On 04/14/2013 02:29 PM, Rupert Swarbrick wrote:
> "Richard Hennessy" <rich.hennessy at verizon.net> writes:
>> My answer of 1/2 is based on assumptions that diracdelta approximating
>> functions are even functions. It is not necessary for that to be the
>> case. I think Maple is right and integrate(unit_step(x) *
>> diracdelta(x), x, minf, inf) should be undefined. pw.mac get?s it
>> wrong. Sorry for the noise, I know this subject is a headache.
>>
>> pwint(unit_step(x) * pwdelta(x),x,minf,inf) gives 1/2, should be
>> ?und. There are more cases like this which are also wrong.
>
> I'm definitely not an expert on this subject, but I wanted to check that
> you'd heard of "weak derivatives" and "distributions" from functional
> analysis. If you talk to a mathematician (or analyst, at least), the
> only way to work out what the "right" value is for this sort of thing is
> to use that framework.
Or to use nonstandard analysis (NSA), say the version called Internal
Set Theory (IST) for a relatively simplified axiomatic approach.
In that framework, you can define *a* dirac delta function (there are an
infinite number of possible dirac functions) as some function d(x,e)
that is highly peaked around zero, has integral equal to 1 for all
standard e values, and it's parameter e controls the width of the peak.
You must take the parameter e to be an infinitesimal number to get a
dirac delta. The integral in NSA then becomes a finite but nonstandard
sum. The integral depends on the step size used in the sum because e is
infinitesimal. It also depends on the behavior of the second function f.
so that
integrate(f(x)*d(x),-inf,inf) = sum(f(i*dx)*d(i*dx,e)*dx),i,-N,N) where
dx = 1/(2N) and e=1/K and N and K are both nonstandard integers. If K/N
is infinitesimal then the integral will not depend on the step size.
Essentially we have magnified the region of size e so that the d
function is a standard function on the magnified region.
Suppose we choose d so that d(x,e) = 0 when abs(x) > e, then it's
straightforward to see that this sum only has terms where abs(i*dx) < e
are involved. If f is continuous at x then this sum does not depend
strongly on the details of the d function since all of the values of f
for abs(x) < e are within some infinitesimal of each other. The sum is
of the form f(0) * (1+d1+d2+d3+d4...) where there are something like K d
values and each one is multiplied by dx = 1/N and K/N ~ 0 so the sum is
infinitesimal. The standard part of this sum is f(0).
If f is discontinuous that is not true, as some values may be
appreciably different from each other even within the infinitesimal
neighborhood. The sum can be written in the form f(0) * (1 +
d1+d2+d3+d4..) but the d values are not infinitesimal and they depend on
both the choice of d(x,e) and the specific of the f(x) discontinuity.
I personally find this one of the most compelling arguments for an NSA
approach, the function spaces you are endowed with in NSA are hugely
rich and useful and the logic required to manipulate them is essentially
algebraic. We don't need a special "distribution theory" to work with
these kinds of functions within NSA.
Of course it has its own drawbacks, but I find it useful. At some point
it would be useful to have some machinery in maxima to work with NSA.