Well, in Physics we use this very often, also in combination with Fourier series, and very much like a function. But of course we keep in mind that it is actually a short pulse with short -> 0
@stayawayfrommrrogers5 жыл бұрын
I would totally enjoy a series on distributions.
@imgayasheck5955 жыл бұрын
Dr peyam you make me love dual spaces because they are very versatile
@johannesh76105 жыл бұрын
Man, functions of functions are more and more fun the more of your videos I see about them!
@tylershepard42695 жыл бұрын
This is very neat. In electrical engineering we study impulse responses a lot because it contains one frequency. In fact summing all cosine overtures at the same amplitude results in a periodic dirac delta “functions.” Even though it’s quite useful for this reason, you never see an actual impulse. A voltage will follow a very sharp Gaussian envelope instead. Even more so, that’s how ultra wide and works. We transmit sharp Gaussian envelopes to transmit over a wide variety of frequencies. I know this was long, but it’s amazing how useful it is!
@smg00035 жыл бұрын
U just solved a physics problem. Two slit experiment: - when wave passes through a slit it must slow to virtually zero so energy goes sideways as fast as u can't obviously blink an eye at, to relieve pressure. However slit has walls so side bounce gangs up between sides then once through other side it has forward + sideways movement Cheers mate
@Tomaplen5 жыл бұрын
everyone: Yeah i know its not a function its obvious etc etc.. only Dr Peyam: Here is why booom
@jaronfeld1235 жыл бұрын
This one was seriously amazing. I loved the Dual Space work a lot. I work in Riemann Geometry and Sub-Manifolds so I love the linear algebra+analysis combinations. This was fun!
@RalphDratman5 жыл бұрын
This is so much fun! I have never seen such joy in connection with mathematics. Thank you Peyam.
@shafihaidery8489 ай бұрын
such a life saver, thank you so much
@dgrandlapinblanc5 жыл бұрын
Hard but always interesting. Thank you very much.
@정대영-l1e5 жыл бұрын
Thank you for this video. Now I got some points of distribution and how linear algebra becomes powerful.
@imgayasheck5955 жыл бұрын
I'm just going to say this before watching: MEASURE
@Murat-fi6ff5 жыл бұрын
Thank u Payem👏👏👏
@kevin3265205 жыл бұрын
Wonderful video! I have always hated this "function" when it was introduced in my physic classes, but after I learn more about math, I see that its nature is actually that of a functional. This made me more comfortable when thinking of Dirac Delta. Your video gives me a clear understanding of what it is, at least in its basic definition. Thanks a lot. Just some quick note, at 17:27 you talk about taking open interval to deal with the problem if the chosen x being ±1. I wouldn't suggest that since it might give rise to situations that you don't get a finite number after integration, e.g. g(x) = 1/(x-1). You may just assume that x, while not being 0, is also not ±1. You can deal with that with continuity, like what you were going to do in 20:15.
@LobsterInSuit5 жыл бұрын
Very nice video, I hope you keep talking about functional analysis! It would be interesting to touch some elements like tempered distributions and Schwartz spaces!
@mehdisi91945 жыл бұрын
Very nice video. Thank you dr.peyam so much. If possible, make a video about distributions.
@youssefbenhachem9935 жыл бұрын
Fantastic !
@patryk_495 жыл бұрын
Can you make a video about inverse laplace transform of 1 or derivative of dirac delta function?
@drpeyam5 жыл бұрын
I don’t know anything about the Laplace transform! And the derivative is just the functional T(f) = f’(0)
@johannesh76105 жыл бұрын
@@drpeyam The Fourier-Transform of the delta function is just constant 1. But doing the inverse transformation is actually a non converging integral which integrates / sums up the unit circle infinitely often. Still, it makes much sense to say it is still the delta function. I don't know the exact difference to a Laplace transform
@Jose_Hunters_EWF_Remixes5 жыл бұрын
Would a less restrictive condition on f hat suffice? For example, could we consider the case of measurable functions on [-1,1]? My impression would be "yes".
@Bombelus5 жыл бұрын
20:20 - On the left side you've just proved that "d != f hat", but on right side showing what "d = f hat" looks like. I take it that the latter "f hat" is different from the former. Also, isn't the latter "f hat" just a non-continuous function (which happens zero everywhere but at zero)?
@Jose_Hunters_EWF_Remixes5 жыл бұрын
So, in the finite dimensional case, V and V* are isomorphic as vector spaces? Aren't all n-dimensional vector spaces isomorphic (as vector spaces over the same field) to all other n-dimensional vector spaces?
@drpeyam5 жыл бұрын
In finite dimensions, they are! It’s in infinite dimensions that they aren’t
@shandyverdyo76885 жыл бұрын
I still don't understand what is the defference between "a function" and "a distribution" dr? :'(
@Aviationlover-belugaxl5 жыл бұрын
Shandy Verdyo Distributions are functionals from functions p(x) (maybe with certain properties) to a number. All functions have a distribution form but the reverse is not true. The Dirac delta is an example of this.
@mike4ty45 жыл бұрын
It's not a bad question: in fact, distributions _are_ a kind of function. They just aren't functions that take _real numbers_ for their input. Rather, they are functions which take _other functions_ as input - they are an example of so-called "higher-order functions" or "functionals". When we say "the Dirac delta function is 'not a function'", what we mean is that it is not a function _that takes in a real number_ like the functions studied in first-year Calculus, and cannot be made into such. But we want to kind of find some clever way to treat it like it were, because it's too damn useful when thought of that way. And that's what distributions + a little bit of added machinery let us do. They aren't the only way, however. Measures are another, and perhaps a much more profound approach that actually allows us to come very close to the naive intuition of an "infinitely thin and infinitely tall spike with an 'infinity' of 'just the right size' to have an area of 1 underneath it", is the use of _non-standard analysis_ . However, this last approach requires a deep dive into mathematical logic to make work, but some consider the payoff worth the cost.
@newtonnewtonnewton15875 жыл бұрын
Nice D peyam السلام عليكم ورحمة الله
@michalbotor5 жыл бұрын
dr peyam i was wondering whether you're familiar with the theory of distributions as conceived by the polish mathematicians: jan mikusiński and roman sikorski in the 50s? essentially they understood distributions being to functions, what cantor understood real numbers being to the rational numbers, namely the limits of some suitable equivallent sequences of the latter. it goes as follows: 1. let C(a, b) be a space of continous functions on the interval (a, b), where -oo
@chirayu_jain5 жыл бұрын
Nice
@cameronspalding97924 жыл бұрын
With the delta function: can it only be said to belong to L1
@ShubhayanKabir4 жыл бұрын
Idk why but I just start by liking the video. 😅
@MrGeorge18965 жыл бұрын
So the anti derivative of δ(x) is h(x) with h(x)=0 for x=0 (constant omitted), ok? So we can find a solution for y'"+y=δ(x) with y(x)=A*sin(x)+B*cos(x)+h(x)*sin(x) when we consider δ(x)*f(x)=0 if f(0)=0 and δ(x)*f(0) if f(0) is not 0 (and f(x) somewhat nice). So d/dx(h(x)*sin(x))=h(x)*cos(x)+δ(x)*sin(x)=h(x)*cos(x) and so d²/dx²(h(x)*sin(x))=-h(x)*sin(x)+δ(x)*cos(x)=-h(x)*sin(x)+δ(x)*1. While it is weird it surprisingly works...
@koenth23595 жыл бұрын
Aaaah your necktie!
@cameronspalding97924 жыл бұрын
The delta functions are the eigenfunctions of the position operators
@leonardromano14915 жыл бұрын
And then there's people who square Dirac deltas lol. In QFT we sometimes identify delta^n with V^(n-1)*delta where V is the system volume.
@anonymaus81915 жыл бұрын
No, it's not a function. Continuous functions do not form a Hilbert space. It's not Cauchy complete.
@D-Bar4 жыл бұрын
From what I understand, there is no continuous function whose hat is the Dirac delta. But what about discontinuous functions? You used continuity in the proof, but why can’t the Dirac delta, while not continuous, be a discontinuous function?
@drpeyam4 жыл бұрын
Still not because if f is a function and f = 0 almost everywhere then the integral of f should be 0, but that’s not true for Dirac
@D-Bar4 жыл бұрын
Dr Peyam Oh that’s right. Thank you for replying!
@xy94395 жыл бұрын
10:55 😂
@blackpenredpen5 жыл бұрын
Adàlia Ramon Hahaha!!
@RalphDratman5 жыл бұрын
Yes that is funny!
@aleksandervadla48405 жыл бұрын
Could someone solve this equation?: pi•x^2=pi^(x^2) There are 4 solutions. It is easy to SEE that two of them are 1 and -1, but I cant find the two remaining solutions.....
@554558172963123456785 жыл бұрын
I don't think that equation can be solved analytically - Try solving it numerically instead?
@MrGeorge18965 жыл бұрын
@@55455817296312345678 if you allow the product logarithm function to be used it can be solved analytically. en.wikipedia.org/wiki/Lambert_W_function