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[MUG] Re: Dirac delta
| [MUG] Re: Dirac delta |
|
Author: Maple User Group
Posted: Wed, 7 Aug 2002 16:24:10 -0400 (
|
>> From: Maple User Group "maple_gr"
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Date: Tue, 6 Aug 2002 15:13:43 -0400 (EDT)
From: Fred Chapman "fwchapma"
To: "maple-list"
Subject: Dirac delta
By definition, Dirac's delta distribution satisfies (in Maple notation)
infinity
/
|
| f(x) Dirac(x) dx = f(0)
|
/
-infinity
In your case, f(x) is implicitly defined as the Heaviside function H(x),
because
infinity infinity
/ /
| |
| Dirac(x) dx = | H(x) Dirac(x) dx
| |
/ /
0 -infinity
where
{ 0 x < 0
H(x) = {
{ 1 0 < x
The value of your integral is therefore H(0). The problem is that the
definition of H(0) is itself ambiguous! Some people define H(0) = 1 so
that H(x) is right-continuous at x = 0; some people define H(0) = 0 so
that H(x) is left-continuous at x = 0; some people define H(0) = 1/2,
which is the average (arithmetic mean) of the right-sided and left-sided
limits 1 and 0.
I recommend that you make an arbitrary choice for the value of H(0),
defining it to be always 0 or always 1/2 or always 1 according to what
makes the most sense in the context of your larger problem. Whatever
value you assign to H(0) is also the value of your original integral.
You can't avoid making an arbitrary choice, but you can apply your choice
consistently throughout your work, which is the important thing here.
For a practical introduction to distribution theory, I recommend Bernard
Friedman's books: "Principles and Techniques of Applied Mathematics"
(Dover, 1990) and "Lectures on Applications-Oriented Mathematics" (Wiley,
1991). For a more in-depth theoretical treatment, I recommend A.H.
Zemanian's "Distribution Theory and Transform Analysis: An Introduction
to Generalized Functions, with Applications" (Dover, 1987).
---------------- http://www.scg.uwaterloo.ca/~fwchapma/ ----------------
Frederick W. Chapman, Ph.D. Student UW Office: Math & Computer 5162
Department of Applied Mathematics UW Phone: (519) 888-4567 x6672
University of Waterloo E-Mail: "fwchapman"
Waterloo, Ontario, N2L 3G1, Canada Curriculum Vitae: see home page
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Date: Tue, 6 Aug 2002 17:05:11 -0700 (PDT)
From: Robert Israel "israel"
To: "maple-list"
Subject: Dirac delta
The answer is that the question is not well-defined.
The Dirac delta is a Borel probability measure on the real line
with a unit mass at the origin. So you can integrate this measure
over any Borel set, in particular an interval from 0 to infinity.
But when you said "between 0 and infinity" you didn't specify whether
the interval is [0,infinity) or (0,infinity), i.e. whether or not 0
is included in the set. The integral over [0,infinity) is 1, the
integral over (0,infinity) is 0. Mathematica's answer of 1/2 is
not the integral over any set (which doesn't mean it's necessarily
wrong - you could interpret it as integrating a function H(x) that is
0 for x < 0, 1/2 for x = 0 and 1 for x > 0).
Robert Israel "israel"
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
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Date: Wed, 7 Aug 2002 15:41:27 +0200 (CEST)
From: "Richard B. Kreckel" "kreckel"
To: "maple-list"
Subject: Dirac delta
It is not defined. Both Maple and Mathematica define it in some
way they consider meaningful or elegant. Mathematica obviously defines it
to be symmetric.
You may "represent" the Dirac delta function as any ordinary sharply
peaked function (such that their integral is normalized to unity) and take
the limit where the peak gets infinitely sharp. Take step functions for
instance, you may chose a step sitting just right of zero, or just left of
zero or centered around zero. From the point of view of distribution
theory it does not matter, that representation is at best a visualization
anyway.
> Does anyone understand this?. Can somebody tell me where I can I find
> a formal proof using distribution theory whether this integral is
> 1,1/2 or zero?.
There is no such thing.
Regards
-richy.
--
Richard B. Kreckel
"Richard.Kreckel"
<http://wwwthep.physik.uni-mainz.de/~kreckel/>
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From: "Daniel Willard" "willardd"
To: "maple-list"
Subject: Dirac delta
Date: Wed, 7 Aug 2002 09:51:59 -0400
If your delta function is "delta(x)" then Mathematica is considered correct.
If the limit as epsilon goes to zero gets all the way to zero, then 0 is an
acceptable answer. Question of open versus closed interval.
| >> From: Herman Jaramillo "HJaramillo"
|Dear Maple Users:
|
|When I compute the integral of the Dirac delta function between 0 and
|infinity using Maple 7, I get "1". When I do it using Mathematica I
|get "1/2".
|
|When I try to do it using distribution theory by choosing a test
|function which has bounded support on the positive real line and using
|the limit from the left and the right of the interval of integration
|(epsilon, infinity) as epsilon goes to 0 I get that the integral
|should be "0".
|
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