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On Wikipedia, the definition of the dirac delta function is given as:

enter image description here

Suppose I have a function where at two points, the function goes to infinity. Given that the distance between the two points is $a$, if I take $a$ tends to 0, will I get a Dirac delta function? That is, when the two spikes "superimpose". The domain is [0,a] and the function is zero everywhere in between the two end points.

enter image description here

Korra
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  • If both the points you mention satisfy the integral and area is 1 under them individually then together they would have area 2, so technically no, it won't be a Dirac delta. – abkds Aug 25 '19 at 06:10
  • Do you mean a function like $f(x):=\dfrac{1}{x^2(x-a)^2}$ ? I would be tempted to say that you are complicating things, because already $f(x):=\dfrac{1}{x^2}$ isn't assimilable to Dirac's $\delta$... – Jean Marie Aug 25 '19 at 06:12
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    The wikipedia article is a mess, written by poor phisicists. If $f$ is $L^1$ with $C = \int_{-\infty}^\infty f(x)dx$ then $n f(n.)$ converges to $C\delta$ is the sense of distributions, in particular for any $\phi$ continuous and bounded then $\lim_{n \to \infty} \int_{-\infty}^\infty n f(nx) \phi(x)dx=C \phi(0)$. In any case you can't think to $\delta$ in term of its "values". It is outside of the set of functions in the same way that real numbers aren't rational numbers. – reuns Aug 25 '19 at 06:18
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    The Dirac delta function is a mess, written by poor physicists. $\leftarrow$ Fixed that for you. –  Aug 25 '19 at 06:20
  • @abkds, I don't understand how the area will be two. I am taking the distance between the two points as 0. Isn't $area=length×breadth$? – Korra Aug 25 '19 at 06:24
  • I hope you agree with me @MatthewDaly given the first formula they give is $\delta(x-\alpha)=\frac{1}{2\pi} \int_{-\infty}^\infty dp\ \cos (px-p\alpha) $ – reuns Aug 25 '19 at 06:24
  • The Dirac delta distribution is the limit in the sense of distributions of any sequence of functions $f_n$ such that $\lim_{n \to \infty} \int_{-\infty}^\infty f_n(x) \phi(x)dx= \phi(0)$ for any $\phi$ smooth bounded or compactly supported. That's the only one way to think about it. – reuns Aug 25 '19 at 06:25
  • Korra, if you are saying that the value is infintite over the entire interval between those values, then you are correct. @abkds was probably assuming that you were talking about a graph with two "spikes". –  Aug 25 '19 at 06:27
  • @Matthew Daly, abkds was assuming correct. But what I am saying is, if the two spikes are brought "on top of each other", then how is the area 2? Should I include a diagram to show what I mean in the question. – Korra Aug 25 '19 at 06:31
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    You're getting exactly to the heart of why mathematicians fight this concept. The height of the spike when they're superimposed is $2\infty$, so the area should be 2. But $2\infty=\infty$, so the area should be 1. The truth is that the area under the zero function with one discontinuity is 0 no matter how high that discontinuity is. But this concept seems to keep the physicists and engineers productive, God bless them. –  Aug 25 '19 at 06:37
  • @Matthew Daly I am asking regarding a physics problem. The problem will be much easier if what I assume above is correct. – Korra Aug 25 '19 at 07:08
  • You might have better luck asking at the Physics SE. I honestly don't know if you're right or if you should divide your answer by two. They'd probably appreciate the context better than we would. –  Aug 25 '19 at 07:12
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    @Matthew Daly oh no, my question would just be put as off topic. Thanks for the response though. – Korra Aug 25 '19 at 07:30

2 Answers2

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The Dirac delta as function takes continuous functions as arguments and returns their value at $x=0$. That is, another name would be "evaluation operator". In still another characterization, it is the unit of convolution, $f*\delta=f$.

If you want to think in real functions, you can only sensibly define approximations of the behavior of the delta distribution. A robust class of such approximations is given by function sequences $\phi_n\ge 0$ with $\int_{\Bbb R}\phi_n dx=1$ and $\phi_n(x)\to0$ for $x\ne 0$. Examples are $\phi_n(x)=n\phi_1(nx)$ where $\phi_1$ satisfies the above conditions.

A rectangular pulse of width $a$ and height $1/a$ located around the origin (meaning the center converges to zero for $a\to 0$) would also satisfy these conditions, as $a$ goes to zero, $1/a$ goes to infinity, and the shape of the graph stays rectangular.

What does not work is a function that is not integrable, and not a function in the strict sense. Your description of a function jumping to infinity and staying there for a positive length is such a non-function.

Manipulation of the Dirac delta as function follows intuitive rules, $\lim_{a\to 0}\delta_a+\delta_0=2\delta_0$. What you can not do is start with the informal characterization of infinity peaks, as it is not precise enough. An infinity peak can correspond to $0\delta$, $\delta$, $3\delta$ or something without any useful finite results.

Lutz Lehmann
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Suppose a sequence $\delta_n$ of functions is proposed to distributionally satisfy $\delta_n\stackrel{n\to\infty}{\to}\delta$ (it has to be distributional convergence; it can't be pointwise, because $\delta$ isn't a true function). For $f$ a Schwartz function,$$f(0)=\int_{\Bbb R}\delta(x)f(x)dx=\lim_{n\to\infty}\int_{\Bbb R}\delta_n(x)f(x)dx.$$Your proposal is of the form $\delta_n=\infty\sum_{i=0}^1[x=ia_n],\,a_n\to0$, in terms of the Iverson bracket. Unfortunately, this gives$$\int_{\Bbb R}\delta_n(x)f(x)dx=0\implies\lim_{n\to\infty}\int_{\Bbb R}\delta_n(x)f(x)dx=0.$$What you should look for is nascent delta functions of the form $\delta_n(x):=ng\left(nx\right)$ with $g$ a pdf, so$$\lim_{n\to\infty}\int_{\Bbb R}\delta_n(x)f(x)dx=\lim_{n\to\infty}\int_{\Bbb R}g(y)f(y/n)dy=\int_{\Bbb R}g(y)\left[\lim_{n\to\infty}f(y/n)\right]dy=f(0)\int_{\Bbb R}g(y)dy=f(0).$$

J.G.
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  • No, my function doesn't satisfy the first condition and it's just these two points I am talking about. Points at the start and end of my domain [0,a] only at those points it is $\inf$ and is zero everywhere in between – Korra Aug 25 '19 at 09:09
  • @Korra I know. That's why I wrote $\infty\sum_i[x=ia]$ instead of $\infty[0\le x\le a]$. – J.G. Aug 25 '19 at 12:44
  • Then I don't understand your first statement – Korra Aug 25 '19 at 14:18