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A257096
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Decimal expansion of I3(u,v) = A248897/AG3(u,v) for u=1, v=2.
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3
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7, 2, 4, 2, 3, 5, 6, 3, 3, 8, 0, 0, 9, 7, 1, 4, 2, 9, 5, 3, 8, 9, 2, 3, 3, 3, 1, 1, 1, 1, 5, 0, 1, 8, 3, 8, 3, 3, 0, 9, 7, 6, 3, 4, 4, 6, 8, 3, 2, 9, 5, 5, 3, 0, 4, 9, 8, 9, 2, 4, 7, 6, 0, 7, 2, 5, 1, 1, 4, 3, 5, 6, 4, 7, 3, 6, 3, 5, 5, 8, 5, 5, 2, 3, 5, 8, 4, 6, 2, 2, 3, 9, 6, 1, 3, 9, 4, 0, 3, 8, 9, 3, 8, 5, 4
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OFFSET
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0,1
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COMMENTS
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For positive u and v, AG3(u,v) is defined as the common limit of u_k, v_k such that u_0=u, v_0=v, u_(k+1)=(u_k+2*v_k)/3, v_(k+1)=(v_k*(u_k*u_k+u_k*v_k+v_k*v_k)/3)^(1/3). Since the iterative algorithm is similar to that for AGM, AG3 is sometimes referred to as "cubic AGM".
An alternative definition of I3(u,v) is by means of the definite integral I3(u,v) = Integral[x=0,inf](x/((u^3+x^3)*(v^3+x^3)^2)^(1/3)).
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LINKS
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FORMULA
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Equals Integral[x=0,inf](x/((1+x^3)*(8+x^3)^2)^(1/3)).
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EXAMPLE
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0.724235633800971429538923331111501838330976344683295530...
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MATHEMATICA
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RealDigits[ NIntegrate[(x/((1 + x^3) (8 + x^3)^2)^(1/3)), {x, 0, Infinity}, AccuracyGoal -> 111, WorkingPrecision -> 111]][[1]] (* Robert G. Wilson v, Apr 16 2015 *)
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PROG
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(PARI) I3(u, v)={my(an=u+0.0, bn=v+0.0, anext=0.0, ncyc=0,
eps=2*10^(-default(realprecision)));
while(1, anext=(an+2*bn)/3;
bn=(bn*(an*an+an*bn+bn*bn)/3)^(1/3); an=anext;
ncyc++; if((ncyc>3)&&(abs(an-bn)<eps), break));
return((2*Pi/(3*sqrt(3)))/an); }
a = I3(1, 2)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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