On Thu, Apr 19, 2012 at 03:12:34PM -0800, Katrina Bennett wrote:
> Dear R Help,
>
> Sorry I wasn't more clear before. Here is another crack at this.
>
> What I am still trying to do is estimate the point on a line when the
> slope changes or asymptotes. I have found some similar postings
> talkin
Katrina:
>
> Is there a better way to then pick off the change points or find the
> asymptotes of a function in R?
Basic calculus? (if parametrically specified).
Otherwise, probably best to consult your local statistician (or maybe
a numerical analyst).
-- Bert
>
> Thank you.
>
> Katrina
>
>
Dear R Help,
Sorry I wasn't more clear before. Here is another crack at this.
What I am still trying to do is estimate the point on a line when the
slope changes or asymptotes. I have found some similar postings
talking about this but no answers.
https://stat.ethz.ch/pipermail/r-help/2003-Januar
Hi David, thanks for the reply.
This is not a homework problem, although it may sound like one :) I
was trying to provide a reproducible example of what I am trying to
do.
The problem is something I am trying to work on for my PhD program.
I've been using the nls() function to derive a self-start
On Apr 19, 2012, at 4:41 AM, Katrina Bennett wrote:
Hi all,
I would like to find the x position of an two asymptotes.
Here is a sample of what I would like to do:
x <- seq(1, 153,, 153)
a <- 85
m <- 65
s =-1.5
fn <- function (x, a, m, s) { a * (exp((m - x)/s) * (1/s))/((1 +
exp((m - x)/s)))^
Hi all,
I would like to find the x position of an two asymptotes.
Here is a sample of what I would like to do:
x <- seq(1, 153,, 153)
a <- 85
m <- 65
s =-1.5
fn <- function (x, a, m, s) { a * (exp((m - x)/s) * (1/s))/((1 +
exp((m - x)/s)))^2 }
plot.deriv1 <- fn(1:153, a, m, s)
I can find the mi
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