Fractional derivatives

(just for fun …)

… …

Complex Systems Summer School, Santa Fe

Tom Carter http://cogs.csustan.edu/˜ tom/

June 20, 2001

How would one define:

\frac{d^{a}(f(t))}{dt^{a}}

For an arbitrary real (or even complex) number a?

Note: we probably at least want our new definition to agree with the old one wherever it can, and to have:

\frac{d^{a}}{dt^{a}} (\frac{d^{b}(f(t))}{dt^{b}} ) = \frac{d^{a+b} (f(t))}{dt^{a+b}} .

Consider this:

If

F (s) = ∫_{−∞}^{∞} f(t)e^{−2πist}dt

is the Fourier transform of the function f(t), then we have

f(t) = ∫_{−∞}^{∞} F (s)e^{2πist}ds.

(The inverse transform of the transform of a function is the original function back again.)

Hence, taking derivatives with respect to t of each side, we have

f′(t) = ∫_{−∞}^{∞} 2πisF (s)e^{2πist}ds.

Taking another derivative, we have

f′′(t) = ∫_{−∞}^{∞}(2πis)^{2}F (s)e^{2πist}ds,

and, in general, for the n^{th} derivative,

f^{(n)}(t) = ∫_{−∞}^{∞}(2πis)^{n}F (s)e^{2πist}ds.

Homework: Check this.

So, how about if we define:

\frac{d^{a}f(t)}{dt^{a}} = ∫_{−∞}^{∞}(2πis)^{a}F (s)e^{2πist}ds.

Note that for any positive whole number n, this definition agrees with the traditional definition of the derivative.

Quick sanity checks: Do we have

\frac{d^{0}f(t)}{dt^{0}} = f(t)?

Do we have

\frac{d^{a}}{dt^{a}} (\frac{d^{b}(f(t))}{dt^{b}} ) = \frac{d^{a+b} (f(t))}{dt^{a+b}} ?

What about negative a? We have that if G(s) is the Fourier transform of f′(t), then

G(s) = 2πisF (s)

(homework: check this …) and hence:

\frac{d^{−1}f′(t)}{dt^{−1}} = ∫_{−∞}^{∞}(2πis)^{−1}G(s)e^{2πist}ds
= ∫_{−∞}^{∞}(2πis)^{−1}(2πis)F (s)e^{2πist}ds
= ∫_{−∞}^{∞} F (s)e^{2πist}ds
= f(t).

In other words, for nice functions (there may be details to worry about), we have roughly:

\frac{d^{−1}f}{dt^{−1}} = ∫ f

(i.e., \frac{d^{−1}}{dt^{−1}} acts like the integral …). (Note: this means we also got fractional integrals for free, right? :-)

Homework question: How could we interpret

\frac{d^{i}f(t)}{dt^{i}} ?

More homework:

1. Is this operation linear? I.e., do we have:

\frac{d^{a}(c ∗ f(t))}{dt^{a}} = c ∗ \frac{d^{a}f(t)}{dt^{a}}

and

\frac{d^{a}(f(t) + g(t))}{dt^{a}} = \frac{d^{a}f(t)}{dt^{a}} + \frac{d^{a}g(t)}{dt^{a}}?

2. How does this operation work with products? I.e., what can we say about

\frac{d^{a}(f(t) ∗ g(t))}{dt^{a}} ?

Note: If your response to all this is

“Cool!!!!!”

then you have the makings of a mathematician!

(As you might expect, I won’t go into what it means if your response is “But what is it good for?” – and I’m confident by now that nobody’s response is “hunh?” …)

:-)