Talk:Integral transform

message: what is the discrete analog of the integral transform?  the summation transform? - Omegatron 14:55, Sep 30, 2004 (UTC)
: in a way Fourier series is a discrete analogy, in that it maps to a discrete span of functions. — Preceding unsigned comment added by 83.233.144.116 (talk) 23:12, 3 October 2006 (UTC)
thread_title:  Discrete analog? 

message: (Disclaimer: I am not a native english speaker, so the following remark might be false.) One can find the word "integral transform" as well as "integral transformation" in the literature. I tried to figure out the appropriate meanings with the following result: These words should be used as in "The Laplace transformation L associates to a function f its Laplace transform Lf". If this is correct, the page should be adjusted. Th. Bliem. --134.95.214.156 08:57, 17 October 2005 (UTC)
thread_title:  Transform vs. Transformation 

message: I find the section on orthogonality to be useless or incorrect.  It speaks of basis functions, which are undefined and not obviously related at all to general integral transforms.  It refers to the kronecker delta, when I think it means dirac; but it would still be wrong because it should be delta(y-5)*delta(x-3) - not any scaling.  If people agree, we should just get rid of it.
:I agree, it is worded incorrectly, I have updated it. Jackminardi (talk) 09:53, 25 December 2011 (UTC)
replies: text: I never knew that the basis functions of integral transforms had to be orthogonal, can you give a reference? How are the basis functions of the Laplace transform orthogonal? Classwarz (talk) 10:28, 24 January 2012 (UTC)
thread_title:  Section on Orthogonality 

message: I found the exposition here extremely clear, and the links very helpful.  It gave me context I needed for digital signal processing without overwhelming me with the mathematics.
Dana Good
71.112.107.85 16:18, 28 March 2007 (UTC)
thread_title:  Helpful for a layperson 

message: I don't pretend to be an expert on this subject, but in "A symmetric kernel is one that is unchanged when the two variables are permuted." couldn't 'permuted' be changed to 'swapped', or something else implying that their order doesn't matter? Using the word permuted seems a little excessive when there's only two possible permutations. I may be misinterpreting this, of course (in which case the sentence should probably be clarified!) --Jonnty (talk) 07:09, 27 December 2009 (UTC)
thread_title:  Overcomplicated wording 

message: I have added the wavelet transform to the list, nevertheless I am not a real expert and am not entirly sure this is mathematically a intergral transfomr, nevertheless would appriciate input from someone with better knowledge. I ' have added it because I think it had to be added to be more complete.  —Preceding unsigned comment added by 146.175.108.218 (talk) 13:21, 4 January 2010 (UTC)
thread_title:  Wavelet transform 

message: The link Integral operator redirects to this entry, but integral transforms are only very particular integral operators: integral operators are simply maps between function spaces defined by means of a integral of one or more dimension. An ideal entry about integral operators should describe the ideas of Vito Volterra, Ivar Fredholm and necessarily describe also nonlinear operators (note the irony of this last redirected link :D ). --Daniele.tampieri (talk) 19:50, 5 June 2010 (UTC)
thread_title:  Wrong redirection 

message: This stuff is pretty abstract, but there is a simple, very visual analogy to so-called setup moves on Rubik's cube. For example, there is a simple algorithm for switching two sets of two cublets in the Rubik's cube, but it only works if the cublets are in the right place. So you use a setup move that moves the cublets to the (relative) positions, execute the algorithm, and "undo" the setup moves to get the desired result. — Preceding unsigned comment added by 62.143.45.154 (talk) 01:17, 21 November 2013 (UTC)
replies: text: The analogy you give is more in the realm of group theory (see Rubik's Cube group). Setup moves are simply group elements (i.e. moves) that can be expressed in the form . To me, it seems a bit misleading to compare setup moves to integral transforms, though, since the idea behind an integral transform is that you move into a different mathematical space (e.g. in the case of the Laplace transform, you move from the space of real functions to the space of complex functions) where finding the solution is easier. In the case of setup moves, we don't move into a distinct space, find the solution in that space, and then move the solution back to the original space; rather, we are simply taking advantage of the fact that the particular effect that we want to induce in the cube is more easily done by moving into a certain cube state (performing the  move), then we induce the state (perform ), and then we move back to the original state (perform ). They are more akin to commutators in this regard (which are of the form ). Perhaps more succinctly, integral transforms are a form of homomorphism; setup moves could be seen as an endomorphism, but this is not a particularly useful observation, in my opinion. —JivanP (talk) 23:29, 26 August 2019 (UTC)
thread_title:  Rubik's Cube Analogy 

message: Should it not be  T(f)(u) = \int \limits_{t_1}^{t_2} K(t, u)\, f(t)\, dt  instead of  T(f(u)) = \int \limits_{t_1}^{t_2} K(t, u)\, f(t)\, dt  ?
The transform takes the function f, which is a function of t, and transforms it to T(f), which is a function of u. So it should be T(f)(u). Right?
The way it is at the moment, it looks like you get f, evaluate it at u, and then take T of the the result. Which is wrong for several reasons. --AndreRD (talk) 07:50, 15 March 2015 (UTC)
thread_title:  T(f(u)) 

message: The page repeatedly talks about an integral transform acting on equations, but this seems wrong to me. Don't integral transforms act on functions, not equations?
NoahSD (talk) 04:21, 26 February 2018 (UTC)
thread_title:  Functions instead of equations? 

message: What would be a typical domain of f? Does it have to be a one-dimensional function if K is symmetric, since K seems to depend on one time variable and "space" variable u in the domain of f.  — Preceding unsigned comment added by 129.94.177.30 (talk) 05:22, 12 June 2019 (UTC)
replies: text: See Green's functions for an example of an application of integral transforms to multivariate functions; in particular, they are useful as the kernel of an integral transform for finding solutions to partial differential equations in the space–time domain with given boundary conditions. For a problem only in the space domain (not in the time domain), we could have the Green's function , where  is the 3D coordinate being integrated over, and  is the 3D coordinate in which the solution to the differential equation (i.e. the function which results from the integral transform) is given. In general, if the function  which has  parameters is being transformed by the kernel  into a function  which has  parameters, then  requires  parameters, since  of these will be integrated over to yield . —JivanP (talk) 17:37, 14 June 2019 (UTC)
replies: text:  To respond to your question about the notion of a symmetric kernel: the term is only formally defined in this article when the kernel has two arguments/parameters. Whilst I am not sure how the term is used in the literature with regard to multivariate functions, I would assume that, in particular, it is used when  and  each have the same arity, and the kernel is unchanged when the parameters being integrated over are interchanged with the other parameters. To go back to our example with the Green's function , I assume one would say  is symmetric iff   for all  and . However, it is conceivable that one would use the term symmetric in a different manner, and ultimately one should refer to the author's definition. —JivanP (talk) 17:43, 14 June 2019 (UTC)
thread_title:  Domain of f and K 

message: It seems that this was invented by Euler before Fourier, so I understand from several sources.
So, I think this needs to be updated on this page.  Pls. confirm 
https://link.springer.com/article/10.1007/BF00348586 Nelagnelag (talk) 12:06, 19 August 2019 (UTC)
replies: text: The very page you link to states: "It would, however, be reading too much into that earlier work [by Euler] to see it as contributing to the theory of the integral transform." In particular, the discovery that periodic functions, such as those used as initial conditions for the wave equation and heat equation, can be represented as a sum of sine waves is generally attributed to Euler (e.g. see page 5 of this PDF; page 189 as it appears in the document itself). This discovery contributed to the devising of the Fourier series, but came a few decades before the Fourier transform and Laplace transform were devised, which in turn paved the way for the more general theory of integral transforms. One can certainly say that Euler's earlier work served as the spark for integral transform theory, but it was hardly a major contribution to integral transform theory as a whole, and especially not to our modern eyes, where the theory is now much broader than it was at the end of the 18th century. For more info on the contributions made by Euler, d'Alembert, and the Bernoulli family in this regard, please see Fourier series#History. See Laplace transform#History for more of Euler's influence. —JivanP (talk) 23:07, 26 August 2019 (UTC)
thread_title:  History 

message: Looking at the path integral formulation:
:\psi(x,t) = \int_{-\infty}^\infty \psi(x',t') K(x,t; x', t') dx'.
This seems wrong, since there seems to be a t' that is not integrated away.  — Preceding unsigned comment added by Jeffsink (talk • contribs) 21:33, 8 January 2021 (UTC)
thread_title:  Path integral formulation wrong? 

message: The section on the path integral links to a StackExchange page, which in turn only links back to Wikipedia. Surely this does not count as an acceptable source by Wiki's standards, right?  — Preceding unsigned comment added by 129.93.184.39 (talk) 21:13, 11 February 2021 (UTC)
thread_title:  Feynman Integral?