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Re: NonlinearRegress and numerical functions...
*To*: mathgroup at smc.vnet.net
*Subject*: [mg20180] Re: [mg20132] NonlinearRegress and numerical functions...
*From*: BobHanlon at aol.com
*Date*: Sun, 3 Oct 1999 21:07:44 -0400
*Sender*: owner-wri-mathgroup at wolfram.com
Larske,
If definition of g, use Evaluate so that you only solve the DE once, rather
than each time g is called
g[t1_] :=
Evaluate[y[t] /. DSolve[{y'[t] == -y[t] 3, y[0] == 6}, y[t], t][[1]] /.
t -> t1];
Fit won't be very accurate for four data points, use more
data = {0, 1, 2, 5, 10, 20, 50, 100};
fit = Transpose[{data, g[data]}];
There is no need for a numerical solution (NDSolve) in definition of f. Also,
use Evaluate as above.
f[t1_, a_, b_] :=
Evaluate[y[t] /. DSolve[{y'[t] == -y[t] a, y[0] == b}, y[t], t][[1]] /.
t -> t1];
Needs["Statistics`NonlinearFit`"]
You need to fit to "fit" not "data"
BestFitParameters /. NonlinearRegress[fit, f[t, a, 6], {t}, {a}]
{a\[Rule]3.}
BestFitParameters /. NonlinearRegress[fit, f[t, a, b], {t}, {a,b}]
{a\[Rule]3.,b\[Rule]6.}
Bob Hanlon
In a message dated 10/2/1999 8:47:21 AM, loke at ic.chalmers.se writes:
>I've got a big problem making Mathematica realize that my function is
>numerical. The following problem illustrates it in a rather simplified
>way:
>
>g[t1_] := y[t] /. DSolve[{y'[t] == -y[t] 3, y[0] == 6}, y[t], t][[1]] /.
>t -> t1
>
>data = {0, 10, 20, 100};
>
>fit = Transpose[{data, g[data]}];
>
>f[t1_, a_,b_] := (y[t] /.NDSolve[{y'[t] == -y[t] a, y[0] == b}, y[t],
>{t, 0, 100}][[1]]) /.t -> t1
>
>NonlinearRegress[data, f[t, a, 6], {t}, {a, {2.9, 3.2}, 2, 4}]
>
>
>It should be rather straightforward to solve this, but NonlinearRegress
>seem to evaluate the function 'f' first and the results:
>
>NDSolve::"ndnum": "Encountered non-numerical value for a derivative at
>\
>
>\!\(t\) == \!\(2.680466916142905`*^-274\)."
>
>etc....
>
>I've tried to use Unevaluated and Hold but nothing works!
>
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