# Does Fractional Calculus define the derivative over the Weirstrass Function?

I recently read this paper on defining the fractional derivative for the Wierstrass function. This seems very interesting since derivatives over fractals are generally not well defined. Yet, this paper seems to good to be true, as it doesn't really elaborate on any possible uses or meanings for this result.

Specifically, I'm having trouble understanding the actual implications of this result. In the paper, it states that for all derivatives less than a certain number, the derivative exists, and that for all derivatives greater than a certain order, the derivative doesn't exist, yet it doesn't elaborate on whether or not any of the existing derivatives have useful properties.

My question is this, does this "derivative" over the Wierstrass function have any real physical meaning? More concretely, can I think of the derivative as measuring the slope of the function, or is this just a mathematical trick without application? (Feel free to present a third option). If it has applications, I'd like to know what they are. If not, what was the point in defining this derivative?

What I already know: I'm not asking about the intuition for the fractional derivative, I'm asking about the intuition/meaning of this specific derivative over this specific fractal. So, I'd appreciate it if you don't link/refer me to other papers that don't specifically talk about intuition for this derivative and fractal.

• I barely glanced at the paper but this is what I guess is the point. The weierstrass function is famously continuous everywhere but also undifferentiable everywhere. However, fractional calculus allows you to differentiate the function but you must use a fractional using a noninteger order up to some integer value. The reason is because the RL definition smooths things out with integration first. As for meaning of the actual derivative... I'd say it's just an integral transform. – abnry May 15 '15 at 13:53
• @nayrb you're passing the buck. What's the meaning of the integral transform? – Zach466920 May 15 '15 at 14:41
• @NonStandard I'd say before discussing the physical meaning of the derivative it would be handy to prescribe some physical meaning to the Weirstrass function itself. Do you have some in mind? – Andrew May 16 '15 at 6:05
• @Andrew (I'm not non-standard by the way) you literally ignored the question already on the table, so once again you're passing the buck... – Zach466920 May 16 '15 at 15:09
• @Andrew having that cleared up. The weirstrass function is a fractal. Considering that fractals are prevalent through nature this is physical enough. However, if you don't know already, fractals are self similar objects. For instance a lightning bolt and a weirstrass function are both fractals. In fact the first could model the latter. – Zach466920 May 16 '15 at 15:13

A (fractional) derivative is defined as $\dfrac{d^a}{dx}$, where $a$ is any number. The usual values of $a$ are integers and these form the usual derivatives, fractional derivatives consider the cases where $a$ is not an integer.

To begin with, we need to visualize what this means. $\dfrac{d }{dx}$ represents the gradient of a curve, and $\dfrac{d^2 }{dx}$ represents the rate of change of this gradient. And so we need to come up with some meaning for say halfway between these two derivatives, namely $\dfrac{d^{3/2} }{dx}$. This may or may not have any meaning, and should be considered as an abstract concept.

However there is some point to the fractional derivative, especially when integral derivatives do not exist. We can use the fractional derivative to tend to an integral derivative, and might write:

$$\dfrac{d^k}{dx} = \lim_\limits{a\to k}\dfrac{d^a}{dx}, \;\;k\in\mathbb{Z}$$

This works because the fractional derivative, as in the Weierstrass function, can sometimes be defined, whereas the integral derivatives are not, and so we have a useful tool for obtaining a 'limit value' for an integral derivative.

As an answer to the question for an actual value for the derivative, we can reference the paper posted in the first line of the original question, and use $(15)$. If we substitute $q=1$ we get:

$$\dfrac{dW_\lambda(t)}{dt}=\sum_\limits{k=1}^\infty \lambda^{(s-1)k}\cos(\lambda^kt)$$

because $d^0/dx$ is 'no derivative' and leaves the function unaffected, and so is just $cos(\lambda^tx)$.

This is consistent with differentiating $(1)$ from the paper.

• Comments are not for extended discussion; this conversation has been moved to chat. – Daniel Fischer May 20 '15 at 19:36