Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've recently been studying the concept of taking fractional derivatives and antiderivatives, and this question has come to mind: If a first derivative, in Cartesian coordinates, is representative of the function's slope, and the second derivative is representative of its concavity, is there any qualitative relationship between a 1/2 derivative and its original function? Or a 3/2 derivative with its respective function?

share|improve this question
    
    
Thanks @Amzoti, for the resource. This may reveal my ignorance, but I wasn't really aware of a subfield "fractional-calculus (like I said, this may simply reveal a certain degree of new-bie ignorance on my part!) –  amWhy Feb 22 '13 at 0:13
    
@AmWhy: I am also ignorant on the subject - so found that nice paper to try and put context on it - and it sounds like it is a relatively new area - since the paper is from 2007. Regards –  Amzoti Feb 22 '13 at 0:20
    
+ And the question is a good question! Not to mention my lack of familiarity with fractional-calculus, the question of understanding what that means, intuitively, is a question very worth the asking! –  amWhy Feb 22 '13 at 0:23
    
The topic shows up in Laplace transform as a curiosity. $L[(-t)^n f(t)]= F^{(n)}(s)$. Now presumably you can take $n=1/3$ for example. –  Maesumi Feb 22 '13 at 4:53

3 Answers 3

up vote 4 down vote accepted

(Non integer) fractional derivatives are non-local so the 1/2 derivative can not have a local meaning like tangent or curvature but would have to take into account the properties of the curve over a large extent (boundary conditions). See http://en.wikipedia.org/wiki/Fractional_calculus#Nature_of_the_fractional_derivative

In addition some of the applications of fractional derivations (http://en.wikipedia.org/wiki/Fractional_calculus#Applications) show the physical meaning is non-local.

Why is the fractional derivative non-local? Integrating a function is not unique or local, because it depends the values of the function over the entire range of integration. The generalization to fractional derivatives unifies differential and integral operators into one Differintegral operator. Whole derivatives are both unique and local. The apparent paradox of fractional derivatives being non-local is actually the natural case, just as integration is non-local.

There is a neat demonstration of calculating the fraction derivative from first principles as a limit in this article http://mathpages.com/home/kmath616/kmath616.htm and it shows how in the whole derivative case the extra terms zero out, but in the fractional case they do not.

share|improve this answer
    
Very interesting, but why must the fractional derivative be non-local? The Wikipedia article doesn't seem to give a justification. –  Jonathan Feb 26 '13 at 22:58
    
I added to my answer to say why it is non-local –  Michael Smith Feb 27 '13 at 3:42
    
Just because it's non-local doesn't mean it doesn't have a geometric interpretation. Also, is there a concept of $\frac{d^b}{da^b} \frac{d^a}{dx^a}$? –  ChickenGod May 23 '13 at 4:37
    
You are right it can have a geometric interpretation just as the single integral can be interpreted as the area under the curve. And it is not a local geometric property as the tangent is. I am having trouble reading the small print in your question and if you are asking can you have arbitrary real numbers in the power of the derivative operator then I believe the answer is yes. Actually I have read you can even put complex numbers in there! –  Michael Smith May 23 '13 at 12:30

Your question is an interesting one. Obviously you are looking for the presentation for this particular theory. As other responders have already stated, the fractional derivative is usually defined in academic papers by a fractional integral. To find the "fractional derivative slope" of a function it is simply a matter of assigning a value to the unknown in the function representing the fractional derivative derived from the fractional derivative. It should be noted that this value will not usually equal the functions normal derivative value, and that normal value is the actual rate of increase of the function. Therefore, the line tangent to the function created by the fractional derivative is different from the whole number value derivative of the function. The different "slope" line created by the fractional derivative can be advantageous in solving certain problems, such as the vector of a particle under different forces, in that particular case, if the forces can be expressed in terms of a function and it's fractional derivative, it greatly simplifies a problem that could involve solving a system of differential equations by simply taking fractional derivatives. Perhaps you are thinking of the definition of a derivative: lim x-> <>X (F(x + <>x) - F(x))/<>x I remember reading a math text in which fractional derivatives were defined in terms of a limit instead of a fractional integral. Unfortunately I have forgotten the author and title. However, defining a fractional derivative in terms of a limit is much more advantageous than in terms of an integral since not all fractional integrals may be integrated. Once again it all depends on the theories presentation in both the case of a limit and integral definition.

share|improve this answer

There several approaches to fractional derivatives. I use the Grunwald-Letnikov derivative and its generalizations to complex plane and the two-sided derivatives. However, most papers use what I call "walking dead" derivatives: the Riemann-Liouville and Caputo. If you want to start, don't loose time with them. There are some attempts to give interpretations to the FD: Prof. Tenreiro Machado and also Prof. Podlubny. The best interpretation in my opinion is the system interpretation: there is a system (linear) called differintegrator with transfer function H(s)=s^a, for Re(s)>0 (forward, causal case) or Re(s) < 0 (backward, anti-causal case). The impulse response of the causal system is t^(a-1)/gamma(-a).u(t) where u(t) is the Heaviside function. Send me a mail and i'll send you some papers mdo@fct.unl.pt

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.