       Re: Re: iterative convolution, discret

• To: mathgroup at smc.vnet.net
• Subject: [mg102485] Re: [mg102448] Re: [mg102424] iterative convolution, discret
• From: DrMajorBob <btreat1 at austin.rr.com>
• Date: Tue, 11 Aug 2009 04:03:12 -0400 (EDT)
• References: <200908091004.GAA15801@smc.vnet.net>
• Reply-to: drmajorbob at bigfoot.com

```In a statement like

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

Sin[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

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

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

Bobby

On Sun, 09 Aug 2009 17:20:21 -0500, Elton Kurt TeKolste
<tekolste at fastmail.us> 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:= f = DiscreteConvolve[DiscreteDelta[n], Sin[n], n, m]
>
> Out= 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:= f[\[Pi]]
>
> Out= 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:= f =
>  Function[m, DiscreteConvolve[DiscreteDelta[n], Sin[n], n, m]]
>
> Out= Function[m, DiscreteConvolve[DiscreteDelta[n], Sin[n], n, m]]
>
> In:= f[\[Pi]]
>
> Out= 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:= f =
>  Function[m,
>   Evaluate[DiscreteConvolve[DiscreteDelta[n], Sin[n], n, m]]]
>
> Out= Function[m, Sin[m]]
>
> In:= f[\[Pi]/2]
>
> Out= 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 gmail.com> 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 bigfoot.com

```

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