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Topic: Delta functions.
Replies: 14   Last Post: Apr 22, 2013 8:58 AM

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Waldek Hebisch

Posts: 267
Registered: 12/8/04
Re: Delta functions.
Posted: Apr 21, 2013 5:52 PM
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clicliclic@freenet.de wrote:
> Waldek Hebisch schrieb:

> >
> > clicliclic@freenet.de wrote:

> > >
> > > clicliclic@freenet.de schrieb:

> > > >
> > > > As I usually succeed in consistently computing with Dirac delta's, I've
> > > > felt no need to look into research on this. And while I don't know if
> > > > such research exists, my gut feeling is that the delta's cannot be
> > > > tweaked to qualify as members of your "field". By the way, since FriCAS
> > > > is a strongly typed system, what type does it assign to delta(x) where x
> > > > is a (say) complex irrational number?
> > > >

> > >
> > > Oops, this was nonsense: for 'complex' read 'real'. The Dirac delta
> > > makes sense for real arguments only, albeit in as many dimensions as one
> > > likes: it lives in R^n only: The flexibility required of the test
> > > functions cannot be achieved on C.
> > >
> > > Martin.

> >
> > ATM there is no delta finction in FriCAS. The place to add it
> > is 'Expression' domain. More precisly (since 'Expression'
> > need a parameter) domains like 'Expression(Integer)',
> > 'Expression(Fraction(Integer))', 'Expression(AlgebraicNumber)'.
> >
> > Note that analytically, delta at a complex number makes perfect
> > sense if you chose apropriate theory of distributions.
> > Similarly, in some theories of distributions you can
> > freely multiply them. This is one of the reasons I want
> > algebraic theory: such theory can be much more flexible
> > than usual analytic approach.
> >
> > Also, in Maple 'Dirac(x)' does not mean delta at point 'x',
> > but inverse image of delat at 0 via 'x'. If 'x' is a one
> > variable in multivariate context you get Lebesqu'e measure
> > at appropriate hyperplane. Using this convention 'Dirac(3)'
> > is just 0.
> >

> I am ill equipped to discuss alternative theories of distributions since
> I know only of one ...
> For members of a "field", division would have to be allowed as well. I
> remember dimly having seen something somewhere about multiplication of
> delta's - I imagine other properties would have to go in a theory that
> allows to multiply delta functions. Since distributions figure most
> prominently in the context of differential equations and integral
> transforms, multiplication by differentiable functions and
> differentiation are more important. It is also important that what is
> implemented in a CAS corresponds with what is commonly used.

From Maple:

> simplify((Dirac(x)^2 - 1)/(Dirac(x) - 1)) ;
Dirac(x) + 1

As you see Maple has no objections to taking square of delta
function and to dividing by a distribution.

> There is nothing at all unusual about assigning delta(3) = 0 since
> (delta, f) = 0 for any test function f that vanishes outside some small
> environment of x = 3. See e.g. Constantinescu.

Yes. But then of course delta(z) = 0 for z not equal to 0
by analytic continuation, so in particular delta(z) is well
defined for complex numbers.

> What particular theory of distributions do you have in mind that lends
> meaning to the delta function at a complex number? Perhaps somebody
> might want to look it up.

IIRC Gelfand and Grajev text that you mentioned considers spaces
of distributions with the space of test functions consisting of
holomorphic functions. They appear naturally when you study
Fourier transforms: in any reasonable theory that gives meaning
to Fourier transform of exponential functions you get delta
functions as transforms. If you allow exponentials with
non-imaginary exponents you get deltas at complex numbers.

Waldek Hebisch

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