Another way to define fractional derivatives

Posted on by John

There are many ways to define fractional derivatives, and in general they coincide on nice classes of functions. A long time ago I wrote about one way to define fractional derivatives using Fourier transforms. From that post:

Here’s one way fractional derivatives could be defined. Suppose the Fourier transform of f(x) is g(ξ). Then for positive integer n, the nth derivative of f(x) has Fourier transform (2π i ξ)n g(ξ). So you could take the nth derivative of f(x) as follows: take the Fourier transform, multiply by (2π i ξ)n, and take the inverse Fourier transform. This suggests the same procedure could be used to define the nth derivative when n is not an integer.

Here’s another way to define fractional derivatives that doesn’t use Fourier transforms, the Grünwald-Letnikov derivative. It’s a fairly direct approach.

The starting point is to note that for positive integer n, the nth derivative can be written as

f^{(n)}(x) = \lim_{h\to 0} \frac{(\Delta^n_h f)(x)}{h^n}

where

(\Delta_h f)(x) = f(x) - f(x - h)

and Δn iterates Δ. For example,

(\Delta_h^2 f)(x) = (\Delta_h (\Delta_h) f)(x) = f(x) - 2 f(x-h) + f(x - 2h).

In general, for positive integer n, we have

(\Delta^n_h f)(x) = \sum_{k=0}^\infty (-1)^k {n \choose k} f(x - kh).

We could set the upper limit of the sum to n, but there’s no harm in setting it to ∞ because the binomial coefficients will be zero for k larger than n. And in this form, we can replace n with the integer n with any positive real number α to define the αth derivative. That is, the Grünwald-Letnikov derivative of f is given by

f^{(\alpha)}(x) = \lim_{h\to 0} \frac{(\Delta^\alpha_h f)(x)}{h^\alpha} = \lim_{h\to 0} h^{-\alpha} \sum_{k=0}^\infty (-1)^k {\alpha \choose k} f(x - kh).

See these notes for the definition of binomial coefficients for possibly non-integer arguments and for an explanation why for integer n the coefficients are eventually zero.

Notice that fractional derivatives require non-local information. Ordinary derivatives at a point are determined by the values of the function in an arbitrarily small neighborhood of that point. But notice how the fractional derivative, as defined in this post, depends on the values of the function at an evenly spaced infinite sequence of points. If we define fractional derivatives via Fourier transform, the non-local nature is more apparent since the Fourier transform at any point depends on the values of the function everywhere.

This non-local feature can be good or bad. If you want to model a phenomena with non-local dependence, fractional differential equations might be a good way to go. But if your phenomena is locally determined, fractional differential equations might be a poor fit.

Related post: Mittag-Leffler functions

Update: Yet another way to define fractional derivatives, this time via fractional integrals, which can be defined in terms of ordinary integrals

3 thoughts on “Another way to define fractional derivatives”

  2. Haniffa Nasir

    The grunwald-Letnikov and Riemann-Liuville definitions are said to be equivalent in many books.
    But, the R-L derivative of constant is not zero while G-L derivative of constant is zero.
    How are these two equivalent?

  3. Babak Farhang

    Very interesting! But I don’t understand why fractional derivatives as defined here require “non-local” information. While f(x-kh) is not as “local” as f(x-h), if I understood right, only finitely many k’s count, and h -> 0, besides.

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