Fractional Calculus has many applications! One such application is expressed in the first paper I published where fractional calculus is used in the application of gas solubility. In this paper, we solve Fractional Differential Equations (FDEs). What I will go over in this article is simpler, more intuitive, and are the building blocks for the Bhatti polynomials used to estimate the solutions of FDEs, the Fractional Derivatives of Polynomials.

Other applications of Fractional Calculus can be found in this paper.

Fractional derivatives don't have a clear geometric/physical interpretation as regular derivatives have. For instance, the derivative of a function that describes the position of an object would describe it's velocity but the fractional derivative isn't as clear.

As you may or may not know, polynomials are expressions that consists of only addition, subtraction, and multiplication operations. Here is one way of expressing them, using summation notation:

$$\sum_i^n a_i x^i=a_n x^n + a_{n-1} x^{n-1}+\dots+a_2 x^2 + a_1 x + a_0$$

To find the derivative of a polynomial you can use the definition of derivatives:

$$\frac{d}{dx}f(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$$

Note: I'm not going to do a rigorous derivations but intuitive ones.

Now the derivative of a polynomial can be acquired using the power rule:


If you need a proof for the power rule check this page out.

The next step in our path is to get the equation for the $n$th derivative . We can interpolate this by creating a small table of derivatives and derivative of derivatives.

$n$ $\frac{d^n x^m}{dx^n}$
$0$ $x^m$
$1$ $mx^{m-1}$
$2$ $m (m-1) x^{m-2}$
$3$ $m(m-1)(m-2)x^{m-3}$
$\vdots$ $\vdots$
$n$ $m(m-1)(m-2)\dots(m-n+1)x^{m-n}$

We can now put down the products as a division of products:

$$\frac{d^n x^m}{dx^n}=\frac{m(m-1)(m-2)\dots1}{(m-n)(m-n-1)(m-n-2)\dots1} x^{m-n}$$

Which, by definition, we can represent using factorials:

$$\frac{d^n x^m}{dx^n}=\frac{m!}{(m-n)!} x^{m-n}$$

Now that we have a function for the $n$-th derivative of a polynomial, we can continue by making the function continuous by using the Gamma function where:


We finish when the function becomes:

$$\frac{d^n x^m}{dx^n}=\frac{\Gamma(m+1)}{\Gamma(m-n+1)} x^{m-n}$$

Let's try it out with the half-derivative of $x^2$:

\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}} x^2&=\frac{\Gamma(2+1)}{\Gamma(2-\frac{1}{2}+1)} x^{2-\frac{1}{2}} \\
\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}} x^2&=\frac{\Gamma(3)}{\Gamma(\frac{5}{2})} x^{\frac{3}{2}} \\
\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}} x^2&=\frac{2}{\frac{3\sqrt{\pi}}{4}} x^{\frac{3}{2}} \\
\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}} x^2&=\frac{8}{3\sqrt{\pi}} x^{\frac{3}{2}}

Let's look at a plot of that function

We can figure out If our derivation and the half-derivative is correct by confirming by calculating Caputo's fractional derivative. Caputo's fractional derivative is as follows:

$$\frac{d^\nu}{dx^\nu} f(t)=\frac{1}{\Gamma(n-\nu)}\int_0^t (t-u)^{(n-\nu-1)}f^{(n)}(u)du \quad (n-1)<\nu<n$$

We can select $n=\lceil{\nu}\rceil=1$, set $f(t)=t^2$, and $\nu=\frac{1}{2}$:

\frac{d^\frac{1}{2}}{dx^\frac{1}{2}} t^2&=\frac{1}{\Gamma(1-\frac{1}{2})}\int_0^t (t-u)^{(1-\frac{1}{2}-1)} \frac{d}{du} (u^2) du \\
\frac{d^\frac{1}{2}}{dx^\frac{1}{2}} t^2&=\frac{2}{\Gamma(\frac{1}{2})}\int_0^t \frac{u}{\sqrt{t-u}} du

In the integral, lets substitute $s=t-u$ and $ds=-du$:

\int\frac{u}{\sqrt{t-u}}&=\int \frac{t-s}{\sqrt{s}} ds \\
&= \int \bigg( \frac{t}{\sqrt{s}}-\sqrt{s}\bigg) ds \\
&= \int \sqrt{s} ds - t \int \frac{1}{\sqrt{s}} ds \\
&= \frac{2s^{\frac{3}{2}}}{3}-2t\sqrt{s}+C \\
&= \frac{2(t-u)^{\frac{3}{2}}}{3}-2t\sqrt{t-u}+C \\
\int_0^t\frac{u}{\sqrt{t-u}}&=\frac{2(t-u)^{\frac{3}{2}}}{3}-2t\sqrt{t-u}\Big|_ {u=0}^{u=t} \\
&= \frac{4t^{\frac{3}{2}}}{3}

Then we place the integral back in our equation:

And simplified:

To make this equation the same as the derived equation, you would just switch $t\rightarrow x$.