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Re: radicals: zeroequivalence [was: Error in Maple V Release 4]
Posted:
Dec 6, 1996 6:17 AM


Bill Dubuque <wgd@martigny.ai.mit.edu> writes: >> : Consider f := x^3  12*x^2 + 20*x + 59 >> : Using solve(f,x), Maple returns 3 imagionary solutions. This is incorrect >> : since all odd degree polynomials must have at least one real solution. >> Mathematica also returns the imaginaries without simplification, likewise
As another datapoint, MuPAD returns the 3 real roots directly from solve(f,x);
I cannot, however, seem to get MuPAD to plug the values back in and simplify them down to 0, even though they're verifiably numerically 0.
Chris
 Here it is if you care:
>> f := x^3  12*x^2 + 20*x + 59;
2 3 20 x  12 x + x + 59
>> PRETTY_PRINT:=FALSE: >> solve(f,x); {(3^(1/2)*7^(1/2)*56/9)^(1/3)*cos(PI*1/3 + atan(108^(1/2)*84541^(1/2)*1/59\ 4)*(1/3)) + (3^(1/2)*7^(1/2)*56/9)^(1/3)*cos(PI*(1/3) + atan(108^(1/2)*8\ 4541^(1/2)*1/594)*1/3) + 4, 3^(1/2)*(3^(1/2)*7^(1/2)*56/9)^(1/3)*sin(PI*1/\ 3 + atan(108^(1/2)*84541^(1/2)*1/594)*(1/3)) + (3^(1/2)*7^(1/2)*56/9)^(1/\ 3)*cos(PI*1/3 + atan(108^(1/2)*84541^(1/2)*1/594)*(1/3))*(1/2) + (3^(1/2\ )*7^(1/2)*56/9)^(1/3)*cos(PI*(1/3) + atan(108^(1/2)*84541^(1/2)*1/594)*1/\ 3)*(1/2) + 4, 3^(1/2)*(3^(1/2)*7^(1/2)*56/9)^(1/3)*sin(PI*1/3 + atan(108\ ^(1/2)*84541^(1/2)*1/594)*(1/3)) + (3^(1/2)*7^(1/2)*56/9)^(1/3)*cos(PI*1/\ 3 + atan(108^(1/2)*84541^(1/2)*1/594)*(1/3))*(1/2) + (3^(1/2)*7^(1/2)*56\ /9)^(1/3)*cos(PI*(1/3) + atan(108^(1/2)*84541^(1/2)*1/594)*1/3)*(1/2) + \ 4}



