aakashl

(subject)[object]

Fractional Calculus

Calculus is the manipulation of one basic operator: the derivative or $\frac{\mathrm{d} }{\mathrm{d} x}$ . This operator operates on functions and by repeatedly applying it, you can get higher order derivatives. It’s inverse operator is known as the integral. Similar to matrix operators which have eigenvalues and eigenvectors, this operator also has eigenvalues and eigenfunctions. The eigenfunction is the function which only goes through some scalar change when acted on by the operator. This scalar that the function is scaled by is called the eigenvalue of the eigenfunction. For the derivative operator, the eigenfunctions would be any function of the form $e^{ax}$ because $\frac{\mathrm{d} }{\mathrm{d} x}e^{ax}=a^n e^{ax}$ . The eigenvalue would be $a$ . This means applying the derivative operator to this function $n$ times only results in multiplying the original function by its eigenvalue $n$ times. It is explicitly written below.

$\frac{\mathrm{d} }{\mathrm{d} x}e^{ax}=a^ne^{ax}$

This means the half derivative of the exponential function, in some sense, is $\sqrt{a}e^{ax}$ . Using this, one can even find fractional derivatives of trigonometric functions by writing them as complex exponentials. The half derivative of sine, for example, is shown below.

$\left ( \frac{\mathrm{d} }{\mathrm{d} x} \right )^{0.5}\textup{sin}(x)=\left ( \frac{\mathrm{d} }{\mathrm{d} x} \right )^{0.5}\left ( \frac{e^{ix}-e^{-ix}}{2i} \right )$

$=\frac{\sqrt{i}e^{ix}-\sqrt{-i}e^{-ix}}{2i}$

$=\frac{e^{i(x+\frac{\pi}{4})}-e^{-i(x+\frac{\pi}{4})}}{2i}$

$=\textup{sin}\left ( x+\frac{\pi}{4} \right )$

In fact, a general formula for both the derivatives of sine and cosine are given below using eigenfunctions.

$\left ( \frac{\mathrm{d} }{\mathrm{d} x} \right )^n \textup{sin}(x)=\textup{sin}\left ( x+\frac{\pi}{2} \right )$
$\left ( \frac{\mathrm{d} }{\mathrm{d} x} \right )^n \textup{cos}(x)=\textup{cos}\left ( x+\frac{n\pi}{2} \right )$
This seems to only work for trigonometric and exponential functions. However, using a Fourier transform, one can rewrite any function in terms of exponentials. The Fourier transform of any function $f(x)$ is shown below.

$f(x)=\int_{-\infty}^{\infty}F(\omega)e^{i\omega x}\textup{d}\omega$
A general method can then be created for the $n^{th}$ derivative of a function $f(x)$ as shown below.

$\left ( \frac{\mathrm{d} }{\mathrm{d} x} \right )^nf(x)=\left ( \frac{\mathrm{d} }{\mathrm{d} x} \right )^n\int_{-\infty}^{\infty}F(\omega)e^{i\omega x}d \omega$

$=\int_{-\infty}^{\infty}\left ( \frac{\mathrm{d} }{\mathrm{d} x} \right )^nF(\omega)e^{i\omega x}d \omega$

$=\int_{-\infty}^{\infty}F(\omega)(i\omega)^ne^{i\omega x}d \omega$

$\mathcal{F}^{-1}[F(\omega)(i\omega)^n]$

$\mathcal{F}^{-1}[\mathcal{F}[f(x)](i\omega)^n]$

This can also be done with Laplace transforms which would give the relation $( \frac{\mathrm{d} }{\mathrm{d} x})^n f(x)=\mathcal{L}^{-1}[\mathcal{L}[f(x)](s)^n]$ .

Through a similar process, one can find out how to calculate fractional Fourier transforms. The eigenfunction of the fractional Fourier transforms is actually the Gaussian so $\mathcal{F}[e^{ax^2/2}]=\sqrt{a}e^{ax^2/2}$ . One can put any function in terms of Gaussians using Hermite transforms.

There are actually much better ways of calculating fractional derivatives through fractional integrals and even fractional Fourier transforms which are used for more rigorous calculations in the field of fractional calculus but these are just some neat methods to do so.

If you want to learn more or see where I learned this from, watch Ahmed Isam.

aakashl

April 29, 2016

Math

3 responses to “Fractional Calculus”

zimbles

May 2, 2016 at 6:24 am

What an interesting post, I never heard of ‘half-derivatives’, yet it makes the connection between exponentials, trigonometry and calculus so clear.

Can you do abstract algebra with the calculus operator? With functions as elements. Using the ‘calculus-operator’ d/dx, can you make a group? It seems like e^x would be kind of like an ‘identity-element’. However I don’t think associativity or closure holds. It confuses me a bit.

LikeLiked by 1 person

Reply
1. aakashl
  
  May 2, 2016 at 4:01 pm
  
  One idea to think about is forming a group of all operators such that two composed together would form another operator. The group of operators would consist of all orders of derivatives and all orders of integrals so each element has an inverse and the identity is simply, in some sense, 1. When this group is applied in regards to an exponential function, in some sense, they all become the identity or 1 because they all act as identity operators. The operators in the group then change based on the function in question. This is similar to the reflections and rotations in a dihedral group that vary based on the polygon that they are being applied to. So, in some sense, yes. You can think of derivatives in the context of abstract algebra but the idea of using the exponential as an identity may be misled due to the fact that this implies that 2 different elements of the group can be operated on to form e^x which is not the case. I guess it is possible to assume these two elements are d/dx and e^x but this adds a lot of complications as to what the entire group would actually be and where the inverses lie. It is definitely a very interesting thought though!
  
  LikeLiked by 1 person
  
  Reply
  1. zimbles
    
    May 3, 2016 at 4:11 am
    
    Thank you for making it a bit more clear! I look forward to your posts in the future.
    
    LikeLike

Fractional Calculus

3 responses to “Fractional Calculus”

Leave a Reply Cancel reply