n=0 f n(z) converges, we say that f n (z) = lim N s N(z) (37)

Transcription

1 10 Power Series A fuctio defied by a series of fuctios Let {f } be defied o A C The formal sum f is called a series of fuctios If for ay z A, the series f (z) coverges, we say that f coverges o A I this case, we deote f (z) = lim N s N(z) (37) where s N (z) := N f (z) f is called coverget uiformly o A if {s N } N=1 coverges uiformly If the series does ot coverges, we say that it diverges By(37),theproblemforcovergece ofseriesoffuctiocabereduceditotheproblem for covergece of sequece of fuctios The we get immediately the followig Theorem 101 (i) (Cauchy criterio) f coverges uiformly o A ǫ > 0, 0 > 0 such that f j (z) < ǫ, m 0, z A j=m (ii) If f are cotiuous, ad f coverges uiformly, the f is cotiuous Theorem 102 (Weierstrass M-Test) If there exist real umbers M > 0 such that f (z) M, N, z A, ad if M coverges, the f coverges absolutely ad uiformly o A Theorem 103 If f are cotious o a ope subset D C ad f coverges uiformly o compact subsets of D, the f = f is cotiuous o D Power series We have defied complex-valued fuctio f(z) I case f(z) is a power series, it ca be writte as i,j a ijz i z j Let us cosider a class of complex-valued fuctio: a power series j a jz j (without z k terms), where a C More eerally, we ca cosider a complex-valued fuctio a (z a) where a C ad a C For simplicity, we assume a = 0 here Cosider the followig three statemets: Let z 0 C 58

2 (i) a z 0 coverges (ii) lim a z 0 = 0 (iii) lim a z 35 0 < We have (i) = (ii) = (iii) (38) Lemma 104 If ay of (i)(ii)(iii) holds, the a z coverges absolutely ad uiformly o compact subsets of ( z 0 ) 36 Proof: Assume z 0 0 ad (iii) holds = By (iii), M > 0 such that a z0 M, Otherwsie, suppose k, k with a k z k 0 k, which implies: a subsequece {a k z k 0 } + The lim a z 0 =, a cotradictio To show: Fix ay r, 0 < r < z 0, we have a umber b with 0 < b < 1, a z b z r, where0 < b < 1 (39) (Assumig this is true, the by Weierstrass M-Test (Theorem 112), we are doe) I fact, whe z r, a z a r = a z 0 Sice r z 0 < 1 ad M 1 1, (39) is proved By Lemma above, we defie { R := sup z 0 r z 0 M( r ) = (M 1/ r ) z 0 z 0 } a z0 coverges 35 Here lim a is the upper limit that equals to the maximum of the limits of its all coverget subsequeces 36 It geeralizesthe fact: ifaseriesofrealumbers a x covergeswhe x = c, the the seriescoverges absolutely ad uiformly o proper closed subiterval of [ c, c ] 59

3 which is called the radius of covergece By (38), R has other two formulas: R = sup { z 0 lim a z 0 = 0} = sup { z 0 lim a z 0 < } R could be 0, fiite positive umber, or + If R = 0, a z is ot coverget except z = 0 If 0 < R <, a z is a coverget power series, hece a cotiuous fuctio, o the disk (R) If R = +, a z coverges at ay z C Remark: 1 All above tell us that a power series a z is a well-defied fuctio o the disk (R) 2 If R = +, we call this fuctio a etire fuctio 3 I case 0 < R < +, we ca prove late that there is at least oe poit i the boudary (R) at which the fuctio a z diverges We also ca show late that for ay 0 < R < +, give ay subset S (R), there exists a power series f(z) = a z with R as its radius of covergece such that at ay poit i (R) S, f(z) coverges, ad at ay poit i S, f(z) diverges We ca show that for ay z (R), outside of the closed disk, a z diverges 4 Geometrically, for a give series, its radius of covergece is the upper limit of the radius of disks (r) i which the series coverges Theorem 105 (Cauchy-Hardamard theorem) For ay power series a z, its radius of covergece is equal to 1 R = lim a I mathematics, the Cauchy - Hadamard theorem was published i 1821 by Cauchy, but remaied relatively ukow util Hadamard rediscovered it i 1888 For the proof of this theorem, see Rudi, Priciple of Mathematical Aalysis, p

4 [Example] If j=0 a z where a = 2 +( 2), the Recall The so that R = 1 2 a = 2 +( 2) = 2 +( 2) lim = e limlog = e lim 1 log = 1 lim a = lim 2 +2 = 2 Sometimes, Cauchy-Hadamard formula could have differet simpler versio: Theorem 106 If lim a a +1 exists (fiite or + ), the R = lim a a +1 I Calculus, we cosider a real-valued power series f(x) = a x where a are real umbers, defied domia of covergece ( R, R) ad radius of covergece R We may regard this fuctio f(x) is the restrictio of a complex-valued power series f(z) = a z The radius of covergece of f(z) is the same as the oe of f(x) This exlais why f(x) may coverge or diverge at the poit R ad R Expoetial Fuctio e z := z, first defied by Euler i 1747, is called the! expoetial fuctio Here 0! = 1 ad! = 1 2 ( 1) e z has the followig properties: 1 The radius of covergece R = + I fact, R = lim a a +1 = lim = + The e z is defied o C 2 Whe z = x R, the restrictio e x is the usual real expoetial fuctio 3 e z+w = e z e w, z,w C I fact, we ca compute e z e w = ( z )( w j ) ( ) z j w j =! j! j! ( j)! j=0 j=0 1 ( ) = z j w j (z +w) = = e z+w! j! j=0 61

5 ( ) 4 We have e z = e z by the power series defiitio 5 e x = lim (1+ x ), x R 6 Euler s formula: e ix = cos x+i si x, x R (40) 7 From Euler s formula, takig cojugate: e ix = cos x i si x so that cos x = eix +e ix, si x = eix e ix 2 2i which ca be used to defie complex-valued fuctio cos z ad si z Hece e iz = cos z +isi z, ad cos z = cos z := eiz +e iz, si z := eiz e iz 2 2i ( 1) (2)! z2, si z = ( 1) (2+1)! z2+1 Here the formulas are similar to the real case: cos x = ( 1) (2+1)! x2+1 ( 1) (2)! x2 ad si x = We shall see the differece betwee the real case ad the complex case I the real case, cos x 1 ad si x 1; i the complex case, cosz 1 ad si z 1 are o loger true 8 We ca verify: cos 2 z +si 2 z = 1 cos(z +w) = cos z cos w si z si w, si(z +w) = si z cos w+cos z si w We ca also defie ta z := si z cos z, cot z := cos z si z 62

6 Remarks for hyperbolic trigoometric fuctios Hyperbolic fuctios were itroduced i the 1760s idepedetly by Vicezo Riccati ad Joha Heirich Lambert Riccati used Sc ad Cc (ie, sius circulare) to refer to circular fuctios ad Sh ad Ch (ie, sius hyperbolico) to refer to hyperbolic fuctios Lambert adopted the ames but altered the abbreviatios to what they are today The hyperbolic cosie ad hyperbolic sie are defied by cosh x = ex +e x, sih x = ex e x 2 2 Sice cosh 2 x sih 2 x = 1, we may regarded it as a parametric fuctio from R to R 2 : x ( cosh x,sih x ) whose image is the hyperbola: x 2 y 2 = 1 By the way, the sie ad cosiefuctioscaberegardeditasaparametricfuctiofromrtor 2 : x ( cosx,sihx ) whose image is the circle: x 2 +y 2 = 1 Now we ca use complex umber to defie cosh z := (e z +e z )/2 ad sih z := (e z e z )/2 similarly Recallig the formulas oe immediately obtais cos z = eiz +e iz, si z = eiz e iz, 2 2 cos(iz) = cosh z, si(iz) = i sih z It meas that oe ca freely go back ad forth betwee circular ad hyperbola fuctios i the realm of complex umbers, while oe ca oly observe the formal aalogies betwee them i the realm of real umbers I other words, i complex aalysis, the distictio betwee these two classes of fuctios disappears Negative, fractioal ad complex expoets The moder symbolism for powers of umbers was itroduced by Reé Descartes i his La géomtrie, Paris, 1637 For example, he used aa or a 2 to represet multiplicatio of a ad itself The symbolism aa was preferred by Descartes, Huyges, Rah, Kersey, Wallis, Newto, Halley, Rolle, Euler, while a 2 was preferred by Leibiz, Ozaam ad David Gregory 37 Negative ad fractioal expoetial otatios has bee suggested by Chuquet, Stevo ad others The moder symbolism is due to Wallis ad Newto 37 F Cajori, History of the expoetial ad logarithmic cocepts, America Mathematical Mothly, vol 20,(1913) 2, p36 63

7 I his Arithmetica ifiitorum, Oxford, 1656, Wallis uses positive itegral expoets ad speaks of egative ad fractioal idices But he does ot actually write a 1 for 1/a, or a 3 2 for a 3 He speaks of the series 1/ 1, 1/ 2, 1/ 3 etc, as havig the idex The moder fractioal expoet was first itroduced by Newto i the aoucemet of his Biomial theorem, iveted by him some time aroud 1665, i terms of moder otatio, (x+y) r = k=0 ( ) r x r k y k k Whe r is a o-egative iteger, the biomial coefficiets for k > r are zero, so the above is the stadard Biomila formula,, ad there are at most r +1 ozero terms For other values of r, the series has ifiitely may ozero terms, at least if x ad y are o-zero Newto also cosider egative fractioal expoets I particular, (1+x) r = k=0 ( ) r x k k I November, 1676, Leibiz collected some of his results o a sheet of paper; he uses here the otatio x 3,x 1 3 The discovery of the expoetial fuctio e z allows us to defie complex expoet e z over e After we discussed how to defie the logarithmic fuctio log z, by usig the idetity b = e log b, we are able to defie complex expoet over eve a complex umber: a z = e log az = e z log a 64