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Numerators of bivariate Taylor expansion of the incomplete elliptic integral of the first kind.
1

%I #37 Apr 29 2016 17:06:59

%S 1,0,1,0,-4,9,0,16,-180,225,0,-64,3024,-12600,11025,0,256,-48960,

%T 529200,-1323000,893025,0,-1024,785664,-20275200,110602800,-196465500,

%U 108056025,0,4096,-12579840,749548800,-8072064000,28605376800,-39332393100,18261468225

%N Numerators of bivariate Taylor expansion of the incomplete elliptic integral of the first kind.

%C Table has only rows for odd h because all coefficients for even h are zero:

%C ===|==========================================================================

%C h\s| 0 1 2 3 4 5 6

%C ---|--------------------------------------------------------------------------

%C 1 | 1

%C 3 | 0 1

%C 5 | 0 -4 9

%C 7 | 0 16 -180 225

%C 9 | 0 -64 3024 -12600 11025

%C 11 | 0 256 -48960 529200 -1323000 893025

%C 13 | 0 -1024 785664 -20275200 110602800 -196465500 108056025

%C 15 | 0 4096 -12579840 749548800 -8072064000 28605376800 39332393100

%C 17 | 0 -16384 201314304 -27298252800 553339987200 -3514374864000 9125115199200

%C ...

%C Conjecture:

%C If t(h,s) is any term of the previous table after the first column (s>0), then:

%C t(h,s) = -( (2*s)^2*t(h-2,s) - (2*s-1)^2*t(h-2,s-1) ), with t(1,0) = 1, t(h,0) = 0 for h>1 and t(h,s) = 0 for odd h = 1..2*s-1. Some example:

%C t(11,3) = -((2*3)^2*t(9,5) - (2*3-1)^2*t(9,2)) = -(36*(-12600) - 25*3024) = 529200;

%C t(17,5) = -((2*5)^2*t(15,5) - (2*5-1)^2*t(15,4)) = -(100*(28605376800) - 81*(-8072064000)) = -351437486400.

%C Also:

%C t(h,1) = (-1)^(h/2+1/2)*A000302(h/2-3/2) for h>1;

%C t(h,2) = (-1)^(h/2-1/2)*(16*t(h-2,2)+9*2^(h-5)) for h>3.

%H Francesco Franco, <a href="/A272102/b272102.txt">Rows n = 1..14 of triangle, flattened</a>

%F F(m,phi) = Int_{theta=0..phi} 1/sqrt(1-m*sin^2 theta) d theta.

%F F(m,phi) = Sum_{n=1,3,5,7,9,...} ( Sum_{s=0..(n-1)/2} a((n-1)/2,s) * m^s)*phi^n/n!.

%e F(m,phi) = phi + m*phi^3/3! + (-4*m+9*m^2)*phi^5/5! + (16*m-180*m^2+225*m^3)*phi^7/7! + (-64*m+3024*m^2-12600*m^3+11025*m^4)*phi^9/9! + ...

%e so the first row (order phi^1) is a(1,1)=1 for the coefficient of phi,

%e the second row (order phi^3) is a(2,0)=0 for the missing coefficient of m^0*phi^3, and a(2,1)=-1 for the coefficient of m^1*phi^3/3!.

%p an := proc(m, n, s) local f: f := coeftayl(EllipticF(sin(phi), m^(1/2)), phi=0, n); coeftayl(f*n!, m=0, s) ; end: nmax := 28 ; for n from 1 to nmax by 2 do for s from 0 to (n-1)/2 do printf("%d, ", an(m, n, s)) ; od ; od;

%t a[1, 0] = 1; a[n_, s_] := SeriesCoefficient[EllipticF[phi, m], {phi, 0, n}, {m, 0, s}]*n!; Table[a[n, s], {n, 1, 17, 2}, {s, 0, n/2}] // Flatten

%Y Cf. A120362.

%K sign,tabl

%O 1,5

%A _Francesco Franco_, Apr 23 2016

%E Missing 0 added by _Bruno Berselli_, Apr 29 2016