[R] numericDeriv

[Florent Bresson]

Effectively, it's much better
thanks
--- Prof Brian Ripley <ripley at stats.ox.ac.uk> a
écrit :

> On Wed, 16 Nov 2005, Florent Bresson wrote:
> 
> > I have to compute some standard errors using the
> delta
> > method and so have to use the command
> "numericDeriv"
> > to get the desired gradient. Befor using it on my
> > complicated function, I've done a try with a
> simple
> > exemple :
> >
> > x <- 1:5
> > numericDeriv(quote(x^2),"x")
> >
> > and i get :
> >
> > [1]   1   8  27  64 125 216
> > attr(,"gradient")
> >     [,1] [,2] [,3] [,4] [,5] [,6]
> > [1,]  Inf    0    0  NaN    0    0
> > [2,]    0    0    0  NaN    0    0
> > [3,]    0  Inf    0  NaN    0    0
> > [4,]    0    0    0  NaN    0    0
> > [5,]    0    0  Inf  NaN    0    0
> > [6,]    0    0    0  NaN    0    0
> >
> > I don't understand the result. I thought I will
> get :
> >
> > [1]   1   8  27  64 125 216
> > attr(,"gradient")
> >     [,1]
> > [1,]  1
> > [2,]  4
> > [3,]  6
> > [4,]  8
> > [5,]  10
> > [6,]  12
> >
> > The derivative of x^2 is still 2x, isn't it ?
> 
> and (1:5)^2 is still
> 
> [1]  1  4  9 16 25
> 
> !
> 
> Try
> 
> > x <- as.numeric(1:5)
> > numericDeriv(quote(x^2),"x")
> 
> since the author of numericDeriv has forgotten some
> coercions.
> 
> -- 
> Brian D. Ripley,                 
> ripley at stats.ox.ac.uk
> Professor of Applied Statistics, 
> http://www.stats.ox.ac.uk/~ripley/
> University of Oxford,             Tel:  +44 1865
> 272861 (self)
> 1 South Parks Road,                     +44 1865
> 272866 (PA)
> Oxford OX1 3TG, UK                Fax:  +44 1865
> 272595
>