William Sit <wyscc <at> cunyvm.cuny.edu> writes: Hi, William. Thanks for your answer.
> The a = %%A0 is the real root of the denominator, which then factors as: > 2*s**3 - 3*s + 4 = (s - a)(2*s**2 + 2*a*s + 2*a**2 -3) I didn't know that %%A0 was the real root of the denominator. When I execute inverseLaplace, it only show the packages loading but not what %%A0 is. Ought I to configurate something in order that Axiom automatically shows the roots values? > The coefficients of t in the exponential functions are the roots of > the quadratic factor based on the quadratic formula. Axiom should have > said what %%A0 is. This Axiom's solution doesn't seem equal to the solution that I obtained by partial fraction decomposition using the approximate root value of -1.647426657: f(t) = -1.088456053*exp(-1.647426657*t) + exp(0.823713328*t) * * [ 1.588456054*cos(0.731786132*t - 20.77542511*sin(0.731786132*t) ] In fact, if we replace the approximate root value, the radicands turn into negative value :-S I think that my solution is correct because f(t) is continuous in t = 0 and f(0) ---the exercise request to find f(0)--- can be verified by the initial value theorem. I saw the resolution of cathedra's director and she solved the exercise _only_ using the initial value theorem but I think that it's wrong because it's only true if f(t) is continuous :-| Regards, Daniel _______________________________________________ Axiom-math mailing list Axiom-math@nongnu.org http://lists.nongnu.org/mailman/listinfo/axiom-math