Could this be paradoxial?
Han de Bruijn
supposed to be trivial and harmless:
a = lim(x->a) x
Where the limit is defined as (note the '<>'):
forAll delta > 0 : | x - a | < delta and x <> a
Question: _is_ this statement trivial and harmless indeed, or can it
give rise to paradoxes, when applied without additional conditions ?
What conditions would be involved then ? Or is the above nonsensical
altogether and why?
Han de Bruijn
Gus Gassmann
It is nonsensical, for a number of reasons.
First, limits exist in the context of sequences and functions. Your
situation does not seem to fit either. In both cases you need two
quantifiers (for all epsilon > 0 exist N resp. x ...)
Second, what is x? You say it is a real number, but you clearly intend
it to *vary*.
Third, what is your indented statement even supposed to say? I cannot
decipher it other than conditions on the *number* x. But that is not a
limit.
Fourth, with the above interpretation, you clearly cannot find such a
real *number* x.
So you must have had something else in mind.
David C. Ullrich
<umu...@gmail.com> wrote:
>Let x and a in the real numbers. Then the following statement is
>supposed to be trivial and harmless:
>
> a = lim(x->a) x
>
>Where the limit is defined as (note the '<>'):
>
> forAll delta > 0 : | x - a | < delta and x <> a
Huh? That's not the definition. The definition of
" a = lim(x->a) x" is
for all eps > 0 there exists delta > 0 such that
for all x with |x - a| < delta and x <> a we have |x - a| < eps.
>Question: _is_ this statement trivial and harmless indeed, or can it
>give rise to paradoxes, when applied without additional conditions ?
It's impossible to say for certain that _anything_ cannot "give rise
to paradoxes". But why do you imagine that this definition might do
so?
>What conditions would be involved then ? Or is the above nonsensical
>altogether and why?
Not sure what you're asking - which statement are you referring to?
The statement that "forAll delta > 0 : | x - a | < delta and
x <> a" is the definition of " a = lim(x->a) x" is simply false.
The statement "forAll delta > 0 : | x - a | < delta and
x <> a" is obviously false for any given x and a.
Luckily this has no relevance to the definition of limits.
The statement "a = lim(x->a) x" is harmless and does not
lead to any paradoxes, as far as anyone knows. Of course
it has to be interpreted correctly, which is to say one has
to get the definition straight. And yes, it's trivial to prove
from the definition.
>Han de Bruijn
>
Han de Bruijn
> On Sat, 26 Mar 2011 05:31:57 -0700 (PDT), Han de Bruijn
>
> >Let x and a in the real numbers. Then the following statement is
> >supposed to be trivial and harmless:
>
> > a = lim(x->a) x
>
> >Where the limit is defined as (note the '<>'):
>
> > forAll delta > 0 : | x - a | < delta and x <> a
>
> Huh? That's not the definition. The definition of
> " a = lim(x->a) x" is
>
> for all eps > 0 there exists delta > 0 such that
> for all x with |x - a| < delta and x <> a we have |x - a| < eps.
You're right. Let's choose delta(eps) = eps then and we're done. OK?
> >Question: _is_ this statement trivial and harmless indeed, or can it
> >give rise to paradoxes, when applied without additional conditions ?
>
> It's impossible to say for certain that _anything_ cannot "give rise
> to paradoxes". But why do you imagine that this definition might do
> so?
I'm trying to be cautious once in a while :-)
> >What conditions would be involved then ? Or is the above nonsensical
> >altogether and why?
>
> Not sure what you're asking - which statement are you referring to?
>
> The statement that "forAll delta > 0 : | x - a | < delta and
> x <> a" is the definition of " a = lim(x->a) x" is simply false.
>
> The statement "forAll delta > 0 : | x - a | < delta and
> x <> a" is obviously false for any given x and a.
> Luckily this has no relevance to the definition of limits.
>
> The statement "a = lim(x->a) x" is harmless and does not
> lead to any paradoxes, as far as anyone knows. Of course
> it has to be interpreted correctly, which is to say one has
> to get the definition straight. And yes, it's trivial to prove
> from the definition.
Your comments have been very helpful. Thanks!
Han de Bruijn
Han de Bruijn
wrote:
Not quite. I've just mixed up a delta and an eps (where delta = eps).
See the comment by David Ullrich, for a _correct_ definition as well.
Han de Bruijn
Han de Bruijn
Hmm .. You wrote as the definition of a = lim(x->a) x :
for all eps > 0 there exists delta > 0 such that
for all x with |x - a| < delta and x <> a we have |x - a| < eps.
There indeed _exists_ a delta , namely delta = eps . Therefore:
[ a = lim(x->a) x ] if and only if
[ for all eps > 0 and x <> a we have |x - a| < eps ]
Which is _exactly_ my Ansatz. Not?
Han de Bruijn
Jesse F. Hughes
> Hmm .. You wrote as the definition of a = lim(x->a) x :
>
> for all eps > 0 there exists delta > 0 such that
> for all x with |x - a| < delta and x <> a we have |x - a| < eps.
>
> There indeed _exists_ a delta , namely delta = eps . Therefore:
>
> [ a = lim(x->a) x ] if and only if
> [ for all eps > 0 and x <> a we have |x - a| < eps ]
>
> Which is _exactly_ my Ansatz. Not?
You actually wrote something else. You wrote
forAll delta > 0 : | x - a | < delta and x <> a
I assume you meant to say this holds for all x, in which case it is of
course false (since it asserts that x <> a).
But it's not clear what you're after. Since we can prove
a = lim(x->a) x, it trivially follows that
a = lim(x->a) x if and only if <insert any true statement here>.
--
"How can people [philosophers] talk like that? Acting as if they're
/glad/ they don't know things! Finding out more and more things they
don't know! It's like children proudly coming to show you a full
potty!" -- Terry Pratchett, /Small Gods/
Ilmari Karonen
>
> Hmm .. You wrote as the definition of a = lim(x->a) x :
>
> for all eps > 0 there exists delta > 0 such that
> for all x with |x - a| < delta and x <> a we have |x - a| < eps.
>
> There indeed _exists_ a delta , namely delta = eps .
Indeed. so far so good.
> Therefore:
>
> [ a = lim(x->a) x ] if and only if
> [ for all eps > 0 and x <> a we have |x - a| < eps ]
Um, no. Substituting delta = eps into the definition you quoted above
does not give that. What it does give is
for all eps > 0
for all x with |x - a| < eps and x <> a we have |x - a| < eps,
which is of course trivially true (since it is of the form "for all
eps, x: (P and Q) => P", which is a logical tautology).
--
Ilmari Karonen
To reply by e-mail, please replace ".invalid" with ".net" in address.
Mistress Helios
David C. Ullrich
Yes, I agreed that it's true and trivial that lim_{x->a} x = a.
You've given the trivial proof.
I have no idea what your point is.
>Han de Bruijn
David C. Ullrich
Delete the "yes" I just posted - I hadn't looked carefully
at what you wrote (when I saw the "delta = eps" I
assumed you had the proof right, since the proof
is trivial and that's part of it.)
The statement " for all eps > 0 and x <> a we have |x - a| < eps"
is obviously false.
What's obviously true is _this_:
(*)
For all x with |x-a| < eps and x <> a we have |x-a|<eps.
And _that_ is what shows, by letting delta = eps in the
definition, that lim_{x->a} x = a.
>Han de Bruijn
Han de Bruijn
> Han de Bruijn <umum...@gmail.com> writes:
>
> > Hmm .. You wrote as the definition of a = lim(x->a) x :
>
> > for all eps > 0 there exists delta > 0 such that
> > for all x with |x - a| < delta and x <> a we have |x - a| < eps.
>
> > There indeed _exists_ a delta , namely delta = eps . Therefore:
>
> > [ a = lim(x->a) x ] if and only if
> > [ for all eps > 0 and x <> a we have |x - a| < eps ]
>
> > Which is _exactly_ my Ansatz. Not?
>
> You actually wrote something else. You wrote
>
> forAll delta > 0 : | x - a | < delta and x <> a
>
> I assume you meant to say this holds for all x, in which case it is of
> course false (since it asserts that x <> a).
> But it's not clear what you're after. Since we can prove
> a = lim(x->a) x, it trivially follows that
>
> a = lim(x->a) x if and only if <insert any true statement here>.
It has no sense to reveal what I'm after until my formalism has become
exactly right. Which obviously is not the case yet.
Han de Bruijn
Gus Gassmann
> It has no sense to reveal what I'm after until my formalism has become
> exactly right. Which obviously is not the case yet.
Just a suggestion: If you have trouble writing down a simple concept
such as a limit, it seems you are ill-equipped to tackle whatever it
is you really are after. I'm just sayin'.
Han de Bruijn
wrote:
Oh well, I'm not at all impressed by your arrogance. I'm just sayin'.
I'm not a talented musician. I'm not a talented mathematician.
And yet I'm making music. And yet I'm doing mathematics.
BTW, try to deny that I've been well-equipped to tackle e.g. this:
http://hdebruijn.soo.dto.tudelft.nl/jaar2011/steiner3.pdf
But you're too much of an arrogant twit to admit, huh ..
Han de Bruijn
Han de Bruijn
> On Sat, 26 Mar 2011 12:42:47 -0700 (PDT), Han de Bruijn
>
>
>
>
>
Next attempt to make sense.
Definition. a and b are real numbers in:
a # b if and only if forAll eps > 0 : | a - b | < eps
Proposition: # is an equivalence relation (equality).
Proof: a # a : trivial
if (a # b) then (b # a) : trivial
if (a # b) and (b # c) then (a # c)
because | a - b | < eps1 and | b - c | < eps2 gives
| a - c | <= | a - b | + | b - c | < eps1 + eps2
and take eps1 = eps2 = eps/2 .
Trivial perhaps, yes. But is _this_ correct?
Han de Bruijn
Gus Gassmann
Look, do you really not understand that
forAll eps > 0 : | a - b | < eps
implies a = b?
Han de Bruijn
wrote:
Maybe you do not intend to be kind, but thanks anyway.
And what if forAll eps > 0 : | a - b | < eps
and a <> b ?
Han de Bruijn
Jesse F. Hughes
>> Look, do you really not understand that
>>
>> forAll eps > 0 : | a - b | < eps
>>
>> implies a = b?
>
> Maybe you do not intend to be kind, but thanks anyway.
>
> And what if forAll eps > 0 : | a - b | < eps
> and a <> b ?
So, you are not discussing the classical reals here?
If not, then it's not altogether obvious what the various functions
involved here are. But, assuming | a - b | = | b - a |,
| a - a | = 0 and | a - c | <= | a - b | + | b - c |, then it seems to
me that # is an equivalence relation.
Do you have a particular context in mind? Some notion of constructive
real analysis, say?
--
"I told her that I loved her.
She said she loved me too.
Neither one was lying,
Yet it wasn't true." -- Del McCoury Band
Tim Golden BandTech.com
wrote:
> > | a - c | <= | a - b | + | b - c | < eps1 + eps2
> > and take eps1 = eps2 = eps/2 .
>
> > Trivial perhaps, yes. But is _this_ correct?
>
> Look, do you really not understand that
>
> forAll eps > 0 : | a - b | < eps
>
> implies a = b?
I agree with Gus here on the logical implication as a sensibility that
the
a < > x
criterion betrays. I think there is also a larger statement here on
the structure of mathematical argumentation. For instance we could
declare that any restriction placed upon values 'a' or 'x' after they
have been declared as generally in the real numbers as a conflict
based on a conflict of freedom; the first statement declaring more
freedom than the more restrictive secondary statements.
I think that we have to admit that the structure of mathematical
argumentation tends to narrow down freedom as the statements
accumulate. This is nearly an informational requirement, and the
ability to raise the freedom of a previously more restrictive
statement appears to us as a larger conflict than its inverse. If for
instance I were to declare a variable 'd' to be in the reals, and then
later raise its freedom to a complex value then we should admit that
the reasoning which I am trying to engage is operant. Sadly, this does
go on within mathematics, so that this subject is alive. This
awareness feeds the dissatisfaction of the square root of minus one,
and also feeds my own discontent with abstract algebra and its
derivation of complex numbers from polynomials with real coefficients.
So, to sum up my own position on your own construction:
it comes down to a matter of definition, and you have offered up a
more restrictive definition of a limit that the standard definition.
The dissection of a series of mathematical statements generally
proceeds from most general to less general via the accumulation of
statements, and so your own restriction if it were to come in a grand
ordering would follow after the definition of limit and does appear to
have some essence worth thought. I like it. I can offer another even
more restrictive form by imposing
0.5 delta
as the actual difference between x and a, but I would try not to
confuse this restriction with the definition of limit. This offering
fits in that tendril of restriction in agreement with your own, but
again I would not attempt to replace your own with it as equivalent.
- Tim
William Hughes
If a#b then for all n, |a-b| < 1/n
In the standard reals this means a-b=0.
If your reals have non zero infinitesimals, then
|a-b| is an infinitesimal.
Unless you do something really weird, #
is an equivalence relation.
Yes there is a (generalized) sequence a_1=0 ... a_w0 = 1
such that a_i#a_i+1, but the sequence is not indexed by
a countable ordinal so you cannot conclude a_1 # a_w0.
- William Hughes
Gus Gassmann
Then you are no longer talking about real numbers. Perhaps you
intended to show that such reals a and b do not exist?
Han de Bruijn
Oh well, think I can answer this question in the classical way myself.
Suppose that | a - b | = delta then take eps = delta/2 to get
| a - b | > eps , which is not forAll eps .
So classical mathematics is consistent in that a = b and a =/= b
cannot be true at the same time. There is no "approximate equality"
in the sense that | a - b | < eps forAll eps or some such.
So far, so good.
Han de Bruijn
Han de Bruijn
But then I don't understand limits anymore ..
The exact definition of a limit is as follows. Let f(x) be a function
defined on an interval that contains x = a . Then we say that
lim(x->a) f(x) = L
iff for every number eps > 0 there is some number delta > 0 such that
| f(x) - L | < eps whenever | x - a | < delta
But according to previous postings, the fist part means that f(x) = L
( because it is equivalent to forAll eps : | f(x) - L | < eps ) while
all positive values of delta are present in | x - a | < delta too;
therefore it follows that x = a as well. The net result is:
x = a therefore f(x) = L or f(a) = L
Note, however, that f(x) is possibly not defined at x = a . So we may
add the _additional_ condition x <> a , according to standard theory.
But this is contradictory to x = a and f(a) = L . Quite confusing.
Han de Bruijn
Han de Bruijn
Example.
sinc(x) = sin(x)/x
Then for x <> 0 and from a picture: cos(x) < sin(x)/x < 1
So | sin(x)/x - 1 | < 1 - cos(x) < x^2/2 < eps
if | x | < sqrt(2.eps) for | x - 0 | < delta
ForAll eps > 0 , hence forAll delta > 0 because delta = sqrt(2.eps)
Therefore x = 0 and sinc(0) = 1 .
No separate definition for sinc(0) is thus needed.
Han de Bruijn
Brian Chandler
<snip>
> But then I don't understand limits anymore ..
>
> The exact definition of a limit is as follows. Let f(x) be a function
> defined on an interval that contains x = a . Then we say that
>
> lim(x->a) f(x) = L
>
> iff for every number eps > 0 there is some number delta > 0 such that
>
> | f(x) - L | < eps whenever | x - a | < delta
Yes, I believe this is correct.
> But according to previous postings, the fist part means that f(x) = L
> ( because it is equivalent to forAll eps : | f(x) - L | < eps ) while
> all positive values of delta are present in | x - a | < delta too;
> therefore it follows that x = a as well.
But this doesn't make sense. What is this "first part"? Write out the
above definition carefully:
lim(x->a) f(x) = L iff the following is true:
for every number eps > 0 there is some number delta > 0 such that ...
(Q)
((( | x - a | < delta implies | f(x) - L | < eps )))
The quantification bit (Q) applies to the rest of the statement: (if x
is within delta of a, then f(x) is within epsilon of L ).
Your attempt to pull out part of this and apply part of the
quantification amounts to a quantifier swap. (QD, in other words.)
> The net result is:
>
> x = a therefore f(x) = L or f(a) = L
So this is wrong.
> Note, however, that f(x) is possibly not defined at x = a . So we may
> add the _additional_ condition x <> a , according to standard theory.
Indeed, f(a) may be defined, and different from lim x->a f(x). So the
extra condition is necessary, and checking a Real Book (Apostol, p 61)
we spot that his version is (modifying the symbols to match):
[Apostol]
whenever 0 < | x - a | < delta, then | f(x) - L | < epsilon
HTH
Brian Chandler
Han de Bruijn
Let f(x) not be a function per se but we call it a "thingie".
The thingie f is continuous at x = a iff for every number eps > 0
there is some number delta > 0 such that
| f(x) - f(a) | < eps whenever | x - a | < delta
As proved (?) earlier, this is equivalent with:
x = a therefore f(x) = f(a)
With other words: continuous thingies are _functions_.
Han de Bruijn
Gus Gassmann
That's why I said you don't know what you are doing. The "whenever" in
| f(x) - L | < eps whenever | x - a | < delta
contains a /quantifier/ that is missing in any of the other attempts
you made.
Han de Bruijn
wrote:
> Han de Bruijn wrote:
>
> <snip>
>
> > But then I don't understand limits anymore ..
>
> > The exact definition of a limit is as follows. Let f(x) be a function
> > defined on an interval that contains x = a . Then we say that
>
> > lim(x->a) f(x) = L
>
> > iff for every number eps > 0 there is some number delta > 0 such that
>
> > | f(x) - L | < eps whenever | x - a | < delta
>
> Yes, I believe this is correct.
>
> > But according to previous postings, the fist part means that f(x) = L
> > ( because it is equivalent to forAll eps : | f(x) - L | < eps ) while
> > all positive values of delta are present in | x - a | < delta too;
> > therefore it follows that x = a as well.
>
> But this doesn't make sense. What is this "first part"? Write out the
> above definition carefully:
>
> lim(x->a) f(x) = L iff the following is true:
>
> for every number eps > 0 there is some number delta > 0 such that ...
> (Q)
> ((( | x - a | < delta implies | f(x) - L | < eps )))
>
> The quantification bit (Q) applies to the rest of the statement: (if x
> is within delta of a, then f(x) is within epsilon of L ).
>
> Your attempt to pull out part of this and apply part of the
> quantification amounts to a quantifier swap. (QD, in other words.)
I don't know what "QD" means. Please take a look at my example though,
which is repeated here for convenience.
<quote>
Example.
sinc(x) = sin(x)/x
Then for x <> 0 and from a picture: cos(x) < sin(x)/x < 1
So | sin(x)/x - 1 | < 1 - cos(x) < x^2/2 < eps
if | x | < sqrt(2.eps) for | x - 0 | < delta
ForAll eps > 0 , hence forAll delta > 0 because delta =
sqrt(2.eps)
Therefore x = 0 and sinc(0) = 1 .
No separate definition for sinc(0) is thus needed.
</quote>
Really don't get it. The limit is completed. So for All eps has become
a _fact_. And we have seen that sinc(0) = 1 follows then. But if All
eps has become completed, then also All delta has become completed and
it follows that x = 0 because it cannot be otherwise if | x | < delta.
> > The net result is:
>
> > x = a therefore f(x) = L or f(a) = L
>
> So this is wrong.
>
> > Note, however, that f(x) is possibly not defined at x = a . So we may
> > add the _additional_ condition x <> a , according to standard theory.
>
> Indeed, f(a) may be defined, and different from lim x->a f(x). So the
> extra condition is necessary, and checking a Real Book (Apostol, p 61)
> we spot that his version is (modifying the symbols to match):
>
> [Apostol]
> whenever 0 < | x - a | < delta, then | f(x) - L | < epsilon
>
> HTH
Hope That Helps?
Han de Bruijn
Han de Bruijn
Sure quantifier dislexia may be one of my shortcomings .. The point is
that I find it not clear _conceptually_ what the difference is between
expressions of the form | a - b | < eps as such - which simply means
that a = b (if for all eps) and the _same_ expression within limits.
The more confusing because these limits are considered as _completed_,
so eps and delta indeed "have got all values", upon completion.
Han de Bruijn
Gus Gassmann
> Sure quantifier dislexia may be one of my shortcomings .. The point is
> that I find it not clear _conceptually_ what the difference is between
> expressions of the form | a - b | < eps as such
It is not an expression as such, but rather you say, implicitly or
explicitly:
forAll a, b real numbers { a = b iff for all eps > 0 : |a - b| < eps }
This means that you *fix* the a and b *before* you start varying the
epsilon.
You also wrote:
> > The exact definition of a limit is as follows. Let f(x) be a function
> > defined on an interval that contains x = a . Then we say that
> > lim(x->a) f(x) = L
> > iff for every number eps > 0 there is some number delta > 0 such that
> > | f(x) - L | < eps whenever | x - a | < delta
Quibble, but probably important: The function is f, not f(x). So the
first line of your quoted passage should read: Let f be a function
defined on an *open* interval that contains the point a. (You must
also make sure that a is not an endpoint of the interval, which might
get you in trouble --- e.g., f = sqrt and a = 0.)
forAll eps > 0 exist delta > 0 such that *FOR ALL* x {if |x - a| <
delta then |f(x) - L| < eps}
Note that in the curly braces here the x is allowed to vary (because
of the forAll that I emphasized for you), whereas above the a and b
were fixed.
You're a programmer, right? Perhaps you can figure out that in the
first case Equals can be thought of as a method that has two
parameters a and b, i.e., you call it as
Equals(a, b);
and expect to see a return value of true or false, but *without*
changing the parameter values a and b.
In the second case you have a method
Limit(f,a,L);
which returns true or false, but note that the x is set *inside* the
method in the second case.
Hope that helps.
Han de Bruijn
wrote:
Thank you very much for this elaborate clarification.
Han de Bruijn
David C. Ullrich
(*) | f(x) - L | < eps whenever | x - a | < delta
No, that is _not_ the definition of "lim(x->a) f(x) = L". It
is the definition of "f is continuous at a", which is the
same as "lim_{x->a} f(x) = f(a)".
>But according to previous postings, the fist part means that f(x) = L
It means that f(a) = L, yes.
>( because it is equivalent to forAll eps : | f(x) - L | < eps ) while
>all positive values of delta are present in | x - a | < delta too;
>therefore it follows that x = a as well. The net result is:
>
> x = a therefore f(x) = L or f(a) = L
>
>Note, however, that f(x) is possibly not defined at x = a .
No, that's not true. (*) implies that f _is_ defined at x = a.
It's true that "lim_{x->a} f(x) = L" allows the possibility
that f is not defined at x = a. But (*) is _not_ the same
as "lim_{x->a} f(x) = L".
> So we may
>add the _additional_ condition x <> a , according to standard theory.
>But this is contradictory to x = a and f(a) = L . Quite confusing.
Confusing to you because you're insisting that the definition
of one thing is the definition of another thing.
You claim you're trying to be careful and get things straight.
You need to be a lot more careful - if you see the definition
of "zebra" and conclude that it applies to tomatoes it's
pretty clear that you're going to remain confused.
>
>Han de Bruijn
Han de Bruijn
> On Tue, 29 Mar 2011 01:11:29 -0700 (PDT), Han de Bruijn
[ .. snip redundancy .. ]
> >But then I don't understand limits anymore ..
>
> >The exact definition of a limit is as follows. Let f(x) be a function
> >defined on an interval that contains x = a . Then we say that
>
> > lim(x->a) f(x) = L
>
> >iff for every number eps > 0 there is some number delta > 0 such that
>
> (*) | f(x) - L | < eps whenever | x - a | < delta
>
> No, that is _not_ the definition of "lim(x->a) f(x) = L". It
> is the definition of "f is continuous at a", which is the
> same as "lim_{x->a} f(x) = f(a)".
>
> >But according to previous postings, the fist part means that f(x) = L
>
> It means that f(a) = L, yes.
>
> >( because it is equivalent to forAll eps : | f(x) - L | < eps ) while
> >all positive values of delta are present in | x - a | < delta too;
> >therefore it follows that x = a as well. The net result is:
>
> > x = a therefore f(x) = L or f(a) = L
>
> >Note, however, that f(x) is possibly not defined at x = a .
>
> No, that's not true. (*) implies that f _is_ defined at x = a.
>
> It's true that "lim_{x->a} f(x) = L" allows the possibility
> that f is not defined at x = a. But (*) is _not_ the same
> as "lim_{x->a} f(x) = L".
Huh ? Should I rather say:
| f(x) - L | < eps whenever 0 < | x - a | < delta
-----------------------------^^^
> > So we may
> >add the _additional_ condition x <> a , according to standard theory.
> >But this is contradictory to x = a and f(a) = L . Quite confusing.
>
> Confusing to you because you're insisting that the definition
> of one thing is the definition of another thing.
Oh well, approximately the _same_ thing. Not?
> You claim you're trying to be careful and get things straight.
> You need to be a lot more careful - if you see the definition
> of "zebra" and conclude that it applies to tomatoes it's
> pretty clear that you're going to remain confused.
Yeah, I need to be a lot more careful, I know, I know ..
Han de Bruijn
Han de Bruijn
Yes, I *am* a programmer. So let's play that game of "doing the limit"
in the programmer's way. We define an eps and ask for a delta that
guarantees | f(x) - L | < eps for | x - a | < delta . Each turn we
make the game more "difficult" by halving eps .
program proef;
procedure test;
var
eps,x : double;
begin
eps := 0.1;
while true do
begin
x := sqrt(2*eps);
Writeln('| sin(x)/x - 1 | = ',(1 - sin(x)/x):8:6,
' < eps = ',eps:8:6,' for x = ',x:8:6,' < delta');
eps := eps/2;
Readln;
end;
end;
begin
test;
end.
Output:
| sin(x)/x - 1 | = 0.033002 < eps = 0.100000 for x = 0.447214 < delta
| sin(x)/x - 1 | = 0.016584 < eps = 0.050000 for x = 0.316228 < delta
| sin(x)/x - 1 | = 0.008313 < eps = 0.025000 for x = 0.223607 < delta
| sin(x)/x - 1 | = 0.004161 < eps = 0.012500 for x = 0.158114 < delta
| sin(x)/x - 1 | = 0.002082 < eps = 0.006250 for x = 0.111803 < delta
| sin(x)/x - 1 | = 0.001041 < eps = 0.003125 for x = 0.079057 < delta
| sin(x)/x - 1 | = 0.000521 < eps = 0.001563 for x = 0.055902 < delta
| sin(x)/x - 1 | = 0.000260 < eps = 0.000781 for x = 0.039528 < delta
| sin(x)/x - 1 | = 0.000130 < eps = 0.000391 for x = 0.027951 < delta
| sin(x)/x - 1 | = 0.000065 < eps = 0.000195 for x = 0.019764 < delta
| sin(x)/x - 1 | = 0.000033 < eps = 0.000098 for x = 0.013975 < delta
| sin(x)/x - 1 | = 0.000016 < eps = 0.000049 for x = 0.009882 < delta
| sin(x)/x - 1 | = 0.000016 < eps = 0.000049 for x = 0.009882 < delta
| sin(x)/x - 1 | = 0.000008 < eps = 0.000024 for x = 0.006988 < delta
| sin(x)/x - 1 | = 0.000004 < eps = 0.000012 for x = 0.004941 < delta
| sin(x)/x - 1 | = 0.000002 < eps = 0.000006 for x = 0.003494 < delta
| sin(x)/x - 1 | = 0.000001 < eps = 0.000003 for x = 0.002471 < delta
| sin(x)/x - 1 | = 0.000001 < eps = 0.000002 for x = 0.001747 < delta
| sin(x)/x - 1 | = 0.000000 < eps = 0.000001 for x = 0.001235 < delta
--
| sin(x)/x - 1 | = 0.000000 < eps = 0.000000 for x = 0.000000 < delta
Runtime error 207 at 00402909
I've included the last line for the sake of honesty: the program will
run into a "division by zero" at last. Despite your efforts, I still
find it difficult to comprehend that | sin(x)/x - 1 | is not exactly
zero (?) and | x | would be not exacly zero, such that sin(x)/x = 1
and x = 0 and that's it. Note that the current Runtime error is due
to limited precision of a machine. Within the machinery of mathematics
as an exact discipline, such an error would not occur (due to infinite
precision). So AT THE END OF THE GAME, I expect all eps and all delta
being done and so x = 0 and sin(0)/0 = 1 . (I _know_ it's "illegal"
what I'm writing up here. But the above output provokes me to do so.)
Why oh why am I so difficult to convert ..
Han de Bruijn
Gus Gassmann
(Convert to what?)
Because you (intentionally?) pick difficult functions. The function f
defined by
f(x) = [sin(x)]/x
is simply not defined at 0. You can *extend* the function by making a
definition at 0, but then the question is, what should f(0) be? Any
number is equally valid a priori. And when you evaluate the function
you have to check:
if (x != 0) f(x) = sin(x)/x else f(x) = ...
Han de Bruijn
function sinc(x : double) : double;
var
f : double;
begin
if x = 0 then f := 1 else f := sin(x)/x;
sinc := f;
end;
But never mind. It only cures the computation error.
Do you really think that it is possible for sinc(x) to _ever_ assume
another value for x = 0 than sinc(0) = 1 , in any application which
is _outside_ pure mathematics? As a physicist by education, I'm pretty
sure that such a thing will never happen. Repeat: NEVER. This in fact
is the reason why I'm questioning all this. I think that there is not
an a priori reason indeed. But maybe there is an a posteriori reason,
whatever that may turn out to be. The objective of this thread is to
sort out material for such an a posteriori argument, if ever ..
Han de Bruijn
Brian Chandler
Quantifier Dyslexia. One of your problems.
> <quote>
>
> Example.
>
> sinc(x) = sin(x)/x
>
> Then for x <> 0 and from a picture: cos(x) < sin(x)/x < 1
OK. (Seems to be right: close to x=0 sin(x) curves downward, and
therefore sin(x)/x is more than the derivative of sin(x), which equals
cos(x). )
> So | sin(x)/x - 1 | < 1 - cos(x) < x^2/2 < eps
> if | x | < sqrt(2.eps) for | x - 0 | < delta
Well, as you have already remarked, you need to say that x is not 0,
since sin(0)/0 is undefined.
I understand as far as "So | sin(x)/x - 1 | < 1 - cos(x) " -- after
that you lose me. All you need to do (assuming you are trying to show
that the limit as x->0 is 1) is to present a way of finding a delta
for any epsilon. And actually delta = epsilon will do. For if x <
epsilon (let's write e), then
| f(x) - 1 | < 1 - cos(x) < 1 - cos(e) <= e
because 1 - cos(x) is a function mapping 0 to 0, and having slope less
than 1, as long as x < pi/2
So we have an epsilon for any delta (with the usual conditions), and
we have shown that
lim x->0 sin(x)/x = 1
Now what are you doing...
> ForAll eps > 0 , hence forAll delta > 0 because delta =
> sqrt(2.eps)
Let's skip this line...
> Therefore x = 0 and sinc(0) = 1 .
Huh? What on earth does it mean? x is a variable... I have no idea how
you think you get this line.
> No separate definition for sinc(0) is thus needed.
Huh?? No, sinc(0) is undefined, since sinc(0) is already defined as
sin(0)/0, which is undefined.
> </quote>
>
> Really don't get it. The limit is completed. So for All eps has become
> a _fact_.
No, you _really_ don't get it. You have gone from a statement of (more
or less) the normal definition of a limit to mystical claims that a
limit is "completed" -- meaningless, but I think a common
misunderstanding of the expression "in the limit" is "when you get to
[whatever it is that doesn't exist]". Then you claim that a
quantification has become a fact??? "For all epsilon" cannot be a
"fact" in any normal sense.
You really need to go back to the very beginning...
> And we have seen that sinc(0) = 1 follows then. But if All
> eps has become completed, then also All delta has become completed and
> it follows that x = 0 because it cannot be otherwise if | x | < delta.
So this is not even wrong -- it's nonsense.
> > HTH
>
> Hope That Helps?
Yes. Well done.
Brian Chandler
Gus Gassmann
I do not know about sinc, but I see functions of the form
f(x) = 0 if x = 0
f(x) = 100 + 10x if x > 0
all the time. (Look up "fixed charge problem".)
Mike Terry
news:aa1dee67-b580-40df...@d19g2000yql.googlegroups.com...
as you realise later, this should be "whenever 0 < | x - a | < delta"
>
> But according to previous postings, the fist part means that f(x) = L
> ( because it is equivalent to forAll eps : | f(x) - L | < eps ) while
> all positive values of delta are present in | x - a | < delta too;
> therefore it follows that x = a as well.
Well this has obviously gone horribly wrong somewhere! I think the problem
is in how you're interpreting nested qualifiers, and what variables are
decided at each level of quantification. Here's the definition again a bit
more spaced out with some comments (I use // to prefix comments):
lim(x->a) f(x) = L // note - this expression *appears* to contain
// x, but that's just a feature of the notation.
// Logically the statement says something about
// the function f, the number a, and the number
L,
// so f, a, and L are fixed for the remainder of
// the definition. The "x" is really not there
at
// all - it should more logically be written as
// "(limit of f at a) = L"
iff
// the definition...
ForAll eps > 0
// (note: from now on we treat eps as fixed but at this point
// x is not yet mentioned)
// [3]
Exists delta > 0 // ... there is a delta > 0 that makes
// the remaining nested bit of the
definition
// true...
// NOTE: FIRST eps is selected, and THEN
delta
// must be found. So delta typically can
depend
// the eps
// (note from now on we treat both eps and delta as fixed)
// [1]
ForAll x satisfying 0 < | x - a | < delta
// (now in what follows x is considered fixed...)
// [2]
| f(x) - L | < eps // finally! :) a statement that is
// obviously true or false.
So you seem to have focussed on [2], at which point we consider x, eps and
delta as all fixed ("decided" might be better), and noted | x - a | < delta,
which is true at this level of the definition, but you are viewing it as x
being fixed before delta is chosen, whereas in fact x is undefined until the
quantification at [1]. *If* x were fixed outside of the eps quantification
[3] it would be correct to deduce x=a. (This is what Brian meant about you
swapping quantifiers - you've taken "x" that is only decided at [2] and
treated it as though it were decided at [3]...)
If this isn't all clear just by carefully looking at the nested definition,
maybe think of it as a two-player game:
1) the game starts with f, L and a, as givens
2) FIRST player-one chooses (and reveals to player-two) an eps > 0
3) THEN it's player-two's turn to choose a delta > 0
(note at this point player-two knows eps, but not x...)
4) THEN it's player-one's turn again - now she must choose x satisfying
0 < | x - a | < delta
5) The game's over! Who's won:
Player two has won if | f(x) - L | < eps, otherwise player one has
won.
Well I won't bother copywriting this game, as I can't see it catching on,
but it encpasulates the limit definition. With sensible play, player two
will always win if "(limit of f at a) = L" is true, whereas player one wins
if it is false. If the game doesn't make sense, try actually playing it
with your f(x) = sin(x)/x, a = 0, L = 1, and decide whether player one or
two always wins with best play...
So within the context of the game, does it make sense to claim that x = a ?
Clearly not, but that is effectively what you're doing in the context of the
continuity definition...
Hope this helps,
Mike.
Virgil
<aa1dee67-b580-40df...@d19g2000yql.googlegroups.com>,
Han de Bruijn <umu...@gmail.com> wrote:
> But then I don't understand limits anymore ..
>
> The exact definition of a limit is as follows. Let f(x) be a function
> defined on an interval that contains x = a . Then we say that
>
> lim(x->a) f(x) = L
>
> iff for every number eps > 0 there is some number delta > 0 such that
>
> | f(x) - L | < eps whenever | x - a | < delta
Actually, that is NOT the correct definition, as your supposed
definition requires the function to be DEFINED at the point x = a, a
which is NOT required by the true definition.
Thus, for example, yours does NOT work for lim(x -> 0) sin(x)/x = 1.
Nor for the usual definition of a derivative at a point as,
f'(a) = lim(h -> 0) (f(a+h) - f(a))/h
For the truly correct definition, it is only necessary that x = a be a
limit point of the domain of the function f(x).
Or, equivalently be a member of the closure of the domain.
--
Han de Bruijn
Plz be more specific, because can't find the function you wrote about.
Han de Bruijn
Han de Bruijn
<news.dead.person.sto...@darjeeling.plus.com> wrote:
> "Han de Bruijn" <umum...@gmail.com> wrote in messagenews:aa1dee67-b580-40df...@d19g2000yql.googlegroups.com...
Thanks Mike! But it would help even more if you put a constrain of 70
characters wide on the lines you're posting, because Google groups is
scrambling everything beyond. Here is a helpful tool for doing this:
1234567890123456789012345678901234567890123456789012345678901234567890
:-)
Han de Bruijn
Han de Bruijn
> In article
> <aa1dee67-b580-40df-9f67-32eefb645...@d19g2000yql.googlegroups.com>,
Or, in layman's terms:
iff for every number eps > 0 there is some number delta > 0 such that
| f(x) - L | < eps whenever 0 < | x - a | < delta
Right? (must have learned all this in school, but it has become rusty)
Han de Bruijn
Han de Bruijn
wrote:
> On Mar 29, 11:45 am, Han de Bruijn <umum...@gmail.com> wrote:
[ .. snip things done .. ]
> > Do you really think that it is possible for sinc(x) to _ever_ assume
> > another value for x = 0 than sinc(0) = 1 , in any application which
> > is _outside_ pure mathematics? As a physicist by education, I'm pretty
> > sure that such a thing will never happen. Repeat: NEVER. This in fact
> > is the reason why I'm questioning all this. I think that there is not
> > an a priori reason indeed. But maybe there is an a posteriori reason,
> > whatever that may turn out to be. The objective of this thread is to
> > sort out material for such an a posteriori argument, if ever ..
>
> I do not know about sinc, but I see functions of the form
>
> f(x) = 0 if x = 0
> f(x) = 100 + 10x if x > 0
>
> all the time. (Look up "fixed charge problem".)
Suppose we have measurements (x_1,y_1),(x_2,y_2),(x_3,y_3)..(x_n,y_n).
One of the possible standard procedures is to try a Least Squares best
fit with a straight line:
y = a.x + b where a and b are calculated with an L.S. procedure.
An "ideal" best fit - but not really - would be discontinuous though:
(x - x_1)(x - x_2)(x - x_3) .. (x - x_n)
y = x.---------------------------------------- where defined, and:
(x - x_1)(x - x_2)(x - x_3) .. (x - x_n)
y = y_1 for x = x_1
y = y_2 for x = x_2
y = y_3 for x = x_3
..
y = y_n for x = x_n
Nobody is going to adopt the latter function as a mathematical model.
Han de Bruijn
Gus Gassmann
How can you even *think* that this refutes my point (or even addresses
it remotely)?
Jesse F. Hughes
> Thanks Mike! But it would help even more if you put a constrain of 70
> characters wide on the lines you're posting, because Google groups is
> scrambling everything beyond. Here is a helpful tool for doing this:
>
> 1234567890123456789012345678901234567890123456789012345678901234567890
It would also be nice if everyone involved would trim quoted material.
Quoting 268 lines just to comment on the length of a few of them is
unnecessary. I don't mean to pick on Han here -- had others trimmed
quotes, there wouldn't have been 268 lines for him to quote.
It just doesn't make sense that everyone has to page through so many
irrelevant citations.
--
Jesse F. Hughes
"Until the patch is released, Microsoft said computer users should be
careful not to visit unfamiliar Web sites."
-- CNN article on the latest MS vulnerability.
Brian Chandler
> On Mar 29, 4:13 pm, Gus Gassmann <horand.gassm...@googlemail.com>
> wrote:
> > On Mar 29, 10:43 am, Han de Bruijn <umum...@gmail.com> wrote:
> > > Yes, I *am* a programmer. So let's play that game of "doing the limit"
> > > in the programmer's way. We define an eps and ask for a delta that
> > > guarantees | f(x) - L | < eps for | x - a | < delta . Each turn we
> > > make the game more "difficult" by halving eps .
> >
> > > program proef;
> >
> > > procedure test;
> > > var
> > > eps,x : double;
> > > begin
> > > eps := 0.1;
> > > while true do
> > > begin
> > > x := sqrt(2*eps);
> > > Writeln('| sin(x)/x - 1 | = ',(1 - sin(x)/x):8:6,
> > > ' < eps = ',eps:8:6,' for x = ',x:8:6,' < delta');
> > > eps := eps/2;
> > > Readln;
> > > end;
> > > end;
Right, you've written a loop with no chance of exit, so it's pretty
obvious that eventually some computing resource will be exhausted, and
there will be some sort of crash. But this is a representation of an
ideal (mathematical) sequence which has no end. In mathematics, when
there is no end, you never get to it* either. (* 'it' has no existent
referent, so this isn't a properly formed way of saying it. But for
brighter students, saying "endless" should be enough.)
> > > Why oh why am I so difficult to convert ..
> >
> > (Convert to what?)
> >
> > Because you (intentionally?) pick difficult functions.
I think that exchange misses the point. You are pretending to talk
about mathematics, but actually you are talking about modelling
physical processes (or something).
> function sinc(x : double) : double;
> var
> f : double;
> begin
> if x = 0 then f := 1 else f := sin(x)/x;
> sinc := f;
> end;
>
> But never mind. It only cures the computation error.
There was no computation error. See above.
> Do you really think that it is possible for sinc(x) to _ever_ assume
> another value for x = 0 than sinc(0) = 1 , in any application which
> is _outside_ pure mathematics?
Are you asking a question about mathematics or physics? In any event
it is extremely odd to talk about a function "assuming" different
possible values. A function is defined to have the values it is
defined to have. The function sind(x) I define to be the same as
sinc(x) except that sind(0) = 3.7. This is a function in mathematics.
> As a physicist by education, I'm pretty
> sure that such a thing will never happen.
Well, yawn. If you are claiming something about physics, why not post
it in a physics group.
Brian Chandler
Tim Little
> It just doesn't make sense that everyone has to page through so many
> irrelevant citations.
In my newsreader, quotes show up in a different colour. If the whole
first page of a message is in that colour (especially if most of the
lines have multiple quote prefixes) then I usually just hit the "next
message" key.
If all of the people involved are too lazy to trim quotes, they're
probably too lazy to come up with anything worth reading.
--
Tim
Jimmy John
<65fe4ac5-df03-45be...@s11g2000yqh.googlegroups.com>,
Han de Bruijn <umu...@gmail.com> wrote:
How is a function which is not defined at ANY of the desired points
qualify as "ideal"?
Jimmy John
<5ac3266f-ab3b-431b...@k7g2000yqj.googlegroups.com>,
Han de Bruijn <umu...@gmail.com> wrote:
Only right if one also assumes certain things about the domain of the
function in question.
Jesse F. Hughes
> On 2011-03-30, Jesse F. Hughes <je...@phiwumbda.org> wrote:
>> It just doesn't make sense that everyone has to page through so many
>> irrelevant citations.
>
> In my newsreader, quotes show up in a different colour.
In mine also.
> If the whole first page of a message is in that colour (especially if
> most of the lines have multiple quote prefixes) then I usually just
> hit the "next message" key.
>
> If all of the people involved are too lazy to trim quotes, they're
> probably too lazy to come up with anything worth reading.
For some reason, there are a few regular posters who buck that trend.
Their original contributions are worth reading, but they rarely bother
to trim quoted material.
--
"Flowers in the Attic" was based on a true story. [...] HOW is it OK
to just butcher such an awesome piece of work? It's like passing
Pokemon off as the Mona-Lisa; sick and entirely wrong.
-- An Amazon reviewer pissed that the movie didn't include incest
Han de Bruijn
wrote:
This brings us to the core of the matter. Is the (yes / no) existence
of completed infinities a mathematics or a physics question? Once you
have decided _that_, I will choose my group.
Han de Bruijn
Brian Chandler
As usual, your question doesn't make much sense. What does the "(yes /
no) existence
of completed infinities" refer to?? Arguably there are two questions:
(1) In mathematics, are there "completed infinities"?
(2) In physics, are there "completed infinities"?
I can't personally understand what question (1) is supposed to mean,
but it seems to me that in mathematics, if anyone defines
(mathematically!) what a "completed infinity" is, then any mathematics
relating to this is perfectly valid. Mathematics isn't about answering
philosophical questions like "Do completed infinities _really_ exist?"
Again, I've no idea what a "completed infinity" would be in physics
either. Consider that the universe might or might not be infinite in
extent; current theories seem to suggest it is not, but at least
conceptually it is perfectly plausible that the real number line might
be a model of a line in some direction with my desk as 0, extending in
two opposite directions without ever ending. There would then be an
infinite set of "metre marks" on this line. But extending without ever
ending does not get you to a place called "completed infinity", to
mention but one non-existent possible interpretation of your mystery
phrase. Or whatever.
Brian Chandler
Han de Bruijn
wrote:
> Han de Bruijn wrote:
> > On Mar 30, 5:12 pm, Brian Chandler <imaginator...@despammed.com>
> > wrote:
> > > Well, yawn. If you are claiming something about physics, why not post
> > > it in a physics group.
>
> > This brings us to the core of the matter. Is the (yes / no) existence
> > of completed infinities a mathematics or a physics question? Once you
> > have decided _that_, I will choose my group.
>
> As usual, your question doesn't make much sense. What does the "(yes /
> no) existence
> of completed infinities" refer to?? Arguably there are two questions:
>
> (1) In mathematics, are there "completed infinities"?
I protest against the use of infinite magnitude as something completed
which is never permissible in mathematics. Infinity is merely a way of
speaking, the true meaning being a limit which certain ratios approach
indefinitely close, while others are permitted to increase without
restriction.} ( C.F. Gauss [in a letter to Schumacher, 12 July 1831] )
> (2) In physics, are there "completed infinities"?
No. Period.
> I can't personally understand what question (1) is supposed to mean,
> but it seems to me that in mathematics, if anyone defines
> (mathematically!) what a "completed infinity" is, then any mathematics
> relating to this is perfectly valid. Mathematics isn't about answering
> philosophical questions like "Do completed infinities _really_ exist?"
Oh yeah, Carl Friedrich Gauss wasn't _really_ a mathematician, huh?
> Again, I've no idea what a "completed infinity" would be in physics
> either. Consider that the universe might or might not be infinite in
> extent; current theories seem to suggest it is not, but at least
> conceptually it is perfectly plausible that the real number line might
> be a model of a line in some direction with my desk as 0, extending in
> two opposite directions without ever ending. There would then be an
> infinite set of "metre marks" on this line. But extending without ever
> ending does not get you to a place called "completed infinity", to
> mention but one non-existent possible interpretation of your mystery
> phrase. Or whatever.
Blah, blah. "Infinitum Actu Non Datur" is a perfectly clear statement.
Old philosphers did perfectly understand what it means, but in modern
times, we've learned backwards, obviously (in more than one respect).
Han de Bruijn
David C. Ullrich
For heaven's sake. No, if you actually want to get these
things straight then i'ts not approximately the same
thing. You claim to be confused about something.
The confusion is because you have a definition wrong.
I give you the right definition. You reply with another
wrong definition. And you're still confused.
Jeez.
See, this stuff _does_ work. But it doesn't work
if you do it wrong. And if, when you're talking about A,
you use the definition of B, that's a big problem.
If you think that using the definition of B in place
of the definition of A is "approximately the same
thing" it's simply hopeless.
Han de Bruijn
> On Tue, 29 Mar 2011 06:29:32 -0700 (PDT), Han de Bruijn
>
> Jeez.
Yes, Jeez, from:
http://archives.math.utk.edu/visual.calculus/1/definition.6/index.html
Definition. The limit of f(x) as x approaches a is L
lim(x->a) f(x) = L
if and only if, given e > 0 , there exists d > 0 such that
0 < |x - a| < d implies that |f(x) - L| < e .
Is that - finally - correct? Or is everybody contradicting each other?
Jimmy John
<284281bf-6747-4b92...@l18g2000yqm.googlegroups.com>,
Han de Bruijn <umu...@gmail.com> wrote:
That definition will work in many, possibly even most, cases, provided
one makes suitable assumptions about the domain of the function and the
domain of the variable, but would not be acceptable in any but
elementary calculus texts.
Jimmy John
<f78af707-5289-4c88...@i14g2000yqe.googlegroups.com>,
Han de Bruijn <umu...@gmail.com> wrote:
> On Mar 31, 10:02 am, Brian Chandler <imaginator...@despammed.com>
> wrote:
> > Han de Bruijn wrote:
> > > On Mar 30, 5:12 pm, Brian Chandler <imaginator...@despammed.com>
> > > wrote:
> > > > Well, yawn. If you are claiming something about physics, why not post
> > > > it in a physics group.
> >
> > > This brings us to the core of the matter. Is the (yes / no) existence
> > > of completed infinities a mathematics or a physics question? Once you
> > > have decided _that_, I will choose my group.
> >
> > As usual, your question doesn't make much sense. What does the "(yes /
> > no) existence
> > of completed infinities" refer to?? Arguably there are two questions:
> >
> > (1) In mathematics, are there "completed infinities"?
>
> I protest against the use of infinite magnitude as something completed
> which is never permissible in mathematics. Infinity is merely a way of
> speaking, the true meaning being a limit which certain ratios approach
> indefinitely close, while others are permitted to increase without
> restriction.} ( C.F. Gauss [in a letter to Schumacher, 12 July 1831] )
A lot has changed in the last 180 years.
>
> > (2) In physics, are there "completed infinities"?
>
> No. Period.
Actually we do not know for sure. The universe, for all we know, may be
infinite, and is certainly complete in itself.
>
> Blah, blah. "Infinitum Actu Non Datur" is a perfectly clear statement.
And a lot has changed since Latin was the dominant language of science.
> Old philosphers did perfectly understand what it means, but in modern
> times, we've learned backwards, obviously (in more than one respect).
We still know what it means, but are no longer sure that it has to be
the case.
We no longer regard the earth as flat nor the Earth as necessarily the
center of the universe and a number of similar old beliefs, so why
accept "Infinitum Actu Non Datur" as necessary without a proof
satisfying modern standards.
FredJeffries
> On Mar 31, 10:02 am, Brian Chandler <imaginator...@despammed.com>
> wrote:
>
> > Han de Bruijn wrote:
> > > On Mar 30, 5:12 pm, Brian Chandler <imaginator...@despammed.com>
> > > wrote:
> > > > Well, yawn. If you are claiming something about physics, why not post
> > > > it in a physics group.
>
> > > This brings us to the core of the matter. Is the (yes / no) existence
> > > of completed infinities a mathematics or a physics question? Once you
> > > have decided _that_, I will choose my group.
>
> > As usual, your question doesn't make much sense. What does the "(yes /
> > no) existence
> > of completed infinities" refer to?? Arguably there are two questions:
>
> > (1) In mathematics, are there "completed infinities"?
>
> I protest against the use of infinite magnitude as something completed
> which is never permissible in mathematics. Infinity is merely a way of
> speaking, the true meaning being a limit which certain ratios approach
> indefinitely close, while others are permitted to increase without
> restriction.} ( C.F. Gauss [in a letter to Schumacher, 12 July 1831] )
For those able to read German, the 1831 correspondence with Schumacher
which prompted the above oft-quoted passage may be found in Band 8 of
his
collected works beginning at page 210
http://resolver.sub.uni-goettingen.de/purl?PPN236010751
and go to page 216:210 in the dropdown
The passage itself appears on page 222:216, second paragraph
See also William C. Waterhouse "Gauss on Infinity" Historia
Mathematica
Volume 6, Issue 4, November 1979, Pages 430-436
Abstract: In opposing the use of completed infinity in mathematics,
Gauss was
making a valid criticism of one particular kind of argument. His
celebrated
statement has no connection with the set theory to which it was later
applied.
David C. Ullrich
Yes. Curiously it's exactly the same as what I said several
posts ago.
>Or is everybody contradicting each other?
??? Who in this thread, except you, has said that the definition
is anything else?
Gus Gassmann
> On Mar 31, 1:30 am, Han de Bruijn <umum...@gmail.com> wrote:
> > I protest against the use of infinite magnitude as something completed
> > which is never permissible in mathematics. Infinity is merely a way of
> > speaking, the true meaning being a limit which certain ratios approach
> > indefinitely close, while others are permitted to increase without
> > restriction.} ( C.F. Gauss [in a letter to Schumacher, 12 July 1831] )
>
> For those able to read German, the 1831 correspondence with Schumacher
> which prompted the above oft-quoted passage may be found in Band 8 of
> his
> collected works beginning at page 210
>
> http://resolver.sub.uni-goettingen.de/purl?PPN236010751
> and go to page 216:210 in the dropdown
This is fascinating stuff. Thanks very much for sharing.
Han de Bruijn
> In article
> <284281bf-6747-4b92-ba92-16bea5e62...@l18g2000yqm.googlegroups.com>,
> Han de Bruijn <umum...@gmail.com> wrote:
[ .. snip redundancy .. ]
> >http://archives.math.utk.edu/visual.calculus/1/definition.6/index.html
>
> > Definition. The limit of f(x) as x approaches a is L
>
> > lim(x->a) f(x) = L
>
> > if and only if, given e > 0 , there exists d > 0 such that
> > 0 < |x - a| < d implies that |f(x) - L| < e .
>
> > Is that - finally - correct? Or is everybody contradicting each other?
>
> That definition will work in many, possibly even most, cases, provided
> one makes suitable assumptions about the domain of the function and the
> domain of the variable, but would not be acceptable in any but
> elementary calculus texts.
Kerm .. !! Are you kidding? Because I don't find it funny anymore.
The domain of the variable is an open interval in the reals. OK?
The domain of the function is an open interval in the reals. OK?
> Or is everybody contradicting each other?
Han de Bruijn
Jesse F. Hughes
> On Apr 1, 12:35 am, Jimmy John <Ji...@john.sci> wrote:
>> In article
>> <284281bf-6747-4b92-ba92-16bea5e62...@l18g2000yqm.googlegroups.com>,
>> Han de Bruijn <umum...@gmail.com> wrote:
>
> [ .. snip redundancy .. ]
>
>> >http://archives.math.utk.edu/visual.calculus/1/definition.6/index.html
>>
>> > Definition. The limit of f(x) as x approaches a is L
>>
>> > lim(x->a) f(x) = L
>>
>> > if and only if, given e > 0 , there exists d > 0 such that
>> > 0 < |x - a| < d implies that |f(x) - L| < e .
>>
>> > Is that - finally - correct? Or is everybody contradicting each other?
>>
>> That definition will work in many, possibly even most, cases, provided
>> one makes suitable assumptions about the domain of the function and the
>> domain of the variable, but would not be acceptable in any but
>> elementary calculus texts.
>
> Kerm .. !! Are you kidding? Because I don't find it funny anymore.
I'm not sure what Jimmy John means either.
Perhaps he's concerned that f(x) may be undefined close to x in which
case |f(x) - L| is not defined. In that case, it's easy enough to fix
it by either
(1) interpreting |f(x) - L| < e is false if f(x) is not defined or
(2) simply replacing the consequent by "f(x) is defined and
|f(x) - L| < e".
Perhaps Jimmy John should be more explicit on the problems of the above
definition. (Note that Han is only defining limits of functions R -> R,
as far as I can tell, so the fact that this is not as general as the
definition of limit in point-set topology is surely not a problem.)
--
Jesse F. Hughes
"Basically there are two angry groups. I am a harsh force of
one. Against me is a society of mathematicians. So far it's been a
draw." -- JSH gives another display of keen insight.
Jimmy John
<c15442e5-f798-41d7...@j13g2000yqj.googlegroups.com>,
Han de Bruijn <umu...@gmail.com> wrote:
Then provided that the overlap and the limit point, a, is in the closure
of each, it should work.