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The delta function is not a function between real numbers, but a construct only to be used as a limit expression of the functional defined by various models for an approximation of this delta function with the pertinent function being integrated alongside with it.
Thank you very much for the most obvious explanation, but what is the intuition of the integral of the product of a function and Dirac distribution? I mean what does such integral express, mathematically and physically?
The Dirac distribution is the Fourier transform of unity and a special case of convolution, where A*f=g, g(x)=d(x-y). f(y)dy , if we imagine the gravitational interaction as a function of g(x) and the electromagnetic interaction as a function of f(y), then these forces (i.e. the lines of force) only interact when x is equal to y ( the Dirac impulse).
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The definition of Dirac delta in this video is wrong. Ali Grami, in Introduction to Digital Communications, 2016 3.5.9 Dirac Delta or Unit Impulse Function The Dirac delta or unit impulse function is a singularity function, and defined mathematically to provide a very useful tool for representing a physical phenomenon that occurs in an extremely short period of time, which is too short to be measured, and with an extremely large amplitude. It is an even function and the total area under it is unity. The Dirac delta or unit impulse function δ(t) is defined by having zero amplitude everywhere except at t=0 where it is infinitely large (unbounded).
It is a generalization of a function, in this case a functional. The mapping that relates variables for a functional is not necessarily the same as a traditional function (although your comment hints that you understand it isn't a standard function), so it's not fair to call his definition "wrong" if you're not specifying this difference of mapping between a standard function and another definition that generalizes functions to functionals. You're arguing against convention rather than utility at that point. Neither of you are wrong. You can certainly argue that a function is just a limiting case of a functional over a given vector space, but when most people talk about functionals they are referring to a mapping of a vector space V over a field F onto a field F (a function with a function as the input). A traditional function is the limiting case where the input itself is constant.
Since I never saw it discussed, I must protest against the fact that f(x)δ(x)=f(0)δ(x). What I mean is, this property is true, but it is also true for every a in the domain of f. That is f(x)δ(x)=f(a)δ(x) for every constant a we want, it doesn't have to be 0. Also f(x)δ(x)=δ(x), since at x=0, we have infinity=infinity, while at x different from 0, we'll have 0=0. So these functions are all equal. And so, it would follow that the integral of f(x)δ(x) with the property that the integral of δ(x) is 1, is any value of f we want, and that's utter non-sense. Would you care to explain how am I wrong here? I found 0 exaplanations regarding this obvious inconsistency.
The way I personally like to think of the delta function is δ(x) = 1/dx if x = 0, and δ(x) = 0 otherwise. That way, if you do the integral ∫ f(x)δ(x − a)dx over ℝ, you first integrate over ℝ\{a}, and since for all x in ℝ different than a, δ(x − a) = 0, you get 0 for that whole portion of the integral. What's left to integrate over is the singleton {a}. Summing over a single element is like getting rid of the integral sign and simply plugging-in x = a. So in the end we have f(a)δ(0)dx = f(a)dx/dx = f(a).
consider the functions g, h defined by : g(x) = f(x)δ(x) h(x) = f(a)δ(x) for any a fixed and f a random function if those are equal then their integral from -∞ to +∞ must be too so ∫f(x)δ(x)dx = f(a)∫δ(x)dx so f(0) = f(a) which is absurd as we chose f any function unless a=0. We can do the same to disprove f(x)δ(x) = δ(x) By integrating we get f(0) = 1 for any function f
No, you don't, because the integral sign is actually an abuse of notation. What you do is not an integral operation but the application of a functional to a test function.
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This is not only true for the delta function! Any function f(x) shifts to the positive x-direction, if you subtract x0 from the argument: f(x-x0). And if you add x0 to the argument, f(x+x0), then the function shifts to the negative x-direction. Plot the quadratic functions f(x) = x^2 and f(x-x0) = (x-x0)^2 then you will see that f(x-x0) = (x-x0)^2 is shifted to the positive direction. Feel free to join the polls I'm doing in the community tab. There you can vote for the next video topic: ua-cam.com/users/universaldenker-physicscommunity If you regularly visit my channel and like my videos, I would be happy about 1 x coffee ☕ per month: ua-cam.com/users/universaldenker-physicsjoin
Support me if you want me to create more theoretical physics videos: ua-cam.com/users/universaldenker-physicsjoin
Do you want to solve every physics problem? Try my eBook!
✅ Contains many useful physics formulas.
✅ Understandable for everyone, because it contains no vectors and integrals.
✅ Formulas are colored and visualized.
✅ Perfect for high school and undergraduate physics students.
Click on the following link to get the book:
en.universaldenker.org/physics-equations-book
I'm extremely grateful I came across this video. The level of clarity in this video is insane! Thanks man, you're amazing.
The best demonstration of Dirac Delta Function on the net. Very Nice.
Possibly the best explanation for the Dirac delta function that you can find. Thank you for your hard work.
This was one of the best and most succinct explanations I saw on yt. Congrats and please do a video on green's functions!
Thanks, will do it at some point! Please join the polls in community tab, which video I should make next.
You make 14 yo understand this, Thank you so much ❤
I like your definition of δ(0) = 1, as it highlights the δ-function's role as a filter.
This actually is undefined..
@@rauldurand The Dirac Delta Function is always a pointed subject!
The delta function is not a function between real numbers, but a construct only to be used as a limit expression of the functional defined by various models for an approximation of this delta function with the pertinent function being integrated alongside with it.
The best explanation!
So beautifully explained and visualized❤️
i never comment but this video is amazing universaldenker. thank you so much.
Since i wacthed ur video on eletromagnetism i knew i found a gold mine. Have been binging since.
Quality animation with superb explanation!
wow Muhammad, thank you for your comment!
Thank you very much for the most obvious explanation, but what is the intuition of the integral of the product of a function and Dirac distribution? I mean what does such integral express, mathematically and physically?
What an amazing explanation! Thank you!
Hey andres! Thank you for your comment!
Exactly the video I needed to see right now. Beautifully done!!!
nice! Thank you very much 👍
Such an awesome explaination! Thank you very much, Sir.
Thank you for your kind comment!
The Dirac distribution is the Fourier transform of unity and a special case of convolution, where A*f=g, g(x)=d(x-y). f(y)dy , if we imagine the gravitational interaction as a function of g(x) and the electromagnetic interaction as a function of f(y), then these forces (i.e. the lines of force) only interact when x is equal to y ( the Dirac impulse).
Great video. Clearly explained. Thanks.
Thank you!
There's a detailed discussion on the Delta Function in Ch. 2 of "Green's Functions with Applications" by Dean G. Duffy.
These videos are gold, I have no idea why they don't get more attention
Thank you very much! Maybe that will change with time!
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Thanks a lot
Thanks
Thank you too for the comment!
Great video!
Thank you very much! 🙂
Videos excelentes, muy didáctica sus explicaciones.
gracias!
The definition of Dirac delta in this video is wrong. Ali Grami, in Introduction to Digital Communications, 2016
3.5.9 Dirac Delta or Unit Impulse Function
The Dirac delta or unit impulse function is a singularity function, and defined mathematically to provide a very useful tool for representing a physical phenomenon that occurs in an extremely short period of time, which is too short to be measured, and with an extremely large amplitude. It is an even function and the total area under it is unity. The Dirac delta or unit impulse function δ(t) is defined by having zero amplitude everywhere except at t=0 where it is infinitely large (unbounded).
It is a generalization of a function, in this case a functional. The mapping that relates variables for a functional is not necessarily the same as a traditional function (although your comment hints that you understand it isn't a standard function), so it's not fair to call his definition "wrong" if you're not specifying this difference of mapping between a standard function and another definition that generalizes functions to functionals. You're arguing against convention rather than utility at that point. Neither of you are wrong.
You can certainly argue that a function is just a limiting case of a functional over a given vector space, but when most people talk about functionals they are referring to a mapping of a vector space V over a field F onto a field F (a function with a function as the input). A traditional function is the limiting case where the input itself is constant.
Wow.. so helpful
Thank you :)
thank you soo much
Thank you for your feedback, prachiti!
Love it!
Thank you very much, Christian!
Awesome video! Thank you!
Awesome!!
Thank you!
Since I never saw it discussed, I must protest against the fact that f(x)δ(x)=f(0)δ(x). What I mean is, this property is true, but it is also true for every a in the domain of f. That is f(x)δ(x)=f(a)δ(x) for every constant a we want, it doesn't have to be 0. Also f(x)δ(x)=δ(x), since at x=0, we have infinity=infinity, while at x different from 0, we'll have 0=0. So these functions are all equal. And so, it would follow that the integral of f(x)δ(x) with the property that the integral of δ(x) is 1, is any value of f we want, and that's utter non-sense. Would you care to explain how am I wrong here? I found 0 exaplanations regarding this obvious inconsistency.
The way I personally like to think of the delta function is δ(x) = 1/dx if x = 0, and δ(x) = 0 otherwise. That way, if you do the integral ∫ f(x)δ(x − a)dx over ℝ, you first integrate over ℝ\{a}, and since for all x in ℝ different than a, δ(x − a) = 0, you get 0 for that whole portion of the integral. What's left to integrate over is the singleton {a}. Summing over a single element is like getting rid of the integral sign and simply plugging-in x = a. So in the end we have f(a)δ(0)dx = f(a)dx/dx = f(a).
consider the functions g, h defined by :
g(x) = f(x)δ(x)
h(x) = f(a)δ(x) for any a fixed and f a random function
if those are equal then their integral from -∞ to +∞ must be too
so ∫f(x)δ(x)dx = f(a)∫δ(x)dx
so f(0) = f(a) which is absurd as we chose f any function unless a=0.
We can do the same to disprove f(x)δ(x) = δ(x)
By integrating we get f(0) = 1 for any function f
at ur examples 7:40, dont u have to integrate first before u set in your value. I mean cos x integrated is sinx and then u place pi inside right?=
No, you don't, because the integral sign is actually an abuse of notation. What you do is not an integral operation but the application of a functional to a test function.
You save me from going mad
Not gone into madness/10, would watch again
Nice!
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Still dont quite understand why shifitng in positive x direction, the delat function become delta(x-x0) why the -ve i never understand that
This is not only true for the delta function! Any function f(x) shifts to the positive x-direction, if you subtract x0 from the argument: f(x-x0). And if you add x0 to the argument, f(x+x0), then the function shifts to the negative x-direction. Plot the quadratic functions f(x) = x^2 and f(x-x0) = (x-x0)^2 then you will see that f(x-x0) = (x-x0)^2 is shifted to the positive direction.
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That the value is ¥|∆√°
语言不通好难受，好想听懂，那么多优秀的英文讲解
❤🔥
thanks
Thanks for such an amazing explanation!
Thank you for your feedback, Hwayeon!
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