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.

>

> > Barry

Perhaps 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.