Elliptic functions: Introduction course

1 Elliptic functions: Introduction courseVladimir G. TKACHEVD epartment of Mathematics, Royal Institute of TechnologyLindstedtsv agen 25, 10044 Stockholm, Swedenemail: tkatchevContentsChapter 1. Elliptic integrals and Jacobi s theta Elliptic integrals and the AGM: real Lemniscates and elastic Euler s addition Theta functions: preliminaries24 Chapter 2. General theory of doubly periodic Periods of analytic Existence of doubly periodic Liouville s The Weierstrass function (z) Modular forms51 Bibliography613 CHAPTER 1 Elliptic integrals and Jacobi s theta Elliptic integrals and the AGM: real Arclength of an ellipse with major and minor arcs 2aand2band eccentricitye:= (a2 b2)/a2 [0,1), ,x2a2+y2b2= is the arclength`(a;b) of the ellipse, as a function ofaandb?]

2 There are two easyobservations to be made:(1)`(ra;rb) =r`(a;b), because rescaling by a factor r increases the arclength by thesame factor;(2)`(a;a) = 2 a, because we know the circumference of a course , is transcendental so it is debatable how well we understand it! 1 2 112 Figure +y24= 1 The total arclength is four times the length of the piece in the first quadrant, where wehave the relationsy=b 1 (x/a)2,y (x) = xba21 1 (x/a) we obtain`(a, b) = 4 a0 1 +y 2(x)dx=substitutingz=x/a= 4a 10 1 ez21 z2dx== 4a 101 ez2 (1 ez2)(1 z2) is an example of anelliptic integral of the second The simple do we compute the period of motion of a simplependulum?

3 Suppose the length of the pendulum isLand the gravitational constant be the angle of the displacement of the pendulum from the vertical. The motion ofthe pendulum is governed by a differential equation (t) = gLsin (t).In basic calculus and physics classes, this is traditionally linearized to (t) = gL (t), 0,so that the solutions take the form (t) =Acos t+Bsin t, = obtain simple harmonic motion with frequency and period 2 / .We shall consider the nonlinear equation, using a series of substitutions. First, note thatour equation integrates to12 2 2cos = constAssume that the pendulum has a maximal displacement of angle = ; then ( ) = 0 sowe have12 2= 2(cos cos ),and thus, = 2(cos cos ).

4 We take positive square root before the maximal displacement isachieved. Integrating again,we obtain t= 0d 2(cos cos )=12 0d sin2 2 sin2 2sin 2, =sin 2sin 2,e= sin2 2 [0,1),6we obtain t= 0dz (1 z2)(1 ez2).At maximal displacement = we have = 1, so the first time where maximal displacementoccurs is given byT4=1 10dz (1 z2)(1 ez2),whereTis theperiodof the oscillation (which is four times the time needed to achieve themaximal displacement). These are examples ofelliptic integrals of the first , we should point out that actually computing the function (t) involves invertingthe function 0dz (1 z2)(1 ez2).]

5 The arithmetic-geometric mean arithmetic-geometric meanof two numbersaandbis defined to be the common limit of the two sequences{an} n=0and{bn} n=0determined by the algorithma0=a,b0=ban+1=an+bn2,bn+1= anbn,n= 0,1,2.. ,( )wherebn+ 1 is always the positive square root thata1andb1are the respective arithmetic and geometric means ofaandb,a2andb2the corresponding means ofa1andb1, etc. Thus the limitM(a, b) := limn an= limn bn( )really does deserve to be called the arithmetic-geometric mean (AGM) ofaandb. Thisalgorithm first appeared in papers of Euler and Lagrange (sometime before 1785), but itwas Gauss who really discovered (in the 1790s at the age of 14) the amazing depth of thissubject.

6 Unfortunately, Gauss published little on the AGM during his positive real numbers. Then the limits in ( ) do existand will assume thata b >0, and we let{an} n=0and{bn} n=0be as in ( ).The usual inequality between arithmetic and geometric means,an+bn2 anbn1By May 30th, 1799, Gauss had observed, purely computationally, that1M(1, 2)and2 10dt 1 t4agreed to at least eleven (!) decimal places. He commented in his diary that this result will surely openup a whole new field of analysis a claim vindicated by the subsequent directions of nineteenth-centurymathematics.

7 The inverse of the above (indefinite) integral is the lemniscate sine, a function Gauss studiedin some detail. He had recognized it as a doubly periodc function by the year 1800 andhence had anticipatedone of the most important developments of Abel and Jacobi: the inverse of algebraic 2. GAUSS Carl Friedrich(1777-1855)immediately implies thatan bnfor alln 0. Actually, much more is true: we havea1 a2 .. an an+1 .. bn+1 bn .. b1 b0( )and0 an bn 2 n(a b).( )To prove ( ), note thatan bnandan+1 bn+1implyan an+bn2=an+1 bn+1= anbn bn,and ( ) follows. Frombn+1 bnwe obtainan+1 bn+1 an+1 bn= 2 1(an bn),and ( ) follows by induction.

8 From ( ) we see immediatelythat limn anand limn bnexist, and ( ) implies that the limits are equal. Thus, we can use ( ) to define the arithmetic-geometric meanM(a, b) we list the simple properties of the 1:M(a, a) =a;Fact 2:M(a, b) =M(b, a);Fact 3:M(a,0) = 0;Fact 4:M(a, b) =M(a1, b1) =M(a2, b2) =..;Fact 5:M( a, b) = M(a, b);Fact 6:M(a, b) =M(a+b2, ab).In particular, the latter relation leads us toM(1, x) =M(1 +x2, x),which shows that the AGMf(x) :=M(1, x) is a solution to the following functional equationf(x) =1 +x2f(2 x1 +x).Our next result shows that the AGM is not as simple as indicated bywhat we have doneso far.

9 We now get our first glimpse of the depth of this (Gauss, 1799).Letaandbare positive reals. Then1M(a, b)=2 /20d a2cos2 +b2sin2 Proof before, we assume thata b >0. LetI(a, b) denote the above integral,and set =M(a, b). Thus we need to proveI(a, b) = 2 .The key step is to show thatI(a, b) =I(a1, b1).( )Let us introduce a new variable such thatsin =2asin a+b+ (a b) sin2 .( )Note that 0 2corresponds to 0 2. To see this we consider the functionf(t) :=2ata+b+ (a b) (t) = 2a(a+b (a b)t2)(a+b+ (a b)t2)2 2ab(a+b+ (a b)t2)2>0which means thatf(t) increasing in [0,1].

10 On the other hand,f(0) = 0,f(1) = 1,which yields our , we note thatd a2cos2 +b2sin2 =d a21cos2 +b21sin2 .( )Indeed, one can find from ( )cos =2 cos a21cos2 +b21sin2 a+b+ (a b) sin2 ( )and it follows (by straightforward manipulations) that a2cos2 +b2sin2 =aa+b (a b) sin2 a+b+ (a b) sin2 .( )Then ( ) follows from these formulas by taking the differential of ( ).Iterating ( ) gives usI(a, b) =I(a1, b1) =I(a2, b2) =.. ,so thatI(a, b) = limn I(an, bn) =I( , ) = 2 ,9since the functions1 a21cos2 +b21sin2 converge uniformly to the constant function1.