**kvanston@muddcs.UUCP (Kathryn Vanstone)** (05/01/85)

*** REPLACE THIS LINE WITH NOTHING *** 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!"

**mvm9745@acf4.UUCP (Michael V. Mascagni)** (05/03/85)

It appears that the gamma function is unknown to this mathematician. To derive the gamma function from the usual multiplicative definition on the integers, use the Euler product. It is not hard to show that the gamma function is analytic on a real segment, and then by analytic continuation to the complex plane with negative integral points deleted. If one would like to stick to the integers, one can be made to look unnaturally restricted, since the complex field is determined by the ring of integers ... it is the algebraic closure of the completion of the field of fractions of the ring of integers. A.P. Mullhaupt