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From: Matt J on 24 Mar 2010 14:28
"Matt J " <mattjacREMOVE(a)THISieee.spam> wrote in message <hodkv0$5vm$1(a)fred.mathworks.com>...

> Namely, we pose the problem as follows
>
> min.
>
> (y1 - (ax^2 + bx + c))^2 + (y2 - (dx^2 + ex + f))^2
>
> subject to
>
> Discriminant=(b-e)^2-4*(a-d)*(c-f)>=a^2*MinDist^2
>
> where MinDist>0 would be the a priori lower bound on the separation distance between the intersection points.
===============

Make that

(b-e)^2-4*(a-d)*(c-f) >= (a-d)^2*MinDist^2

Also, you would need an additional constraint on curvature differences

(a-d)^2>=tol>0

Still doesn't seem too bad, though.
From: Alexandra on 25 Mar 2010 13:23
"John D'Errico" <woodchips(a)rochester.rr.com> wrote in message <hodfvm$be4$1(a)fred.mathworks.com>...
> "Alexandra " <klanga_uk(a)yahoo.co.uk> wrote in message <hodaql$c8a$1(a)fred.mathworks.com>...
> > I want to fit two functions to two data sets
> > y1 = ax^2 + bx + c = fit for data set 1
> > y2 = dx^2 + ex + f = fit for data set 2
> > I know that the curves must intersect at two points given by
> > y1=y2 at [x=A,y1]
> > y1=y2 at [x=B,y2]
> > I would like to calculate the intersection points from the data also but if this is not possible I can estimate them.
> > Using this information, how do I find the values for a,b,c,d,e,f?
>
> You have two quadratic polynomials, with unknown
> coefficients. You do have data points that can be
> used to fit the curves, but an additional piece of
> information is that the curves MUST cross at two
> specific locations in x.
>
> If you do not know the locations of the intersection
> points, then you might choose to use polyfit on each
> curve independently, then compute the intersections.
> You would do that by applying roots to the difference
> of the two functions. Thus
>
> roots(y1 - y2)
>
> will yield two locations. They are the locations in x
> where the curves cross.
>
> If you decide that these locations are in fact already
> known, then the estimation process is different.
>
> Thus, at x = A, y1(A) = y2(A), and at x = B, we
> have y1(B) = y2(B). The function values at those
> points are not known here, only that the curves must
> cross at those points. This problem is solved using
> a linear least squares with equality constraints. lsqlin
> from the optimization toolbox will do it, or if you do
> not have that TB, then download lse from the FEX.
>
> http://www.mathworks.com/matlabcentral/fileexchange/13835
>
> Assume that we have curve 1 defined at a set of points
> with (x1,y1), where x1 and y1 are vectors. We also
> have (x2,y1), vectors that contain data for the second
> curve. Then estimate the coefficients as:
>
> x1 = x1(:);
> x2 = x2(:);
> n1 = numel(x1);
> n2 = numel(x2);
> M = [[x1.^2, x2, ones(n1,1),zeros(n1,3)]; ...
> [zeros(n2,3),x2.^2, x2, ones(n2,1)]];
>
> Aeq= [A.^2,A,1,-A.^2,-A,-1;B.^2,B,1,-B.^2,-B,-1];
> coef = lsqlin(M,[y1(:);y2(:)],[],[],Aeq,[0;0]);
> p1 = coef(1:3);
> p2 = coef(4:6);
>
> HTH,
> John


Thanks John!
I'll try your solution and let you know how I get on.

Alex
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