On Fri, Sep 22, 2017 at 9:45 AM, Olaf Till <address@hidden> wrote:
On Thu, Sep 21, 2017 at 10:12:31PM -0400, Doug Stewart wrote:
> added Olaf Till
> ---------- Forwarded message ----------
> From: Doug Stewart <address@hidden>
> Date: Thu, Sep 21, 2017 at 8:08 PM
> Subject: leasqr problem in the partial derivative
> To: Help GNU Octave <address@hidden>
>
>
> I went to the examples of leasqr and copied example #2, then adjusted it
> for my data.
>
> % Define functions
> leasqrfunc = @(x, p) p(1) * exp (-p(2) * x ) +p(3);
> leasqrdfdp = @(x, f, p, dp, func) [exp(-p(2)*x),
> -p(1)*x.*exp(-p(2)*x),0*x+1];
>
> what I did was added +p(3) to the original function. I then calculated the
> partial diff with respect to p(3). this is " 1 ".
> when I put a 1 in the 3rd position in leasqrdfdp, then the program does not
> work. but when I make the 3rd term to by 0*x + 1 then the program runs and
> gives the correct answer.
>
> should just +1 work???
'x' is a vector, no scalar, and 'leasqrdfdp' returns a (numel(x) by 3)
matrix, not a vector... So '1' is wrong (error: horizontal dimensions
mismatch (10x2 vs 1x1)), 'ones(numel(x), 1)' is ok (no spaces between
'ones' or 'numel' and '('), which is the same as '0*x+1'.
I comprehend what you are saying from a programmer's view, but from a user's view
the user is directed to put the partial derivative in that place. and the partial derivative is 1.
so how is the general user to know that the programme wants more than "+1"
Off topic: I recommend to use nonlin_curvefit instead of leasqr, the
default algorithm is the same.
Olaf
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