why the Wiles's FLT, Appel&Haken's 4Color Mapping, Hales's KPP are fakes #1000 Correcting Math
Archimedes Plutonium
I do not have much time for "people talk" in this book, but I should
give it some comment,
because, if nothing else, most people do not know math and are scared
and afraid of
math and about the only thing they know of mathematics other than add
or multiply is
the stories about people in math. They know little to nothing of math
proofs or math
results. So I should comment on math people talk, for the reason that
the nonmathematician
knows more about math people talk than they know about math itself.
Now why do I say that Wiles's FLT is a fake, and Hales's Kepler
Packing is a fake
and Appel & Haken's 4 Color Mapping is a fake?
The first sentence in Wiles's FLT should have been to well-define
Finite versus Infinity.
He never has a precision definition of what finite versus infinite
means, and so in the
pythagorean triples of a^n + b^n = c^n for n>2, Wiles is never clear,
nor Fermat himself,
clear as to what are Finite numbers to fit into that equation and
what
are Infinite numbers.
So if the first sentence in the Wiles alleged proof had been "the
border between Finite
and Infinity is 10^700, then Mr. Wiles could have launched a valid
math proof, because then,
and only then would he have well-defined what a finite number is
versus an infinite number.
The Wiles's fake proof of FLT is no better, nor any different from
Wiles proving that
"all numbers greater than 2 are peachy". That is a nonsense
statement,
but the FLT, without
a clear precise definition of Finite versus Infinite is a nonsense
conjecture.
The first sentence in Hales's Kepler Packing fake proof should have
been the same as
Mr. Wiles first sentence, where Mr. Hales well-defines Finite versus
Infinity. He could have
done this by saying the border between finite 3D Euclidean space is
10^700. Then the task
of the proof was to show that the hexagonal closed packing is the
maximum density. But since we have a border, the hexagonal closed
packing is not the maximum packing density
and that because of the border, there needs some adjustment of
packing
at the border itself.
In various posts of mine I described this as oblong packing at the
borderline.
In the recent Poincare conjecture fake proof, here again, the first
sentence would have been to
define with precision, what it means to be finite versus infinite in
the small scale, of between points. Does mathematics have a continuum
of points? It looks as though math does not and that that Euclidean
plane geometry postulates are inconsistent with a continuum of
points.
That there must be gaps between any two points as we go smaller and
smaller. So that in the
Poincare Conjecture, same as FLT and Kepler Packing, a precision
definition of Finite versus
Infinity such as 10^700 would say that there are no more points in
geometry less than 10^-700.
That means the Poincare Conjecture is false in mathematics. Unlike
the
Kepler Packing that needs adjustments for its conclusion, or like the
FLT that is true for a borderline of say 10^700,
unlike them, the Poincare Conjecture is flat out totally false. And
it
is a hypocrisy to read how
some group of mathematicians were able to foist such a fake. It is
understandable that FLT
and KPP needed adjustments to be true, but in the Poincare Conjecture
the conjecture is totally false and can never be salvaged. It is why
the PC has gone on so long without ever being proven-- because it has
no truth to it at all.
Now the irony of this last fake proof, is that the borderline between
countries in the 4 Color Mapping problem of Appel & Haken, that fake
proof does not require them, for their first sentence to well-define
Finite versus Infinity, although it would not hurt to be their first
sentence. But the irony of 4Color Mapping is that the error is the
ignoring of countries's borderline. To well define Finite versus
Infinity we must state a borderline. But in the 4Color
Mapping, it is the borderline between countries that is tossed out
the
window, and we are expected to believe and accept that two countries
can be without borders. Once anyone reviewing the 4Color Mapping,
insists that the borders be installed, would immediately recognize
that the mapping is a 2Color Mapping is sufficient to color all maps,
because the
border is one color and the interior is the second color. In fact,
Information theory should have a theorem and proof that all
information can be stored with 2 colors or 2 digits or 2 things. The
4Color Mapping, because it expects the countries to toss out their
borderlines is not a mathematics problem but a psychology problem of
visual perspective. Here again, if the
4Color Mapping is a math problem then so is the ridiculous "all
countries are peachy countries
a math problem".
Gauss, a long time ago warned people, that when math has failings,
the
failure is usually due to some definition that is ill-defined or ill-
conceived, and until that lousy definition is corrected, it is not
mathematics. In the above examples, although the conjectures looked
like real math
problems. All of them had huge errors of precision definition of
Finite versus Infinite and borderlines.
The Riemann Hypothesis and the Goldbach Conjecture are unproveable,
until and unless
the first sentence in these two conjectures reads: Finite versus
Infinity is defined at the
border of 10^618.
Archimedes Plutonium
http://www.iw.net/~a_plutonium/
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
Archimedes Plutonium
surface area go to
infinity as the poles go to infinity. The proof simply relies on a
construction of a cone that is
thinly shaped, like a needle that fits inside the pseudosphere poles
and since the volume of
a cone is 1/3(pi)R^2h, that the h goes to infinity means the volume is
infinite.
So I want to track down where the error in math history occurred.
Where someone published
a bogus proof of the volume and surface area of pseudosphere. I hear
tell the bogus proof
relies on a Series convergence of 1+1/2 + 1/4 + 1/8+...
So I looked in Wolfram's MathWorld for some reference to the
historical proof of pseudosphere volume:
--- quoting from http://mathworld.wolfram.com/Pseudosphere.html ---
REFERENCES:
Fischer, G. (Ed.). Plate 82 in Mathematische Modelle aus den
Sammlungen von Universitäten und Museen, Bildband. Braunschweig,
Germany: Vieweg, p. 77, 1986.
Geometry Center. "The Pseudosphere." http://www.geom.umn.edu/zoo/diffgeom/pseudosphere/.
Gray, A.; Abbena, E.; and Salamon, S. Modern Differential Geometry of
Curves and Surfaces with Mathematica, 3rd ed. Boca Raton, FL: CRC
Press, pp. 477 and 480, 2006.
JavaView. "Classic Surfaces from Differential Geometry: Pseudo
Sphere." http://www-sfb288.math.tu-berlin.de/vgp/javaview/demo/surface/common/PaSurface_PseudoSphere.html.
Steinhaus, H. Mathematical Snapshots, 3rd ed. New York: Dover,
p. 251, 1999.
Wells, D. The Penguin Dictionary of Curious and Interesting Geometry.
London: Penguin, pp. 199-200, 1991.
--- end quoting from http://mathworld.wolfram.com/Pseudosphere.html
---
The above hints to me that Beltrami, who is credited with the first
discovery of the pseudosphere, had no role in determining the volume
or surface area. Although Beltrami
would have known of that Series.
So I need to find out if the proof is a recent one such as in the 20th
century.
Is there some sort of mathematical tome that records who proved what
and when?
Archimedes Plutonium
http://en.wikipedia.org/wiki/Tractrix
And they show the tractrix graph of (4,0) or a pseudosphere of radius
4
Now that would mean a sphere of radius 4 has volume of approx
4/3(pi)R^3
or 4(4^3) = 256
Now the volume of that tractrix made into pseudosphere would be an
enclosed
volume inside the sphere can be approximated as the two cones of
radius 4
and approx equal to 1/3(pi) R^2h where R is 4 and h is 4 so approx
4^3= 64
and since there are two cones approx the enclosed pseudosphere we have
an interior volume of 128 for the pseudosphere.
Now all we need to account for is the exterior volume of the
pseudosphere and
that volume cuts the surface of the enclosed sphere giving an exit-
circle of
radius 1. Notice in the graph that at y= 4 and -4 that the x component
is 1 so
a radius of 1 for the exit-circle of the pseudosphere.
Now we do the Needle-Cone accounting of volume. This cone has radius
1. So that
eliminates the 1/3(pi)R^2h to be (1)h. So for this Needle-Cone we only
need to go out
to a distance of 64 for the height of this cone and thus give us a
volume of 64. Now there
are two pseudosphere poles and both of them have volume 64, having
traveled 68 units
from the center of the sphere and pseudosphere.
Now this does not agree with the Wikipedia page on pseudosphere which
claims the pseudosphere volume never goes beyond 1/2 the volume of the
sphere of identical radius.
Here we contradicted Wikipedia's entry of pseudosphere volume for that
pseudosphere of radius 4 equals the volume of the sphere of radius 4
at a distance of 68 from the center of the
sphere. And because the pseudosphere goes to infinity, the volume of
pseudosphere of radius 4 is also infinite, since the height of the
Needle Cone is infinity.
So I wonder, is everyone in the mathematics community, asleep at the
watchtower?
Archimedes Plutonium
Archimedes Plutonium wrote:
> Alright, so on Wikipedia they give a page on the Tractrix:
>
> http://en.wikipedia.org/wiki/Tractrix
>
> And they show the tractrix graph of (4,0) or a pseudosphere of radius
> 4
>
> Now that would mean a sphere of radius 4 has volume of approx
> 4/3(pi)R^3
> or 4(4^3) = 256
>
> Now the volume of that tractrix made into pseudosphere would be an
> enclosed
> volume inside the sphere can be approximated as the two cones of
> radius 4
> and approx equal to 1/3(pi) R^2h where R is 4 and h is 4 so approx
> 4^3= 64
> and since there are two cones approx the enclosed pseudosphere we have
> an interior volume of 128 for the pseudosphere.
Now that is a very poor approximation of the interior enclosed
pseudosphere
inside the sphere of radii 4 for both. The experiment that was
performed earlier
in this thread shows that the pseudosphere volume enclosed in sphere
is
approx 25% of the sphere volume whereas above I got 50%. And we can
see
why that is a poor approximation because the sides of the cone of
radius 4 based
on the plane of the equator of the sphere extends far outside the
sides of the
pseudosphere. In the Wikipedia drawing of tractrix (4,0) we can see
that a better
approx of the interior pseudosphere volume would be where we used the
equatorial
plane radius of the pseudosphere as radius of 2, not 4 so that the
cone fits inside
the pseudosphere and does not go outside the pseudosphere.
With the cone formula for volume, 1/3(pi)R^2h we would then have the
interior
volume of pseudosphere as (4)4 and since there are two of these cones
we
would have volume of 16x2= 32. Now a interior pseudosphere volume of
32 enclosed
by sphere of volume 256 is not 25%, because we have not captured the
volume of the
pseudosphere that ranges from 2 to 4 on the x axis. So let us assume
that the volume
of this interior pseudosphere is 25% of the sphere volume and thus
would be a total
interior pseudosphere volume of 64.
>
> Now all we need to account for is the exterior volume of the
> pseudosphere and
> that volume cuts the surface of the enclosed sphere giving an exit-
> circle of
> radius 1. Notice in the graph that at y= 4 and -4 that the x component
> is 1 so
> a radius of 1 for the exit-circle of the pseudosphere.
>
> Now we do the Needle-Cone accounting of volume. This cone has radius
> 1. So that
> eliminates the 1/3(pi)R^2h to be (1)h. So for this Needle-Cone we only
> need to go out
> to a distance of 64 for the height of this cone and thus give us a
> volume of 64. Now there
> are two pseudosphere poles and both of them have volume 64, having
> traveled 68 units
> from the center of the sphere and pseudosphere.
Now here I need 256-64 = 192 volume for the poles that extend out of
the sphere
so that the pseudosphere volume equals the sphere volume of radii 4.
So I divide
that by two because of two poles gives me 192/2 = 96
The same formula applies as above for the Needle Cone, only the
distance traveled
is not 64 for the height of the Needle Cone but is 96.
So that for a sphere of radius 4 and volume 256, a pseudosphere of
radius 4 and distance
of 96 + 4 = 100 from the center of the sphere has equal volume with
the sphere itself.
So who is Wikipedia and MathWorld joking around with, when they claim
the volume of the
pseudosphere of identical radius to a sphere has a maximum of 1/2 the
sphere volume?
Who are they joking with, because the Needle Cone construction shows
that the volume
of pseudosphere is infinite.
>
> Now this does not agree with the Wikipedia page on pseudosphere which
> claims the pseudosphere volume never goes beyond 1/2 the volume of the
> sphere of identical radius.
>
> Here we contradicted Wikipedia's entry of pseudosphere volume for that
> pseudosphere of radius 4 equals the volume of the sphere of radius 4
> at a distance of 68 from the center of the
> sphere. And because the pseudosphere goes to infinity, the volume of
> pseudosphere of radius 4 is also infinite, since the height of the
> Needle Cone is infinity.
>
> So I wonder, is everyone in the mathematics community, asleep at the
> watchtower?
>
Or I should say, asleep at the switch.
My program in this thread is to show that Finite versus Infinity is
precision defined by
stating the border between Finite versus Infinity and the pure
mathematics has a natural
border given by the pseudosphere versus sphere. There is one and only
one, or a unique large number, such as 10^618 where the volume and
surface area of a sphere and pseudosphere
equal each other and equal this unique large number. In my next post
I should work out the
surface area of this pseudosphere of radius 4.
Archimedes Plutonium
I think the best approximation of volume would be a concatenation of
truncated
or the sawed off tips of cones stacked together all having a height of
1 unit
from that Wikipedia graph of tractrix (4,0)
For the pseudosphere poles a concatenation of truncated cones would
get a more
accurate volume and surface area, but essentially since the height
goes to infinity
the volume goes to infinity.
Let me start on the Surface Area using that tractrix graph (4,0) as
shown
in Wikipedia.
The right circular cone surface area is (pi)R^2 + (pi)Rs where s is
the slant height
and where s is known as the hypotenuse of a right triangle cross
section.
The height in our example is infinity so the slant height is infinity.
Now some are going to complain and argue that I cannot construct a
solo cone into that pseudosphere pole that goes to infinity. And I
would point out to them that no matter how
far down the line one goes on a pole, and although that diameter of
the pseudosphere
pole gets smaller and smaller, that I can still maintain a very narrow
cone that keeps
marching along to infinity.
Now for my example of the tractrix of (4,0) the sphere of radius 4 has
surface area:
4(pi)R^2 which gives us approx 192 sq units.
the enclosed part of the pseudosphere inside the sphere is
approximated by a radius of 2
cone surface area which yields (pi)Rs which is (pi)4(5.6) where the
slant height is about
5.6 so the interior pseudosphere enclosed by the sphere has surface
area approx 67.2
times two for both sides is 134.4 sq. units
So much of a distance on the pseudosphere poles do we have to travel
in order to have
equal volume with the sphere surface area? Well that would be 192 -
134.4 =57.6
and that divided by two for the two poles is 28.8. The radius of the
exiting-circle of pseudosphere is 1 so for (pi)Rs , I would need a 3s
= 28.8 or s = 9.6
That 9.6 is square root of R^2 + h^2 and since radius is 1 then h is
9.5
So the pseudosphere distance from the center of the sphere out to 4 +
9.5 = 13.5
distance, does the surface area of the pseudosphere equals the surface
area of the
sphere of radius 4 itself. But according to Wikipedia and MathWorld
there is no more surface area for it should be finite and equal to the
sphere, but here we show that there is an infinity
of surface area for pseudosphere of radius 4.
Again, I ask, where has the proof of this Pseudosphere volume as 1/2
of the associated
sphere of same radius and equal in surface area of that same sphere?
Where have those
erroneous conclusions come from and who did the erroneous proof? I
doubt it was Beltrami,
although it may have been?
Apparently the error filled proof never considered tiling the
pseudosphere with cones.
Archimedes Plutonium
I was correct with
volume and surface area of pseudosphere is infinite or whether the
math literature with MathWorld, encyclopedias and Wikipedia were
correct by saying the pseudosphere volume is
1/2 the sphere volume of identical radius and equal to sphere surface
area of identical radius.
Alright, I hope this argument and reasoning will be the end of all
such doubt.
All of us in mathematics accepts that the volume of a sphere of radius
R is
4/3(pi)R^3 and surface area is 4(pi)R^2
And for a right circular cone the volume of radius R is 1/3(pi)R^2h
and the surface
area is (pi)R^2 + (pi)Rs with s the slant height.
None of us doubts that, and all of us agree those are true beyond
question.
Now, tell me what the volume is on a right circular cone if the height
is infinity
and the base is of radius 10^1000^10000? Instinctively, and without
hesitation
nearly every one of us will immediately say the volume is infinite. So
do you see
where I can go with this? Do you see that no matter what the "small
size" of the
radius is, except infinity, that the infinity for the height of the
cone makes the volume
equal to an infinite volume.
So somewhere in the history of mathematics when it came time to prove
the volume and
surface area of pseudosphere, everything went kaput and heywire and
all the fuses seemed
to have gone up in ash.
Noone bothered with the above style of convincing argument.
Yes the poles of the pseudosphere rapidly decline and get smaller and
smaller. But since those poles go to infinity, that Euclidean geometry
of cones also go to infinity and the cone widths can be
smaller than the width of the pseudosphere pole widths. That we can
insert the cone as a sort of needle all the length to infinity with
this infinitely long needle-cone.
The errors of the formal proof that claimed the volume and area of
pseudosphere as finite, never raised a argument of a infinitely long
cone. Why they never did this? Maybe just blind spotted. They probably
used a concatenation of cones of different widths and then they
applied a Series of 1+1/2+1/4+1/8+... that was irrelevant and false
accounting just as the
hotel paradox I discussed earlier.
So if this argument does not convince those sitting on the fences, I
do not have anything more
simple of a argument than this.
If you still are unconvinced, would only mean that you deny the
existence of very thin cones
and the formula for the volume of cones. But you know, everytime I get
a splinter in me or use a needle to mend and sew my work clothes, I am
using one of those cones.
Archimedes Plutonium
Archimedes Plutonium wrote:
> Alright, someone called me and said he was on the fence as to whether
> I was correct with
> volume and surface area of pseudosphere is infinite or whether the
> math literature with MathWorld, encyclopedias and Wikipedia were
> correct by saying the pseudosphere volume is
> 1/2 the sphere volume of identical radius and equal to sphere surface
> area of identical radius.
>
>
> Alright, I hope this argument and reasoning will be the end of all
> such doubt.
>
> All of us in mathematics accepts that the volume of a sphere of radius
> R is
> 4/3(pi)R^3 and surface area is 4(pi)R^2
>
> And for a right circular cone the volume of radius R is 1/3(pi)R^2h
> and the surface
> area is (pi)R^2 + (pi)Rs with s the slant height.
>
> None of us doubts that, and all of us agree those are true beyond
> question.
>
> Now, tell me what the volume is on a right circular cone if the height
> is infinity
> and the base is of radius 10^1000^10000? Instinctively, and without
Sorry, a big error here that should have been negative exponents but
then
there is a notation confusion because 10^-1000^-10000 may mean several
different things. So I should change that to read 10^-10000000 in the
original followed
by a (sic) sign.
Archimedes Plutonium
Archimedes Plutonium wrote:
(snipped)
> >
> > Now, tell me what the volume is on a right circular cone if the height
> > is infinity
> > and the base is of radius 10^-10000000 (sic) ? Instinctively, and without
> > hesitation
> > nearly every one of us will immediately say the volume is infinite. So
> > do you see
> > where I can go with this? Do you see that no matter what the "small
> > size" of the
> > radius is, except infinity, that the infinity for the height of the
> > cone makes the volume
> > equal to an infinite volume.
> >
Hold on a moment, for I can do more to make it easy to grasp that the
pseudosphere
volume and surface area go to infinity.
I can marshall the use of Series, in a correct manner though, not the
incorrect manner.
I believe the error of Series is when you apply them to discontinuous
geometry of
disconnected cylinders that are graduated into smaller sizes of
applying
1+ 1/2 + 1/4 + 1/8+... when the geometry is continuous and has 1/3,
and 1/5 and 1/6 etc etc.
So that in a mathematical proof, if a geometry is continuous and you
try to apply a discontinuous Series, you abandoned logical
consistency.
So let me show that the Pseudosphere volume goes to infinity by
applying a true Series to
a continuous geometrical figure of the right circular cone. And let me
use the graph of the tractrix (4,0) as shown in Wikipedia for the
poles as an exit-circle of 1. I will prove that the
one pole, alone, has infinite volume.
We know the Series 1 + 1/2 + 1/3 + 1/4 + 1/5 + (or 1/n in
general) . . . that this
series diverges to infinity.
Now let me display the pole of the pseudosphere on the x-axis with its
circle radius
of 1 at x = 4, its exit circle radius. And parallel to this
pseudosphere pole, let me draw
a needle-cone that goes to infinity on the x-axis. Let me define a
concept of x-axis width
as the width of the pseudosphere pole when x=5, x=6, etc etc and the
same concept of
width of needle cone at x=5, x=6, etc etc. And measure the width in
terms of radius.
So that at x=4 the pseudosphere width is 1. And now I construct a
needle cone whose
width is going to be much smaller than the pseudosphere width along
the x-axis.
Now I assign the divergent series 1 + 1/n to each segment of the
needle cone such that
the second segment has 1/2 the volume of the original, and the next
segment has 1/3 the
volume of the original, and the next segment 1/4 volume of original.
Now the volume of this needle cone is infinite and the Series concurs
with the infinite volume.
And the segments along the needle cone, all fit inside the
pseudosphere pole all along the x-axis and since the Series is
divergent means that the pseudosphere pole is infinite in volume
also.
So there I outlined a true proof of how to use Series in determining
volume.
Archimedes Plutonium
Using the pseudosphere formed from the tractrix of the graph (4,0)
http://en.wikipedia.org/wiki/Tractrix
We can estimate the radius of the exit-circle of the pseudosphere
poles as
it exits the enclosing sphere.
That graph (4,0) indicates a radius of 1 for the pseudosphere pole and
the
circumference of enclosing sphere is 2(4)(pi) = 25. And so a
percentage of
the enclosing sphere's circumference to its radius of pseudosphere
pole
is 1/25 or 4%. So that given any general sphere enclosing of
pseudosphere
the exiting pole will be about 4% of the circumference.
This is useful in estimating whether the B-number (border between
Finite versus
Infinity) is 10^618. In the example of the graph of pseudosphere (4,0)
as given
in Wikipedia, that would use a B-number of 256 since the volume is
4^4, and we can
begin to understand that 256 cannot be the B-number because the volume
and surface
area of the sphere versus pseudosphere with radii of 1, 2, 3, 4 cannot
be equal.
That we need a large number such as 10^618 where the poles of the
pseudosphere
are the lionshare of the volume and surface area so that 10^309 and
10^206 as radii
can equilibrate the volume and surface area for the sphere versus
pseudosphere.
Now someone is quick to ask the question, so what? So what if the
volume and surface
area equal the sphere versus sphere when the upper boundary is 10^618.
And the answer
to that nagging perplex question is that since a pseudosphere
stretches to infinity and since a
sphere volume is always known as finite, that when this happens
naturally in mathematics that the infinite expanse of pseudosphere
volume and surface area are equal to that of finiteness
of sphere, then that truly is a boundary set up within mathematics
itself. That Finite meets
Infinity at the B-number, and it is the number for which mathematics
has Aristotelian Logic where the answers by numbers below the B-number
are reliable, dependable and not something of Quantum logic duality.
Above the B-number, mathematics is untrustworthy for it
has duality issues.
Archimedes Plutonium
that should read "sphere versus pseudosphere", not sphere versus
sphere
Archimedes Plutonium
> area equal the sphere versus pseudosphere (sic) when the upper boundary is 10^618.
> And the answer
> to that nagging perplex question is that since a pseudosphere
> stretches to infinity and since a
> sphere volume is always known as finite, that when this happens
> naturally in mathematics that the infinite expanse of pseudosphere
> volume and surface area are equal to that of finiteness
> of sphere, then that truly is a boundary set up within mathematics
> itself. That Finite meets
> Infinity at the B-number, and it is the number for which mathematics
> has Aristotelian Logic where the answers by numbers below the B-number
> are reliable, dependable and not something of Quantum logic duality.
> Above the B-number, mathematics is untrustworthy for it
> has duality issues.
>
So, using that fact that the exit circle is going to be about, roughly
4% of the
sphere circumference.
Now we ask, is 10^618 the likely B-number, where the volume and
surface area
of a sphere and pseudosphere of identical radius are equal and are
within that range
of numbers from 0 to 10^618?
In volume we have R^3 so that the volume must be from the radius of
10^206
and for surface area we must have R^2 so that the radius must be
10^309
Computing the volume and surface area of the sphere of those radii is
straightforward,
but the pseudosphere is what all the mess is about.
So the exiting-circle radius for 10^206 and 10^309 radii is 4% of the
circumference.
Basically, with these large numbers it is going to be 10^206 and
10^309 and then the
volume and surface area is all about that longer stretch from 10^206
and 10^309 to
that of 10^618. It is all about 10^618 where the 10^206 and 10^309 are
practically
insignificant.
Sure, those volumes and surface area are huge ones, but the problem
reduces to how
much do the poles of the pseudosphere hold in terms of volume and
surface area upon
exiting the spheres of radii 10^206 and 10^309 travelling to 10^618 ??
In the sphere of radii 4 as shown in Wikipedia we can instantly sense
that no equilibration is
going to occurr for radii of 1,2, 3, 4 because there is not enough of
the poles of the pseudosphere to be the "leading acting role in the
drama" . With small radii, the sphere enclosing pseudosphere has the
major acting role, but at these large numbers of 10^618,
the leading acting role is all about the poles of the pseudosphere.
Now I do have some good estimates of percentages that I should be able
to roughly estimate if 10^618 is the B-number of mathematics.
Good estimates:
(1) exiting-circle radius is 4% of sphere circumference
(2) pseudosphere enclosure by sphere volume is 25% of sphere enclosure
volume
(3) pseudosphere enclosure by sphere surface area is 50% of sphere
enclosure surface area
And I have the Needle Cone as a model. I may have to use a triangle in
3rd dimension, whether called a wedge or what, as a model. For
basically, at 10^618, it all becomes that stretch of the pseudosphere
poles that matters.
Archimedes Plutonium
Archimedes Plutonium wrote:
(snipped)
> Good estimates:
>
> (1) exiting-circle radius is 4% of sphere circumference
> (2) pseudosphere enclosure by sphere volume is 25% of sphere enclosure
> volume
> (3) pseudosphere enclosure by sphere surface area is 50% of sphere
> enclosure surface area
>
> And I have the Needle Cone as a model. I may have to use a triangle in
> 3rd dimension, whether called a wedge or what, as a model. For
> basically, at 10^618, it all becomes that stretch of the pseudosphere
> poles that matters.
>
Let me try only the volume. Using that estimate that at 10^206 radius
for
sphere and pseudosphere that 25% of the sphere volume will be the
pseudosphere
volume that resides inside the sphere. The sphere volume is 4x 10^618.
So the enclosed portion of the pseudosphere
is of volume 1 x 10^618. That means I have to retreive 3 x 10^618
volume from the two
poles of the pseudosphere or 1.5 x 10^618 volume from each
pseudosphere pole.
Old Math would have said I could only retrieve another 25% from the
poles and be done with it. But Old Math never had a fake proof as to
pseudosphere volume.
To proceed, I suspect only a concatenation of truncated cones can
proceed here and in that
concatenation the formula for a concatenated truncated cones retains
the height of 10^618
but it no longer retains the R^2. I do not know if anyone has
assembled a formula for volume of concatenated truncated cones. Have
to look that up.
Archimedes Plutonium
But Old Math only had a fake proof as to
pseudosphere volume, where they used cylinders to model.
Archimedes Plutonium
be blamed almost
totally on the picking of a bad model. I suspect from other posters
who talk about using
cylinders into the psuedosphere poles, that cylinders were used for
the Old Math proof that
concludes volume is only 1/2 sphere of identical radius. If the proper
model of a concatenation
of truncated cones had been used, it would have been realized that the
Series 1 + 1/n as
divergent was the end result. That the truncated cones concatenated
bounded below the volume of the pole goes to infinity so the pole
itself must have infinite volume.
Surely, I do not believe anyone in Old Math is going to say that if
you concatenate truncated cones out to infinity of Old Math that you
have a finite volume, nor a finite surface area. Is there anyone,
please raise your hand, in Old Math that believes a concatenation of
truncated cones infinitely long has finite volume?
And it is far easier in a proof, to have a correspondence of 1, then
1/2, then 1/3, then 1/4, then 1/5 with truncated cones concatenated
rather than with a concatenation of cylinders as was done in the the
Old Math fake proof. The cones fit more snugly, even they they are a
lower bound, but we can do the upper bound with concatenated truncated
cones.
Now I have to see if anyone has some formula for volume of truncated
cone concatenation.
Archimedes Plutonium
Just curious, and the title says it all.
Archimedes Plutonium
concatenation of truncated
cones. Which leads to the Series 1/n, the divergent series that goes
to infinity, and so the
volume of pseudospheres goes to infinity.
But my curiosity just lead me to question at what entry of the Series
1 + 1/2 + 1/3+ . . .
at what entry does that Series reach 10^618?
I am going to poke a wild blue guess. I am going to guess that at the
618! entry of the
1/n Series is it equal to 10^618. I say that because the factorial is
the fastest number
operator to gain large numbers, whereas the 1/n Series is an infinite
extending Series using small terms.
Has anyone explored this channel before?
Archimedes Plutonium
Archimedes Plutonium wrote:
> I am on the topic of pseudosphere poles and the modeling by a
> concatenation of truncated
> cones. Which leads to the Series 1/n, the divergent series that goes
> to infinity, and so the
> volume of pseudospheres goes to infinity.
>
> But my curiosity just lead me to question at what entry of the Series
> 1 + 1/2 + 1/3+ . . .
> at what entry does that Series reach 10^618?
>
> I am going to poke a wild blue guess. I am going to guess that at the
> 618! entry of the
Instead a wild blue yonder boo boo. I think PhD candidates should be
able
to use the formal term of "boo boo" in their dissertation and their
commencment
speech.
What I meant to say is that I suspect at the (10^618)! the enter
number of 10^618 being
factorialized is where the Series 1/n equals 10^618. I say that
because if 1+1/2+1/4+...
series converges to 2 and it takes infinity to get 2 and it takes till
what to get to 1.999..99
with 618 digits of 9s? In this case would it take 618! entries of 1/n
in order to reach
1.999..99 which has 618 digits of 9s?
Archimedes Plutonium
There is not much in the math literature about when a divergent Series
adds up to a specified
sum. Looks as though the interest in Series was placed on convergent
Series and the divergent Series were ignored.
There is a Series, I do not recall at the moment in which the
denominator is a factorial.
And which this denominator factorial sequence converges. Then again
there is a Series
in which we take the divergent Series and square the denominator and
it becomes
convergent.
I do not see any theorems in mathematics talking about how many
entries of a Series like
1/n where the string equals a preconceived specific number. Of course,
in the Series 1/n,
the second entry equals 1.5 and the third entry equals 1.8333... What
I want to know is what
entry does 1/n equal to 10^618 ? Is it the 10^1236 entry or 618!
entry?
Archimedes Plutonium
are infinite in volume.
We know that the function 1/x is a hyperbola in first quadrant. We
know that the pseudosphere
poles, formed from the revolution of the tractrix, although not equal
to 1/x is bounded by 1/x.
Since the Series 1/n is divergent and since the tractrix is bounded
below by 1/x. And since
1/n Series is infinite, we conclude that the pseudosphere is infinite
in volume. In other words the pseudosphere is hyperbolic and if
hyperbolic functions go to infinity, the
pseudosphere goes to infinity.
Now I did some searching to see if some broad formula was ever devised
that tells us the
approx summation of a Series 1/n at a entry distance. The question of
how many entries
for 1/n for the sum to equal 10^618. It looks like Abel did some work
in this topic, but unrelated
to what I seek.
It maybe the case, that the number 10^618 is unique to the
pseudosphere versus sphere where the volume and surface area are equal
given identical radius. But perhaps also, unique,
in that 10^618 maybe the only number in the Series 1/n where 10^618 is
10^1236 entries or
perhaps 618! entries long. A uniqueness of the number 10^618 with the
number of entries in Series 1/n.
Archimedes Plutonium
and is the modeling of
the pseudosphere pole. It is in the bathtub that small rubber drain
plug that is curved on its sides to fit waterproof. Now imagine a
concatenation of these drain plugs all along the stretch
of the drain, and as the drain gets more and more narrow, so do the
drain plugs. Since it takes
an infinite number of these drain plugs, each having some volume, the
pseudosphere pole
is infinite in volume.
Archimedes Plutonium
time I ever heard
of a digamma function:
--- quoting from (where some of the format did not take)
http://mathworld.wolfram.com/HarmonicSeries.html
The sum of the first few terms of the harmonic series is given
analytically by the th harmonic number
(6)
(7)
where is the Euler-Mascheroni constant and is the digamma
function.
The only values of for which is a regular number are , 2, and 6
(Havil 2003, pp. 24-25).
The number of terms needed for to exceed 1, 2, 3, ... are 1, 4, 11,
31, 83, 227, 616, 1674, 4550, 12367, 33617, 91380, 248397, ...
(Sloane's A004080; DeTemple and Wang 1991). Using the analytic form
shows that after terms, the sum is still less than 20. Furthermore,
to achieve a sum greater than 100, more than terms are needed!
Written explicitly, the number of terms is
--- end quoting ---
MathWorld mentions some conventional mixup as to the digamma function
where some use the factorial, others something different.
Anyway, the above shows that since 10^3 is 1.1 x 10^434, that my first
instinct that 10^618
is probably going to require 10^618factorial terms. Now that is a huge
large number.
And I wonder if it is a unique huge large number in that the entries
for the Series 1/n for
10^618 is the only number where 10^618 = 10^618factorial terms.
So then the question becomes, is the B-number to be taken as 10^618 or
is it to be taken
as 10^618factorial?
Archimedes Plutonium
Archimedes Plutonium wrote:
> Alright, someone pointed me into the proper direction, and the first
> time I ever heard
> of a digamma function:
>
> --- quoting from (where some of the format did not take)
> http://mathworld.wolfram.com/HarmonicSeries.html
>
You look at the asymptotes of the Digamma function or polygamma
function
and you wonder, immediately, is there ever a asmyptotic graph with
finite area
under the curve? Then you want a proof that all asymptotes have
infinite area
under the curve.
Finally, you realize the tractrix of the 2D of a pseudosphere has two
asymptotes
and thus the area is infinite and the volume would be infinite in 3D.
Nice, easy
breeze proof.
Then there is the more practical proof, the kind I like most of all.
We construct a
concatenation of truncated cones and fill the poles of the
pseudosphere. There will still
be alot of gaps and spaces where the truncated cones miss. It is a
compacting or
bathtub rubber plugs. Those stoppers in the bathtub that keep the
water in. Those are
truncated cones. So if we pile on ever smaller truncated cones that
goes to infinity,
we will have a lower bound of volume and surface area of pseudosphere.
And it is infinite
volume and surface area since the height is infinite. In fact we can
lower bound the concatenated truncated cones with one long continuous
cone whose height is infinite
and thus volume and surface area infinite.
But back to this idea of asymptotes. I cannot think of one such entity
that stretches
to infinity yet whose area under the curve is finite.
I cannot think of one single asymptotic function with finite area
under the curve.
So who is the author of the proof that the pseudosphere surface area
and volume
are finite? Was it Beltrami?
P.S. Apparently most in the math community has been ordered to not to
reply to any
AP posts, or that there is some blocking of reply posts. Who puts the
"blocking tag" on a
sci.math poster? So we have a reply-moderator in
sci.math that coerces those who reply to AP. They coerce repliers
through email and by
blocking. And this is what happens when mathematics has a huge blunder
in it-- pseudosphere has infinite volume and infinite
area. And where the math community, rather than admit their error and
clean up their mistake, instead, blacklists the author who found the
error and pretend as if no error ever existed.
I mean, well, that is really funny for the history of mathematics that
throughout the entire
20th century every mathematician believed the pseudosphere volume and
area were finite, which would be akin to thinking that throughout the
20th century, every mathematician thought
the primes were finite, akin, would you not agree.
When a big mistake occurs in math, everyone shuts up, because they
hate to think that all
their teaching has to corrected or modified. As my Wikipedia entry
says about me --" made much of modern day math obsolete"
Well, yes, if I make your math understanding obsolete, of course you
are going to shut up
and not reply to my posts. But what you really should be doing is
admitting and then fixing
the mistake.
Archimedes Plutonium
Archimedes Plutonium wrote:
(snipped)
> When a big mistake occurs in math, everyone shuts up, because they
> hate to think that all
> their teaching has to corrected or modified. As my Wikipedia entry
> says about me --" made much of modern day math obsolete"
>
> Well, yes, if I make your math understanding obsolete, of course you
> are going to shut up
> and not reply to my posts. But what you really should be doing is
> admitting and then fixing
> the mistake.
>
Well, I have two Wikipedia entries and the one I was attempting to
quote above is
this one--
--- quoting Wikipedia ---
Archimedes Plutonium (born Ludwig Poehlmann in 1950, raised as Ludwig
Hansen, legally changed his name to Ludwig van Ludvig, then Ludwig
Plutonium, then later changed it to Archimedes Plutonium) - noted for
his many posts about his own theories of physics, mathematics, and
stock market investing, and in particular his "Plutonium Atom
Totality" theory, which posits that the universe is a giant plutonium
atom and that galaxies are "dots" in the electron dot cloud of this
atom.[3][4] He states that the cosmic atom must be plutonium in order
to explain the values of the mathematical constants e and π, and
certain physical constants such as the fine structure constant and
proton-to-electron mass ratio.[citation needed] He also claimed to
have invented a new decimal number notation that leads to proofs/
disproofs of the prime number theorem, Poincare conjecture, Goldbach
conjecture, Fermat's Last Theorem, and the Riemann hypothesis, and
which will render current methods of mathematics obsolete.[citation
needed] As of 2009, Plutonium has authored over 20,000 unique postings
to dozens of science newsgroups such as "sci.physics", "sci.math",
"sci.chem", and "sci.bio.misc".[citation needed]
--- end quoting ---
When someone comes along and points out the error of Euclid's
Infinitude of Primes proof and how the correcting of that error leads
to a quick and easy Infinitude of Twin Primes proof, well,
that does not embroil the math community.
But when someone comes along and points out that the math community
had believed in a finite volume and area for a pseudosphere for over a
century of this mistake, and that the solving of this mistake is to
have a borderline between Finite and Infinity, is a colossal change
in mathematics, and really does make current math obsolete.
And when someone does that to mathematics, they clamm-up. When instead
they should be
vocal and seeking to fix the mistake.
Well I have a little time in this 1000+ page book for "people talk".
Over the phenomenon
of sci.math postings, where it is freedom of speech. But there is what
can be called,
unseen coercion that readers and future readers would not understand
had gone on, unless I spoke of it.
You can see it in LWalk's various posts to me where he apologizes to
those who are harrassing and coercing LWalk. They harrass and coerce
LWalk into not replying to any of
my posts. They do it mostly via email, with threats. And you see it in
LWalk's post where he
gracefully refers to an apology to those that want no reply made by
LWalk.
When I first arrived at sci.math in 1993, I also received threatening
emails from those
self appointed dictators who threatened me if I replied to a specific
poster. So alot of
coercion goes on behind the scenes of sci.math itself. And hopefully,
someday, some of these
people who have been coerced to not to reply will come forth and post
who their coercion bully
was, and what threats they made. Back in 1993, the threats I usually
received was a threat to
email bomb me. That is, fill my email box with junk if I continued to
reply to someone.
Archimedes Plutonium
Archimedes Plutonium wrote:
(snipped)
>
> When I first arrived at sci.math in 1993, I also received threatening
> emails from those
> self appointed dictators who threatened me if I replied to a specific
> poster. So alot of
> coercion goes on behind the scenes of sci.math itself. And hopefully,
> someday, some of these
> people who have been coerced to not to reply will come forth and post
> who their coercion bully
> was, and what threats they made. Back in 1993, the threats I usually
> received was a threat to
> email bomb me. That is, fill my email box with junk if I continued to
> reply to someone.
Let me comment especially on one vicious form of bullying and coercion
of sci.math
that is unseen by sci.math. I know this because I was witness to this
bullying
circa 1994 from someone in Texas. They would email me to not to reply
to a certain poster,
because that poster was, according to this bully, that poster was
undergoing psychriatric
medical treatment for which the bully went into in-depth details in
the email. So, partly a prank, partly a act of
bullying, and fully a liar. And how did I respond.
Well, I am the kind of person like the Missouri licence plate "Show
me".
I was undaunted and unfazed by this punk bully coercion attempt, and
continued to reply
to whomever I wanted to, whenever I wanted to.
But that sort of shenanigans goes on all the time and especially in
these sci newsgroups.
I take the habit, and most people should follow this rule, that an
email from someone you
do not know is 99% a prank. In fact, I rarely open up email from
unknown sources. In 1994 I was more inclined to open emails, now I
rarely do. Most emails are like the telemarketing
phone call we have no time for.
So I think the bully email to get people to stop replying to other
people in the sci newsgroups
is a problem that should be addressed. Perhaps the Sci.Math Faq should
make a paragraph mention that there are silly oaf bullies who have
nothing better to do than to email participants
in sci newsgroups insisting you not
reply to some posters. It is these bullies that need medical
attention.
I think the Sci.Math FAQ should mention that some posters are bullies
that email you and try
to suppress your freedom of speech.
Archimedes Plutonium
http://mathworld.wolfram.com/HarmonicSeries.html
We see how Oresme proved that the Series 1/n diverges by means of
grouping the terms
as in
1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) +. . .
So that each successive grouping is at least 0.5 or greater. That is
the whole
of the proof.
Now let us do the same for the pole of a pseudosphere that comes
poking out of the
enclosing sphere with its exit-circle. Let us mold a truncated cone
whose volume is
1 and will fit like a bathtub plug into the exit circle hole of the
pseudosphere pole and
the fit is only tight at the narrow end so that there is some gaps
along the side of the
plug for it is not a contiguous intersection of truncated cone with
the sides of the
pseudosphere pole. Now we mold another rubber truncated cone with
volume 1/2
and whose narrowest end is contiguous with the next stretch of the
pole. We thence
have begun a concatenation of truncated cones to fit inside the
pseudosphere pole.
Now we mold two more bathtub plugs whose volume is 1/3 and 1/4 and fit
them after
the 1/2 plug. And there is nothing to stop us from doing this to
infinity. So the volume
of the pseudosphere pole is infinite volume.
So one really has to wonder how in the world anyone came up with a
cylinder modelling
that yielded a finite volume using the convergent Series 1 + 1/2 + 1/4
+ . . . It sort of reminds me of Statistics work, where field
researchers go out and look for what they would like the
conclusions to be and ignore anything that is opposed to what their
end conclusion is wanted to be. Sure we can take a pseudosphere pole
and mold cylinders of volumes 1 then 1/2, then
1/4 and say that they are upper bounds of the entire pole, and then
falsely conclude the pseudosphere is finite. Just like in the Hotel
Paradox of false accounting of $27 + $2 + where is the missing $1 to
make $30, whereas the true accounting is $25 +$3 + $2 = $30. So by
using the convergent Series for the pseudosphere is a false
accounting. Just like if someone told you to take a thermometer and go
over there and measure the distance between A and B, and then take the
Convergent Series and and measure the volume of a pseudosphere.
So Oresme's proof that the harmonic Series 1/n is divergent is an
elegant proof. And so elegant that it easily is the same proof that
the pole of a pseudosphere is infinite in volume.
Archimedes Plutonium
Again, looking at this website of MathWorld:
http://mathworld.wolfram.com/HarmonicSeries.html
where it talks about the fact that at 10^3 requires 1.1 x 10^434 terms
of the harmonic
series 1/n.
That provides another proof that the pseudosphere is not as others
say-- 1/2 the volume
of the sphere of radius 10. Because here we are going to show a
contradiction, because the
sphere of radius 1 has volume of about 4 and half of that is 2. Now we
construct truncated cones as bathtub plugs and we are able to specify
the volume of each truncated cone to be the terms of the series 1/n,
so that after summing 10^434 terms we have
a volume of each pseudosphere pole that exits the enclosing sphere as
1000 volume and since there are two such poles means we have 2000
volume from just the exiting poles and
2000 is far larger of a volume than the 4 for the enclosing sphere of
identical radius to the pseudosphere.