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Tony Orlow (aeo6)
science forum Guru
Joined: 24 Mar 2005
Posts: 4069
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 Posted: Thu Jun 22, 2006 5:37 pm Post subject:
Infinite Induction and the Limits of Curves
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Hi All -
It has been my position that the method of inductive proof is valid, not only
for all finite natural n, but for the infinite case as well, given certain
precautions. An equality proven inductively, such as f(n)=g(n), always holds in
the infinite case, but with inequalities we must be careful that the difference
that causes the inequality does not have a limit of 0 as n->oo, or the proof
does only hold for the finite case. Otherwise, if we can inductively prove that
f(n)>g(n) for all n greater than some finite m, and that f(n)-g(n) does not
have a limit of 0 as n->oo, then we can say that f(n)>g(n) for all infinite n
as well. This has important implications.
While discussing this in Calculus XOR Probability, Chas offered a
counterexample to infinite induction which became an interesting discussion. It
involved a staircase function from (0,0) to (1,1), in the limit as the number
of steps approached oo, and its equivalence to the diagonal line with those
endpoints, the contradiction coming in the form of two different arc length
measures for the "same" curve. Well, this led me to devise an improved
definition of the curve and demonstrate the difference between the two curves
using this definition. I was then challenged to find another curve which WAS
equivalent to the staircase in the limit given my definition, which I devised,
with interesting results, especially regarding the use of infinitesimals. I
wrote up a little paper on the topic which can be viewed at:
http://www.people.cornell.edu/pages/aeo6/Induction/Limits.htm
Please enjoy. I look forward to all comments, either here or in email (link at
bottom of page). I'll update the page with any pertinent ideas or comments.
Thanx. Have a nice day!
--
Smiles,
Tony |
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Virgil
science forum Guru
Joined: 24 Mar 2005
Posts: 5536
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| Posted: Thu Jun 22, 2006 7:10 pm Post subject:
Re: Infinite Induction and the Limits of Curves
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In article <MPG.1f049f5a354e71af98ad6f@newsstand.cit.cornell.edu>,
Tony Orlow <aeo6@cornell.edu> wrote:
| Quote: | Hi All -
It has been my position that the method of inductive proof is valid, not only
for all finite natural n, but for the infinite case as well, given certain
precautions.
|
There is a valid form of transfinite induction on well ordered sets, but
TO's is not valid, as it requires things which contradict the standard
properties of limits, such as the commutativity of all limit processes.
| Quote: | An equality proven inductively, such as f(n)=g(n), always holds in
the infinite case
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Only in TOmatics.
| Quote: |
While discussing this in Calculus XOR Probability, Chas offered a
counterexample to infinite induction which became an interesting
discussion. It involved a staircase function from (0,0) to (1,1), in
the limit as the number of steps approached oo, and its equivalence
to the diagonal line with those endpoints, the contradiction coming
in the form of two different arc length measures for the "same"
curve.
|
This is one of the cases in which two limit processes do not commute,
the limit of lengths of approximating polygons as the number of segments
increases and the limiting set of points as the number of steps
increases. The result depends on which limit is taken first.
| Quote: | Well, this led me to devise an improved definition of the
curve
A different definition, granted, but it is only in TO's eyes that there |
is any improvement seen.
| Quote: | and demonstrate the difference between the two curves using
this definition. I was then challenged to find another curve which
WAS equivalent to the staircase in the limit given my definition,
which I devised, with interesting results, especially regarding the
use of infinitesimals. I wrote up a little paper on the topic which
can be viewed at:
http://www.people.cornell.edu/pages/aeo6/Induction/Limits.htm
|
As TO's theory of infinitesimals does not agree with any of the standard
theroies allowing infinitesimals which all work within standard
axiomatic system, and TO has no axiomatic system in which his system can
be shown to exist, he is merely hawking vaporware again.
| Quote: |
Please enjoy. I look forward to all comments, either here or in email
(link at bottom of page). I'll update the page with any pertinent
ideas or comments.
Thanx. Have a nice day!
|
TO says in the first paragraph of
http://www.people.cornell.edu/pages/aeo6/Induction/Limits.htm
" Thus, proving inductively that, for all n greater than 2, 2*n<n^2, is
taken to be a valid proof for all finite n, but is not considered valid
for infinite n. This would contradict transfinite set theory"
That a proof does not consider certain cases is not evidence that those
cases are false, merely that they have not been proven true, so limiting
attention to finite cases says nothing either for or against infinite
cases.
Since TO's paper starts with a false claim ( that standard finite
induction says anything at all about "infinite" cases), anything and
everything based on that false assumption is at best suspect and at
worst horribly false. |
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cbrown
science forum Guru
Joined: 02 Jun 2005
Posts: 371
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Han de Bruijn
science forum Guru
Joined: 18 May 2005
Posts: 1285
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| Posted: Fri Jun 23, 2006 8:52 am Post subject:
Re: Infinite Induction and the Limits of Curves
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cbrown@cbrownsystems.com wrote:
Yes. The picture is missing.
Han de Bruijn |
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Tony Orlow (aeo6)
science forum Guru
Joined: 24 Mar 2005
Posts: 4069
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| Posted: Fri Jun 23, 2006 2:06 pm Post subject:
Re: Infinite Induction and the Limits of Curves
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cbrown@cbrownsystems.com said:
That was fixed shortly after you tried it. Try again. the link should now be
simply to "staircase.GIF".
--
Smiles,
Tony |
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Tony Orlow (aeo6)
science forum Guru
Joined: 24 Mar 2005
Posts: 4069
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| Posted: Fri Jun 23, 2006 2:14 pm Post subject:
Re: Infinite Induction and the Limits of Curves
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Han de Bruijn said:
Ack! Still? I looked at it from home last night on dialup.....
Looking at the html code now, it seems it has some conditional code pointing to
the image you mention, which I had missed. I had to change the file name in two
places apparently, since I used Word to make the html file. I'll have to get
something better together soon for doing web pages. Can you see if it works
now?
Sorry, and thanks!
--
Smiles,
Tony |
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David R Tribble
science forum Guru
Joined: 21 Jul 2005
Posts: 1005
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| Posted: Fri Jun 30, 2006 7:00 pm Post subject:
Re: Infinite Induction and the Limits of Curves
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Tony Orlow wrote:
| Quote: | While discussing this in Calculus XOR Probability, Chas offered a
counterexample to infinite induction which became an interesting
discussion. It involved a staircase function from (0,0) to (1,1), in
the limit as the number of steps approached oo, and its equivalence
to the diagonal line with those endpoints, the contradiction coming
in the form of two different arc length measures for the "same"
curve.
|
Virgil wrote:
| Quote: | This is one of the cases in which two limit processes do not commute,
the limit of lengths of approximating polygons as the number of segments
increases and the limiting set of points as the number of steps
increases. The result depends on which limit is taken first.
|
Yes, which was the whole point of Chas's example, that you can't
assume the limits work out the same.
Tony's "better definition of the limit of a curve" tries to explain the
apparent paradox by redefining the stairstep, so that instead of
finding the limit of delta_x+delta_y (which is 2), he is finding the
limit of the diagonal at each step. This is obviously a circular
argument, since he's using the diagonals of the steps to find
the diagonal of the square. But that is exactly the opposite of
the whole point of the example. |
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Tony Orlow (aeo6)
science forum Guru
Joined: 24 Mar 2005
Posts: 4069
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| Posted: Fri Jun 30, 2006 7:10 pm Post subject:
Re: Infinite Induction and the Limits of Curves
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David R Tribble said:
| Quote: | Tony Orlow wrote:
While discussing this in Calculus XOR Probability, Chas offered a
counterexample to infinite induction which became an interesting
discussion. It involved a staircase function from (0,0) to (1,1), in
the limit as the number of steps approached oo, and its equivalence
to the diagonal line with those endpoints, the contradiction coming
in the form of two different arc length measures for the "same"
curve.
Virgil wrote:
This is one of the cases in which two limit processes do not commute,
the limit of lengths of approximating polygons as the number of segments
increases and the limiting set of points as the number of steps
increases. The result depends on which limit is taken first.
Yes, which was the whole point of Chas's example, that you can't
assume the limits work out the same.
Tony's "better definition of the limit of a curve" tries to explain the
apparent paradox by redefining the stairstep, so that instead of
finding the limit of delta_x+delta_y (which is 2), he is finding the
limit of the diagonal at each step. This is obviously a circular
argument, since he's using the diagonals of the steps to find
the diagonal of the square. But that is exactly the opposite of
the whole point of the example.
|
As usual, you missed the point. I was comparing the steps of the staircase with
the corresponding segments of the diagonal. I did not create a limit of the
staircase which WAS the diagonal. I demonstrated a salient difference between
the two, using an equally valid sequence of pairs representing xy offsets of
each segment, rather than a set of xy coordinates of each point on the curve. I
further went on to show a different curve definition which WAS the same in the
limit as the staircase, additionally demonstrating the difference between
infinitesimals and subinfinitesimals. Do you actually read the thing, or are
you just going by memory from our conversations in Calculus XOR Probability?
http://www.people.cornell.edu/pages/aeo6/Induction/Limits.htm
--
Smiles,
Tony |
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David R Tribble
science forum Guru
Joined: 21 Jul 2005
Posts: 1005
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| Posted: Fri Jun 30, 2006 7:36 pm Post subject:
Re: Infinite Induction and the Limits of Curves
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David R Tribble said:
| Quote: | Tony's "better definition of the limit of a curve" tries to explain the
apparent paradox by redefining the stairstep, so that instead of
finding the limit of delta_x+delta_y (which is 2), he is finding the
limit of the diagonal at each step. This is obviously a circular
argument, since he's using the diagonals of the steps to find
the diagonal of the square. But that is exactly the opposite of
the whole point of the example.
|
Tony Orlow wrote:
| Quote: | As usual, you missed the point. I was comparing the steps of the staircase with
the corresponding segments of the diagonal. I did not create a limit of the
staircase which WAS the diagonal. I demonstrated a salient difference between
the two, using an equally valid sequence of pairs representing xy offsets of
each segment, rather than a set of xy coordinates of each point on the curve. I
further went on to show a different curve definition which WAS the same in the
limit as the staircase, additionally demonstrating the difference between
infinitesimals and subinfinitesimals.
|
Well, you assume the value of sqrt(2) in deriving your limit, which is
also the result you're looking for, so it seems circular to me.
And your whole premise about "infinite" induction remains as flawed
as ever:
"However, since Peano's axiom of induction is presented with the
other axioms for the natural numbers, and since natural numbers
are all taken to be finite, this form of proof is only considered
valid for the finite case, that is, over any finite number of
successions, but not for any iterations beyond those corresponding
to the finite naturals."
Your "in the finite case" premise is simply wrong. Induction over
the naturals already covers an infinite set, and there are no
"iterations beyond the finite naturals" in the Peano system. All of
this has been pointed out to you countless times. |
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Tony Orlow (aeo6)
science forum Guru
Joined: 24 Mar 2005
Posts: 4069
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| Posted: Fri Jun 30, 2006 7:47 pm Post subject:
Re: Infinite Induction and the Limits of Curves
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David R Tribble said:
| Quote: | David R Tribble said:
Tony's "better definition of the limit of a curve" tries to explain the
apparent paradox by redefining the stairstep, so that instead of
finding the limit of delta_x+delta_y (which is 2), he is finding the
limit of the diagonal at each step. This is obviously a circular
argument, since he's using the diagonals of the steps to find
the diagonal of the square. But that is exactly the opposite of
the whole point of the example.
Tony Orlow wrote:
As usual, you missed the point. I was comparing the steps of the staircase with
the corresponding segments of the diagonal. I did not create a limit of the
staircase which WAS the diagonal. I demonstrated a salient difference between
the two, using an equally valid sequence of pairs representing xy offsets of
each segment, rather than a set of xy coordinates of each point on the curve. I
further went on to show a different curve definition which WAS the same in the
limit as the staircase, additionally demonstrating the difference between
infinitesimals and subinfinitesimals.
Well, you assume the value of sqrt(2) in deriving your limit, which is
also the result you're looking for, so it seems circular to me.
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I use the normal metric to state the x and y offsets of each segment, that
being sqrt((x2-x1)^2+(y2-y1)^2), which for (x1,y1,x2,y2)=(0,1,0,1), gives sqrt
(2) in the diagonal segments. That is, if the x,y offset is {n,n} we use this
metric to ascertain that the length of that segment is sqrt(2)*n. In the
staircase, if each step spans this segment, it is divided into two segments
{0,n} and {n,0}, and the sum of the lengths is n+n=2n. There is no *assumption*
of sqrt(2) beyind assuming that you are familiar with the Pythagorean Theorem.
| Quote: |
And your whole premise about "infinite" induction remains as flawed
as ever:
"However, since Peano's axiom of induction is presented with the
other axioms for the natural numbers, and since natural numbers
are all taken to be finite, this form of proof is only considered
valid for the finite case, that is, over any finite number of
successions, but not for any iterations beyond those corresponding
to the finite naturals."
Your "in the finite case" premise is simply wrong. Induction over
the naturals already covers an infinite set, and there are no
"iterations beyond the finite naturals" in the Peano system. All of
this has been pointed out to you countless times.
|
The Peano system makes no reference to finite or infinite values, either as a
value within the set or as a measure pertaining to its size. Do you really not
see how accepting this notion leads immediately to a rich formulaic comparison
of an infinite number of ordered infinities? Look again.
--
Smiles,
Tony |
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david petry
science forum Guru
Joined: 18 May 2005
Posts: 503
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| Posted: Fri Jun 30, 2006 7:51 pm Post subject:
Re: Infinite Induction and the Limits of Curves
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Tony Orlow wrote:
| Quote: | Hi All -
It has been my position that the method of inductive proof is valid, not only
for all finite natural n, but for the infinite case as well, given certain
precautions. An equality proven inductively, such as f(n)=g(n), always holds in
the infinite case,
|
Hmm. I wonder if the following proof works in your mathematical
system. It doesn't work in ZFC.
Theorem: The rationals are equinumerous with the irrationals.
Proof. Let U be [0,1) (i.e. the unit interval including 0
but not 1). I will show that the rationals in U are
equinumerous with the irrationals in U.
Let T be an enumeration of the rationals in U starting
with 0 (e.g. T = {0, 1/2, 1/3, 2/3, 1/4 ...} ) and let T_n
be the set containing the first 'n' elements of T.
Let S_n be the set of pieces (connected components) left over
after removing the elements of T_n from U. For example,
S_3 = { (0, 1/3), (1/3, 1/2), (1/2, 1) }
In the limit as n -> oo, T_oo = {x | x in U and x is rational},
and S_oo = { {x} | x in U and x is irrational}
But for all n, the number of elements in T_n is exactly
equal to the number of elements in S_n, and hence, taking
the limit as n -> oo, we conclude that the number of elements
in T_oo equals the number of elements in S_oo QED |
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Dave L. Renfro
science forum Guru
Joined: 29 Apr 2005
Posts: 570
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| Posted: Fri Jun 30, 2006 8:08 pm Post subject:
Re: Infinite Induction and the Limits of Curves
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Virgil wrote:
| Quote: | This is one of the cases in which two limit
processes do not commute, the limit of lengths
of approximating polygons as the number of segments
increases and the limiting set of points as the
number of steps increases. The result depends
on which limit is taken first.
|
David R Tribble wrote:
| Quote: | Yes, which was the whole point of Chas's example,
that you can't assume the limits work out the same.
|
For what it's worth, arc length, while not continuous,
is "half-way" continuous in this context:
sci.math, "Strange objcet...", 29 September 2002
http://groups.google.com/group/sci.math/msg/71efa9c214d3f721
Dave L. Renfro |
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Virgil
science forum Guru
Joined: 24 Mar 2005
Posts: 5536
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| Posted: Sat Jul 01, 2006 3:09 am Post subject:
Re: Infinite Induction and the Limits of Curves
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In article <MPG.1f0f412c7358b96f98ada2@newsstand.cit.cornell.edu>,
Tony Orlow <aeo6@cornell.edu> wrote:
| Quote: | As usual, you missed the point.
There are myriad points that TO misses in almost every post. |
| Quote: | I was comparing the steps of the staircase with the corresponding
segments of the diagonal. I did not create a limit of the staircase
which WAS the diagonal.
|
You should have. By any mathematical standards it is.
| Quote: | I demonstrated a salient difference between
the two, using an equally valid sequence of pairs representing xy offsets of
each segment, rather than a set of xy coordinates of each point on the curve.
|
That might be equally valid in TO's promised future system, if it ever
gets completed, but not in any system now extant.
| Quote: | I
further went on to show a different curve definition which WAS the same in
the
limit as the staircase, additionally demonstrating the difference between
infinitesimals and subinfinitesimals.
|
None of which exist in any standard geometry.
| Quote: | Do you actually read the thing
|
Only far enough to see that it doesn't work. |
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Tony Orlow (aeo6)
science forum Guru
Joined: 24 Mar 2005
Posts: 4069
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| Posted: Mon Jul 03, 2006 7:01 pm Post subject:
Re: Infinite Induction and the Limits of Curves
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Dave L. Renfro said:
| Quote: | Virgil wrote:
This is one of the cases in which two limit
processes do not commute, the limit of lengths
of approximating polygons as the number of segments
increases and the limiting set of points as the
number of steps increases. The result depends
on which limit is taken first.
David R Tribble wrote:
Yes, which was the whole point of Chas's example,
that you can't assume the limits work out the same.
For what it's worth, arc length, while not continuous,
is "half-way" continuous in this context:
sci.math, "Strange objcet...", 29 September 2002
http://groups.google.com/group/sci.math/msg/71efa9c214d3f721
Dave L. Renfro
|
Hi David -
Thanks for your comment. I still am not sure what is meant by "semicontinuous".
The staircase is continuous in that there are no gaps, locationwise. In the
sense that the direction of the curve changes instantaneously where riser meets
tread, it is not continuous. Is that related? Did you read the referenced
page? It offers a different definition of the diagonal and staircase that shows
a clear difference in the limit, and then introduces a vertical sawtooth curve
which leans as it thins, until it IS the staircase in the limit by this new
definition. There are minor differences between the two which may be
disregarded, as they are on the scale of the square of an infinitesimal, but
given that stipulation, the concept works, as far as I can tell. Of course, the
whole point was to defend the notion of normal inductive proof holding in the
infinite case, undr certain conditions, and was meant to refute a
counterexample to that notion. Have a nice day!
http://www.people.cornell.edu/pages/aeo6/Induction/Limits.htm
--
Smiles,
Tony |
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Tony Orlow (aeo6)
science forum Guru
Joined: 24 Mar 2005
Posts: 4069
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| Posted: Mon Jul 03, 2006 7:01 pm Post subject:
Re: Infinite Induction and the Limits of Curves
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david petry said:
| Quote: | Tony Orlow wrote:
Hi All -
It has been my position that the method of inductive proof is valid, not only
for all finite natural n, but for the infinite case as well, given certain
precautions. An equality proven inductively, such as f(n)=g(n), always holds in
the infinite case,
Hmm. I wonder if the following proof works in your mathematical
system. It doesn't work in ZFC.
Theorem: The rationals are equinumerous with the irrationals.
|
Hi David -
I'd be interested in the exact reason why this doesn't work in ZFC. Is there a
particular point where it breaks, or does it simply contradict the conclusion
of another proof in ZFC?
| Quote: |
Proof. Let U be [0,1) (i.e. the unit interval including 0
but not 1). I will show that the rationals in U are
equinumerous with the irrationals in U.
Let T be an enumeration of the rationals in U starting
with 0 (e.g. T = {0, 1/2, 1/3, 2/3, 1/4 ...} ) and let T_n
be the set containing the first 'n' elements of T.
Let S_n be the set of pieces (connected components) left over
after removing the elements of T_n from U. For example,
S_3 = { (0, 1/3), (1/3, 1/2), (1/2, 1) }
|
Okay. I notice that for n sequential and separate points, there will be n-1
segments between them, but you are including 1 as an n+1th element, so indeed
you have n segments in [0,1) for n rationals in order. I can see that.
| Quote: |
In the limit as n -> oo, T_oo = {x | x in U and x is rational},
and S_oo = { {x} | x in U and x is irrational}
|
This part seems a little iffy. In all finite cases, the segment must include
rationals as well as irrationals, since between any two rationals is another
rational. Of course, between any two rationals must also lie an irrational, I
think. I think that uultimately the question becomes whether there must always
be a rational between any two irrationals. If you can prove this, as well as
there beign an irrational between any two rationals, then I think the case is
made, that the number line alternates between rationals and irrationals.
However, I don't think this is provable.
| Quote: |
But for all n, the number of elements in T_n is exactly
equal to the number of elements in S_n, and hence, taking
the limit as n -> oo, we conclude that the number of elements
in T_oo equals the number of elements in S_oo QED
|
Like I said, I think there is an assumption that each of these segments will
boil down to a single irrational number, though that's not provable in the
finite case. I am not sure it's provable in the infinite case either. There
seems to be something missing in that area. I think it's interesting, but I'm
not quite convinced it's right.
My thoughts on the rationals vs. the irrationals run as follows. There are an
infinite number of reals in the unit interval. Call this number Big'un. Over
the infinite real line, there are an equal number of unit intervals (extending
into infinite values). This gives Big'un^2 hyperreals. This corresponds to the
Big'un x Big'un matrix of hyperrationals, but that matrix includes the vast
majority as duplicate values. In fact, for every unique value in Q, there are a
countably infinite set of equivalent fractions. In my mind, this vast majority
of the rational fractions which are quantitatively redundant corresponds to the
vast majority of reals which are not rational, in agreement with standard
thinking. Of course, I cannot quantify the relative sizes of those sets at this
point, so my thinking on that matter is not complete.
--
Smiles,
Tony |
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