Talk:Differintegral

message: Moved here from initialized fractional calculus;
A certain oddity about the differintegral should be pointed out.  If the differintegral is "uninitialized", then although:
:\mathbb{D}^q\mathbb{D}^{-q} = \mathbb{I}
(That is, Dq is the left inverse of D-q.), the converse is not necessarily true.
:\mathbb{D}^{-q}\mathbb{D}^q \neq \mathbb{I}
But hold on, the ship isn't sunk yet!  Let's take a look at integral calculus, to get a better idea of what's happening. First, let's integrate, then differentiate, using the arbitrary function 3x2+1:
:d\left[\int (3x^2+1)dx\right]/dx = d[x^3+x+c]/dx = 3x^2+1
Well, that was pretty straightforward, and it worked.  Now, what happens when we exchange the order of composition?
:\int [d(3x^2+1)/dx]dx = \int 6xdx = 3x^2+c
Hmmm... I think it's fairly obvious what that integration constant is.  Even if it wasn't obvious, we would simply use the initialization terms such as f'(0) = c, f' '(0)= d, etc.  If we neglected those initilization terms, the last equation would fail our test.  This is exactly the problem that we encountered with the differintegral.  If the differintegral is initialized properly, then the composition holds.  The problem is that in differentation, we lose state information, as we lost the c in the first equation. (see dynamical systems).
In fractional calculus, however, since the operator has been fractionalized and is thus continuous, an entire complimentary function is needed, not just a constant or set of constants.
:{}_a\mathbb{D}^q_tf(x)=\frac{1}{\Gamma(n-q)}\frac{d^n}{dx^n}\int_{a}^{t}(x-\tau)^{n-q-1}f(\tau)d\tau + \Psi(x)
Charles Matthews 06:41, 22 Apr 2004 (UTC)
thread_title: Oddity

message: This page needs a definition of the differintegral at the top of the page. All the equations need to define what all the variables inside it mean. The article on fractional calculus doesn't give decent enough perspective for someone to understand this page. Like.. what do the superscripts mean? Fresheneesz 17:22, 26 November 2005 (UTC)
replies: text: The notation is an extension of the notation for differential.  Understanding calculus is a prerequisite for understanding fractional calculus.  If you understand calculus, you're familiar w/the notation, thus if you're not familiar w/the notation, you don't understand calculus, and thus you won't be able to understand fractional calculus regardless of whether or not we explain the notation. Kevin Baastalk 14:44, 17 June 2009 (UTC)
thread_title: Definitions and explanation

message: Does the composition rule hold always without exception? Consider this:
:\mathbb{D}^{\frac{5}{6}}\mathbb{D}^{2}(t)=\mathbb{D}^{\frac{17}{6}}(t)=\frac{1}{\Gamma(-\frac{5}{6})t^{\frac{11}{6}}}
However, it is clear that
:\mathbb{D}^{2}(t)=0--
Therefore
:\mathbb{D}^{\frac{5}{6}}\mathbb{D}^{2}(t)=0
--88.101.219.136 08:26, 4 July 2007 (UTC)
replies: text: Yeah, there are some issues like that with composition.  It's discussed to some extent in the literature, but I don't think it's made clear in the article. Kevin Baastalk 15:10, 9 June 2009 (UTC)

text: I tried explaining here. To give a short answer to this particular example: when you differentiated to the 5/6th order, you were integrating.  And when you integrate, you've got to remember to add the " + C".  The literature might not always have that taken into account and the equations on here might thus not be accurate in that respect (verifiability, not truth) - but that's what's happening. I went into some depth about  " + C" and composition, esp. in relation to fractional calculus in the link I just provided.  If you can understand it (i think it needs some work), it might help answer your question in the general case.  Kevin Baastalk 16:33, 9 June 2009 (UTC)
thread_title:  Composition rule 

message: First, this is at least 50% a duplicate of the more fundamental article on fractional calculus. Since it is also a "low-priority" article, perhaps it should be merged?
Second, on neither of these pages is there any discussion of application. I understand that it has application at least in analyzing so-called "meta-materials" (e.g. optical media with negative refractive index) - it would be awesome if I could read about this here; I've only heard hints of it from those in the know.  208.120.110.46 (talk) 03:14, 7 April 2008 (UTC)
thread_title:  Merger? and applications 

message: Which of the following is clearer?
Option 1:
Option 2:
Personally, I find the latter version preferable.  (What does the phrase "does not define a separate function" mean, for example?)  If there is a dispute about this, then I suggest that it be escalated by placing an  tag on the article and an appropriately neutral thread at WT:WPM, rather than reverting good-faith edits to the article.  Thanks, Le Docteur (talk) 00:51, 24 October 2009 (UTC)
replies: text: Option 2 is obviously clearer. Kevin Baastalk 15:11, 17 March 2010 (UTC)
thread_title:  Which is clearer? 

message: you'll notice that if you integrate an n dimensional function (i mean orthogonal dimensions i.e. x,y,z) over an m dimensional region, it will be an m-th order integration that results in an n-m dimensional function.
:I(F(x{n)),R{x{m}}) =  f(x{n-m})
as differentiation is the opposite of integration, the reverse should also hold.  that is:
:d(f(x{n-m)),R{x{m}}) =  F(x{n})
where the d is an m-th order differentiation.  and you'll notice it requires a region of differentiation.  this "region of differentiation" (my own wording, i know) is the "differentiation with respect to".  I'm speaking kind of roughly here, but my point is that you have to consider that:
# mth order integration of an n-dimensional function over an m-dimensional region results in an n-m dimensional function (namely, the dimensions in the integrated function but not the region)
# and so the inverse operation, differentiation, necessarily has the reverse relations, if composition is to hold.
# i.e. if going one direction requires a region of integration, then going the other direction requires something like the reverse.
# when applying this to fractional order orthogonal integrations/differentiations, the region of integration/differentiation is necessarily of fractional dimension (fractal) and thus so is the result and/or the function being integrated/differentiated.
# and furthermore there is the issue i discussed (again, roughly) here.
of all the literature i've read the closest that comes to addressing these mathematical issues is a paper by a division of NASA called "initialized fractional calculus".  yet neither it, or any of these issues, are mentioned anywhere in the article. Kevin Baastalk 15:34, 17 March 2010 (UTC)
thread_title:  spatial relations and composition of differintegration w/orthogonal spaces 

message: can someone put the chain rule of differintegration in the basic formal properties section?
Kevin Baastalk 19:47, 2 June 2010 (UTC)
thread_title:  chain rule? 

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thread_title:  External links modified