Hetware
Posts:
148
Registered:
4/13/13


Re: Is (t^29)/(t3) defined at t=3?
Posted:
Oct 19, 2013 11:40 PM


On 10/19/2013 4:52 PM, quasi wrote: > Hetware wrote: >> >> Let me propose a test question: >> >> Given >> >> f(x) defined over R, > >> limit[f(s), s>x] exist over R, > >> and g(x) = limit[f(s), s>x] >> >> can g(x) be used to find the value of f(x) for all x in R? > > No, not unless it is also given that f is continuous. >
On 10/19/2013 7:34 AM, Peter Percival wrote:> Hetware wrote: > >> If we define f(x) as >> >> f(x) = x/x and c[f(x)], > > That's cheating. You've been told at least eight million times that you > may define f with a formula, but you cannot put "and c[f(x)]" in the > definition. If c[f(x)] is true it has to be proved _from_ the definition. > > > [snip] >> I believe this is a profound observation, closely related to Russell's >> Paradox. > >
The definition of c[f(x)] was given as follows:
Let c[f(t)] be the assertion that for every real number t there is some real number y such that f(t) = y, and also that for every real number t limit[f(s), s>t] where s is a real number in the neighborhood of t. In poor man's notation:
c[f(t)] => [y = f(t) y in R, all t in R] AND [limit[f(s), s>t] = f(t)]
'=>' is to be read 'implies'.
> For example, define f by > > f(x) = 1 if x != 0 > > f(0) = 2 > > Then > > f(x) is defined for all x in R, > > and limit (s > x) f(s) exists for all x in R, > > but if g is defined by > > g(x) = limit (s > x) f(s) > > then g(x) = 1 for all x in R, hence > > g(0) != f(0) > > so f,g are not the same function. > > Now suppose f(0) is known to exist, but the value of f(0) is > not explicitly specified. Then f(0) could potentially be any > real number. In particular, in the absence of more information, > you can't conclude that f(0) = g(0). > > On the other hand, suppose your proposed question is reworded > in the following way ... > > Given: > > f(x) is defined and continuous for all x in R.
I believe we can drop the "defined" since it is implied by "continuous".
> Then if g is defined by > > g(x) = limit (s > x) f(s) > > can g(x) be used to determine the value of f(x) for all x in R? > > The answer is yes  in fact, f(x) = g(x) for all x in R. > > quasi >
What is your opinion of the definition given as follows:
Let f(x) be continuous over R, and f(x) = x/x for all x in R where x/x is a determinate form?
I believe f(x) to be consistently and uniquely defined for all x in R, under that specification.
Clear the previous definition of f(x).
Elsewhere in this thread, I recently posted a development similar to the following.
Let P(f) be the proposition /f(x) is continuous for all x in R/.
P(f) => f(x) = limit[f(s), s > x] such that f(x) in R for all x in R.
Let D(f) mean f(x) = x/x for all x in R where x/x is a determinate form.
So D(f) does not defined f(0). That is not to say f(0) cannot be defined. Since x/x is only indeterminate for x = 0, we shall focus our attention on x = 0 and its neighborhood.
Now assert P(f) on f(x) satisfying D(f).
P(f) => f(x) = limit[x/x, s > x] such that f(x) in R for all x in R. P(f) => f(x) = 1 for all x in R. P(f) is true.
That is to say, f(x) provides a means of determining a suitable value for all x in R under the assertion that f(x) is continuous. In particular, f(0) = 1.
Next assert !P(f) on f(x) satisfying D(f).
!P(f) => [(f(x) != limit[f(s), s > x]) or (f(x) not in R) or (limit[f(s), s > x] not in R)] for some x in R.
Since we know x = 0 is the only value for x where f(x) is not given explicitly by D(f), f(0) must be the case where !P(f).
I'm going to have to review this in the morning. I was already suffering from focus fatigue when I went to buy my 3 12oz. bottles of Sneaky Pete, Imperial IPA.

