Dear Ravi,
the inaccuracy seems to creep in when powers are calculated. Apparently,
some quite general function is called to calculate the squares, and one
can avoid the error by reformulating the example as follows:
rosen <- function(x) {
n <- length(x)
x1 <- x[2:n]
x2 <- x[1:(n-1)]
sum(100*(x1-x2*x2)*(x1-x2*x2) + (1-x2)*(1-x2))
}
x0 <- c(0.0094, 0.7146, 0.2179, 0.6883, 0.5757, 0.9549, 0.7136, 0.0849, 0.4147,
0.4540)
h <- c(1.e-15*1i, 0, 0, 0, 0, 0, 0, 0, 0, 0)
xh <- x0 + h
rx <- rosen(xh)
Re(rx)
Im (rx)
I don't know which arithmetics are involved in the application you
mentioned, but writing some auxiliary function for the calculation of
x^n when x is complex and n is (a not too large) integer may solve some
of the numerical issues. A simple version is:
powN <- function(x,n) sapply(x,function(x) prod(rep(x,n)))
The corresponding summation in 'rosen' would then read:
sum(100*powN(x1-powN(x2,2),2) + powN(1-x2,2))
HTH,
Martin
Ravi Varadhan wrote:
Dear All,
Hans Borchers and I have been trying to compute "exact" derivatives in R using the idea of complex-step derivatives that Hans has proposed. This is a really, really cool idea. It gives "exact" derivatives with only a minimal effort (same as that involved in computing first-order forward-difference derivative).
Unfortunately, we cannot implement this in R as the "complex arithmetic" in R
appears to be inaccurate.
Here is an example:
#-- Classical Rosenbrock function in n variables
rosen <- function(x) {
n <- length(x)
x1 <- x[2:n]
x2 <- x[1:(n-1)]
sum(100*(x1-x2^2)^2 + (1-x2)^2)
}
x0 <- c(0.0094, 0.7146, 0.2179, 0.6883, 0.5757, 0.9549, 0.7136, 0.0849, 0.4147,
0.4540)
h <- c(1.e-15*1i, 0, 0, 0, 0, 0, 0, 0, 0, 0)
xh <- x0 + h
rx <- rosen(xh)
Re(rx)
Im (rx)
# rx = 190.3079796814885 - 12.13915588266717e-15 i # incorrect imaginary part
in R
However, the imaginary part of the above answer is inaccurate. The correct
imaginary part (from Matlab) is:
190.3079796814886 - 4.66776376640000e-15 i # correct imaginary part from Matlab
This inaccuracy is serious enough to affect the acuracy of the compex-step
gradient drastically.
Hans and I were wondering if there is a way to obtain accurate "small" imaginary part for complex arithmetic.
I am using Windows XP operating system.
Thanks for taking a look at this.
Best regards,
Ravi.
____________________________________________________________________
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology
School of Medicine
Johns Hopkins University
Ph. (410) 502-2619
email: rvarad...@jhmi.edu
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--
Dr. Martin Becker
Statistics and Econometrics
Saarland University
Campus C3 1, Room 206
66123 Saarbruecken
Germany
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