Talk:Dirac delta function/Archive 1

Archive 1

Archive 2

Graphic/Animation

The animation of the Gaussian limit is not normalized -> Incorrect. 128.200.93.188 (talk) 21:26, 16 April 2009 (UTC)

Definite integrals of the delta function

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?

—The preceding unsigned comment was added by 217.158.106.142 (talk • contribs) 22:36, 22 November 2002 (UTC2)

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)

---

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)

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)

---

Would anyone please put something in the article about the value of

$\int _{0}^{\infty }\delta (x)dx?$

I was taught that if you for some reason need to evaluate such an expression in a physical problem, you need to state which convention you use (0, 1, or 1/2). But I was educated as a physicist, who are known to treat mathematics from a practical point of view. :-) Han-Kwang 21:01, 5 June 2006 (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 ?

-- < OBAKA E >

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

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

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

$\int _{a}^{b}f(t)dt=b-a$ ,

where integral means improper Riemann integral.

Let $\phi$ be a function $\mathbb {Q} -\{0\}\to \mathbb {R} -\{-\infty ,+\infty \}$ ,   and let $g$ be a distribution $\mathbb {R} \to \mathbb {R}$ :

$g(t)={\begin{cases}\phi (t)&(t\in \mathbb {Q} -\{0\})\\\delta (t)&(t\in (\mathbb {R} -\mathbb {Q} )\cup \{0\})\end{cases}}$ .

For every $x\in \mathbb {R}$ ,

$\int _{-\infty }^{x}g(t)dt=h(x)$ ,

where $h:\mathbb {R} \ni t\mapsto {\frac {1+{\rm {sgn}}t}{2}}\in \mathbb {R}$ .

—The preceding unsigned comment was added by 219.49.2.14 (talk • contribs) 13:45, 13 April 2006 (UTC2)

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)

this is the way to define it when δ is the Dirac delta measure. Eliding x may make that more clear. $\int f\,\mathrm {d} \delta =f(0)$--MarSch 14:46, 13 April 2006 (UTC)

The subsection with that equation ("The delta function as a measure") is very short and a quarter of it is about distributions instead of measures. Should the subsection be there at all? Is there any satisfactory way of thinking about the delta function as a measure? How do you get the $\int _{-\infty }^{0}f(x)\,\mathrm {d} \delta (x)=\int _{0}^{\infty }f(x)\,\mathrm {d} \delta (x)=f(0)/2$ property? Perhaps by setting measure of {0} to 1/2 and then adding two infinitesimals around 0 so that the measure of {-infinitesimal, +infinitesimal} is 1/2, and the measure of the rest is 0, and extend f's domain accordingly? Surely there's some standard way of doing this (unless this subsection is original research) and it'd be nice if it were explained. The scaling property too - it's given as a "helpful identity", but as far as I can see it does not follow from this measure definition, which I take as an indication that the definition is lacking. (intuitively it follows from the delta function being an approximation of a very thin box and scaling the axis changes the area of the box which should be reflected in the approximation, but that's motivation for the definition, not a definition) 82.103.198.180 23:31, 14 July 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)

maybe you should look up two topics? --MarSch 14:43, 13 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, much more so than that "delta measure". I therefore propose that Dirac delta measure be merged into Dirac delta function. --Zvika 18:15, 4 April 2006 (UTC)

Dirac delta is also widely used and not incorrect. I think we should merge to there. -MarSch 12:59, 5 April 2006 (UTC)

That article is quite silly. It never even defines the Dirac measure! The measure is defined here though, so I don't even know if there's anything to merge. Maybe can just be made into a redirect here. (And I favor the name dirac delta function over the slightly name Dirac delta). -lethe ^{talk +} 15:08, 5 April 2006 (UTC)

Keep an open mind I've never heard of a dirac measure either, but it sounds right as rain to me. And I've formally studied the Dirac delta function from the perspective of Papoulis's The Fourier integral and its applications (McGraw Hill, 1962). --Firefly322 (talk) 12:26, 11 March 2008 (UTC)

Unit impulse function??

If the article on the unit impulse function is correct, then to say that "The Dirac delta function (is) sometimes referred to as the unit impulse function" is wrong. The integral of a delta function is 1, whereas the integral of the unit impulse function would be infintesimal. What should we do about that? Fresheneesz 05:00, 19 April 2006 (UTC)

The term unit impulse directly redirects here, which I don't think it is appropriate. At least unit impulse should be directed into unit impulse function I mean disambiguation because Kronecker delta is more appropriate for discrete domain unit impulse. —Preceding unsigned comment added by Lielei (talk • contribs) 21:16, 9 December 2007 (UTC)

That impulse function is defined on the integers, so an "integral" is really a summation. If you sum that bad boy, you do indeed get 1. So I don't think there's a contradiction. -lethe ^{talk +} 05:05, 19 April 2006 (UTC)

Yes an integral is a sum, but its a weighted sum. Each infintesimal part is multiplied by the function at that point, and thats added up. If you only have one infinitesimal, the product of those is infintesimal. There is a contradiction. You don't indeed get 1. At least... I don't understand how you could. Fresheneesz 07:49, 19 April 2006 (UTC)

An integral over the reals can be thought of as a sum of infinitesimals. That is, it's the supremum of a sum of smaller and smaller bits. But when you're working over the integers, there is no "smaller and smaller". There is no infinitesimal. There's just a sum. So the integral of the function x² over the integers from 0 to 3 is just 0+1+4+9 = 16. Similarly, the integral over the integers of the impulse function on the integers will just be 1. -lethe ^{talk +} 08:01, 19 April 2006 (UTC)

I believe one of the properties of the dirac delta function is that its width is infinitely small but the area underneath is equal to 1. (It is infinitely skinny but infinitely tall, but the one constant is that its integral is 1--this is the reason the sifting property exists.) -Msa11usec 03:55, 20 April 2006 (UTC)

It's true that the dirac delta can be approximated by functions with decreasing width, increasing height, and constant area of 1. Thank you for the synopsis. -lethe ^{talk +} 04:07, 21 April 2006 (UTC)

I can believe that "The Dirac delta function (is) sometimes referred to as the unit impulse function", so we should deal with that reality. But linking to unit impulse function is not currently a solution, because that is an article about the Kronecker delta. A possible solution is to convert that article into a disambiguity page that points to both Kronecker delta and Dirac delta function. --Bob K 16:25, 19 April 2006 (UTC)

Looks like you did that. But in response to Lethe, an integral is a sum over a continuous set, an integral doesn't exist on integers. An integral is always an infinite sum, no matter what interval. But sums of integers are not always infinite, and are never continuous. In any case, your example with integers doesn't apply anyway, because the Dirac delta function is a function of a continuous set. Fresheneesz 08:06, 21 April 2006 (UTC)

Let me summarize for you the definition of an integral, maybe that will help. An integral of a function over a set is the greatest of the sums of the heights of the function over small bits times the size of the small bits. When considering real functions, one uses the Lebesgue measure, which in the limit mutliplies by vanishingly small bits. When considering functions on the integers, one uses the counting measure, which simply counts the number of points in the small bits. The latter kind of integral is simply a sum over the integers, and the Dirac delta functional on that measure space is just the Kronecker delta, what someone around here is calling the unit impulse function. In short, the definition of an integral allows you to integrate over the reals or the integers, using the appropriate measure. Your comment that integrals are always over continuous sets and that integrals don't exist over the integers is simply not true. Integrals do exist over the integers, where they're known as sums. -lethe ^{talk +} 08:18, 21 April 2006 (UTC)

I had the notion that an integral was a type of sum - the fact that they have different symbols implies (of course doesn't prove) that they are in fact different. In anycase, I could hypothetically agree with your definition, but this would mean that the "area under the curve" approach to an integral no longer works. Fresheneesz 22:03, 21 April 2006 (UTC)

That's true, it's hard to interpret the integral of a function over the integers as an area. However, in modern mathematics, the integral is not defined in terms of area. On the contrary, area is usually defined in terms of integrals. -lethe ^{talk +} 23:47, 21 April 2006 (UTC)

Maybe it would be useful to re-frame this discussion as a mis-communication due to overloading of the word integration (and the disambiguation page isn't much help!). You probably don't need me to tell you this, but I have a few idle minutes to waste. One person is using a broader, more general, definition than the other person. But indeed, "integration", is commonly used in the more specific sense of $\mathbb {R}$ and Lebesgue measure. In fact I would dare to suggest that that is its most common usage (in a mathematical context). Anyhow, it's a safe bet that the intuitively-pleasing association with "area under the curve" will be around longer than Wikipedia. Fagetaboutit, and have a great weekend everybody. --Bob K 00:04, 22 April 2006 (UTC)

Bob is right. Fresh, the most common usage of the word "integral" is like you describe. Area under a continuous real curve. However, there is a commonly used rigorous definition which allows the definition of delta functions in such a way as that the unit impulse function you're complaining about is perfectly rigorous. This is why my point was originally, and is still, there is no contradiction. -lethe ^{talk +} 01:33, 22 April 2006 (UTC)

By the way, your explanation inspired me to read Integral, which led me to Calculus. If Wikipedia is correct (always a big if), Calculus appears to be limited to the more specific meaning of integral. And the Integral article begins with the statement: "This article deals with the concept of an integral in calculus." So where is the proper place for your information/explanation? My instinct is to add a small section to Integral and get rid of the leading caveat. The only alternative I can think of is to rename Integral (e.g. "Integral calculus") and create a new article for your information. (What would it be called?) Then Integral would become a disambiguation page. --Bob K 14:08, 22 April 2006 (UTC)

This is a small problem, since the two different definitions of "integral" have nontrival differences. If the integral of the Kronecker delta function is 1, then it should be explained why that is, and what the connotations are of such a thing. If the statement is true.. but useless, then it probably shouldn't be in the article. In any case, the Kronecker delta article has nothing on its integral - but it would be very interesting to note what the significance of an integral over integers is. Fresheneesz 03:11, 23 April 2006 (UTC)

There aren't two different definitions of "integral". There is only one definition, but there are two choices of measure: the Lebesgue measure on the reals, and the counting measure on the integers. The fact that the Kronecker delta satisfies this property is already discussed in the article, where it's referred to as the sifting property, but if you would like the article to mention integration, I can certainly try to accommodate. I've added a comment to the article Kronecker delta. How do you like it? -lethe ^{talk +} 04:11, 23 April 2006 (UTC)

Oh, my mistake. Heh, well.. I don't understand your revision - it doesn't clarify anything for me : ) . But I don't know enough about either sifting properties .. or integration (apparently).. or delta fuctions to be able to help. I was just trying to point out what a thought was an inconsistancy. It's all fine with me now. Fresheneesz 11:04, 23 April 2006 (UTC)

If you would like to understand how summation is an example of integration, first you should learn what a measure is, and then how integrals are defined for arbitrary measure. This is standard material for a first semester graduate class in real analysis. Wikipedia has all the relevant material as well. -lethe ^{talk +} 04:21, 26 April 2006 (UTC)

I also think we're in pretty good shape. The change to Kronecker delta looks good to me... short but sweet with a link for those who would like to know more. My only concern is that when I turn to Wikipedia to learn the definition of integral, I find the calculus definition, which is the [limited] common definition, not the [general] mathematical one. Effectively there are two definitions. I think it is fine to emphasize the common definition, but the integral article is probably remiss not to mention the real definition somewhere in the fine print. If it had done so, this discussion would have been either unnecessary or much more efficient. --Bob K 12:01, 23 April 2006 (UTC)

Is it really unit?

Another question about calling it a "unit impulse function": Is it really called that? I've learned about a unit impulse function that is called unit because it jumps to 1 at time 0. Any old impulse function is a multiple of the unit impulse function. In that case, I don't think the dirac delta function is a unit impulse function - but rather an infinite impulse function. Fresheneesz 03:39, 26 April 2006 (UTC)

Also, delta function links here - but doesn't "delta function" have a more general meaning (ie. a delta function is 0 everywhere except at x=0) ?

I would say that a unit impulse function is called unit because the integral (using the appropriate measure) is 1. The integral is the only distinguishing feature of $\delta (x)\,$ and $\delta (2x)\,$. They both have infinite height. For me the term infinite impulse function implies a Dirac delta, because the concept of infinite is irrelevant to the concept of Kronecker delta. --Bob K 04:04, 26 April 2006 (UTC)

Those are the only meanings I know of for delta function. My thought is to make delta function a disambiguation page that goes to both places. --Bob K 04:04, 26 April 2006 (UTC)

I don't use the term "unit impulse function", nor am I familiar with that usage, though I do have a vague impression that it's common with engineers. I note that the Dirac delta function is an identity element under convolution. -lethe ^{talk +} 04:18, 26 April 2006 (UTC)

Neither do I, and I am an EE/DSP type. However, Google-searches for the exact phrase "unit impulse" found matches at 29 Wiki articles and 70,000 web sites. So it seems to be serving a useful purpose. I spent a little time perusing the hits and came away with the impression that most people associate it with the Dirac delta, probably because of the word "impulse". In fact the original unit impulse function article was simply a redirect to here. That was changed 14-Apr-2006 into an article about the Kronecker delta (except the name was omitted). Then on 20-Apr I moved its contents into the Kronecker delta article, and changed it into a disambiguation page, pointing back to here. But actually it seems that the original article (a simple redirect to Dirac delta) is probably the one in agreement with most of the user world. Shall we revert? --Bob K 12:32, 26 April 2006 (UTC)

If I may vote on my own question, I vote to not revert, because all DSP engineers use the term impulse response, even if they prefer delta function to unit impulse. So the word impulse is firmly entrenched in DSP-land, where it refers to the Kronecker delta. --Bob K 12:49, 26 April 2006 (UTC)

I'm not clear on what reversion you are considering. You want Dirac delta to be a disambiguation? I don't support that. I think Dirac delta should go here, and Kronecker delta should be separate. -lethe ^{talk +} 20:39, 26 April 2006 (UTC)

I agree with that. The reversion I was considering is the Unit_impulse_function article... full circle... back to its original incarnation: "unit impulse function, take 1" --Bob K 23:01, 26 April 2006 (UTC)

Oh, I see. I'm having a hard time making myself care about whether unit impulse function is a redirect or a disambig, so whatever you like is fine with me. -lethe ^{talk +} 23:12, 26 April 2006 (UTC)

If both are refered to as unit impulse functions, then I vote no revert. Fresheneesz 03:13, 27 April 2006 (UTC)

Unit impulse now redirects to Delta function which points to both Kronecker delta and Dirac delta function. --Bob K 14:56, 27 April 2006 (UTC)

It really is a unit

We engineers really do refer to the "Dirac delta" as the "unit impulse": Defined as the (normalized) product of the amplitude and the time (t-axis) in the limit as t --> 0 (thus the amplitude --> infinity). It's used in communication theory in particular to "sample a wave". Hence the comb function or "shah" function is sampling at a fixed interval of time, ad infinitum. This is what a DSP does, and when it does the process creates the opportunity for aliasing, thus the need for low-pass filtering before the sampling-device. The Fourier transform of the unit impulse is a straight line across the spectrum indicating noise of equal amplitude at all frequencies: Easy to see with a discrete Fourier transform that you can put on your Excel spreadsheet in about 15 minutes. The phrase "Kronecker delta" I have stuck in my head for some reason and went hunting in my old text books but haven't found a reference yet, but I did find references to "the Dirac delta", if anybody wants me to add them as references lemme know. What I read here seems accurate.

I added the notation to the derivative of the sigmoid function, an absolutely fascinating function. The derivative -- as pointed out in the sigmoid page -- is also a nice function plus it is a "hump" that, in the limit and multiplied by a constant, makes a perfectly-good (mathematically-continuous) impulse-function. 1/wvbaileyWvbailey 14:42, 11 June 2006 (UTC)

I am not disagreeing with your description of sampling, but I think it can bear some clarification.

• A DSP (more specifically an analog-to-digital converter) produces samples of a waveform at regular increments of time. The samples themselves are not a function of continuous time, and their continuous [time] Fourier transform is undefined.
• To use that analysis tool, a continuous-time function is contrived conceptually (not actually nor numerically) by using the samples to modulate the "teeth" of a Dirac comb function, which does have a continuous-time Fourier transform.
  • The transform of the modulated comb is related to the transform of the original waveform in a very insight-filled way, which leads to an understanding of aliasing and ways to mitigate it, such as lowpass filtering and/or increasing the sample-rate.
  • It also reveals that the sample-rate can be unnecessarily high. Undersampling, which causes aliasing, is not a reversible operation. Oversampling is inefficient/wasteful, but it is also reversible, meaning that no information is actually lost (see sampling theorem).
• The continuous-time Fourier transform of the modulated comb can be mathematically simplified (reduced) to a numerical calculation (that a computer can do) using just the original samples (without the infinite-valued delta functions).
  • That special case of the continuous-time Fourier transform is called discrete-time Fourier transform (DTFT). But it is a continuous-frequency function, which means that a computer cannot evaluate it at every frequency (because it is a continuum).
  • The discrete [frequency] Fourier transform (DFT) is a formula for computing regularly-spaced values (i.e. at discrete frequencies) of the DTFT function (which is always periodic).
  • The fast Fourier transform (FFT) is an algorithm for computing the DFT very efficiently.

--Bob K 15:48, 12 June 2006 (UTC)

A holding place for this reference: I finally found the definition of "Kronecker delta" in my perusing of Kreider, et. al. An Introduction to Linear Analysis, Addison Wesley, Reading Mass, 1966. In the following, the bold-face indicates these are vectors and the * should be a dot-product:

Eq. 7.31: xi*xj = 0 whenever i <> j

Eq. 7.32: xi*xj = 1 whenever i = j

"For economy of notation when discussing orthonormal sets, Eqs. (7.31) and (7.32) are frequently combined by writing

"xi*xj =δij, where δij = {0 if i <>j, 1 if i = j }

"The symbol introduced here is called the Kronecker delta" (Kreider p.268-269)

Thus it is the diagonal vector of 1's in all 0's otherwise. wvbaileyWvbailey

So are you suggesting that Kronecker delta is just another name for an identity matrix? --Bob K 20:57, 13 June 2006 (UTC)

Yes. The Kronecker Delta is the unit matrix. See next section below. wvbaileyWvbailey 13:38, 14 June 2006 (UTC)

I did. And it does not say that the Kronecker delta is a matrix. When the elements of a square matrix are denoted by the Kronecker delta, the matrix (not the Kronecker delta) is an identity matrix. --Bob K 16:11, 14 June 2006 (UTC)

You are entitled to your opinion. But I've re-read it and can't see your point of view. Point me to a paper-document source (not a website) that documents this and clarifies it for me and for others (i.e. a math text book, etc).wvbaileyWvbailey 16:24, 14 June 2006 (UTC)

You expect me to "prove" a negative for you? And if I can't find a book that states the Kronecker delta is not a sandbar at the mouth of the Mississippi River, are you also going to conclude that it is? --Bob K 17:03, 14 June 2006 (UTC)

My request is legitimate. I want to see your sources so I can understand what you are asserting, your point of view. Either put up or shut up and by the way, I find your tone abrasive. wvbaileyWvbailey 17:10, 14 June 2006 (UTC)

• You cited "next section below" as the source for the assertion: "The Kronecker Delta is the unit matrix". I read that source, and it does not justify the assertion. That section (written by you) is the only source I need for that observation. My only other statement is: "When the elements of a square matrix are denoted by the Kronecker delta, the matrix (not the Kronecker delta) is an identity matrix." I think we agree on the positive part of that statement. --Bob K 20:43, 14 June 2006 (UTC)

I have given you a source you can verify. Library <== operative word. I'm waiting for you quote me an alternate source that I can verify. I'm agnostic about the definition of "kronecker delta". Whatever the definition is: is. But so far you haven't provided me with an alternate source. You haven't offered an alternate definition. I'm waiting. I'm patient.wvbaileyWvbailey 22:38, 14 June 2006 (UTC)

• I do not owe you an alternate definition. I have merely observed that your source does not support your conclusion, in case you care. --Bob K 04:17, 15 June 2006 (UTC)

I on the other hand find your tone (put up or shut up) very soothing. You should become a diplomat. -lethe ^talk+ 20:25, 14 June 2006 (UTC)

You can also find that information in the Wikipedia article Kronecker delta. -lethe ^{talk +} 19:17, 13 June 2006 (UTC)

Clearly Mr. Lethe likes to intrude into little cat-fights where his opinion is not welcome.wvbaileyWvbailey 22:38, 14 June 2006 (UTC)

Yeah but the difference is: this is a verifiable source, not the rubbish that passes for references on that page. Actually there aren't any references there. This dirac delta page isn't much better with respect to references. wvbaileyWvbailey 19:32, 13 June 2006 (UTC)

Another difference is that a lot more people have easy access to Wikipedia than those who have your reference on their bookshelf. Anyhow, why all the noise? Why don't you just quietly add your reference to the Kronecker delta article and be done with it? --Bob K 20:57, 13 June 2006 (UTC)

Alternate definitions, sources

copied from above to keep the continuity: A holding place for this reference: I finally found the definition of "Kronecker delta" in my perusing of Kreider 1966, Kreider et. al. An Introduction to Linear Analysis, Addison Wesley, Reading Mass, 1966. In the following, the bold-face indicates these are vectors and the * should be a dot-product:

Eq. 7.31: xi*xj = 0 whenever i <> j

Eq. 7.32: xi*xj = 1 whenever i = j

"For economy of notation when discussing orthonormal sets, Eqs. (7.31) and (7.32) are frequently combined by writing

"xi*xj =δij, where δij = {0 if i <>j, 1 if i = j }

"The symbol introduced here is called the Kronecker delta" (Kreider p.268-269)

Thus it is the diagonal vector of 1's in all 0's otherwise. wvbaileyWvbailey

So are you suggesting that Kronecker delta is just another name for an identity matrix? --Bob K 20:57, 13 June 2006 (UTC)

That would seem to be what Kreider suggests. But there is no more to be found in his text of 773 pages of dense math. I found a reference Topper 1962 that states this explicitly:

"1.5 The unit matrix.

"We already have a matrix which corresponds in matrix algebra to the number zero in the algebra of numbers. We now need a matrix 1 to take the place of the number unity. 1 must have the property that 1A = A for every A, whenever the product on the left exists. Consider the diagonal matirx 1m of order m x m whose diagonal elements are all equal to unity.

[drawing of 1m = unit matrix]

" The elements of this matrix are usually denoted by the Kronecker delta δik, which is such that

δik = 0 (i<>k), δii = 1

"Then

[1m*A]ik = from i=1 to m Σ(δik*aik) = δik*aik = aik [* is just regular multiplication]

"so that 1m*A=A for any matrix 1m*A of order m x n. The matrix 1m is called the unit matrix of order m, and there are unit matrices of all (square) orders. [etc]"(her italics and boldface: Topper, p. 19)

Noble 1969 defines the Kronecker delta this way:

"For an orthonormal set we have

(ur, us) = δrs

where δrs is the Kronecker delta [his italics], which is unity if r = s and zero if r <>s.

" We have already met these ideas in Chapter 9 [Eigenvalues and Eigenvectors]. The main difference is that we are now talking about any abstract vector space instead of the space of n x 1 column vectors. As an example of a type that we have not met before, consider

ur = 2^1/2 sin rΠt, 0 <=t <=1, r = 1, 2, 3, ...

the inner product being that defined ... with w(t) = 1 [weight]. We have

(ur, us) = 2*integral from 0 to 1 [sin rΠt sin sΠt dt] = integral from 0 to 1 { cos (r - s)Πt - cos(r + s)Πt } dt

"It is left to the reader to show that this is unity if r = s and zero if r<>s, so that the ur form an orhtonormal set. note that the set contains and infinite number of vectors."

" We now give a theorem which generalizes results proved in Chapter 9 for n x 1 column vectors..."(p. 487)

In his chapter The impulse symbol δ(x), Bracewell 1965 defines the Kronecker delta in exactly the same way as Kreider. On page 97, problem 18:

"18 The Kronecker delta is defined by

δij = {0 if i <>j, 1 if i = j }

"show that it may be expressed as a null function of i - j as follows:

δij = δ^0(i - j)" (p. 97)

He defines a "null function" as:

"Null functions are known chiefly for having Fourier transforms which are zero, while not themselves being identically zero. By definition, f(x) is a null function if

integral from a to b (f(f(x) dx) = 0

for all a and b.... Null functions arise in connection with the one-to-one relationship between a function and its transform, a relationship defined by Lerch's theorem, which says that if two functions f(x) and g(x) have the same transform, then f(x) - g(x) is a null function.

"An example of a null function ... is

δ^0(x) = { 0 if x<>0, 1 x=0 }

"under the ordinary rules of integration, the integral of δ^0(x)is certainly zero.(Bracewell p. 82, 83)

While we're at it, Carlson 1968 defines "the unit impulse (also called Dirac impulse or delta function)"(p. 45) in terms of convolution:

"... a function with unit area satisfying

δ(u) = { 0 at u<>0, infinity at u=0) (2.42)

"More precisely it should be staated that d(u) has the property

integral from -infinity to +infinity (g(u)*δ(u - u0)du = g(u0) (2.43)

"where g(u) is any regular function continuous at u0. As a special case of (2.42) we have the more familiar relation

integral from -infinity to +infinity (δ(u)du) = integral from 0- to 0+ (δ(u)du)= 1

In words, an imulse has unit are or weight [his italics] concentrated at the point where its argument is zero and one elsewhere [his italics]. For most purposes we can also say that d(u - u0) is located at u=u0 and is zero everywhere else. Thus Ad(u-u0), an impulse of weight A, is graphically represented by ... [drawing here of an arrow of height "A"]

" Strictly speaking, impulses are not functions in the usual sense. Consequently, (2.42) and (2.43) -- or any expression containing impulses -- require a standard for interpretation. the usual convention is to replace δ(u) by a a unit-area pulse of finite amplitude and nonzero duration, the pulse shape being relatively unimportant. Operations involving δ(u) are then carried out by the finite pulse, after which we consider the limit as the duration approaches zero.

" Of the many possible shapes which become impulses in the limit, we shall have use for the gaussian, rectangular, sinc and sinc^2 pulses [his italics].... Further detailed treatment of impulses and the associated concept of generalized functions can be found in the literature, e.g., Bracewell (1965, chap. 5) and Lighthill (1958)." (Carlson, p. 45-46)

in yet another book that describes the Dirac Delta: I have found Lighthill 1962 referenced Jordan and Balmain 1950,1968. Also he references Papoulis 1962. See below.

In Jordan and Balmain 1950, 1968, section

1.06 The Dirac Delta

"... Under the name of "unit impulse" the Dirac delta has ben widely used in circuit theory to represent a verys short pulse of high amplitude...detailed discussions of the properties of the Dirac delta may be found in the books referenced at the end of the chapter [Lighthill 1962 and Popoulis 1962].

" The Dirac delta at the point x = xo is designated by δ(x - xo) and at x = 0 it is designated by δ(x). It has the property

integral from a to b [δ(x - xo)]dx = 1 if xo is in (a, b), =0 if xo is not in (a, b)

" Thus the delta behaves as if it were a very sharply peaked function of unit are; in fact, the detlat may be represented rigoursly as a vanishingly thin Gaussian function of unit area..."

" Another fundamental property may be stated as follows:

integral from a to b [f(x)δ(x - xo)]dx = f(xo) if xo is in (a, b), =0 if xo is not in (a, b)

"Under the integral operation, the delta has the property of selecting the value of the function f(x) at the point xo; thus the delta is a kind of mathematical "sampling" device or gating operation.

" The derivative of the Dirac delta [sketch in Fig. 1-12] is useful in representing charge diopolses and electric double layers [etc.]" (Jordan and Balmain, pages 18-19)

The Dirac delta function is defined in Cunningham 1965, p. 148 as the following:

"Let us consider the step function f(x0 which is defined as follows

f(x) = 1/ε, -ε/2 <= x <= ε/2

f(x) = 0, otherwise

" The area enclosed by such a function and the x-axis will clearly be unity (see Fig. 8.7) [drawing of the step function with base from -ε/2 to +ε/2 and height 1/ε]. If now we consider the limit of the function f(x) as e-->0 we obtain the so-called Dirac delta function which as the property δ(x) = 0 provided x<>0,δ(0)=infinity, together with the condition that

integral from -inf to +inf δ(x)dx = 1

"in order that the 'area under the curve' remains unity. Clearly this is not a function in the usuaal sense of the word but pyscial parallels are legion."(Cunningham, p. 148)

There is a lot more in Cunningham -- including a Theorem of 5 parts that look very much like Carlson in the sense of defining/using the delta function in convolution (p. 149 ff); most of this is directed to complex variable theory.

Cannon 1967 defines the "unit impulse" this way:

"A unit impulse-- denoted by δ(t - τ) ... is basically a mathematical device obtained by reducing the width of a real impulse until it occupies a time which is very short compared to characteristic time constants of the system being studied, but at the same time maintaining the magnitude (are) of the impulse unity. Formally, the unit impulse is defined by

δ(t - τ)= lim Δt-->0 of {a[u(t - τ) - u(t - τ + Δt)) with aΔt = 1"(Cannon p. 211)

References:

Topper, A. Mary, Matrix Theory for Electrical Engineers, George G. Harrap ,London -- Addison-Wesley Publishing CompanyReading, Reading Mass, 1962.

Cannon, Robert H, Dynamics of Physical Systems, McGraw-Hill, New York, 1967.

Noble, Ben, Applied Linear Algebra, Prentice-hall, Englewood Cliffs, NJ, 1969.

Cunningham, John, Complex Variable Methods in Science and Technology, D. Van Nostrand, London, 1965.

Bracewell, Ron, The Fourier Transform and Its Applications, McGraw-Hill, New York, 1965.

Carlson, Bruce A, Communication Systems: An Introduction to Signals and Noise in Electrical Communication, McGraw-hill, New York,

Jordan, Edward C. and Balmain, Keith G., Electromagnetic Waves and Radiating Systems: Second edition, Prentice-hall, Englewood Cliffs, NJ, 1950, 1968.

The following I have not examined, but are referenced by Jordan and Balmain, and by others:

Lighthill, M.J., Fourier Analysis and Generalized Functions, Cambridge university Press, London, 1960.

Papoulis, A., 'he Fourier Inegral and its Applications, McGraw-Hill Book Company, New York, 1962.

wvbaileyWvbailey 16:00, 14 June 2006 (UTC)

Delta function on more complicated arguments.

Although everything is written here about the properties of delta function(and their proofs) can be found in the "External links" and any standard textbook on this subject, I wasn't able to find something about the n-dimensional generalized scaling property ---> $\int _{V}f(\mathbf {r} )\,\delta (g(\mathbf {r} ))\,d^{n}r=\int _{\partial V}{\frac {f(\mathbf {r} )}{|\mathbf {\nabla } g|}}\,d^{n-1}r$

I would be grateful if someone could supply a reference about this property.

StefanosNikolaou 15:33, 9 November 2006 (UTC)

Please could someone supply a reference? —Preceding unsigned comment added by 130.226.56.2 (talk) 11:01, 18 May 2009 (UTC)

Derivative of Dirac delta function

Does the Dirac delta function have a derivative? --Abdull 14:37, 23 January 2007 (UTC)

Yes - see Scienceworld entry. This article doesn't have a section on that, but it needs one, definitely. When you become an expert, why don't you add to the article? PAR 16:21, 23 January 2007 (UTC)

In short, it has a distributional derivative Lavaka 23:30, 12 April 2007 (UTC)

The Scienceworld article has some very useful equations which would enhance the wikipedia article, eg,

$x\,\delta '(x)=-\delta (x)$

Fourier Representation justification

Since the dirac delta function is a tempered distribution, we can define it's Fourier Transform, which, as the article states, is 1. Hence, it seems OK at first to define the delta function as the Inverse Fourier Transform of 1, which is what the article states:

$\int _{-\infty }^{\infty }1\cdot e^{-i2\pi kt}\,dt=\delta (k)$

My question: why is this true? For nice functions $g$ (say, in the Schwartz Space, or in L^1), we have, where $F$ represents the Fourier Transform, and $F^{-1}$ represents the Inverse Fourier Transform, the following fundamental result

$F^{-1}(F(g))=g$

but as far as I know, this is not true for distributions. So, how to justify the first formula? Maybe the author of that formula can tell me where they got it from? --Lavaka 23:37, 12 April 2007 (UTC)

It really is a function

The Dirac delta does I believe satisfy the Wikipedia definition of a function, so the page here was edited to provide consistency. - Lee

No, the Dirac delta is not a function. If it were a true function, with values of zero except the origin, where the value would be infinite, such a thing would have a Lebesgue integral of zero, and not equal to 1, as it happens. Oleg Alexandrov (talk) 05:07, 2 May 2007 (UTC)

Well, it's a real-valued function (in fact, a bounded, linear functional) defined on a space of test functions, it's just not a function defined on the domain of those test functions. :-) Sullivan.t.j 05:54, 2 May 2007 (UTC)

Well, if you put it that way, then yeah. :) Oleg Alexandrov (talk) 06:07, 2 May 2007 (UTC)

It just occurred to me that the difference in domains might have been the source of the confusion. Maybe, maybe not. Sullivan.t.j 07:12, 2 May 2007 (UTC)

$\delta (0)=?$

It is commonly said in introductory texts that the delta function has value $\infty$ at the origin and value zero elsewhere. However this is not true. We can construct a delta sequence $\{\delta _{n}\}$ which converges pointwisely to zero at the origin simply by forcing every sequence member to have value zero at the origin. Frigoris 14:54, 1 July 2007 (UTC)

Fourier Transform

1610 Hello Sir, my name is Joni.

I want to ask: What is the Fourier Transform of δ(Хо,Уо)? Thanks.

Joni NTUST-Taiwan email address:m9602801@mail.ntust.edu.tw —Preceding unsigned comment added by 140.118.123.226 (talk) 13:48, 16 October 2007 (UTC)

what is the differance?

what is the difference between those symbole delta for functions?

δn(r) and δ(r) —Preceding unsigned comment added by Greeniq (talk • contribs) 21:30, 17 February 2008 (UTC)

$\delta _{a}(x)\,$ represents a function of x that approaches $\delta (x)\,$ in the limit as parameter $a\,$ approaches to zero. There are many such functions, and $\delta _{a}(x)\,$ is just a general notation for any of them.

--Bob K (talk) 01:32, 18 February 2008 (UTC)

so when I have function like this (δn(r)- δ(r))how can I solve such a function? the general function I have is

-Δφ(r)=(e/ε)(δn(r)- δ(r))? —Preceding unsigned comment added by Greeniq (talk • contribs) 09:18, 18 February 2008 (UTC)

It does not appear that you are referring to the article. This talk page is only for the article.

--Bob K (talk) 12:50, 18 February 2008 (UTC)

You may ask your question on Reference_desk/Science or Reference_desk/Mathematics which would be better places for your quesion. - Justin545 (talk) 03:02, 18 March 2008 (UTC)

Laplace transform

Given inverse Laplace transform of $\cos(as)$ is incorrect. This can be easily verified by substituting the sum of delta functions into Laplace transform definition. As can be seen the point $t=i*a $ does not belong to the interval $(0,\infty)$. Thus, the integral should be equal to zero. Vladimir1954 (talk) 23:38, 23 February 2008 (UTC)

UNITs

Can anybody tell me : what are the units of delta func? For example: $[\delta (\hbar \omega -E_{g})]=?$

Is its unit in "energy"?

Thanks a lot !!!

                     An Asian boy

notation

Is there really the need to use the notation $\delta _{a}(x)$ for the representations ("nascent deltas")? This looks like delta centered at a, quite popular notation. It's a bit confusing. JWroblewski (talk) 08:04, 24 June 2008 (UTC)

Egorov's theory

An outline should be added of the properties of the delta function in Egorov's theory of generalised functions. In this theory there are several deltas, all of which have the properties of the "usual" delta. Egorov's theory is important because it allows multiplications, powers, and indeed almost any functions of deltas. The theory is moreover very near to Dirac's original formulation and to the use of delta functions in physics.

See

- Yu. V. Egorov, A contribution to the theory of generalized functions, Russ. Math. Surveys (Uspekhi Mat. Nauk) 45(5), 1–49 (1990).

- Yu. V. Egorov, Generalized functions and their applications, in P. Exner and H. Neidhardt, eds., Order, Disorder and Chaos in Quantum Systems: Proceedings of a conference held at Dubna, USSR on October 17–21, 1989 (Birkhäuser Verlag, Basel, 1990).

- A. S. Demidov, Generalized Functions in Mathematical Physics: Main Ideas and Concepts (Nova Science Publishers, Huntington, USA, 2001), with an addition by Yu. V. Egorov. —Preceding unsigned comment added by 81.172.158.10 (talk) 15:45, 8 July 2008 (UTC)

Broken External Link

The link at the bottom "Integral and Series Representations of the Dirac Delta Function" goes to an inaccessable page. I get a page saying "You do not have access to this file." This link should either be fixed or removed. 165.230.20.142 (talk) 16:46, 22 October 2008 (UTC)

How the Dirac delta is a generalization of the Kronecker delta

I have a few comments about the new section. I think the title is a bit long, perhaps "Relation to Kronecker delta"? Then we invoke a lot of machinery to describe the relationship. One say more simply that and avoid discussing eigenvalues of operators.

$\sum _{k=1}^{n}\delta _{ik}v_{k}=v_{i}\qquad \int \delta (x-x_{0})f(x)\,dx=f(x_{0})$.

We are assuming various things about the operator to turn the sum into an integral, we should really point out what they are. Lastly, the notation seems a bit off. The article claims ξ' both as a complex number and a vector, but I think this could be cleared up by indexing eigenvalues and eigenvectors. Thenub314 (talk) 12:23, 23 December 2008 (UTC)

Dimensional analysis?

The phrase "In terms of dimensional analysis, this definition of δ(x) implies that δ(x) has dimensions reciprocal to those of dx." seems to be somewhat irrelevant. I'm guessing it's come from the fact that the definite integral results in a dimensionless quantity, but this is true of all definite integrals (unless units are specifically invoked). Can we safely remove this sentence? Oli Filth^{(talk|contribs)} 12:12, 4 January 2009 (UTC)

Definition of the Dirac Delta

WRONG!! THIS IS THE CORRECT DEFINITION:

$\delta (x)={\begin{cases}\ 0,&x\neq 0\\\int _{-\infty }^{\infty }\delta (x)\,dx=1.\end{cases}}$

When x=0, the Dirac function is undefined, NOT equal to infinity!!! In some cases when x=0, infinity works, but sometimes (as in the sampling theory) one finds that a value of one is helpful. —Preceding unsigned comment added by 216.207.242.34 (talk) 22:00, 4 March 2009 (UTC)

To be technically correct, both definitions are wrong because neither defines an actual function on the real number line. The definition you refer to in the article is merely a heuristic definition useful for remembering properties of the dirac distribution (but makes no sense mathematically). I think maybe your are referring to scaling of the Dirac distribution. --67.193.128.233 (talk) 01:34, 9 March 2009 (UTC)

Explanation of table removal

I have removed the section containing the table, which was restored by User:PAR with the additional implication that I was "destroying this section", which I can only imagine means removing those items from the table that were already treated in context above. I do feel that perhaps to avoid further misunderstandings, I should explain the removal of the table here.

• First of all, it is completely mathematically trivial to write down a nascent delta function. However, some at least have identifiable importance in various areas of mathematics such as partial differential equations, probability theory, and Fourier analysis. The appropriate format is to build context for the examples in the text, rather than lump them all together in a table.

• Second, the following paragraph appeared in the section under discussion, as though it communicated some deep fact about nascent delta functions:

Note: If η(ε,x) is a nascent delta function which is a probability distribution over the whole real line (i.e. is always non-negative between -∞ and +∞) then another nascent delta function η_φ(ε, x) can be built from its characteristic function as follows:

$\eta _{\varphi }(\epsilon ,x)={\frac {1}{2\pi }}~{\frac {\varphi (1/\epsilon ,x)}{\eta (1/\epsilon ,0)}}$

where

$\varphi (\epsilon ,k)=\int _{-\infty }^{\infty }\eta (\epsilon ,x)e^{-ikx}\,dx$

is the characteristic function of the nascent delta function η(ε, x). This result is related to the localization property of the continuous Fourier transform.

Now, this is really just a special case of the fact that f(x/ε)/ε is a nascent delta function for any f of total integral 1, and this already appears in the subsection on approximations to the identity.

• Third, the material on probability distributions had already been integrated elsewhere into the section.

What remained of the table was the sinc-squared function and the derivative of the sigmoid function. I have no idea in what context these appear, or indeed if they were merely constructed as examples for the naive purposes of enlarging the table, but it seems better not to have them at all in light of my first point above than to run the risk of being accused of committing "slow destruction". Sławomir Biały (talk) 03:56, 16 June 2009 (UTC)

Unless the section is completely wrong, it would be better to move it to the talk page: in this way, disagreements between different editors can be resolved faster and more constructively. This is a fairly standard course of action for chunks of articles that seem irrelevant, and serves other useful purposes: some material may turn out to be useful, after all; if later identical or similar material is readded, there is a transparent record of why it had been removed in the first place, etc. At the moment, I cannot even follow your arguments, since they refer to the non-existing context. Arcfrk (talk) 13:34, 16 June 2009 (UTC)

There is a link to the removed content in the first line of my comment. I have also included it in the box below for easier navigation. Sławomir Biały (talk) 13:41, 16 June 2009 (UTC)

Table of nascent delta functions
The following discussion has been closed. Please do not modify it.
One often imposes symmetry or positivity on the nascent delta functions. Positivity is important because, if a function has integral 1 and is non-negative (i.e., is a probability distribution), then convolving with it does not result in overshoot or undershoot, as the output is a convex combination of the input values, and thus falls between the maximum and minimum of the input function. Some nascent delta functions are: $\eta _{\epsilon }(x)={\frac {1}{\epsilon {\sqrt {\pi }}}}\mathrm {e} ^{-x^{2}/\epsilon ^{2}}$ Limit of a normal distribution $\eta _{\epsilon }(x)={\frac {1}{\pi }}{\frac {\epsilon }{\epsilon ^{2}+x^{2}}}={\frac {1}{2\pi }}\int _{-\infty }^{\infty }\mathrm {e} ^{\mathrm {i} kx-|\epsilon k|}\;dk$ Limit of a Cauchy distribution $\eta _{\epsilon }(x)={\frac {e^{-|x/\epsilon |}}{2\epsilon }}={\frac {1}{2\pi }}\int _{-\infty }^{\infty }{\frac {e^{ikx}}{1+\epsilon ^{2}k^{2}}}\,dk$ Cauchy φ (see note below) $\eta _{\epsilon }(x)={\frac {{\textrm {rect}}(x/\epsilon )}{\epsilon }}={\begin{cases}{\frac {1}{\epsilon }},&-{\frac {\epsilon }{2}}<x<{\frac {\epsilon }{2}}\\0,&{\mbox{otherwise}}\end{cases}}={\frac {1}{2\pi }}\int _{-\infty }^{\infty }{\textrm {sinc}}\left({\frac {\epsilon k}{2\pi }}\right)e^{ikx}\,dk$ Limit of a rectangular function $\eta _{\epsilon }(x)={\frac {1}{\pi x}}\sin \left({\frac {x}{\epsilon }}\right)={\frac {1}{2\pi }}\int _{-1/\epsilon }^{1/\epsilon }\cos(kx)\;dk$ Limit of the sinc function (or Fourier transform of the rectangular function; see note below) $\eta _{\epsilon }(x)=\partial _{x}{\frac {1}{1+\mathrm {e} ^{-x/\epsilon }}}=-\partial _{x}{\frac {1}{1+\mathrm {e} ^{x/\epsilon }}}$ Derivative of the sigmoid (or Fermi-Dirac) function $\eta _{\epsilon }(x)={\frac {\epsilon }{\pi x^{2}}}\sin ^{2}\left({\frac {x}{\epsilon }}\right)$ Limit of the sinc-squared function $\eta _{\epsilon }(x)={\frac {1}{\epsilon }}A_{i}\left({\frac {x}{\epsilon }}\right)$ Limit of the Airy function $\eta _{\epsilon }(x)={\frac {1}{\epsilon }}J_{1/\epsilon }\left({\frac {x+1}{\epsilon }}\right)$ Limit of a Bessel function $\eta _{\epsilon }(x)={\begin{cases}{\frac {2}{\pi \epsilon ^{2}}}{\sqrt {\epsilon ^{2}-x^{2}}},&-\epsilon <x<\epsilon \\0,&{\mbox{otherwise}}\end{cases}}$ Limit of the Wigner semicircle distribution (This nascent delta function has the advantage that, for all nonzero $a$, it has compact support and is continuous. It is not smooth, however, and thus not a mollifier.) $\eta _{\epsilon }(x)={\frac {\Psi (x/\epsilon )}{\int _{-\infty }^{\infty }\Psi (x/\epsilon )\,dx}}\qquad \Psi (x)={\begin{cases}e^{-1/(1-|x|^{2})}&{\text{ if }}|x|<1\\0&{\text{ if }}|x|\geq 1\end{cases}}$ This is a mollifier: Ψ is a bump function (smooth, compactly supported), and the nascent delta function is just scaling this and normalizing so it has integral 1. Note: If η(ε,x) is a nascent delta function which is a probability distribution over the whole real line (i.e. is always non-negative between -∞ and +∞) then another nascent delta function ηφ(ε, x) can be built from its characteristic function as follows: $\eta _{\varphi }(\epsilon ,x)={\frac {1}{2\pi }}~{\frac {\varphi (1/\epsilon ,x)}{\eta (1/\epsilon ,0)}}$ where $\varphi (\epsilon ,k)=\int _{-\infty }^{\infty }\eta (\epsilon ,x)e^{-ikx}\,dx$ is the characteristic function of the nascent delta function η(ε, x). This result is related to the localization property of the continuous Fourier transform. There are also series and integral representations of the Dirac delta function in terms of special functions, such as integrals of products of Airy functions, of Bessel functions, of Coulomb wave functions and of parabolic cylinder functions, and also series of products of orthogonal polynomials.

So, just to reiterate more succinctly for those who may have trouble following my above arguments:

• The first paragraph in the deleted section already appears elsewhere in the article in a better context.
• The second-to-last paragraph is saying something that is a trivial special case of the general construction in a very convoluted way.
• The entries of the table itself have been, over the past few days, almost entirely integrated into the text for better context.

--Sławomir Biały (talk) 15:09, 16 June 2009 (UTC)

---

Please comment here on the recent and unnecessary creation of the new article List of nascent delta functions. As indicated above, all of this material already appears in context in the article. Our own manual of style tends to favor list incorporation, rather than standalone lists. Also, it is a stupid endeavor to attempt to give a "list of nascent delta functions" that attempts to be at all comprehensive. The article already indicates the many ways in which they occur "naturally", and it is plainly obvious that any attempt to list them is futile at best, and potentially misleading at worst. 173.75.158.194 (talk) 15:30, 4 November 2009 (UTC)

I disagree with the removal of this table. I think the above remarks are all very well able for an able and knowledgeable mathematician, but a wikipedia article should be aimed at improving the understanding of those who are less able, and a list of examples (not intended to be comprehensive) is a useful adjunct. I have also found it a handy reference in the past. RQG (talk) 11:36, 29 November 2009 (UTC)

Each and every one of the examples is still in the article, just not in the form of a table. The only difference is that the examples are now organized (and that requires text) and given appropriate context. As a matter of fact, there are now more examples than there ever were in the table, because neither the plane wave decomposition, the Cauchy integral, nor the fundamental solution of the wave equation, were mentioned in the table. Anyway, physicists who need particular nascent delta functions (for whatever reason) usually get them as fundamental solutions. So a more useful organization of that sort of thing would be in an article about those. Sławomir Biały (talk) 15:01, 29 November 2009 (UTC)

The is something wrong with the following statement

In the section Dirac delta function#Composition with a function the article states "... provided that g is a continuously differentiable function with g′ nowhere zero." A few line down it then states

This distribution satisfies

$\delta (g(x))=\sum _{i}{\frac {\delta (x-x_{i})}{|g'(x_{i})|}}$

where x_i are the real roots of g(x) (which are all simple by the restriction on the derivative of g).

Now this is of course correct, but it is somewhat of an odd statement since the condition that g'(x) != 0 everywhere implies that there is at most one root to begin with. I guess that the proper condition meant was that g'(x) != 0 at any of the roots of g. Anybody care to confirm this? (TimothyRias (talk) 08:49, 25 June 2009 (UTC))

Actually, the correct condition is that |g'(x)| must be nonzero everywhere, for otherwise the composition δ(g(x)) requires much more effort to define. Taking this into account, the expression on the right-hand side should be either zero (the 'empty sum') or the singleton sum, as you have noted. Sławomir Biały (talk) 00:09, 26 June 2009 (UTC)

too many directions

Only readers who already know about can understand the article! Otherwise there is no chance…

I think that the best way is giving up the"unformal/heuristic" definitions.

Can I suggest to:

1) Introduce delta in the Lebesgue measure context

2) Then define it as a Radon measure

3) briefly explain why/how these two definitions coincide (approximate identity and so on)

4) Give the distribution point of view: as a Radon mesure, delta is a distribution of order 0.

NB: the writing $\delta (x)$ has NO MEANING. Only $\delta (\{x\})$ gets one in a measure context. Otherwise one has to write $\delta (\phi )$ or $\langle \delta ,\phi \rangle$ (some would do $\langle \phi ,\delta \rangle$ ;) )

Regards —Preceding unsigned comment added by 83.199.27.48 (talk) 20:25, 3 July 2009 (UTC)

Doesn't the article already do precisely what you think it should? Granted, the approximate identity comes later in the article. But I think that is as it should be, because there are some differences between the weak versus distributional meaning of the approximate identities. Sławomir Biały (talk) 02:41, 4 July 2009 (UTC)

Also, I disagree with your proposal to abandon the heuristic definition. This is how non-mathematicians typically define the delta function, and the article would be incomplete without this heuristic definition. Even Gel'fand and Shilov, on the first page of their seminal treatise, start with this. Also, while you may think that it is absolutely incorrect to write δ(x) under any circumstances, actually a lot of highly respected mathematicians and other scientists do indeed write it this way. See for instance the aforementioned work. Sławomir Biały (talk) 02:59, 4 July 2009 (UTC)

Who shall read the article: Gelfand or Mr Smith? That's why I think that, after an unformal overview, precise definitions should be given. Then you can introduce approximate identities and explain how the formal settings meet the first intuition. Thus, you can give the two points of view into keeping rigorous.

The article should be a readable introduction, not an abstract targeted for experts who already master unformal notations. —Preceding unsigned comment added by 83.199.27.48 (talk) 12:24, 4 July 2009 (UTC)

But as far as I can tell, the article does give precise definitions immediately after the informal overview. Also, you seem to be saying two contradictory things: one that the article is too informal, and the other that the article is so rigorous as to be completely inaccessible. For these two reasons, I don't see how one should edit based on your critique. Sławomir Biały (talk) 13:57, 4 July 2009 (UTC)

Your presentation is messy: you begin by some casual writings, then you claim that the Dirac is a measure, then a distribution without really linking the two parts: the point does not emerges.(Cf. L Schwartz-Analysis III) And why don't you mention the weak-* convergence?

$\{f_{n}\}$ is a sequence of approximate identities then tends w-* to $\delta$

(i.e for every test-function and so on)

Writings like $\int \delta (x)\phi (x)\mathrm {d} x$ are just conventions. They are not necessary and make the equalities heavier than they should be. like:

$\int _{-\infty }^{\infty }f(x)\,\delta \{dx\}=f(0)$

instead of :

$\int _{\mathbb {R} }f\mathrm {d} \delta =f(0)$

?

I'm not an expert: I only try to imagine how a BA student whould react. I bet that I'm not the only one to think this way. Maybe some people else could give their point of view… —Preceding unsigned comment added by 83.199.27.48 (talk) 14:35, 4 July 2009 (UTC) --83.199.27.48 (talk) 14:37, 4 July 2009 (UTC)

The notation $\textstyle {\int _{-\infty }^{\infty }f(x)\,\delta \{dx\}=f(0)}$ is fairly standard for the Lebesgue integral with respect to a measure (see, for instance, Feller's textbook on probability theory). It is not as popular, perhaps, as the notation $\textstyle {\int _{\mathbb {R} }f\mathrm {d} \delta }$, but this clashes with the (quite standard) notation for the Stieljes integral that is used in the same section, which is the reason I settled on the former notation instead of the latter. Sławomir Biały (talk) 14:59, 4 July 2009 (UTC)

With regard to your former point on weak-* convergence, I do mention this on several occasions in the relevant section. One does need to be careful about which weak-* topology is being used: for instance the weak-* topology on the space of measures (to which I have used the standard term vague topology) is different from the weak-* topology on the space of distributions. Finally, I do actually link the two notions in the "Definitions" section. Sławomir Biały (talk) 15:07, 4 July 2009 (UTC)

To quote from the article:

In the context of measure theory, the Dirac measure gives rise to a distribution by integration. Conversely, equation (2) defines a Daniell integral on the space of all compactly supported continuous functions φ which, by the Riesz representation theorem, can be represented as the Lebesgue integral of φ with respect to some Radon measure.

--Sławomir Biały (talk) 15:09, 4 July 2009 (UTC)

I have added a paragraph that attempts to put the link between the measure and distribution formulations up front in a more accessible manner. Sławomir Biały (talk) 15:26, 4 July 2009 (UTC)

I suggest to order the definitions like this:

The Dirac δ function is the borel measure that only loads the singleton {0}, namely: $\delta (\{0\})=1,\,\delta (Q)=0\quad ({\text{Q is a segment/closed cube which does not contains 0}})$

Fix A a Borel set. A straightforward computation now shows that $\delta$ is indeed a Borel measure, the one such that:

- if A contains 0: $\delta (A)=1{\text{ i.e }}\int {1}_{A}\mathrm {d} \delta =1$

- if not: $\delta (A)=0{\text{ i.e }}\int {1}_{A}\mathrm {d} \delta =0$

Since any Borel function f is the pointwise limit of simple functions, we reach:

$\int f\mathrm {d} \delta =f(0)$

With a 'radonian' point of view, we set: $\delta :{\mathcal {C}}_{c}(\mathbf {R} ^{n})\rightarrow \mathbf {\mathbf {R} } ,\,f\rightarrow f(0)$.

Let be $K$ a compact: the restriction of a such linear form on ${\mathcal {C}}_{K}(\mathbf {R} ^{n})$ is clearly continuous (hence, δ is a Radon measure), of norm equal to 1:

$\sup \left\{|f(0)|:\,\|f\|_{\infty }=1\right\}=1$

We now can see δ as distribution of order 0:

$\delta :{\mathcal {D}}\rightarrow \mathbf {R} ,\,\phi \rightarrow \phi (0)$ is a linear form such that:

$\forall K{\text{ compact}}:\,|\delta (\phi )|\leqslant \|\phi \|_{0}\quad (\phi \in {\mathcal {D}}_{K})$

(Recall that $\|.\|_{0}$ denotes the supremum norm in the distributions context)

Applying the definition of a distribution support, we conclude that ${\text{spt }}\delta =\{0\}$ is compact. Hence $\delta$ is tempered

Remark: If we put $T_{H}(\phi )\!:=\int H\phi \quad (\phi \in {\mathcal {D}})$ - H denotes the Heaviside function, we obtain

$T_{H}'(\phi )=-T_{H}(\phi ')=-\int H\phi '=-[\phi ]_{0}^{\infty }=\phi (0)=\delta (\phi )$

This way we have the chain: Borel->Radon/linear form->distribution.

On a second step w* convergences can be mentioned without loss of clarity.

I suggest to mention Radon-Nikodym theorems at the end, with probability issues. —Preceding unsigned comment added by 83.199.27.48 (talk) 20:22, 4 July 2009 (UTC)

While I do see a few details in the above that could perhaps be mentioned in the article (e.g., that the delta measure is a Borel measure of total variation 1, and that it is a tempered distribution), I still don't see how the article is in any way at odds with the spirit of what the article tries to accomplish. Indeed, the overall structure of measure -> linear form -> distribution is certainly there, which is what you seem to continue to suggest is missing. Perhaps another set of eyes could be helpful here. Sławomir Biały (talk) 20:39, 4 July 2009 (UTC)

indeed it would be —Preceding unsigned comment added by 83.199.27.48 (talk) 21:22, 4 July 2009 (UTC)

Continuous with compact support?

Why is the the integral of a function f with respect to the Dirac measure at x not defined for all measurable f? At the moment, the text says this holds for continuous f with compact support, which seems unnecessarily restrictive. — Preceding unsigned comment added by 131.111.185.68 (talk) 09:58, 6 November 2013 (UTC)

The (admittedly somewhat unsatisfactory) answer is that this is the smallest class of functions needed to define delta as a Radon measure. But actually, since all subsets of $\mathbb {R}$ are measurable with respect to the measure, every function is measurable with respect to the measure, and the integral of f is always just f(0). (I think this might be confusing to some readers, since often in analysis functions are identified if they differ only on a set of Lebesgue measure zero, but since delta is not absolutely continuous with respect to the Lebesgue measure, such an identification can no longer be enforced. For instance, the integral with respect to delta of an f in L^1(R) is not well-defined.) Sławomir Biały (talk) 13:09, 6 November 2013 (UTC)

Hyperfunction

I'm not following the discussion closely, so forgive me for any possible misunderstanding. But, at least to me, the article currently looks perfectly fine (organizationally speaking). The notions somehow differ from what I'm used to, but that's really not an issue. Also, I know of no mainstream definitions Dirac delta other than one by measures or distributions (maybe tempered distributions to be more precise?). I know there is the definition of Dirac in terms of Sato's hyperfunction, but this I don't think is standard. Personally, I think seeing generalized functions as limits of some nicely behaving functions is very problematic. I know Yoshida defines Sobolev spaces in this way; as the completion of some space of test functions. This doesn't work quite well in practice because there are too many choices of convergence and test functions. (Yeah, this reminded me: the article probably should relate the theory of Sobolev spaces to Dirac delta; not sure how, though. See below) I also note that the article probably spend too much spaces just for the definitions of delta, which isn't really an interesting topic after all. -- Taku (talk) 00:49, 6 July 2009 (UTC)

I didn't carefully read the article, so maybe I just missed, but there are a couple of things I think it should mention:

• Poisson summation formula in terms of delta
• How the Cauchy integral formula relates to delta
• The structure theorem for distributions, which is formulated in terms of derivatives of delta; in particular, a distribution with finite support is completely characterized by
• Delta in the representation theory: when you're dealing with group algebra, convolutions (not necessarily commutative?) involve delta.
• A better exposition of the relationship to fundamental solution; that is, why and how solving Pu = f is equivalent to solving it when f is delta
• A spherical version of the representation of delta
• $\delta (x_{1},x_{2},...x_{n})=\delta (x_{1})\delta (x_{2})...\delta (x_{n})$ (Don't know how to put this rigorously.)
• Plane-wave decomposition of the delta:

$\delta (x)=\int _{S^{n-1}}{\omega (\xi ) \over \langle x,\xi \rangle }$

(where $\omega$ is a volume form of the sphere) or something like it:) It's probably due to F. John.)

-- Taku (talk) 01:25, 6 July 2009 (UTC)

I agree with most of the points you make. Actually the biggest serious omission is a proper treatment of delta in more than one variable. This problem is endemic to most of the articles here on Fourier analysis. The article Fourier transform, for instance, was clearly written with the case of one variable primarily in mind and higher dimensions only secondarily so. (I have already attempted to change this somewhat, as have other mathematics editors.)

Fundamental solutions, are already mentioned. But, since this was what emerged out of the current article structure, it is in connection with "nascent" delta functions, which is a somewhat artificial context. Perhaps the next big task will be to fork material out into a more natural section on this topic. Sławomir Biały (talk) 02:52, 6 July 2009 (UTC)

The connection to hyperfunctions, mentioned above, could be made more explicit based on the article linked at the bottom of the page. Tkuvho (talk) 15:41, 18 July 2010 (UTC)

Laugwitz 1989?

Someone seems very keen on including a reference to Laugwitz (1989) in the "overview" section. To me, this is totally out of place. For one thing, the text

Cauchy's formula for a unit-impulse, infinitely tall, infinitely narrow delta function defined in terms of infinitesimals is reproduced by D. Laugwitz (1989), p. 230.

Doesn't fit with the surrounding paragraphs, which mentions no such formula at all. Secondly, here we are discussing Cauchy, Kirchoff, Dirac, and now Laugwitz?! The name-drop is totally jarring. I also do not think this adds anything in the way of meaningful content that is helpful for an "Overview" section. I am willing to hear reasons for the inclusion of this, but I will remove it again if I am not convinced. Please refer to WP:BRD: the editor who added this was WP:BOLD, but then reverted. Now the onus is on him or her to discuss the merits of inclusion. Sławomir Biały (talk) 12:02, 4 February 2010 (UTC)

We are not discussing Laugwitz. We are discussing Cauchy. The first paragraph in this section states that Cauchy considered sequence of taller and taller unit-impulse functions. I would like to see a source for that. In reality, Cauchy wrote down formulas for unit-impulse delta function explicitly defined in terms of infinitesimals. Note that the book you mentioned is a scientific, not a historical, monograph. The article I referred to is published in a leading historical journal. It is a more reliable source on Cauchy. Tkuvho (talk) 12:11, 4 February 2010 (UTC)

I fail to see how being a scientific rather than exclusively historical monograph disqualifies something as a source for the history of a subject. It is the only monograph (that I am aware of) that gives a detailed historical account of the Dirac delta function. It is much more detailed than the source that you are keen on citing, because it includes not only Cauchy's approach, but Poisson's, Kirchoff's, Helmholtz's, etc. Anyway, I have now included the reference you are keen on putting there in a manner that does not interrupt the flow of the text. Hopefully this compromise will be suitable. Sławomir Biały (talk) 12:23, 4 February 2010 (UTC)

You might want to consult L"utzen, J.: The prehistory of the theory of distributions. Studies in the History of Mathematics and Physical Sciences 7 Springer-Verlag, New York-Berlin, 1982. The idea that a book (van der Pol and Bremmer) written half a century ago is the last word on the subject is a little far-fetched. Unfortunately, my library does not have it. Could you tell me what they say about Cauchy exactly? As I mention below, the current version is unacceptable, as it misrepresents both Cauchy and Laugwitz. Tkuvho (talk) 12:28, 4 February 2010 (UTC)

Thank you, I will do that. Most of the van der Pol and Bremmer text is available through google books, if your library lacks it. Sławomir Biały (talk) 12:37, 4 February 2010 (UTC)

I will try to look it up. I would be very surprised if it is as detailed on Cauchy as Laugwitz's monumental 50-page text, which as I mentioned is published in a leading historical periodical. The page should contain a reference to Cauchy's infinitesimal delta function, not an ahistorical fantasy about his ideas about sequences of functions. Tkuvho (talk) 12:42, 4 February 2010 (UTC)

I see you have latched on to the word "sequences" as objectionable. Perhaps this could be changed to something more satisfactory. Sławomir Biały (talk) 12:52, 4 February 2010 (UTC)

How about mentioning the infinitesimal formula? That would have the advantage of being both more concise and more accurate. Tkuvho (talk) 13:06, 4 February 2010 (UTC)

revert by Bialy

The following material was reverted twice by Bialy without awaiting outcome of discussion on this page:

Cauchy's formula for a unit-impulse, infinitely tall, infinitely narrow delta function defined in terms of infinitesimals is reproduced by D. Laugwitz (1989), p. 230. Tkuvho (talk) 12:17, 4 February 2010 (UTC)

The current version states that "The idea of using a sequence of functions to approximate a unit impulse goes back to the early 19th century, and was considered by Augustin Louis Cauchy", with reference to Laugwitz. This is incorrect on the counts. Laugwitz did not say Cauchy's idea was to use a sequence of functions. Instead, he reproduces a formula (not an idea) giving an infinitesimal delta function in Cauchy. Do you have a source for the current claim about Cauchy, with a citation in Cauchy? Tkuvho (talk) 12:22, 4 February 2010 (UTC)

I reverted only once. The second time, I moved the reference to against Cauchy, per your insistence in the above thread, and cut extraneous details that were irrelevant to the purpose of an "Overview" section. Sławomir Biały (talk) 12:27, 4 February 2010 (UTC)

They are not extraneous at all. On a page dealing with the dirac delta function, it is not extraneous to note that Cauchy defined it in 1827. Tkuvho (talk) 12:29, 4 February 2010 (UTC)

Interesting comment, given that no version under discussion even mentions this. Rather the version under discussion appears to give credit to Laugwitz for reproducing Cauchy's formula (which the article does not actually state) in 1989, which seems like fairly extraneous information to me. Sławomir Biały (talk) 12:40, 4 February 2010 (UTC)

I re-read the current version, and it still states that Cauchy talked about sequences of functions, with a reference to Laugwitz. Cauchy did not talk about a sequence but rather a single infinitesimal delta function, and Laugwitz was clear on this point, since he reproduces Cauchy's explicit definition in terms of infinitesimals. Tkuvho (talk) 12:45, 4 February 2010 (UTC)

Looking at the section of Laugwitz on Cauchy, the delta function appears in his derivation of the Fourier inversion formula as a limit of regularized integrals. This does seem to be consistent with the text as written. Sławomir Biały (talk) 13:05, 4 February 2010 (UTC)

Which formula are you looking at? Laugwitz mentions that Cauchy states explicitly that alpha and epsilon are infinitesimals, and writes down a single formula, not a sequence. Here epsilon is irrelevant as Laugwitz mentions later in the article. The height of Cauchy's delta function is 1/alpha. Tkuvho (talk) 13:08, 4 February 2010 (UTC)

Cauchy before hyperreals

(moved from User talk:Sławomir Biały)

Hi, Your historical observation at Dirac delta function is perfectly correct: Cauchy came before the hyperreals. As documented at that page, Cauchy defined a Dirac-type function in terms of an infinitesimal, and a hyperreal definition thereof similarly uses infinitesimals. Thus it is inaccurate to say that a Dirac delta function cannot be defined as a true function. You don't have to be an adherent of the hyperreals to recognize an inaccuracy. Tkuvho (talk) 01:38, 9 July 2010 (UTC)

If this is an inaccuracy, it is one that is in nearly every reliable source. Anyway, I see nothing wrong with adding a separate referenced section on hyperreals, but cryptic parenthetical remarks about them in the midst of intuitive discussions are not helpful. Sławomir Biały (talk) 11:49, 9 July 2010 (UTC)

OK, thanks. Perhaps the solution is to add a brief section toward the end containing an intuitive discussion of a unit impulse, infinitely narrow, infinitely tall Dirac delta function; mention Cauchy's use of such functions in 1827; present an example in the context of a modern theory of an infinitesimal-enriched continuum; and give a bibiliography discussing infinitesimal Dirac delta fuctions starting from the 1960s. Tkuvho (talk) 13:46, 9 July 2010 (UTC)

This sounds like a good idea, except I'd recommend some caution in how Cauchy's exact conception is addressed. Cauchy explicitly stated that for him "infinitely small" meant "tending to zero". So one widely-held view (for instance, Boyer's History of the Calculus) is that Cauchy did not use literal infinitesimals (as he equally rejected "actual infinities"). Laugwitz notes both of these facets early on in his previously cited paper, but then he appears to reject this historiographical viewpoint. Anyway, it is evidently not clear that Cauchy's conception of infinitesimals squares with the modern one introduced by Robinson et al. Certainly Robinson and presumably other infinitesimal analysts have no qualms in attributing infinitesimal results to Cauchy, but this seems to rely on a literal reading of "infinitely small" that is not directly supported by Cauchy's own conception of a limit. Anyway, with appropriate care to source the history observing WP:NPOV, I see no problem in what you propose. Sławomir Biały (talk) 13:43, 10 July 2010 (UTC)

You are right, some of these issues need to be clarified before the page is modified. What, in your opinion, is Cauchy's definition of an infinitesimal? And what sources is your opinion of Robinson based upon? Tkuvho (talk) 21:29, 10 July 2010 (UTC)

In the Cours d'Analyse, Cauchy states that a variable is infinitely small if the magnitude of the variable is taken to decrease to zero. So it is not essentially different from other real-valued variables, except in the understanding that it tends to zero. This interpretation is supported by Boyer's analysis in The History of the Calculus p. 278. By opinion of Robinson is, I admit, based on a rather cursory skim of part of the previously referenced paper by Laugwitz, who in turn refers to Robinson's Nonstandard analysis text. But Robinson's text does not appear to verify Laugwitz's point, at least not straightforwardly. Rather Robinson gives what I would consider to be a fairly correct account of Cauchy's view of infinitesimals, but then explicitly reinterprets Cauchy's writings in a nonstandard setting. So I was, I think, mistaken in my original post. Sławomir Biały (talk) 14:09, 11 July 2010 (UTC)

Good, I am glad we agree on Robinson. Maybe we can try to agree on Cauchy, because Boyer's is one of many, many interpretations, differing from ones that preceded him as well as ones that followed him. So what did Cauchy say exactly? He said that a variable tending to zero "becomes" an infinitesimal. Is that a fair assessment? A recent study by K. Brating in Archive for history of exact sciences paraphrases Cauchy's definition by saying that a Cauchy infinitesimal is generated by a null sequence. Is that a fair assessment? Tkuvho (talk) 14:23, 11 July 2010 (UTC)

I think "becomes infinitely small" is the exact translation. But it's a stretch for me to read this ontologically and conclude that Cauchy thought of an infinitesimal as being some definite quantity apart from the limiting process (which "generates an infinitesimal" seems to suggest). There does appear to be consensus in the literature at least on the point that for Cauchy an infinitely small quantity was an ordinary real variable that is regarded as tending to zero. What there seems to be confusion about is the ontology: is there an "actual infinitesimal" that underlies this limiting process, or does infinitely small simply mean the manner in which the variable is treated? Lakatos' 1978 "Cauchy and the continuum" seems to have a very interesting and sophisticated take on this dilemma. He breaks down two historical trends into viewing Cauchy as either an "inarticulate Weierstrass" or an "inarticulate Robinson". I'm curious what, if any, the relevance of Lakatos' view is to our discussion. You are probably already familiar with this paper, and in a better position to answer than I. Sławomir Biały (talk) 15:26, 11 July 2010 (UTC)

I am not sure which phrase you are translating as "becomes infinitely small". The term "infinitely small" is Cauchy's exclusive term for infinitesimals. He never uses the term "infinitesimal" itself, even though it does have a French equivalent. Other authors have used it, such as L. Carnot, but not Cauchy. My reading of Cauchy is that in 1821 he defined them in terms of variable tending to zero, where the variable is understood to be a discrete variable. Thus, he gives an example which is a slight perturbation of the sequence 1, 1/2, 1/3, 1/4, etc. Note that Carnot defined infinitesimals the same way. Meanwhile, in 1823 Cauchy gave a definition in terms of a variable tending to zero, where the implied variable varies continuously. So your remark is correct with regard to the 1823 definition. Incidentally, Lakatos's 1978 text was written in the 60s (published inedited form by Cleave in 1978). Lakatos changed his mind a few times about Cauchy's infinitesimals. I personally find Laugwitz's 1989 text far more scholarly and reliable. My immediate reaction to "gut W" or "gut R" is scepticism. Why can't we take Cauchy and his definitions at face value? I am interested in the fact that Cauchy never says that a null sequence IS an infinitesimal, but only BECOMES infinitesimal. Tkuvho (talk) 16:08, 11 July 2010 (UTC)

At face value, as you say, Cauchy doesn't use the term infinitesimal, merely the term "becomes infinitely small" as a way of saying "variable tending to zero"—whether the limit is discrete or continuous doesn't seem to be especially a concern for Cauchy. Sławomir Biały (talk) 16:53, 11 July 2010 (UTC)

Again, I am not aware of Cauchy using the expression "becoming infinitely small" anywhere. Are you? Cauchy uses the expression "an infinitely small quantity" in exactly the same sense as Carnot when he uses the expression "infinitesimal", since Cauchy and Carnot give an identical definition of the respective term. Tkuvho (talk) 17:12, 11 July 2010 (UTC)

I naturally assumed that you were referring above (with the first use of "becomes" in the thread) to Cauchy's first use of the term infiniment petite in the 1821 Cours: "On dit qu'un quantite variable devient infiniment petite, lorsque sa valuer numerique decroit indefiniment de manier a converger vers la limite zero." Sławomir Biały (talk) 17:56, 11 July 2010 (UTC)

I see what you mean. What I was pointing out is that Cauchy is using "infiniment petite" here as shorthand for "quantite infiniment petite", in other words as a noun. I have the impression that most commentators agree that Cauchy is not envisioning some hitherto unmentioned "infinitesimal values" that the variable takes on its way to zero. Rather, the sequence (i.e. variable) itself is/becomes an infinitesimal. The meaning of "becoming" here is a fascinating subject, particularly since Cauchy obviously did not have access to the set-theoretic framework we take for granted today. This may not be relevant for the purposes of wiki pages, though. If we agree about the meaning of Cauchy's definition of infinitesimal, perhaps we can use it as a basis for what to write at this page. Tkuvho (talk) 18:17, 11 July 2010 (UTC)

Thanks, that clarifies things for me. I think perhaps our views of Cauchy are not so radically different as I had once thought. Sławomir Biały (talk) 00:57, 12 July 2010 (UTC)

OK then, we can mention that Cauchy used an infinitesimal $\alpha$ to write down a unit impulse, infinitely tall and narrow Dirac-type delta function $\delta _{\alpha }$ satisfying $\int F\delta _{\alpha }=F(0)$ in a number of articles in 1827. Cauchy defined an infinitesimal in 1821 in terms of a sequence tending to zero. Namely, such a null sequence becomes an infinitesimal in Cauchy's and Carnot's terminology. We can mention that modern set-theoretic approaches allow one to define infinitesimals via the ultrapower construction where a null sequence becomes an infinitesimal in the sense of an equivalence class modulo a suitable ultrafilter, and perhaps mention an article or two giving a bibliography on modern Dirac delta functions in the context of an infinitesimal-enriched continuum. Tkuvho (talk) 01:49, 12 July 2010 (UTC)

Well, let's see how it looks. Sławomir Biały (talk) 03:49, 12 July 2010 (UTC)

I was thinking of pretty much copying the material as it appears above, some like this:

Infinitesimal delta functions

Cauchy used an infinitesimal $\alpha$ to write down a unit impulse, infinitely tall and narrow Dirac-type delta function $\delta _{\alpha }$ satisfying $\int F(x)\delta _{\alpha }(x)=F(0)$ in a number of articles in 1827, see Laugwitz (1989). Cauchy defined an infinitesimal in 1821 (Cours d'Analyse) in terms of a sequence tending to zero. Namely, such a null sequence becomes an infinitesimal in Cauchy's and Carnot's terminology.

Modern set-theoretic approaches allow one to define infinitesimals via the ultrapower construction, where a null sequence becomes an infinitesimal in the sense of an equivalence class modulo a relation defined in terms of a suitable ultrafilter. The article by ... contains a bibliography on modern Dirac delta functions in the context of an infinitesimal-enriched continuum such as the hyperreals. Tkuvho (talk) 04:36, 12 July 2010 (UTC)

Two obvious problems

I saw that the article was nominated for a good article. I haven't read the entire thing, so I can't really give a proper review... but I notice two obvious problems that should be fairly easy to sort out before someone does review it properly.

• The sentence "Informally, it is a generalized function representing an infinitely sharp peak bounding unit area: a 'function' δ(x) that has the value zero everywhere except at x = 0 where its value is infinitely large in such a way that its total integral is 1." is overly long and contains too many redundant words. I know sometimes you need to explain what you have just said, but the redundancy doesn't seem to achieve that. How about "Informally, it is a generalized function that has the value zero everywhere except at x = 0 where its value is infinitely large in such a way that its total integral is 1."?
• The equations are numbered in an odd way. How about numbering them in numerical order, starting with 1? I don't really know how this aspect of wiki code works, but I think it would be a search and replace job to make everything right.

Yaris678 (talk) 17:37, 11 July 2010 (UTC)

I have changed the lead sentence in line with my first point. Yaris678 (talk) 11:11, 13 July 2010 (UTC)

I notice that Sławomir Biały has fixed the equation numbering with this change. Thanks. Yaris678 (talk) 11:56, 13 October 2010 (UTC)

Representations of the delta function

This section seems a bit weak to me. It should be connected with expressions involving sums of eigenfunctions, for example:

$\delta (x-\xi )=2\sum _{1}^{\infty }\sin n\pi x\ \sin n\pi \xi \ .$

$\delta (x-\xi )={\frac {2}{\pi }}\int _{0}^{\infty }\ \sin kx\sin k\xi \ dk\ .$

These are generated from the completeness relations for various sets of functions, for example eigenfunctions, and also are closely related to Green's functions:

$\delta (x-\xi )=-{\frac {1}{2\pi }}\oint G_{\lambda }(x,\ \xi )\ d\lambda \ .$ Brews ohare (talk) 15:44, 12 July 2010 (UTC)

For an inadequate discussion see Resolution of identity and this. A very particular example:

$\delta (x-x')=\oint _{C}\psi _{+}(x,\ k)\psi _{-}(x',\ k)\ dk\ ,$

is found here. A basic intro is here and here. Brews ohare (talk) 16:13, 12 July 2010 (UTC)

I added a section on eigenfunction representations, but have not dealt with Green's functions. Brews ohare (talk) 15:08, 13 July 2010 (UTC)

Nice. I've tried to make it a little less formal. Green's functions and fundamental solutions should be treated together, and I agree this is a major omission. However, for what it's worth, I think an explicitly "spectral" approach to them may be unwarranted. There already is some scattered discussion of fundamental solutions of evolution equations, but there needs to be some discussion to tie this together. Sławomir Biały (talk) 15:50, 13 July 2010 (UTC)

You've tightened this up nicely. Can you add some explanation or sources or links to guide the reader to "converges in the distributional sense"? Brews ohare (talk) 16:08, 13 July 2010 (UTC)

Identity removed

I have reorganized the material to what seems to be a more logical order (e.g., discussing the distribution derivative of the delta function in the section that treats derivatives). In the process, I removed the identity

${\frac {d}{dx}}\log x={\frac {1}{x}}+i\pi \delta (x).$

As stated, it doesn't make sense because 1/x is ambiguous. I had initially changed this to a principal value, but I think that's wrong. Rather it seems to be getting at Sokhostky's formula, which I will add a short section on. Sławomir Biały (talk) 16:33, 13 July 2010 (UTC)

McMahon reference

Reference 29 to McMahon appears to be misplaced as it is simply a general intro to the delta function and doesn't refer to the rectangular approximation nor to statistical usage. Brews ohare (talk) 13:28, 14 July 2010 (UTC)

I replaced this reference with this one. Brews ohare (talk) 14:46, 14 July 2010 (UTC)

Inner products with a weight function

Here is a possible re-write of a portion of the "Resolutions of the identity" section:

...the expression for f can be rewritten as:

$f(x)=\int _{D}\,d\xi \ w(\xi )\ \left(\sum _{n=1}^{\infty }\ \varphi _{n}(x)\varphi _{n}^{*}(\xi )\right)f(\xi ).$

where provision is made for a weight function w(x) in the inner product. The right-hand side converges to f in the L² sense. It need not hold in a pointwise sense, even when f is a continuous function. Multiplying both sides by the weight function:

$w(x)f(x)=\int _{D}\,d\xi \ w(\xi )f(\xi )\ \left(w(x)\sum _{n=1}^{\infty }\ \varphi _{n}(x)\varphi _{n}^{*}(\xi )\right).$

It is common to abuse notation and write

$w(x)f(x)=\int \ d\xi \ w(\xi )f(\xi )\ \delta (x-\xi ),$

resulting in the representation of the delta function:

$\delta (x-\xi )=w(x)\sum _{n=1}^{\infty }\ \varphi _{n}(x)\varphi _{n}^{*}(\xi ).$

Although this looks superficially OK, the symmetry in x and ξ is lost, with the result that integral of the delta function over x is 1, but the integral over ξ is not 1.

Can this be fixed up and inclusion of a weight function made part of the article? Brews ohare (talk) 16:28, 14 July 2010 (UTC)

Should the first formula be

$w(x)f(x)=\sum _{n=1}^{\infty }\int _{D}\,d\xi \ w(\xi )\ \left(\ \varphi _{n}(x)\varphi _{n}^{*}(\xi )\right)f(\xi )?$

By the way I think the sum should go outside the integral- it doesn't converge except in a weak sense. Then the last formula would be

$\delta (x-\xi )={\frac {w(\xi )}{w(x)}}\sum _{n=1}^{\infty }\ \varphi _{n}(x)\varphi _{n}^{*}(\xi ).$

Seems to me that the initial formula you wrote would only work if $\varphi _{n}$ was an o.n. basis with respect to the weighted inner product. Do you have a source? Holmansf (talk) 17:11, 14 July 2010 (UTC)

Holmansf: The source Davis & Thompson is typical. As you point out, the discussion is limited to the case:

$\int _{D}\ d\xi \ w(\xi )\ \phi _{n}(\xi )\phi _{m}^{*}(\xi )=\delta _{nm}\ ,$

with {$\phi _{n}$} a complete set. That established,

$f(x)=\sum _{n=1}^{\infty }\ \phi _{n}(x)\int _{D}\ d\xi w(\xi )\phi _{n}^{*}(\xi )f(\xi )\ .$

Thus, one finds:

$w(x)f(x)=\sum _{n=1}^{\infty }\ w(x)\ \phi _{n}(x)\int _{D}\ d\xi w(\xi )\phi _{n}^{*}(\xi )f(\xi )\ .$

I'd guess that in most cases of interest the sum and the integration can be interchanged. I do not know the exact technical requirements for that to be OK, but it is done all the time in physics texts without comment. Assuming some cases exist where interchange is OK, one finds:

$h(x)=\int _{D}\ d\xi \ \sum _{n=1}^{\infty }\ \ w(x)\phi _{n}(x)\phi _{n}^{*}(\xi )h(\xi )\ ,$

with h(x) = w(x)f(x), suggesting:

$\delta (x-\xi )=\sum _{n=1}^{\infty }\ \ w(x)\phi _{n}(x)\phi _{n}^{*}(\xi )\ .$

The source does not make this identification as they treat the case where w(x) = 1 in talking about the delta function, and do not use the delta function when w(x) ≠ 1. As you can see, this result cannot be true because of the asymmetry in the two arguments, as also is true of your suggested alternative.

Maybe the correct approach is to say the delta function is like any other function so that:

$(f,\delta (x-x_{0}))=\int _{D}\ d\xi \ w(\xi )\ f(\xi )\delta (\xi -x_{0})\ .$

With the δ-function:

$\delta (\xi -x_{0})=\sum _{n=1}^{\infty }\ \ \phi _{n}(x_{0})\phi _{n}^{*}(\xi )\ ,$

this produces:

$(f,\delta (x-x_{0}))=\sum _{n=1}^{\infty }\ \ \phi _{n}(x_{0})c_{n}=f(x_{0})\ ,$

as required.

Any suggestions? Brews ohare (talk) 19:44, 14 July 2010 (UTC)

Here is a sample re-write of this section. Please comment. Brews ohare (talk) 20:16, 14 July 2010 (UTC)

The formula I wrote is correct despite your reservations about asymmetry. It's actually easy to see it's correct since if you take any smooth function $F(x,\xi )$ such that $F(x,x)=1$ for all x, then $F\delta =\delta$. $F(x,\xi )=w(\xi )/w(x)$ is such a function.

Some of the formulas you have in your rewrite are not quite correct. Also, you should use some notation like $L_{w}^{2}(D)$ to indicate you are in a weighted space. Personally, I'd leave the section as is, and then add a comment about weighted spaces at the end.

Also, I'd make the example much shorter. You could put in the final formula giving the delta as a sum, but I think the details you have should go in Fourier-Bessel series.

The sum should go outside the integral. Otherwise it's just a formal notation and not an actual integral. Holmansf (talk) 21:36, 14 July 2010 (UTC)

I lean in the direction of doing all this using the Dirac measure, although I am as yet unclear just how to travel that path. This may help as may this. Brews ohare (talk) 01:15, 15 July 2010 (UTC)

I always think of the delta function as a distribution, and am not really sure of the advantage of regarding it as a measure. I am moving the sum outside the integral in accordance with my comments above. Holmansf (talk) 02:16, 15 July 2010 (UTC)

GA Review

This review is transcluded from Talk:Dirac delta function/GA1. The edit link for this section can be used to add comments to the review.

Reviewer: tb240904 (talk) 01:53, 26 July 2010 (UTC)

Criteria taken from Wikipedia:Good_article_criteria

1. Well-written

(a) the prose is clear and the spelling and grammar are correct

no problems found

(b) it complies with the manual of style guidelines for lead sections, layout, words to watch, fiction, and list incorporation

Overall:

2. Factually accurate and verifiable

(a) it provides references to all sources of information in the section(s) dedicated to the attribution of these sources according to the guide to layout

(b) it provides in-line citations from reliable sources for direct quotations, statistics, published opinion, counter-intuitive or controversial statements that are challenged or likely to be challenged, and contentious material relating to living persons-science-based articles should follow the scientific citation guidelines

Could do with some more inline citations

(c) it contains no original research

Overall:

3. Broad in its coverage

(a) it addresses the main aspects of the topic

(b) it stays focused on the topic without going into unnecessary detail (see summary style)

Goes into detail but I can't see anything that's obviously unnecessary.

Overall:

4. Neutral

It represents viewpoints fairly and without bias.

no problems

Overall:

5. Stable

It does not change significantly from day to day because of an ongoing edit war or content dispute.

several editors working on this with no edit warring

Overall:

6. Illustrated, if possible, by images

(a) images are tagged with their copyright status, and valid fair use rationales are provided for non-free content

File:Dirac_distribution_PDF.svg - CC Attribution Share Alike

File:Dirac_function_approximation.gif - Public domain, released by author

(b) images are relevant to the topic, and have suitable captions

File:Dirac_distribution_PDF.svg - relevant, with caption.

File:Dirac_function_approximation.gif - relevant, with caption.

Overall:

7. Conclusion

This article may not meet the following points of the good article criteria:

• 2(b)

Overall: Article is on hold Overall: Article failed 01 October 2010

--tb240904 ^{Talk Contribs} 17:23, 1 October 2010 (UTC)

8. Notes for the Editor

• More inline citations

• Separate notes and citations

  This appears to have been already done. Sławomir Biały (talk) 20:19, 24 August 2010 (UTC)

  The notes section expands on the information given in the article (i.e. notes) and gives references for individual material (i.e. references). There is then a reference section which gives details on the books referenced. --tb240904 ^{Talk Contribs} 17:32, 1 October 2010 (UTC)

Reviewer: tb240904 ^{Talk Contribs} 01:53, 26 July 2010 (UTC)

What's the status on the review? No updates on this page in nearly a month. Wizardman _{Operation Big Bear} 04:57, 18 August 2010 (UTC)

I have been out of town, and apparently the only one around with an interest in getting this GA to go through. Give me a week or so, and I will bring it into shape. Sławomir Biały (talk) 21:07, 24 August 2010 (UTC)

This review has been going on for far too long. All editors involved should make every effort to close this review as soon as possible. Also, keep in mind that we have special citation guidelines for math-related articles. Edge3 (talk) 15:14, 9 September 2010 (UTC)

Agreed, it should be closed one way or the other within the next few days. Wizardman _{Operation Big Bear} 17:53, 13 September 2010 (UTC)

I have left another note on the reviewer's talk page. Jezhotwells (talk) 19:48, 27 September 2010 (UTC)

OK, as the reviewer is not responding and as I personally don't feel capable of reviewing this. I am going to close it as a failed nomination now. Jezhotwells (talk) 01:59, 29 September 2010 (UTC)

I have renominated, with a timestamp one hour later to allow this keep its place in the queue. Jezhotwells (talk) 02:27, 29 September 2010 (UTC)

I appologise for not responding but my home computer is currently unable to connect to the internet and I was unable to access my user talk page due to my school's firewall. I only had concerns about the references/notes sections but I see the article has been reviewed by another editor and passed. --tb240904 ^{Talk Contribs} 17:32, 1 October 2010 (UTC)

Application to quantum mechanics

Application to quantum mechanics

==Application to quantum mechanics==

We give an example of how the delta function is expedient in quantum mechanics. A set {φ_n} of orthonormal wave functions is complete in the space L² of square-integrable functions if any wave function ψ can be expressed as a combination of the φ_n:

$\psi =\sum c_{n}\phi _{n},$

with $c_{n}=\langle \phi _{n}|\psi \rangle$. Complete orthonormal systems of wave functions appear naturally as the eigenfunctions of the Hamiltonian (for bound states) in quantum mechanics that measures the energy levels, which are called the eigenvalues. The set of eigenvalues, in this case, is known as the spectrum of the Hamiltonian. In bra-ket notation, as above, this equality implies that

$\delta =\sum |\phi _{n}\rangle \langle \phi _{n}|.$

For other observables in quantum mechanics, the set of eigenfunctions may be continuous rather than discrete. An example is the position observable, Qψ(x) = xψ(x). The spectrum of the position observable (in one dimension) is the entire real line, and is called a continuous spectrum. However, unlike the Hamiltonian, the position operator lacks proper eigenfunctions. The conventional way to overcome this shortcoming is to widen the class of available functions by allowing distributions as well: that is, to replace the Hilbert space of quantum mechanics by an appropriate rigged Hilbert space. In this context, the position operator has a complete set of eigen-distributions, labeled by the points y of the real line, given by

$\phi _{y}(x)=\delta (x-y).$

The eigenfunctions of position are denoted by $\phi _{y}=|y\rangle$ in Dirac notation, and are known as position eigenstates. Although they are not physical particle states, they are fundamental in quantum mechanics.

Similar considerations apply to the eigenstates of the momentum operator, or indeed any other observable P with a complete set of eigen-distributions. What this means is that there is a set {φ_y} of functions, labeled by y in some subset Ω of the complex plane such that

$P\phi _{y}=y\phi _{y}$

and, for any test function ψ,

$\psi (x)=\int _{\Omega }c(y)\phi _{y}(x)\,dy$

where

$c(y)=\langle \psi ,\phi _{y}\rangle$.

If the eigenvectors are normalized so that

$\langle \phi _{y},\phi _{y'}\rangle =\delta (y-y')$

in some sense, then there is a resolution of the identity

$\delta =\int _{\Omega }|\phi _{y}\rangle \,\langle \phi _{y}|\,dy$

where the operator-valued integral is understood in the weak sense.

The delta function also has many more specialized applications in quantum mechanics, such as the delta potential models for a single and double potential well.

I moved to here the section Application to quantum mechanics because it has so many problems and is misinformation. For example, in the following equations from the section

$\delta =\sum |\phi _{n}\rangle \langle \phi _{n}|$

and

$\delta =\int _{\Omega }|\phi _{y}\rangle \,\langle \phi _{y}|\,dy$

the left hand sides should be 1, not $\delta$. And if these equations are corrected they lose their purpose in the context.

In my opinion, it would be best to completely rewrite the section. --Bob K31416 (talk) 19:16, 3 September 2010 (UTC) P.S. Instead of generating error-filled OR, I would suggest using sources. For example, Dirac, Paul A. M. (1967). Principles of Quantum Mechanics (revised 4th ed.). London: Oxford University Press. p. 58. ISBN 978-0198520115. or college textbooks on quantum mechanics. --Bob K31416 (talk) 20:46, 3 September 2010 (UTC)

I guess I assume some responsibility for some of the errors in the section. This was my own effort to rewrite what was here before in a way that actually made sense, although I do apologize for the errors that I may have introduced in my carelessness along the way. I support the idea of rewriting from sources, but the emphasis that most sources have often makes it very difficult to write material on the delta function itself, as it is seldom the focus. It seems to me the basic assumption in quantum mechanics is that it is possible to normalize the eigenstates of a self adjoint operator (with a continuous spectrum) so that

$\langle x|y\rangle =\delta (x-y).$

One can then resolve any state as a superposition of these eigenstates by the resolution of the identity. This is, at least, what it seems to me that the earlier revision had intended to convey, and it was this sense that I had intended to preserve. My own carelessness in the exposition notwithstanding, this particular conclusion is surely not "original research". Whether this bears enough relevance to this particular article is a valid concern, though. Sławomir Biały (talk) 22:39, 3 September 2010 (UTC)

I've attempted again to improve the section, hopefully with fewer errors this time. Constructive feedback, as always, is most appreciated. Sławomir Biały (talk) 01:17, 4 September 2010 (UTC)

Why did you use an undefined quantity "I" instead of 1? What is the purpose of the two corresponding equations in discussing the delta function? --Bob K31416 (talk) 03:52, 4 September 2010 (UTC)

I is fairly standard notation for the identity operator, and we already use this in the section on Resolutions of the identity. The notation "1" might be taken to refer to the function f(x) = 1 in the rigged Hilbert space. Sławomir Biały (talk) 10:52, 4 September 2010 (UTC)

And your response to my other question re purpose is....? --Bob K31416 (talk) 13:40, 4 September 2010 (UTC)

The description already in the Resolutions of the identity section seems apropos. The operator identity

$I=\int |x\rangle \langle x|\,dx$

is equivalent to the normalization

$\langle x|y\rangle =\delta (x-y)$

by applying each equality to a pair of test functions. Sławomir Biały (talk) 14:18, 4 September 2010 (UTC)

<outdent>That doesn't appear to be quite true. If you feel they are equivalent, try deriving each from the other. However, from looking at the "Resolutions of the identity" subsection of the Hilbert space theory section and the source there " Davis & Thomson 2000, Equation 8.9.11, p. 344" , I think I see what the original forms of the equations at the top of this Talk section were trying to do.

$\delta (x-\xi )=\sum _{n=1}^{\infty }\ \varphi _{n}(x)\varphi _{n}^{*}(\xi ).$

I'm not sure of the similar expression for the continuous index case, which isn't mentioned in the "Resolutions of the identity" subsection. Is there a source for the continuous index case? --Bob K31416 (talk) 16:12, 4 September 2010 (UTC)

Sorry, my mistake. The eigenvectors need to be suitably complete in order to show equivalence: every compactly supported function on the spectrum must have the form $\phi (x)=\langle \phi |x\rangle$ for some $|\phi \rangle$. I think this is true automatically for bounded self-adjoint operators by the Gelfand-Naimark theorem. I don't know about the unbounded case, which is the primary case of importance here. Sławomir Biały (talk) 21:50, 4 September 2010 (UTC)

I'm not sure how to put my feelings about the situation at this article. I think I'll just leave. I do recognize that you are trying but I think there is a fundamental problem here. Anyhow, good luck. --Bob K31416 (talk) 00:59, 5 September 2010 (UTC)

Some remarks:

• Quantum mechanics does not deal exclusively with the Hilbert space L². Spin systems often have finite dimensional Hilbert spaces. For this reason it is not immediately clear how the given example is situated in physics. Some attempt to setup up the physical context (e.g. a free particle) could be made.
• The Hamiltonian itself may have a continuous spectrum. (For example, the Hamiltonian of a free particle has a continuous spectrum) This makes the sentence: "For other observables in quantum mechanics, the set of eigenfunctions may be continuous rather than discrete." a bit weird.
• Strictly speaking, neither position nor momentum operators are self-adjoint, because their adjoint is not well-defined. Within this context a looser definition of self-adjoint applies.
• This appearance of the delta function in QM is hardly the most notable (although it is important.) At least equally important is the appearance of the Dirac delta in the canonical commutation relations, and the role of the Dirac delta in the definition of Green's functions. (The latter is actually an application of the Delta function that is important in nearly any field of physics.
• It would greatly help if the section were based of a RS. Using Dirac's book might not be the best idea though.TimothyRias (talk) 11:40, 6 September 2010 (UTC)

Thanks for the remarks. Three of these issues are fairly easily addressed. I have attempted to set up the physical context for your first remark. Also, in the second remark, An earlier revision said that this was the Hamiltonian of a bound state, but that somehow didn't make it into the version you saw: I have restored it. For the third bullet, there is a standard way of treating self-adjoint unbounded operators (provided they are densely defined). Usually (at least in my experience) "self-adjoint" means that the operator can be unbounded, whereas "hermitian" implies boundedness (or do I have it backwards?) At any rate, I have added an appropriate link to the unbounded operator article.

I have a question regarding your fourth point. Presumably you mean the canonical commutation relations of field theory (i.e., "second quantization")? Sławomir Biały (talk) 13:20, 6 September 2010 (UTC)

Tweaks look good. The phrase "Hamiltonian of a bound state" is a bit peculiar. I guess it is supposed to convey the idea that the eigenstates of the Hamiltonian are bound states. Wouldn't "Hamiltonian of a bound system" be better? As for your last question, yes I was referring to the canonical commutation relations of a quantum field theory, as used in canonical quantization. (I really hate the term "second quantization" since it is somewhat misleading.) As for the standard usage of either the term self-adjoint of Hermitian for a densely defined self-adjoint operator: Don't know, physicists generally use the term Hermetian in this context (and simply ignore any subtleties relating to this issue).TimothyRias (talk) 13:38, 6 September 2010 (UTC)

Edits to lead

I agree with some of the edits to the lead, but disagree with others. For one thing, the lead sets up a contrast between the informal description of the delta function as a quantity that is zero at every nonzero point, but with total integral one, and its actual definition as a measure or distribution. The article discusses this in great detail. I have added a reference to this effect. It is important to note that Dirac's 1927 textbook is not the ultimate authority on the delta function: much has been done in both mathematics and physics to lay this on a more satisfactory foundation. Sławomir Biały (talk) 11:06, 4 September 2010 (UTC)

I think the 1st edition of Dirac's book was in 1930 and the last edition that Dirac revised was in 1967.

Here's an excerpt from Dirac's discussion of the delta function on p.58 in the 1967 edition of his book.

"δ(x) is not a function of x according to the usual mathematical definition of a function, which requires a function to have a definite value for each point in its domain..."

The corresponding sentence of mine that you removed[4] was

"The δ function is not a mathematical function according to the usual definition because it doesn't have a definite value when its parameter is zero."

In your edit summary you wrote, "I don't agree with this edit. Lack of a definite value is not essential ...".

It appears that you disagree with Dirac's comment that the definition of a function requires that a function have a definite value for each point in its domain. Could you give a source and the excerpt from the source that contains the idea, that you seem to have, that the definition of a function does not require a function to have a a definite value for each point in its domain? Thanks. --Bob K31416 (talk) 14:28, 4 September 2010 (UTC)

I'm not disagreeing that functions (as conventionally defined in most areas of mathematics) have values at each point of the domain. What I'm disagreeing with is that this is at all relevant to the discussion. Elements of the space L¹, for instance, while commonly called "integrable functions", are actually equivalence classes of functions that are mutually equal almost everywhere. Thus such "functions" indeed lack a definite value at any given point of the domain. Such functions have a well-defined integral, and in fact have the aforementioned property: that if they are zero at every point save one, then they must have total integral zero. (An example of an L¹ function from physics is the Heaviside function on the interval [−1,1]. In practice, it seldom matters how (or even if) the Heaviside function is defined at the origin.) However, whereas these L¹ functions fit neatly into conventional integration theory, the delta function actually requires a different formalism altogether to put on a proper foundation. The latter point is more substantial from the point of view of measure theory and distribution theory. So, while Dirac's observation that the delta function is not a function because it lacks a specific value at one point of the domain may be technically correct, it emphasizes the wrong aspect of the theory from the point of view of the modern parallel developments of integration theory and the theory of distributions. Sławomir Biały (talk) 15:53, 4 September 2010 (UTC)

It's not clear what your position is regarding the definition of a function. You began by saying, "I'm not disagreeing that functions (as conventionally defined in most areas of mathematics) have values at each point of the domain." You continue with remarks that don't seem to be consistent with this beginning statement. With the phrase "(as conventionally defined in most areas of mathematics)" perhaps you feel that there is a definition of function that does not require a function to have a definite value for each point in its domain?

As you noted there are mathematical objects that have the word function in their names that don't have a definite value for each point in their respective domains, but please note that like the delta function, this does not mean they are functions according to the usual definition. My edit made it clear at the beginning, and in a way that coordinated well with the previous sentence, that the delta function is not a function according to the usual definition, and was simply reporting what the source correctly stated. The next paragraph of the lead in that previous version of the article that was reverted, was there before my edit and satisfies the other subjects that you brought up. Regards, --Bob K31416 (talk) 17:30, 4 September 2010 (UTC)

The fact that the delta function is of "indeterminate value" at a particular point misses the mark. The integral of a function whose value at a single point is unspecified (or even is infinite) and which is zero everywhere else is well-defined, and the result is zero. Strictly speaking, it doesn't even make sense to talk about the "value" of the delta function at any point: it is a distribution (mathematics) pure and simple, not a pointwise object at all. Compare this situation with actual functions and actual integrals: simply lacking a value at a point does not turn a function into a more mysterious object, it simply changes the domain of the function (and not its integral). Sławomir Biały (talk) 20:10, 4 September 2010 (UTC)

It doesn't look like we're getting anywhere in this discussion. --Bob K31416 (talk) 20:27, 4 September 2010 (UTC)

To put it another way, the delta function isn't just "not a function", it isn't even a locally integrable function. Sławomir Biały (talk) 20:45, 4 September 2010 (UTC)

That brings to mind an issue regarding the definition of the delta function. Its defined in terms of an integral that does not exist, according to the definition of an integral, since the integrand does not have a definite value at x=0. Perhaps that was your point all along? --Bob K31416 (talk) 21:49, 4 September 2010 (UTC)

Well yes. Something like: if the integral existed, then it would necessarily be zero by the definition of the integral (e.g., Lebesgue integral). Sławomir Biały (talk) 21:52, 4 September 2010 (UTC)

I was thinking of the definition of the Riemann integral with regard to it not existing, rather than being zero. --Bob K31416 (talk) 22:22, 4 September 2010 (UTC)

P.S. I don't readily have available the Vladimirov reference that appears in the 2nd paragraph of the lead. Could you give the excerpt from that reference that supports the point that the integral is zero? --Bob K31416 (talk) 22:38, 4 September 2010 (UTC)

After motivating the definition of the Delta function as a limit of step functions, Vladimirov on page 64 says:

We shall first take the density to be point limit of the sequence of average densities f_ε, that is,

$\delta (x)=\lim _{\epsilon \to 0}f_{\epsilon }(x)={\begin{cases}+\infty &\mathrm {if} \ x=0\\0&\mathrm {if} \ x\not =0\end{cases}}$ (1)

It is naturally required that the integral of the density δ over any volume G should give the mass of this volume, that is,

$\int _{G}\delta (x)\,dx={\begin{cases}1&\mathrm {if} \ 0\in G\\0&\mathrm {if} \ 0\not \in G\end{cases}}$

But, by virtue of (1), the left-hand side of this equation is always equal to zero if the integral is taken to be improper.

Vladimirov then proceeds to define generalized functions, and distinguishes between regular and singular generalized functions: the regular ones being those that come from locally integrable functions. Ultimately, on page 74, he presents a more careful argument that the delta function is not regular which proceeds by showing under the assumption that δ is regular, that is equal to zero almost everywhere, but that this contradicts the definition of the delta function. Sławomir Biały (talk) 12:58, 6 September 2010 (UTC)

Disambiguation really needed

I have noticed that a hatnote directing readers to the Kronecker delta has been added twice to this article. I am inclined to removed this, per WP:RELATED. The article already amply covers the Kronecker delta in a section. Is there any real concern that someone will have arrived here in error? Sławomir Biały (talk) 13:10, 11 October 2010 (UTC)

There's a huge problem in Wikipedia with technical articles being too technical and confusing for non-expert users. Check out WP:TECHNICAL and Wikipedia:Lead section#Introductory text and you'll see some info on how to make technical articles better. In technical articles it's vitally important that the intro cover the most pertinent issues and be directed towards the non-expert, who makes up the vast majority of the readers of the article. This includes what a topic is (explained, if at all possible, in a way that makes sense to the non-expert), why a topic is important, what it's useful for, and any other considerations relevant to the non-expert. In this case it's very common for non-expert readers to confuse the Dirac delta and Kronecker delta, and thus this needs to go in the lead. It doesn't have to be a hatnote -- imo even better might be a short section that actually explains the difference. In fact, I just changed the article correspondingly.

Now you might object that we already have a section comparing the two. The problem is (1) this section is way down towards the end rather than near the beginning; and (2) it's phrased in a way that will make no sense to a non-expert user. Furthermore, there's nothing wrong with having the comparison mentioned twice. In fact, it's probably exactly the right thing. A Wikipedia article should follow the inverted pyramid structure of a newspaper article, with the most important info first, summarized in the lead and then explained in more detail later on. The lead should stand on its own -- see WP:LEAD. Benwing (talk) 04:06, 12 October 2010 (UTC)

The tone of your reply suggests that I am making an argument that I never did. Nowhere in WP:MTAA or WP:LEAD does it say that we should populate our technical articles with irrelevant hatnotes. I have no problem mentioning the Kronecker delta in the lead, although an entire paragraph on the subject is disproportionate to the coverage in the article (and even level of interest and importance of the Kronecker delta vis-a-vis the Dirac delta). So I've trimmed this down somewhat and put it in what seems to be a natural place.

Regarding making technical articles more accessible, I think we should focus there on the big picture. The Kronecker delta is of strictly peripheral importance: even if it is more "accessible", moving that content upwards without a view for how that enhances an understanding of the Delta function is not likely to lead to improvement. Instead someone needs to take a stab at improving the introductory section, with a care to sourcing it properly and including only content that is part of the "standard" treatment of the subject. Some pseudo-mathematical examples might be helpful as well, with illustrations showing limits of Gaussians or hat functions tending to the delta function. If I recall, the Strichartz text referenced in the article has a fairly good introductory account, and might be useful as a source for writing such a section. Then a good source for the history needs to be found. At one point, I had suggested van der Pol and Bremmer, which has a rather good account of the early history. I'm told that Lutzen's "Prehistory of the theory of distributions" might be better. Probably any textbook on distributions will give some account of the more recent history. Thirdly, someone needs to focus on writing an Applications section, including the current application to quantum mechanics, and the stublike application to probability, as well as including applications to other diverse areas like Green's functions and partial differential equations. Once all this has been done, hopefully a natural article structure will emerge that will facilitate greater accessibility. Sławomir Biały (talk) 12:03, 12 October 2010 (UTC)

I agree: Kronecker delta is peripheral here. Tkuvho (talk) 13:39, 12 October 2010 (UTC)

the integral of the Kronecker delta is equal to zero?

This phrase from the introduction is misleading. Usually Kronecker delta is useful in the case of a finite domain, where one should be "integrating" with respect to point measures. Tkuvho (talk) 13:33, 12 October 2010 (UTC)

I agree that one should not think of the Kronecker delta on a continuous domain. The fact that the integral of such a function over a continuous domain is zero emphasizes the dissimilarity of the Kronecker delta and Dirac delta rather than their fundamental similarity: in some sense the Dirac delta is the proper generalization of the Kronecker delta from a discrete domain to a continuous one. Sławomir Biały (talk) 11:27, 13 October 2010 (UTC)

Informally?

This article starts: "The Dirac delta function, or δ function, is (informally)..." At this point, one would hope most readers would either roll their eyes, start laughing, or just stop reading. Wiki is at best "informal" to begin with - this isn't a mathematics journal or textbook. If this definition is informal in an already informal environment - what is it? Well, being wiki I could tell you what it is, but some children might read this in the future - let's just say it rythms with nit and wit. —Preceding unsigned comment added by 192.158.61.139 (talk) 20:48, 13 January 2011 (UTC)

What could rhyme with nit and wit that could possibly describe a definition in such an offensive way? genuinely curious here. — Preceding unsigned comment added by 60.184.245.143 (talk) 15:23, 25 January 2012 (UTC)

Bullshit. Bill Wvbailey (talk) 23:14, 25 January 2012 (UTC)

dirac delta function == sum of exponentials

Could someone please check this formula ?

$\delta (x)={\frac {1}{2\pi }}\sum _{n=-\infty }^{\infty }e^{inx}.$

I would suppose the RHS gives rise to a whole comb of delta-functions: Setting x=2*pi definitely does not give zero. Benjamin.friedrich (talk) 12:50, 7 February 2011 (UTC)

This gives the delta function on the circle rather than the real line (ie x is in [-pi,pi]). If x is regarded as a variable on the real line, then this is the periodization of the Dirac delta, which is is the Dirac comb. The Poisson summation formula is relevant as well. Sławomir Biały (talk) 13:28, 7 February 2011 (UTC)

I have added a statement that hopefully clarifies this assertion in the text. It is clear from the context that it intends to refer to functions on the circle. Sławomir Biały (talk) 13:58, 7 February 2011 (UTC)

Composition with a function example slightly confusing

Hi all, excuse me if I make any blunders - this is my first attempt at creating a discussion section.

We are told the change of variables shown holds " provided that g is a continuously differentiable function with g′ nowhere zero. " However the subsequent example sets $g(x)=x^{2}-\alpha ^{2}$ which does not obey the second requirement that the first differential is nowhere zero.

This is remedied by the fact that the requirement for change of variables need only be true over the domain of integration, not necessarily the whole real line as shown in the equation

$\int _{\mathbf {R} }\delta {\bigl (}g(x){\bigr )}f{\bigl (}g(x){\bigr )}|g'(x)|\,dx=\int _{g(\mathbf {R} )}\delta (u)f(u)\,du$

In the example, the function $g(x)$ need only satisfy the above requirements in the neighbourhood of α and -α. I suggest replacing $\mathbf {R}$ with an arbitrary domain such as $\mathbf {\Omega }$ in order to clarify this, but please correct me if you believe I'm wrong.--Gabs (talk) 15:17, 9 June 2011 (UTC)

The definition does not require that the derivative of g be nonzero, only that the roots be simple. This is only used to motivate writing down the definition. Sławomir Biały (talk) 15:25, 9 June 2011 (UTC)

I'm not sure I understand you. The change of variables method requires that the derivative is nowhere zero on the domain of integration - this is what the article states and references. In practise, if the derivative is zero at a finite number of points then we split the domain of integration around these points.Gabriel Rosser (talk) 10:09, 10 June 2011 (UTC)

From the article,

It is natural therefore to define the composition δ(g(x)) for continuously differentiable functions g by

$\delta (g(x))=\sum _{i}{\frac {\delta (x-x_{i})}{|g'(x_{i})|}}$

where the sum extends over all roots of g(x), which are assumed to be simple.

This is referenced to Gelfand and Shilov, who also define the composition in this way. This definition doesn't involve change of variables in an integral. That was just to motivate writing this definition down. Sławomir Biały (talk) 11:45, 10 June 2011 (UTC)

OK, I understand. Thanks for explaining. Gabriel Rosser (talk) 13:00, 10 June 2011 (UTC)

Layman

It's not difficult to give 'the gist' of what the dirac delta function is to the layman, but the first sentence introduces difficulties by using the term 'generalised function'. So no way can the layman get the gist. I will attempt to give the gist in a new first paragraph, which will be there by the time you read this. — Preceding unsigned comment added by Mal (talk • contribs) 09:17, 16 January 2012 (UTC)

What you added was incorrect. I have removed it. We should, of course, try to simplify the treatment as much as possible, but we shouldn't tell lies in the process. The term "generalized function" is technically correct, and yet it also conveys the right sense to a layman. Sławomir Biały (talk) 10:00, 16 January 2012 (UTC)

Most laymen would have no idea what a generalised function is. If you doubt this, try asking your nearest and dearest (if he/she doesn't have a science degree!) S/he might also have trouble with "function" - but s/he might understand the first paragraph of the "function" article.Mal (talk) 16:03, 17 January 2012 (UTC)

Well that's most unfortunate. The concept of a function is included in elementary school education throughout much of the world. But this is irrelevant. Obviously one can't hope to get anything out of the article without knowing what a function is. The Dirac delta function is a "generalized function" (forget that this has a technical definition for a moment). To someone who understands the concept of "function", we should expect that such a person also understands the implications of the English word "generalized". Sławomir Biały (talk) 16:41, 17 January 2012 (UTC)

The first sentence contains "The Dirac delta function, or δ function, is (informally) a generalized function depending on a real parameter such that it is zero for all values of the parameter except when the parameter is zero". This is not true, the sinc(x/a)/a function being a counter example. I agree, we should not tell lies in the process, and I wonder if this is too big a lie. PAR (talk) 01:12, 17 January 2012 (UTC)

It's not a lie. The delta function is zero at every nonzero point. (More precisely, its support is the single point at the origin.) The counterexample of sinc(x/a)/a doesn't tend pointwise to zero, but it does tend weakly to zero away from the origin. The only "lie" here is one that hasn't been committed, namely the naive belief that the pointwise limit is relevant. Sławomir Biały (talk) 01:23, 17 January 2012 (UTC)

One way of conveying what the delta function is intuitively would be to refer to the way Dirac described it. This way we don't commit ourselves to mathematical accuracy, and also give an idea of the intuition behind it. I don't have Dirac's book in front of me but I assume he spoke of a function that's zero everywhere except the origin, is infinite at the origin, and integrates to 1. If we say anything of this sort we would have to attribute it explicitly to Dirac. Tkuvho (talk) 13:02, 17 January 2012 (UTC)

This description appears in numerous sources, including Dirac, Gelfand and Shilov, Vladimirov, etc. It doesn't need to be specifically attributed to anyone. It is commonly accepted that the delta function is a measure supported at the origin with total mass one. Sławomir Biały (talk) 13:07, 17 January 2012 (UTC)

What I was pointing out is that we could distinguish between initial pre-mathematical intuitions, on the one hand, and mathematical implementations (of which there are several, including "true" functions), on the other. If we choose to present the former in the lede, then there is no need to mention "generalized" since Dirac did not do it. Also, there is no need to suppress "infinite at the origin", since Dirac presumably mentioned it, and also because it helps to give an intuitive idea of the delta function. Tkuvho (talk) 13:12, 17 January 2012 (UTC)

The definition is on p.58 of TPQM, and Tkuvho is correct. Mal (talk) 16:03, 17 January 2012 (UTC)

I just checked this reference. It agrees exactly with how we describe it in the article. Nowhere does Dirac appear to say "equal to zero everywhere except the origin, where it is infinite". Moreover, Dirac is careful to avoid calling it a function. He calls it an "improper function". The standard term nowadays is "generalized function." Sławomir Biały (talk) 16:51, 17 January 2012 (UTC)

Is the argument over whether to use the term "generalized function" in the lede? I think it should stay. Even if you don't know the technical definition of a generalized function the term still gets across the idea that the delta is like a function in some ways, but not really a function. The business about the delta as a function on the hyperreals is obscure and should not be a consideration for the first few sentences of the intro IMHO.Holmansf (talk) 16:49, 17 January 2012 (UTC)

The title of this talk section is "Layman". To a layman, the idea that

$\lim _{a\rightarrow 0}{\frac {\sin(x/a)}{xa}}=\delta (x)$

and

$\delta (1)=0$

means that

$\lim _{a\rightarrow 0}{\frac {\sin(1/a)}{a}}=0$

which is false, the limit does not exist. The error is in the statement $\delta (1)=0$. The delta function should NOT be thought of as necessarily zero at every nonzero point. To simply blow off the problem by saying "pointwise convergence is irrelevant" is sloppy. To say that its support is the single point at the origin is NOT the same as saying it is equal to zero everywhere else. If f(x)=0 except f(0)=1, then it has no support on the real line. This does not mean that f(x) is zero everywhere. The delta function is "code" for a more complicated concept. The delta function has no real meaning except when used inside an integral. To say it is supported nowhere except at zero immediately invokes the concept of integration. To say it is zero except at the origin does not. Its like talking about $\lim _{x\rightarrow \infty }1/x=0$. There is no such real number as infinity, the whole statement is "code" for a more complicated limiting process which involves only real numbers. PAR (talk) 16:53, 19 January 2012 (UTC)

This post really veers off from anything that might possibly concern a layman. Rigorous formal definitions are given in the appropriate section, for someone to whom such issues might be a worry. I agree that talking about pointwise properties of the delta function is simply "code" to mean where it is supported, but I think you'll find that (outside of very formalistic writings) it is a code that is generally used to talk about the support of a distribution, if not to rigorously define it. Sławomir Biały (talk) 17:05, 19 January 2012 (UTC)

Slawomir Bialy is correct - the issues you're discussing would only concern someone who has spent some considerable time thinking about the delta function, not a general layman. But since you brought it up, I believe the mistake in the above "layman's" argument is actually a misunderstanding of the limit. It is not a pointwise limit, but rather in the sense of distributions. One can also say for example $\lim _{a\rightarrow \infty }e^{iax}=0$, but this doesn't mean the distribution zero cannot be said to be pointwise equal to zero. Certainly any distribution can be evaluated pointwise in the complement of its Singular support, and for the delta function that is everywhere except the origin. Holmansf (talk) 18:42, 19 January 2012 (UTC)

I read through this entire section, and although I'm not quite sure what the disagreement is, I think part of it is due to confusing several distinct objects as if they were the same:

1. The δ function, a non-existent function on the real line which is equal to 0 everywhere except possibly x=0, and which integrates to 1 on the entire real line.
2. The δ functional (distribution), which as such is not a function on the reals, so we cannot even ask whether it is zero at, e.g., x=2.
3. The point mass probability measure on the reals located at the origin. Again, this is not a function, so we cannot ask whether it is zero at x=2.

The latter two bullets do not give a definition of the δ function; they give examples of things that do exists although the δ function does not.

By analogy, if I accept that dragons do not exist, but are simply fictionalized crocodiles, and I accept that crocodiles cannot breathe fire, I can still assert that dragons can breathe fire. Similarly, even though the δ function does not exist, it is still 0 everywhere except possibly the origin. — Carl (CBM · talk) 22:53, 19 January 2012 (UTC)

Yes, I think it would be useful to clarify what precisely is being debated. My understanding is that the original proposal was to rewrite the lede in some way that does not include the term generalized function. Personally, I think generalized function should be used there, but I don't like the use of the word "parameter" and would change the first sentence to:

"The Dirac delta function, or δ function, is a generalized function on the real number line that is zero everywhere except at zero, and when integrated over the whole line is equal to one."

Holmansf (talk) 00:45, 20 January 2012 (UTC)

I also think that is better than the "parameter" sentence. — Carl (CBM · talk) 03:29, 20 January 2012 (UTC)

Looks good to me. Sławomir Biały (talk) 03:31, 20 January 2012 (UTC)

I'm not forgetting about the layman - I'm just saying that to the layman, the three equations I wrote will seem inconsistent. Yes, the first limit equation is "code" for something more complicated, not to be interpreted as if δ was a function. The second equation is not code for anything, it is simply wrong, it assigns a specific value to the delta function. The third equation therefore does not follow. So how do we present this to the layman? I think we do that by saying that thinking of the delta function as zero everywhere, infinity at the origin, and the integral of the delta function is unity is a good way to think of the delta function to begin with, but that further study will show that this is only a first step. I think it is wrong to tell or imply to a layman that this is the final definition of a delta function. Then present the sinc function as a counterexample to illustrate that fact. Adding the term "generalized", which changes the meaning of "equal" is not something that we should expect a layman will immediately get. I'm in favor of saying something like this in the introduction:

The Dirac delta function, or δ function, is (informally) a generalized function depending on a real parameter. It may be thought of as being zero for all values of the parameter except at zero, at which point it is infinite, and its integral over the parameter from −∞ to ∞ is equal to one. This interpretation is useful for many situations, but not appropriate for all situations. The full definition of the delta function rests upon its behavior upon integration, and thinking of it as having a specific value as a function of its parameter is sometimes counterproductive.

This will not call upon a layman to understand and probably misinterpret the unmentioned subtleties of "equal" with regard to a generalized functions. PAR (talk) 06:13, 20 January 2012 (UTC)

Your proposal is likely to be more confusing to readers rather than less so, for the common objection that it talks around the subject without defining it (even informally). Moreover, I know of no reliable sources that are concerned with this issue. Indeed, Dirac includes no such equivocal language. Laurent Schwartz (commonly credited as the founder of modern distribution theory) introduces it in the same way as Dirac, as do Gelfand and Shilov (two more founders of the modern theory of distributions). Do you have sources of similar pedigree that justify rewriting the lead to avoid the particular bogeyman that you are worried about?

Moreover, it has already been argued by multiple editors that the lead sentence is not misleading. The simplest solution to the issue you pose is not to write $\delta (x)=\lim _{a\to 0}{\frac {\operatorname {sinc} (x/a)}{\pi x}}$. It is this expression that is misleading to a layman, since the limit involved is a weak limit rather than a pointwise limit. I hope it's clear to you that the delta function is not a pointwise limit of any sequence of functions. So the misleading thing here is to believe that it is. Furthermore, as a weak limit, it is zero away from the origin. So it's also misleading to suggest that it might not be. Sławomir Biały (talk) 11:50, 20 January 2012 (UTC)

This language has exactly the problem I was talking about. The full definition of the δ function is that it is zero everywhere except the origin, and integrates to 1. (Of course no such function exists.) The full definition of the δ distribution is based on how it integrates with test functions, but the δ distribution is not the same thing as the δ function. — Carl (CBM · talk) 12:23, 20 January 2012 (UTC)

My engineering texts agree with Carl, it's very simple: "an ideal impulse has finite area but "zero" duration (it occurs entirely within the interval 0 < t < 0+). It must therefore have "infinite" height. . . on one ground or another, impulses are specifically exempted from the category of "finite disturbances" in the spirit of Table VI, p. 202. So, for exactly the same reasons, are the derivatives of steps." Cannon 1967:217. I.e. if integrated from -infinity to +infinity the integral produces an ideal step-function at time t=0. And vice versa, the derivative of an ideal step function is the delta function. The one sided Laplace transform of an impulse function is 1. Bill Wvbailey (talk) 14:40, 20 January 2012 (UTC)

Ok, I am going to go out on a limb here, tell me where I go wrong. There is no such thing as a "delta function", the notation $\delta (x)$ has no meaning except inside of an integral, and whenever you see it alone in an equation, it is "code" for describing the behavior of that equation upon integration. The behavior of the delta function upon integration is $\int _{-\infty }^{\infty }f(x)\delta (x)dx=f(0)$ and thats the entire definition. Ok, thats a bit informal, better is the exact definition of the Dirac measure. Furthermore, the idea that the "delta function" has a particular value at a particular point makes no sense given this definition. When you say $\delta (1)=0$ you are misleading the layman into thinking that it is some kind of point-wise statement applying to some function, when in fact it is a limiting statement about the integral of $\int f(x)\delta (x)dx$ over an interval about 1.

When you say

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

you are really saying

$\int _{-\infty }^{\infty }\lim _{a\rightarrow 0}{\frac {{\textrm {sinc}}(x/a)}{a}}\varphi (x)\,dx=\varphi (0)$

Now, I admit, there is trouble. This is certainly true for certain restrictions on $\varphi (x)$, like it has to be continuous almost everywhere, I think. So in that sense, the sinc definition of the delta function MAY be misleading if you are thinking of L^2 integrable functions. I say "may" because I don't know what happens if you admit L^2 integrable $\varphi$. Is it misleading for Reimann integrable functions? I don't know.

When you say that the sinc definition of the delta function is a "weak limit", I interpret that to mean that the above integral equation is true almost anywhere for any $\varphi$ in L^2. Is this what you mean? I also admit that I do not know whether that is true or not.

The bottom line is that the sinc definition may very well be misleading, but the statement that $\delta (1)=0$ is also misleading, to the layman. Its a statement about the support of the Dirac measure, which, as Carl says, is not the same as the delta function, which, as Carl correctly says, does not exist. PAR (talk) 06:28, 27 January 2012 (UTC)

I'm just not seeing it. The article says that the delta function "is zero everywhere except at zero", which is a true statement about the delta function. If you're worried that it isn't a function, it agrees with the zero function away from the origin in a standard way: see Distribution (mathematics)#Functions as distributions. As far as I can tell, the the equation you seem to have a problem with, $\delta (1)=0$, doesn't appear anywhere in the article. The article is very careful to not refer to the pointwise value of the delta function at all. In the second paragraph of the lead, it explains that we need measures to make the (admittedly informal) statement of the first paragraph rigorous. Sławomir Biały (talk) 11:35, 27 January 2012 (UTC)

I meant that to a layman, $\delta (1)=0$ is a special case of "is zero everywhere except at zero". That statement "is zero everywhere except at zero" is misleading to a layman. Rather than say it "is equal" to zero, say it may be thought of as being (pointwise) equal to zero, but that ultimately this idea is not productive. Then we can go into the more complicated measure theory meaning of "is equal". PAR (talk) 05:03, 28 January 2012 (UTC)

Surely this is no more misleading than referring to elements of L² as "functions". Sławomir Biały (talk) 11:15, 28 January 2012 (UTC)

Certainly true. So let's not mislead the layman by making either statement.

I think of a function as being defined point by point, and I think of elements of L² as functions with an equivalence class imposed on them. This equivalence is not the same as pointwise equivalence, and that is the whole problem here. A layman sees an equal sign and assumes pointwise equivalence. In the very first sentence of the article, the article uses "equals" in the sense of an L² equivalence class, without ever explaining the difference to the layman, and that is what I see as misleading. PAR (talk) 10:06, 29 January 2012 (UTC)

The δ function is pointwise equal to 0 everywhere except at the origin, though. That is one of its two defining properties, after all. The first sentence of the article isn't about L² equivalence, nor is it about the δ measure or δ distribution. The point of the first sentence is to give the definition of the δ function.

By comparison, imagine if we had an article on dragons that first said that a dragon is a flying lizard the breathes fire, and then went on to give examples of actual animals that might have inspired the myth of dragons. Of course there is not really a fire-breathing flying lizard, but we would still need to say that is what a dragon is. The δ function is a similar mythological beast. — Carl (CBM · talk) 12:37, 29 January 2012 (UTC)

Note that the fire-breathing variety of the delta function is definitely infinite at the origin. This should be mentioned in the lede. Tkuvho (talk) 12:57, 29 January 2012 (UTC)

Since that first sentence has three sources, I'd be interested to know what they say about that before changing the sentence that is cited to them. The first sentence of the overview gives a different idea of what the graph would look like. I don't think that the value at 0 is worth adding a second sentence to the lede, because it seems to be a somewhat delicate issue. But if the sources already in use say that the function is infinite then I would add it to the first sentence. — Carl (CBM · talk) 13:12, 29 January 2012 (UTC)

They don't. Sławomir Biały (talk) 13:22, 29 January 2012 (UTC)

I don't really accept the whole "fire breathing" argument. Since no one seems to agree about anything here, we should at least be able to agree with what reliable sources have to say about the delta function. I've already pointed out, all of the sources I checked initially gave an informal definition that was some variant on "zero everywhere with integral one". Not one of them said "infinite at the origin" (as Tkuvho would like the article to say). Not one of them seemed to feel that this heuristic was misleading in the way that PAR thinks it is. I checked the following very high-quality sources: Dirac, Laurent Schwartz (discoverer of distributions), Gelfand-Shilov (discoverers of distributions), Bracewell (standard signal processing text), Arfken and Weber (standard text for engineers and physicists). Now, I think I have given very good reasons (to PAR) that the informal definition is not misleading. This is supported by the presence of the same informal definition in many many high quality sources. But "zero everywhere except at the origin, where it is infinite" is misleading and misses the key point altogether. Some sources, like Vladimirov (standard text for mathematical physics) emphasize this very issue. Sławomir Biały (talk) 13:22, 29 January 2012 (UTC)

Thanks for looking up the sources, I really appreciate it. I think the source of "infinite at the origin" may be that, as everyone here knows, if the function was finite at the origin, it would (more) clearly not have a positive integral, and because if we think of the function as representing an impulse it would have to have infinite magnitude and zero duration. Of course, as everyone here also knows, the definition for nonzero points already implies having a zero integral, which is why the naive δ function doesn't exist, regardless what the value at the origin is supposed to be. — Carl (CBM · talk) 13:36, 29 January 2012 (UTC)

ok, then go back to the sinc function problem. Its not pointwise equal to zero everywhere except at the origin, yet upon integration it behaves as a delta function for a certain set of pointwise test functions. As I understand the delta function, $\int _{a}^{b}f(x)\delta (x)\,dx$ equals f(0) if [a,b] contains zero, zero otherwise. That is the entire definition. Forget being zero everywhere and infinite at the origin. It has no real meaning as a pointwise function. You can profitably visualize it as that pointwise function sometimes, but that visualization will lead you astray in certain cases. Thats what we should tell the layman in the lead. The integral definition (more formally stated than above) will not, but thats something that needs to be developed. PAR (talk) 17:26, 29 January 2012 (UTC)

Slawik wrote above "But "zero everywhere except at the origin, where it is infinite" is misleading and misses the key point altogether. Some sources, like Vladimirov (standard text for mathematical physics) emphasize this very issue." What does Vladimirov say exactly about this issue? Tkuvho (talk) 13:22, 30 January 2012 (UTC)

Vladimirov first defines the delta function to be density corresponding to a point mass of mass one situated at the origin. He approximates by a sequence of indicator functions, and then computes the pointwise limit, which has total integral zero. That is, the (Lebesgue or improper) integral of the function

$f(x)={\begin{cases}0&x\not =0\\\infty &x=0\end{cases}}$

is zero. So it is misleading in the extreme to define the delta function this way, since it is inconsistent with the requirement that the delta function have total mass one. For this reason, he emphasizes that the appropriate limit is not the pointwise limit, but the weak limit instead. Sławomir Biały (talk) 13:39, 30 January 2012 (UTC)

I think we all agree about the mathematical point. I was wondering about precisely what Vladimirov described as being "misleading in the extreme" if he indeed used such terminology. Tkuvho (talk) 13:47, 30 January 2012 (UTC)

His precise words were "The contradiction here shows that the point limit of the sequence $f_{\epsilon }$ as $\epsilon \to 0$ cannot be taken as the density $\delta (x)$". He arrived at this conclusion because the function that is "zero everywhere but the origin where it is infinite" has total integral zero, which of course contradicts the key property of the delta function. Sławomir Biały (talk) 14:15, 30 January 2012 (UTC)

Thanks. I found an even more useful quote in an article by Aguirregabiria et al: "Notice also that, contrary to the intuitive idea that delta(x) is infinite at the origin, a sequence converging to delta(x), that is, corresponding to *R g(u)du51 with R5(2`,`), will behave at the origin as g g(0), which may converge to 1`, but also to 0, or even to 2`." This is a good candidate for the lede: we can mention that, contrary to the intuitive idea that delta(x) is infinite at the origin, the nascent functions can be assigned arbitrary values there, and give a reference to Aguirregabiria et al. The 2001 article is entitled "delta-function converging sequences". Tkuvho (talk) 14:38, 30 January 2012 (UTC)

There may be a place lower in the article for that, but I think it goes into too much detail for the lede. The point of the lede is just to give a general summary, not to get into fine details. At the same time, we need to be careful to distinguish the δ function from the δ distribution; Vladimirov seems to conflate them, or to only talk about the δ distribution. — Carl (CBM · talk) 14:52, 30 January 2012 (UTC)

Carl, why are you being difficult? Being infinite at the origin is a common "fire-spitting" intuition about the delta function, and should be addressed in the lede. If it were not a common intuition, Vladimirov and Aguirregabiria would not have addressed the issue. Besides, some of the most obvious nascent functions will indeed display such behavior at the origin. Moreover, for every fixed "small" interval [-e,e], the supremum of the nascent function will necessarily tend to infinity. In this sense, the intuition is not so far off the mark. Tkuvho (talk) 16:53, 30 January 2012 (UTC)

I don't think we should emphasize the "value" at the origin at all. The intuitive description should stand as it is, since this is how the vast majority of reliable sources describe the delta function. The fact that there are sources saying that the delta function is not infinite at the origin seems to further bolster this point that it should not be described this way. Sławomir Biały (talk) 19:27, 30 January 2012 (UTC)

In exactly the same spirit, we should not emphasize the "value" anywhere else either. Then we are left with the fact that there is no "delta function" and that every reference to a "delta function" is a veiled reference to the delta distribution. Then we are left with the problem of how to introduce the "delta function" to the layman. Again, I say we should describe it as zero everywhere, infinite at the origin, such that the integral of f(x) over [a,b] is zero if [a,b] does not contain zero, f(0) if it does. Then we can say this is an oversimplification that is a good way to start but can lead to trouble, and then go on to explain from there. PAR (talk) 16:19, 1 February 2012 (UTC)

This is not a bad summary, but it may be better to refer to the bit about "infinite at the origin" as "an intuition that is not entirely correct", which is sourceable. On the other hand, this isn't going anywhere as we have two editors supporting the proposal and two opposing. Perhaps someone else can comment. Given the difficulty of the material, the more we say about motivating intuitions, the better, even if ultimately some of them have to be given up. Remember, a concrete mathematical implementation is always an impoverished version of the original pre-mathematical insight, as Weyl put it. Tkuvho (talk) 16:27, 1 February 2012 (UTC)

Sorry to keep pointing this out, but this is not how reliable sources introduce the delta function. The "infinite at the origin" bit might be sourceable to poor quality sources, but the majority of high quality sources do not define it in this way at all. I think we should stick to what the best sources say, which is what Dirac says, which is what the article says. Sławomir Biały (talk) 11:43, 2 February 2012 (UTC)

I have added the following sentence, that should hopefully be a suitable compromise: "The delta function is sometimes thought of as an infinitely high, infinitely thin spike at the origin, with total area one under the spike, and physically represents an idealized point mass or point charge." Sławomir Biały (talk) 11:55, 2 February 2012 (UTC)

Thanks, that sounds fine. As far as reliability of sources is concerned, there is no reason to believe Aguirregabiria is a poor source and not a solid scientist. A brief check at mathscinet reveals a number of texts of his. Tkuvho (talk) 13:00, 2 February 2012 (UTC)

Does he have a book or something? I can't find a citation. The sources I have mentioned (Dirac, Schwartz, Gelfand-Shilov, Bracewell, Arfken-Weber) all have thousands of citations, and are considered to be canonical texts. From the Aguirregabiria source that I see in AJP, he seems to argue against the point that the delta function should be thought of as "infinite at the origin". Have I missed something? Sławomir Biały (talk) 13:10, 2 February 2012 (UTC)

No, you haven't missed anything. I have been precisely suggesting that we refer to an intuition that is not entirely correct according to this source (and I imagine others). I also mentioned that fact that if authors find it necessary to mention this point, it must be indeed a common intuition. Also, hyperreals delta functions do take infinite values at suitable infinitesimal points. Tkuvho (talk) 13:53, 2 February 2012 (UTC)

--- re the proposed "intuitive" wording, above, for non-mathematicians: Good point about Bracewell as a source. If you want a firm source, with an elegant, "intuitive" interpretation, here's what Bracewell 1965:69ff has to say. I haven't looked at this in 40 years, and yet it comes back as clear and in focus as the day I first read it as an engineering student; it will be highly intuitive to an engineer or physicist or a calculus student. Notice the notion of a "measuring equipment", and his definition dodges the problems of the simplistic infinite integral of a point at zero. The treatment is quite nuanced:

(He references, via a book by van der Pol and Bremmer, Hermite, Caucy, Poisson, Kirchoff, Helmhotz, Kelvin and Heaviside as "historical sources"):

"It is convenient to have notation for intense unit-area pulses so brief that measuring equipment of a given resolving power is unable to distinguish between them and even briefer pulses. This concept is covered in mechanics by the term "impulse". . . The notation δ(x), which was subsequently [subsequent to Heaviside for the derivative of the unit step function cf footnote 2 p. 69] introduced into quantum mechanics by Dirac₃, is now in general use [₃ P. A. M. Dirac, The Principles of Quantum Mechanics, 3rd ed., Oxford University Press, Oxford, 1947]. The underlying concept permeats physics. Point masses, point charges, point sources, concentrated forces, line sources, surface charges, and the like are familiar and accepted entities in physics. Of course, these things do not exist. . . .However, the impulse symbol δ(x) [footnote refers to it as a "generalized function" or "symbol"] does not represent a function in the sense in which the word is used in analysis (to stress this fact Dirac coined the term "imporper function"), and the above integeral [the classical one in the article that everyone's been fussing about] is not a meaningufl quantity until some convention for interpreting it is declared. . . ." (page 69-70)

He then defines δ(x) as the limit, as τ-->0, of the integral from -infinity to +infinity with respect to x of: 1/τ times the rectangle function Π(x/τ). He draws this notion on page 72 with rectangle functions of decreasing width as 1/τ gets taller, shows how the step function's slope increases as τ --> 0. And then goes on to extend exactly the same notion to triangle functions, sinc functions, Gaussian function, plus some other more fussy oscillatory functions, each with slightly different properties (cf pages 69-74). Bill Wvbailey (talk) 18:15, 2 February 2012 (UTC)

Thanks for this. Elaborating on physical motivation is a very good idea, I think. I am a little less enthusiastic than you are about how effective the passage above is. He mentions some important physical applications, but does not really explain how they tie in with the Dirac delta. His explanation in terms of the integral is the usual one, with a partilar choice of (discontinuous) nascent function, which is already mentioned in the lede. Tkuvho (talk) 09:24, 3 February 2012 (UTC)

I hunted in the article for Bracewell's definition but not knowing what a "nascent function" means I guess I missed it. (The article is a bit overwhelming for someone not a mathematician). I realized after posting the above that the animation of the spike is the same as Bracewell's drawings. RE physical motivation for a "delta" function (a pulse so brief it cannot be measured) Bracewell gives it in the sentences that follow what I quoted above, and in a subsequent example of a low-pass filter; I add an example of plucking a wire:

" . . . Of course, these things do not exist. Their conceptual value stems from the fact that the impulse response -- the effect associated with the impulse (point mass, point charge, and the like) -- may be indistinguishable, given measuring equipment of specified resolving power, from the response due to a physically realizable pulse. It is then a conveinece to have a name for pulses which are so brief and intense that making them any briefer and more intense does not matter." (p. 69)

He then continues with an example of a lowpass filter whacked with a voltage impulse (standard operating procedure for engineers curious about the natural response of a system -- hit it with an impulse):

" . . .it is readily observable, as the applied pulses are made briefer and brier, that the response setles down to a definite form. It is also observable that the form of the response is then independent of the input pulse shape, be it rectangular, triangular, or eve a pair of pulses. This happens because the high-frequency components, which distinguish the different applied pulses, produce negligible response. . . . the details [of the pulse] are irrelevant; it is necessary only that they be brief enough. Since the response may be scrutinized with an oscilloscope of the highest precision and time resolution, we must, of course, be prepared to keep the applied pulse duration shorter than the minimum set by the quality of the measuring instrument. The impulse symbol enables us to make abbreviated statements about arbitrarily shaped indefinitely brief pulses." (p. 71)

Finally (tho this is probably in the article, somewhere), he presents the Maclaurin series of the function e^(-1/x)H(x)(where H(x) is the "nascent function" above), and he claims (without a demonstration, actually) that all the terms vanish except "the remainder" that does not go away -- it contains all the "information", as it were.

This was just an example of impulse "functions" in engineering, what's important is the above: Finally, RE convolution integrals and natural responses of systems: another physical explanation of the lowpass filter hit by an impulse is that the impulse contains all frequency components with amplitudes of 1 (like white noise, only not persistent) or in real life, usually a gaussian distribution rolling off slowly to 0 at infinity (pink noise). Thus an impulse sounds like a click. Whatever the system's natural response is, the convolution of the impulse with the (unknown) response will reveal it. This is especially useful for resonant circuits: Example: Guitarists and pianists rely on this fact -- that you hit or pluck a string (in its center) and it vibrates afterwards with its natural (fundamental) frequency. The first major engineering project I worked on was to "pluck" (impact) a wire using an electromagnet and a capacitive discharge (i.e. an impulse). It didn't work for short wires without sending a pulse so intense it ran the danger of exploding the coal mines where this equipment would be used -- solution: lots of little pulses spaced out at different intervals to create a spread-spectrum. (I can source this, if necessary). Anyway, those are the sorts of physical motivations behind "impulse functions". Bill Wvbailey (talk) 15:25, 3 February 2012 (UTC)

I'm not sure how much of this belongs in this article, since there is already an article impulse response (that would actually benefit from some attention in my opinion). Whether this article should mention the impulse response more prominently than it does already, I don't really have an opinion on. Sławomir Biały (talk) 15:47, 3 February 2012 (UTC)

I struck the above as a distraction. The question is whether further development along Bracewell's line of attack would help a layman: RE concerning the idealization of the nature of events below the "resolution of instrumentation", ie the discovery that the shape of an "event" (e.g. the exact mathematical equation) doesn't matter beyond a certain "resolution" because no instruments can detect exactly what's happening. Is this idealization just lucky heuristic, a discovery about the behavior of the world, not provable, or is it derivable/provable from deeper physics? The article does poke at the idea with the decent example of the baseball hitting the bat, but it's not very prominent or developed, and nothing in the pertainent paragraph is sourced (now we have a pinpoint source). Maybe all this paragraph needs is a tidbit re instrumentation resolution and the reference to Bracewell. The historical is not bad (the source mentioned there -- van der Pol -- is the same as noted in Bracewell). It just needs something more specific about why Dirac up with the notion. Bill Wvbailey (talk) 22:58, 3 February 2012 (UTC)

Indeed Bracewell's heuristic is quite elegant, but as far as the lead goes, his view doesn't really fit in with how the article is currently written. Maybe a paragraph in the Overview section would be a good place to start. In the long run, a well thought-out "Physical motivation" section is definitely lacking. Regarding Dirac, in my view he was a mathematician first and a physicist second, so his view as expounded in TPQM is more like that of the early parts of the article in its present form, although there have obviously been subsequent refinements to Dirac's ideas that deserve elaboration. Sławomir Biały (talk) 03:03, 4 February 2012 (UTC)

RE augmenting the "Overview" and then adding a "Physical motivation section": I agree.

RE slightly different motivation in free-body mechanics: My old physics text that introduced us to "particles in a box" i.e. quantum mechanics has nothing about Dirac. My classical physics text (Resnick and Halliday 1966:214) gives a brief discussion of impulses (using the baseball and bat analogy) that is somewhat similar to Bracewell; but with application to free-body mechanics their motivation for use of a δ-"function" is slightly different: "During the collision we can safely ignore [gravity] in determining the change in motion of the ball; the shorter the duration of the collision the more likely this is to be true".

RE generalized functions: this is discussed nicely in Bracewell p. 87ff.

RE references: In my communications-theory text (Carlson 1968:44ff) the definitions of the δ-"function" are similar to those of Bracewell, but written differently to include 4 possible functions in the limit as candidates: the gaussian, the rectangular function, the sinc function and the sinc² function. Carlson references Bracewell 1965 and Lighthill 1958 [see below]. Toward the end of the chapter he again discusses the theoretical behind the heuristic (cf p. 88):

"A satisfactory mathematical formulation of theory of impulses has been evolved along these lines [i.e. limit as tau --> 0 of the integral that I wrote above] and is expounded in the books of Lighthill¹ and Friedman². Lighthill credits Temple with simplifying the mathematical presentation; Temple³ in turn credits the Polish mathematician Mikusinski⁴ with introducing the presentation in terms of sequences in 1948. Schwartz's two volumes⁵ on the theory of distributions unify "in one systematic theory a number of partial and special techniques proposed for the analytical inerpretation of 'imporper' or 'ideal' fucntions and symbolic methods⁶. // the idea of sequences was curernt in physical circles before 1948, however⁷.

• 1: M. J. Lighthill, "An introduction to Fourier Analsys and Generalised Functions", Cambridge University Press, Cambridge UK 1958.
• 2: B. Friedman, "Principles and Techniques of Applied Mathematics," John wiley & Sons, New York, 1956.
• 3: G. Temple, Theories and Appliations of Generalized Functions, J. Lond. Math. Soc., vol. 28, p. 181, 1953.
• 4 J. G.-Mikusinski, Sur la methode de generalisation de Laurent Schwartz et sur la convergence faible, Fundamenta Mathematicae, vol. 35, p. 235, 1948.
• 5 L. Schwartz, "Theorie des distributions," vols. 1 and 2, Herman & Cie, Paris, 1950 and 1951.
• 6 Temple, op. cit. p. 175
• 7 B. van der Pol, Discontinuous Pheonmena in Radio Communication, J. Inst. Elec. Engrs, vol. 81, p. 381, 1937.
• 8 Schwartz, op. cit. p. 22.

I am curious about Dirac's motivation. I'm thinking it has to do with sampling theory and the distortions caused by the "aperature effect", when applied to particle waves (cf Carlson 1968:285). I'll sign off for now re "Layman"; maybe someday I'll come back to it, or maybe somebody ambitious can use some of the above. Bill Wvbailey (talk) 16:52, 4 February 2012 (UTC)

Sorry to add to this lengthy and old section, but I think there is a mistake above: "Again, I say we should describe it as zero everywhere, infinite at the origin, such that the integral of f(x) over [a,b] is zero if [a,b] does not contain zero, f(0) if it does." I would replace "f(0)" with 1, since nowhere is the former claimed to be the definition of the Dirac delta, while the latter is claimed in multiple locations. And, concerning the original topic of this section, I am sympathetic to the non-mathematician, but starting with an integration-based definition is more confusing for such a person than starting with the more intuitive yet essentially erroneous point-wise "function" definition. David Spector (talk) 19:26, 26 July 2013 (UTC)

real

The claim that "The Dirac delta is not a true function, as no function has these properties" is factually incorrect. It would be accurate to say that no real function has these properties. Tkuvho (talk) 14:18, 17 January 2012 (UTC)

Well, yes, we are talking about Lebesgue integrable functions from the reals to something. Making the codomain be the set of complex numbers would not make any difference. I don't see that the claim is incorrect, could you explain what you mean? — Carl (CBM · talk) 14:22, 17 January 2012 (UTC)

A Lebesgue integrable function f has a natural hyperreal extension *f by the transfer principle. There exist "true" hyperreal functions which, when integrated against *f, will have the effect of the delta function on f. Thus, the delta distribution can be represented by a genuine function, if one extends the domain. Slawek feels that elaborating on the hyperreal viewpoint in this early section is inappropriate, and he may be right. What I am discussing is a separate issue, namely, that the sentence as stated is incorrect: it is possible to represent the dirac delta by a function, though not over the real domain. Adding a one-word clarification would not be counter to wiki undue weight policies. Tkuvho (talk) 14:27, 17 January 2012 (UTC)

The construction claimed at PlanetMath using hyperreals [5] is also "factually incorrect", IMHO, because the Dirac delta must have the property that it is zero everywhere except at one point x=0, while the function constructed on PlanetMath is nonzero on some non-zero values. In particular, if the function constructed there is integrated on $(-\infty ,0)\cup (0,\infty )$ (the hyperreal line minus exactly one point) the answer must still be 1, but it should be 0 if we were integrating the Dirac delta function. So at best we can represent some of the properties of the Dirac delta using nonstandard methods, but we do not seem capture all of them. But this is a digression.

In any case the due weight for nonstandard analysis in this article is extremely low, and I agree with Slawek that we do not need to add "real" to the paragraph in question. Is there any example of a graduate-level analysis book that is not directly about nonstandard analysis that uses nonstandard analysis to treat the Dirac delta function? It might be worth mentioning the hyperreal analogy in a separate paragraph, though. — Carl (CBM · talk) 14:41, 17 January 2012 (UTC)

You are right, you can't have it zero everywhere, though you can arrange for it to be infinitesimal at every real point (except 0). What I was referring to is the defining property of the delta function, namely that it produces the value of f at 0 when integrated against f. Apart from the hyperreal issue, this is something that should be emphasized in the lede or at the very least early on in the article. I withdraw my claim that the current statement is factually incorrect: I was mentally substituting the latter property. Tkuvho (talk) 14:51, 17 January 2012 (UTC)

The next subsection, "as a measure", states: "no true function for which the property $\int _{-\infty }^{\infty }f(x)\delta (x)\,dx=f(0)$ holds." This claim does not appear to be correct, as discussed above. I would suggest adding "real" here. Tkuvho (talk) 14:56, 17 January 2012 (UTC)

We could make similar corrections to almost every article on real analysis (e.g. "there is no number that is greater than 0 and less than 1/n for every natural number n" and every other statement that relies on the Archimedean property). I don't think that is generally worthwhile; the context of these articles is real analysis, and in that context the statements are correct. Moving to the context of hyperreals changes the meanings of all the symbols, so that they no longer mean what they used to. Thus statements that start out correct can become incorrect.

By analogy, in an article on topology, we could say that the rational numbers have $2^{\aleph _{0}}$ subspaces, even though our article on locale theory says there are $2^{2^{\aleph _{0}}}$ sublocales of the rational numbers. The statement in topology is not trying to talk about locale theory, though; statements in topology that are correct might nevertheless become incorrect if they are re-interpreted to be statements about locale theory. — Carl (CBM · talk) 16:31, 17 January 2012 (UTC)

Unlike the theory of real numbers per se, there is nothing about the Dirac delta or any of its numerous applications in physics that stipulate that it must be handled in the context of the real numbers specifically. Statements about cardinality are certainly about the real numbers. Statements about the Dirac delta are not. Certainly there weren't to Cauchy. Tkuvho (talk) 17:58, 17 January 2012 (UTC)

I don't think it would be out of line to add a subsection to the definitions section about this hyper real construction. However, I don't think it should be overly stressed. What do you think of a wording like this: "The Dirac delta is not a function in the traditional sense as no function defined on the real numbers has these properties. Most sources rigorously define the Dirac delta function as either a distribution or a measure. It is also possible to construct the delta function as a function on the hyper reals" I would shy away from the wording "The Dirac delta function is not a real function ..." as I think most people would parse this to "The Dirac delta function is not a real valued function ... ." You could then add a section about the hyper real construction after the other definitions. Holmansf (talk) 17:13, 17 January 2012 (UTC)

I think Carl makes a convincing case that the hyperteal construction does not belong in the Definitions section, since it's not actually a definition of the same thing. Perhaps it could be mentioned at the end of the article as a related construction though. Sławomir Biały (talk) 17:25, 17 January 2012 (UTC)

I certainly think mentioning the hyperreal construction in a section at the end is worthwhile, as it is an interesting result. I also like the wording suggested by Holmansf for the lede. I would also try to avoid the wording "true function"; "function in the traditional sense" is much more clear to me. — Carl (CBM · talk) 18:12, 17 January 2012 (UTC)

Break

The infinitesimal section currently refers only to a bibliography by Yamashita. Is anyone willing to look at the original papers by Todorov to see what he actually says? Sławomir Biały (talk) 17:10, 19 January 2012 (UTC)

Which Todorov paper are you referring about? There is a paper of his in AMM as I recall in the 1990s but there were already several articles about the hyperreal Dirac delta by that time in the literature. Tkuvho (talk) 08:07, 20 January 2012 (UTC)

I mean the ones referenced by Yamashita. He's a bit vague on the details, but my impression is that others had only "approximated" distributions by smooth hyperreal functions, while Todorov was able to eliminate the "infinitely near" relation. Anyway, some kind of proper clear reference is highly desirable, whether it be to Todorov or someone else. Sławomir Biały (talk) 11:16, 20 January 2012 (UTC)

Actually, I take it back. You reference the bibliography inline, but there is another paper by Yamashita appearing in the footnote that contains more details. Sławomir Biały (talk) 11:19, 20 January 2012 (UTC)

Doing it "up to an infinitesimal" is already in Cauchy! You can start with any bump function, and use a sequence of rescalings to obtain an internal hyperreal function which evaluates to F(0) up to an infinitesimal at every real continuous function F. I recall there was a paper by a japanese in the 1960s, Giorello in the 1970s, and others before Todorov. Actually, it is surely already in Abraham Robinson, since the implementation of the Dirac delta is one item that Heyting singled out as a particular achievement of non-standard analysis. Tkuvho (talk) 12:58, 22 January 2012 (UTC)

That may be so, but clear inline citations should be added to the relevant section then. The current reference to an incomplete bibliography that references none of what you just mentioned seems poorly selected. Sławomir Biały (talk) 01:56, 23 January 2012 (UTC)

I have to revert myself again. If one starts with a bump function f with compact support and rescales it to obtain a sequence of "nascent" delta functions <f_n>, then the associated internal function [f_n] will indeed vanish at every nonzero real point, as I originally claimed. Therefore we obtain a function defined on the hyperreals which vanishes at every real point such that, when integrated against, will have the effect of the Dirac delta function. This was my original remark: it is not entirely correct to assert that "there is no function with these properties". It is only correct to assert that there is no real function with these properties. Sorry about the confusion. Tkuvho (talk) 15:16, 1 February 2012 (UTC)

The hyperreal delta function thus constructed will take infinite values for suitable infinitesimal values of the input. Tkuvho (talk) 15:17, 1 February 2012 (UTC)

Note that Yamashita does cite an earlier source: G. Takeuti, Proc. Jpn. Acad. 38. 414 (1962). Also, non-standard delta functions are indeed mentioned in Robinson's 1966 book. Tkuvho (talk) 15:03, 15 March 2012 (UTC)

Error under Representations of the delta function - Probabilistic Considerations

The formula under Representations of the delta function - Probabilistic Considerations is wrong. The formula given is

$\eta _{\varepsilon }(x)=\delta _{n}(x)=2n\ {\textrm {rect}}(nx/2)={\begin{cases}2n,&-{\frac {1}{n}}<x<{\frac {1}{n}}\\0,&{\text{otherwise}}.\end{cases}}$

Which is correctly copied from the reference (Aratyn & Rasinariu 2006), but the book has a mistake in, they say that 2n(1/n - (-1/n)) = 1 which is just wrong, it's 4. The correct formula should just be

$\eta _{\varepsilon }(x)={\frac {1}{\varepsilon }}\ {\textrm {rect}}\left({\frac {x}{\varepsilon }}\right)={\begin{cases}{\frac {1}{\varepsilon }},&-{\frac {\varepsilon }{2}}<x<{\frac {\varepsilon }{2}}\\0,&{\text{otherwise}}.\end{cases}}$

I don't know if this counts as "original research" though so I thought I should mention it here before adding it... — Preceding unsigned comment added by 129.11.69.169 (talk) 12:15, 6 March 2012 (UTC)

It's fine by me if you want to make the change. It also seems strange that we would change to a sequence involving n rather than ε. It's not a big issue, but if you can find another reference that would be ideal. Sławomir Biały (talk) 12:25, 6 March 2012 (UTC)

I just noticed Saichev and Woyczyński mention it - they're already listed in the references! Up to a factor of 2 anyway, but that's consistent with η_ε(x) = η₁(x/ε)/ε — Preceding unsigned comment added by 94.192.37.97 (talk) 10:08, 7 March 2012 (UTC)

8th of August 2012

Hello, in the Definition it is said, that the Heavyside step function is 1 for $x\geq 0$, otherwise 0. below that, the characteristic function is said to be in the range $(-\infty ,0]$.

Hasn't this to be $[0,\infty )$

I'm a bit confused, since it appears several times, so it might be on purpose.

Thank you for clearance.

Best

Phil. — Preceding unsigned comment added by UnameAlreadyTaken (talk • contribs) 22:23, 7 August 2012 (UTC)

Where do you see the characteristic function of (-∞,0]? Sławomir Biały (talk) 00:47, 8 August 2012 (UTC)

Elementary representation of the function

Is this function not easily representable as the elementary function ${\frac {0}{x}}$? Pokajanje|Talk 04:22, 18 May 2013 (UTC)

No, it's not. The delta function, as noted in the article, must be thought of as a distribution rather than a function. But see distribution (mathematics) for how to multiply a distribution by a scalar. In particular, zero times any distribution gives the zero distribution. So as a distribution $0/x=0$, even though as a function the left-hand side is not defined at 0. Sławomir Biały (talk) 11:58, 18 May 2013 (UTC)

No, it's not. Your proposed function is everywhere zero except at x=0, where it is undefined. In most contexts, the expression ${\frac {0}{0}}$ can have any value at all, and so it is not a function. To clarify, the Dirac delta is properly defined only in the context of integration. Thus, the Dirac delta should be thought of as similar in usage to an infinitesimal like $dx$. It has no existence by itself in the conventional sense. The Dirac delta is a generalized function and not a function at all. David Spector (talk) 19:45, 26 July 2013 (UTC)

not $x\delta '(x)f(x)=-\delta (x)f(x)$

this does not seem to be correct! Have a look at: http://www.wolframalpha.com/input/?i=integrate+from+-inf+to+inf+f%28x%29*x*delta%28x%29%27%3D-integrate+from+-inf+to+inf+f%28x%29*x*delta%28x%29 . A restriction has to be made!!!!!--92.202.98.107 (talk) 19:32, 28 November 2013 (UTC)

You made an input error in Wolfram Alpha. Please don't continue to insert your addition. Sławomir Biały (talk) 20:41, 28 November 2013 (UTC)

I do not see an input error. Show me what is wrong. In my opinion the "property" does not hold for an "additional test function!". You can also try to apply partitial integration... i dont see why this property should hold in general... So now show me the claimed input error! (In my opinion there is none) — Preceding unsigned comment added by 92.202.98.107 (talk) 20:50, 28 November 2013 (UTC)

Wolfram showed that f(0)=0 if this property holds for all test functions f. This is for sure not true. So this claimed "property" allows no further test function. Please provide a source for your claim.--92.202.98.107 (talk) 20:56, 28 November 2013 (UTC)

Ok, then why would you think that what is written in the article implies that

$\int x\delta '(x)f(x)=-\int x\delta (x)f(x)$

?? I agree that this expression is incorrect, but it's certainly not what the article asserts. Sławomir Biały (talk) 21:03, 28 November 2013 (UTC)

Good that you agree. For sure the article would state $\int x\delta '(x)f(x)dx=-\int x\delta (x)f(x)dx$ without a further restriction because because the two *real* properties just above this one are general! The pseudo property for the test function x arises from partial integration. Therefore it holds for the test function x. In some sense it is interesting (for example I just searched for something like that), but it should be clear that this (pseudo) property of the delta function is not general (it is not a real property of the delta function since it only holds for one specific test function - or the test functions that fulfill f(0)=0).--92.202.98.107 (talk) 21:14, 28 November 2013 (UTC)

0).

What your wolfram alpha link shows is that $x\delta '(x)\not =x\delta (x)$. I agree with this, and it is not inconsistent with the article, which asserts that $x\delta '(x)=\delta (x).$ Do you not see the difference? Sławomir Biały (talk) 21:16, 28 November 2013 (UTC)

I am soo sorry for the inconvenience!!!!! http://www.wolframalpha.com/input/?i=integrate+from+-inf+to+inf+f%28x%29*x*delta%28x%29%27%3D-integrate+from+-inf+to+inf+f%28x%29*delta%28x%29 --92.202.98.107 (talk) 21:26, 28 November 2013 (UTC)

Here is the diff for all my changes: https://en.wikipedia.org/w/index.php?title=Dirac_delta_function&diff=583710621&oldid=577041997 --92.202.98.107 (talk) 21:40, 28 November 2013 (UTC)  Done

Application:Solving differential equations

The Dirac delta function is used to find the impulse response when taking a Laplace transform https://en.wikipedia.org/wiki/Impulse_response. — Preceding unsigned comment added by 66.37.66.96 (talk) 01:19, 21 July 2015 (UTC)

Can we rename it to 'Dirac delta distribution' ?

We say all the time to the students that $\delta$ isn't a function, so it would be nice if the wikipedia article wasn't named... Dirac delta function ! I hope you agree with that ? Can someone do the renaming properly ?

78.196.93.135 (talk) 21:13, 13 July 2016 (UTC)

The guideline for article naming is WP:COMMONNAME. The term "Delta function" is much more common than "Delta distribution", as an immediately recognizable name for the subject of the article. Sławomir Biały (talk) 23:49, 13 July 2016 (UTC)

You are proving you are wrong, the most common in mathematics is of course "Dirac delta distriАbution" while in physics it is probably "Dirac delta function" (anyway Dirac is always mentioned). This is clearly a mathematical article, and letting Dirac delta function in the introduction will let it referenced at 1st on google. — Preceding unsigned comment added by 78.196.93.135 (talk) 17:39, 15 July 2016 (UTC)

The most common in mathematics is certainly not "Dirac delta distribution". For example the Springer Encyclopedia of mathematics article is also entitled "delta function". Indeed, most sources in mathematics as well as physics and engineering use the term "delta function" quite happily. The multivolume authoritative treatise by Israel Gelfand and Georgiy Shilov, for example, uses "delta function". Sławomir Biały (talk) 16:09, 16 July 2016 (UTC)

It appears the article should be renamed. The first sentence of the article even clarifies that the Dirac delta distribution (or whatever we should call it) is not technically speaking a function. Somebody should find a good mathematical source that references it and rename the article as it is currently misleading. Jbeyerl (talk) 14:40, 16 July 2016 (UTC)

What is the basis for the proposed renaming? The subject of the article is almost universally known as the "delta function". Calling it "delta distribution" is very likely to be unfamiliar to most readets. Sławomir Biały (talk) 16:09, 16 July 2016 (UTC)

It seems to fail the precision guideline at Wikipedia:Article_titles because it purports that it is a function. Indeed as far as I know experts usually call it the "Dirac delta function" or "delta function", but their intended audience are typically other experts that know what they mean. Wikipedia should be more accessible to non-experts. The term is a colloquialism or perhaps even an idiom. What about this, leave the title the same but change the first sentence to "In mathematics, the Dirac delta function, or δ function, is a colloquialism used to describe the distribution on the real number line that is zero everywhere except at zero, with an integral of one over the entire real line." Jbeyerl (talk) 21:57, 16 July 2016 (UTC)

Yes experts call it the "delta function", students call it the "delta function", engineers call it the "delta function", physicists call it the "delta function". That's precisely why the title of the article ought to be "delta function". It's the overwhelmingly widely accepted term for the subject of the article. WP:COMMONNAME. Sławomir Biały (talk) 22:45, 16 July 2016 (UTC)

The second paragraph of the article describes the sense in which the delta function is not a function. I don't think the reader is served by calling it a "colloquialism". Anyway, this lacks a reference. It is typically called a "generalized function" in the literature. This has the virtue of being true, well-sourced, and also clarifying the sense in which "function" is appropriate. As I said, the sense in which it is not a function are clearly explained in the second paragraph of the lead, with much more detail in the body of the article for those readers that wish to read further. Sławomir Biały (talk) 12:09, 17 July 2016 (UTC)

This article is titled function but is not about a function hence the very first sentence should make clear what it's actually about. I can't think of a better title, but as you even pointed me toward WP:COMMONNAME, it specifically says "Ambiguous or inaccurate names for the article subject, as determined in reliable sources, are often avoided even though they may be more frequently used by reliable sources." The best that seems doable in this case is clarifying what it means. Jbeyerl (talk) 13:41, 17 July 2016 (UTC)

Please read the second paragraph of the lead of the article. Please discuss and get consensus before reinstating these edits. You are "fixing" nonexistent problems with the article. It is made very clear that the delta function is a generalized function, and not a true function in the strict sense. We do this in a manner supported by very high-quality references, representing mainstream scholarship in the area. Sławomir Biały (talk) 14:27, 17 July 2016 (UTC)

The original point the anon editor pointed out brought out what does appear to be an issue. It's not clear to a novice that a generalized function is not a function. So the article reads as quite confusing to nonexperts. Jbeyerl (talk) 10:43, 18 July 2016 (UTC)

I think we should assume that a novice wishing to find out about the delta function will read more than the first sentence of the article. The entire second paragraph of the article concerns the delta function's lack of functionhood. Calling it an "idiom" is not helpful to a novice, and is borderline WP:OR because it is not supported by the cited sources, which include Gel'fand and Shilov, Dirac, and Schwartz. And I struggle to see how this edit, which is ungrammatical, clarifies anything. Finally, I have said to get consensus first, and you've not done that. Sławomir Biały (talk) 10:49, 18 July 2016 (UTC)

The common name of the thing is the Dirac delta function, and we can't change that. Don't read it as two separate word but as one name. The Sierpinski sponge is not really a sponge. Do you mean we should we rename that article too? For the fine print, in another article it is referred to as a "formal device" with a well-defined distribution backing up its formal behavior rigorously. "Idiom" does not sound right. Could "formal device" be used in descriptive terms here? YohanN7 (talk) 11:56, 18 July 2016 (UTC)

I don't see anything wrong with the current formulation. "Generalized function" already signals that it is not quite a function. The second paragraph then elucidates in what sense this is not a function. Note that the vast majority of readers will not have a precise notion of what a function is (mathematically speaking) in the first place. Introducing woolly terminology like "idiom" or "formal device" will (at best) serve to add confusion, and are certainly not helpful.TR 12:29, 18 July 2016 (UTC)

For a precise description, "generalized function/distribution/continuous linear functional/whatever applies" is fine with me too of course. But the Dirac delta function is USED as a formal device. Few physicists and no engineers bother about what it really is, they just compute using the rules (which say it can formally be treated as a function) of the device. YohanN7 (talk) 12:43, 18 July 2016 (UTC)

I don't think "formal device" is appropriate for the first sentence. One thing that Schwartz notes is that the delta function was quickly adopted into the operational calculus (which we might call the "calculus of formal devices that engineers use"), and along with all of the methods of the operational calculus was viewed with suspicion by most of the mathematicians of the early 20th century. Sławomir Biały (talk) 12:51, 18 July 2016 (UTC)

I've added a little bit on this perspective to the second paragraph. Probably it could be improved. Sławomir Biały (talk) 13:03, 18 July 2016 (UTC)

I didn't mean to be used in the first sentence. It was just an idea, and now reading the lead (something I should perhaps have done in the first place), it is pretty much all there, especially after your edits. YohanN7 (talk) 13:05, 18 July 2016 (UTC)

My 2p as an uninvolved math editor. The term "delta function" is standard in the literature, so it's a good name for the article. "Delta distribution" is not as good because "delta function" can also refer, interchangeably, to the point-mass Dirac measure. I do think the lede should spend more time on the measure-theoretic representation, and not begin with a sentence about the distribution. It might be better to say that the delta function is a hypothetical function with certain properties; it turns out that such a function cannot exist, but similar objects do exist, are widely used, and are called the "delta function". Good luck, — Carl (CBM · talk) 14:57, 18 July 2016 (UTC)

The current second paragraph discusses the non-functionhood of the delta function. Is this what you mean? I don't see any reasonable way to pack the content of the second paragraph into the first paragraph. I think a basic rule of thumb is not to rush into things. It is not a problem if the first sentence misses some nuances. Sławomir Biały (talk) 15:32, 18 July 2016 (UTC)

I mean that, in my opinion, the first paragraph jumps the gun by mentioning the distribution interpetation but not the measure-theoretic one. The term "delta function" can be interpreted as a distribution or as a measure; the first paragraph instead says that it is a distribution. The section "Definitions" does a good job with both definitions, but then the "properties" section assumes we are back to the distribution interpretation, with no comment to that effect. — Carl (CBM · talk) 16:40, 18 July 2016 (UTC)

Well, strictly speaking the measure and the distribution are two different objects, and the term "delta function" really means the distribution rather than the measure. Indeed, there is a separate article Dirac measure. Really, the "distribution" is actually a Radon measure; but this isn't generally what people think of when they say "Dirac measure". They're thinking of the "geometrical" point mass at the origin, rather than its action on test functions obtained by integration. This should perhaps be clarified in the "Definition" section more than it already is. Sławomir Biały (talk) 16:59, 18 July 2016 (UTC)

This is the difference in our perspectives. In my mind, the term "delta function" refers ambiguously to both the distribution and the geometrical point-mass measure, by referring directly to neither, allowing the reader to interpret it either way in many situations, such as in the integral $\int _{-1}^{1}\delta (x)\,dx=1$. In some contexts "delta function" would almost always mean the distribution, but in others it would almost always mean the geometrical measure. If the goal of this article is to describe only the distribution, then I suppose it might make more sense to rename it to a different title and make the delta function article a disambiguation page. — Carl (CBM · talk) 20:27, 18 July 2016 (UTC)

I don't think this is a view that is widely held in the literature. I have not seen in the literature a definition resembling the following: "The delta function is the measure that assigns to a set A of real numbers the integer 1 if 0 is in A and 0 otherwise". One would always say "delta measure" (or "point measure", "unit mass", etc.) in that case. Typically, "Delta function" refers exclusively to the distribution, although I'm willing to be proven wrong by some references of a suitably high quality. Sławomir Biały (talk) 20:52, 18 July 2016 (UTC)

I guess, you could find something like this: "In particular, the delta function is evidently positive, and therefore, it is in fact a measure (in contrast to its derivative)". Boris Tsirelson (talk) 20:58, 18 July 2016 (UTC)

True enough, but this does not really support CMBS contention that distributions are over-emphasized in the article at the expense of measures. It is a theorem that the delta function is the distribution associated to a measure. But one does not usually refer to the measure itself (a function on a sigma algebra of a specific kind) as "delta function". The article discusses distributions and measures. I think giving measures more significance along the lines of what CBM is suggesting does not seem in line with NPOV. Sławomir Biały (talk) 21:21, 18 July 2016 (UTC)

Locally finite (signed) measures are (canonically) embedded into Schwartz distributions, as (for instance) integer numbers are (canonically) embedded into real numbers. Who bothers to specify, is 25 treated as integer or real (unless we do numerics)? Boris Tsirelson (talk) 20:47, 18 July 2016 (UTC)

A quote: “Like the alligator pear that is neither an alligator nor a pear and the biologist’s white ant that is neither white nor an ant, the probabilist’s random variable is neither random nor a variable.” S. Goldberg “Probability: an introduction”, Dower 1986, p. 160. (Alligator pear = avocado; white ant = termite.)

Such is the life. Indeed, also "delta function" is not a function. Boris Tsirelson (talk) 20:00, 18 July 2016 (UTC)

In addition: "finite measure" is a measure, but "signed measure", "vector measure" and "finitely additive measure" are (generally) not measures. Boris Tsirelson (talk) 20:41, 18 July 2016 (UTC)

---

This is a maths article, and I'm quite sure all the people who approve Dirac delta function are indeed not mathematicians, but engineers or physicists instead. In maths we say all the time that $\delta$ is a DISTRIBUTION, and its name is depending on the context : Dirac delta or Dirac delta distribution. In french, they renamed it to fr:Distribution_de_Dirac Please also understand it is different to the Kronecker delta, a function this time. This is addressed in particular to @Sławomir Biały: you said your opinion, let the other participate too, tks. 78.196.93.135 (talk) 03:52, 7 August 2016 (UTC)

You are mistaken. CBM and myself are both mathematicians. I wrote most of the article, that you think is a "maths article". But it's not really important what we think. It's important what reliable sources call it. The Springer Encyclopedia of Mathematics calls is the "delta function". Israel Gel'fand and Georgiy Shilov call it the "delta function" in their authoritative treatise "Generalized functions". Laurent Schwartz calls it "la function de Dirac". Nelson Dunford and Jacob Schwartz call it the δ-function in their "Linear operators" text. These are all mathematical works, written by the 20th century's most distinguished authorities on the theory of distributions. So, they cannot simply be dismissed as the works of physicists and engineers. Finally, the neutral point of view policy strongly urges against dismissing other points of view in this discussion. So, to get a crude estimate of how common "delta function" is versus "delta distribution" in the literature, we can compare Google scholar hits: "Dirac delta distribution" versus "Dirac delta function". Thus "Dirac delta function" is almost 20 times more common than "Dirac delta distribution" in the scholarly literature. The scholar search without the word "Dirac" shows a similar 20-fold preference of "function" over "distribution": "delta distribution" versus "delta function". While this certainly has engineering and physics mixed with mathematical sources, I've already demonstrated that even exceptionally high mathematical sources quite happily call it the "Dirac delta function". Sławomir Biały (talk) 11:30, 7 August 2016 (UTC)

you said your opinion : follow the wikipedia guideline, rely on old textbooks, some of them written before distributions had been even invented (~1950). Whereas our is : this is a maths article, and in mathematics the un-official guideline is to use the technical term whenever it is possible, since maths articles are mainly read by maths/science students. Here, the problem is that students discover $\delta (x)$ with the Fourier series, 2-3 years before they study the distributions, so there are 2-3 years while it is very possible they think that $\delta (x)$ is a function, and this is very error prone (see any maths forum, you'll see what I mean). The last but not the least argument, is that wikipedia is becoming the standard in science, i.e. the name you choose today for a wikipedia article will be the name tomorrow's teachers will use in class. 78.196.93.135 (talk) 23:26, 16 August 2016 (UTC)

I understand that your opinion is that the article should be renamed to "Dirac delta distribution". However, opinions of random people on the internet carry very little weight, and Wikipedia is not a democracy. We go by reliable sources, which I have given several, and Wikipedia guidelines, which clearly say we should use the more common term and I have demonstrated that sources overwhelmingly favor "function" in this context. Unless arguments are based on our policies and guidelines, they are likely to be disregarded. Sławomir Biały (talk) 23:42, 16 August 2016 (UTC)

Also, for some reason IP wants to disregard from consideration books written by the originators of the theory of distributions (Schwartz, Gel'fand). A much more recent work is Michael E. Taylor's three-volume work "Partial differential equations", wherein he refers to the delta function. Sławomir Biały (talk) 12:01, 19 August 2016 (UTC)

@ Ip: If you look at the top of this page you'll see that this article is within the scope of physics and engineering, hence not a pure math article. So even if your arguments had been correct (and they aren't) for a pure math article, this isn't such an article. YohanN7 (talk) 08:41, 17 August 2016 (UTC)

Obligatory google ngram test for "Dirac delta function" vs. "Dirac delta distribution" for a WP:COMMONNAME discussion. No contest in this case.TR 10:48, 17 August 2016 (UTC)

Ok then look also at the results for "dirac delta"+"is not a function" compared to "dirac delta" you'll find 70000 vs 500000 results. In other words, your choice is error prone, and a very bad idea from the scientific point of view. I'll conclude by saying wikipedia is the 1st result on google, so the name on wikipedia is the name that teachers and students will use for the 30 years to come...

And @YohanN7 : if you want to say that my arguments aren't correct, you have to explain why (and from what point of view). In particular you have to explain if you explain why the Dirac delta is a distribution and not a function (which is quite a good reason for not naming this article Dirac delta function). 78.196.93.135 (talk) 02:33, 14 November 2016 (UTC)

Riemannian metrics aren't metrics. They are Riemannian metrics. No problem there, right? There is no particular inherent problem with delta functions not being functions either. It is simply what the things are called. YohanN7 (talk) 09:58, 14 November 2016 (UTC)

The article very carefully explains why the delta function is not a function. It would need to do this whether the title of the article was "dirac delta function" or "dirac delta distribution", for two reasons. First, because it is overwhelmingly called "delta function" in the literature, it is necessary to convey this because of readers arriving here by typing "delta function" into the search bar. Secondly, this is a notable aspect of the distribution that a large number of high quality sources discuss. Furthermore, if we were to change title to "Dirac delta distribution", most readers would arrive here by a redirect, and we would need to spend more time, not less, explaining that the subject of the article is usually called the "Dirac delta function", in addition to discussion about why it is not a function. Sławomir Biały (talk) 11:14, 14 November 2016 (UTC)

Note that your comments make no sense what-so-ever with regard to ngram (For example, ngram does not support a "+" join, and returns only relative values, not absolute one). I think you are mistaking it with a standard google search. Ngram simply compares the occurrences of specific phrases in the printed literature over time. It is a great tool for determining which of alternative phrases (or spellings of the same word) is more common (and how this may have changed in recent history).TR 11:46, 14 November 2016 (UTC)

The "not a function" also refers to that it is not a real-valued function of a real variable. I think it is correct to say that it can be modeled as a perfectly acceptable (set-theoretic) function from the reals to the extended reals. The extended reals can certainly be modeled as a set. Begin with defining positive infinity to be any set except a real number (reals can be modeled as sets), e.g. the set of the real numbers itself, which certainly isn't a real number, thus put ∞ ≡ ℝ. (Define negative infinity if desired, define order, topology and algebraic operations in the nearly obvious ways if desired.) The result is that the delta function is a function. This is not very useful, but is i m o worth mentioning here on the talk page since so many get so involved in this issue. YohanN7 (talk) 12:36, 15 November 2016 (UTC)

That's actually not the case. The resulting extended real function would be measurable, and it would have Lebesgue integral equal to zero, not one. Sławomir Biały (talk) 13:05, 15 November 2016 (UTC)

You miss my point. It is a function, but it isn't very useful. Obviosuly, the Lebesgue integral with Lebesgue measure isn't what we have. Just a set-function. YohanN7 (talk) 13:10, 15 November 2016 (UTC)

But this set function would in no way model the delta function. That's a "function" whose integral is one that is equal to zero away from the origin. There is no function (in the set theoretical sense) with this property, extended real-valued or otherwise. Sławomir Biały (talk) 14:53, 15 November 2016 (UTC)

Point taken. But equally well, can't you put the quotation marks around "integral" (or call it a Dirac delta integral)?. YohanN7 (talk) 15:20, 15 November 2016 (UTC)

@YohanN7: @Sławomir Biały: Here in the comments, example of a mathematical student who thinks that the Dirac delta is a function because you on wikipedia don't understand what I'm saying. The introduction of the article is very unclear : "From a purely mathematical viewpoint, the Dirac delta is not strictly a function". Why not simply say that the Dirac delta is NOT a function ? 78.196.93.135 (talk) 21:26, 30 May 2017 (UTC)

(re to 78.196.93.135) Yes, why not? Done. YohanN7 (talk) 07:15, 31 May 2017 (UTC)

I'll take Gelfand, Shilov, and Schwartz over random anonymous people on the internet, thanks. The second paragraph already elaborates on the sense in which it is not a function, and how to make the idea of the Dirac delta function rigorous. Anyone who reads the paragraph and gets the idea that the Delta function is actually a proper mathematical function has issues with reading comprehension that are well beyond our scope to fix. Sławomir Biały (talk) 22:04, 30 May 2017 (UTC)

Obviously you didn't understand Gelfand, Shilov, and Schwartz. How do you explain everything I wrote ? How do you explain that many students think it is a function ? How do you explain I'm coming here to complain ? You have issues understanding something as simple as wikipedia is used as a main mathematical ressource by millions of students, and we have to make it clear that $\delta$ is not a function. Also I said it 5 times : we understood what you had to say, no need to repeat it. And when reading YohanN7's comment about $\delta$ being a function from $\mathbb {R}$ to $\mathbb {R} \cup \{\infty \}$ I'm really worried. No $\delta$ is not a function. It is a measure, a generalized-function or a distribution. The most useful definition in Fourier analysis is as the linear operator

$\int _{-\infty }^{\infty }\delta (t)\varphi (t)dt{\overset {def}{=}}\lim _{\epsilon \to 0^{+}}\int _{-\infty }^{\infty }{\frac {1_{|t|<\epsilon }}{2\epsilon }}\varphi (t)dt=\lim _{\epsilon \to 0^{+}}{\frac {1}{2\epsilon }}\int _{-\epsilon }^{\epsilon }\varphi (t)dt$ which reduces to $\int _{-\infty }^{\infty }\delta (t)\varphi (t)dt=\varphi (0)$ whenever $\varphi$ is continuous at $0$. 78.196.93.135 (talk) 00:25, 31 May 2017 (UTC)

I think perhaps it is your own reading comprehension that is a problem if you believe that people in this discussion think that the delta function is really a mathematical function, or if you think that the article gives this impression either. Also, the belief that this is the "most useful definition in Fourier analysis" is naive and wrong. I suspect it was copied from some textbook without understanding the wider mathematical context. Sławomir Biały (talk) 01:05, 31 May 2017 (UTC)

"Dirac delta function" is nothing but one of Oddities of mathematical terminology. Wikipedia mirrors the world. See also List of types of functions#More general objects still called functions. Boris Tsirelson (talk) 08:52, 31 May 2017 (UTC)