Do different methods of calculating fractional derivatives have to be equal?

Do different methods of calculating fractional derivatives have to be equal? Or do they sometimes end up differently?

An example would be nice, and if possible, an explanation as too why such formulas can disagree with one another would be exceptional.

The main reason behind this is because I noted that if we could take the fractional derivative of a function through its Taylor series, this would imply that $\frac{d^q}{dx^q}e^x-\frac{d^{q+1}}{dx^{q+1}}e^x={x^{-q-1}\over\Gamma(-q)}$, which tends to $0$ as $q$ tends to become a whole number, but still, this goes against what I would expect.

And of course, explanation on the role of constants of integration would be nice.

• I'm not sure my understanding is correct. Are you talking about different methods for computing the same fractional derivative or different definitions from the fractional derivative, such as Weyl derivative, Marchaud derivative, Liouville derivative, etc. – Upax Jan 1 '17 at 11:15
• @Upax The second one, different definitions for computing, say, the half order derivative of the same function. – Simply Beautiful Art Jan 1 '17 at 12:39
• Doesn't your last line simply implicate that Taylor series are not applicable to fractional derivatives? – Han de Bruijn Jan 2 '17 at 11:04
• @HandeBruijn That's what I found out quite a while after posting this question. As you can see, its a little old, but for the most part, I am still interested in the questions and would like an answer. – Simply Beautiful Art Jan 2 '17 at 12:40
• In general, the Riemann-Liouville and Caputo derivatives would give different results; you are after all composing two operators that do not necessarily commute... – J. M. is a poor mathematician Jan 3 '17 at 21:02

The number of different kinds of fractional derivatives are many, from using laplace and fourier transforms, to integral transform definitions, to finite differences. Some are equivalent but some are different. See my answer here for a rundown. One of issues of fractional calculus, as I see it, is that there is no equivalent Bohr-Mollerup theorem that gives the most appropriate extension of differentiation like the theorem does for factorials.

But I want to make it clear that even in one particular definition, the Riemann-Liouville definition, we have to choose a reference point of integration in order to compute a fractional derivative or integral. In a bit I'll give you a concrete example.

The big jump from regular derivatives to fractional derivatives is that fractional derivatives are non-local operators. That is, $f'(c)$ is well defined and its value unchanged however $f$ is defined on $R \setminus B_{\delta}(c)$.

You are already aware of a non-local operator, which is itself a non positive integer derivative: integration, with $q=-1$. Consider the operator $[D^{-1} f](x) = \int_{-\infty}^x f(t) dt$. At $x=c$, any change in the value of $f$ in some positive measure set in $(-\infty, c)$ will potentially change the value of $[D^{-1} f](x)$. But if you look closely, we have a choice in the lower bound of $-\infty$. We could have just as well chosen $0$ or $-1$, or $1$.

Now you might think that this isn't a big deal because the difference is just a constant. Well, for integration that's true, but it gets more complicated with fractional differentiation.

The R-L definition of a fractional derivative is:

$$[D_a^{q}]f(x) = \frac{d}{dx^n}\frac{1}{\Gamma(n-q+1)} \int_a^x (x-t)^{1-q+n} f(t) dt$$ where $n=\max([q], 0)$ is the smallest nonnegative integer greater than $q$. (The reason we need to do these gymnastics is because the generalization of repeated integration has convergence issues for $q$ negative. So we integrate up fractionally and differentiate integerally to get a net differentiation of $q$.)

So now here is your example. Picking $a=-\infty$, one can compute using the R-L definition that

$$D_{-\infty}^q e^{bx} = b^q e^{bx},$$ for $b > 0$, while on the other hand $$D_{0}^q e^{bx} = x^{-q} E_{1, 1-q}(bx)$$

where $$E_{a,b}(x) = \sum_{k=0}^\infty \frac{x^k}{\Gamma(ak+b)}$$ is the Mittag-Leffler function. It would be a good exercise to compute these.

But actually, a simpler example to work out would be just to compute $$D_{a}^q 1 = \frac{(x-a)^{-q}}{\Gamma(1-q)},$$ and realize the dervitative cannot be equal for different choices of $a$ even up to a constant and more so, even up to a polynomial for non-integer $q$.

The question might now be this: Instead of making equivalence classes of functions modulo constants to make integration unique, can we make some new equivalence class of functions modulo some condition to make fractional differentiation unique? I am not sure this is possible, but it is worth investigating.

• I'm going to leave this bounty open for a while, but you've hit most of the points I wanted to get covered. The last paragraph of yours is of particular interest, which I noticed when I tried the following:$$D_a^q0=?$$While attempting to keep the operators linear. You can see I'd have some problems with this, and so I kind of came to the conclusion that even when considering fractional differentiation via one definition, it cannot be unique due to $D_a^q0$. :-( – Simply Beautiful Art Jan 1 '17 at 20:30
• With respect to integer differentiation and integration, you have to consider functions equivalent up to polynomials. With fractional derivatives, perhaps you need to consider functions modulo terms of the form $\sum x^a_k$ for $a_k$ real. But I truly don't know. Maybe if you limit yourself to analytic functions and work via the series representations you can show this. – abnry Jan 1 '17 at 21:16
• I am not quite sure of your problem with $D_a^q 0$. Can you elaborate? – abnry Jan 1 '17 at 21:17
• Oh, I've studied up on the subject for quite a while (ok, maybe not that long, but long enough) to see some crazy things. And I was thinking that if the operation were linear, than our uniqueness problem reduces down as follows:$$D_a^q[f(x)+0]=D_a^qf(x)+D_a^q0$$And so we have that thing... – Simply Beautiful Art Jan 1 '17 at 21:19
• I'm not sure, but whether or not $D_a^q0=0$ may depend on the definition we are using. And you are right, I was thinking indefinite, since choosing indefinite fulfills the necessary "polynomials" you need to check the functions you have. – Simply Beautiful Art Jan 1 '17 at 21:26

Consider fractional integrals. More than one version of the fractional integral exist. We can consider: \begin{equation} _{c} D_{x}^{-v} f(x)= \frac{1}{\Gamma(v)} \int_{c} ^{x} (x-t)^{v-1} f(t) dt \end{equation} the Riemann definition, \begin{equation} _{x} W_{\infty}^{-v} f(x)= \frac{1}{\Gamma(v)} \int_{x} ^{\infty} (x-t)^{v-1} f(t) dt \end{equation} the Weyl definition, and \begin{equation} _{-\infty} D_{x}^{-v} f(x)= \frac{1}{\Gamma(v)} \int_{-\infty} ^{x} (x-t)^{v-1} f(t) dt \end{equation} the Liouville definition. If we set $c=0$ into the Riemann definition we get the Riemann-Liouville fractional integral. These are not the only definition that exist. For Riemann and Weyl definition we assume $Re \{v\}>0$ and the class of functions to which $f(x)$ belongs is not necessarely the same of the other definitions. If for instance for the Riemann definition which consider a function $f(x)$ which is constant, than the Riemann definition and the Riemann-Liouville definition are meaningfull, but this is not the case for the the Liouville and Weyl definition. The situation is similar when considering fractional derivative. More than one formulation exists, such as The Riemann-Liouville derivative, the Grünwald-Letnikov derivative, the Marchaud derivative. The Marchaud derivative for periodic functions coincides with that of Weyl. The Liouville-Grünwald derivative and the strong Marchaud derivative concide for periodic functions in $L_{2 \pi}^{p}$ with $1\leq p < \infty$. From a purely mathematical point of view, all these definitions are acceptable, even if are not fully equivalent. But since I'm not a mathematicians, I would like to stress the point of view of an engineer. From a practical popint of view we should look at a formulation which is consistent with practical cases. For instance the Grünwald–Letnikov derivative leads to a correct generalization of the current linear systems theory. The Grünwald–Letnikov definition does not assume any bound on the domain of $f(x)$. If we assume that $f(t)=0$ for $t<0$, the derivative imposes causality. Let us apply Riemann and Liouville integral definitions to the function $f(t)=e^{k t}$. We have: \begin{equation} _{0} D_{x}^{-v} e^{k x}= \frac{1}{\Gamma(v)} \int_{0} ^{x} (x-t)^{v-1} e^{k t} dt= k^{-v} e ^{k x} \left(1-\frac{\Gamma(v,k x)}{\Gamma(v)} \right) \end{equation} with $x>0$, and $\Gamma(v,k x)$ the incomplete Gamma function. When using the Liouville definition: \begin{equation} _{-\infty} D_{x}^{-v} e^{k x}= \frac{1}{\Gamma(v)} \int_{-\infty} ^{x} (x-t)^{v-1} e^{k f} dt= k^{-v} e ^{k x} \end{equation} Using the Liouville fractional integral definition the function e kx van- ishes for $k > 0$ at the lower boundary of the integral domain. Consequently there is no additional contribution. with $x>0$ and $k>0$. When considering the fractional derivative according to the Grünwald-Letnikov definition we have: \begin{equation} D_a ^q f(x) = e ^{-j a q} \lim_{h \to 0} \frac{\sum_{k=0} ^n (-1) \binom{q}{k}f(x- k h)}{|h|^q} \end{equation}
where h is a complex number, with $a \in (-\pi,\pi]$. To understand the above formula, assume that x is a time and that h is real, and $a$ is zero or $\pi$. If $a=\pi$ only the present and future values are used, while if $a=0$ only the present and past values are used. This means that if we look at above expression as a linear system, the second case is causal, while the first is not.For $a=0$ we get the so called forward Grünwald–Letnikov derivative, while for $a=\pi$ we have the backward Grünwald–Letnikov derivative. It should be noted that causality makes sense only if we are using the forward or backward derivatives and if $x$ is real. In the case of the constant function $f(x)=0$ for $x \in \mathbb{R}$ and $q \in \mathbb{R} / \mathbb{Z}$ we can write \begin{equation} D_a ^q f(x) = \lim_{h \to 0} \frac{\sum_{k=0} ^{\infty} (-1) \binom{q}{k}}{h^q}= \begin{cases} 0,& \text{if } q> 0\ \\ \infty ,& \text{if } q< 0\ & \end{cases} \end{equation}
As a matter of fact, consider the partial sum of the series: \begin{equation} \sum_{k=0} ^{n} (-1) \binom{q}{k} = (-1) ^n \binom{q-1}{n} = \frac{1}{\Gamma(1-q)} \frac{\Gamma(1+n-q)}{\Gamma(1+n)} \end{equation} As $n \rightarrow \infty$ e quotient of two gamma functions has a known limiting behaviour: \begin{equation} \lim_{n \to \infty}\frac{\Gamma(1+n-q)}{\Gamma(1+n)}= \frac{1}{n^q} \end{equation} So, the q order fractional derivative of 1 is the null function. If $q < 0$, the limit is infinite. So, there is no fractional primitive of a constant. However, this does not happen if q is a negative integer.