Deeper than primes - Continuation

[doronshadmi]

Let us examine this notion:

"For an infinite expanse of circles of radius r, the length is unbounded."

This is true only if there are infinitely many r's > 0 (also in that case the "host" space is not the sum of many "hosted" sub-spaces, exactly as an actual continuum is not a puzzle of sub-parts).

This is false if r = 0 (we have no more than a one element with r = 0).


"The circles cover the entirety of a straight line."

This is true only if there are infinitely many r's > 0 that touch each other (also in that case the "host" space (the straight line) is not the sum of many "hosted" sub-spaces (segments), exactly as an actual continuum is not a puzzle of sub-parts).

This is false if r = 0 (we have no more than a one element with r = 0).


"As r approaches 0, the circles grow smaller and in a sense are packed more tightly, but they continue to have an infinite expanse, covering the entirety of the straight line."

This is true only if there are infinitely many r's > 0 that touch each other, and there are infinitely many r's > 0 exactly because the r's approach (not actually reach) 0 (r's > 0) (again, the "host" space (the straight line) is not the sum of many "hosted" sub-spaces (segments), exactly as an actual continuum is not a puzzle of sub-parts).


"The limit as r approaches 0, then, is not a single point, but the entire line."

This is true since r's approach (not actually reach) 0 (r's > 0) (and again, the "host" space (the straight line) is not the sum of many "hosted" sub-spaces (segments), exactly as an actual continuum is not a puzzle of sub-parts).

This is false if r = 0 (we have no more than a one element with r = 0).

--------------------------------

An actual continuum (represented by a straight line in the following diagram) is not a collection of sub-parts (whether they are represented as many diameters of touching semi-circles (as shown in the diagram) or a collection of points (where in this case, there are many points as long as there are many diameters, where any given diameter is at AND not at the location of its end-points, and the straight line is at AND not at the domains of the sub-parts along it, because the "host" space (the straight line) is not the sum of many "hosted" sub-spaces):

Persons that are using their verbal_symbolic AND visual_spatial mind's skills, can get that.

Furthermore, they can get http://www.internationalskeptics.com/forums/showpost.php?p=7796635&postcount=219 .

 

[The Man]

Let us examine the following notion:

"As r approaches 0, the circles grow smaller and in a sense are packed more tightly, but they continue to have an infinite expanse, covering the entirety of the straight line. The limit as r approaches 0, then, is not a single point, but the entire line."

1. "r approaches 0" is different than "r is 0".

Certainly as "approaches" "is different than" than, well, "is".

2. If "r approaches 0" then and only then a collection of r (weather it is finite or not) covering the entirety of the straight line, in the sense of the sum of a given length, but even in this case the sum of a given length does not have of the power of the continuum of being one and only one non-composed (atomic) element, exactly as a puzzle of many "hosted" sub-spaces does not have the power of the continuum of the "host" space (the "host" space is not the sum of "hosted" sub-spaces).

You have consistently failed to demonstrate any use of your purported " power of the continuum of being one and only one non-composed (atomic) element" other than taking up decades of your life. Can "being one and only one non-composed (atomic) element" also be discontinuous?

3. "packed more tightly" has nothing to do with r = 0 (which is not less than totally packed).

So your "totally packed" is not "packed more tightly" than what you would consider "less than totally packed"?

4. "The limit as r approaches 0", is not actually be at the limit, and we are talking on not less than actually be at the limit, that if actually fulfilled, the result is exactly one and only one totally curved (totally packed) element, (no matter if the initial state is involved with finite or infinite collection of r>0 objects).

So now your claiming your line is completely covered or "(totally packed)" by points? What 'totally packs' your "one and only one totally curved (totally packed) element other than just BS?

In order to really get 1 to 4 ( or http://www.internationalskeptics.com/forums/showpost.php?p=7823487&postcount=253 ) verbal-symbolic AND visual-spatial skill's must be used.

As is the case with any visual representation of a spoken language.

 

[epix]

Or some other point. How the shrinking circles stay touching would need to be defined precisely; however, having the two shrinking circles moving towards each other equally and symmetrically is probably the most natural scenario, in which case the central point at the limit would be [2,0].

For two circles, there is indeed the third option, but that opens the door to infinitely many possibilities, and I can't see any mathematical instrument that would make the animation possible apart from an algorithm that would run the graphics. Moreover, the tangent point that is supposed to become the single point is defined by the circumferences of both circles. At r=0, there is no circumference, the defining terms cease to exist and therefore there is no point. That's why there is no real solution to the circle at r=0. The proposition made by Gosh is just intuitively conceptual.

 

[jsfisher]

Doron, instead of bounding into new directions to re-demonstrate your poor reading comprehension skills, how about addressing this:

So, Doron, is he right or wrong? Do 1 and 0.999... represent the same number? Is 1/7 rational? At what point does an infinite linear array of shrinking marbles switch from having infinite extent to being just a point?

 

[doronshadmi]

The associations among "host" "hosted" spaces is resulted by discontinuous forms of summed "hosted" spaces, that do not have the power of continuum of the "host" space (an actual continuum ("host" space) is not a puzzle of sub-spaces ("hosted" spaces)).

We have not less than Cross-contexts ("host" space) AND Context-dependent ("hosted" spaces) framework, which is known if verbal_symbolic AND visual_spatial skills are used.

 

[doronshadmi]

Persons that are using only verbal_symbolic skills or only visual_spatial skills, suffer from poor understanding.

 

[epix]

Let's say we start with N circles of radius r where N is some integer. The length of the linear array of those (touching) circles is 2Nr. As the circles shrink, r approaches 0 and the length, 2Nr also approaches 0. At the limit, the circles collapse to a single point.

For an infinite expanse of circles of radius r, the length is unbounded. The circles cover the entirety of a straight line. As r approaches 0, the circles grow smaller and in a sense are packed more tightly, but they continue to have an infinite expanse, covering the entirety of the straight line. The limit as r approaches 0, then, is not a single point, but the entire line.

I'm not entirely clear on the description regarding r, but there seems to be only two options:

A) lim. r → 0
B) r = 0

Plugging both options into 2Nr returns

A') 2Nr > 0
B') 2Nr = 0

So in the finite case, for the collapsing process that creates the singularity, r = 0 is the necessary requirement.

For the infinite case where N → ∞, the A and B options return

A'') 2Nr = 2
B'') 2Nr = 0

The conclusion is that for r = 0, the singularity exists for both finite and infinite cases; and for lim. r → 0 the singularity doesn't exist for both finite and infinite cases.

The algebra of the limit case is based on

If n → ∞ then n * 1/n = 1.

 

[The Man]

Persons that are using only verbal_symbolic skills or only visual_spatial skills, suffer from poor understanding.

Again, if your fantasy "persons" suffer so then perhaps you should have better fantasies.

 

[jsfisher]

I'm not entirely clear on the description regarding r, but there seems to be only two options:

A) lim. r → 0
B) r = 0

Plugging both options into 2Nr returns

A') 2Nr > 0
B') 2Nr = 0

No. Let L(r) = 2Nr, the length of our chain of N circles. Then we have

A') Limit of L(r → 0) = 0, and
B') L(r=0) = 0.

...
For the infinite case where N → ∞, the A and B options return

A'') 2Nr = 2
B'') 2Nr = 0

No. It is still the limit as r approaches 0 that needs to be explored. For the infinite case, you are starting off with an infinite supply of circles, not a finite quantity that grows to infinity.

A'') limit L(r → 0) is unbounded
B'') L(r = 0) is undefined.

 

[doronshadmi]

Limited persons understand creativity as fantasy.

For example, they are not aware of http://www.internationalskeptics.com/forums/showpost.php?p=7826122&postcount=261 .

 

[jsfisher]

Why do you dodge these simple questions, Doron?

So, Doron, is he right or wrong? Do 1 and 0.999... represent the same number? Is 1/7 rational? At what point does an infinite linear array of shrinking marbles switch from having infinite extent to being just a point?

 

[epix]

No. Let L(r) = 2Nr, the length of our chain of N circles. Then we have

A') Limit of L(r → 0) = 0, and
B') L(r=0) = 0.

That's right. For some reason (LOL), I forgot to evaluate the limit to zero. On the other hand, the concept limit of the calculus may not be applicable in this particular case, coz there are no functions that would handle the shrinking. Maybe that went through my mind when I wrote > instead of = in A'. But one should stick with the intuitive concept and investigate the possibilities under the assumption that the standard concept of limit applies.

No. It is still the limit as r approaches 0 that needs to be explored. For the infinite case, you are starting off with an infinite supply of circles, not a finite quantity that grows to infinity.

A'') limit L(r → 0) is unbounded
B'') L(r = 0) is undefined.

The expression N → ∞ doesn't imply that a finite quantity grows to infinity - it is used to indicate that N doesn't represent finite quantity. If you stick with L= f(r) = 2Nr where r is the variable and

lim.[r → 0] = 0

then

2Nr = 2 (assuming you disregard the non-real result for the circle function.)

That's because, you need to first multiply N and r where N→ ∞ and (lim. r→0) = (1/N).

Can you write down the intermediate steps for the A'' case limit L(r → 0) is unbounded for L = f(r) = 2Nr?

 

[jsfisher]

The expression N → ∞ doesn't imply that a finite quantity grows to infinity - it is used to indicate that N doesn't represent finite quantity.

Ok, but you are still treating it like a limit.

If you stick with L= f(r) = 2Nr where r is the variable and

lim.[r → 0] = 0

then

2Nr = 2 (assuming you disregard the non-real result for the circle function.)

How do you get that? L(r) is unbounded for all r > 0. Since L(r) is unbounded everywhere near r = 0, the limit as r approaches 0 is similarly unbounded.

 

[epix]

Ok, but you are still treating it like a limit.

How do you get that? L(r) is unbounded for all r > 0. Since L(r) is unbounded everywhere near r = 0, the limit as r approaches 0 is similarly unbounded.

Well, there is the issue of what you get when you multiply "infinitely large number" - in this case N - with "infinitely small number" - in this case r. If you take into the consideration that

http://www.mathsisfun.com/calculus/limits-infinity.html

and let N approach infinity and also let r approach zero, then there exists a product Nr as a part of 2Nr. The radius is approaching zero at reciprocal rate of N approaching infinity.

Here is the well-known limit

e = lim[x→ ∞] (1 + 1/x)^x
where 1/x is approaching zero and x is approaching infinity. The formula needed to be evaluated and the result was e - the base of natural logarithms.

Isn't it true that (r→ 0) = lim[N→ ∞] 1/N?

So from the above, the product Nr = N * 1/N = 1.

And so 2Nr = 2. In other words, if you keep shrinking the radius of an infinite numbers of touching circles in a linear array, then as r approaches zero, the lenghth of the array approaches 2. When r = 0, the length is undefined.

 

[jsfisher]

Well, there is the issue of what you get when you multiply "infinitely large number" - in this case N - with "infinitely small number" - in this case r. If you take into the consideration that

[qimg]http://img43.imageshack.us/img43/5627/limit1.png[/qimg]
http://www.mathsisfun.com/calculus/limits-infinity.html

and let N approach infinity and also let r approach zero, then there exists a product Nr as a part of 2Nr. The radius is approaching zero at reciprocal rate of N approaching infinity.

There's a problem right there. You have decided that N and r are approaching there limits together. They are not coupled; N does not get bigger as r gets smaller.

You could express it as a sequence of two limits, if you wish, but with N → ∞ having precedence over r → 0. I.e.

lim[r → 0] { lim[N → ∞] 2Nr }

Be that as it may, even if we do work from a case where N and r are coupled as you seem to prefer but is separate from the original I proposed, we would still need to agree on the exact coupling (you've used N = 1/r, but it could just as easily be N = 16,777,216/r), but we'd still end up contradicting the conclusion of Ghosh: The circles don't reduce to a single point.

 

[epix]

There's a problem right there. You have decided that N and r are approaching there limits together. They are not coupled; N does not get bigger as r gets smaller.

You could express it as a sequence of two limits, if you wish, but with N → ∞ having precedence over r → 0. I.e.

lim[r → 0] { lim[N → ∞] 2Nr }

Be that as it may, even if we do work from a case where N and r are coupled as you seem to prefer but is separate from the original I proposed, we would still need to agree on the exact coupling (you've used N = 1/r, but it could just as easily be N = 16,777,216/r), but we'd still end up contradicting the conclusion of Ghosh: The circles don't reduce to a single point.

It's the matter of the initial condition and the rate of approach to given limits. If you start off with infinite number of circles having radius r=1, for example, then the result would be completely different. If you go by the standard concept of limits, then you can actually arrive at important constants, like the base of natural logarithms, for example, so I sort of automatically used it.

I found a parametric equations plotter on the web, and so I could see what's going on when the case is finite and the hypothetical singularity is supposed to materialize at [0,0]. That 2Nr part showed up in the equations pretty quickly. (L is the length of the array.)

 

[doronshadmi]

One can choose some arbitrary touching point between left and right infinitely many circles along an unbounded straight line.

They are considered as circles only if each one of them has r > 0.

If all r's = 0 we actually left with the arbitrary chosen point, which is located on the unbounded straight line.

So Dr. Ghosh is right, and no verbal_symbolic manipulations (that do not distinguish between "approach" and "reach") can changes that.

 

[jsfisher]

One can choose some arbitrary touching point between left and right infinitely many circles along an unbounded straight line.

They are considered as circles only if each one of them has r > 0.

If all r's = 0 we actually left with the arbitrary chosen point, which is located on the unbounded straight line.

So Dr. Ghosh is right, and no verbal_symbolic manipulations (that do not distinguish between "approach" and "reach") can changes that.

Doron, there was no need for you to again demonstrate you poor reading comprehension skills and your lack of understanding of limits. These are well-established. Please, instead, respond to these ever so simple questions:

So, Doron, is he right or wrong? Do 1 and 0.999... represent the same number? Is 1/7 rational? At what point does an infinite linear array of shrinking marbles switch from having infinite extent to being just a point?

 

[epix]

So Dr. Ghosh is right, and no verbal_symbolic manipulations (that do not distinguish between "approach" and "reach") can changes that.

Things may not be that straightforward. The email was sent to IJPAM by Dr. Kaushik Ghosh.

Yours sincerely,

Dr. Kaushik Ghosh,

Department of Physics,
ST. Xavier's College,
30, Mother Teresa Sarani,
Kolkata - 700016, India.

But the Department of Physics is headed by some other Dr. Gosh.

Head of the Department
Dr. Subhankar Ghosh (SGH)
M.Sc., Ph.D.
Associate Professor

Maybe Dr. Kaushik Ghosh is just a teacher there.

I showed the picture of St.Xavier's College

to Prof. Jones and he said that he was familiar with that place. He also said that the building has some underground structure that houses Department of Cardiology where the controversial "kali ma" (approach and reach) procedure is taught. So it is possible that Dr. Kaushik Ghosh is also teaching there:

 

[doronshadmi]

If a given community of scholars do not distinguish between "approaches" to a given value and "reaches" a given value, then their framework is not well-established, no matter how many verbal_symbolic manipulations are written in vast quantity of their refereed journals, along the years.