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From the Wikipedia page, dirac delta function is described such as:

Description 1: "whose value is zero everywhere except at zero" or

if you scroll down a little bit more, you will find:

Description 2:

$ \delta(x)= \begin{cases} 0&x \neq 0\\ \infty&x=0 \end{cases} $

While reading this, it made me believe that such function(even though it's not a function), its value is only at origin and anywhere else, it's $0$. This is what the above says.

Then, It also shows one of the family of function that "successfully" mimics the behaviour:

$\displaystyle\delta_b(x) = \lim_{b \to 0} \frac{1}{|b| \sqrt{\pi}} e^{-(\frac{x}{b})^2}$

Question: note that the following function doesn't correctly obey the description 1 or either description 2. If you calculate $\delta_b(x)$ at any $x>0.0000001$, it's zero, but just use $x=0.00000001$(extra zero), and the limit returns $\infty$. So we just found $x$ that is not $0$, but for which $\delta$ is not $0$. Could you explain which description is correct or what's going on ? I'm not good at math, so would appreciate the clear/easy explanation.

Math Attack
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Giorgi
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  • The limit is obviously zero for any $x \ne 0$. This is because $e^{-x^2/b^2}$ will go to $0$ a lot faster than $1/|b|$ could ever go to $\infty$ (each as $b \to 0$). Your logic for why this should not be the case is incredibly unclear and you should do some further effort to look into it and justify your conclusion, because right now it's hand-waving nonsense at best. – PrincessEev Nov 23 '23 at 21:52
  • What you observe is just a consequence of your calculator or whatever you use to "calculate" the limit. It is zero – Lorago Nov 23 '23 at 21:53
  • Though yours does have this property (as they say above, your observation is a numerical artifact), it's actually not necessary. For instance, $\sin(Rt)/(\pi t)$ in the limit $R\to\infty$ serves perfectly well as a smooth form of the delta function. See here for more info. – spaceisdarkgreen Nov 23 '23 at 22:01
  • Really ? I checked it on wolframalpha. I wonder why that is, maybe it doesn't have that many decimals support. – Giorgi Nov 23 '23 at 22:01

1 Answers1

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The meaning of the Dirac delta is not the same as what you are used to with "standard functions" (the Dirac delta is a distribution in fact), in fact the definition $$\delta(x)=\begin{cases}\infty&x= 0\\0&x\neq 0\end{cases}$$ it is not very useful for application purposes.

The most useful property of the Dirac delta is the fact that: $$\int_{-\infty}^{\infty}\delta(x)\mathrm{d}x=1$$ However, the meaning of the delta should not be sought in the sense of the functions where you said (for example) $f(5)=\pi$ but it must be found in relation to another function (generally called "test function")

The delta property comes into play when you do operations like this

$$\int_{-\infty}^{\infty}\delta(x-x_0)f(x)\mathrm{d}x=f(x_0)$$

From here you see that the use of delta is to sample functions when doing integrals (you don't evaluate the $\delta$ at a point like in functions but you evaluate other functions by means of integrals)

The fact that it is presented to you as the limit of $$\delta_b(x) = \lim_{b \to 0} \frac{1}{|b| \sqrt{\pi}} e^{-(\frac{x}{b})^2}$$ is just a construction to make you see that the function tends to $0$ everywhere except $0$ and in parallel you have that the area subtended by the function always remains $1$.

Math Attack
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  • Thanks for the answer. is it my mistake that for $x= 0.00000001$, it's truly 0 and not $\infty$ as wolframalpha gives out ? – Giorgi Nov 23 '23 at 22:04
  • WolframAlpha and calculators are not infallible. If you want to be sure, actually do the work by hand. – PrincessEev Nov 23 '23 at 22:05
  • @GiorgiLagidze to me it says $0$, but in general the delta is always $0$ if the argument inside it is different from $0$ https://www.wolframalpha.com/input?i=delta%2810%5E%28-8%29%29 – Math Attack Nov 23 '23 at 22:06
  • Okay, just to be sure 100%, $\delta_b(x) = \lim_{b \to 0} \frac{1}{|b| \sqrt{\pi}} e^{-(\frac{x}{b})^2}$ is really 0 for EVERY value of $x$, doesn't matter how small $x$ is, right ? and it's only $\infty$ at $x=0$. All good ? By hand calculations, this is what I get. – Giorgi Nov 23 '23 at 22:07
  • "Distribution" is the accurate way of seeing it. The thing that gave me helpful intuition was thinking of a point particle eg a concentration of mass at a point (also in terms of discrete probability distributions). There is a fuzzing of the distinction between discrete and continuous. – Mark Bennet Nov 23 '23 at 22:08
  • @GiorgiLagidze Correct, $\delta_\infty(x)\to 0$ for $x\neq 0$ and $\delta_\infty(x)\to\infty$ for $b=0$ – Math Attack Nov 23 '23 at 22:09
  • @MathAttack can you click on that ? https://www.wolframalpha.com/input?i=limit+calculator&assumption=%7B%22F%22%2C+%22Limit%22%2C+%22limit%22%7D+-%3E%220%22&assumption=%7B%22F%22%2C+%22Limit%22%2C+%22limitfunction%22%7D+-%3E%221%2F%28%5Csqrt%7B%CF%80%7D%7Cx%7C%29e%5E%7B-%280.00000001%2Fx%29%5E2%7D%22&assumption=%22FSelect%22+-%3E+%7B%7B%22Limit%22%7D%2C+%22dflt%22%7D – Giorgi Nov 23 '23 at 22:10
  • @GiorgiLagidze Wolfram is a calculator that works in finite precision, if he sees an extremely high number he considers it infinite (in the same way if he sees very small numbers he might consider them 0) – Math Attack Nov 23 '23 at 22:11
  • Got it. It seems like my number is not that small. it's $10^-8$, but i guess it fails. hmm. – Giorgi Nov 23 '23 at 22:12
  • @GiorgiLagidze Keep in mind that very often when working with Wolfram (especially if you deal with complex numbers) they could return 2 results at the same time: the first of the type $z=4.312312+7.312\cdot 10^{-24}i$ and the second of the type $z=4.312312$. The first result is the calculated one which is calculated in finite precision and gives you a very small imaginary part, the second is the Wolfram interpretation which considers it $0$. Example: try $\tan(3+10^{-50} i)$ – Math Attack Nov 23 '23 at 22:16
  • @MathAttack Hey. Wanted to ask something. What really drives me crazy about this concept is that some say that it's infinite at $x=0$ and 0 everywhere else, but also some explain as limit and narrow spike. If it's a narrow spike, it can't be that it's zero everywhere, because close to $x=0$, values won't be 0. So which one is it ? – Giorgi Nov 25 '23 at 11:11