(c562) circumference_ellipse(a,b)/2/%pi 2 b 2 a elliptic_ec(1 - --) 2 a (d562) ----------------------- %pi is clearly a legitimate mean(a,b), satisfying an identity: (c563) (circumference_ellipse(1-x,1+x)/2/%pi = rhs(d448)) 2 (1 - x) 2 (x + 1) elliptic_ec(1 - --------) 2 (x + 1) (d563) ----------------------------------- = %pi 2 2 2 (2 elliptic_ec(x ) - (1 - x) (x + 1) elliptic_kc(x )) ------------------------------------------------------- %pi (c564) taylor(%,x,0,12) 2 4 6 8 10 12 x x x 25 x 49 x 441 x (d564)/T/ 1 + -- + -- + --- + ----- + ------ + ------- + . . . = 4 64 256 16384 65536 1048576 2 4 6 8 10 12 x x x 25 x 49 x 441 x 1 + -- + -- + --- + ----- + ------ + ------- + . . . 4 64 256 16384 65536 1048576 (i.e. 2F1(-1/2,-1/2,1,x^2)) so it is >= arithmetic >= geometric means. Both the "aem" and the "egm" converge quadratically, like the agm, but to what? --rwg PS, John Brillhart challenged me to find the discriminant of the nth Legendre polynomial discriminant. I hope he doesn't mind my ducking and throwing open the question...
[>PS, John Brillhart challenged me to find the discriminant of the nth Legendre [>polynomial discriminant. I hope he doesn't mind my ducking and throwing open [>the question... # Here are the first few values. |\^/| Maple 9.5 (SGI MIPS UNIX) ._|\| |/|_. Copyright (c) Maplesoft, a division of Waterloo Maple Inc. 2004 \ MAPLE / All rights reserved. Maple is a trademark of <____ ____> Waterloo Maple Inc. | Type ? for help.
interface(prettyprint=0); 1 for i to 12 do d[i]:=discrim(orthopoly[P](i,x),x); od; d[1] := 1 d[2] := 3 d[3] := 135/4 d[4] := 23625/16 d[5] := 260465625/1024 d[6] := 11371668721875/65536 d[7] := 7888446990683634375/16777216 d[8] := 21776965089186101310140625/4294967296 d[9] := 15330043202319289712414934678515625/70368744177664 d[10] := 43033523436556282747812223470803609794921875/1152921504606846976 d[11] := 1927983533652930855481078826533672813447199742802734375/ 75557863725914323419136 d[12] := 344795425404576705797704088613314881404393134165837604181728515625/ 4951760157141521099596496896 for i from 2 to 12 do d1[i] := d[i]/d[i-1]; od; d1[2] := 3 d1[3] := 45/4 d1[4] := 175/4 d1[5] := 11025/64 d1[6] := 43659/64 d1[7] := 693693/256 d1[8] := 2760615/256 d1[9] := 703956825/16384 d1[10] := 2807136475/16384 d1[11] := 44801898141/65536 d1[12] := 178837328943/65536 for i from 3 to 12 do d2[i] := d1[i]/d1[i-1]; od; d2[3] := 15/4 d2[4] := 35/9 d2[5] := 63/16 d2[6] := 99/25 d2[7] := 143/36 d2[8] := 195/49 d2[9] := 255/64 d2[10] := 323/81 d2[11] := 399/100 d2[12] := 483/121 for i from 3 to 12 do guess[i] := (2*i-3)*(2*i-1)/(i-1)^2; od; guess[3] := 15/4 guess[4] := 35/9 guess[5] := 63/16 guess[6] := 99/25 guess[7] := 143/36 guess[8] := 195/49 guess[9] := 255/64 guess[10] := 323/81 guess[11] := 399/100 guess[12] := 483/121 quit
I blithered
(c562) circumference_ellipse(a,b)/2/%pi
2 b 2 a elliptic_ec(1 - --) 2 a (d562) ----------------------- %pi
is clearly a legitimate mean(a,b), ...
Guilt by nonassociation: (c138) em(a,b):=circumference_ellipse(a,b)/2/%pi$ (c139) dfloat(em(em(2,4),em(3,1))) (d139) 2.62751721836387d0 (c140) dfloat(em(em(2,1),em(3,4))); (d140) 2.62733046926582d0 Also, Department of Redundancy Department:
... discriminant of the nth Legendre polynomial discriminant. Sorry for the pollution. But thanks (Gary and Dick) for the solution. --rwg BLITHER HILBERT SPACE - PEACH BLISTER
participants (2)
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gfee＠cecm.sfu.ca -
R. William Gosper