R help - Dec 2003 - NUMERIC DERIVATE

UseRs,

I used the optim function

valor.optim <- optim(c(1,1,1),logexp1,method
="BFGS",control=list(fnscale=-1),hessian=T);

and I want to calculate the derivates,

    psi1<-valor.optim$par[1]
    psi2<-valor.optim$par[2]
    psi3<-valor.optim$par[3]
    
    a0=exp(psi1);
    a1=exp(psi2)/(20+exp(psi2)+exp(psi3));
    a2=exp(psi3)/(20+exp(psi2)+exp(psi3))
       
   
deriv.psi1<-numericDeriv(a0,c("psi1","psi2","psi3"));
   
deriv.psi2<-numericDeriv(a1,c("psi1","psi2","psi3"));
   
deriv.psi3<-numericDeriv(a2,c("psi1","psi2","psi3"));

but I found,
>
deriv.psi1<-numericDeriv(a0,c("psi1","psi2","psi3"));
> deriv.psi1
[1] 0.038384
attr(,"gradient")
     [,1] [,2] [,3]
[1,]    0    0    0

>
deriv.psi2<-numericDeriv(a1,c("psi1","psi2","psi3"));
> deriv.psi2
[1] 0.05754
attr(,"gradient")
     [,1] [,2] [,3]
[1,]    0    0    0


deriv.psi3<-numericDeriv(a2,c("psi1","psi2","psi3"));
>  deriv.psi3
[1] 0.93315
attr(,"gradient")
     [,1] [,2] [,3]
[1,]    0    0    0


The derivates are zero. Why?

On Wed, 17 Dec 2003, Savano S. Pereira wrote:
> UseRs,
>
> I used the optim function
>
> valor.optim <- optim(c(1,1,1),logexp1,method
> ="BFGS",control=list(fnscale=-1),hessian=T);
>
> and I want to calculate the derivates,  [ ... snip ... ]
>
> but I found,  [ ... snip ... ]
>
> The derivates are zero. Why?
Do you suppose it could have something to do with the
condition for finding an optimum ?

-  tom blackwell  -  u michigan medical school  -  ann arbor  -