That approximate identity function gn(x) reminds me a lot of the delta function (distribution, for technicality), the one shaped like an infinite spike. If you multiply the delta function (centered at a) with the function you want and integrate from -infinity to +infinity, you just get the original function evaluated at a

@kieransquared

2 жыл бұрын

Great observation! Approximate identities approximate the dirac delta distribution, which is an identity for convolution.

@daleamon2547

Жыл бұрын

In EE we called it the Dirac Delta. Heavily used in transforms.

@elkinmontoya9640

Жыл бұрын

@@kieransquared Would Dirac's deltas be the (unique) limit of all of them? Does this expression even make sense?

@yakov9ify

Жыл бұрын

@@elkinmontoya9640 you can see this limit defined in multiple ways. First you can define it in the space of distributions of locally integrable functions (functionals satisfying some properties), I believe in this case the convergence is unique but I'd have to double check. You can also think of these as probability measures, in which case this converges is what we call "weak convergence" of probability measures, in which case the limit is definitely unique.

this video is so great :). Its helping me with my class. I also participated in the SOME competition last summer.

As a recent Real Analysis student, it was excellent to watch this video. Convolutions certainly are very powerful.

Great video! I never knew about the connection between polynomial approximations and convolution. Taking this video a step further, it was neat to see how in the limit, these polybomial approximations behave just like a delta function, which is consistent with how my signal processing classes define convolution!

As with a lot of things the notion of convolution and its mathematical expressions relation to mixing can be made clear by moving to the realm of algebra and looking at what a convolution means in the context of a group. It's nice to see people take a less algebraic approach, i really liked this video.

Absolutely amazing video! Subscribed.

Great video!! You have a gift at teaching.

Hey (kieran)² ! I really enjoyed the video but I believe some constructive criticism could be appretiated. My main point of criticism for this video is some inconsistency in the redundancy and lack of explanation in some sense. For example, I dont believe explaining that an integral is the area below a curve is neccesary for this video since I doubt that anyone that has heard about convolutions is that unfamiliar with calculus. On the other hand, I think that some relevant information after 8 minutes has been glossed over. Other than that, the video style is great and removing the sutle humming noise in the background would add a cherry on top of your overall video quality. I am looking forward to your next videos and you've just earned yourself another subscriber :)

very good video, done very well, keep it up

Nice video, very clearly explained !

Hella UNDERRATED video. Needs a higher view count

Really liked this video, great job! Do you have any suggestions for resources that go in-depth on all the facts whose proofs you glossed over?

@kieransquared

2 жыл бұрын

Yes! Walter Rudin’s Principles of Mathematical Analysis goes over pretty much the same proof of the Weierstrass theorem, although with unfortunately not many details. This goes over it in more detail: bondmatt.files.wordpress.com/2009/09/weierstrass2-01.pdf

excellent

Very good. What happened to the contest?

A point about the continuous average: With the normal discrete average, we ask "Suppose all of these data points had the same value. What must that value be, if it keeps the same total as the original data?". So what we have is a sum over all a_i points of the original data, equaling a sum over all b_i points of the uniform data, which in turn equals N * b, where b = b_i for all i, and N is the number of points of both data sets. Then, to get the value of b, we divide the sum over a_i by the value of N. This is the average formula we're familiar with. Now, with the continuous average, we ask "Suppose this function were constant. What must its value be, if it keeps the same area as the original function (on this interval)?". So we have some shape given by the function over the interval, and we want to know the necessary height of a rectangle with the same base, and area of that shape. Of course, that will be the area divided by the length of the base, which is the integral of the function over the interval, divided by the length of the interval.

Nice! never think before about the moving average as a convolution with the step function. I look in Wikipedia about the Weierstrass Approximation Theorem and there are mentioned the use of the Bernsteins' Polynomial for the convolution kernel on closed intervals, but those look different from the polynomials Qn(x) you have used on your explanation: ¿How are called this specific family of polynomial Qn(x)=c_n (1-x^2)^n?

I really wish the video shows some examples of what this convolution with Q_n (besides Gaussians) looks like! For example for the square wave or triangle waves.

Beautiful concepts! Now i only wonder why these 4 properties are sufficient for the approximation property. And like always when it comes to convolution i just wonder how this relates to fourier transforms. Thanks:)

I wonder about more examples of approximate identities

Isn't it a problem that Q_n you defined at the end of the video is not really a polynomial? It is a polynomial on [-1,1] and is zero outside. Is the convolution of Q_n with f a polynomial?

@kieransquared

2 жыл бұрын

Since we’re approximating f by polynomials on a closed interval, we don’t need Q_n to be a polynomial everywhere. Still, the definition of convolution and some messy computations should verify that f convolved with Q_n is a polynomial everywhere, provided f is integrable. Q_n(t-x) is a polynomial in x (which is defined for all x) so it comes out of the integral and then the integral in t just gives you the coefficients of the polynomial.

@amirzandieh2793

2 жыл бұрын

@@kieransquared I see. Yes I agree that the integral in t should give the coefficients but it seems to me that the limits of integration will depend on x because of the way Q_n was defined, specifically I think the limits will be x-1 and x+1. So the coefficients will depend on the antiderivative of f at points x-1 and x+1 which varies as a function of x. Am I missing something?

@tdtrtgrytftuih4486

Жыл бұрын

Amir, you’re not missing something, you’re right. I arrive at the same conclusion with different logic: Its pretty simple to show, for instance, that if f(x)=e^x and f*g is well defined then f*g is some constant (namely (f*g)(0)) times e^x, meaning it is not a polynomial. In this case g=Q_n does make this well defined and so you indeed get an exponential, not a polynomial. Your story seems to be falling apart, Kieran.

2:00 I don't think that theorem is complete, since there are continuous functions that are not differentiable at some or all of their points. One should speak of continuous functions and with numbers of derivatives of the same order as each term of the polynomial. Well, speaking of the video, it was excellent.👍

@kieransquared

2 жыл бұрын

You can have a sequence of polynomials converging to a nondifferentiable function, the limit of a sequence of polynomials doesn’t need to itself be differentiable. For example, try to compute a sequence of polynomials converging to |x| using the formula I give later in the video. In fact, not every limit of a sequence of polynomials is even continuous - for example, x^n on [0,1] converges to zero when x is in [0,1) and 1 when x = 1. This is why uniform convergence is an important conclusion of the Weierstrass approximation theorem, because a uniform limit of continuous functions IS continuous (hence x^n does not converge uniformly).

@pablogh1204

2 жыл бұрын

@@kieransquared Thanks for the clarification and I hope you keep uploading videos like this one. 👍

@kieransquared

2 жыл бұрын

@@pablogh1204 Of course! Glad you enjoyed it :)

Ok genius iam also approaching the problem same way like you I don't use matheMatical way my question is so simple because LTI depends on convulution here's my question below Convulution is nothing but stacks and scales the input that's why the input to an amplifier is stacked and scaled or amplified but in filter design it attenuate frequency so I don't know how it regret certain frequency by stacking and scaling the input if possible Please some one explain to me