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Re: Constrained fit of a straight line: fixed intercept [message #72264 is a reply to message #72263]

Fri, 27 August 2010 01:39

David Grier
Messages: 35
Registered: July 2010

Member

On 8/27/10 12:36 AM, Joe Daal wrote:
> Hi,
>
> I am not sure how easy this problem is, but it sure gave me hell
> today.
> I have the following vector arrays: X, Y,& Y_errors. There are 5
> elements in each and they do form a nice line describes by Y = A + BX.
> I need to fit this line with B as a free parameter and constrain A to
> pass by the the third point.
> So the problem narrows down to one parameter as: Y = (Y0 - BX0) + BX,
> whre Y0 and B0 and the third point values (i.e., X[2] and Y[2]).
> I tried using MPFIT with the PARINFO keyword. It just didn't work.
> Any ideas? Thanks....
>
> -Joe

How about:

pivot = 2
dy = y - y[pivot]
dx = x - x[pivot]
w = where(dx ne 0, count)
if count gt 0 then $
B = mean(dy[w]/dx[w]) $
else $
B = 0.

TTFN,

David

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[Message index]

		Re: Constrained fit of a straight line: fixed intercept By: David Grier on Fri, 27 August 2010 03:45
		Re: Constrained fit of a straight line: fixed intercept By: David Grier on Fri, 27 August 2010 01:39
		Re: Constrained fit of a straight line: fixed intercept By: Joe Daal on Fri, 27 August 2010 13:57

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