Beginner really needs help with Maxima's basic concepts

Hello Leo,

*Thanks a lot for your detailed reply* and accept my apologies for my late
follow-up.

Your code is very useful indeed and I understand most of it save for the
syntax of the solve function as in here:
*peq : solve(p='p,exp(r*T));*

I figured out from the documentation that 'p prevents the evaluation of p.
Now what does it mean to have an equation defined as:
*"the evaluated value of p"="the non evaluated value of p"?? *

I expected something like:
*"the evaluated value of p"=0

*This is what I cannot work out from the documentation...

Julien.

2009/11/11 Leo Butler <l.butler at ed.ac.uk>

>
>
> On Tue, 10 Nov 2009, Julien Martin wrote:
>
> < Hello,
> <
> < I am working on the binomial model and my example is based on that but
> really my question relates to Maxima's basic concepts.
> <
> < Here is my example:
> <
> < I have %Delta and p defined as follows:
> <
> < %Delta:(fu-fd)/(S0*u-S0*d); (1)
> <
> < p:(exp(r*T)-d)/(u-d); (2)
> <
> < I need to have the following expression (3) simplified by replacing the
> %Delta with the expression given above in (1) and expressed in terms of p as
> defined above
> < in (2)...
> <
> < f:S0*%Delta*(1-u*exp(-r*T))+fu*exp(-r*T); (3)
> <
> < ...so that I would ideally get this:
> <
> < f:exp(-r*T)*(p*fu + (1-p) * fd); (4)
> <
> < Is this possible? Can someone please help?
> <
> < Julien.
> <
> <
>
> Julien,
> I think the key to the problem is to model your own
> manipulations. To subst p into f, you need to solve for %e^(r*T)
> in terms of p:
>
> (%i2) f:S0*%Delta*(1-u*exp(-r*T))+fu*exp(-r*T);
>
> (%o2) %Delta*S0*(1-u*%e^-(r*T))+fu*%e^-(r*T)
> (%i3) %Delta:(fu-fd)/(S0*u-S0*d);
>
> (%o3) (fu-fd)/(u*S0-d*S0)
> (%i4) p:(exp(r*T)-d)/(u-d);
>
> (%o4) (%e^(r*T)-d)/(u-d)
> (%i5) peq : solve(p='p,exp(r*T));
>
> (%o5) [%e^(r*T) = p*u-d*p+d]
> (%i6) f;
>
> (%o6) %Delta*S0*(1-u*%e^-(r*T))+fu*%e^-(r*T)
> (%i7) rpeq : 1/peq;
>
> (%o7) [%e^-(r*T) = 1/(p*u-d*p+d)]
> (%i8) subst(rpeq,f);
>
> (%o8) %Delta*(1-u/(p*u-d*p+d))*S0+fu/(p*u-d*p+d)
> (%i9) subst(rpeq,''f);
>
> (%o9) (fu-fd)*(1-u/(p*u-d*p+d))*S0/(u*S0-d*S0)+fu/(p*u-d*p+d)
> (%i10) rat(%);
>
> (%o10) ((fu-fd)*p+fd)/(p*u-d*p+d)
> (%i11) subst(rhs(peq[1])=lhs(peq[1]),%);
>
> (%o11) ((fu-fd)*p+fd)*%e^-(r*T)
>
> This final expression is almost what you want.
>
> (%i12) collectterms(expand(part(%o11,1)),fu,fd);
>
> (%o12) fu*p+fd*(1-p)
> (%i13) %*part(%o11,2);
>
> (%o13) (fu*p+fd*(1-p))*%e^-(r*T)
>
> --
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>