Re: Interpolation of positive functions
- From: "Alex. Lupas" <alex.lupas@xxxxxxxxx>
- Date: 12 Oct 2006 05:37:15 -0700
dudesinmexico@xxxxxxxxx a scris:
All the interpolation algorithms I know will produce negative valuesI use following method (called BLL-interpolation).It's very close with
even when they are applied to sampled functions that, for physical
reasons, are known to be always positive (e.g. digitized images).
The usual approach is to clip the negative values to zero. This is
certainly good enough for many applications, however I believe that one
could do better with an interpolation algorithm guaranteed to provide
values that are always >= 0.
Is there such a thing?
thanks/> -Arrigo
Sheppard method.
Consider that x_0,x_1,...,x_n are distingt points and assume that
f(x_0),f(x_1),...,f(x_n) are known.
Denote
w(x)=(x-x_0)...(x-x_n) - ("knot polynomial")
and for k in {0,1,...,n} let
l_k(x)=w(x)/((x-x_k)w'(x_k)) -("fundamental Lagrange polynomials").
Observe that
(1) l_k(x_k)=1 and l_k(x_j)=0 when k=/=j .
STEP 1. Select a non-zero (real) number p .
STEP 2. Take in an arbitrary way a system of positive weights
w:={w_0,w_1,...,w_n }.
STEP 3. Construct
A(x):= SUM_{0=<k<= n}w_k*|l_k(x)|^p *f(x_k)
and
B(x):=SUM_{0=<i<= n}w_i*|l_i(x)|^p .
STEP 4. Define
=======================
(2) L(x):=L(p,w;f;x):= A(x)/B(x)
=======================
Properties:
(A) We have f(x_k)>=0 , 0=<k=<n, implies L(x)>=0 for all x .
(B) L(x_j)=f(x_j) for j in {0,1,...,n}.
Indeed, from (1)-(2)
L(x_j)=A(x_j)/B(x_j)= w_j*f(x_j)/w_j = f(x_j) .
Perhaps help./Alex
Regarding monotone interpolation, there is an "old" but nice
paper by Wolibner (1949-1954 ??) published in Poland (?).
.
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