Fractional calculus

In calculus we consider the idea \displaystyle f'(x) = \lim_{\Delta x \rightarrow 0} \frac{f(x+\Delta x) - f(x)}{\Delta x} where it is assumed that for certain f(x) such a limit is defined. An alternative notation is \displaystyle f'(x) = \frac{df(x)}{dx} or \displaystyle f'(x) = \left(\frac{d}{dx}\right) f(x). We may consider \displaystyle \frac{d}{dx} as an operator, let \displaystyle \mathfrak{D}_x = \frac{d}{dx}. Then f'(x) = \mathfrak{D}_x f(x). We can also say that f''(x) = \mathfrak{D}_x f'(x) = \mathfrak{D}_x ^2 f(x).

As simple example, say that f(x) = x^3 then we obtain that

(1.1) \displaystyle \mathfrak{D}_x x^3 = \lim_{\Delta x \rightarrow 0} \frac{(x+\Delta x)^3 - x^3}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac{x^3 + 3 x^2 \Delta x + 3 x \Delta^2 x + \Delta^3 x - x^3}{\Delta x}

therefore

(1.2) \displaystyle \mathfrak{D}_x x^3 = \lim_{\Delta x \rightarrow 0} \left( 3 x^2 + 3 x \Delta x + \Delta^3 x\right) = 3 x^2

Thus \mathfrak{D}_x x^3 = 3 x^2 as is well-known. Now we may wonder, what is \mathfrak{D}_x^k x^y where k is a positive integer and n is any integer, it is clear that

(2) \mathfrak{D}_x^k x^n = \mathfrak{D}_x^{k-1} n x^{n-1} = \mathfrak{D}_x^{k-2} n (n-1) x^{n-2} = \cdots = n (n-1) \cdots (n-k+1) x^{n-k}

So we may write

(3) \displaystyle \mathfrak{D}_x^k x^n = \frac{n!}{(n-k)!} x^{n-k}

Now we use the definition of \Gamma and we know that \Gamma(n+1) = n! so we can extend the relation as

(4) \displaystyle \mathfrak{D}_x^z x^y = \frac{\Gamma{(y+1)}}{\Gamma(y-z+1)!} x^{z-y}

We now have a definition for the fractional operator \mathfrak{D}^z, as acting on x^y. Another example, consider \mathfrak{D}_x^{1/2} x^4 then we obtain

(5.1) \displaystyle \mathfrak{D}_x^{1/2} x^4 = \frac{\Gamma(5)}{\Gamma(9/2)} x^{7/2}

Now we continue and we have

(5.2) \displaystyle \mathfrak{D}_x^{1/2} \frac{\Gamma(5)}{\Gamma(9/2)} x^{7/2} = \frac{\Gamma(5)}{\Gamma(9/2)} \frac{\Gamma(9/2)}{\Gamma(4)} x^3 = 4 x^3

Therefore

(5.3) \mathfrak{D}_x^{1/2} \mathfrak{D}_x^{1/2} x^4 = 4 x ^3

as expected as \mathfrak{D}_x^{1/2} \mathfrak{D}_x^{1/2} x^4 = \mathfrak{D}_x x^4 = 4 x^4. similar we find that

(6) \mathfrak{D}_x^z \exp(ax) = a^z \exp(ax)

We now can consider something like \mathfrak{D}_x^z \cos(ax) and we find that

(7.1) \mathfrak{D}_x^z \cos(ax) = \mathfrak{D}_x^z \frac{1}{2} \left( \exp(\textbf{i}ax) + \exp(-\textbf{i}ax) \right) so

(7.2) \mathfrak{D}_x^z \cos(ax) = \frac{1}{2} \left\{ (\textbf{i}a)^z \exp(\textbf{i}ax) + (-\textbf{i}a)^z \exp(-\textbf{i}ax) \right\} then

(7.3) \begin{array}{rcl} \mathfrak{D}_x^z \cos(ax) &=& \frac{1}{2} a^z \left\{ \left( \cos(z\pi) + \textbf{i} \sin(z\pi) \right) \left(\cos(ax) + \textbf{i} \sin(ax)\right) \right. \\&& \hspace{2em} + \left. \left(\cos(z\pi) - \textbf{i} \sin(z\pi) \right) \left(\cos(ax) - \textbf{i} \sin(ax)\right) \right\} \end{array} therefore

(7.4) \mathfrak{D}_x^z \cos(ax) = \cos(z\pi) \cos(ax) - \sin(z\pi) \sin(ax) = \cos(ax - z\pi)

Similar we have \mathfrak{D}_x^z \sin(ax) = \sin(ax - z\pi), for fractional calculus we have

(8.1) \displaystyle \mathfrak{D}_x^z x^y = \frac{\Gamma{(y+1)}}{\Gamma(y-z+1)!} x^{z-y}
(8.2) \mathfrak{D}_x^z \exp(ax) = a^z \exp(ax)
(8.3) \mathfrak{D}_x^z \cos(ax) = \cos(ax - z\pi)
(8.4) \mathfrak{D}_x^z \sin(ax) = \sin(ax - z\pi).

Now using the operators \mathfrak{D}_x^z, we may also consider some operator like

(9) \displaystyle \mathfrak{D}_{x;f(z)} = \int_0^\infty dz f(z) \mathfrak{D}_x^z.

The \exp(ax) is then fun part, as we have

(10.1) \displaystyle \mathfrak{D}_{x;f(z)} \exp{ax} = \int_0^\infty dz f(z) \mathfrak{D}_x^z \exp{ax} thus

(10.2) \displaystyle \mathfrak{D}_{x;f(z)} \exp{ax} = \left( \int_0^\infty dz f(z) a^z \right) \exp{ax}

Note the form \mathfrak{D}_{x;f(z)} \mathfrak{e} = \mathfrak{v} \mathfrak{e} – thus \mathfrak{e} are eigenvectors and \mathfrak{v} are eigenvalues.

Fractional calculus is an extension for normal calculus. The purpose of this post is simply to give some idea about fractional calculus. Have fun!

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