> From: Gabriel Dos Reis <[EMAIL PROTECTED]>
>> Paul Schlie <[EMAIL PROTECTED]> writes:
> | > Gabriel Dos Reis wrote:
> | > You probably noticed that in the polynomial expansion, you are using
> | > an integer power -- which everybody agrees on yield 1 at the limit.
> | >
> | > I'm tlaking about 0^0, when you look at the limit of function x^y
> |
> | Out of curiosity, on what basis can one conclude:
> |
> | lim{|x|==|y|->0} x^y :: lim{|x|==|y|->0} (exp (* x (log y))) != 1 ?
>
> The issue is not whether the limit of x^x, as x approaches 0, is 1 not.
> We all, mathematically, agree on that.
>
> The issue is whether the *bivariate* function x^y has a defined limit
> at (0,0). And the answer is unambiguously no.
> Checking just one path does NOT suffice to assrt that the limit
> exists. (However, that might suffice to assert that a limit does not
> exist).
>
> I'm deeply burried somewhere in the middle-west deserts and I have no
> much reliable connection, so I'll point you to the message
>
> http://gcc.gnu.org/ml/gcc/2005-03/msg00469.html
>
> where I've tried to taint this discussion with some realities from what
> standard bodies think on the 0^0 arithmetic, and conterexample you can
> check by yourself.
>
> | As although it's logarithmic decomposition may yield intermediate complex
> | values, and may diverge prior to converging as they approach their limit,
> | it seems fairly obvious that the expression converges to the value of 1
>
> You've transmuted the function x^y to the function x^x which is a
> different beast. Existing of limit of the latter does not imply
> existance of limit of the former. Again check the counterexamples in
> the message I referred to above.
Thank you. In essence, I've intentionally defined the question of x^y's
value about x=y->0 as a constrained "bivariate" function, to where only
the direction, not the relative rate of the argument's paths are ambiguous,
as I believe that when the numerical representation system has no provision
to express their relative rates of convergence, they should be assumed to be
equivalent; as the question of a functions value about any static point such
as (0,0) or (2,4) etc., is invalid unless that point is well defined within
it's arguments path; where if it is, then the constrained representation is
equally valid, but not otherwise (as nor is the question).
Therefore in other words, the question of an arbitrary function's value
about an arbitrary static point is just that, it's not a question about a
functions value about an arbitrary point which may or may not be intersected
by another function further constraining it's arguments.
Therefore the counter argument observing that x^y is ambiguous if further
constrained by y = k/ln(x), is essentially irrelevant; as the question is
what's the value of x^y, with no provision to express further constraints
on it's arguments. Just as the value of (x + y) if further constrained by
y = x, about the point (1,2) would be both ambiguous and an irrelevant to
the defined value of (x + y) about (1,2).
I believe things are being confused by a misinterpretation of the meaning
of what a limit about an infinite boundary truly means; as although most
understand that lim{x->1; y->2} implies convergence about the static valid
point (1,2) where it would be obvious that if x and y were further
constrained such that (1,2) were invalid, then so too would be the question;
just as lim{x->0; y->0} should be equally treated. But it's being abused by
those who don't understand that just because all of a function's arguments
may approach a given set of values eventually, if they do not
simultaneously, then that set of values does not lie in the function's path,
therefore irrelevant; just as (0,0) does not lie in y = k/ln(x)'s path,
therefore an invalidating simultaneous constraint; as otherwise it would be
valid to argue that (0,0) lies on y = x, y = x + 1, y = x + 2, ...
simultaneously, which is more obviously false, making it more apparent that
it's important to differentiate the static points they imply, from the
infinite boundaries they simultaneously abstractly represent in the form of
0 ~ lim{->0} :: lim{->1/inf}, and inf ~ lim{->inf} :: lim{->1/->0}.
Very long story short, it seems clear that:
f(a,b) :: lim{v->1/inf) f(a+/-v,b+/-v)
about any static point, when defined independently of any other arbitrary
constraints.
