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Topic: Is (t^2-9)/(t-3) defined at t=3?
Replies: 166   Last Post: Oct 30, 2013 9:41 AM

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Hetware

Posts: 148
Registered: 4/13/13
Re: Is (t^2-9)/(t-3) defined at t=3?
Posted: Oct 19, 2013 11:40 PM
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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.


Date Subject Author
9/28/13
Read Is (t^2-9)/(t-3) defined at t=3?
Hetware
9/28/13
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Michael F. Stemper
9/28/13
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scattered
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Hetware
9/28/13
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quasi
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Hetware
9/28/13
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quasi
9/28/13
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Peter Percival
9/29/13
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quasi
9/28/13
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Hetware
9/28/13
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Richard Tobin
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Hetware
9/28/13
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9/29/13
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Hetware
10/6/13
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Hetware
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Hetware
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10/13/13
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fom
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10/14/13
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