Beginner really needs help with Maxima's basic concepts

Julien Martin a ?crit :
> Thanks a lot to you too Eric!
>
> I am also experiencing difficulties with the syntax of your:
> *s:solve([F-f,P-p],[F,u,d]); 
> ss:s[2][1];
>
> *Can you please describe in English what the above two line mean - if 
> and when you have the time to do so of course :-)
>
You want to find the value of f in terms of other quantities except u 
and d. So the idea is to solve a system of equations whose unknowns are 
f,u,d.

But you defined f already, it is no more an indeterminate, so this 
symbol cannot be used as an indeterminate of the system of equations. 
You need another symbol, I used  F  but you may use  'f  as well or 
anything new.
For the same reason, since you want to express f in terms of p (not only 
the components of p but the whole expression p) you need asymbol for p 
which is not already assigned. I used P.

So I want to find the value of F,u,d (in terms of fu,fd,S0,P, exp(r*T)) 
knowing that F-f  and P-p are zero.
Hence our equations are F-f=0, P-p=0 and the unknowns are F,u,d. This is 
the meaning of *s:solve([F-f,P-p],[F,u,d]);*

xmaxima answers [ sol1,sol2 ] where the first solution corresponds to 
u=d (not interesting) and the real solution is sol2. This second 
solution is a triplet [F=..., u=..., d=...] and the first term of this 
triplet is the equality F = ((fu-fd)*P+fd)*%e^-(r*T) you're looking for.
This is the meaning of ss:s[2][1]    : take the first term of the second 
solution.

HTH

Eric


> Thanks in advance,
>
> Julien.
>
> 2009/11/11 reyssat <eric.reyssat at math.unicaen.fr 
> <mailto:eric.reyssat at math.unicaen.fr>>
>
>     Julien Martin a ?crit :
>
>         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.
>         ------------------------------------------------------------------------
>
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>          
>
>     Hello,
>     since you want to eliminate u and d in f, and express it in terms
>     of the other parameters only, you can solve in terms of f,u,d :
>
>     (%i2) %Delta:(fu-fd)/(S0*u-S0*d);
>     (%o2) (fu-fd)/(u*S0-d*S0)
>
>     (%i3) p:(exp(r*T)-d)/(u-d);
>     (%o3) (%e^(r*T)-d)/(u-d)
>
>     (%i4) f:S0*%Delta*(1-u*exp(-r*T))+fu*exp(-r*T);
>     (%o4) (fu-fd)*S0*(1-u*%e^-(r*T))/(u*S0-d*S0)+fu*%e^-(r*T)
>
>     (%i5) s:solve([F-f,P-p],[F,u,d]);  ss:s[2][1];
>     (%o5) [[F = %r1,u = %e^(r*T),d = %e^(r*T)],
>          [F = ((fu-fd)*P+fd)*%e^-(r*T),u = %r2,d =
>     -(%e^(r*T)-%r2*P)/(P-1)]]
>     (%o6) F = ((fu-fd)*P+fd)*%e^-(r*T)
>
>     which is almost your expected answer (need to collect coefficient
>     of fd).
>
>     Eric
>
>