[R] Find position of asymptote

[Katrina Bennett]

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-starting logistic
function to calculate snowmelt duration in Alaskan watersheds.

Here's a snippet of the nls function calculation **not reproducible**
#year 2000
print("year 2000")
Asym.2000 <- mean(na.omit(dat.cl.2000$dat)[1:10])
na.dat.2000 <- which(!is.na(dat.cl.2000$dat))
loess.smooth.2000 <- loess.smooth(dat.cl.2000$x[na.dat.2000],
dat.cl.2000$dat[na.dat.2000], span=0.20)
diff.loess.smooth.2000 <- which(diff(loess.smooth.2000$y) ==
(min(diff(loess.smooth.2000$y))), arr.ind=T)
val.in.jday.2000 <- loess.smooth.2000$y[diff.loess.smooth.2000]
xmid.2000 <- val.in.jday.2000
scal.2000 <- (min(na.omit(dat.cl.2000$dat)) -
max(na.omit(dat.cl.2000$dat))) / (max(na.omit(dat.cl.2000$x)) -
min(na.omit(dat.cl.2000$x))) * 10
fit.dat.2000 <- nls(dat ~ SSlogis(x, Asym, xmid,scal), data =
dat.cl.2000, start=list(Asym=round(Asym.2000), xmid=round(xmid.2000),
scal=scal.2000), control = list(maxiter = 500, warnOnly = TRUE),
trace=TRUE)

What I want to be able to do however, is to determine the start and
end points (dates) of the asymptote of the curve when the snowmelt
initiates.

Prior to this I was using the second derivative mins and maxes.

#init <- optimize(f=fn.sslogis.deriv2, interval=c(1:mid.point), a=a,
m=m, s=s)$minimum #this is the start of the snowmelt season
#term <- optimize(f=fn.sslogis.deriv2, interval=c(mid.point:153), a=a,
m=m, s=s, maximum=TRUE)$maximum #this is the end of the snowmelt
season

However, this cuts the melt too short.

Any assistance on this non-hw problem would be appreciated.



Sincerely,
Katrina

On Thu, Apr 19, 2012 at 4:34 AM, David Winsemius <dwinsemius at comcast.net> wrote:
>
> 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)))^2 }
>> plot.deriv1 <- fn(1:153, a, m, s)
>>
>> I can find the midpoint/minimum of this function easily. However, I
>> can not find the start and end of the two asymptote position (along
>> the x).
>
>
> This is not a homework help-line.
>
>
>> mid.point <- optimize(f=fn.sslogis.deriv1, interval=c(1:153), a=a,
>> m=m, s=s)$minimum
>>
>> Thanks for your assistance,
>>
>> Katrina
>
>
> --
>
> David Winsemius, MD
> West Hartford, CT
>



-- 
Katrina E. Bennett
PhD Student
University of Alaska Fairbanks
International Arctic Research Center
930 Koyukuk Drive, PO Box 757340
Fairbanks, Alaska 99775-7340
907-474-1939 office
907-385-7657 cell
kebennett at alaska.edu


Personal Address:
UAF, PO Box 752525
Fairbanks, Alaska 99775-2525
bennett.katrina at gmail.com