On Sat, Oct 1, 2011 at 5:28 AM, Casper Ti. Vector
<caspervec...@gmail.com> wrote:
> Example:
>
>> f <- function(x) { 1 + 2 * log(1 + 3 * x) + rnorm(1, sd = 0.5) }
>> y <- f(x <- c(1 : 10)); y
> [1] 4.503841 5.623073 6.336423 6.861151 7.276430 7.620131 7.913338 8.169004
> [9] 8.395662 8.599227
>> nls(x ~ a + b * log(1 + c * x), start = list(a = 1, b = 2, c = 3), trace =
>> TRUE)
> 37.22954 : 1 2 3
> Error in numericDeriv(form[[3L]], names(ind), env) :
> Missing value or an infinity produced when evaluating the model
> In addition: Warning message:
> In log(1 + c * x) : NaNs produced
>
> What's wrong here? Am I handling this problem in the wrong way?
> Any suggestions are welcome, thanks :)
>
Its linear given c so calculate the residual sum of squares using lm
(or lm.fit which is faster) given c and optimize over c:
set.seed(123) # for reproducibility
# test data
x <- 1:10
y <- 1 + 2 * log(1 + 3 * x) + rnorm(1, sd = 0.5)
# calculate residual sum of squares for best fit given c
fitc <- function(c) lm.fit(cbind(1, log(1 + c * x)), y)
rssvals <- function(c) sum(resid(fitc(c))^2)
out <- optimize(rssvals, c(0.01, 10))
which gives:
> setNames(c(coef(fitc(out$minimum)), out$minimum), letters[1:3])
a b c
0.7197666 2.0000007 2.9999899
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