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RE: Applying Math 6

  • To: mathgroup at smc.vnet.net
  • Subject: [mg82341] RE: [mg82274] Applying Math 6
  • From: "David Annetts" <davidannetts at aapt.net.au>
  • Date: Thu, 18 Oct 2007 04:47:58 -0400 (EDT)
  • References: <200710170747.DAA13615@smc.vnet.net>

Hi Johan,

> I need to fit x-y data to the equation below and also need to 
> determine the derivative.
> Please help.
> 
> y=exp[-(a*ln(x))/bc]^d + exp[-(a*ln(x)/be]^f
> 
> This equation is used in adsorption studies and is normally 
> referred to as the Dubinin-Izotova equation.

The first thing would be to read at least guide/MathematicalFunctions in the
online help (assuming 6.0.x).

Your model is

	mdl = Exp[-(a*Log[x])/bc]^d + Exp[-(a*Log[x])/be]^f

and we can get its derivative wrt x using

	D[mdl, x] // Simplify

Check that this is OK by integrating

	Integrate[%, x]//Simplify

As to your problem ....

We can generate some noisy test data (then plot them) using

	data = Table[{x, RandomReal[{.01, .01}] x + mdl} /. {a -> 1, be ->
.5, bc -> 2, d -> 2, f -> 3}, {x, 1, 10, .1}];
	ListPlot[data]

I don't know whether these parameter values are realistic; I guess that's
your field.

Standard nonlinear least-squares fitting is accomplished using

	fit = FindFit[data, mdl, {a, bc, be, d, f}, x]

and we can compare the original data with the modelled data using

	Show[{
	  ListPlot[data],
	  Plot[mdl /. fit, {x, 1, 10}]
	  },
	 PlotRange -> All]

There is a nonlinear least square fitting package that we can load

	Needs["NonlinearRegression`"]

usage is similar to FindFit, and 

	nlfit = NonlinearRegress[data, mdl, {a, bc, be, d, f}, x]

gives exactly the same answers as FitFit but with more diagnostic
information.

If we know starting parameters, then 

	nlf = NonlinearRegress[data, mdl, {{a, 1}, {bc, 2}, {be, .5}, {d,
2}, {f, 3}}, x, 
		RegressionReport -> {BestFit, PredictedResponse}]

and this also gives the parameters found by FindFit (compare BestFit /. nlf
with mdl /. fit).

We can compare our models with our data using

	Show[{
	  ListPlot[data],
	  Plot[mdl /. fit, {x, 1, 10}],
	  ListPlot[Transpose[{Range[1, 10, .1], PredictedResponse /. nlf}],
Joined -> True, PlotStyle -> Red]
	  },
	 PlotRange -> All]

Straightforward, and powerful.  But only after learing a bit of syntax.

Regards,

Dave.



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