From: Uzuner, Tolga <tolga.uzuner_at_csfb.com>

Date: Fri 06 May 2005 - 08:20:57 EST

b.mtr <- matrix(1, (r - 1), n)

*#------------------------------------------------------------------------
*

*# 1 Calculate the derivative formula given in 'GENERAL APPROACH' section.
*

*# -- The first order derivatives are stored in the matrix a.mtr[k,i],
*

*# where the indexing variables k for rows(1 to r), i for columns
*

*# (1 to n), r is the number of iterations, and n is the number of
*

# variables.

#-------------------------------------------------------------------------

}

if(show) {

}

}

a.mtr[length(a.mtr)]

}

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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Fri May 06 08:29:50 2005

Date: Fri 06 May 2005 - 08:20:57 EST

Hi,

I have been trying to do numerical differentiation using R.

I found some old S code using Richardson Extrapolation which I managed to get to work.

I am posting it here in case anyone needs it.

########################################################################

richardson.grad <- function(func, x, d=0.01, eps=1e-4, r=6, show=F){

*# This function calculates a numerical approximation of the first
*

# derivative of func at the point x. The calculation

# is done by Richardson's extrapolation (see eg. G.R.Linfield and

J.E.T.Penny

*# "Microcomputers in Numerical Analysis"). The method should be used if
**# accuracy, as opposed to speed, is important.
**#
**# * modified by Paul Gilbert from orginal code by XINGQIAO LIU.
**# CALCULATES THE FIRST ORDER DERIVATIVE
**# VECTOR using a Richardson extrapolation.
**#
**# GENERAL APPROACH
**# -- GIVEN THE FOLLOWING INITIAL VALUES:
**# INTERVAL VALUE D, NUMBER OF ITERATIONS R, AND
**# REDUCED FACTOR V.
**# - THE FIRST ORDER aproximation to the DERIVATIVE WITH RESPECT TO Xi IS
**#
**# F'(Xi)={F(X1,...,Xi+D,...,Xn) - F(X1,...,Xi-D,...,Xn)}/(2*D)
**#
**# -- REPEAT r TIMES with successively smaller D and
**# then apply Richardson extraplolation
**#
**# INPUT
**# func Name of the function.
**# x The parameters of func.
**# d Initial interval value (real) by default set to 0.01*x or
**# eps if x is 0.0.
**# r The number of Richardson improvement iterations.
**# show If T show intermediate results.
**# OUTPUT
**#
*

# The gradient vector.

v <- 2 # reduction factor. n <- length(x) # Integer, number of variables. a.mtr <- matrix(1, r, n)

b.mtr <- matrix(1, (r - 1), n)

# variables.

#-------------------------------------------------------------------------

h <- abs(d*x)+eps*(x==0.0)

for(k in 1:r) { # successively reduce h for(i in 1:n) { x1.vct <- x2.vct <- x x1.vct[i] <- x[i] + h[i] x2.vct[i] <- x[i] - h[i] if(k == 1) a.mtr[k,i] <- (func(x1.vct) - func(x2.vct))/(2*h[i]) else{ if(abs(a.mtr[(k-1),i])>1e-20) # some functions are unstable near 0.0 a.mtr[k,i] <- (func(x1.vct)-func(x2.vct))/(2*h[i]) else a.mtr[k, i] <- 0 } } h <- h/v # Reduced h by 1/v.

}

if(show) {

cat("\n","first order approximations", "\n") print(a.mtr, 12)

}

*#------------------------------------------------------------------------
**# 1 Applying Richardson Extrapolation to improve the accuracy of
**# the first and second order derivatives. The algorithm as follows:
**#
**# -- For each column of the 1st and 2nd order derivatives matrix a.mtr,
**# say, A1, A2, ..., Ar, by Richardson Extrapolation, to calculate a
**# new sequence of approximations B1, B2, ..., Br used the formula
**#
**# B(i) =( A(i+1)*4^m - A(i) ) / (4^m - 1) , i=1,2,...,r-m
**#
**# N.B. This formula assumes v=2.
**#
**# -- Initially m is taken as 1 and then the process is repeated
**# restarting with the latest improved values and increasing the
**# value of m by one each until m equals r-1
**#
**# 2 Display the improved derivatives for each
**# m from 1 to r-1 if the argument show=T.
**#
*

# 3 Return the final improved derivative vector.

#-------------------------------------------------------------------------

for(i in 1:(r - m)) b.mtr[i,]<- (a.mtr[(i+1),]*(4^m)-a.mtr[i,])/(4^m-1)

# a.mtr<- b.mtr

# a.mtr<- (a.mtr[2:(r+1-m),]*(4^m)-a.mtr[1:(r-m),])/(4^m-1)

if(show & m!=(r-1) ) { cat("\n","Richarson improvement group No. ", m, "\n") print(a.mtr[1:(r-m),], 12) }

}

a.mtr[length(a.mtr)]

}

## try it out

richardson.grad(function(x){x^3},2)

########################################################################################

Regards,

Tolga Uzuner

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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Fri May 06 08:29:50 2005

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