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From: kvanston@muddcs.UUCP (Kathryn Vanstone)
Newsgroups: net.puzzle,net.math
Subject: RE: the derivative (actually the continuity) of x!
Message-ID: <318@muddcs.UUCP>
Date: Tue, 30-Apr-85 18:15:38 EDT
Article-I.D.: muddcs.318
Posted: Tue Apr 30 18:15:38 1985
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Organization: Harvey Mudd College, Claremont, CA
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I know this is old news, but I'm new to the system, and it has taken 
me some time to get organized.  

While listening to the discussion on the derivative of x! I heard to
false statements, one which grates me as a mathematician and another
which is a minor point.

The first: >A necessary and sufficient condition for differentiability
 is that the function is continuous.

I consider the function abs(x) to be continuous, and I dare anyone to 
find its derivative at 0.  For another example, look at x**1/3.

The second: >X! is not continuous.

X! is continuous (taken as a function defined on the non-negative 
integers) by the strict definition of continuity at a point x. (For any 
epsilon > 0 there exists a delta > 0 (try delta = 0.5) such that for all
y (in the set of positive integers) abs(x-y) < delta (which in this case
means that x=y) implies that abs(x! - y!) < epsilon (which is true since
x = y)

Now I will be the first to agree that the above is a bit trivial, and
it does not change the fact that the x! (defined as above) is not
differentiable at any point.

One last comment:  A general mathematicians answer to the question:

"In what space is the function defined?"

					Kathryn Van Stone
			
{allegra!scgvaxd | ucla-cs}!muddcs!kvanston

"Led go by dose, your hurdig be!"