On 07/12/2012 01:39 AM, Duncan Murdoch wrote:
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
On Jul 11, 2012, at 20:34 , 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
)
--
Best, JN
On 07/12/2012 06:00 AM, r-help-requ...@r-project.org wrote:
From: Jonas Stein n...@jonasstein.de
To: r-h...@stat.math.ethz.ch
Subject: [R] nls problem: singular gradient
Message-ID: e8h0d9-ao4@news.jonasstein.de
Content-Type: text/plain
Why fails
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.
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)
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
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,
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
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