Your model is singular. Varying m and log(l) have the same
effect: log(ir) = log(k) + m * log(l) * ox
Also with plinear you don't specify the linear coefficients but
rather an X matrix whose coefficients represent them:
If we use this model instead:
ir = k * exp(m * ox)
Then:
> mod0 <- lm(log(ir) ~ ox)
> mod0
Call:
lm(formula = log(ir) ~ ox)
Coefficients:
(Intercept) ox
2.199743 0.003835
> nls(ir ~ exp(m * ox), start = list(m = coef(mod0)[2]), algorithm = "plinear")
Nonlinear regression model
model: ir ~ exp(m * ox)
data: parent.frame()
m .lin
0.003991 9.091758
residual sum-of-squares: 0.3551
Number of iterations to convergence: 3
Achieved convergence tolerance: 5.289e-07
On Dec 24, 2007 9:04 AM, <[EMAIL PROTECTED]> wrote:
> I'm trying to fit a function y=k*l^(m*x) to some data points, with
> reasonable starting value estimates (I think). I keep getting "singular
> matrix 'a' in solve".
>
> This is the code:
>
> ox <- c(-600,-300,-200,1,100,200)
> ir <- c(1,2.5,4,9,14,20)
> model <- nls(ir ~ k*l^(m*ox),start=list(k=10,l=3,m=0.004),algorithm="plinear")
> summary(model)
> plot(ox,ir)
> testox <- seq(-600,200,length=100)
> k <- 10
> l <- 3
> m <- 0.004
> testir <- k*l^(m*testox)
> lines(testox,testir)
>
> Any thoughts?
> Thanks!
>
> --
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