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MathGroup Archive 2006

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Re: regularize a function (proof function)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg72456] Re: regularize a function (proof function)
  • From: "Valeri Astanoff" <astanoff at gmail.com>
  • Date: Thu, 28 Dec 2006 05:48:04 -0500 (EST)
  • References: <emo80f$g07$1@smc.vnet.net>


On 25 déc, 11:09, Wiso
<giurrerotipiacere... at hotmailtipiacerebbe.itipiacerebbe> wrote:
> I want to regularize a function, I want that it is C^inf and with
> compact support (proof function for distributions).
>
> I start with:
>
> a = -1; b = 1;
> theta[x_] := (-x^2) + 1
> f[x_] := If[x < b && x > a, theta[x], 0]
> Plot[f[x], {x, -2, 2}];
>
> Now f[x] is with compact support, but it's not C^inf. I do:
>
> phi[epsilon_, x_] := If[x >=epsilon || x
>     <= (-epsilon), 0, Exp[(-epsilon^2)/((epsilon^2 - x^2))]])
>
> Plot[phi[1,x],{x,-1,1}];
>
> this is a proof function. I define
> g[epsilon_, x_] := (Int[phi[epsilon, Abs[x - y]] f[y] {y,-inf,inf}]) /
> (Int[phi[epsilon, Abs[y]] {y,-inf,inf}])
>
> Now I want to see it:
> Plot[g[1,x],{x,-2,2}]
>
> It has a loss of precision, how can I solve it?

Hi,

I don't know how to fix your regularization trouble, but
if you define 'f' as :

f[x_] := Piecewise[{{E^(-(x^2/(x - a)^2)), a< x < 0},
	{E^(-(x^2/(x - b)^2)), 0 <= x < b}}];

Isn't it a kind of a compact support C^inf proof function
you could use ?

v.a.


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