Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

For $f(x)=x$, the half-derivative of $f$ is $$\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}} x = 2 \sqrt{\frac{x}{\pi}} \;.$$ Is there some geometric interpretation of (Q1) this specific derivative, and, (Q2) of the half-derivative more generally? I have read that fractional derivatives are nonlocal, but it seems strange to me that integral derivatives can be described in terms of local geometry only, while fractional derivatives cannot be so described. This would suggest an odd discontinuity between, say, $d^{1}$ and $d^{1.01}$. This seems especially at odds with the many applications of fractional derivatives, which (superficially) suggests continuity should reign.

I'd appreciate someone clearing up my elementary confusions—Thanks in advance!

Addendum (5Jan14). @AlexR. found this geometric interpretation of the fractional integral in Richard Herrmann's book, Fractional Calculus: An Introduction for Physicists,World Scientific, 2011:

share|improve this question
Well, my view of fractional derivatives is from it's kernel (by using Fourier transform), though this has nothing to do with geometric interpretation. For integral derivatives,like $\frac{d^k}{dx^k}$, the Schwartz kernel is $\delta_0^{(k)}$, which measures only on a point, however, when considering the fractional derivative,like $\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}$ on $\mathbb{R}$, the respective kernel is $|x-y|^{-1-\frac{1}{2}}$, which is nonlocal anymore. –  shanlin Jan 4 at 2:43
I haven't found fractional powers of the first derivative $d/dx$, which is indefinite, to be of much use. Fractional powers $\Delta^\alpha$ of the (analyst's) Laplacian $\Delta = -\sum_{i=1}^n \frac{\partial}{\partial x_i^2}$, on the other hand, are much more useful (note that the Laplacian is positive definite, in contrast to the first order operator.) The square root of the Laplacian has a natural geometric interpretation as the Dirichlet-to-Neumann operator for the upper half-plane. –  Terry Tao Jan 4 at 2:46
Also, fractional derivatives depend continuously on the exponent $\alpha$ (at least in the distributional topology), as can be seen on the Fourier side. The non-locality disappears as $\alpha$ approaches a natural number due to denominators such as $\Gamma(-\alpha)$ that appear in the formulae for the kernel away from the origin (which is something like $\frac{1}{\Gamma(-\alpha)} |x-y|^{-1-\alpha}$ in one dimension). –  Terry Tao Jan 4 at 2:49
One obstruction to an obvious geometric interpretation is that the ordinary derivative can be regarded as generalizing to, for example, the exterior derivative, but I'm not aware of a generalization of the half-derivative this broad without extra structure (fractional powers of the Laplacian require a Riemannian metric, for example). –  Qiaochu Yuan Jan 4 at 6:00
A correction to my previous comment: the first order operator $d/dx$ is indefinite on the real line (the spectral variable $\xi$ can be either positive real or negative real), and so it is not natural to consider fractional powers of this operator (one has to arbitrarily choose a branch cut for $\xi^\alpha$). But if one is working on the half-line instead, then the spectral variable $\xi$ now naturally lives on the upper half-plane (Fourier-Laplace transform) and one now has a canonical interpretation of $\xi^\alpha$. So fractional powers of $d/dx$ are reasonable in half-line settings. –  Terry Tao Jan 4 at 16:54
show 5 more comments

2 Answers 2

up vote 31 down vote accepted

A possible mechanical interpretation of the half-derivative can be given in terms of Abel's solution to a classical problem from the calculus of variations (the tautochrone problem).

Let there be a heavy particle which is constrained to slide without friction along the curve $y=y(t)$ in uniform gravity to its lowest point. Then, given a function $T(y)$ that specifies the total time of descent for a given starting height what is an equation of the curve that yields this result?

The principle of conservation of energy implies that the distance $S=S(t)$ travelled by the particle along the curve from the initial height $y_0$ satisfies the equation $$\left(\frac{dS}{dt}\right)^2=2g(y_0-y).$$ This is equivalent to the integral equation $$T(y_0)=\frac{1}{\sqrt{2g}}\int_0^{y_0}\frac{1}{(y_0-y)^{1/2}}\frac{dS}{dy}dy.$$

The r.h.s. of the latter equation is nothing else but the Riemann–Liouville fractional integral of $f=\pi^{1/2}(2g)^{-1/2}dS/dy$, i.e. $$D^{-\alpha}f(x)=\frac{1}{\Gamma(\alpha)}\int_0^x (x-y)^{\alpha-1}f(y)dy$$ of the order $\alpha=1/2$.

The solution to Abel's integral equation $dS/dy$ can be now interpreted (up to a constant factor) as the half-derivative of $T=T(y_0)$.

enter image description here

share|improve this answer
And what is the heuristic interpretation of this... interpretation? –  Qfwfq Jan 4 at 8:35
Beautiful!! Thanks for the lucid lesson! –  Joseph O'Rourke Jan 4 at 13:31
add comment

Interference of traveling waves can be used to model the action of fractional integro-derivatives.

The FIDs can be regarded as an interpolation of the integral derivatives of the function acted upon--to be more precise, a sinc function interpolation, as in the Whittaker-Shannon interpolation formula, of the integral derivatives, properly normalized .

The FID can be expressed as

$$$$ $\displaystyle\frac{d^{\beta}}{dx^\beta}f(x)=\lim_{\:r \rightarrow 1^{-}}\frac{1}{2\pi i}\displaystyle\oint_{|z-x|=r \cdot x}f(z)\frac{\beta!}{(z-x)^{\beta+1}}dz=FP\displaystyle\int_{0}^{x}f(z)\frac{\;\;\;\;\;(x-z)^{-\beta-1}}{(-\beta-1)!} dz$, $$$$

where $x>0$, $(z-x)=r \cdot x e^{i\theta}$ in the contour integral, $-\pi<\theta<\pi$, $FP$ denotes a Hadamard-type finite part, and $\beta$ is real,

or, with a change of variable and notation and using $\binom{0}{\beta}=\frac{1}{\beta!(-\beta)!}=\frac{\sin (\pi \beta)}{\pi \beta}$, formally

$$$$ $\displaystyle D^{\beta} f(x)=FP\displaystyle\int_{0}^{x}f(x-z)\frac{\;\;\;\;\;z^{-\beta-1}}{(-\beta-1)!} dz=FP\displaystyle\int_{0}^{x}\frac{\;\;\;\;\;z^{-\beta-1}}{(-\beta-1)!}exp(-zD)\; dz\; f(x)$ $$$$

$=\displaystyle\sum_{n=0}^{\infty } (-1)^n \frac{f^{(n)}(x)}{n!}\frac{1}{(-\beta-1)!}\frac{x^{n-\beta}}{n-\beta}=\frac{\beta!}{x^{\beta}}\sum_{n=0}^{\infty }\frac{x^{n}}{n!}f^{(n)}(x)\frac{sin(\pi(n-\beta))}{\pi(n-\beta)}.$ $$$$

Summarizing, we have an interpretation of a FID as the sinc fct. interpolation

$$$$ $$\displaystyle \frac{x^{\beta}}{\beta!}D^{\beta} f(x)=\sum_{n=0}^{\infty }\frac{x^{n}}{n!}f^{(n)}(x)\frac{sin(\pi(n-\beta))}{\pi(n-\beta)}.$$ $$$$

For $f(x)=x$ and $\beta=\frac{1}{2}$, this gives $$$$ $$\displaystyle \frac{x^{\frac{1}{2}}}{(\frac{1}{2})!}D^{\frac{1}{2}} x=x\;[\frac{sin(\pi(-\frac{1}{2}))}{\pi(-\frac{1}{2})}+\frac{sin(\pi(\frac{1}{2}))}{\pi(\frac{1}{2})}]=x\;\frac{4}{\pi},$$


$$\displaystyle D^{\frac{1}{2}} x= {\frac{1}{2}}!\;\frac{4}{\pi}x^{\frac{1}{2}}=\;\frac{2}{\sqrt{\pi}}x^{\frac{1}{2}}=\frac{x^{\frac{1}{2}}}{(\frac{1}{2})!}.$$ $$$$

Construct traveling pulses from plane waves over an inverse wavelength ($k=1/\lambda$) bandwidth of $k_0$: $$$$ $$\displaystyle \frac{x}{k_0}\int_{-\frac{k_0}{2}}^{\frac{k_0}{2}}cos(2\pi k(u-vt+n/k_0)) dk=x\frac{sin(\pi k_0(u-vt+n/k_0))}{\pi k_0(u-vt+n/k_0)}$$ $$$$

where $n=0$ or $1$, and $u, v, t,$ and $x$ are the spatial coordinate, wave velocity, time, and wave amplitude, respectively.

The interference of the the pulses at $u=0$ for $t=\frac{\beta}{k_0 v}$ is

$$$$ $$\displaystyle x\;[\frac{sin(\pi(-\beta))}{\pi(-\beta)}+\frac{sin(\pi(1-\beta))}{\pi(1-\beta)}]=\frac{x^\beta}{\beta!}D^{\beta} x.$$ $$$$

For $f(x)=x^m$, $m+1$ traveling sinc pulses are required with the amplitude of the $n$'th pulse being $\binom{m}{n}x^m$.

For $f(x)=x^\alpha$, the interpolation formula gives, for $Re(\alpha)>-1$,

$$$$ $$\displaystyle \frac{x^{\beta}}{\beta!}D^{\beta} x^\alpha=x^\alpha\displaystyle\sum_{n=0}^{\infty }\binom{\alpha }{n}\frac{\sin (\pi (\beta -n))}{\pi (\beta -n)}=\binom{\alpha }{\beta}x^\alpha,$$


$$\displaystyle D^{\beta} \frac{x^\alpha}{\alpha!}=\frac{x^{\alpha-\beta}}{(\alpha-\beta)!}.$$ $$$$


1) $x^\alpha$ is an eigenfunction of the operator $x^\beta D^\beta$, so we can expect the inverse Mellin transform to aid in characterizing the operator's action on various functions. In fact, this interpolation perspective is consistent with the discussion in MO79868 (What does Mellin inversion really mean?) of the Mellin transform as providing an interpolation of the coefficients of the Taylor series of a function (in this case, the function $w(z)=H(x-z)f(x-z)$ where $H(x)$ is the Heaviside step function), a perspective on Ramanujan's Master Formula / Theorem.

2) All of this, of course, is related to the theory of Green's functions and impulse response functions in signal processing. And, the final example concerns essentially the beta function, which is related to scattering amplitudes in string theory.)

3) It seems a little misleading to distinguish between "locality" and "non-locality" based on the singularities of the gamma fct. From the contour integral rep, the distinction is maybe better understood as the difference between pole singularities and branch cuts for the "Cauchy" kernel.

share|improve this answer
The Laplace transform provides the appropriate spectral interpretation of this FID, not the Fourier. Look at the difference between the respective convolution theorems and between the transforms of the Heaviside step function. –  Tom Copeland Feb 11 at 6:02
For an axiomatic derivation of the series expressions for the fractional integroderivatives see The Role of Salvatore Pincherle in the Development of the Fractional Calculus by Mainardi and Pagnini. Harold Davis in The Theory of Linear Operators ... (Principia Press, 1936) gives a detailed history of differently defined FIDs. –  Tom Copeland Mar 14 at 12:01
add comment

Your Answer


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.