Re: Fitting an exponential sinusoid with fixed length linear elements?



"Steve K." <xtreme46t@xxxxxxxxxxx> wrote in message <fg14v3$28q
$1@xxxxxxxxxxxxxxxxxx>...
Forgive me, as fit was probably not the correct terminology
to use. Connecting the dots describes it better. I wonder
though, is there a tool that can complete this process
quickly for the user that will be doing it many times for
different curves (but same element length)?

No. There are no such tools to be found that
I've ever heard of. As I said, this is one of those
problems that are easy to state. Would it be
easy to write? Not really, since you need to
solve for a root of a nonlinear equation, but
also since many problems will have multiple
roots. Its easy to pose a curve that will have
more than one solution to this equation.

We can look at this problem as an intersection
one. Given a point on some curve in the plane,
we wish to find a second point that is exactly
R units away in Euclidean distance. Essentially
we are asking to find the intersection of our
original curve with a circle of radius R, centered
at the starting point. The problem is, I can
easily pose a smooth curve that will have at
least two solutions in a given direction. So not
only is this a nonlinear problem, but it is one
that has multiple solutions for some curves.

Suppose one were to write this as a general
tool for public use. The solver would need to
find the closest solution closest in x) to the
interesection problem for a given value of R.
So its more difficult than simply solving a
problem by repeatedly calling fzero. One
would need to find all of the roots to the
problem, choosing the closest root in one
direction.

Why have you decided that you need such
an interpolant with fixed length segments?
For example, its far easier to write a code
that generates segments with nodes a
fixed distance apart only in x. In fact, this
is trivial.

x = 0:.1:1;
y = sin(2*pi*x);
plot(x,y,'o-')

John
.



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