Distributions: Convolution product

In summary, P1 tried to solve a problem where H*δ'=δ and P2 has problems with unclearityes in the F[δ*δ] equation. P1 found that F[δ*δ]= C F[u*v]F[v] where C is a constant and T1 is the transform of the delta distribution. P2 found that <δ*δ,F[φ]>=<δ,<T1,φ>=<F[φ](0)=∫φe^0dx=∫φdx=∫1φdx=<T1,φ> which proves that
  • #1
DIrtyPio
18
0
So I have some problems and I tried to resolve them, I also have the results so I can check them but I'm curious if I made them good.

P1: (H*δ)'=?, where H is the heavisede distrobution and δ is diracs distributin. So I tried liek this : <(H*δ)',φ>=-<H*δ,φ'>=-<δ,<H,φ'>>, <H,φ'>=∫φ'dx=φ => -<δ,φ>=<(H*δ)',φ> => (H*δ)'=δ. This is the result given so I think it's OK.

P2:Here is my real problem, here I have doubts-unclearityes so it looks like this :
F[δ*δ]=?, Where F[] is the Fourier transform. My solution: <F[δ*δ],φ>=<δ*δ,F[φ]>=
<δ,<δ,F[φ]>, <δ,F[φ]>=F[φ](0)=∫φe^0dx=∫φdx=∫1φdx=<T1,φ> =>
<δ,<δ,F[φ]>=<δ,<T1,φ>> So I by definition <δ,φ>=φ(0) so this is<T1,φ>(0) but I don't really know what tis means... does it mean <T1,φ>? Because if it means this then the solution given by the book is correct and is T1, but if it is not or even if it is could anyone tell me why is it like this ? Because in ∫φdx(0) I don't really see the reason.
 
Physics news on Phys.org
  • #2
P1:

Something weird is going on when you are removing the convolutions here:
DIrtyPio said:
-<H*δ,φ'>=-<δ,<H,φ'>>

I'm not sure what you're doing, but on the face of it, it seems false. <H,φ'> is a number, but there it is used as if it is a function. There may be some subtle notation issue here but I don't see it. Could you please elaborate on what you are thinking of when you do this step?

When I look at -<H*δ,φ'>, the part that immediately jumps out to me is that H*δ can be rewritten as just H (the delta function is the identity under convolution), and if you do that substitution then the result falls out.

P2:Here is my real problem, here I have doubts-unclearityes so it looks like this :
F[δ*δ]=?, Where F[] is the Fourier transform. My solution: <F[δ*δ],φ>=<δ*δ,F[φ]>=
<δ,<δ,F[φ]>, <δ,F[φ]>=F[φ](0)=∫φe^0dx=∫φdx=∫1φdx=<T1,φ> =>
<δ,<δ,F[φ]>=<δ,<T1,φ>> So I by definition <δ,φ>=φ(0) so this is<T1,φ>(0) but I don't really know what tis means... does it mean <T1,φ>? Because if it means this then the solution given by the book is correct and is T1, but if it is not or even if it is could anyone tell me why is it like this ? Because in ∫φdx(0) I don't really see the reason.

The same issue about convolutions as part a arises, so I don't really see what you're doing.

The way I would go about attacking this is to use 2 facts. Fact 1: F[u*v] = C FF[v] where the constant C is (2pi)^(d/2) or whatever depending on your Fourier transform convention. I think this "fact" is often even use as the definition for the convolution of distributions that have locally integrable Fourier transforms. Fact 2: F[δ] = 1/C. Combining these 2 facts you can solve the problem.

For intuition as to why F[δ] = const: the Fourier transform takes functions bunched up at zero and spreads them out, and takes functions that are spread out and bunches them up at zero, while keeping the norm the same. The happy medium is a gaussian of standard width, whose Fourier transform is itself. For skinnier and skinnier functions, their Fourier transform get wider and wider. For the most ultimately skinny "function" - the delta distribution - the Fourier transform is the widest function - a constant. This result can be proven rigorously using the definitions (<F[δ],v>=<δ,F[v]>=F[v](0)= 1/C int v = <1/C,v> for v in the Schwarz space)
 
Last edited:
  • #3
-<H*δ,φ'>=-<δ,<H,φ'>> this is the definiton of the convolution product in a distribution, as I know( www.emis.de/journals/NSJOM/Papers/23_1/NSJOM_23_1_013_027.pdf ). But you are talking abot F[δ] which I did not write anywhere. As you see my problem is that i can't compute <δ,<T1,φ>> , where as you could figure out T1<=>f(x)=1 .
 
  • #4
Oh I see. They are somewhat abusing the notation in the document. They use the shorthand notation:

"<(f*g)(x),φ(x)>" = ∫ f(y) ∫ g(x)φ(x+y) dxdy := "<f(y),<g(x),φ(x+y)>>"

I really don't like the way they are doing it since it is easy to get confused about what is a function and what is a number, and what is the argument of a function vs a parameter. But OK, let's do it their way. Then the step in question is:

<δ(y),<T1(x),φ(x+y)>> = <δ(y), ∫ T1(x)φ(x+y) dx> = (∫ φ(x+y) dx)(0) = ∫φ(x+0) dx = <T1,φ>
 
  • #5
Thank you very much.
 

1. What is a convolution product?

A convolution product is a mathematical operation that combines two functions to create a third function. It is used to describe the distribution of a sum of two independent random variables. It can also be thought of as a way to modify the probability distribution of a random variable by passing it through a filter or modifying function.

2. How is a convolution product calculated?

The convolution product is calculated by integrating the product of two functions over all possible values of the variables. In probability theory, this is equivalent to finding the probability distribution of the sum of two independent random variables by taking the convolution of their individual probability distributions.

3. What is the significance of the convolution product in statistics?

The convolution product is significant in statistics because it allows for the description of complex distributions in terms of simpler distributions. It is also used in signal processing to analyze and manipulate signals. Additionally, it plays a key role in the central limit theorem, which states that the sum of a large number of independent random variables will tend towards a normal distribution.

4. Can the convolution product be applied to continuous and discrete distributions?

Yes, the convolution product can be applied to both continuous and discrete distributions. In the case of continuous distributions, the convolution integral is used. For discrete distributions, the convolution sum is used. However, the general principles and properties of the convolution product remain the same for both types of distributions.

5. What are some real-world applications of the convolution product?

The convolution product has many real-world applications, including in signal processing, probability theory, and statistics. It is used to model the sum of independent random variables in fields such as finance, engineering, and physics. It is also used in image and audio processing to enhance and filter signals. Additionally, the convolution product is used in machine learning algorithms to combine multiple features or inputs into a single output.

Similar threads

  • Calculus and Beyond Homework Help
Replies
16
Views
1K
  • Topology and Analysis
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
3K
  • Calculus and Beyond Homework Help
Replies
4
Views
979
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
20
Views
3K
  • Advanced Physics Homework Help
Replies
28
Views
3K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
Back
Top