Semigroup property for convolution

In summary: Laplace_transform#Properties_and_theoremsIn summary, the conversation discussed the task of proving the semigroup property with respect to convolution for a family of functions defined as $$f_{a}(x) = \frac{1}{a \pi} \cdot \frac{1}{1 + \frac{x^{2}}{a^{2}}}$$ The conversation touched on using the Fourier transform and integration techniques to prove the property, ultimately leading to the solution of using partial fraction decomposition or the residue theorem to simplify the convolution integral. Sources were provided for future reference.
  • #1
AVBs2Systems
39
24
Summary: Show that for this family of functions the following semigroup property with respect to convolution holds.

Hi.

My task is to prove that for the family of functions defined as:

$$
f_{a}(x) = \frac{1}{a \pi} \cdot \frac{1}{1 + \frac{x^{2}}{a^{2}} }
$$
The following semigroup property with respect to convolution holds:
$$
f_{a}(x) * f_{b}(x) = f_{a + b}(x)
$$
I suspect it has to do with their Fourier transforms, my intuition tells me that:
$$
F_{a}( j \omega) \cdot F_{b}( j \omega) = F_{a \cdot b}( j \omega)
$$
Am I on the right track, so far?

A and B are nonzero positive.
 
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  • #2
How is the convolution defined? What do you have to show for the semigroup property?
 
  • #3
The convolution is defined on $$ l^{1}(\mathbb{Z}) $$ and $$ l^{1}(\mathbb{R} ) $$ spaces.
Respectively, for continuous and discrete (absolutely integrable and absolutely summable functions):
$$
x[n] * y[n] = \displaystyle \sum_{k \in \mathbb{Z} } x[k] \cdot y[n - k]
$$
And
$$
x(t) * y(t) = \displaystyle \int_{-\infty}^{\infty} x(\tau) \cdot y(t - \tau) \,\,\,\, \text{d} \tau
$$
You have to show that for the family of the functions below:
$$
f_{a}(x) = \dfrac{1}{a \pi} \cdot \dfrac{1}{ 1 + \dfrac{x^{2}}{a^{2}} }
$$
Forms a semigroup, under convolution:
$$
f_{a}(x) * f_{b}(x) = f_{a + b}(x)
$$
I am on the path, to use the foiurier transform. Please see my intuition and reasoning for this, in my first post.
 
  • #4
It could be that the Fourier transform helps, but I don't see it yet. You can use the formula for ##\mathcal{F}(f*g)## but you have to undo the transformation again, and ##\mathcal{F}(f)\cdot \mathcal{F}(g)## doesn't look much better. Why don't you calculate the integrals?
 
  • #5
Hi

The Fourier transform of:
$$
\mathcal{F} \bigg\{ \frac{1}{a \pi} \cdot \frac{1}{1 + { \frac{x}{a} }^{2} } \bigg \} =\frac{1}{ a^2 \pi} \cdot \sqrt{ \frac{\pi}{2} } \cdot e^{-| \frac{\omega}{a} |}
$$
My idea is that, since, the Fourier transform of the convolution results in a mulitplication of their individual Fourier transforms:
$$
\mathcal{F} (f_{a} * f_{b} ) = \mathcal{F}(f_{a}) \cdot \mathcal{F}(f_{b} )

$$
Our original claim was that:
$$
f_{a} * f_{b} = f_{a + b}
$$
So, if we multiply their Fourier transforms, we get:
$$
\frac{1}{ a^2 \pi} \cdot \sqrt{ \frac{\pi}{2} } \cdot e^{-| \frac{\omega}{a}| } \cdot \frac{1}{b^2 \pi} \cdot \sqrt{ \frac{\pi}{2} } \cdot e^{-| \frac{\omega}{b} |} = \frac{1}{(ab\pi)^{2} } \cdot \frac{\pi}{2} \cdot e^{ - ( \frac{\omega}{a} + \frac{\omega}{b} )}
$$
This equation above, is simply a given, due to convolution and frequency multiplication duality.

My question is, how can I positively show that:
$$
\mathcal{F} \bigg \{ f_{a + b} \bigg \} = \frac{1}{(ab\pi)^{2} } \cdot \frac{\pi}{2} \cdot e^{ - ( \frac{\omega}{a} + \frac{\omega}{b} )}
$$
If I can poisitively show that, then I can complete my proof.-
 
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  • #6
Where is the integral? If we rewrite
$$
f_a(x)=\dfrac{1}{a\pi} \cdot \dfrac{1}{1+\frac{x^2}{a^2}}= \dfrac{1}{\pi}\cdot \dfrac{a}{a^2+x^2}
$$
then this function says tangent to me. E.g. ##\dfrac{d}{dx} \operatorname{arctan}(x)=\dfrac{1}{1+x^2}## Now if we fold ##f_a*f_b## then we could get rid of the integral. How did you remove it?
 
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  • #7
AVBs2Systems said:
$$\mathcal{F} \bigg\{ \frac{1}{a \pi} \cdot \frac{1}{1 + { \frac{x}{a} }^{2} } \bigg \} =\frac{1}{ a^2 \pi} \cdot \sqrt{ \frac{\pi}{2} } \cdot e^{-| \frac{\omega}{a} |}$$
Where did you get this from? It looks wrong to me.
 
  • #8
strangerep said:
Where did you get this from? It looks wrong to me.
Scaling property and linearity property. Note a and b are greater than zero.

The Fourier transform of
$$
\mathcal{F} \bigg \{ \frac{1}{1 + {(\frac{x}{a})}^{2} } \bigg \} = \frac{1}{|a|}\cdot \sqrt{ \frac{\pi}{2} } \cdot e^{-|\frac{\omega}{a} | }

$$
Hence:
$$
\frac{1}{a \pi} \cdot \mathcal{F} \bigg \{ \frac{1}{1 + {(\frac{x}{a})}^{2} } \bigg \} = \dfrac{1}{a^2 \pi} \cdot X( \frac{\omega}{a} )

$$
 
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  • #9
That's not a sensible answer. You have an "##a##" on the RHS (of 1st equation), but not on the LHS.
Show your work in more detail.

(Btw, shouldn't this be in one of the homework forums?)
 
  • #10
In case you will end up in an integration formula, and I still don't see how to avoid this, the substitutions ##x=a\tan \varphi## and ##t-x=b\tan \vartheta## could be helpful.
 
  • #11
fresh_42 said:
[...] I still don't see how to avoid [integration formula]
Maybe I'm missing something but,... are you sure you're not overthinking this?

Imho, if the OP could get his/her Fourier transform expression right (i.e., use the scaling rule correctly), then I think the original problem becomes trivial.
 
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  • #12
I am sure if you search for "proof of semigroup property of convolution for Cauchy distribution" you'll get something hopefully meaningful.
 
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  • #13
MathematicalPhysicist said:
-----
I submitted both the approaches, one with the incorrect Fourier transform and the other with the convolution integral. The convolution integral, reduced using partial fraction decomposition or the residue theorem (neither of which I was able to produce as a solution) reduces it to the form required.

Thank you everyone for your assistance.
$$ \textbf{Sources for the solution} $$
## \text{For future reference, if anyone needs it:} ##1. https://math.stackexchange.com/ques...ion-of-cauchy-density-function-int-infty-inft2. https://www.tandfonline.com/doi/abs/10.1080/00029890.1985.11971537
 
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What is the semigroup property for convolution?

The semigroup property for convolution is a mathematical concept that states that the convolution of two functions is associative. This means that the order in which the functions are convolved does not matter, and the result will be the same.

Why is the semigroup property important in convolution?

The semigroup property is important in convolution because it allows for more efficient and simplified calculations. It also allows for the use of Fourier transforms, which are powerful tools in signal processing and other fields.

How is the semigroup property proven in convolution?

The semigroup property can be proven using mathematical induction, where the base case is two functions and the induction step involves adding a third function. The proof involves manipulating the convolution integral and showing that it is equal regardless of the order of the functions.

Can the semigroup property be extended to more than two functions?

Yes, the semigroup property can be extended to any number of functions. This is known as the semigroup property for multi-convolution. The proof is similar to the two-function case and involves mathematical induction.

What are the practical applications of the semigroup property for convolution?

The semigroup property for convolution has many practical applications in fields such as signal processing, image processing, and data analysis. It is used to efficiently process and manipulate signals and images, and to perform various types of filtering and enhancement. It is also used in the study of differential equations and other mathematical models.

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