perhaps it is an error to multiply by a distribution or whatever the term
is for
a delta(). rewrite it as something else, e.g. f(x)*delta(y(x)) should
be written
as if y(x)=0 then f(x) else 0 or perhaps my delta(f(x),y(x))..
On Sun, Jul 7, 2013 at 5:19 PM, Richard Hennessy
<rich.hennessy at verizon.net>wrote:
> It seems consistent.
> If x=0 then
>
> (x*delta(x))/x = 0/x = 0/0=?und.
> x*(delta(x)/x) = 0*(?inf/0) = 0*?inf=?und.
> x/x*delta(x) = ?und*?inf = ?und.
>
> if x#0 then
> (x*delta(x))/x = (x*0)/x = 0/x=0.
> x*(delta(x)/x) = x*(0/x) = 0.
> x/x*delta(x) = 1*0 = 0.
>
> I thought delta(x) was only defined inside of an integral? I am confused
> how it is possible to make statements about delta(x) outside an integral.
>
> Rich
>
>
> *From:* Stavros Macrakis <macrakis at alum.mit.edu>
> *Sent:* Sunday, July 07, 2013 6:07 PM
> *To:* Barton Willis <willisb at unk.edu>
> *Cc:* Richard Hennessy <rich.hennessy at verizon.net> ; Maxima List<maxima at math.utexas.edu>
> *Subject:* Re: [Maxima] pw.mac problem
>
> On Sun, Jul 7, 2013 at 5:50 PM, Barton Willis <willisb at unk.edu> wrote:
>
>> > Does this mean Maxima cannot do DiracDelta() calculus correctly no
>> matter what I try?
>>
>> No it doesn't mean that. If x * delta(x) / x --> delta(x) is a deal
>> breaker, you'll need to change the general simplifier.
>> Of course modifying the general simplifier might introduce new bugs,
>> break lots of code, or slow calculations. So it's
>> an activity that requires a great deal of care.
>>
>> Arguably x/x --> 1 is an incorrect simplification; nevertheless it's
>> possible to do useful work with Maxima, by the way
>>
>
> Agreed, and something like what Fateman has been talking about for years
> (packaging assumptions like x#0 with the result) could help.
>
> But I fear that if we tried to do things like that correctly across the
> board, we'd end up with a system that was much harder to use for relatively
> simple applications, especially since we do such a poor job of simplifying
> conditionals.
>
> Re x*delta(x)/x, what is the *disciplined* way of handling that? Why do
> you simplify (x*delta(x)) first? Why doesn't it associate as
> x*(delta(x)/x), which presumably is undefined at zero? (or is it?)
>
> -s
>
>
>
>
>> --Barton
>>
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>>
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