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