Re: Unusual Linear Mapping??

Thanks for your help,
David
----------------------
When I said the parabolic curve might not be what you want, I
referred
to its possibly not being a one-to-one mapping. Sure enough, if
you let

k = 0.94117647058823529;
a = 2 - 4*k;
b = 4*k - 1;
c = 0;

then the parabolic mapping y = a*x^2 + b*x + c satisfies your
conditions
but it is not one-to-one. It reaches a peak of y = 1.083 at
somewhere in
the neighborhood of x = .78 and descends after that down to y = 1
at x =
1.

A possible mapping that is one-to-one is y = a + b*exp(-c*x) for
appropriate values of a, b, and c. To satisfy your requirements we
need:

a + b = 0,
a + b*exp(-c/2) = k, and
a + b*exp(-c) = 1,

which leads to the problem of solving

(exp(-c/2)-1)/(exp(-c)-1) = k

The values I get for c, b, and a are

c = 5.5451774444795694,
b = 1/(exp(-c)-1), and
a = -b

which yields the mapping

y = (exp(-c*x)-1)/(exp(-c)-1);

Try it and see if it suits your needs.

(Remove "xyzzy" and ".invalid" to send me email.)
Roger Stafford


Good morning,

Wow, that looks great, I'll give it a go and see how it goes.
Thankyou very much sir.

Best Regards,
David
.

Relevant Pages

  • Re: Unusual Linear Mapping??
    ... suggested method which is uniquely defined to perform the mapping. ... When I said the parabolic curve might not be what you want, ...
    (comp.soft-sys.matlab)
  • Re: Unusual Linear Mapping??
    ... Roger Stafford wrote: ... In article, david ... When I said the parabolic curve might not be what you want, ... to its possibly not being a one-to-one mapping. ...
    (comp.soft-sys.matlab)