Re: visualizing a non-potential

[David Bowman <dbowman@TIGER.GEORGETOWNCOLLEGE.EDU>]

Regarding Bob. S.'s comment made on 27 APR 03 in his discussion with
John D.:

> This may be misleading.  Any continuous, scalar function of position can be
> subjected to the gradient operator and generate a vector function of position.
> The original scalar is then a genuine potential function for the vector.

This is not true in general.  Just because a function is continuous
is no reason to presume it must be differentiable.  It is quite easy
to come up with functions that are continuous everywhere and yet are
differentiable nowhere.  Consider the 1-d example of f(x) defined as
a rapidly convergent Fourier series with a very sparse set of
harmonics:  f(x) = SUM{n = 0, infinity| sin((4^n)*x)/2^n} .  This
series is easily seen to be rapidly convergent and the sum is easily
seen to be continuous everywhere, and yet it is differentiable
nowhere.  If we attempt to take its gradient we do not "generate a
vector function of position".  Rather we get a wildly non-convergent
series that does not represent any function at all.

David Bowman