Formulas for the approximation of the complete Elliptic Integrals

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    a  r   X   i  v  :   1   1   0   4 .   4   7   9   8  v   1   [  m  a   t   h .   G   M   ]   2   5   A  p  r   2   0   1   1 Formulas for the approximation of the complete Elliptic IntegralsNikos Bagis Department of InformaticsAristotle University Thessaloniki Greece.e-mail: nikosbagis@hotmail.gr Abstract In this article we give evaluations of the two complete el-liptic integrals  K   and  E   in the form of Ramanujans type- π formulas. The result is a formula for Γ(1 / 4) 2 π − 3 / 2 with accu-racy about 120 digits per term. keywords:  elliptic functions; singular modulus; Ramanujan; Leg-endre functions; evaluations; constants 1 Elliptic singular moduli It is known that (see [1],[3]) K  ( x ) =    π/ 20 dθ   1 − x 2 sin 2 ( θ )=  π 2 2 F  1  12 ,  12;1; x 2   (1)is the complete elliptic integral of the first kind. The function  k r  is called ellipticsingular moduli and defined from the equation K    1 − k 2 r  K  ( k r ) = √  r  (2)Also it is known that if   r ∈ Q ∗ + , the  k r  are algebraic numbers.For  r  ∈ N  we set  K  [ r ] =  K  ( k r ).  K  [ r ] could be expressed in terms of productsof Γ functions, algebraic numbers and powers of   π . In time it became obviousthat the best way to express the numbers  K  [ r ] most concisely was to use thefunction b (  p ) = Γ 2 (  p )Γ(2  p )   tan(  pπ ) (3)It is also known that if   N   =  n 2 m , where  n  and  m  are positive integers then K  [ n 2 m ] =  M  n ( m ) K  [ m ] ,  (4)where  M  n ( m ) is algebraic. The following formulas for some  M  n ( m ) are known. M  2 ( m ) = 1 +  k ′ m 2 (5)1  27 M  43 ( m ) − 18 M  23 ( m ) − 8(1 − 2 k 2 m ) M  3 ( m ) − 1 = 0 (6)(5 M  5 ( m ) − 1) 5 (1 − M  5 ( m )) = 256 k 2 m (1 − k 2 m ) M  5 ( m ) (7)These formulas for finding  K  [4 r ],  K  [9 r ] and  K  [25 r ] depend only on knowing  k r .Also we consider the complete elliptic integral of the second kind, which is E  ( x ) =  π 2 2 F  1  − 12  ,  12;1; x 2   (8)and related with  K  ( x ) from the relation E  ( k r ) =  K  ( k r ) √  r   π 3 K  ( k r ) 2  − a ( r )  +  K  ( k r ) .  (9)The function  a ( r ) is called elliptic alpha function (see [4]).We will use the elliptic functions theory to evaluate values of   K  ( k r ) and  E  ( k r )in high precision using Ramanujan’s type- π  formulas, but now the constant willbe not  π  but1 πb  14   = Γ  14  2 π 3 / 2  (10)the precision of the application formula, which is our more interesting result inthis paper is an about 120 digits per term.Our methods consists Legendre functions, and we not use the function  a ( r ). 2 Legendre polynomials and the formula The Legendre  P   function is defined by P  µν   ( z ) = 1Γ(1 − ν  )  z  + 11 − z  ν/ 22 F  1  − µ,µ  + 1;1 − ν  ; 1 − z 2   (11)Set φ ( z ) =  2 F  1  ( − µ,µ  + 1;1 − ν  ; z ) =   z 1 − z  ν/ 2 Γ(1 − ν  ) P  µν   (1 − 2 z )Then derivating  φ  we have φ ′ ( z ) = 12(1 − z ) z   z 1 − z  ν/ 2 Γ(1 − ν  ) ×× [( − 1 − µ  +  ν   + 2(1+  µ ) z ) P  µν   (1 − 2 z )+ (1 +  µ − ν  ) P  1+ µν   (1 − 2 z )] (12)If we assume that ∞  n =0 ( − µ ) n  (1 +  µ ) n (1 − ν  ) n z n n !( αn  +  β  ) =  g  (13)2  then βφ ( z ) +  αzφ ′ ( z ) =  g From (11),(12) and (13) we have Theorem 1. If  α  = 2( − 1+  z ) − 1 − µ  +  ν   + 2 z  + 2 µz  (14)then ∞  n =0 ( − µ ) n  (1 + µ ) n (1 − ν  ) n n !  z  n ( αn +1) =( − 1 − µ + ν  )   z 1 − z  ν/ 2 Γ(1 − ν  ) P  1+ µν  (1 − 2 z  ) − 1 − µ + ν   + 2( µ + 1) z   (15) It is known (see [1]), that P  ( − 1 / 2)0  (1 − 2 z ) =  2 F  1  12 ,  12;1; z   (16)hence if we set  µ  = − 3 / 2 and  ν   = 0, then we have Proposition 1. ∞  n =0  32  n  − 12  n ( n !) 2  ( k r ) 2 n  − 4(1 − k 2 r ) n  + 1 − 2 k 2 r   = 2 K  ( k r ) π  = 2 ϑ 23 ( q  ) (17)where  k ′ r  =   1 − k 2 r ,  q   =  e − π √  r .The result of the Proposition 1 is not trivial since the  ϑ 3 -function can be eval-uated from the identity ϑ 3 ( q  ) = ∞  n = −∞ q  n 2 (18)in which the two constants  e  and  π  involved.We know that k 4 r  = 1 − k ′ r 1 +  k ′ r (19)Hence K  [16 r ] = 1 +  k ′ 4 r 2  K  [4 r ] = 1+  k ′ 4 r 21 +  k ′ r 2  K  [ r ]But k ′ 4 r  =   1 − k 24 r  =   1 −  1 − k ′ r 1 +  k ′ r  2 =   (1 +  k ′ r ) 2 − (1 − k ′ r ) 2 1 +  k ′ r =3  Also from (22) we have K  [6400] = 18   1 + 2 √  2  p 1 / 4 2 + √   p  + 2 7 / 8   p 1 / 4 2 + √   p  1 / 4  2 K  [100]But it is known that K  [100] = 4 + 2 √  5 + √  2(3 + 2 · 5 1 / 4 )80  b  14  hence we get an about 120 digits per term formula for  1 π b (1 / 4): 18  4 + 2 √  5 + √  2(3 + 2 · 5 1 / 4 )  − 1   1 + 2 √  2  p 1 / 4 2 + √   p  + 2 7 / 8   p 1 / 4 2 + √   p  1 / 4  − 2 ×× ∞  n =0  32  n  − 12  n ( n !) 2  ( w ) 2 n  − 2(1 − w 2 ) n − w 2 + 1 / 2   =Γ  14  π 3 / 2  : ( a ) The evaluation of   E  ( k r ) /π  follows if we use the formula P  1 / 2 (1 − 2 z ) = 2 π [2 E  ( z ) − K  ( z )] ,  (25)Then one can arrive with the same method as in Proposition 1, to Proposition 2. 4 E  ( k r ) π  = 2 K  ( k r ) π  + ∞  n =0  12  2 n ( n !) 2 ( k r ) 2 n [4(1 − k 2 r ) n  + 1 − 2 k 2 r ] (26)5
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