Ben Bolker bolker at ufl.edu
Sat Oct 20 14:18:50 CEST 2007

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>
> dear list
>   I am student M.S. statistics in department statistics . I am working in
> the function "nls" in the [R 2.3.1] with 246 data and want to fit a model
> to vectors( v and u ) but I have
> a problem to use it
>
>

Thanks for providing a reproducible example.  Your version of R is
_really_ old:
you should probably update (or ask you administrator to update) to make sure
that all possible bugs etc. are fixed.

u=c(5.000000e-13,2.179057e+03,6.537171e+03,1.089529e+04,1.525340e+04,
1.91151e+04,2.396963e+04,2.832774e+04,3.268586e+04,3.704397e+04,
4.140209e+04,4.576020e+04,5.011831e+04,5.447643e+04)
v=c(8.382562e-01,4.090868e+02,1.311053e+03,2.124143e+03,3.365494e+03,
2.138903e+03,7.687774e+03,1.028396e+04,1.004186e+04,2.059798e+04,
1.438464e+04,2.861373e+04,2.294919e+04,2.807701e+04)

## always a good idea to look at the data
plot(v~u)

nls(v~c0+.5(a^2)*(u^2),start=list(c0=0,a=.3))

## you need a * between ".5" and "(a^2)"
m2 = nls(v~c0+.5*(a^2)*(u^2),start=list(c0=0,a=.3))
coef(n2)

## you can also fit this (quadratic regression) with lm
m3 = lm(v~I(u^2))
## converting the coefficient
sqrt(2*coef(m3))

## you need to use sensible starting conditions!
##   one way to get is to look at the plot and
##    take a guess.  Another would be to assume
##   c0=0 and fit a linear model to log(v) as a function
##   of log(u)

nls(v~c0+a*(u^s),start=list(c0=0,a=1,s=2))
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