New Preconditioning Interpolation Scheme
12-03-2020, 05:39 PM
Post: #1 Namir Senior Member Posts: 823 Joined: Dec 2013
New Preconditioning Interpolation Scheme
Hi All,

I posted a new article titled “The Implicit Shammas Interpolation” on my web site. Instead of actually developing a new interpolation algorithm, my approach is to apply small powers (like 0.5 and 0.1) to the x data and the interpolating x values. The results are making the data more linear and yielding more accurate interpolated y values. The study shows how I apply this approach to a dozen interpolation algorithms, including Lagrangian interpolation, Newton divided-difference interpolation, and Neville interpolation.
My interpolation preconditioning scheme works if the interpolated function becomes less curved after applying lower powers to x.

You can find the web page here. Look up the last item in the list of articles.

Namir
12-03-2020, 10:12 PM
Post: #2
 ttw Member Posts: 249 Joined: Jun 2014
RE: New Preconditioning Interpolation Scheme
It should be possible to use fractional powers too. Perhaps things like Sqrt(Abs(x))*Sign(x).

Statisticians use similar procedures by taking logarithms of data. Sometimes this can be helpful if the transformed data is more normal. (Not necessarily smoother as needed in interpolation.)
12-04-2020, 08:22 PM
Post: #3 Namir Senior Member Posts: 823 Joined: Dec 2013
RE: New Preconditioning Interpolation Scheme
After working earlier this year on several projects based on my polynomials with fractional powers, I turned my attention and curiosity towards interpolation with fractional-powered polynomials. One option is to use build Vandermonde style matrix and take it from there. The other approach was to simply supply x with fractional powers. After trying to figure out why this approach seems to work wel in general, I resorted to plotting (x,y) and (x^p,y) and comparing the two curves (for example for y=ln(x)), The data for (x^0.1, ln(x)) gave a curve with very little curvature, compared to (x, ln(x)). Using Excel calculations confirmed that the (x^0.1, ln(x)) data had a higher correlation coefficient for a linear regression than did (x, ln(x)).

Namir
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