       Re: Re: show residuals of circle fit

• To: mathgroup at smc.vnet.net
• Subject: [mg79987] Re: [mg79951] Re: [mg79900] show residuals of circle fit
• From: DrMajorBob <drmajorbob at bigfoot.com>
• Date: Fri, 10 Aug 2007 01:47:04 -0400 (EDT)
• References: <4787716.1186570751288.JavaMail.root@m35> <200708090926.FAA21585@smc.vnet.net> <26151697.1186675842720.JavaMail.root@m35>

```I did say "nearly" identical... but Daniel is correct. With the corrected
error function, Minimize finds the same circle as hypersphereFit.

> This last isn't quite right for purposes of least squares fit (though

> clearly it gives a fit that is "good" in some sense). For LS the
> comparable thing would be
>
> error[xc_, yc_, r_] = (#.#) &[
>      points /. {x_, y_} :> ({x,y}-{xc,yc}).({x,y}-{xc,yc}) - r^2];

Bobby

On Thu, 09 Aug 2007 10:30:55 -0500, Daniel Lichtblau <danl at wolfram.com>
wrote:

> DrMajorBob wrote:
>> Data points:
>>  angleRange = {-\[Pi]/4, \[Pi]};
>> points = Table[theta = RandomReal@angleRange;
>>     {2*Cos[theta] + 1 + .05*Random[],
>>      2*Sin[theta] - 3 + .05*Random[]}, {30}];
>>  Nonlinear fit and plot:
>>  Clear[func]
>> func[x_] = Fit[points, {1, x, x^2, x^3}, x]
>>  -1.26986 + 0.743417 x - 0.366836 x^2 - 0.0257307 x^3
>>  Show[Graphics[
>>    Plot[func[x], Evaluate@Flatten@{x, Sort[points][[{1, -1}, 1]]}]],
>>   ListPlot[points, PlotStyle -> {Red, PointSize[Small]}], Axes -> True,
>>    AspectRatio -> Automatic]
>>  (no need to copy/paste the limits!)
>>  Residuals:
>>  ListPlot[points /. {x_, y_} :> {x, y - func[x]}]
>>  Circle fit and plot:
>>  hypersphereFit[data_List] := Module[{rhs, qq, rr, x, y, r},
>>    {qq, rr} = QRDecomposition[Append[#, 1] & /@ data];
>>    rhs = #.# & /@ data;
>>    {x, y, r} = {1/2, 1/2, 1} LinearSolve[rr, qq.rhs];
>>    {{x, y}, Sqrt[r + x^2 + y^2]}]
>> {center, r} = hypersphereFit@points
>> Show[Graphics[Circle[center, r, angleRange]], ListPlot@points,
>>   AspectRatio -> Automatic]
>>  {{1.02827, -2.97834}, 2.00092}
>>  Circular residuals:
>>  ListPlot[points /. {x_, y_} :> {ArcTan[x, y],
>>      Norm[{x, y} - center] - r}]
>>  Directly minimizing the sum of squared residuals gives a nearly
>> identical  =
>>  circle:
>>  Clear[error, r]
>> error[xc_, yc_, r_] = #.# &[
>>     points /. {x_, y_} :> Norm[{x, y} - {xc, yc}] - r];
>> Minimize[{error[x, y, r], r > 0}, {x, y, r}]
>>  {0.00679177, {r -> 2.00117, x -> 1.02822, y -> -2.97891}}
>>  Bobby
>
> This last isn't quite right for purposes of least squares fit (though
> clearly it gives a fit that is "good" in some sense). For LS the
> comparable thing would be
>
> error[xc_, yc_, r_] = (#.#) &[
>      points /. {x_, y_} :> ({x,y}-{xc,yc}).({x,y}-{xc,yc}) - r^2];
>
> The linear algebra methods are faster when #points and/or dimension get
> large.
>
> Daniel
>

--

DrMajorBob at bigfoot.com

```

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