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Re: Re: 'NonlinearFit` confusion

  • To: mathgroup at smc.vnet.net
  • Subject: [mg49977] Re: [mg49895] Re: [mg49844] 'NonlinearFit` confusion
  • From: "Janos D. Pinter" <jdpinter at hfx.eastlink.ca>
  • Date: Mon, 9 Aug 2004 04:29:15 -0400 (EDT)
  • References: <200408041446.KAA20105@smc.vnet.net> <200408051321.JAA05885@smc.vnet.net> <6.1.2.0.1.20040807114330.04939518@pop.hfx.eastlink.ca> <Pine.OSX.4.58.0408071231350.28763@heisenberg> <6.1.2.0.1.20040807151006.049bd348@pop.hfx.eastlink.ca> <Pine.OSX.4.58.0408071618220.28780@heisenberg>
  • Sender: owner-wri-mathgroup at wolfram.com

Yas,

[my] famous last words on the subject: nothing will make a local search 
method global - except a clairvoyant user who knows where to start the search.

Of course, good insight and modeling practice are always essential, but the 
issue of error surface multi-extremality still neds to be addressed. Hence, 
nonlinear model fitting often (typically) needs a global scope search 
method such as NMinimize, MathOptimizer or MathOptimizer Pro.

Where is that beer?

Cheers,
JDP





At 07:06 PM 8/7/2004, you wrote:
>Hi,
>Yes OK, I agree with your last sentence, but I haven't skipped any data
>points. I have only used what's relevant. On the other hand you have now
>introduced n>1, hills, valleys, a ball, friction and gravity. This sounds
>like an enjoyable discussion but I'd prefer to do that over a beer or two.
>In any case, let's go back to the original problem for the moment, bearing
>in mind that I'm not interested in what might or might not happen, but
>given a problem, what's the solution?.
>I would like to see FindFit come to a solution with so many
>points as posed in the original problem. A computer is a fantastic GIGO
>device. I had a short go, using Simulated Annealing, NelderMead methods,
>Gradien etc, but came to realise that the problem as posed needed some
>thought. What am I fitting to?! Again, the point here remains. I don't
>need to make FindFit start somehwre in the middle. The first point is a
>good starting point, so as long as the number of points used to fit to the
>function are representative of the function! I haven't left out data
>points. I have simply chosen what is physically relevant and used a
>fitting algorithm to fit model function to the data. The
>Levenberg-Marquardt algorithm is good enough. And it leads to the expected
>solution. Even with n>1 dimensions as you say, if there is only one valley
>and you roll down the hill you will end up in the valley!! You don't need
>a million points to work out that you are in the valley. There is of
>course more to it than that, and of course some common sense must come
>into play some time.
>
>Cheers
>Have a good one
>Yas
>
>
>
>
>On Sat, 7 Aug 2004, Janos D. Pinter wrote:
>
> >
> > Yas,
> >
> > in everyday terms, multi-extremality can be visualized by a 'hilly' region
> > where different starting points will lead to different valleys. (Imagine a
> > rolling ball started from different points in this region, assuming gravity
> > and friction). FindFit will end up at the right solution only if started in
> > the right valley (region of attraction). In n>1 dimensions it is a problem
> > to take a good look at the error fct, for most of us. :-)
> >
> > To skip points could (will) bring in another element of subjectivity,
> > leading to point subset-dependent results.
> >
> > Cheers,
> >
> > Janos Pinter
> > www.pinterconsulting.com
> >
> >
> >
> > At 02:47 PM 8/7/2004, you wrote:
> > >G'day,
> > >Yes, but I have no idea what multi-extremality means in English. You can
> > >use FindFit as well. But here, one just needs to take a good look at the
> > >model function to quickly realise where the sample points should be and
> > >how many.
> > >It is always an assumption by most that when fitting data that more
> > >points are better. And its usually forgotten that all you need
> > >are sufficient data points to smoothly sample the model function.
> > >Obviously in this case the model function can be smoothly sampled through
> > >a single period. Perhaps that in itself is too much.
> > >I think Dr Bob has already pointed out something about fitting, non-exact
> > >and science. Which leaves the physical intepretation of the function used
> > >to model the data!
> > >You may have also noticed that I didn't skip data points. I simple took
> > >the first 40 or so.
> > >
> > >Best Regards
> > >Yas
> > >
> > >
> > >  On Sat, 7 Aug 2004, Janos D. Pinter wrote:
> > >
> > > >
> > > > Yas,
> > > >
> > > > skipping some of the data will not eliminate the potential
> > > > multi-extremality of a nonlinear model-fitting problem (but can 
> make the
> > > > solution less stable). As Paul Abbott noted, either one needs a
> > > > 'sufficiently good' starting point for successful local
> > > > fitting/optimization, or needs to apply global optimization methods 
> to find
> > > > the best numerical fit.
> > > >
> > > > Regards,
> > > > Janos Pinter
> > > >
> > > >
> > > > At 10:21 AM 8/5/2004, you wrote:
> > > > >Hi,
> > > > >
> > > > >Why not take less points, rather than more??
> > > > >
> > > > >In[480]:=
> > > > >Clear[datay,datax,data1]
> > > > >datay = Table [6 + 2Sin[3 + x], {x, -10Pi, 10Pi, 0.05}];
> > > > >datax = Table [x, {x, -10Pi, 10Pi, 0.05}];
> > > > >data1 = Table[{datax[[i]], datay[[i]]}, {i, 1, Length[datay]}];
> > > > >Length[data1]
> > > > >First[data1]
> > > > >
> > > > >Out[484]=
> > > > >1257
> > > > >
> > > > >Out[485]=
> > > > >{-31.4159,6.28224}
> > > > >
> > > > >In[511]:=
> > > > >Clear[a,b,c,d,e]
> > > > >sinFit1=NonlinearFit[Take[data1,40], c + a *Sin[d + e *x], x, {a, c,
> > > > >d,e}]
> > > > >
> > > > >Out[512]=
> > > > >6.\[InvisibleSpace]+2. Sin[3.\[InvisibleSpace]+1. x]
> > > > >
> > > > >
> > > > >In[513]:=
> > > > >Clear[y,datay,datax,a,b,c,d,e]
> > > > >datay = Table [6 + 2Sin[3 + 5 x], {x, -10Pi, 10Pi, 0.05}];
> > > > >datax = Table [x, {x, -10Pi, 10Pi, 0.05}];
> > > > >data2 = Table[{datax[[i]], datay[[i]]}, {i, 1, Length[datay]}];
> > > > >Length[data2]
> > > > >First[data2]
> > > > >
> > > > >Out[517]=
> > > > >1257
> > > > >
> > > > >Out[518]=
> > > > >{-31.4159,6.28224}
> > > > >
> > > > >In[519]:=
> > > > >y=a*Sin[d+e x]+c;
> > > > >sinFit2=NonlinearFit[Take[data2,40],y, x, {a,c, d,e}]
> > > > >
> > > > >Out[520]=
> > > > >6.\[InvisibleSpace]+2. Sin[122.381\[InvisibleSpace]+5. x]
> > > > >
> > > > >
> > > > >DisplayTogether[
> > > > >    Plot[sinFit1, {x, -10Pi, -8 Pi}, PlotStyle -> Blue],
> > > > >    Plot[sinFit2, {x, -10Pi, -8 Pi}, PlotStyle -> Red],
> > > > >
> > > > >    ListPlot[Take[data1, 40], PlotStyle -> {Blue, PointSize[0.011]}],
> > > > >    ListPlot[Take[data2, 30], PlotStyle -> {Red, PointSize[0.011]}],
> > > > >    Prolog -> {
> > > > >        {
> > > > >          Blue,
> > > > >          Text["sinFit1", {-27.5, 7}, {-1, 0},
> > > > >            TextStyle -> {FontFamily -> "Times",
> > > > >                FontWeight -> "Bold",
> > > > >                FontSize -> 18
> > > > >                }
> > > > >            ]
> > > > >          },
> > > > >
> > > > >        {
> > > > >          Red,
> > > > >          Text["sinFit2", {-26.9, 6}, {-1, 0},
> > > > >            TextStyle -> {FontFamily -> "Times",
> > > > >                FontWeight -> "Bold",
> > > > >                FontSize -> 18
> > > > >                }
> > > > >            ]
> > > > >          }
> > > > >        },
> > > > >    ImageSize -> 800,
> > > > >    Frame -> False,
> > > > >    AxesOrigin -> {-31.5, 4.0}
> > > > >
> > > > >    ]
> > > > >
> > > > >Cheers
> > > > >Yas
> > > > >
> > > > >
> > > > >
> > > > >On Aug 4, 2004, at 9:46 AM, Klingot wrote:
> > > > >
> > > > > > I'm trying to fit a sinusoidal function to my data using
> > > > > > 'NonlinearFit' but it's exhibiting rather odd behaviour. Please 
> see my
> > > > > > examples below:
> > > > > >
> > > > > > EXAMPLE (1).
> > > > > >
> > > > > > **** As a test, I created a list of data from a function of the 
> form y
> > > > > > = 6 + 2Sin[3 + x] with:
> > > > > >
> > > > > > datay = Table [6 + 2Sin[3 + x], {x, -10Pi, 10Pi, 0.05}];
> > > > > > datax = Table [x, {x, -10Pi, 10Pi, 0.05}];
> > > > > > data = Table[{datax[[i]], datay[[i]]}, {i, 1, Length[datay]}];
> > > > > >
> > > > > > **** Then tested to see whether NonlinearFit would correctly deduce
> > > > > > the equations parameters with:
> > > > > >
> > > > > > NonlinearFit[data, c + a Sin[d + e x], x, {a, c, d,e}]
> > > > > >
> > > > > > **** As expected, it gave me '6.+ 2. Sin[3. + 1. x]' ... exactly as
> > > > > > one would expect :)
> > > > > >
> > > > > >
> > > > > > EXAMPLE (2).
> > > > > >
> > > > > > **** Second test, I modified the equation by multiplying x by 
> 5, ie.
> > > > > > y = 6 + 2Sin[3 + 5x]:
> > > > > >
> > > > > > datay = Table [6 + 2Sin[3 + 5 x], {x, -10Pi, 10Pi, 0.05}];
> > > > > > datax = Table [x, {x, -10Pi, 10Pi, 0.05}];
> > > > > > data = Table[{datax[[i]], datay[[i]]}, {i, 1, Length[datay]}];
> > > > > >
> > > > > > ***** and applied the NonlinearFit as before:
> > > > > >
> > > > > > NonlinearFit[data, c + a Sin[d + e x], x, {a, c, d,e}]
> > > > > >
> > > > > > ***** but this time I get a wildly innacurate result:    6.000165 +
> > > > > > 0.025086 Sin[0.0080308 - 0.247967 x]
> > > > > >
> > > > > > Specifically, the parameters 'a', 'd' and 'e' are all completely in
> > > > > > error by orders of magnitude.
> > > > > >
> > > > > > I tried extending the range of the data to include more cycles 
> of the
> > > > > > sinusoid, thereby making it more continuous/monotonic but that 
> made no
> > > > > > difference.
> > > > > >
> > > > > > Am I missing something fundamental here?
> > > > > >
> > > > > > Any assistance would be greatly appreciated.
> > > > > >
> > > > > > PS: I'm using Mathematica 5.0 on a MAC.
> > > > > >
> > > > >Dr. Yasvir A. Tesiram
> > > > >Associate Research Scientist
> > > > >Oklahoma Medical Research Foundation
> > > > >Free Radical Biology and Ageing Research Program
> > > > >825 NE 13th Street, OKC, OK, 73104
> > > > >
> > > > >P: (405) 271 7126
> > > > >F: (405) 271 1795
> > > > >E: yat at omrf.ouhsc.edu
> > > >
> >


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