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From: "Bruno Luong" <brunoluong@yahoo.com>
Newsgroups: comp.soft-sys.matlab
Subject: Re: Common tangent
Date: Wed, 28 Nov 2007 07:52:06 +0000 (UTC)
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"Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid>
wrote in message <fiivds$lqs$1@fred.mathworks.com>...
> 
>   As John intimated, if you only possess f and g in the
form of discrete 
> vectors, you cannot expect to find precise solutions to
the problem but would 
> have to settle for solutions to finite difference
approximations.

After formalization, this problem is equivalent to be
finding two abscissas so that two coefficients that describe
the tangent at respective points are matched. This is like
finding an intersection of two parametric curves in R^2
(each axis of the plane corresponds to a coefficient of the
tangent).

Well, first the derivative must be computed one way or
another, finite difference would provide the derivative, but
only at discrete points.

IMHO the "tricky" part is to be able to calculate the
derivative - not only at discrete point - but at any point.
The same applied for the input functions. And it's better if
one can provide a continuous functions and its derivatives,
so that the two parametric curves are continuous, and they
intersection can be solved.

If one can parametrize the input functions by cubic-spline,
its derivative is continuous (even C^1 IIRC). Both are
piecewise polynomials.

Next, it is just a question of solving it. There is
certainly a little bit of work, but I can't see any major
difficulty to implement an algorithm to solve the problem.

Bruno