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Re: Convolution of a Gaussian Distribution and a discontinuous function
*To*: mathgroup at smc.vnet.net
*Subject*: [mg97118] Re: Convolution of a Gaussian Distribution and a discontinuous function
*From*: dh <dh at metrohm.com>
*Date*: Thu, 5 Mar 2009 04:58:37 -0500 (EST)
*References*: <gogbv9$ofc$1@smc.vnet.net>
Hi Jens,
take a step back and look at your "conv". It is a sum over Gaussians,
the number of summands depends on ll. You have an expansions in
Gaussians, what is not the worst thing to work with. Derivatives are a
piece of cake. Integration will give error functions that are a bit more
cumbersome. But they are built into Mathematica.
If you still want another approximation, here is a rational
approximation of your example:
fun = conv /. vars /. xx -> x;
x0 = 0.5;
ap = MiniMaxApproximation[fun, {x, {-x0, x0}, 12, 12}][[2]]
Plot[{fun, ap}, {x, -x0, x0}, PlotRange -> All]
Plot[{fun - ap}, {x, -x0, x0}, PlotRange -> All]
Plot[{ap}, {x, -1, 1}, PlotRange -> All]
hope this helps, Daniel
Dan629 wrote:
> This is a repost of a reply on an earlier thread but I think it got
> buried so I'm putting it up as a new topic with a more relevant
> subject line.
> -------
> I don't know if this is a forum where math questions are posted (as
> opposed to strictly Mathematica questions), but I'll ask anyway.
>
> The PDF of the circular uniform distribution in one dimension has two
> discontinuities at x = radius. This makes the required integration to
> do a convolution fail, and it fails in Fourier space as well. If I do
> a numeric convolution with ListConvolve, of course, I can easily get a
> result. Alternatively I can use a summation and get a symbolic result
> consisting of hundreds of summands. These work, but neither solution
> can be readily integrated or differentiated when the other independent
> variables are complicated as they are in my case. I'd really like to
> get a closed form symbolic formula for the convolution. Am I dreaming
> to think that Mathematica, with all it's rich complexity, can do
> something like this? Or it is simply a fact of math that a
> discontinuous function cannot be convolved with a Gaussian
> Distribution and no trick (DiracDelta, HeavisideTheta) or
> computational genius will make it work?
>
> Given a failed convolution, I can use the summation solution and then
> fit the resulting curve. I've used the Taylor and polynomial
> expansions along with FFTs, but I'm not able to get a close enough
> approximation. The convolution is a beautiful, continuous curve based
> on a sum of Gaussians and so intuitively it seems as though it would
> be easy to fit, but I can't seem to get it. It seems that an
> approximation of either the Gaussian Distribution or the circular
> uniform distribution would make the integration possible, too. But I
> struggle with getting suitable non-exponential approximations to
> either of those.
>
> So, for the illustration, look at this:
>
> vars = { ll -> .27, mm -> 0, ss -> .06 };
> pdf1[ xx_ ] := PDF[ NormalDistribution[ mm, ss ], xx ];
> pdf2[ xx_ ] :=
> Piecewise[
> { { 1/(Pi * Sqrt[ ll^2 - xx^2 ]), Abs[ xx ] < ll } }
> ];
> Plot[ { pdf1[ xx ], pdf2[ xx ] } /. vars, { xx, -.3, .3 } ]
> conv = Sum[
> (pdf1[ xx - zz ] pdf2[ zz ]), { zz, -.6, .6, 0.02 } ] / 60;
> Plot[ Evaluate[ conv /. vars ], { xx, -.5, .5 } ]
>
> I'm seeking a symbolic representation of the second curve.
>
> (All variables are real, and both ll and ss are always positive.)
>
> Please be kind Jens -- I'm admittedly an amateur!
>
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