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Since I have just started studying Laplaces Transform and convolutions, some questions came up to my mind and one them is:

(1) These guys (Laplace, Fourier and so on...) had only created their transforms? I mean, could anyone (you and/or I, for example) create an integral transform? If so, what are the requirements for it?

My professor said once that convolution was invented; not discovered such as addition or multiplication, for instance. Unfortunately, I had not had the chance to ask him (1)

Thanks

Mr. N
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    Regarding convolution being invented: The concept of a shift- invariant linear operator is fundamental and unavoidable. But shift-invariant linear operates are the same thing as convolution operators. – littleO Oct 26 '19 at 00:42
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    You could "create" one, but the goal of creating one is for it to be useful. – cmk Oct 26 '19 at 01:39

2 Answers2

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Well, frankly speaking, you can do whatever you want.

For example, I could define the function $f_{a,b,c}(x)$ to be a solution to the ODE $$y''=a\sin(bxy')+cy,$$ but just defining the solution $y=f_{a,b,c}(x)$ does not really give me anymore useful information about the function itself.

When you come up with a definition, it can be whatever you want, but that doesn't mean that it's relevant or useful.

Edit:

Keep in mind, that although something you invent may not have current relevance or use, that may not be the case in the future. Keep playing around and experimenting with definitions and seeing where that gets you. Who knows, maybe you'll find something important, interesting, or just straight up fun. For example, I recently messed around with a function similar to $$f(x)=\exp\left[\int_0^x \ln\sin\pi t\ dt\right]$$ and it got me an infinitude (literally) of infinite product identities involving $\exp(\mathrm G/\pi)$ where $\mathrm G$ is Catalan's constant. See here for more info. I am currently writing a paper on these products and will provide a link once it is published.

clathratus
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    Thanks a lot. I was thinking the same, but not so sure. – Mr. N Oct 26 '19 at 02:43
  • You're welcome :) See the edit – clathratus Oct 26 '19 at 03:14
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    I have already seen that!! Your picture got my attention (hahahaha) and then I went to your profile. Just astonishing! However I'm gonna read it paying more attention and do not forget to provide the link, please. – Mr. N Oct 26 '19 at 13:41
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As other people suggested, you can define any transform you want. The question you need to answer, "why are you defining that?".

In the case of the Fourier transform, the definition of the transform is inspired by Fourier series. Thus, it makes sense to why somebody would want to consider that. After such a transform is introduced, it is then natural to ask for various properties this transform has. The Fourier transform has plenty of very nice properties that make it very valuable in many instances.

The Laplace transform is very similar to the Fourier transform. It is almost a "real"-version of the Fourier transform, so it is again natural to consider it. It also comes with its own nice properties, which is not surprising since it is so similar to the Fourier transform.

There are a few other transforms. For instance, there is something called the Hankel transform. Why was it invented? Because people realized that the Laplace/Fourier transform can be used, among many other things, to solve certain differential equations. Inspired by this, certain mathematicians thought of what other analogue transforms would aid in solving various important differential equations. The Hankel transform was introduced to specifically deal with Bessel-like differential equations, as they are frequently studied in physics.

Necessity drives innovation. Thus, before you go on your own path to invent a new transform, the question you should ask is, "what can my transform be used for solving". Otherwise, if it is an uninspired random transform, then it will probably not be useful in any circumstance and just sit there alone by itself without any need to study it.

  • Thanks for answering me! I got the idea. I will check this Hankel one more closely, since it's brand new for me, never heard about it. – Mr. N Oct 26 '19 at 13:54