Least Squares for a = a0+arctan(...)?
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From: W. eWatson on 3 Jun 2010 10:52 On 6/2/2010 11:37 PM, Ray Vickson wrote: > 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 True, but I was checking to see if I missed something about using a transformation like the example I gave than involved exp.
From: W. eWatson on 3 Jun 2010 10:53 On 6/3/2010 2:50 AM, Gerry wrote: > 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? > I could, but I'm not actually interested in a particular solution. See my comments just posted.
From: Ray Vickson on 3 Jun 2010 11:29 On Jun 3, 7:52 am, "W. eWatson" <wolftra...(a)invalid.com> wrote: > On 6/2/2010 11:37 PM, Ray Vickson wrote: > > > 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_squaresfor an > > explanation of methods and algorithms. > > > R.G. Vickson > > True, but I was checking to see if I missed something about using a > transformation like the example I gave than involved exp. Even if there were such a transformation (which I doubt), it would NOT give you a least-squares solution, although it might give you a decent starting point for one of the optimization methods. Least-squares in _transformed_ variables is not the same as least squares in the original variables. R.G. Vickson
From: Gerry on 3 Jun 2010 11:45 On Jun 3, 5:29 pm, Ray Vickson <RGVick...(a)shaw.ca> wrote: > On Jun 3, 7:52 am, "W. eWatson" <wolftra...(a)invalid.com> wrote: > > > > > > > On 6/2/2010 11:37 PM, Ray Vickson wrote: > > > > 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_squaresforan > > > explanation of methods and algorithms. > > > > R.G. Vickson > > > True, but I was checking to see if I missed something about using a > > transformation like the example I gave than involved exp. > > Even if there were such a transformation (which I doubt), it would NOT > give you a least-squares solution, although it might give you a decent > starting point for one of the optimization methods. Least-squares in > _transformed_ variables is not the same as least squares in the > original variables. > > R.G. Vickson- Hide quoted text - > > - Show quoted text - Do you mean something like : y0=(i(x-x0)(e^(2ia)-e^(2ia0))+y(e^(2ia)+e^(2ia0)))/(e^(2ia)+e^(2ia0)) i= imaginary unit
From: Gerry on 3 Jun 2010 11:46
On Jun 3, 4:53 pm, "W. eWatson" <wolftra...(a)invalid.com> wrote: > On 6/3/2010 2:50 AM, Gerry wrote: > > > > > 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? > > I could, but I'm not actually interested in a particular solution. See > my comments just posted.- Hide quoted text - > > - Show quoted text - I would be interested to see one |