On 11/07/2012 11:04 AM, Jonas Stein wrote:
Why fails nls with "singular gradient" here? I post a minimal example on the bottom and would be very happy if someone could help me.
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
Kind regards, ########### # define some constants smallc <- 0.0001 t <- seq(0,1,0.001) t0 <- 0.5 tau1 <- 0.02 # generate yy(t) yy <- 1/2 * ( 1- tanh((t - t0)/smallc) * exp(-t / tau1) ) + rnorm(length(t))*0.01 # show the curve plot(x=t, y=yy, pch=18) # prepare data dd <- data.frame(y=yy, x=t) nlsfit <- nls(data=dd, y ~ 1/2 * ( 1- tanh((x - ttt)/smallc) * exp(-x / tau2) ), start=list(ttt=0.4, tau2=0.1) , trace=TRUE) # get error: # Error in nls(data = dd, y ~ 1/2 * (1 - tanh((x - ttt)/smallc) * exp(-x/tau2)), : # singular gradient
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