0 suc tat if z S ad 0 < z c < δ, te f(z) L < ɛ. If te domai S is ubouded, we say tat f as a limit L as z approaces, ad we write L = lim z f(z), if for every ɛ > 0 tere exists a positive umber B suc tat if z S ad z B, te f(z) L < ɛ. Aalogously, if S R, we say lim x f(x) = L if for every ɛ > 0 tere exists a real umber B suc tat if x S ad x B, te f(x) L < ɛ. Ad we say tat lim x f(x) = L if for every ɛ > 0 tere exists a real umber B suc tat if x S ad x B, te f(x) L < ɛ. Fially, for f : (a, b) C a fuctio of a real variable, ad for c [a, b], we defie te oe-sided (left ad rigt) limits of f at c. We say tat f as a left ad limit of L at c, ad we write L = lim x c 0 f(x), if for every ɛ > 0 tere exists a δ > 0 suc tat if x (a, b) ad 0 < c x < δ te f(x) L < ɛ. We say tat f as a rigt ad limit of L at c, ad write L = lim x c+0 f(x), if for every ɛ > 0 tere exists a δ > 0 suc tat if x S ad 0 < x c < δ te f(x) L < ɛ. Te first few results about limits of fuctios are ot surprisig. Te aalogy betwee fuctios avig limits ad fuctios beig cotiuous is very close, so tat 85

2 86 IV. DIFFERENTIATION, LOCAL BEHAVIOR for every elemetary result about cotiuous fuctios tere will be a compaio result about limits of fuctios. THEOREM 4.1. Let c be a complex umber. Let f : S C ad g : S C be fuctios. Assume tat bot fad g ave limits as x approaces c. Te: (1) Tere exists a δ > 0 ad a positive umber M suc tat if z S ad 0 < z c < δ te f(z) < M. Tat is, if f as a limit as z approaces c, te f is bouded ear c. (2) lim(f(z) + g(z)) = lim f(z) + lim g(z). z c z c z c (3) (4) If lim z c g(z) 0, te lim(f(z)g(z)) = lim f(z) lim g(z). z c z c z c f(z) lim z c g(z) = lim z c f(z) lim z c g(z), (5) If u ad v are te real ad imagiary parts of a complex-valued fuctio f, te u ad v ave limits as z approaces c if ad oly if f as a limit as z approaces c. Ad, lim f(z) = lim u(z) + i lim v(z). z c z c z c Exercise 4.1. (a) Prove Teorem 4.1. HINT: Compare wit Teorem 3.2. (b) Prove tat lim x c f(x) = L if ad oly if, for every sequece {x } of elemets of S tat coverges to c, we ave lim f(x ) = L. HINT: Compare wit Teorem 3.4. (c) Prove te aalog of Teorem 4.1 replacig te limit as z approaces c by te limit as z approaces. Exercise 4.2. (a) Prove tat a fuctio f : S C is cotiuous at a poit c of S if ad oly if lim x c f(x) = f(c). HINT: Carefully write dow bot defiitios, ad observe tat tey are verbetim te same. (b) Let f be a fuctio wit domai S, ad let c be a limit poit of S tat is ot i S. Suppose g is a fuctio wit domai S {c}, tat f(x) = g(x) for all x S, ad tat g is cotiuous at c. Prove tat lim x c f(x) = g(c). Exercise 4.3. Prove tat te followig fuctios f ave te specified limits L at te give poits c. (a) f(x) = (x 3 8)/(x 2 4), c = 2, ad L = 3. (b) f(x) = (x 2 + 1)/(x 3 + 1), c = 1, ad L = 1. (c) f(x) = (x 8 1)/(x 6 + 1), c = i, ad L = 4/3. (d) f(x) = (si(x) + cos(x) exp(x))/(x 2 ), c = 0, ad L = 1. Exercise 4.4. Defie f o te set S of all ozero real umbers by f(x) = c if x < 0 ad f(x) = d if x > 0. Sow tat lim x 0 f(x) exists if ad oly if c = d.

3 IV. DIFFERENTIATION, LOCAL BEHAVIOR 87 (b) Let f : (a, b) C be a complex-valued fuctio o te ope iterval (a, b). Suppose c is a poit of (a, b). Prove tat lim x c f(x) exists if ad oly if te two oe-sided limits lim x c 0 f(x) ad lim x c+0 f(x) exist ad are equal. Exercise 4.5. (Cage of variable i a limit) Suppose f : S C is a fuctio, ad tat lim x c f(x) = L. Defie a fuctio g by g(y) = f(y + c). (a) Wat is te domai of g? (b) Sow tat 0 is a limit poit of te domai of g ad tat lim y 0 g(y) = lim x c f(x). (c) Suppose T C, tat : T S, ad tat lim y d (y) = c. Prove tat lim f((y)) = lim f(x) = L. y d x c REMARK. We we use te word iterior i coectio wit a set S, it is obviously importat to uderstad te cotext; i.e., is S beig tougt of as a set of real umbers or as a set of complex umbers. A poit c is i te iterior of a set S of complex umbers if te etire dis B ɛ (c) of radius ɛ aroud c is cotaied i S. Wile, a poit c belogs to te iterior of a set S of real umbers if te etire iterval (c ɛ, c + ɛ) is cotaied i S. Hece, i te followig defiitio, we will be careful to distiguis betwee te cases tat f is a fuctio of a real variable or is a fuctio of a complex variable. THE DERIVATIVE OF A FUNCTION Now begis wat is ordiarily tougt of as te first mai subject of calculus, te derivative. DEFINITION. Let S be a subset of R, let f : S C be a complex-valued fuctio (of a real variable), ad let c be a elemet of te iterior of S. We say tat f is differetiable at c if f(c + ) f(c) lim 0 exists. (Here, te umber is a real umber.) Aalogously, let S be a subset of C, let f : S C be a complex-valued fuctio (of a complex variable), ad let c be a elemet of te iterior of S. We say tat f is differetiable at c if f(c + ) f(c) lim 0 exists. (Here, te umber is a complex umber.) If f : S C is a fuctio eiter of a real variable or a complex variable, ad if S deotes te subset of S cosistig of te poits c were f is differetiable, we defie a fuctio f : S C by f f(x + ) f(x) (x) = lim. 0 Te fuctio f is called te derivative of f. A cotiuous fuctio f : [a, b] C tat is differetiable at eac poit x (a, b), ad wose te derivative f is cotiuous o (a, b), is called a smoot fuctio o

4 88 IV. DIFFERENTIATION, LOCAL BEHAVIOR [a, b]. If tere exists a partitio {a = x 0 < x 1 <... < x = b} of [a, b] suc tat f is smoot o eac subiterval [x i 1, x i ], te f is called piecewise smoot o [a, b]. Higer order derivatives are defied iductively. Tat is, f is te derivative of f, ad so o. We use te symbol f () for te t derivative of f. REMARK. I te defiitio of te derivative of a fuctio f, we are iterested i te limit, as approaces 0, ot of f but of te quotiet q() = f(c+) f(c). Notice tat 0 is ot i te domai of te fuctio q, but 0 is a limit poit of tat domai. Tis is te reaso wy we ad to mae suc a big deal above out of te limit of a fuctio. Te fuctio q is ofte called te differetial quotiet. REMARK. As metioed i Capter III, we are ofte iterested i solvig for uows tat are fuctios. Te most commo suc problem is to solve a differetial equatio. I suc a problem, tere is a uow fuctio for wic tere is some id of relatiosip betwee it ad its derivatives. Differetial equatios ca be extremely complicated, ad may are usolvable. However, we will ave to cosider certai relatively simple oes i tis capter, e.g., f = f, f = f, ad f = ±f. Tere are various equivalet ways to formulate te defiitio of differetiable, ad eac of tese ways as its advatages. Te ext teorem presets oe of tose alterative ways. THEOREM 4.2. Let c belog to te iterior of a set S (eiter i R or i C), ad let f : S C be a fuctio. Te te followig are equivalet. (1) f is differetiable at c. Tat is, (2) f(c + ) f(c) lim exists. 0 f(x) f(c) lim exists. x c x c (3) Tere exists a umber L ad a fuctio θ suc tat te followig two coditios old: (4.1) f(c + ) f(c) = L + θ() ad (4.2) lim 0 θ() = 0. I tis case, L is uique ad equals f (c), ad te fuctio θ is uique ad equals f(c + ) f(c) f (c). PROOF. Tat (1) ad (2) are equivalet follows from Exercise 4.5 by writig x as c +. Suppose ext tat f is differetiable at c, ad defie L = f f(c + ) f(c) (c) = lim. 0

5 Set Te clearly wic is Equatio (4.1). Also IV. DIFFERENTIATION, LOCAL BEHAVIOR 89 θ() = f(c + ) f(c) f (c). f(c + ) f(c) = L + θ(), θ() = f(c + ) f(c) f (c) f(c + ) f(c) = f (c), wic teds to 0 as approaces 0 because f is differetiable at c. Hece, we ave establised equatios (4.1) ad (4.2), sowig tat (1) implies (3). Fially, suppose tere is a umber L ad a fuctio θ satisfyig Equatios (4.1) ad (4.2). Te f(c + ) f(c) = L + θ(), wic coverges to L as approaces 0 by Equatio (4.2) ad part (2) of Teorem 4.1. Hece, L = f (c), ad so θ() = f(c+) f(c) f (c). Terefore, (3) implies (1), ad te teorem is proved. REMARK. Toug it seems artificial ad awward, Coditio (3) of tis teorem is very coveiet for may proofs. Oe sould remember it. Exercise 4.6. (a) Wat is te domai of te fuctio θ of coditio (3) i te precedig teorem? Is 0 i tis domai? Are tere ay poits i te iterior of tis domai? (b) Let L ad θ be as i part (3) of te precedig teorem. Prove tat, give a ɛ > 0 tere exists a δ > 0 suc tat if < δ te θ() < ɛ. THEOREM 4.3. If f : S C is a fuctio, eiter of a real variable or a complex variable, ad if f is differetiable at a poit c of S, te f is cotiuous at c. Tat is, differetiability implies cotiuity. PROOF. We are assumig tat lim 0 (f(c+) f(c))/ = L. Hece, tere exists a positive umber δ 0 suc tat f(c+) f(c) L < 1 if < δ 0, implyig tat f(c + ) f(c) < ( L + 1)weever < δ 0. So, if ɛ > 0 is give, let δ be te miimum of δ 0 ad ɛ/( L + 1). If y S ad y c < δ, te, tiig of y as beig c +, f(y) f(c) = f(c + ) f(c) < ( L + 1) = y c ( L + 1) < ɛ. (Every y ca be writte as c + for some, ad y c =.) Exercise 4.7. Defie f(z) = z for z C. (a) Prove tat f is cotiuous at every poit of C. (b) Sow tat, if f is differetiable at a poit c, te f (c) = 0. HINT: Usig part (b) of Exercise 4.1, evaluate f (c) i te followig two ways. f (c) = lim c + 1 c 1

6 90 IV. DIFFERENTIATION, LOCAL BEHAVIOR ad f (c) = lim c + i c. i Sow tat te oly way tese two limits ca be equal is for tem to be 0. (c) Coclude tat f is ot differetiable aywere. Ideed, if it were, wat would te fuctio θ ave to be, ad wy would t it satisfy Equatio 4.2? (d) Suppose f : R R is te fuctio of a real variable tat is defied by f(x) = x. Sow tat f is differetiable at every poit x 0. How does tis result ot cotradict part (c)? Te followig teorem geeralizes te precedig exercise. THEOREM 4.4. Suppose f : S R is a real-valued fuctio of a complex variable, ad assume tat f is differetiable at a poit c S. Te f (c) = 0. Tat is, every real-valued, differetiable fuctio f of a complex variable satisfies f (c) = 0 for all c i te domai of f. PROOF. We compute f (c) i two ways. f (c) = lim f(c + 1 ) f(c) 1 is a real umber.. f (c) = lim f(c + i ) f(c) i Hece, f (c) must be 0, as claimed. is a purely imagiary umber. REMARK. Tis teorem may come as a surprise, for it sows tat tere are very few real-valued differetiable fuctios of a complex variable. For tis reaso, weever f : S R is a real-valued, differetiable fuctio, we will presume tat f is a fuctio of a real variable; i.e., tat te domai S R. Evaluatig lim 0 q() i te two differet ways, real, ad pure imagiary, led to te proof of te last teorem. It also leads us to mae defiitios of wat are called partial derivatives of real-valued fuctios wose domais are subsets of C R 2. As te ext exercise will sow, te teory of partial derivatives of real-valued fuctios is a muc ricer teory ta tat of stadard derivatives of real-valued fuctios of a sigle complex variable. DEFINITION. Let f : S R be defied o a set S C R 2, ad let c = (a, b) = + + bi be a poit i te iterior of S. We defie te partial derivative of f wit respect to x at te poit c = (a, b) by te formula f (a, b) = lim x 0 f(a +, b) f(a, b), ad te partial derivative of f wit respect to y at c = (a, b) by te formula f (a, b) = lim y 0 f(a, b + ) f(a, b), weever tese limits exist. (I bot tese limits, te variable is a real variable.)(

7 IV. DIFFERENTIATION, LOCAL BEHAVIOR 91 It is clear tat te partial derivatives of a fuctio arise we we fix eiter te real part of te variable or te imagiary part of te variable to be a costat, ad te cosider te resultig fuctio of te oter (real) variable. We will see i Exercise 4.8 tat tere is a defiite differece betwee a fuctio s beig differetiable at a poit c = (a + bi) i te complex plae C versus its avig partial derivatives at te poit (a, b) i R 2. Exercise 4.8. (a) Suppose f is a complex-valued fuctio of a complex variable, ad assume tat bot te real ad imagiary parts of f are differetiable at a poit c. Sow tat f is differetiable at c ad tat f (c) = 0. (b) Let f = u + iv be a complex-valued fuctio of a complex variable tat is differetiable at a poit c. Prove tat bot partial derivatives of u ad v exist at c = (a, b), ad i fact tat u v (c) + i x x (c) = f (c) ad u (c) + i v y y (c) = if (c). (c) Defie a complex-valued fuctio f o C R 2 by f(z) = f(x + iy) = x iy. Write f = u + iv, ad sow tat bot partial derivatives of u ad v exist at every poit, but tat f is ot a differetiable fuctio of te complex variable z. Te ext teorem is, i part, wat we call i calculus te differetiatio formulas. THEOREM 4.5. Let f ad g be fuctios (eiter of a real variable or a complex variable), wic are bot differetiable at a poit c. Let a ad b be complex umbers. Te: (1) af + bg is differetiable at c, ad (af + bg) (c) = af (c) + bg (c). (2) (Product Formula) fg is differetiable at c, ad (fg) (c) = f (c)g(c) + f(c)g (c). (3) (Quotiet Formula) f/g is differetiable at c (providig tat g(c) 0), ad ( f g ) (c) = g(c)f (c) f(c)g (c) (g(c)) 2. (4) If f = u+iv is a complex-valued fuctio, te f is differetiable at a poit c if ad oly if u ad v are differetiable at c, ad f (c) = u (c) + iv (c). PROOF. We prove part (2) ad leave parts (1), (3), ad (4) for te exercises. We

8 92 IV. DIFFERENTIATION, LOCAL BEHAVIOR ave (fg)(c + ) (fg)(c) f(c + )g(c + ) f(c)g(c) lim = lim 0 0 were we ave used Teorems 4.1, 4.2, ad 4.3. = lim 0 f(c + )g(c + ) f(c)g(c + ) + lim 0 f(c)g(c + ) f(c)g(c) f(c + ) f(c) = lim lim g(c + ) lim 0 f(c) lim 0 g(c + ) g(c) = f (c)g(c) + f(c)g (c), Exercise 4.9. (a) Prove parts (1), (3), ad (4) of Teorem 4.5. (b) If f ad g are real-valued fuctios tat are differetiable at a poit c, wat ca be said about te differetiability of max(f, g)? (c) Let f be a costat fuctio f(z). Prove tat f is differetiable everywere ad tat f (z) = 0 for all z. (d) Defie a fuctio f by f(z) = z. Prove tat f is differetiable everywere ad tat f (z) = 1 for all z. (e) Verify te usual derivative formulas for polyomial fuctios: If p(z) = =0 a z, te p (z) = =1 a z 1. Wat about power series fuctios? Are tey differetiable fuctios? If so, are teir derivatives agai power series fuctios? I fact, everytig wors as expected. THEOREM 4.6. Let f be a power series fuctio f(z) = =0 a z avig radius of covergece r > 0. Te f is differetiable at eac poit z i its ope dis B r (0) of covergece, ad f (z) = a z 1 = a z 1. =0 PROOF. Te proof will use part (3) of Teorem 4.2. Fix a z wit z < r. Coose r so tat z < r < r, ad write α for r z, i.e., z + α = r. Note first tat te ifiite series =0 a r coverges to a positive umber we will call M. Also, from te Caucy-Hadamard Formula, we ow tat te power series fuctio a w as te same radius of covergece as does f, ad ece te ifiite series a z 1 coverges to a umber we will deote by L. We defie a fuctio θ by θ() = f(z + ) f(z) L from wic it follows immediately tat =1 f(z + ) f(z) = L + θ(), wic establises Equatio (4.1). To complete te proof tat f is differetiable at z, it will suffice to establis Equatio (4.2), i.e., to sow tat θ() lim 0 = 0.

9 IV. DIFFERENTIATION, LOCAL BEHAVIOR 93 Tat is, give ɛ > 0 we must sow tat tere exists a δ > 0 suc tat if 0 < < δ te f(z + ) f(z) θ()/ = L < ɛ. Assumig, witout loss of geerality, tat < α, we ave tat f(z + ) f(z) =0 L = a (z + ) =0 a z ) z ) = = =0 a ( =0 ( L =0 a (( ( =0 ) z ) z ) =1 a ( ( =1 ) z ) = ( = a ( =1 =1 ( = a ( =1 =1 ( = a ( =2 =2 ( a =2 =2 ( a =2 =2 ( a =2 =2 1 α 2 a =0 ) z 1 ) ) z 1 ) ) z 1 ) ) z 1 =0 = 1 α 2 a ( z + α) =0 = 1 α 2 a r = M α 2, =0 L =0 a z L L a z 1 =1 =1 ) z 2 ) z α 2 ( ) z α ( ) a z 1 1 so tat if δ = ɛ/ M α 2, te θ()/ < ɛ, weever < δ, as desired. REMARK. Teorem 4.6 sows tat ideed power series fuctios are differetiable, ad i fact teir derivatives ca be computed, just lie polyomials, by differetiatig term by term. Tis is certaily a result we would ave oped was true, but te proof is ot trivial.

10 94 IV. DIFFERENTIATION, LOCAL BEHAVIOR Te ext teorem, te Cai Rule, is aoter otrivial oe. It deals wit te differetiability of te compositio of two differetiable fuctios. Agai, te result is wat we would ave wated, te compositio of two differetiable fuctios is itself differetiable, but te argumet required to prove it is tricy. THEOREM 4.7. (Cai Rule) Let f : S C be a fuctio, ad assume tat f is differetiable at a poit c. Suppose g : T C is a fuctio, tat T C, tat te umber f(c) T, ad tat g is differetiable at f(c). Te te compositio g f is differetiable at c ad (g f) (c) = g (f(c))f (c). PROOF. Usig part (3) of Teorem 4.2, write ad g(f(c) + ) g(f(c)) = L g + θ g () f(c + ) f(c) = L f + θ f (). We ow from tat teorem tat L g = g (f(c)) ad L f = f (c). Ad, we also ow tat θ g () θ f () lim = 0 ad lim = Defie a fuctio () = f(c + ) f(c). Te, by Teorem 4.3, we ave tat lim 0 () = 0. We will sow tat g f is differetiable at c by sowig tat tere exists a umber L ad a fuctio θ satisfyig te two coditios of part (3) of Teorem 4.2. Tus, we ave tat g f(c + ) g f(c) = g(f(c + )) g(f(c)) = g(f(c) + ()) g(f(c)) = L g () + θ g (()) = L g (f(c + ) f(c)) + θ g (()) = L g (L f + θ f ()) + θ g (()) = L g L f + L g θ f () + θ g (()). We defie L = L g l f = g (f(c))f (c), ad we defie te fuctio θ by θ() = L g θ f () + θ g (()). By our defiitios, we ave establised Equatio (4.1) g f(c + ) g f(c) = L + θ(), so tat it remais to verify Equatio (4.2). We must sow tat, give ɛ > 0, tere exists a δ > 0 suc tat if 0 < < δ te θ()/ < ɛ. First, coose a ɛ > 0 so tat (4.3). L g ɛ + L f ɛ + ɛ 2 < ɛ

11 IV. DIFFERENTIATION, LOCAL BEHAVIOR 95 Next, usig part (b) of Exercise 4.6, coose a δ > 0 suc tat if < δ te θ g () < ɛ. Fially, coose δ > 0 so tat if 0 < < δ, te te followig two iequalities old. () < δ ad θ f () < ɛ. Te first ca be satisfied because f is cotiuous at c, ad te secod is a cosequece of part (b) of Exercise 4.6. Te: if 0 < < δ, θ() = L g θ f () + θ g (()) L g θ f () + θ g (()) < L g ɛ + ɛ () = L g ɛ + ɛ f(c + ) f(c) = L g ɛ + ɛ L f + θ f () L g ɛ + ɛ L f + ɛ θ f () < L g ɛ + ɛ L f + ɛ ɛ = ( L g ɛ + L f ɛ + ɛ 2 ), wece θ()/ < ( L g ɛ + L f ɛ + ɛ 2 ) < ɛ, as desired. Exercise (a) Derive te familiar formulas for te derivatives of te elemetary trascedetal fuctios: exp = exp, si = cos,, si = cos, cos = si ad cos = si. (b) Defie a fuctio f as follows. f(z) = cos 2 (z) + si 2 (z). Use part (a) ad te Cai Rule to sow tat f (z) = 0 for all z C. Does tis imply tat cos 2 (z) + si 2 (z) = 1 for all complex umbers z? (c) Suppose f is expadable i a Taylor series aroud te poit c : f(z) = =0 a (z c) for all z B r (c). Prove tat f is differetiable at eac poit of te ope dis B r (c), ad sow tat f (z) = a (z c) 1. =1 HINT: Use Teorem 4.6 ad te cai rule. CONSEQUENCES OF DIFFERENTIABILITY, THE MEAN VALUE THEOREM DEFINITION. Let f : S R be a real-valued fuctio of a real variable, ad let c be a elemet of te iterior of S. Te f is said to attai a local maximum at c if tere exists a δ > 0 suc tat (c δ, c + δ) S ad f(c) f(x) for all x (c δ, c + δ). Te fuctio f is said to attai a local miimum at c if tere exists a iterval (c δ, c + δ) S suc tat f(c) f(x) for all x (c δ, c + δ). Te ext teorem sould be a familiar result from calculus.

12 96 IV. DIFFERENTIATION, LOCAL BEHAVIOR THEOREM 4.8. (First Derivative Test for Extreme Values) Let f : S R be a real-valued fuctio of a real variable, ad let c S be a poit at wic f attais a local maximum or a local miimum. If f is differetiable at c, te f (c) must be 0. PROOF. We prove te teorem we f attais a local maximum at c. Te proof for te case we f attais a local miimum is completely aalogous. Tus, let δ > 0 be suc tat f(c) f(x) for all x suc tat x c < δ. Note tat, if is sufficietly large, te bot c + 1 ad c 1 belog to te iterval (c δ, c + δ). We evaluate f (c) i two ways. First, f (c) = lim f(c + 1 ) f(c) 1 0 because te umerator is always opositive ad te deomiator is always positive. O te oter ad, f f(c 1 (c) = lim ) f(c) 1 0 sice bot umerator ad deomiator are opositive. Terefore, f (c) must be 0, as desired. Of course we do ot eed a result lie Teorem 4.8 for fuctios of a complex variable, sice te derivative of every real-valued fuctio of a complex variable ecessarily is 0, idepedet of weter or ot te fuctio attais a extreme value. REMARK. As metioed earlier, te zeroes of a fuctio are ofte importat umbers. Te precedig teorem sows tat te zeroes of te derivative f of a fuctio f are itimately related to fidig te extreme values of te fuctio f. Te zeroes of f are ofte called te critical poits for f. Part (a) of te ext exercise establises te familiar procedure from calculus for determiig te maximum ad miimum of a cotiuous real-valued fuctio o a closed iterval. Exercise (a) Let f be a cotiuous real-valued fuctio o a closed iterval [a, b], ad assume tat f is differetiable at eac poit x i te ope iterval (a, b). Let M be te maximum value of f o tis iterval, ad m be its miimum value o tis iterval. Write S for te set of all x (a, b) for wic f (x) = 0. Suppose x is a poit of [a, b] for wic f(x) is eiter M or m. Prove tat x eiter is a elemet of te set S, or x is oe of te edpoits a or b. (b) Let f be te fuctio defied o [0, 1/2) by f(t) = t/(1 t). Sow tat f(t) < 1 for all t [0, 1/2). (c) Let t ( 1/2, 1/2) be give. Prove tat tere exists a r < 1, depedig o t, suc tat t/(1 + y) < r for all y betwee 0 ad t. (d) Let t be a fixed umber for wic 0 < t < 1. Sow tat, for all 0 s t, (t s)/(1 + s) t. Probably te most powerful teorem about differetiatio is te ext oe. It is stated as a equatio, but its power is usually as a iequality; i.e., te absolute value of te left ad side is less ta or equal to te absolute value of te rigt ad side.

13 IV. DIFFERENTIATION, LOCAL BEHAVIOR 97 THEOREM 4.9. (Mea Value Teorem) Let f be a real-valued cotiuous fuctio o a closed bouded iterval [a, b], ad assume tat f is differetiable at eac poit x i te ope iterval (a, b). Te tere exists a poit c (a, b) suc tat f(b) f(a) = f (c)(b a). PROOF. Tis proof is tricy. Defie a fuctio o [a, b] by (x) = x(f(b) f(a)) f(x)(b a). Clearly, is cotiuous o [a, b] ad is differetiable at eac poit x (a, b). Ideed, (x) = f(b) f(a) f (x)(b a). It follows from tis equatio tat te teorem will be proved if we ca sow tat tere exists a poit c (a, b) for wic (c) = 0. Note also tat (a) = a(f(b) f(a)) f(a)(b a) = af(b) bf(a) ad (b) = b(f(b) f(a)) f(b)(b a) = af(b) bf(a), sowig tat (a) = (b). Let m be te miimum value attaied by te cotiuous fuctio o te compact iterval [a, b] ad let M be te maximum value attaied by o [a, b]. If m = M, te is a costat o [a, b] ad (c) = 0 for all c (a, b). Hece, te teorem is true if M = m, ad we could use ay c (a, b). If m M, te at least oe of tese two extreme values is ot equal to (a). Suppose m (a). Of course, m is also ot equal to (b). Let c [a, b] be suc tat (c) = m. Te, i fact, c (a, b). By Teorem 4.8, (c) = 0. We ave te tat i every case tere exists a poit c (a, b) for wic (c) = 0. Tis completes te proof. REMARK. Te Mea Value Teorem is a teorem about real-valued fuctios of a real variable, ad we will see later tat it fails for complex-valued fuctios of a complex variable. (See part (f) of Exercise 4.16.) I fact, it ca fail for a complex-valued fuctio of a real variable. Ideed, if f(x) = u(x) + iv(x) is a cotiuous complex-valued fuctio o te iterval [a, b], ad differetiable o te ope iterval (a, b), te te Mea Value Teorem certaily olds for te two real-valued fuctios u ad v, so tat we would ave f(b) f(a) = u(b) u(a) + i(v(b) v(a)) = u (c 1 )(b a) + iv (c 2 )(b a), wic is ot f (c)(b a) uless we ca be sure tat te two poits c 1 ad c 2 ca be cose to be equal. Tis simply is ot always possible. Loo at te fuctio f(x) = x 2 + ix 3 o te iterval [0, 1]. O te oter ad, if f is a real-valued fuctio of a complex variable (two real variables), te a geeralized versio of te Mea Value Teorem does old. See part (c) of Exercise 4.35.

14 98 IV. DIFFERENTIATION, LOCAL BEHAVIOR Oe of te first applicatios of te Mea Value Teorem is to sow tat a fuctio wose derivative is idetically 0 is ecessarily a costat fuctio. Tis seemigly obvious fact is just ot obvious. Te ext exercise sows tat tis result olds for complex-valued fuctios of a complex variable, eve toug te Mea Value Teorem does ot. Exercise (a) Suppose f is a cotiuous real-valued fuctio o (a, b) ad tat f (x) = 0 for all x (a, b). Prove tat f is a costat fuctio o (a, b). HINT: Sow tat f(x) = f(a) for all x [a, b] by usig te Mea Value Teorem applied to te iterval [a, x]. (b) Let f be a complex-valued fuctio of a real variable. Suppose f is differetiable at eac poit x i a ope iterval (a, b), ad assume tat f (x) = 0 for all x (a, b). Prove tat f is a costat fuctio. HINT: Use te real ad imagiary parts of f. (c) Let f be a complex-valued fuctio of a complex variable, ad suppose tat f is differetiable o a dis B r (c) C, ad tat f (z) = 0 for all z B r (c). Prove tat f(z) is costat o B r (c). HINT: Let z be a arbitrary poit i B r (c), ad defie a fuctio : [0, 1] C by (t) = f((1 t)c + tz). Apply part (b) to. Te ext exercise establises, at last, two importat idetities. Exercise 4.13.) (cos 2 + si 2 = 1 ad exp(iπ = 1.) (a) Prove tat cos 2 (z) + si 2 (z) = 1 for all complex umbers z. (b) Prove tat cos(π) = 1. HINT: We ow from part (a) tat cos(π) = ±1. Usig te Mea Value Teorem for te cosie fuctio o te iterval [0, π], derive a cotradictio from te assumptio tat cos(π) = 1. (c) Prove tat exp(iπ) = 1. HINT: Recall tat exp(iz) = cos(z)+i si(z) for all complex z. (Note tat tis does ot yet tell us tat e iπ = 1. We do ot yet ow tat exp(z) = e z.) (d) Prove tat cos 2 z si 2 z = 1 for all complex umbers z. (e) Compute te derivatives of te taget ad yperbolic taget fuctios ta = si / cos ad ta = si / cos. Sow i fact tat ta = 1 cos 2 ad ta = 1 cos 2. Here are two more elemetary cosequeces of te Mea Value Teorem. Exercise (a) Suppose f ad g are two complex-valued fuctios of a real (or complex) variable, ad suppose tat f (x) = g (x) for all x (a, b) (or x B r (c).) Prove tat tere exists a costat suc tat f(x) = g(x) + for all x (a, b) (or x B r (c).) (b) Suppose f (z) = c exp(az) for all z, were c ad a are complex costats wit a 0. Prove tat tere exists a costat c suc tat f(z) = c a exp(az) + c. Wat if a = 0? (c) (A geeralizatio of part (a)) Suppose f ad g are cotiuous real-valued fuctios o te closed iterval [a, b], ad suppose tere exists a partitio {x 0 < x 1 <... < x } of [a, b] suc tat bot f ad g are differetiable o eac subiterval

15 IV. DIFFERENTIATION, LOCAL BEHAVIOR 99 (x i 1, x i ). (Tat is, we do ot assume tat f ad g are differetiable at te edpoits.) Suppose tat f (x) = g (x) for every x i eac ope subiterval (x i 1, x i ). Prove tat tere exists a costat suc tat f(x) = g(x) + for all x [a, b]. HINT: Use part (a) to coclude tat f = g + were is a step fuctio, ad te observe tat must be cotiuous ad ece a costat. (d) Suppose f is a differetiable real-valued fuctio o (a, b) ad assume tat f (x) 0 for all x (a, b). Prove tat f is 1-1 o (a, b). Exercise Let f : [a, b] R be a fuctio tat is cotiuous o its domai [a, b] ad differetiable o (a, b). (We do ot suppose tat f is cotiuous o (a, b).) (a) Prove tat f is odecreasig o [a, b] if ad oly if f (x) 0 for all x (a, b). Sow also tat f is oicreasig o [a, b] if ad oly if f (x) 0 for all x (a, b). (b) Coclude tat, if f taes o bot positive ad egative values o (a, b), te f is ot 1-1. (See te proof of Teorem 3.11.) (c) Sow tat, if f taes o bot positive ad egative values o (a, b), te tere must exist a poit c (a, b) for wic f (c) = 0. (If f were cotiuous, tis would follow from te Itermediate Value Teorem. But, we are ot assumig ere tat f is cotiuous.) (d) Prove te Itermediate Value Teorem for Derivatives: Suppose f is cotiuous o te closed bouded iterval [a, b] ad differetiable o te ope iterval (a, b). If f attais two distict values v 1 = f (x 1 ) < v 2 = f (x 2 ), te f attais every value v betwee v 1 ad v 2. HINT: Suppose v is a value betwee v 1 ad v 2. Defie a fuctio g o [a, b] by g(x) = f(x) vx. Now apply part (c) to g. Here is aoter perfectly reasoable ad expected teorem, but oe wose proof is toug. THEOREM (Iverse Fuctio Teorem) Suppose f : (a, b) R is a fuctio tat is cotiuous ad 1-1 from (a, b) oto te iterval (a, b ). Assume tat f is differetiable at a poit c (a, b) ad tat f (c) 0. Te f 1 is differetiable at te poit f(c), ad f 1 (f(c)) = 1 f (c). PROOF. Te formula f 1 (f(c)) = 1/f (c) is o surprise. Tis follows directly from te Cai Rule. For, if f 1 (f(x)) = x, ad f ad f 1 are bot differetiable, te f 1 (f(c))f (c) = 1, wic gives te formula. Te difficulty wit tis teorem is i provig tat te iverse fuctio f 1 of f is differetiable at f(c). I fact, te first tig to cec is tat te poit f(c) belogs to te iterior of te domai of f 1, for tat is essetial if f 1 is to be differetiable tere, ad ere is were te ypotesis tat f is a real-valued fuctio of a real variable is importat. Accordig to Exercise 3.12, te 1-1 cotiuous fuctio f maps [a, b] oto a iterval [a, b ], ad f(c) is i te ope iterval (a, b ), i.e., is i te iterior of te domai of f 1. Accordig to part (2) of Teorem 4.2, we ca prove tat f 1 is differetiable at f(c) by sowig tat f 1 (x) f 1 (f(c)) lim = 1 x f(c) x f(c) f (c).

16 100 IV. DIFFERENTIATION, LOCAL BEHAVIOR Tat is, we eed to sow tat, give a ɛ > 0, tere exists a δ > 0 suc tat if 0 < x f(c) < δ te f 1 (x) f 1 (f(c)) x f(c) 1 f (c) < ɛ. First of all, because te fuctio 1/q is cotiuous at te poit f (c), tere exists a ɛ > 0 suc tat if q f (c) < ɛ, te (4.4). 1 q 1 f (c) < ɛ Next, because f is differetiable at c, tere exists a δ > 0 suc tat if 0 < y c < δ te f(y) f(c) (4.5). f (c) < ɛ y c Now, by Teorem 3.10, f 1 is cotiuous at te poit f(c), ad terefore tere exists a δ > 0 suc tat if x f(c) < δ te (4.6). f 1 (x) f 1 (f(c) < δ So, if x f(c) < δ, te But te, by Iequality 4.5, f 1 (x) c = f 1 (x) f 1 (f(c)) < δ. f(f 1 (x)) f(c) f 1 (x) c from wic it follows, usig Iequality 4.4, tat as desired. f 1 (x) f 1 (f(c)) x f(c) f (c) < ɛ, 1 f (c) < ɛ, REMARK. A result very lie Teorem 4.10 is actually true for complex-valued fuctios of a complex variable. We will ave to sow tat if c is i te iterior of te domai S of a oe-to-oe, cotiuously differetiable, complex-valued fuctio f of a complex variable, te f(c) is i te iterior of te domai f(s) of f 1. But, i te complex variable case, tis requires a somewat more difficult argumet. Oce tat fact is establised, te proof tat f 1 is differetiable at f(c) will be te same for complex-valued fuctios of complex variables as it is ere for realvalued fuctios of a real variable. Toug te proof of Teorem 4.10 is reasoably complicated for real-valued fuctios of a real variable, te correspodig result for complex fuctios is muc more deep, ad tat proof will ave to be postpoed to a later capter. See Teorem 7.10.

17 IV. DIFFERENTIATION, LOCAL BEHAVIOR 101 THE EXPONENTIAL AND LOGARITHM FUNCTIONS We derive ext te elemetary properties of te expoetial ad logaritmic fuctios. Of course, by expoetial fuctio, we mea te power series fuctio exp. Ad, as yet, we ave ot eve defied a logaritm fuctio. Exercise (a) Defie a complex-valued fuctio f : C C by f(z) = exp(z) exp( z). Prove tat f(z) = 1 for all z C. (b) Coclude from part (a) tat te expoetial fuctio is ever 0, ad tat exp( z) = 1/ exp(z). (c) Sow tat te expoetial fuctio is always positive o R, ad tat lim x exp(x) = 0. (d) Prove tat exp is cotiuous ad 1-1 from (, ) oto (0, ). (e) Sow tat te expoetial fuctio is ot 1-1 o C. (f) Use parts b ad e to sow tat te Mea Value Teorem is ot i ay way valid for complex-valued fuctios of a complex variable. Usig part (d) of te precedig exercise, we mae te followig importat defiitio. DEFINITION. We call te iverse exp 1 of te restrictio of te expoetial fuctio to R te (atural) logaritm fuctio, ad we deote tis fuctio by l. Te properties of te expoetial ad logaritm fuctios are strogly tied to te simplest ids of differetial equatios. Te coectio is suggested by te fact, we ave already observed, tat exp = exp. Te ext teorem, corollary, ad exercises mae tese remars more precise. THEOREM Suppose f : C C is differetiable everywere ad satisfies te differetial equatio f = af, were a is a complex umber. Te f(z) = c exp(az), were c = f(0). PROOF. Cosider te fuctio (z) = f(z)/ exp(az). Usig te Quotiet Formula, we ave tat (z) = exp(az)f (z) a exp(az)f(z) [exp(az)] 2 = exp(az)(f (z) af(z)) [exp(z)] 2 = 0. Hece, tere exists a complex umber c suc tat (z) = c for all z. Terefore, f(z) = c exp(az) for all z. Settig z = 0 gives f(0) = c, as desired. COROLLARY. (Law of Expoets) For all complex umbers z ad w, exp(z + w) = exp(z) exp(w). PROOF OF THE COROLLARY. Fix w, defie f(z) = exp(z + w), ad apply te precedig teorem. We ave f (z) = exp(z + w) = f(z), so we get exp(z + w) = f(z) = f(0) exp(z) = exp(w) exp(z). Exercise (a) If is a positive iteger ad z is ay complex umber, sow tat exp(z) = (exp(z)). (b) If r is a ratioal umber ad x is ay real umber, sow tat exp(rx) = (exp(x)) r.

18 102 IV. DIFFERENTIATION, LOCAL BEHAVIOR Exercise (a) Sow tat l is cotiuous ad 1-1 from (0, ) oto R. (b) Prove tat te logaritm fuctio l is differetiable at eac poit y (0, ) ad tat l (y) = 1/y. HINT: Write y = exp(c) ad use Teorem (c) Derive te first law of logaritms: l(xy) = l(x) + l(y). (d) Derive te secod law of logaritms: Tat is, if r is a ratioal umber ad x is a positive real umber, sow tat l(x r ) = r l(x). We are about to mae te coectio betwee te umber e ad te expoetial fuctio. Te ext teorem is te first step. THEOREM l(1) = 0 ad l(e) = 1. PROOF. If we write 1 = exp(t), te t = l(1). But exp(0) = 1, so tat l(1) = 0, wic establises te first assertio. Recall tat e = lim (1 + 1 ). Terefore, l(e) = l(lim (1 + 1 ) ) = lim l((1 + 1 ) ) = lim l(1 + 1 ) = lim l(1 + 1 ) 1 = lim l(1 + 1 ) l(1) 1 = l (1) = 1/1 = 1. Tis establises te secod assertio of te teorem. Exercise (a) Prove tat e = =0 1!. HINT: Use te fact tat te logaritm fuctio is 1-1. (b) For r a ratioal umber, sow tat exp(r) = e r. (c) If a is a positive umber ad r = p/q is a ratioal umber, sow tat a r = exp(r l(a)). (d) Prove tat e is irratioal. HINT: Let p /q be te t partial sum of te series i part (a). Sow tat q!, ad tat lim q (e p /q ) = 0. Te use Teorem We ave fially reaced a poit i our developmet were we ca mae sese of raisig ay positive umber to a arbitrary complex expoet. Of course tis icludes raisig positive umbers to irratioal powers. We mae our geeral defiitio based o part (c) of te precedig exercise.

19 IV. DIFFERENTIATION, LOCAL BEHAVIOR 103 DEFINITION. For a a positive real umber ad z a arbitrary complex umber, defie a z by a z = exp(z l(a)). REMARK. Te poit is tat our old uderstadig of wat a r meas, were a > 0 ad r is a ratioal umber, coicides wit te fuctio exp(r l(a)). So, tis ew defiitio of a z coicides ad is cosistet wit our old defiitio. Ad, it ow allows us to raies a positive umber a to a arbitrary complex expoet. REMARK. Let te bugles soud!! Now, avig made all te appropriate defiitios ad derived all te relevat teorems, we ca fially prove tat e iπ = 1. From te defiitio above, we see tat if a = e, te we ave e z = exp(z). Te, from part (c) of Exercise 4.13, we ave wat we wat: e iπ = 1. Exercise (a) Prove tat, for all complex umbers z ad w, e z+w = e z e w. (b) If x is a real umber ad z is ay complex umber, sow tat (e x ) z = e xz. (c) Let a be a fixed positive umber, ad defie a fuctio f : C C by f(z) = a z. Sow tat f is differetiable at every z C ad tat f (z) = l(a)a z. (d) Prove te geeral laws of expoets: If a ad b are positive real umbers ad z ad w are complex umbers, ad, if x is real, a z+w = a z a w, a z b z = (ab) z, a xw = (a x ) w. (e) If y is a real umber, sow tat e iy = 1. If z = x + iy is a complex umber, sow tat e z = e x. (f) Let α = a + bi be a complex umber, ad defie a fuctio f : (0, ) C by f(x) = x α = e α l(x). Prove tat f is differetiable at eac poit x of (0, ) ad tat f (x) = αx α 1. (g) Let α = a + bi be as i part (f). For x > 0, sow tat x α = x a. THE TRIGONOMETRIC AND HYPERBOLIC FUNCTIONS Te laws of expoets ad te algebraic coectios betwee te expoetial fuctio ad te trigoometric ad yperbolic fuctios, give te followig additio formulas: THEOREM Te followig idetities old for all complex umbers z ad w. si(z + w) = si(z) cos(w) + cos(z) si(w). cos(z + w) = cos(z) cos(w) si(z) si(w). si(z + w) = si(z) cos(w) + cos(z) si(w).

20 104 IV. DIFFERENTIATION, LOCAL BEHAVIOR cos(z + w) = cos(z) cos(w) + si(z) si(w). PROOF. We derive te first formula ad leave te oters to a exercise. First, for ay two real umbers x ad y, we ave cos(x + y) + i si(x + y) = e i(x+y) = e ix e iy = (cos x + i si x) (cos y + i si y) wic, equatig real ad imagiary parts, gives tat = cos x cos y si x si y + i(cos x si y + si x cos y), cos(x + y) = cos x cos y si x si y ad si(x + y) = si x cos y + cos x si y. Te secod of tese equatios is exactly wat we wat, but tis calculatio oly sows tat it olds for real umbers x ad y. We ca use te Idetity Teorem to sow tat i fact tis formula olds for all complex umbers z ad w. Tus, fix a real umber y. Let f(z) = si z cos y + cos z si y, ad let g(z) = si(z + y) = 1 2i (ei(z+y) e i(z+y) = 1 2i (eiz e iy e iz e iy ). Te bot f ad g are power series fuctios of te variable z. Furtermore, by te previous calculatio, f(1/) = g(1/) for all positive itegers. Hece, by te Idetity Teorem, f(z) = g(z) for all complex z. Hece we ave te formula we wat for all complex umbers z ad all real umbers y. To fiis te proof, we do te same tric oe more time. Fix a complex umber z. Let f(w) = si z cos w + cos z si w, ad let g(w) = si(z + w) = 1 2i (ei(z+w) e i(z+w) = 1 2i (eiz e iw e iz e iw ). Agai, bot f ad g are power series fuctios of te variable w, ad tey agree o te sequece {1/}. Hece tey agree everywere, ad tis completes te proof of te first additio formula. Exercise (a) Derive te remaiig tree additio formulas of te precedig teorem. (b) From te additio formulas, derive te two alf agle formulas for te trigoometric fuctios: si 2 (z) = 1 cos(2z), 2 ad cos 2 (z) = 1 + cos(2z). 2

21 IV. DIFFERENTIATION, LOCAL BEHAVIOR 105 THEOREM Te trigoometric fuctios si ad cos are periodic wit period 2π; i.e., si(z + 2π) = si(z) ad cos(z + 2π) = cos(z) for all complex umbers z. PROOF. We ave from te precedig exercise tat si(z + 2π) = si(z) cos(2π) + cos(z) si(2π), so tat te periodicity assertio for te sie fuctio will follow if we sow tat cos(2π) = 1 ad si(2π) = 0. From part (b) of te precedig exercise, we ave tat 0 = si 2 (π) = 1 cos(2π) 2 wic sows tat cos(2π) = 1. Sice cos 2 + si 2 = 1, it te follows tat si(2π) = 0. Te periodicity of te cosie fuctio is proved similarly. Exercise (a) Prove tat te yperbolic fuctios si ad cos are periodic. Wat is te period? (b) Prove tat te yperbolic cosie cos(x) is ever 0 for x a real umber, tat te yperbolic taget ta(x) = si(x)/ cos(x) is bouded ad icreasig from R oto ( 1, 1), ad tat te iverse yperbolic taget as derivative give by ta 1 (y) = 1/(1 y 2 ). (c) Verify tat for all y ( 1, 1) 1 + y ta 1 (y) = l( 1 y ). Exercise (Polar coordiates) Let z be a ozero complex umber. Prove tat tere exists a uique real umber 0 θ < 2π suc tat z = re iθ, were r = z. HINT: If z = a + bi, te z = r( a r + b r i. Observe tat 1 a r 1, 1 b r 1, ad ( a r )2 + ( b r )2 = 1. Sow tat tere exists a uique 0 θ < 2π suc tat a r = cos θ ad b r = si θ. L Hopital s Rule May limits of certai combiatios of fuctios are difficult to evaluate because tey lead to wat s ow as idetermiate forms. Tese are expressios of te form 0/0, /, 0 0,, 1, ad te lie. Tey are precisely combiatios of fuctios tat are ot covered by our limit teorems. See Teorem 4.1. Te very defiitio of te derivative itself is suc a case: lim 0 (f(c + ) f(c)) = 0, lim 0 = 0, ad we are iterested i te limit of te quotiet of tese two fuctios, wic would lead us to te idetermiate form 0/0. Te defiitio of te umber e is aoter example: lim(1 + 1/) = 1, lim =, ad we are iterested i te limit of (1 + 1/), wic leads to te idetermiate form 1. L Hopital s Rule, Teorem 4.16 below, is our strogest tool for adlig suc idetermiate forms. To begi wit, ere is a useful geeralizatio of te Mea Value Teorem. THEOREM (Caucy Mea Value Teorem) Let f ad g be cotiuous real-valued fuctios o a closed iterval [a, b], suppose g(a) g(b), ad assume

22 106 IV. DIFFERENTIATION, LOCAL BEHAVIOR tat bot f ad g are differetiable o te ope iterval (a, b). Te tere exists a poit c (a, b) suc tat f(b) f(a) g(b) g(a) = f (c) g (c). Exercise Prove te precedig teorem. HINT: Defie a auxiliary fuctio as was doe i te proof of te origial Mea Value Teorem. Te followig teorem ad exercise comprise wat is called L Hopital s Rule. THEOREM Suppose f ad g are differetiable real-valued fuctios o te bouded ope iterval (a, b) ad assume tat f (x) lim x a+0 g (x) = L, were L is a real umber. (Implicit i tis ypotesis is tat g (x) 0 for x i some iterval (a, a + α).) Suppose furter tat eiter lim f(x) = lim g(x) = 0 x a+0 x a+0 or te lim f(x) = lim g(x) =. x a+0 x a+0 f(x) lim x a+0 g(x) = L. PROOF. Suppose first tat lim f(x) = lim g(x) = 0. x a+0 x a+0 Observe first tat, because g (x) 0 for all x i some iterval (a, a + α), g (x) is eiter always positive or always egative o tat iterval. (Tis follows from part (d) of Exercise 4.15.) Terefore te fuctio g must be strictly mootoic o te iterval (a, a + α). Hece, sice lim x a+0 g(x) = 0, we must ave tat g(x) 0 o te iterval (a, a + α). Now, give a ɛ > 0, coose a positive δ < α suc tat if a < c < a + δ te L < ɛ. Te, for every atural umber for wic 1/ < δ, ad every a < x < a + δ, we ave by te Caucy Mea Value Teorem tat tere exists a poit c betwee a + 1/ ad x suc tat f (c) g (c) f(x) f(a + 1/) g(x) g(a + 1/) L = f (c) g L < ɛ. (c) Terefore, taig te limit as approaces, we obtai f(x) f(a + 1/) L = lim f(x) g(x) g(x) g(a + 1/) L ɛ

23 IV. DIFFERENTIATION, LOCAL BEHAVIOR 107 for all x for wic a < x < a + δ. Tis proves te teorem i tis first case. Next, suppose tat lim f(x) = lim g(x) =. x a+0 x a+0 Tis part of te teorem is a bit more complicated to prove. First, coose a positive α so tat f(x) ad g(x) are bot positive o te iterval (a, a + α). Tis is possible because bot fuctios are tedig to ifiity as x approaces a. Now, give a ɛ > 0, coose a positive umber β < α suc tat f (c) g (c) L < ɛ 2 for all a < c < a + β. We express tis absolute value iequality as te followig pair of ordiary iequalities: L ɛ 2 < f (c) g (c) < L + ɛ 2. Set y = a + β. Usig te Caucy Mea Value Teorem, ad te precedig iequalities, we ave tat for all a < x < y implyig tat L ɛ 2 < f(x) f(y) g(x) g(y) < L + ɛ 2, (L ɛ 2 )(g(x) g(y)) + f(y) < f(x) < (L + ɛ )(g(x) g(y)) + f(y). 2 Dividig troug by g(x) ad simplifyig we obtai L ɛ 2 (L ɛ 2 )g(y) g(x) + f(y) g(x) < f(x) g(x) < L + ɛ 2 (L + ɛ 2 )g(y) + f(y) g(x) g(x). Fially, usig te ypotesis tat lim x a+0 g(x) =, ad te fact tat L, ɛ, g(y), ad f(y) are all costats, coose a δ > 0, wit δ < β, suc tat if a < x < a + δ, te (L ɛ 2 )g(y) + f(y) g(x) g(x) < ɛ 2 ad (L + ɛ 2 )g(y) g(x) Te, for all a < x < a + δ, we would ave + f(y) g(x) < ɛ 2. L ɛ < f(x) g(x) < L + ɛ, implyig tat f(x) L < ɛ, g(x)

24 108 IV. DIFFERENTIATION, LOCAL BEHAVIOR ad te teorem is proved. Exercise (a) Sow tat te coclusios of te precedig teorem also old if we assume tat f (x) lim x a+0 g (x) =. HINT: Replace ɛ by a large real umber B ad sow tat f(x)/g(x) > B if 0 < x a < δ. (b) Sow tat te precedig teorem, as well as part (a) of tis exercise, also olds if we replace te (fiite) edpoit a by. HINT: Replace te δ s by egative umbers B. (c) Sow tat te precedig teorem, as well as parts a ad b of tis exercise, old if te limit as x approaces a from te rigt is replaced by te limit as x approaces b from te left. HINT: Replace f(x) by f( x) ad g(x) by g( x). (d) Give a example to sow tat te coverse of L Hopital s Rule eed ot old; i.e., fid fuctios f ad g for wic lim x a+0 f(x) = lim x a+0 g(x) = 0, f(x) lim x a+0 g(x) exists, but lim f (x) x a+0 g (x) does ot exist. (e) Deduce from te proof give above tat if lim x a+0 f (x)/g (x) = L ad lim x a+0 g(x) =, te lim x a+0 f(x)/g(x) = L idepedet of te beavior of f. (f) Evaluate lim x x 1/x, ad lim x 0 (1 x) 1/x. HINT: Tae logaritms. HIGHER ORDER DERIVATIVES DEFINITION. Let S be a subset of R (or C), ad Let f : S C be a fuctio of a real (or complex) variable. We say tat f is cotiuously differetiable o S 0 if f is differetiable at eac poit x of S 0 ad te fuctio f is cotiuous o S 0. We say tat f C 1 (S) if f is cotiuous o S ad cotiuously differetiable o S 0. We say tat f is 2-times cotiuously differetiable o S 0 if te first derivative f is itself cotiuously differetiable o S 0. Ad, iductively, we say tat f is -times cotiuously differetiable o S 0 if te 1st derivative of f is itself cotiuously differetiable o S 0. We write f () for te t derivative of f, ad we write f C (S) if f is cotiuous o S ad is times cotiuously differetiable o S 0. Of course, if f C (S), te all te derivatives f (j), for j, exist d are cotiuous o S 0. (Wy?) For completeess, we defie f (0) to be f itself, ad we say tat f C (S) if f is cotiuous o S ad as ifiitely may cotiuous derivatives o S 0 ; i.e., all of its derivatives exist ad are cotiuous o S 0. As i Capter III, we say tat f is real-aalytic (or complex-aalytic) o S if it is expadable i a Taylor series aroud eac poit c S 0 REMARK. Keep i mid tat te defiitio above, as applied to fuctios wose domai S is a otrivial subset of C, as to do wit fuctios of a complex variable tat are cotiuously differetiable o te set S 0. We ave see tat tis is quite differet from a fuctio avig cotiuous partial derivatives o S 0. We will retur to partial derivatives at te ed of tis capter.

25 IV. DIFFERENTIATION, LOCAL BEHAVIOR 109 THEOREM Let S be a ope subset of R (or C). (1) Suppose W S is a subset of R. Te, for eac 1, tere exists a fuctio i C (S) tat is ot i C +1 (S). Tat is, C +1 (S) is a proper subset of C (S). (2) If f is real-aalytic (or complex-aalytic) o S, te f C (S). (3) Tere exists a fuctio i C (R) tat is ot real-aalytic o R. Tat is, te set of real-aalytic fuctios o R is a proper subset of te set C (R). REMARK. Suppose S is a ope subset of C. It is a famous result from te Teory of Complex Variables tat if f is i C 1 (S), te f is ecessarily complex aalytic o S. We will prove tis amazig result i Teorem 7.5. Part (3) of te teorem sows tat te situatio is quite differet for real-valued fuctios of a real variable. PROOF. For part (1), see te exercise below. Part (2) is immediate from part (c) of Exercise Before fiisig te proof of part (3), we preset te followig lemma: LEMMA. Let f be te fuctio defied o all of R as follows. { 0 x 0 f(x) = p(x)e 1/x x x > 0 were p(x) is a fixed polyomial fuctio ad is a fixed oegative iteger. Te f is cotiuous at eac poit x of R. PROOF OF THE LEMMA. Te assertio of te lemma is clear if x 0. To see tat f is cotiuous at 0, it will suffice to prove tat p(x)e 1/x lim x 0+0 x = 0. (Wy?) But, for x > 0, we ow from part (b) of Exercise 3.22 tat e 1/x > 1/(x +1 ( + 1)!), implyig tat e 1/x < x +1 ( + 1)!. Hece, for x > 0, f(x) = p(x) e 1/x x < ( + 1)!x p(x), ad tis teds to 0 as x approaces 0 from te rigt, as desired.. Returig to te proof of Teorem 4.17, we verify part (3) by observig tat if f is as i te precedig lemma te f is actually differetiable, ad its derivative f is a fuctio of te same sort. (Wy?) It follows tat ay suc fuctio belogs to C (R). O te oter ad, a otrivial suc f caot be expadable i a Taylor series aroud 0 because of te Idetity Teorem. (Tae x = 1/.) Tis completes te proof. Exercise (a) Prove part (1) of Teorem Use fuctios of te form x si(1/x). (b) Prove tat ay fuctio of te form of te f i te lemma above is everywere differetiable o R, ad its derivative as te same form. Coclude tat ay suc fuctio belogs to C (R).

26 110 IV. DIFFERENTIATION, LOCAL BEHAVIOR (c) For eac positive iteger, defie a fuctio f o te iterval ( 1, 1 by f (x) = x 1+1/. Prove tat eac f is differetiable at every poit i ( 1, 1), icludig 0. Prove also tat te sequece {f } coverges uiformly to te fuctio f(x) = x. (See part () of Exercise 3.28.) Coclude tat te uiform limit of differetiable fuctios of a real variable eed ot be differetiable. (Agai, for fuctios of a complex variable, te situatio is very differet. I tat case, te uiform limit of differetiable fuctios is differetiable. See Teorem 7.11.) Exercise (A smoot approximatio to a step fuctio.) Suppose a < b < c < d are real umbers. Sow tat tere exists a fuctio χ i C (R) suc tat 0 χ(x) 1 for all x, χ(x) 1 for x [b, c], ad χ(x) 0 for x / (a, d). (If a is close to b ad c is close to d, te tis fuctio is a C approximatio to te step fuctio tat is 1 o te iterval [b, c] ad 0 elsewere.) (a) Let f be a fuctio lie te oe i te lemma. Ti about te graps of te fuctios f(x c) ad f(b x). Costruct a C fuctio g tat is 0 betwee b ad c ad positive everywere else. (b) Costruct a C fuctio tat is positive betwee a ad d ad 0 everywere else. (c) Let g ad be as i parts (a) ad (b). If j = g +, sow tat j is ever 0, ad write for te C fuctio = 1/j. (d) Examie te fuctio, ad sow tat it is te desired fuctio χ. THEOREM (Formula for te coefficiets of a Taylor Series fuctio) Let f be expadable i a Taylor series aroud a poit c : Te for eac, a = f () (c)/!. f(x) = a (x c). PROOF. Because eac derivative of a Taylor series fuctio is agai a Taylor series fuctio, ad because te value of a Taylor series fuctio at te poit c is equal to its costat term a 0, we ave tat a 1 = f (c). Computig te derivative of te derivative, we see tat 2a 2 = f (c) = f (2) (c). Cotiuig tis, i.e., arguig by iductio, we fid tat!a = f () (c), wic proves te teorem. TAYLOR POLYNOMIALS AND TAYLOR S REMAINDER THEOREM DEFINITION. Let f be i C (B r (c)) for c a fixed complex umber, r > 0, ad a positive iteger. Defie te Taylor polyomial of degree for f at c to be te polyomial T T(f,c) give by te formula: were a j = f (j) (c)/j!. (T(f,c) )(z) = a j (z c) j, j=0 REMARK. If f is expadable i a Taylor series o B r (c), te te Taylor polyomial for f of degree is otig but te t partial sum of te Taylor series for f o B r (c). However, ay fuctio tat is times differetiable at a poit c as a Taylor polyomial of order. Fuctios tat are ifiitely differetiable ave