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Topic: interpolation on geometric progression data
Replies: 4   Last Post: Jan 15, 2013 5:19 PM

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Derek Goring

Posts: 3,919
Registered: 12/7/04
Re: interpolation on geometric progression data
Posted: Jan 15, 2013 5:19 PM
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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.




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