*Dear Barton Willis,
Thank you very much for your easy answer.
It may seem an odd idea to try to compute numerical values
using maxima rather than Octave or C++, but it was only a
small thing embedded in a larger symbolic program.
Best regards,
Esben Byskov
*
Esben Byskov, Ph.D., Dr.Techn.
Professor Emeritus of Structural Analysis
Department of Civil Engineering
Aalborg University
Sohngaardsholmsvej 57
DK-9000 Aalborg
Denmark
Phone: +45 3963 7328
+45 2178 8365
e-mail: eb at civil.aau.dk
esben at annegrete.dk
On 08/16/2013 06:17 PM, Barton Willis wrote:
> If the ultimate goal is a decimal approximation, a purely floating point method is often better than symbolic solution followed by a conversion
> to floating point. Try:
>
> (%i102) allroots(-2*q^3+3*q-1.357008100494576);
> (%o102) [q=0.58757180617091,q=0.82024709886951,q=-1.40781890504042]
>
> (%i104) bfallroots(-2*q^3+3*q-1.357008100494576), fpprec : 28;
> (%o104) [q=5.875718061709101990263852045b-1,q=8.202470988695097658442859279b-1,q=-1.407818905040419964870671132b0]
>
>
> --Barton
>
> ________________________________________
> From: maxima-bounces at math.utexas.edu [maxima-bounces at math.utexas.edu] on behalf of Thomas D. Dean [tomdean at speakeasy.org]
> Sent: Friday, August 16, 2013 10:41
> To: maxima at math.utexas.edu
> Subject: Re: [Maxima] 3rd order equations
>
> On 08/16/13 05:56, Esben Byskov wrote:
>
>> *Maybe somebody can help.
>>
>> I have tried to solve the following rather simple equation using maxima:
>>
>> (%i1) Eq: -2*q^3+3*q-1.357008100494576;
>> (%o1) - 2 q + 3 q - 1.357008100494576
>>
>>
> Eq: -2*q^3+3*q-1357008100494576/1000000000000000;
> load(to_poly_solve)$
> to_poly_solve(Eq,q);
>
> Tom Dean
>
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