Beginner really needs help with Maxima's basic concepts

Hi,
I now have a much better understanding of the concepts and tools used in
order to solve the problem.
Thanks to you two,
Julien.

2009/11/14 Leo Butler <l.butler at ed.ac.uk>

>
>
> On Sat, 14 Nov 2009, reyssat wrote:
>
> < 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]);*
>
> NB. There is no need to solve for u, you could have solved for F and d,
> alone.
>
> (%i21) solve([F=f,P=p],[F,d]);
>
> (%o21) [[F = ((fu-fd)*P+fd)*%e^-(r*T),d = -(%e^(r*T)-u*P)/(P-1)]]
>
> Leo
>
>
> <
> < 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
> < >
> < >
> <
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