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
Sent: Sunday, July 07, 2013 6:07 PM
To: Barton Willis
Cc: Richard Hennessy ; Maxima List
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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