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

 From: William Sit Subject: [Axiom-developer] Re: [Axiom-math] Re: About "%%" Date: Mon, 16 Jan 2006 22:52:15 -0500

```
Daniel wrote:
>
> > 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?

I think it is a bug in Axiom not reporting what the temporary symbol %%A0 is. I
figured that out by constructing the differential equation satisfied
by f(t).

> > 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

Axiom is correct. When the root is substituted into Axiom's solution, I got:
(14)
+-----------------------+
3.5487528461 274440233\|- 2.1420437737 48171979
+
0.7942280262 445812132
*
+-----------------------+
0.5 t\|- 2.1420437737 48171979  + 0.8237133286 6012776746 t
%e
+
- 1.6474266573 202555349 t
- 1.0884560524 89162426 %e
+
+-----------------------+
- 3.5487528461 274440233\|- 2.1420437737 48171979
+
0.7942280262 445812132
*
+-----------------------+
- 0.5 t\|- 2.1420437737 48171979  + 0.8237133286 6012776746 t
%e

This has the form:

c1 * %e**((a+b*%i)*t) + c2 * %e**((a - b*%i)*t) + c3 * %e**(c*t)

where c is the real root of the denominator. If you apply the Euler identity

%e**((a+b*%i)*t) = %e**(a*t)(cos(b*t) + %i*sin(b*t))

%e**((a-b*%i)*t) = %e**(a*t)(cos(b*t) - %i*sin(b*t))

you can rewrite the inverse Laplace transform into the form you have.  Here is
the answer from Mathematica (simplified after removing near-zero imaginary
components in the coefficients):

e^(0.8237133286601279 t)(1.5884560524891629 cos(0.73178613230714 t)
- (10.387712479126234 sin(0.73178613230714t))
- (1.0884560524891655) e^(-1.6474266573202556 t)

So I believe the coefficient of the sine term in your solution is not correct
and should be half its value or -10.387712479126234. The value of f(0) = 0.5.

> 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

William

```