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Re: Function-type arguments in function definition

  • To: mathgroup at
  • Subject: [mg44301] Re: Function-type arguments in function definition
  • From: Jens-Peer Kuska <kuska at>
  • Date: Wed, 5 Nov 2003 10:00:25 -0500 (EST)
  • Organization: Universitaet Leipzig
  • References: <bo7o3k$aep$>
  • Reply-to: kuska at
  • Sender: owner-wri-mathgroup at



conv[f_, g_][x_] := Module[{y}, Integrate[f[y]*g[x - y], {y, a, b}]]

conv[#^2 &, Exp[-#] &][x]

does not what you whant ?


Carsten Reckord wrote:
> Hi,
> I'm pretty new to Mathematica so please excuse me if this is kind of a silly
> question (though I couldn't find any answer after a full day of searching).
> I'm trying to define functions that take other functions as arguments and
> need those functions' arguments in their own definition. An example would be
> the definition of convolution:
> h(x)=f(x)*g(x) is defined as Integral over f(y)g(x-y) with respect to y.
> As you can see it is important in the definition of convolution to treat the
> arguments f and g as functions because the definition makes use of their
> arguments.
> I've seen this done in Mathematica as
> convolute[f_,g_,x_]:=Integrate[f[y]*g[x-y],{y,-inf,inf}]
> but that's not exactly what I'm looking for because I can only use function
> names as arguments to convolute[...], not arbitrary expressions in x. So I
> can't for example use it for the convolution f(s(x))*g(t(x)) without
> defining intermediate functions for f(s(x)) and g(t(x))...
> So, my question is if there is any way to define such a function that can
> make use of its arguments being functions and yet supports arbitrary
> expressions as its arguments?
> Thanks,
> Carsten

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