Talk:Dirac delta function

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From the article:

If you integrate the delta function between ANY limits a and b, then the integral is:

0 if a,b > 0 or a,b < 0
1 if a < 0 < b
0.5 if a = 0 or b = 0

Really? I'm not sure about the last of these lines, the one with value 0.5. Surely this contradicts the "compact support" bit?

Looks right to me -- I remember the delta func being defined as the limit of a sequence of functions, each getting pointier. A pointy function of full integral 1, centred on 0 clearly has a half-integral of 1/2. with Lebesgue integration (which I think is the only thing you can use for the delat function, you can't use Riemann), there's something about the limit of the integrals is the integral of the limits (uniform convergences is probably a requirement too) -- Tarquin 20:44 Nov 22, 2002 (UTC)

The difference is that in Lebesgue integration you really integrate over a set. For Lebesgue measure it doesn't matter whether that set is an open interval or its closure, but for the Dirac measure it does. Thus when you integrate over [0, b] you get 1 and when you intergrate over (0, b) you get zero. There is no way to get 1/2. Just as there is no set of which 0 is half an element. -MarSch 17:57, 5 May 2005 (UTC)

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The sequence of functions that go into the delta function does not necessarily have to be centered at zero. I conjecture the sufficient condition is simply that the center approaches 0 as the function approaches an infinite spike. For example, consider the rectangular function

$f_{\epsilon }(x)=\left\{{\begin{matrix}1/\epsilon ,&{\mbox{if }}(c-1/2)\epsilon <x<(c+1/2)\epsilon \\0,&{\mbox{otherwise}}\end{matrix}}\right.$

This function is centered at cε. For any real c,

$\lim _{\epsilon \to 0}f_{\epsilon }(x)=\delta (x)$

However, the value of the half-integral depends on c.

Cyan 05:58 4 Jul 2003 (UTC)

I don't think I really understand this discussion. It is morally equivalent to trying to convolve the Dirac delta-function with the Heaviside function, no? Which is like trying to specify the value of the Heaviside function at 0. Saying it is 0.5, i.e. halfway up, is sort of the right answer in Fourier theory - but I doubt it is the right way to say it.

Charles Matthews 07:48 4 Jul 2003 (UTC)

I think what you are trying to say is: if you evaluate the Fourier series of a discontinuous periodic function at a discontinuity, it converges to the average of the original function's limiting values at the discontinuity. But all that means is that the original function and the Fourier series representation can disagree at discontinuities.

The delta function can be defined in various ways, e.g. as a measure in measure theory, or as a linear functional, or as an integral satisfying certain properties under a limiting operation. I don't know squat about measure theory or functional analysis, so I go with the third definition:

if

$\lim _{\epsilon \to 0}\int _{-\infty }^{\infty }f_{\epsilon }(x)g(x)dx=g(0)$

we say that $f_{\epsilon }(x)$ is a delta sequence, and for shorthand, we abuse proper notation by writing

$\lim _{\epsilon \to 0}f_{\epsilon }(x)=\delta (x)$

(Lists of delta sequences may be found at [1] and [2].)

Here's the issue: is the following statement true for all delta sequences?

$\lim _{\epsilon \to 0}\int _{0}^{\infty }f_{\epsilon }(x)g(x)dx=0.5\cdot g(0)$

Now, some delta sequences are symmetric about the y-axis, and would yield a half-integral of 0.5*g(0). But other delta sequences, like the one I defined above, are not necessarily symmetric about the y-axis. The half-integral is really indeterminate, because the definition of a delta sequence doesn't constrain it to any particular value.

Cyan 05:18 7 Jul 2003 (UTC)

We Japanese think that

$\int _{-\epsilon }^{0}\delta (x)dx=\int _{0}^{+\epsilon }\delta (x)dx=1/2$ for all $\epsilon >0$

and every image $H(x)$ of Heaviside step function $H:\mathbb {R} \ni x\mapsto H(x)\in H(\mathbb {R} )\subset \mathbb {R}$ is

$H(x)=\int _{-\infty }^{x}\delta (\xi )d\xi ={\frac {1+{\rm {sgn}}x}{2}}=\left\{{\begin{matrix}0&\left(x<0\right)\\1/2&\left(x=0\right)\\1&\left(x>0\right)\end{matrix}}\right.$ .

User:Koiki Sumi 00:00, 15 Sep 2003 (UTC) & 00:30, 18 Sep 2003 (UTC) Who had changed $\mapsto$ for $\longrightarrow$? It has been returned as before.

I believe that my off-center rectangular function (see above) is a counter-example to the idea that

$\int _{-\epsilon }^{0}\delta (x)dx=\int _{0}^{+\epsilon }\delta (x)dx=1/2$ for all $\epsilon >0$. -- Cyan 04:47, 15 Sep 2003 (UTC)

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The indefinite integral of a distribution is another distribution, not a pointwise valued function. So, the whole question about $\int _{0}^{a}\delta (x)dx$ just isn't well-defined. Phys 17:44, 6 Nov 2003 (UTC)

Japanese

Cannot understand the sense of that paragraph... Pfortuny 11:29, 13 Sep 2004 (UTC)

the statement that $\delta (\mathbb {R} )\subset \mathbb {R}$ is probably false. for example, $\delta (0)\not \in \mathbb {R}$. I am going to remove that part. --> You say that $\delta (0)\in \mathbb {C} -\mathbb {R}$, don't you?. (OMAERA WA BAKA KA ?) ---

Anyway, in short, this japanese version (i have never heard this version referred to as "japanese", can someone attest that usage?) simply says that the delta function is the derivative of the Heaviside step function. In other words, let

$\theta (x)={\begin{cases}0&x<0\\1&x\leq 0\end{cases}}$

(this is what is known as the Heaviside step function. It can also be written in terms of the signum function, as is done in the article). Then you can simple define the delta function to be

$\delta (x)={\frac {d\theta }{dx}}$

Your homework: figure out how what the article says is the same thing as what I said. -lethe ^talk

On "The delta function as a probability density function" in the article, most Japanese see that $\delta$ and $h$ must be homeomorphic. -- OMAERA WA BAKA KA ?

Let $f:\mathbb {R} \to \mathbb {R}$ be

$f(x)={\begin{cases}-1&(x\in \mathbb {Q} )\\1&(x\in \mathbb {R} -\mathbb {Q} )\end{cases}}$ .

For every $a,b\in \mathbb {R}$,

$\int _{a}^{b}f(x)dx=b-a$ ,

where integral means improper Riemann integral.

sequences

Well, I just want to add this admittedly pedantic piece of comment: From the point of view of mathematical exactness and completeness it should be emphasized that the parameter a shares the property of all epsilon-small quantities in maths: It is positive! Otherwise, many given formulae are incorrect in the sense that they in fact define sign(a)*delta(x).

The support of the delta function

Ok, maybe the support of the delta function is zero, but consider this:

$\lim _{a\rightarrow 0}{\frac {{\textrm {sinc}}(x/a)}{\pi a}}=\delta (x)$

where sinc(x)=sin(x)/x is the sinc function. This is indeed a delta function, but the reason it yields zero when integrated over any interval not containing zero is NOT because it goes to zero, but because its period of oscillation goes to zero, while its amplitude stays between ±1/x. This means that it goes negative sometimes, and is therefore not a probability density function. Only if you specify that the Dirac delta function be further constrained to be a probability density function can it be thought of as approaching zero except at x=0. That is why I question the support of δ(x) being zero, and the inclusion of the probability infobox in the article. PAR 01:50, 24 Mar 2005 (UTC)

FWIW, I find the probability infobox annoying; these infoboxes would be a lot more useful if they were consolidated onto one page, so that user could actually compare distributions to find the one that they wanted, instead of clawing randomly through various articles hoping to trip over the infobox they wanted. linas 06:11, 26 Jun 2005 (UTC)

Broken HTML table

The html table at the bottom, listing many different kinds of limits, it totally broken on my browser. I notice User:PAR added this maybe 3 months ago. Can I get rid of it? I'd rather do that than try to figure out the bug in the HTML markup. (We shouldn't be using html markup in WP articles anyway). linas 06:11, 26 Jun 2005 (UTC)

I am against removing it. It is not an HTML table, its a wiki table. I have looked at it on three different browsers on two different machines, and it always looks fine. It could be a temporary glitch in the wikipedia software, or it could be a problem with your browser or machine. If you can, try looking with a different browser and/or a different machine. Also wait a while and try again. What is the nature of the problem? I mean what does it look like on your browser? PAR 16:34, 26 Jun 2005 (UTC)

Are you talking about "Some nascent delta functions are:" invisible table? Looks fine here. --MarSch 18:56, 26 Jun 2005 (UTC)

Its still broken. I'm using Konqueror version 3.3.2 which I beleive is the same as what is the main default browser on Macintoshes (Safari). Visually, the formulas overlap with text and each other; the thing is considerably wider than what fits in a standard column width. (My monitor is set for 1600x 2000 and the table is still to wide for that.) linas 20:30, 24 July 2005 (UTC)

Can you get access to a different browser? Try it with that. Really, the table is fine, its broken on your end. PAR 22:19, 24 July 2005 (UTC)

No, its simply the best browser out there. No other web sites are broken, no other WP articles are broken. Just this one. I'm assuming its not my browser, but something about the html that is generated. Maybe it should be run through an HTML checker? linas 03:24, 25 July 2005 (UTC)

 1 - Why run it through an HTML checker when its a Wikipedia table?
 2 - Do you have another browser?

PAR 15:33, 25 July 2005 (UTC)

Konqueror is krap, in my experience.

I just ran this page through HTML validation, and it came out fine. "This Page Is Valid XHTML 1.0 Transitional!" So there's nothing wrong with the code itself. Of course validation doesn't point out visual errors with otherwise legit code.

Can you take a screenshot of your problem?

You should just get Firefox.  :-) - Omegatron 16:02, July 25, 2005 (UTC)

Distribution Notation?

Can someone define what this notation means, or link to page that defines it? I've never seen it before, and its not used on the page about distributions:

As a distribution, the Dirac delta is defined by

$\delta [\phi ]=\phi (0)\,$

—the preceding unsigned comment is by 129.7.57.161 (talk • contribs) 01:07, 8 January 2006 (UTC1)

It means that the delta functional takes a function φ as intput, and returns that function's value at 0 as output. I guess it's the square brackets that are throwing you off? Sometimes physicists like to use square brackets for things that take entire functions as inputs, instead of just numbers. -lethe ^talk 01:15, 8 January 2006 (UTC)

My problem was that I didn't understand what a distribution was. I read the distribution page enough to see that a distribution was a generalized function, and that they did not use the same notation. Reading further, I finally see that a distrubution is a map on the domain of test functions. —the preceding unsigned comment is by 129.7.57.161 (talk • contribs) 16:53, 8 January 2006 (UTC1)

Formal introduction

Is this a typo?: $\int _{-\infty }^{\infty }f(x)\,d\delta (x)=f(0)$

If not, it needs more explanation, IMO. --Bob K 07:51, 23 January 2006 (UTC)

I think it's a case of someone following an infelicitous convention, at best. Michael Hardy 23:12, 31 March 2006 (UTC)

cumulative distribution function

When talking about probability density functions, it is conceivable that there is a reason to prefer a convention that, at points when a jump discontinuity occurs, one defines the value of the function to be halfway between the two one-sided limits. But with cumulative distribution functions rather than probabilbity density functions, that is absolutely incorrect. By the usual convention, one defines the cumulative probability distribution function of a real-valued random variable X by

F(x) = Pr(X ≤ x).

There is also a convention, seldom seen, as far as I can tell, that defines it thus:

F(x) = Pr(X < x).

But either way, picking the halfway point at a jump discontinuity is completely wrong. Michael Hardy 23:16, 31 March 2006 (UTC)

You should probably add something to the article explaining this, as the subtlty of this point will be lost on most readers, leaving the field open for this error to be repeated over and over again ... linas 23:31, 31 March 2006 (UTC)

The Dirac Delta Function in Curvilinear Orthogonal Coordinates

it should be interesting to introduce the three dimensional delta functions, along with it's definition for other orthogonal coordinates (spherical and cylindrical) since it is useful for finding Green's function.

a good reference may be found here: [3]

$\delta ^{3}({\vec {r}}-{\vec {r'}})=\delta (x-x')\delta (y-y')\delta (z-z')$

$\delta ^{3}({\vec {r}}-{\vec {r'}})={\frac {\delta (r-r')\delta (\phi -\phi ')\delta (z-z')}{r}}$

$\delta ^{3}({\vec {r}}-{\vec {r'}})={\frac {\delta (r-r')\delta (\theta -\theta ')\delta (\phi -\phi ')}{r^{2}\sin \theta }}$

Cumulative distribution function

I restored the cumulative distribution function plot because there was no explanation given as to why it was wrong. If its wrong, please give an indication of how it is wrong. PAR 03:57, 5 April 2006 (UTC)

Merge with Dirac delta function

The articles are obviously equivalent, and Dirac delta function is much more complete than Dirac delta measure. I recognize that "function" is technically an incorrect term, but it is still extremely widely used,