Vladimir Bondarenko schrieb:
> Hello computer algebra Buff the Earthling,
>
> You've spotted that we keep preparing for the Ultimate Battle...
> Now you? Ready to fight for our C12-based Human Civilization?
>
> (If not, think again, and get ready!)
>
> The Last Combat against the bad (striped!) Computers is near!
>
> As Field Marshal Suvorov teaches us, Train hard, fight easy!
> (don't give your little gray cells to get too slow... or else
> the wicked (striped!) Computers will rob us!)
>
> So, is there a Valiant Warrior the Simplifier to invent a string
> of CAS commands to squeeze (20+ times) this expression
>
> -(1/120*(30-6*5^(1/2))^(1/2)+1/120*5^(1/2)-1/40)*arctan((8*(30-6
> *5^(1/2))^(1/2)-8*5^(1/2)-8)/(8*(10-2*5^(1/2))^(1/2)+8*3^(1/2)*(
> 5^(1/2)+1)))+(1/120*(30-6*5^(1/2))^(1/2)-1/120*5^(1/2)+1/40)*arc
> tan((8*(30-6*5^(1/2))^(1/2)+8*5^(1/2)+8)/(8*(10-2*5^(1/2))^(1/2)
> -8*3^(1/2)*(5^(1/2)+1)))-(1/120*(30+6*5^(1/2))^(1/2)+1/120*5^(1/
> 2)+1/40)*arctan((8*(30+6*5^(1/2))^(1/2)-8*5^(1/2)+8)/(8*(10+2*5^
> (1/2))^(1/2)+8*3^(1/2)*(5^(1/2)-1)))-(1/120*(30+6*5^(1/2))^(1/2)
> -1/120*5^(1/2)-1/40)*arctan((8*(30+6*5^(1/2))^(1/2)+8*5^(1/2)-8)
> /(8*(10+2*5^(1/2))^(1/2)-8*3^(1/2)*(5^(1/2)-1)))+(1/240*(30+6*5^
> (1/2))^(1/2)+1/240*5^(1/2)+1/80)*arctan(((10-2*5^(1/2))^(1/2)+3^
> (1/2)*(5^(1/2)+1))/((30-6*5^(1/2))^(1/2)-5^(1/2)-1))+(1/240*(30+
> 6*5^(1/2))^(1/2)-1/240*5^(1/2)-1/80)*arctan(((10-2*5^(1/2))^(1/2
> )-3^(1/2)*(5^(1/2)+1))/((30-6*5^(1/2))^(1/2)+5^(1/2)+1))-(1/240*
> (30-6*5^(1/2))^(1/2)-1/240*5^(1/2)+1/80)*arctan(((10+2*5^(1/2))^
> (1/2)+3^(1/2)*(5^(1/2)-1))/((30+6*5^(1/2))^(1/2)-5^(1/2)+1))+(1/
> 240*(30-6*5^(1/2))^(1/2)+1/240*5^(1/2)-1/80)*arctan(((10+2*5^(1/
> 2))^(1/2)-3^(1/2)*(5^(1/2)-1))/((30+6*5^(1/2))^(1/2)+5^(1/2)-1))
> +Pi*(1/24*(15+6*5^(1/2))^(1/2)+7/120*5^(1/2)+1/10)
>
> ?
> Best wishes,
>
> Vladimir Bondarenko
>
It simplifies to
Pi*(Sqrt[6=B7Sqrt[5] + 15]/30 + Sqrt[5]/15 + 1/10)
in a fraction of a second on Derive 6.10.
Martin.
Forgot to "translate" the square roots in my first reply!
|
