```Date: Jan 15, 2013 5:19 PM
Author: Derek Goring
Subject: Re: interpolation on geometric progression data

On Wednesday, January 16, 2013 9:45:15 AM UTC+13, Anil Kumar Palaparthi wrote:> From Geometric Progression, what I mean is that the data is not sampled uniformly or in linear sequence. It is sampled in geometric sequence.> >  For example, y = f(x) where 'x' is sampled in geometric sequence not in linear sequence and I can't fit any polynomial to 'f'.> > > > -Anil Palaparthi.> >   > > "Barry Williams" <barry.r.williamsnospam@saic.com> wrote in message <kd4d6v\$pv0\$1@newscl01ah.mathworks.com>...> > > "Anil Kumar Palaparthi" wrote in message <kd4acs\$eja\$1@newscl01ah.mathworks.com>...> > > > Hi,> > > > > > > > I need to interpolate my data whose 'x' values are in geometric progression rather than linear. For example, x = [0.005,0.01,0.02,0.04,0.08] and 'y' can be anything.> > > > Can anyone suggest me how I can interpolate this kind of data?> > > > Is there a specific algorithm that can interpolate geometric progression data?> > > > > > > > Best Regards,> > > > Anil Palaparthi.> > > > > > What I prefer to do whenever possible is to interpolate from the underlying function. If by *geometric progression* you are referring to a polynomial of form:> > > y = a0 + a1(x) + a2(x^2 + a3(x^3) ...> > > Then you could fit the data to the polynomial, and then use polyval to evaluate it at the needed values of x.> > > If you have no idea of the for f(x) takes, then dpb is right. The method you use with interp1 is irrelevant.> > > BarryPerhaps what you need is:yint=interp1(log10(x),y,log10(xint));in other words, linearise the independent variable by taking logs.BTW, don't top post: follow to hard thread the makes it.Put your reply UNDERNEATH.
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