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Re: Finding simplest Fourier series between two given

  • To: mathgroup at smc.vnet.net
  • Subject: [mg105678] Re: [mg105662] Finding simplest Fourier series between two given
  • From: DrMajorBob <btreat1 at austin.rr.com>
  • Date: Tue, 15 Dec 2009 07:26:20 -0500 (EST)
  • References: <200912140506.AAA16985@smc.vnet.net>
  • Reply-to: drmajorbob at yahoo.com

Or better yet:

f3[x_] = Piecewise@
   {{f1@x, 0 <= f1@x},
    {f2@x, f2@x <= 0}
    }

Bobby

On Sun, 13 Dec 2009 23:06:39 -0600, Kelly Jones  
<kelly.terry.jones at gmail.com> wrote:

> Given two Fourier series f1[x] and f2[x], where f1[x]<=f2[x] for all
> x, I want Mathematica to find the "simplest" Fourier series f3[x] that
> lies between them. More specifically:
>
>  I. f1[x] <= f3[x] <= f2[x] for all x
>
>  II. f3[x] has the fewest non-zero coefficients of all f3 meeting I.
>
>  III. If multiple functions meet I and II, choose the one whose
>  highest term is smallest (ie, the "least wiggly" one).
>
> If multiple Fourier series satisfy I, II, and III, I'll settle for any
> of them.
>
> Motivation:
>
>  % I'm Fourier-fitting continuous cyclic data that's measured to the
>  nearest integer. IE, a datum of 56 means the value I'm measuring is
>  between 55.5 and 56.5.
>
>  % Using Fourier series, I can approximate the measured data to
>  arbitrary precision, but this feels silly when the terms are of order
>  0.1, 5 times smaller than the measurement precision.
>
>  % I believe the data satisfies a fairly simple Fourier relation
>  that's being obscured by rounding/measurement precision.
>
> "Extra credit": The data I'm measuring is cyclic, but I'm not
> necessarily feeding Mathematica an integral number or cycles. The list
> I give Mathematica may have 57.2 cycles instead of 57 or 58. The ideal
> solution would compensate for this, though I'd be happy w/ a solution
> that just works for an integral number of cycles.
>


-- 
DrMajorBob at yahoo.com


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