Displaying 4 results from an estimated 4 matches for "lassp".
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2001 Dec 15
1
fit to spike with exponential decay : optim() question
...differentiable function, I had (stupidly)
'simplified' the problem until it became singular.
Can I do something to make optim() less sensitive to my initial guess? For
this example, I get a lousy solution if I make the initial guess for t0 =
min(t) = 0.05.
Thanks again,
--
Robert Merithew
LASSP, Clark Hall
Cornell University, Ithaca NY
-----------
t <- c(0.05,0.9,1.4,2.38,3.42,5.4,8.31,12.4)
amp <- c(1.0,0.85,7.4, 6.1, 4.95, 3.5, 2.3, 1.5)
spike <- function (x, t) {
b0 <- x[1]
b1 <- x[2]
tau <- x[3]
t0 <- x[4]
temp <- exp((-t+t0)/tau)
(b0 + (b1 *...
2001 Oct 05
1
nls() fit to a lorentzian - can I specify partials?
...lue
})
I think .expr10 should look more like:
Q * sqrt((1 - .expr1^2)^2 + (.expr1 * 1/Q)^2)
Is there simply a problem with the way the output of deriv() is displayed
(missing 2 sets of parentheses)?
Finally: Is there a better way to do my lorentzian fits with R?
thanks,
--
Robert Merithew
LASSP, Cornell University
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2001 Sep 01
2
interpolation and numerical differentiation in R ?
...state physics experiments, and very little of
the analysis is really *statistical*, so it may be that R is the wrong
tool for me.
If anyone can point me toward documentation covering how to do tasks like
the above in R, I'd appreciate it.
thanks,
-Robert Merithew, PhD rdm28 at cornell.edu
LASSP, Cornell University
Ithaca, NY
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Send "info", "help", or "[un]subscribe"
(in the "body", not the subjec...
2002 Jun 04
4
par(xaxp)
I think this is a bug; at least this behavior is not documented in plot
or plot.default.
plot.default resets xaxp, and leaves xaxp reset when it exits:
par(xaxp=c(0,1,4))
print(par("xaxp"))
plot(c(0,1),c(0.2,0.3))
print(par("xaxp"))
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