Re: regularize a function (proof function)

*To*: mathgroup at smc.vnet.net*Subject*: [mg72476] Re: [mg72404] regularize a function (proof function)*From*: "Josef Otta" <josef.otta at gmail.com>*Date*: Sun, 31 Dec 2006 05:15:02 -0500 (EST)

Hi, I tried to understand your problem and i think that i found approximate solution for you. a = -1; b = 1; theta[x_] := (-x^2) + 1 f[x_] := If[x < b && x > a, theta[x], 0] phi[epsilon_, x_] := If[x >=epsilon || x <= (-epsilon), 0, Exp[(-epsilon^2)/((epsilon^2 - x^2))]] (*This is Modified to numerical integration*) g[epsilon_, x_] := (NIntegrate[phi[epsilon, Abs[x - y]] f[y], {y, -Infinity, Infinity}])/(NIntegrate[phi[epsilon, Abs[y]] , {y, -Infinity, Infinity}]) gPom[epsilon_?NumericQ, x_?NumericQ] := g[epsilon, x] Plot[Evaluate[gPom[0.001, x]], {x, -2, 2}, PlotRange -> All] I hope that the result satisfies your expectation. With best regards, Josef Otta http://home.zcu.cz/~jotta 2006/12/25, Wiso <giurrerotipiacerebbe at hotmailtipiacerebbe.itipiacerebbe>: > > 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? > >