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%Slide 1
\title{{\LARGE\bf Fractional derivatives (just for fun \ldots )} \\
% or, \\ Can we compute faster in a multiverse? \\
 \dots \\
 \dots \\
  Complex Systems Summer School, Santa Fe}
\author{Tom Carter
\newline
\newline
http://cogs.csustan.edu/\~\ tom/}
\date{June 20, 2001}


\maketitle


%Slide 2
\sectionhead{How would one define:}
     { \Large
                 $$\frac{d^a(f(t))}{dt^a}$$
}

 {\large
          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:
$$\frac{d^a}{dt^a}\left(\frac{d^b(f(t))}{dt^b}\right) = \frac{d^{a+b}(f(t))}{dt^{a+b}}.$$


%\thepage
\pagedone

%Slide 3
\sectionhead{Consider this:}

  If 
     $$F(s) = \int_{-\infty}^\infty f(t) e^{-2\pi i s t}dt$$
  is the Fourier transform of the function $f(t)$, then we have
  $$f(t) = \int_{-\infty}^\infty F(s) e^{2\pi i s t}ds.$$

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

\pagedone

Hence, taking derivatives with respect to $t$ of each side, we have
$$f'(t) = \int_{-\infty}^\infty 2\pi i s F(s) e^{2\pi i s t}ds.$$

Taking another derivative, we have
$$f''(t) = \int_{-\infty}^\infty (2\pi i s)^2F(s) e^{2\pi i s t}ds,$$
and, in general, for the $n^{th}$ derivative,
$$f^{(n)}(t) = \int_{-\infty}^\infty (2\pi i s)^nF(s) e^{2\pi i s t}ds.$$

Homework:  Check this.

\pagedone

So, how about if we define:
$$\frac{d^af(t)}{dt^a} = \int_{-\infty}^\infty (2\pi i s)^aF(s) e^{2\pi i s t}ds.$$

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

Quick sanity checks:  Do we have
$$\frac{d^0f(t)}{dt^0} = f(t)?$$
Do we have
$$\frac{d^a}{dt^a}\left(\frac{d^b(f(t))}{dt^b}\right) = \frac{d^{a+b}(f(t))}{dt^{a+b}}?$$

\pagedone

What about negative $a$?  We have that if $G(s)$ is the Fourier transform of $f'(t)$, then
$$G(s) = 2\pi i s F(s)$$
(homework:  check this \ldots )\newline
and hence:
\begin{eqnarray*}
     \frac{d^{-1}f'(t)}{dt^{-1}} & = & \int_{-\infty}^\infty (2\pi i s)^{-1}G(s) e^{2\pi i s t}ds\\
                                           & = & \int_{-\infty}^\infty (2\pi i s)^{-1}(2\pi i s)F(s) e^{2\pi i s t}ds\\
                                           & = & \int_{-\infty}^\infty F(s) e^{2\pi i s t}ds\\
                                           & = & f(t).
\end{eqnarray*}

\pagedone

In other words, for nice functions (there may be details to worry about), we have
roughly:
$$ \frac{d^{-1}f}{dt^{-1}}  =  \int f$$
(i.e., $\frac{d^{-1}}{dt^{-1}}$ acts like the integral \ldots ).
(Note: this means we also got fractional integrals for free, right?~:-)

Homework question:  How could we interpret
$$ \frac{d^{i}f(t)}{dt^{i}}\ \ \ ?$$

\pagedone
More homework:
\begin{enumerate}
     \item Is this operation linear?  I.e., do we have:
       $$\frac{d^a(c*f(t))}{dt^a} = c * \frac{d^af(t)}{dt^a}$$
and
     $$\frac{d^a(f(t)+g(t))}{dt^a} = \frac{d^af(t)}{dt^a} + \frac{d^ag(t)}{dt^a}?$$
   \item How does this operation work with products?  I.e., what can we say about
$$\frac{d^a(f(t)*g(t))}{dt^a}?$$
\end{enumerate}

\pagedone

Note:  If your response to all this is

\begin{center} ``Cool!!!!!'' \end{center}

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?'' \dots )

:-)
}
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