numericDeriv

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