Re: Computable functions/reals.

On 21 elo, 23:11, "David C. Ullrich" <dullr...@xxxxxxxxxxx> wrote:
In article
<f5bde8f3-92d0-41f8-909c-2cfa63870...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,



Gc <Gcut...@xxxxxxxxxxx> wrote:
On 21 elo, 18:35, "David C. Ullrich" <dullr...@xxxxxxxxxxx> wrote:
In article
<b3e82287-b4fd-4cf3-9ce3-872506627...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,

Gc <Gcut...@xxxxxxxxxxx> wrote:
On 21 elo, 11:35, David C. Ullrich <dullr...@xxxxxxxxxxx> wrote:
On Wed, 20 Aug 2008 04:24:03 -0700 (PDT), Gc <Gcut...@xxxxxxxxxxx>
wrote:

On 20 elo, 13:41, David C. Ullrich <dullr...@xxxxxxxxxxx> wrote:
On Wed, 20 Aug 2008 02:30:52 -0700 (PDT), Gc <Gcut...@xxxxxxxxxxx>
wrote:

On 20 elo, 01:20, Gc <Gcut...@xxxxxxxxxxx> wrote:
On 20 elo, 00:21, Gc <Gcut...@xxxxxxxxxxx> wrote:

On 19 elo, 23:58, "David C. Ullrich" <dullr...@xxxxxxxxxxx>
wrote:

In article
<2b239c4e-894d-4ed0-bb9e-a6028350d...@xxxxxxxxxxxxxxxxxxxxxxxx
.com
,

Gc <Gcut...@xxxxxxxxxxx> wrote:
On 19 elo, 22:01, "David C. Ullrich" <dullr...@xxxxxxxxxxx>
wrote:
In article
<11869f74-fa93-4870-8833-2dfc56add...@xxxxxxxxxxxxxxxxxxxx
oups
.com>,

Gc <Gcut...@xxxxxxxxxxx> wrote:
On 19 elo, 19:00, "David C. Ullrich"
<dullr...@xxxxxxxxxxx>
wrote:
In article
<f74eca44-7067-4c5e-a633-807a5ab3d...@xxxxxxxxxxxxxxxx
legr
oups.com>,

Gc <Gcut...@xxxxxxxxxxx> wrote:
On 19 elo, 13:31, David C. Ullrich
<dullr...@xxxxxxxxxxx> wrote:
On Tue, 19 Aug 2008 02:42:38 -0700 (PDT), Gc
<Gcut...@xxxxxxxxxxx>

but fine let`s drop this whole
a while.

It's curious that you've stopped saying that the
whole notion of "computable function from R
to R" makes no sense, as of the point when
I gave that link to that wikipedia page (you
remember, the one that you totally misread
as talking about functions defined on the
_computable_ reals).

It In the definition we have "A function f \colon
\mathbb{R} \to
\mathbb{R} is sequentially computable if, for every
computable
sequence \{x_i\}_{i=1}^\infty of real numbers, the
sequence \{f(x_i)
\}_{i=1}^\infty is also computable." So this
definition
depens only on
the computable sequences and that the functions
"behave
well" in the
noncomputable reals.

You're totally hopeless. Even after I say you should
read
it
much more carefully you simply _don't_.

Yes, the definition of "sequentially computable"
depends
only
on f(x) for computable x - what you say about _that_
definition
makes sense.

NOW look at the definition of "computable"! To do
that
you're
going to have to look down a few lines. The
definition of
"computable" does _not_ have the property that it
depends
only
on the restriction of f to the computable reals.

You should read the rest of my post, before calling me
hopeless! I
never said that
that it depens only about computable reals, it depens
also
from the
smoothness of the function on R.

In practice nothing would lost if we restricted
to the computable reals. I don`t still approve that
we
can say that
the function f:R--->R is computable in the point a,
where a is a
noncomputable real. Wikipedia is not the God you
know,

And neither are you. It _is_ curious that when I say
something
about this you insist it makes no sense and
contradicts
the CT
thesis, now when you see something analogous on
Wikipedia
that's weakened to you don't "approve".

Heh. This Wikipedia thing is not even essentially your
original
definition,

I haven't yet decided whether it's equivalent, but the
two
are very closely related. Before saying anything silly
about
that you should stop and recall the other day, when you
seemed to think that the definition I gave must allow
discontinuous functions, just because it didn't mention
the word "continuous".

But It does allow, but not on irrationals :)))
Explain this. Let f be 1 on R/{0} and 0 on 0. Is this
computable by
your definition,

No. I've already explained why not. Twice.

Heh. Explain again:) Let me sketch imprtant parts of this
machine
to
you. If it reads 0 it goes to certain stage and moves right. If
it
then reads blank it halts and prints 0...otherwise when it
finds
first
1 on the input tape it halts and prints 1.

God you can be dense.

What if the input tape has all 0's on it? The machine is supposed
to ouput 0 if and only if _all_ of the input is 0's. But it never
finds out that _all_ of the input is 0's, so it can never output
anything. That function is not computable.

This is I think the fourth time this has been explained to you
(the third explanation from me, one was from someone else).

Eh. Now you are claiming that you decimal expanions are non standard .

Huh????????????????????

How is 0.0000000000... a "nonstandard" decimal?

You Idiot. Look athttp://mathworld.wolfram.com/DecimalExpansion.html
You can`t play with representations against the fact itself.

Relying on mathworld is a bad idea. In mathematics the decimal
expansion of a number (between 0 and 1, say) is an infinite sequence
of digits, such that [etc].

I think there seems to be a difference between decimal reprensation
and decimal expansion but fine. Let`s say your machine prints
0,0000... if the first decimal on the input is 0, otherwise it prints
1.0000000....

So you lied when you said that other post was your last post.

I changed my mind for a while. This is my last post on the probabilty
0.9.


If the machine does what you say then it's not calculating
a function from R to R, because it outputs different things
for the input .0999... and the input 0.1000... .

OK.



Before you complain that now I'm adding new requirements,
I'm not. Go back to the very first thing I said:

" f : R -> R is computable
if there exists a Turing machine which writes
a decimal expansion of f(x) on the output if
given a decimal expansion of x on the input."

The crucial part is can it start in some point all over again, in a
way that
it puts what it previously wrote to the trash can. It can do this at
least if the output is any Cauchy sequence.
When you said that you rather work with Cauchy sequences you didn`t
explicitly say that for every computable cauchy sequence your
computation gives the same function, which is effectively the
requirement for continuity. There are functions on R, where a proper
function gives different value for different Cauchy sequences
convergin to x you know.

Note the words "a decimal expansion of f(x)"
and "a decimal expansion of x". What you just
said does not define a function from R to R;
if you change that by defining f(1/10) = something
it still doesn't work, because whichever you choose
for f(1/10) there is _a_ decimal expansion of 1/10
such that the machine does _not_ output an expansion
of f(1/10).

--
David C. Ullrich

.

Relevant Pages

  • Re: how to prove that a set is complete?
    ... it's a Cauchy sequence of real numbers and therefore it ... "because we construct the reals so that they are complete", ... Assume the Supremum Principle. ...
    (sci.math)
  • Re: ga4- Integers found that form the sqrt(2) or transcendental numbers.
    ... The statement that "reals are the ... >arrives at a quotient of integers. ... and thus has no finite decimal expansion. ... In your answer you require that in my sequence r must be rational to start ...
    (sci.math)
  • Re: Dedekind Cuts, Fundamental Sequences: why?
    ... sequence does not require any reference to a value to which the sequence ... "A sequence converges if and only if it is a Cauchy sequence, ... It is possible to give a definition of convergence in terms of rational ... reals", ...
    (sci.math)
  • Re: Cantor Confusion
    ... Then you're not talking about the reals. ... sequence of rationals converges to a unique real number, ... countable decimal expansion represents a real number. ... And where does Cantor argue this? ...
    (sci.math)
  • Re: Dedekind Cuts, Fundamental Sequences: why?
    ... sequence does not require any reference to a value to which the sequence ... A Cauchy sequence in any complete metric space must converge. ... rationals are not a complete metric space. ... That's true in the reals, ...
    (sci.math)
Loading