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## [Axiom-math] Re: About "%%"

**From**: |
Daniel |

**Subject**: |
[Axiom-math] Re: About "%%" |

**Date**: |
Mon, 16 Jan 2006 18:05:01 +0000 (UTC) |

**User-agent**: |
Loom/3.14 (http://gmane.org/) |

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