In article
<c1f51eeb-ff3c-42f6-9470-2bcb23501a02@[EMAIL PROTECTED]
>,
Vladimir Bondarenko <vb@[EMAIL PROTECTED]
> wrote:
>Hello computer algebra Fighter the Earthling,
>Train hard, fight easy! The Last Battle is coming!
>Go and do***ent our human's intelligence superiority over the
>dreadful (striped!) Computers coming.... maybe driving at our
>beer, cheese and Southern hot sausages, who knows?!
>My carbon-based brethren worldwide, it is only daily training,
>training, and more training that will allow us to curb the
>aggressors, and stop once and for all the advance of the World
>Silicon/Superconductive/Qbit Evil and this way to rescue the
>Universe.... along with our eatables!
>Back to the training exercises...
>Is there an Insensible-To-Fear Human Warrior the Simplifier to
>come up with a succession of CAS commands to squeeze this very,
>very, very much
One reason for computers is that a carbon-based person cannot
even keep all of this straight, while a computer can.
One thing I might do is to see if there are any relation****ps
between the large number of arctangents, including of complex
arguments, and logarithms, which might come from them.
>((-1)^(1/4)*((1-I)*Pi+I*(Pi-ArcTan[4])+I*ArcTan[4]+(2*I)*ArcTan
>[2+(1-I)*Sqrt[2]]-2*ArcTan[2+(1+I)*Sqrt[2]]+I*ArcTan[(-6+2*Sqrt
>[2])/(-1+2*Sqrt[2])] - I*ArcTan[(2+2*Sqrt[2])/(1+2*Sqrt[2])]-I*
>ArcTan[(5+2*Sqrt[2])/(4+2*Sqrt[2])]+I*ArcTan[(5+4*Sqrt[2])/(4+4
>*Sqrt[2])] +I*ArcTan[(6+3*Sqrt[3])/(6+2*Sqrt[3])]-I*ArcTan[(69+
>12*Sqrt[3])/(114 -22*Sqrt[3])]+2*ArcTanh[(14-4*Sqrt[3])/(20-11*
>Sqrt[3])] + (2*I)*ArcTanh[((6+4*I)+2*Sqrt[3])/((3+4*I)-(1+2*I)*
>Sqrt[3])]+(1+I)*Log[-6+7*Sqrt[3]]-I*((-2*I)*Pi+I*(Pi-ArcCot[4])
>+I*(Pi+ArcTan[(-1+2*Sqrt[2])/(-6+2*Sqrt[2])])-Log[17]/2+Log[(-6
>+2*Sqrt[2])^2 +(-1+2*Sqrt[2])^2]/2)+Log[(-6+2*Sqrt[2])^2+(-1+2*
>Sqrt[2])^2]/2 + I*( (-I)*ArcCot[4]+I*ArcTan[(1+2*Sqrt[2])/(2+2*
>Sqrt[2])]-Log[17]/2+Log[(1+2*Sqrt[2])^2+(2+2*Sqrt[2])^2]/2)-Log
>[(1+2*Sqrt[2])^2+(2+2*Sqrt[2])^2]/2+I*(I*ArcTan[(4+2*Sqrt[2])/(
>5+2*Sqrt[2])]+Log[(4+2*Sqrt[2])^2+(5+2*Sqrt[2])^2]/2)-Log[(4+2*
>Sqrt[2])^2 + (5+2*Sqrt[2])^2]/2-I*(I*ArcTan[(4+4*Sqrt[2])/(5+4*
>Sqrt[2])] + Log[(4 + 4*Sqrt[2])^2+(5+4*Sqrt[2])^2]/2)+Log[(4+4*
>Sqrt[2])^2 + (5+4*Sqrt[2])^2]/2+I*(I*ArcTan[(6+2*Sqrt[3])/(6+3*
>Sqrt[3])]+Log[(6+2*Sqrt[3])^2+(6+3*Sqrt[3])^2]/2)+Log[(6+2*Sqrt
>[3])^2 +(6+3*Sqrt[3])^2]/2-I*(I*ArcTan[(114-22*Sqrt[3])/(69+12*
>Sqrt[3])]+Log[(114-22*Sqrt[3])^2+(69+12*Sqrt[3])^2]/2)-Log[(114
>-22*Sqrt[3])^2+(69+12*Sqrt[3])^2]/2))/4
>(* = 0.87041.... *)
> ?
>Best wishes,
>Vladimir Bondarenko
>VM and GEMM architect
>Co-founder, CEO, Mathematical Director
>http://www.cybertester.com/
Cyber Tester, LLC
>http://maple.bug-list.org/
Maple Bugs Encyclopaedia
>http://www.CAS-testing.org/
CAS Testing
>-----------------------------------------------------------------
>"We must understand that technologies
>like these are the way of the future."
>-----------------------------------------------------------------
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@[EMAIL PROTECTED]
Phone: (765)494-6054 FAX: (765)494-0558
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