Anonymous at Tue, 22 Oct 2024 05:08:33 UTC No. 16443634

>>16443631
Your question is nonsensical. Convolution is defined on function spaces, not the reals.

Anonymous at Tue, 22 Oct 2024 05:09:53 UTC No. 16443636

>>16443634
Sorry, but how is this question nonsensicle? I am looking for an integrable and bounded function from the reals to the reals which satisfies an equation involving convolution.

Anonymous at Tue, 22 Oct 2024 05:10:41 UTC No. 16443638

Fixed formatting

Is there a function [math] g: R -> R [/math] such that [math] g^2=g*g [/math], where * denotes convolution? We can take g to be integrable and bounded. This problem is hard because it’s both nonlinear and non-local. Obvious strategies all fail, with simple inequalities failing to rule out a solution and Fourier transforming getting you back to where you started.

Attempting to do power series or other forms of expansion end poorly because what is well behaved for the square term isn’t for the autoconvolution, and vice versa. Simple inequalities show that no such solution can exist which is supported on (1/e, infinity), even over the shared domain of [math] g^2 [/math] and [math] g*g [/math]. In general, no finite support works because the convolution will double it.

It’s worth noting that a solution to the problem can be horizontally rescaled to a solution for [math] g^2= c g*g [/math]. This isn’t the case if we define the problem over the domain (0,1) and take the convolution to be periodic (modulo 1). In that case, Jensen immediately rules out solutions to [math] g^2= c g*g [/math] unless c>1. Beyond that I have no idea. Another way to solve the problem would be to find a function who is its own Fourier transform, and whose square is its own Fourier transform as well. This is evidently difficult. Any new strategies would be greatly appreciated.

Anonymous at Tue, 22 Oct 2024 05:16:08 UTC No. 16443646

>>16443636
So you mean to say you’re looking for a convolution of constant functions? It’s not possible because the functions you’re convoluting must satisfy /int f(x)dx = c, where c is finite. It’s in the definition.

Anonymous at Tue, 22 Oct 2024 05:25:26 UTC No. 16443653

>>16443646
No. Please read the question closely. I am looking for a function for which the operations of squaring and Fourier transforming commute.

Anonymous at Tue, 22 Oct 2024 05:30:02 UTC No. 16443658

>>16443653
I read the question closely. You need to learn how to define functions properly, because, I repeat, the domain of convolution is a function space, not the reals.

Anonymous at Tue, 22 Oct 2024 06:18:23 UTC No. 16443699

>>16443658
>R -> R
are you trolling? my function is g: R->R, a function from the reals to the reals and the convolution of g with itself is g*g. g*g is also a function from the reals to the reals. I am clearly using convolution to map a function to another function.

Anonymous at Tue, 22 Oct 2024 07:00:19 UTC No. 16443735

>>16443699
Nigger, how many times do you have to say this? A convolution is a binary operator on functions. Not on the reals. And its codomain is also a function space. Literally look at the fucking formula, you blind retard.

Anonymous at Tue, 22 Oct 2024 07:09:50 UTC No. 16443744

>>16443699
what he is not telling you is that function spaces are generally defined on an infinite dimensional vector space, if you understood what that meant then you would not be arguing, a convolution of g with itself is basically defined on the cartesian product gxg, not the reals

Anonymous at Tue, 22 Oct 2024 07:31:24 UTC No. 16443751

>>16443744
What does this have to do with the definition? Cartesian products are defined on all sets, not just vector spaces.

Anonymous at Tue, 22 Oct 2024 07:40:48 UTC No. 16443757

>>16443751
its easier to understand cartesian products than inifnite dimensional vector spaces since op doesn't seem to get the difference btn reals and function spaces

Anonymous at Tue, 22 Oct 2024 07:41:06 UTC No. 16443758

>>16443735
please ignore the idiot, he wont be of any help to your problem

Anonymous at Tue, 22 Oct 2024 08:09:43 UTC No. 16443775

Dirac delta function

Anonymous at Tue, 22 Oct 2024 08:12:39 UTC No. 16443778

>>16443758
I am the one explaining to the idiot that he can’t even formulate the problem properly.

Anonymous at Tue, 22 Oct 2024 08:13:33 UTC No. 16443779

>>16443775
not bounded or integrable or even a function

Anonymous at Tue, 22 Oct 2024 08:14:58 UTC No. 16443782

>>16443778
The problem is completely fine and you're a larping retarded pseud. g is a real function (suitable for convolution). conv(g,g) is a real function. g^2 is a real function. What is the problem?

Anonymous at Tue, 22 Oct 2024 08:18:06 UTC No. 16443784

>>16443779
Its integral is one, dickhead. This is a very well defined function used in engineering and physics all the time. Go back to college.

Anonymous at Tue, 22 Oct 2024 08:21:54 UTC No. 16443790

>>16443782
So is there a function such that g times g equals conv(g,g)? Jesus, what a convoluted way of formulating this? g^2 in abstract algebra may denote any binary operator, which is why I was confused. It seemed to me like your question was “is there a function such that it can be convoluted with itself?” Fucking anal guys and their implicit assumptions.

Anonymous at Tue, 22 Oct 2024 08:22:55 UTC No. 16443792

>>16443784
It’s not a function, but a distribution, engineerlet. Its codomain is undefined.

Anonymous at Tue, 22 Oct 2024 08:23:48 UTC No. 16443793

>>16443790
It's not my question, but I understood wtf he was talking about in 0.5 seconds because it's the only thing he could possibly mean.
>anal guys
You're the one having an autistic meltdown here, my friend.

Anonymous at Tue, 22 Oct 2024 08:25:33 UTC No. 16443795

>>16443793
>because it’s the only thing he could mean
no, because g^2 could simply be a shorthand for conv(g,g). Very standard practice in algebra.

Anonymous at Tue, 22 Oct 2024 08:26:22 UTC No. 16443797

>>16443795
Yes but that would be a completely retarded question, wouldn't it? g^2 is also very common shorthand for g to the power of 2.

Anonymous at Tue, 22 Oct 2024 08:26:30 UTC No. 16443798

>>16443792
It fucking works for this thread and you can't admit it because you just can't be wrong.

Anonymous at Tue, 22 Oct 2024 08:28:51 UTC No. 16443801

>>16443798
No it doesn't. OP asked for a function.

Anonymous at Tue, 22 Oct 2024 08:29:29 UTC No. 16443803

>>16443798
It’s not what OP asks for. He asks for a function. The Dirac delta isn’t a function. Wtf is even a square of the Dirac delta in the first place?

Anonymous at Tue, 22 Oct 2024 08:30:15 UTC No. 16443804

>>16443801
It is a function. Dirac function. English much?

Anonymous at Tue, 22 Oct 2024 08:30:50 UTC No. 16443805

>>16443797
>Yes but that would be a completely retarded question, wouldn't it?
Which is why I was confused. I work with too many abstract operators to just assume that we’re talking about good ol’ multiplication.

Anonymous at Tue, 22 Oct 2024 08:31:05 UTC No. 16443806

>>16443804
Unfortunately we're talking about mathematics, not English.

Anonymous at Tue, 22 Oct 2024 08:31:25 UTC No. 16443807

>>16443803
The square of the dirac delta function is itself. Look it up.

Anonymous at Tue, 22 Oct 2024 08:31:35 UTC No. 16443808

>>16443805
OK, thanks for sharing

Anonymous at Tue, 22 Oct 2024 08:32:02 UTC No. 16443810

>>16443804
What’s the codomain, Mr Engineer? Tell me the set it maps to.

Anonymous at Tue, 22 Oct 2024 08:33:01 UTC No. 16443811

>>16443806
Unfortunately for you, it's a well defined function. You need to go back.

Anonymous at Tue, 22 Oct 2024 08:33:41 UTC No. 16443813

>>16443811
So, what's your definition of a function?

Anonymous at Tue, 22 Oct 2024 08:35:15 UTC No. 16443815

>>16443813
Didn’t you read? He thinks it’s a function because it has that name in English. Emf must also be a force according to that retard.

Anonymous at Tue, 22 Oct 2024 08:35:41 UTC No. 16443816

>>16443810
[eqn]\delta(f) = f(0)[/eqn]
It maps any real valued function to a real number by evaluating it at zero.

Anonymous at Tue, 22 Oct 2024 08:36:04 UTC No. 16443817

>>16443815
Emf IS a force. Duh.

Anonymous at Tue, 22 Oct 2024 08:40:22 UTC No. 16443819

>>16443816
So is delta's domain the reals or the function space? And what's delta(0) then? A real number? Holy fuck, you're retarded. Please leave this thread.

Anonymous at Tue, 22 Oct 2024 08:44:38 UTC No. 16443823

>>16443819
The domain of delta is a function space.
delta(0) = 0
if you intepret the 0 in the input as the neutral element of addition in that function space which is also the constant zero function.

Anonymous at Tue, 22 Oct 2024 08:46:01 UTC No. 16443824

>>16443823
OK but then it isn't a real function and doesn't answer the question

Anonymous at Tue, 22 Oct 2024 08:47:15 UTC No. 16443826

>>16443823
Great, so it's not the function OP asks for. He asks a function with the reals as its domain.

Anonymous at Tue, 22 Oct 2024 08:48:56 UTC No. 16443827

The obvious answer would be a Gaussian, since squaring it or convoluting it with itself will give another Gaussian. But the numbers don't add up. Maybe you could invent some function that's a power series of Gaussians or something but I think it's probably impossible.

Anonymous at Tue, 22 Oct 2024 08:51:14 UTC No. 16443831

try defining an arbitrary function, e = conv(f,g) and its lp norm where p=2 then see for what conditions that becomes equal to the lp norm of 1/f*g

Anonymous at Tue, 22 Oct 2024 09:03:57 UTC No. 16443842

>OP is retarded for asking a question more advanced than calc 1 on 4chan
>idiot he's arguing with is retarded for not being able to understand the question
>person who suggested delta function is retarded for missing the part where the range is supposed to be R
>I'm retarded for not having an answer but analysis isn't my thing so fuck you

Anonymous at Tue, 22 Oct 2024 15:54:16 UTC No. 16444268

>>16443807
In general you won't be able to define squares of distributions like this, either as limits in some metric space or even just as an operator which maps other functions to values. In this specific problem, a delta function won't work because the convolution of a delta function with itself has value "delta(0)" at 0, whereas the square has value "delta(0) squared" at 0, which is "bigger." There is no limiting behavior which the delta function solution will get right.

Anonymous at Tue, 22 Oct 2024 15:57:11 UTC No. 16444270

>>16443827
Yes, this was my first guess as well, but as you correctly point out it is hopeless. This is because convolution doubles a gaussian's width (adding two random variables doubles variance), wheras squaring reduces it.

However, I am able to find infinitely many solutions to the linearized version of the problem [math] fg=f*g [/math] by taking f and g to be gaussians with complex variance. Their variance should have amplitude one and differ in phase by [math] 2 \pi /3 [/math]. Unfortunately this solution can not be ported over to the nonlinear case.

Anonymous at Tue, 22 Oct 2024 15:57:43 UTC No. 16444271

>>16443842
thank you for understanding...

Anonymous at Wed, 23 Oct 2024 12:28:15 UTC No. 16445615

>>16443631
Bumping for interest.
I've thought about this for a bit and I believe it's not possible, but I can't think of a convincing argument.
I'll do some numerical attempts later.

Anonymous at Wed, 23 Oct 2024 12:57:59 UTC No. 16445629

>range has to be R
just drop this requirwment

Anonymous at Wed, 23 Oct 2024 17:10:38 UTC No. 16445998

>>16443631
i could find it for a multivariate function, but otherwise you might have to settle with a numerical solution of some sort.

Anonymous at Wed, 23 Oct 2024 17:39:08 UTC No. 16446047

>>16443631
I was thinking you should just use a properly scaled gaussian.
g(x) = exp(-pi*x^2) should work.

Anonymous at Wed, 23 Oct 2024 17:44:04 UTC No. 16446058

>>16443631
yes. g identically 0

Anonymous at Wed, 23 Oct 2024 18:26:25 UTC No. 16446124

>>16446047
Take a Gaussian exp(-a x^2). The square gives you exp(-2a x^2), and the convolution gives you something proportional to exp(-a x^2/2). So it doesn't work.

Anonymous at Wed, 23 Oct 2024 21:43:17 UTC No. 16446398

>>16443631
Let a such that -2=<a=<2, and a=/=0
a
Let b= (arcsin(a/2)+2pi * n)/a, for n in the integers
Let C be any Real number
We get f(x) = cos(ax + C) from -b to b inclusive and 0 everywhere else, satisfies the condition
f(x)^2 = f(x) convoluted with itself.
Quick proof (ill omit the C constant so C=0 but its the same result just more steps):
cos^2(ax)= 1/2 * cos(ax+ax) cos(ax - ax) = (cos(2ax) + 1)/2
Now the integral only has non zero values between -b and b since that's how we defined the function
Integral from -b to b cos(ax)cos(ax-at)dt = integral -b to b 1/2(cos(2ax -at) + cos(at))dt.
We can now seperate the integral (and ill skip to the result but you can plug it in any solver online):
2cos^2(ax)sin(ab)/a = (cos(2ax)+1)sin(ab)/a
Good now we make the square expression equal to the convolution one and we get:
sin(ab)/a=1/2.
The solutions to this equation is our choice of b up above (by isolating b)

Anonymous at Wed, 23 Oct 2024 22:05:25 UTC No. 16446418

>>16446398
Im retarded, disregard entire post

Anonymous at Wed, 23 Oct 2024 22:43:39 UTC No. 16446464

>>16443631
>>16443638

Why are you wasting so much time when it sounds like you basically have all the tools to solve this?

Just integrate both sides of g^2 = g*g and you get
[math] \int g^2 dt = \left(\int g dt\right)^2. [/math]
Can you find a g that solves this? No, not unless you are taking convolution over a periodic unit interval and g is constant. Or if you are discretizing the integrals and you take g to be a Kronecker delta (which is the Fourier transform of the constant periodic case).

If you are considering convolution over the entire real line the answer is simply no.

Anonymous at Thu, 24 Oct 2024 17:20:38 UTC No. 16447536

>>16446464
>Can you find a g that solves this?
Yes, very easily. Even a function that is 1 between 0 and 1 and vanishes otherwise can do that much.

Anonymous at Thu, 24 Oct 2024 17:27:21 UTC No. 16447548

>>16447536
>Even a function that is 1 between 0 and 1 and vanishes otherwise can do that much.
Good, now you understand why I said a constant over a unit interval works. Can you find something less trivial? No

Anonymous at Thu, 24 Oct 2024 18:33:21 UTC No. 16447642

>>16447548
>constant over a unit interval works
You did not actually say that, you were babbling about periodic intervals or something.
Anyway, there's a million such functions.
Just take
[math]
g(x) =x^2 e^{-\frac{32 \pi}{9}x^2},\\
g(x) = \frac{\sin( \pi x)}{x},
[/math]
or any number of other examples. You can make pretty much anything work.

Anonymous at Thu, 24 Oct 2024 19:38:49 UTC No. 16447749

>>16447642
>You did not actually say that, you were babbling about periodic intervals or something.
Consider the convolution integral. If you want to have a constant function over a unit interval, a piecewise function defined over the entire real line doesn't work, you need to have it be periodic.

Anyway I concede your main point that it is easier to find nontrivial function satisfying the integral expression than I thought.

Anonymous at Sat, 26 Oct 2024 09:15:16 UTC No. 16450216

>>16443638
yeah