It seems to work!
Thanks and sorry for the personal message
Francisco
On Tue, 17 Apr 2012 09:09:24 +0200, peter dalgaard wrote
> On Apr 17, 2012, at 06:23 , Francisco Mora Ardila wrote:
>
> > Hi
> >
> > I´m trying to fit a nonlinear model to a derivative of the logistic function
> >
> > y = a
2 1.080
residual sum-of-squares: 50.5
Number of iterations to convergence: 0
Achieved convergence tolerance: 4.818e-07
>
On 04/17/2012 06:00 AM, r-help-requ...@r-project.org wrote:
> Message: 112
> Date: Mon, 16 Apr 2012 23:23:07 -0500
> From: "Francisco Mora Ardila"
>
On Tue, Apr 17, 2012 at 12:23 AM, Francisco Mora Ardila
wrote:
> Hi
>
> I´m trying to fit a nonlinear model to a derivative of the logistic function
>
> y = a/(1+exp((b-x)/c)) (this is the parametrization for the SSlogis function
> with nls)
>
> The derivative calculated with D function is:
>
>>
On Apr 17, 2012, at 06:23 , Francisco Mora Ardila wrote:
> Hi
>
> I´m trying to fit a nonlinear model to a derivative of the logistic function
>
> y = a/(1+exp((b-x)/c)) (this is the parametrization for the SSlogis function
> with nls)
>
> The derivative calculated with D function is:
>
>> l
Hi
I´m trying to fit a nonlinear model to a derivative of the logistic function
y = a/(1+exp((b-x)/c)) (this is the parametrization for the SSlogis function
with nls)
The derivative calculated with D function is:
> logis<- expression(a/(1+exp((b-x)/c)))
> D(logis, "x")
a * (exp((b - x)/c) * (1
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