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# Initialized fractional calculus

mathematics -> fractional calculus -> 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 neccessarily 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, ect. 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.

${}_a\mathbb{D}^q_tf(x)=\frac{1}{\Gamma(n-q)}\frac{d^n}{dx^n}\int_{a}^{t}(t-\tau)^{n-q-1}f(\tau)d\tau + \Psi(x)$

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