Can I tell Maxima somewhere/somehow that %beta is real and positive and
thus make my original question solvable by Maxima?
On Mon, 29 Nov 2010, Barton Willis wrote:
> Ignoring the x = 0 root, you are asking for the real roots of (%beta*x^3-2*%beta*x^2+%beta*x-3*x+2) = 0.
> When %beta = 1 there are three real roots; when %beta = -2, there is one real root:
>
> (%i39) (%beta*x^3-2*%beta*x^2+%beta*x-3*x+2);
> (%o39) %beta*x^3-2*%beta*x^2+%beta*x-3*x+2
>
> (%i40) subst(%beta=1,%);
> (%o40) x^3-2*x^2-2*x+2
>
> (%i41) nroots(%,minf,inf);
> (%o41) 3
>
> (%i44) subst(%beta = -2,%o39);
> (%o44) -2*x^3+4*x^2-5*x+2
>
> (%i45) nroots(%,minf,inf);
> (%o45) 1
>
> Maxima is unable to easily determine the number of real roots. Just looking at solutions
> for which freeof(%i, rectform(xxx)) is true isn't going to work. I think you'll to give
> axima plenty of help.
>
> --Barton
>
> -----maxima-bounces at math.utexas.edu wrote: -----
>
>
>> I'm?trying?the?following:
>> urb(x):=x*ui/(x*(1-x)*%beta+1);
>> deltaurb(x):=ui*x-urb(x);
>> diff(deltaurb(x),x);
>> solve(%,x);
>>
>> How?can?I?make?Maxima?show?only?the?"real"?(not?complex)?solutions?
>>
>>
>> hugo
>>
>