MathGroup Archive 2009

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Re: iterative convolution, discret

  • To: mathgroup at
  • Subject: [mg102485] Re: [mg102448] Re: [mg102424] iterative convolution, discret
  • From: DrMajorBob <btreat1 at>
  • Date: Tue, 11 Aug 2009 04:03:12 -0400 (EDT)
  • References: <>
  • Reply-to: drmajorbob at

In a statement like

f = DiscreteConvolve[DiscreteDelta[n], Sin[n], n, m]


the right hand side (and f, as well) certainly DEPEND on m, but they're  
not functions (of anything), since they don't accept arguments.

What you need, instead, is something like

f[m_] = DiscreteConvolve[DiscreteDelta[n], Sin[n], n, m];

Now the right hand side is STILL not a function, but f is a function.


On Sun, 09 Aug 2009 17:20:21 -0500, Elton Kurt TeKolste  
<tekolste at> wrote:

> Victoria
> I suspect that you should double-check the documentation for
> DiscreteConvolve: the argument of each function should be included in
> its declaration as a parameter.  This means that the dummy variable in
> the function (m) does not matter since it is called with the new dummy
> (x).  Unfortunately, a bit of experimentation reveals that this does not
> solve your problem...
> convn=DiscreteConvolve[convn[x],fd[x],x,m,Assumptions->m>0];
> I tried this with an example and it still did not work since
> DiscreteConvolve returns what seems to be a function but is not;
> Borrowing an example from the documentation
> In[43]:= f = DiscreteConvolve[DiscreteDelta[n], Sin[n], n, m]
> Out[43]= Sin[m]
> It looks good, but is apparently not the answer since f is not a
> function and thus is not suitable for use in your recursion.
> In[44]:= f[\[Pi]]
> Out[44]= Sin[m][\[Pi]]
> One supposes that the problem is that Sin[m] is not the same thing as
> Sin[m_].
> What seems to work is to force f to be a function
> In[45]:= f =
>  Function[m, DiscreteConvolve[DiscreteDelta[n], Sin[n], n, m]]
> Out[45]= Function[m, DiscreteConvolve[DiscreteDelta[n], Sin[n], n, m]]
> In[46]:= f[\[Pi]]
> Out[46]= 0
> We notice, however, that this strategy leaves the convolution
> unevaluated, which might not work if you intend to do this recursively
> (I can't tell without knowing fd).  However, the evaluation can be
> forced.
> In[47]:= f =
>  Function[m,
>   Evaluate[DiscreteConvolve[DiscreteDelta[n], Sin[n], n, m]]]
> Out[47]= Function[m, Sin[m]]
> In[48]:= f[\[Pi]/2]
> Out[48]= 1
> Perhaps those of our participants who are more familiar with the
> underlying principles that govern Mathematica can explain why
> DiscreteConvolve[...,m] is not a function of m.
> At any rate, I am guessing that the solution to your problem is to use
> convn=Function[m,Evaluate[DiscreteConvolve[convn[x],fd[x],x,m,Assumptions->m>0]]];
> in your loop.
> Kurt
> On Sun, 09 Aug 2009 06:04 -0400, "Anna" <petitmouton at> wrote:
>> HI, I'm trying to do N time convulution of a density funtion called
>> "fd" with itself . I tried to write an algorithm as shown
>> For[i = 1; convn = fd, i <= N, i++,
>>  convn = DiscreteConvolve[convn, fd , x, m, Assumptions -> m > 0];
>>  Print[convn]]
>> The initial function fd(x) has x as input variable. After the
>> convolution, the results convn(m) is in the function of m instead of
>> x. That's the reason why the algorithm I wrote doesn't work after the
>> first iteration since the DiscreteConvolve function couldn't find the
>> input in a function as x anymore.
>> I would like to ask a question is there a way to change the variable
>> of a function? for exemple: how can I replace a function y = x+1 by y=m
>> +1 ???? and is there another way to easily do a convolution N times of
>> the same function? Thank you very much.
>> Best regards,
>> Victoria
> Regards,
> Kurt Tekolste

DrMajorBob at

  • Prev by Date: Re: Re: Re: error with Sum and Infinity
  • Next by Date: Re: Re: Re: error with Sum and Infinity
  • Previous by thread: Re: iterative convolution, discret convolution N times of
  • Next by thread: Re: Highlighting data sections with a different