[R] nls problem: singular gradient

[Duncan Murdoch]

On 12-07-11 2:34 PM, Jonas Stein wrote:
>> Take a look at the predicted values at your starting fit:  there's a
>> discontinuity at 0.4, which sure makes it look as though overflow is
>> occurring.  I'd recommend expanding tanh() in terms of exponentials and
>> rewrite the prediction in a way that won't overflow.
>>
>> Duncan Murdoch
>
> Hi Duncan,
> Thank you for your suggestion. I wrote a function "mytanh" and
> nls terminates a bit later with another error message:
>
> Error in nls(data = dd, y ~ 1/2 * (1 - mytanh((x - ttt)/1e-04) * exp(-x/tau2)),  :
>    number of iterations exceeded maximum of 50
>
> How can i fix that?
> Kind regards,
> Jonas
>
> ============================ R CODE STARTS HERE =======
>
> mytanh<- function(x){
>    return(x - x^3/3 + 2*x^5 /15 - 17 * x^7/315)
> }

That looks like it would overflow as soon as abs(x-ttt) got large, just 
like the original.  You might be able to fix it by following the advice 
I gave last time, or maybe you need to rescale the parameters.  In most 
cases optimizers work best when the uncertainty in the parameters is all 
on the same scale, typically around 1.

Duncan Murdoch

>
> t<- seq(0,1,0.001)
> t0<- 0.5
> tau1<- 0.02
>
> yy<- 1/2 * ( 1- tanh((t - t0)/0.0001) * exp(-t / tau1) ) + rnorm(length(t))*0.001
>
> plot(x=t, y=yy, pch=18)
>
> dd<- data.frame(y=yy, x=t)
>
> nlsfit<- nls(data=dd,  y ~  1/2 * ( 1- mytanh((x - ttt)/0.0001) * exp(-x / tau2) ), start=list(ttt=0.5, tau2=0.02) , trace=TRUE)
>
> ============================ R CODE ENDS HERE =======
>