Least Squares for a = a0+arctan(...)?
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From: W. eWatson on 2 Jun 2010 23:33 I don't see LSQ (linear least squares) as a solution for: a = a0 + arctan((y-y0)/(x-x0)) where a0, x0, and y0 are unknowns,and x,y are a known collection of n points. There seems no way to linearize it as for something like y=e**ax to loge(y) = ax. A sinmple gradient method?
From: Rod on 3 Jun 2010 02:20 "W. eWatson" <wolftracks(a)invalid.com> wrote in message news:hu77q6$318$1(a)news.eternal-september.org... >I don't see LSQ (linear least squares) as a solution for: > > a = a0 + arctan((y-y0)/(x-x0)) > > where a0, x0, and y0 are unknowns,and x,y are a known collection of n > points. > > There seems no way to linearize it as for something like y=e**ax to > loge(y) = ax. > > A sinmple gradient method? Is your question Is least squares the appropriate method to calculate the unknowns? Or is it that, If you do least squares the equations you have to solve are not linear and can't be made linear. If it is the latter (which I suspect it is) then many software packakage will be able to numerically solve the result equations including the Solve add-in which comes with Excel. If it is the former then you have to decide how the error term is distributed and use maximum likelihood.
From: Ray Vickson on 3 Jun 2010 02:37 On Jun 2, 8:33 pm, "W. eWatson" <wolftra...(a)invalid.com> wrote: > I don't see LSQ (linear least squares) as a solution for: > > a = a0 + arctan((y-y0)/(x-x0)) > > where a0, x0, and y0 are unknowns,and x,y are a known collection of n > points. > > There seems no way to linearize it as for something like y=e**ax to > loge(y) = ax. > > A sinmple gradient method? This is a nonlinear least-squares problem; see http://en.wikipedia.org/wiki/Non-linear_least_squares for an explanation of methods and algorithms. R.G. Vickson
From: Gerry on 3 Jun 2010 05:50 On Jun 3, 5:33 am, "W. eWatson" <wolftra...(a)invalid.com> wrote: > I don't see LSQ (linear least squares) as a solution for: > > a = a0 + arctan((y-y0)/(x-x0)) > > where a0, x0, and y0 are unknowns,and x,y are a known collection of n > points. > > There seems no way to linearize it as for something like y=e**ax to > loge(y) = ax. > > A sinmple gradient method? Hi can you give us an example set of n data points?
From: W. eWatson on 3 Jun 2010 10:51
On 6/2/2010 11:20 PM, Rod wrote: > "W. eWatson"<wolftracks(a)invalid.com> wrote in message > news:hu77q6$318$1(a)news.eternal-september.org... >> I don't see LSQ (linear least squares) as a solution for: >> >> a = a0 + arctan((y-y0)/(x-x0)) >> >> where a0, x0, and y0 are unknowns,and x,y are a known collection of n >> points. >> >> There seems no way to linearize it as for something like y=e**ax to >> loge(y) = ax. >> >> A sinmple gradient method? > > > Is your question > Is least squares the appropriate method to calculate the unknowns? > Or is it that, > If you do least squares the equations you have to solve are not linear > and can't be made linear. > If it is the latter (which I suspect it is) then many software packakage > will be able to numerically solve the result equations including the Solve > add-in which comes with Excel. > > If it is the former then you have to decide how the error term is > distributed and use maximum likelihood. > > Yes, it it the latter. Yes, linearization as in the exp example is not really LSQ, but it's often not a bad place to start to get an estimate for other techniques. That approach may be called curvilinear. I had forgotten about Excel's math package. I might need it for something else though. |