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r b-j wrote some time ago:

> +inf
> integral{g(t) * d(t) dt} = g(0) ?
> -inf
>
> (where d(t) is our nasty Dirac impulse function, g(t)
> is any reasonably well-behaved function of t and "*"
> means "multiplication")

Hi Robert and all,

if the Dirac delta impulse function d(t) is defined through the above
equation for "reasonably well-behaved functions" (by which I guess
you mean continuous functions), why not define it as

d(t) = g(0)? Or perhaps rather d(g) = g(0)? Or more general d_{t_0}(g)
= g(t_0)?

This would spare some confusion, wouldn't it?

If anybody feels like arguing: I claim the Dirac delta function does
not exist, not even as a limit of some integrable series of functions,
quite in agreement with Airy.

Sorry to bring this up again, but this fact only became clear to me
recently, and I find it quite interesting.

Regards,
Andor

Hi Andor,
Just for fun, I thought I'd reply to this. I haven't followed the thread
until now so if I reiterate something or don't make sense just ignore.


"Andor" <[email protected]> wrote in message
news:[email protected] om...
> r b-j wrote some time ago:
>
> > +inf
> > integral{g(t) * d(t) dt} = g(0) ?
> > -inf
> >
> > (where d(t) is our nasty Dirac impulse function, g(t)
> > is any reasonably well-behaved function of t and "*"
> > means "multiplication")
>
> Hi Robert and all,
>
> if the Dirac delta impulse function d(t) is defined through the above
> equation for "reasonably well-behaved functions" (by which I guess
> you mean continuous functions), why not define it as
>
> d(t) = g(0)? Or perhaps rather d(g) = g(0)? Or more general d_{t_0}(g)
> = g(t_0)?

Sure. We could. I guess it would be a functional, right? This is how it
is "defined" I believe. However, for utilitarian purposes we shorten it to
the simple Dirac delta "function" (fictitious, I agree) and use it's
functional properties.


>
> This would spare some confusion, wouldn't it?
>
> If anybody feels like arguing: I claim the Dirac delta function does
> not exist, not even as a limit of some integrable series of functions,
> quite in agreement with Airy.
>

I like to think of the Dirac Delta "function" as a different tool in
different situations. Sometimes it is helpful to think of it as:

lim y->x of 1/(y-x) * integral(x,y,f(t))dt = f(x)

assuming f is integrable (and I did my calculus correctly) this has similar
properties to the sifting property described above as:
+inf
integral{g(t) * d(t-t_0) dt} = g(t_0) ?
-inf


> Sorry to bring this up again, but this fact only became clear to me
> recently, and I find it quite interesting.
>

I also find it iteresting because it is something that any undergrad
engineer "knows" and can use quite well. However, as we think about its
consequences in different situations where the normal undergrad eduacation
doesn't go, it becomes an enigma. Especially since we have been taught to
throw it around carelessly.

Brandon

> Regards,
> Andor

Andor wrote:
>
> r b-j wrote some time ago:
>
> > +inf
> > integral{g(t) * d(t) dt} = g(0) ?
> > -inf
> >
> > (where d(t) is our nasty Dirac impulse function, g(t)
> > is any reasonably well-behaved function of t and "*"
> > means "multiplication")
>
> Hi Robert and all,
>
> if the Dirac delta impulse function d(t) is defined through the above
> equation for "reasonably well-behaved functions" (by which I guess
> you mean continuous functions), why not define it as
>
> d(t) = g(0)? Or perhaps rather d(g) = g(0)? Or more general d_{t_0}(g)
> = g(t_0)?
>
> This would spare some confusion, wouldn't it?
>
> If anybody feels like arguing: I claim the Dirac delta function does
> not exist, not even as a limit of some integrable series of functions,
> quite in agreement with Airy.
>
> Sorry to bring this up again, but this fact only became clear to me
> recently, and I find it quite interesting.
>
> Regards,
> Andor

Of course a delta doesn't exist; nobody can produce one. It's a handy
thought tool, though. Zero doesn't exist either, by definition, and it
was a long time after people started using numbers that its utility
became evident.

For that matter, one doesn't exist either. Some things like Johansson
blocks come very close, but whatever, it won't be exact.

"Ah", you say! "No object can be size one, but when I'm counting, I know
perfectly well what one means." I'll give you that. Hold on to that
thought and extend it.

Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

"Jerry Avins" <[email protected]> wrote in message
news:[email protected]..
> Andor wrote:
> >
> > r b-j wrote some time ago:
> >
> > > +inf
> > > integral{g(t) * d(t) dt} = g(0) ?
> > > -inf
> > >
> > > (where d(t) is our nasty Dirac impulse function, g(t)
> > > is any reasonably well-behaved function of t and "*"
> > > means "multiplication")
> >
> > Hi Robert and all,
> >
> > if the Dirac delta impulse function d(t) is defined through the above
> > equation for "reasonably well-behaved functions" (by which I guess
> > you mean continuous functions), why not define it as
> >
> > d(t) = g(0)? Or perhaps rather d(g) = g(0)? Or more general d_{t_0}(g)
> > = g(t_0)?
> >
> > This would spare some confusion, wouldn't it?
> >
> > If anybody feels like arguing: I claim the Dirac delta function does
> > not exist, not even as a limit of some integrable series of functions,
> > quite in agreement with Airy.
> >
> > Sorry to bring this up again, but this fact only became clear to me
> > recently, and I find it quite interesting.
> >
> > Regards,
> > Andor
>
> Of course a delta doesn't exist; nobody can produce one. It's a handy
> thought tool, though. Zero doesn't exist either, by definition, and it
> was a long time after people started using numbers that its utility
> became evident.
>
> For that matter, one doesn't exist either. Some things like Johansson
> blocks come very close, but whatever, it won't be exact.
>
> "Ah", you say! "No object can be size one, but when I'm counting, I know
> perfectly well what one means." I'll give you that. Hold on to that
> thought and extend it.

Cool.

Fred

In article [email protected], Andor at
[email protected] wrote on 08/21/2003 08:28:

> r b-j wrote some time ago:
>
>> +inf
>> integral{g(t) * d(t) dt} = g(0) ?
>> -inf
>>
>> (where d(t) is our nasty Dirac impulse function, g(t)
>> is any reasonably well-behaved function of t and "*"
>> means "multiplication")

BTW, i only wrote it (some time ago) because it is the main property of the
Dirac delta "function" (or whatever mathematicians wanna call it) that both
engineers/physicists and pure mathematicians can agree on. that rectangular
or gaussian or whatever shaped pulse that has unit area and a width that
approaches zero in the limit is a fine way to look at it from my POV but the
math guys cry "foul" because, given their rigorous ways of looking at
things, the Dirac Delta Function is not really a function at all. (the name
they give it is "distribution". my response to that is "you can have your
semantics and i'll have mine.")

> Hi Robert and all,

hi Andor and all,

>
> if the Dirac delta impulse function d(t) is defined through the above
> equation for "reasonably well-behaved functions" (by which I guess
> you mean continuous functions),

i dunno if that is either sufficient or necessary. sometimes a fully
continuous function is still nasty (thwarting the sampling of it) and i
think that the operation or definition above will work in for nice functions
that have a few (or countable) nice step discontinuities. as long as g(t)
is defined a 0 or t_0 or whenever the impulse happens.

> why not define it as
>
> d(t) = g(0)? Or perhaps rather d(g) = g(0)? Or more general d_{t_0}(g)
> = g(t_0)?

that's what it is. it's a "sifting" operator or a "sampling" operator.

> This would spare some confusion, wouldn't it?
>
> If anybody feels like arguing: I claim the Dirac delta function does
> not exist, not even as a limit of some integrable series of functions,
> quite in agreement with Airy.

Airy was saying other things, too, that was a bit off-the-wall.

> Sorry to bring this up again, but this fact only became clear to me
> recently, and I find it quite interesting.

the reason for me sticking with the non-rigorous engineering definition and
POV of the Dirac impulse "function" or "operator" or "distribution" or
"whatever" is that it corresponds to a physical concept (sorta what happens
regarding force when two perfectly hard and incompressible billiard balls
collide and the momentum of one is instantly transferred from one to the
other, force is the derivative of momentum and the momentum function has a
step discontinuity at the time of impact). the pure mathematicians would
rather never see the Dirac function unless it was contained in an integral,
but i don't think it creates confusion and i believe that, in the context of
EE and continuous-time linear systems (LTI or LTV), that the concept of an
impulse (with finite area and virtually zero width) that stands all by
itself and is input to the linear system is a useful one that works
pedagogically. the pure math guys don't like it, but the concept works
nonetheless (unless g(t) is a "bad" function).

because, from a rigorous real analysis POV, i expect to eventually lose the
argument sticking with the common engineering definition of the Dirac
impulse function, what i like to do as a "fall-back" is define it to be one
of them limiting pulses (say the rectangular one) with width of 1
femto-second, or if they really complain, a width of 1 Planck Time and with
a constant area of 1 (dimensionless). that is a *real* function and it will
work, to within any finite constraint of accuracy (meaning that all of the
properties will be indistinguishable from the "true" Delta function to
within any finite degree of accuracy), in any physical system. that, to me,
justifies the use of the non-rigorous common engineering definition.

BTW, the other responses to your post are valid also, IMO.

> Regards,

and also to you.

r b-j

Ian Buckner wrote:
>
> "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message
> news:F581b.4259$[email protected]. .
> >
> > "Jerry Avins" <[email protected]> wrote in message
> > news:[email protected]..
> > > Andor wrote:
> > > >
> > > > r b-j wrote some time ago:
> > > >
> > > > > +inf
> > > > > integral{g(t) * d(t) dt} = g(0) ?
> > > > > -inf
> > > > >
> > > > > (where d(t) is our nasty Dirac impulse function, g(t)
> > > > > is any reasonably well-behaved function of t and "*"
> > > > > means "multiplication")
> > > >
> > > > Hi Robert and all,
> > > >
> > > > if the Dirac delta impulse function d(t) is defined through the
> above
> > > > equation for "reasonably well-behaved functions" (by which I
> guess
> > > > you mean continuous functions), why not define it as
> > > >
> > > > d(t) = g(0)? Or perhaps rather d(g) = g(0)? Or more general
> d_{t_0}(g)
> > > > = g(t_0)?
> > > >
> > > > This would spare some confusion, wouldn't it?
> > > >
> > > > If anybody feels like arguing: I claim the Dirac delta function
> does
> > > > not exist, not even as a limit of some integrable series of
> functions,
> > > > quite in agreement with Airy.
> > > >
> > > > Sorry to bring this up again, but this fact only became clear to
> me
> > > > recently, and I find it quite interesting.
> > > >
> > > > Regards,
> > > > Andor
> > >
> > > Of course a delta doesn't exist; nobody can produce one. It's a
> handy
> > > thought tool, though. Zero doesn't exist either, by definition,
> and it
> > > was a long time after people started using numbers that its
> utility
> > > became evident.
> > >
> > > For that matter, one doesn't exist either. Some things like
> Johansson
> > > blocks come very close, but whatever, it won't be exact.
> > >
> > > "Ah", you say! "No object can be size one, but when I'm counting,
> I know
> > > perfectly well what one means." I'll give you that. Hold on to
> that
> > > thought and extend it.
> >
> > Cool.
> >
> > Fred
> >
> >
> >
> So if delta, zero and one don't exist, by simple extension no numbers
> exist...
>
> I think I'll go and lie down for a while.
>
> Regards
> Ian
>
> ;-)

No need. Take solace from Peano. "There is a number. For every number
there is a successor number. ..."

Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

>Subject: Re: Dirac delta
>From: Jerry Avins [email protected]
>Date: 8/22/2003 1:39 PM Eastern Daylight Time
>Message-id: <[email protected]>
>
>Ian Buckner wrote:
>>
>> "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message
>> news:F581b.4259$[email protected]. .
>> >
>> > "Jerry Avins" <[email protected]> wrote in message
>> > news:[email protected]..
>> > > Andor wrote:
>> > > >
>> > > > r b-j wrote some time ago:
>> > > >
>> > > > > +inf
>> > > > > integral{g(t) * d(t) dt} = g(0) ?
>> > > > > -inf
>> > > > >
>> > > > > (where d(t) is our nasty Dirac impulse function, g(t)
>> > > > > is any reasonably well-behaved function of t and "*"
>> > > > > means "multiplication")
>> > > >
>> > > > Hi Robert and all,
>> > > >
>> > > > if the Dirac delta impulse function d(t) is defined through the
>> above
>> > > > equation for "reasonably well-behaved functions" (by which I
>> guess
>> > > > you mean continuous functions), why not define it as
>> > > >
>> > > > d(t) = g(0)? Or perhaps rather d(g) = g(0)? Or more general
>> d_{t_0}(g)
>> > > > = g(t_0)?
>> > > >
>> > > > This would spare some confusion, wouldn't it?
>> > > >
>> > > > If anybody feels like arguing: I claim the Dirac delta function
>> does
>> > > > not exist, not even as a limit of some integrable series of
>> functions,
>> > > > quite in agreement with Airy.
>> > > >
>> > > > Sorry to bring this up again, but this fact only became clear to
>> me
>> > > > recently, and I find it quite interesting.
>> > > >
>> > > > Regards,
>> > > > Andor
>> > >
>> > > Of course a delta doesn't exist; nobody can produce one. It's a
>> handy
>> > > thought tool, though. Zero doesn't exist either, by definition,
>> and it
>> > > was a long time after people started using numbers that its
>> utility
>> > > became evident.
>> > >
>> > > For that matter, one doesn't exist either. Some things like
>> Johansson
>> > > blocks come very close, but whatever, it won't be exact.
>> > >
>> > > "Ah", you say! "No object can be size one, but when I'm counting,
>> I know
>> > > perfectly well what one means." I'll give you that. Hold on to
>> that
>> > > thought and extend it.
>> >
>> > Cool.
>> >
>> > Fred
>> >
>> >
>> >
>> So if delta, zero and one don't exist, by simple extension no numbers
>> exist...
>>
>> I think I'll go and lie down for a while.
>>
>> Regards
>> Ian
>>
>> ;-)
>
>No need. Take solace from Peano. "There is a number. For every number
>there is a successor number. ..."
>
>Jerry
>--
>Engineering is the art of making what you want from things you can get.

I would suggest to anyone who wants a more in depth understanding of the delta
function (distribution) to read Appendix I of Papoulis' book entitled The
Fourier Integral and Its Application (1962).

Scott

Hi Brandon,

> Hi Andor,
> Just for fun, I thought I'd reply to this. I haven't followed the thread
> until now so if I reiterate something or don't make sense just ignore.

OK .

> "Andor" <[email protected]> wrote in message
> news:[email protected] om...
> > r b-j wrote some time ago:
> >
> > > +inf
> > > integral{g(t) * d(t) dt} = g(0) ?
> > > -inf
> > >
> > > (where d(t) is our nasty Dirac impulse function, g(t)
> > > is any reasonably well-behaved function of t and "*"
> > > means "multiplication")
> >
> > Hi Robert and all,
> >
> > if the Dirac delta impulse function d(t) is defined through the above
> > equation for "reasonably well-behaved functions" (by which I guess
> > you mean continuous functions), why not define it as
> >
> > d(t) = g(0)? Or perhaps rather d(g) = g(0)? Or more general d_{t_0}(g)
> > = g(t_0)?
>
> Sure. We could. I guess it would be a functional, right? This is how it
> is "defined" I believe.

Absolutely correct. Let's call the functional D, and in the notation of
linear operators, one can write

D g := g(0) (for continuous function, this is a perfectly valid
definition).

If I want to extend the definition of the operator D to integrable functions
there is a problem: functions of the same class may differ pointwise. A
simple example is the function f(x) = 0 for x in ]0, 1] and f(0) = 1. This
function is in the same L^2 class on [0,1] with the zero function (meaning
Integral[ |f(x) - 0|^2, {x,0,1}] = 0). If I want the operator D to make any
sense on the space of L^2 functions, it must return the same value for
functions in the same class.

So the reason to define this operator via the integral is to kind of take
the "average" value around 0. The advantage is that you have extended the
operator to a larger space of functions (from continuous functions to
integrable ones). It is a common myth that every L^2 integrable function can
be approximated by a series of continuous ones (this myth stems from the
fact that the space of continuous functions lies "dense" in the L^2 space).

> I like to think of the Dirac Delta "function" as a different tool in
> different situations. Sometimes it is helpful to think of it as:
>
> lim y->x of 1/(y-x) * integral(x,y,f(t))dt = f(x)

Again, this equation is only correct for continuous functions. The value
f(x) could be far away from the term on the right of the equal sign if f is
not continuous (but still integrable).

> > Sorry to bring this up again, but this fact only became clear to me
> > recently, and I find it quite interesting.
> >
>
> I also find it iteresting because it is something that any undergrad
> engineer "knows" and can use quite well. However, as we think about its
> consequences in different situations where the normal undergrad eduacation
> doesn't go, it becomes an enigma. Especially since we have been taught to
> throw it around carelessly.
>
> Brandon

Just the experience I had .

> > Regards,
> > Andor

Andor wrote:
>
> Jerry Avins wrote:
> ...

...
>
> >
> > "Ah", you say! "No object can be size one,
>
> Wrong again. I can easily define the size of an object as being "1", with
> the apropriate unit. We do it all the time, that's what units are all about.

Agreed. Every object has size one when the measurement unit is "self".
That's what normalizing is all about.

> I'm not quite sure what this has to do with the Dirac delta though .
>
It has to do with the existence of intangibles, a class that includes
the Dirac delta. :-)

Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

In article [email protected], Andor at [email protected] wrote on
08/24/2003 06:51:

> Jerry Avins wrote:
> ...
>> Of course a delta doesn't exist; nobody can produce one. It's a handy
>
> Hey waitaminute! I think I have to get the definition of "non-existence"
> straight: the Dirac delta does not exist as a limit of integrable functions,
> because any limit of integrable functions does not have the properties of
> the Dirac delta. So, by contradiction, it _cannot_ exist.

that's Fatou's lemma in Real Analysis, right. the integral of a function
that is zero almost everywhere is zero. the thing that bothers me (with
this lemma) is that the limit of the integral of these integrable functions
(not the limit of the functions themselves) *does* exist and is 1 (for the
unit impulse).

In article 3f48a838$[email protected], Andor at [email protected] wrote
on 08/24/2003 08:01:

> robert bristow-johnson wrote:
>
>>> r b-j wrote some time ago:
>>>
>>>> +inf
>>>> integral{g(t) * d(t) dt} = g(0) ?
>>>> -inf
>>>>
>>>> (where d(t) is our nasty Dirac impulse function, g(t)
>>>> is any reasonably well-behaved function of t and "*"
>>>> means "multiplication")
>>
>> BTW, i only wrote it (some time ago) because it is the main property of the
>> Dirac delta "function" (or whatever mathematicians wanna call it) that both
>> engineers/physicists and pure mathematicians can agree on.
>
> We actually agree on most points. All I wanted to state was the fact that if
> the Dirac delta were defined as D g = g(0)
>
> (or if this equation were true:
>>>> +inf
>>>> integral{g(t) * d(t) dt} = g(0)
>>>> -inf
> )
>
> then this defintion would be useful for continuous functions only.

the functions, g(t), don't have to be continuous, just *defined* at 0, for
the latter definition. there could be a step discontinuity at 0 but if g(0)
is defined (not at some undefined meanpoint between g(-epsilon) and
g(+epsilon). what if:

{ 1 for t >= 0
g(t) = {
( 0 for t < 0

? doesn't the functional definition work in that case?

for the definition regarding the limit of the rectangular or whatever
function becoming an impulse, then maybe g(t) has to be a continuous
function. i'm not sure about that.

> I wanted
> to make it clear that the integral was needed to "average" the value of g
> around 0 (read my reply to Brandon) to extend D on the class of integrable
> functions, so the integral is needed in the definition of the Dirac delta
> functional. I think that this is also the reason for all the confusion.

but in the limit, there isn't much around 0 that g() has to be averaged.

> The use of the Dirac delta outside of integrals can be justified as a measure
> (delta(A) = 1 if 0 is element of A, A a subset of the real numbers).

but that isn't an expression of d(t) as a function of t. which is fine.
it's just that i don't understand where your functional "delta()" comes into
play.

> With
> this delta measure, you can well, measure sets and, more importantly,
> define integrals over delta-measurable functions.
>
> Both constructions result in the same thing, but one defines a functional on
> the space of integrable functions, and one defines a measure on subsets of
> the real numbers. This measure (like all normalized measures) can then be
> used to define a distribution for a random variable with a point mass at 0,
> thus the term "distribution".

in other words, a random variable that is a constant (not very random) so
it's p.d.f. is an impulse, right?

> Regards,

and also to you.

r b-j

Jerry Avins wrote:
....
> I'm too ignorant to be bothered. It makes perfect sense to me that the
> integral of a function that is zero a.e. be zero AS LONG AS THE
> EXCEPTIONAL POINTS ARE FINITE. So it doesn't seem contradictory to me
> that the integral of a Dirac delta is a step.

Jerry, we talked about this before . The integral is zero if the
function is zero a.e. The exceptional points can be infinite, and
there can be uncountably many infinite points (ie. as many as there
are real numbers) in the interval where you integrate, as long as the
measure of the set of these uncountably many points is zero.

I wrote:

> > > The use of the Dirac delta outside of integrals can be justified as a measure
> > > (delta(A) = 1 if 0 is element of A, A a subset of the real numbers).
> >
> > but that isn't an expression of d(t) as a function of t. which is fine.
> > it's just that i don't understand where your functional "delta()" comes into
> > play.
>
> d(t) := delta( {t} ) (the argument to delta has to be a subset of
> the reals)
>
> ...
>
> > in other words, a random variable that is a constant (not very random) so
> > it's p.d.f. is an impulse, right?
>
> Correct. And the distribution of this random variable X is the delta
> measure:
>
> P[ X <= t ] = delta( ]-infinity, t] )


On second thoughts, this doesn't get us any further. If the
distribution of this rv is the delta measure, we want that the density
is the Dirac delta. But we can't get such a density, because the delta
measure is not absolute continuous w.r.t the Lebesgue measure. This
just means that you cannot get the value of the point mass back in an
integral with the Lebesgue measure - ie. there is no density function.

So the probability measure has to be split up into a sum of an
integral over an absolute continuous function (the density) and a
signular measure (delta measure) which returns the discrete steps
(like for example all discrete random variables have no density
function). Just one more proof that the Dirac delta does not exist as
a function .

So the name "distribution" for the Dirac delta doesn't come from
probability theory (because the Dirac delta is a "density", so to
speak). I looked it up in a physics book - they define a distribution
as a functional. Since this is a German book, and the use the term
"Distribution" and not "Verteilung" which would be the correct
translation of the probability distribution, I think the two have
nothing in common, except their English names.

Regards,
Andor

Jerry Avins wrote:
....
> I know we did. But I can't pound into my head that while the derivative
> of a unit step is a delta, the integral of a delta isn't a unit step.
> You can see why I so admire Heavyside. I don't believe everything math
> tells me. I know I won't ever be able to dice up a gold sphere, then
> reassemble the pieces in a way that would make me rich. Who shaves the
> barber? :-)

Perhaps alchemists should have studied more measure theory and paradox
decompositions instead of fiddling with Toad's Skin and Bat's Ears ...