Christian Ritz pisze:
> Hi Jarek,
>
> an alternative approach is to provide more precise starting values!
>
> It pays off to realise that it's possible to find quite good guesses
> for some of the parameters in your model function:
>
> t ~ tr+(ts-tr)/((1+(a*h)^n)^(1-(1/n)))
>
> The par
Hi Jarek,
an alternative approach is to provide more precise starting values!
It pays off to realise that it's possible to find quite good guesses
for some of the parameters in your model function:
t ~ tr+(ts-tr)/((1+(a*h)^n)^(1-(1/n)))
The parameters tr and ts correspond to the re
On Jan 23, 2008 3:44 PM, Jarek Jasiewicz <[EMAIL PROTECTED]> wrote:
> Gabor Grothendieck wrote:
> > Try port algorithm. Please provide complete reproducible code and data
> > when posting.
> >
> >
> >> y <- 1/(x <- 1:5)
> >> nls(y~k/x^n, start=list(k=1, n=1), algorithm = "port")
> >>
> > Nonlinear
Gabor Grothendieck wrote:
> Try port algorithm. Please provide complete reproducible code and data
> when posting.
>
>
>> y <- 1/(x <- 1:5)
>> nls(y~k/x^n, start=list(k=1, n=1), algorithm = "port")
>>
> Nonlinear regression model
> model: y ~ k/x^n
>data: parent.frame()
> k n
> 1 1
Try port algorithm. Please provide complete reproducible code and data
when posting.
> y <- 1/(x <- 1:5)
> nls(y~k/x^n, start=list(k=1, n=1), algorithm = "port")
Nonlinear regression model
model: y ~ k/x^n
data: parent.frame()
k n
1 1
residual sum-of-squares: 0
Algorithm "port", converge
Hi!
How to write in model's formula of type:
nls(y~k/(x^n), data=data, start=list(k=1,n=1))
i.e the problem is on x^n, I(x^n) generate error
thanks
Jarek
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