second method proof pseudosphere area with infinity at 9.99..x10^603 Chapt13.40085 Maxwell Equations placing demands on mathematics #658 New Physics #778 ATOM TOTALITY 5th ed
Jun 21, 2012 3:15 PM
On Jun 21, 1:23 am, Archimedes Plutonium <plutonium.archime...@gmail.com> wrote: > I spent the day thinking about the problem and perhaps I was too hasty > in emitting overboard > the Calculus on tractrix or pseudosphere. > > Here is pi to 603 digits rightward: > 3.1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 > 5923078164 0628620899 8628034825 3421170679 8214808651 3282306647 > 0938446095 5058223172 5359408128 4811174502 8410270193 8521105559 > 6446229489 5493038196 4428810975 6659334461 2847564823 3786783165 > 2712019091 4564856692 3460348610 4543266482 1339360726 0249141273 > 7245870066 0631558817 4881520920 9628292540 9171536436 7892590360 > 0113305305 4882046652 1384146951 9415116094 3305727036 5759591953 > 0921861173 8193261179 3105118548 0744623799 6274956735 1885752724 > 8912279381 8301194912 9833673362 4406566430 8602139494 6395224737 > 1907021798 6094370277 0539217176 2931767523 8467481846 7669405132 000 > > Now let me state the question as clearly as I can. We think infinity > border is > 9.99..x 10^603 but have to prove it. The way to prove it is that > pseudosphere > has the same exact area as the respective sphere at infinity. We can > use > the tractrix in 2 dimension which is easier, and in fact the first > quadrant is all > we need. Now to prove that 9.99..x10^603 is the borderline we have to > show that > the tractrix is always smaller in area than the circle. We use unit > circle and unit > tractrix. Of course we only pay attention to the first quadrant. > > So the unit circle at infinity if our number is the border of > infinity, the unit circle area > is that number for pi given above. (Of course 1/4 for 1st quadrant.) > > Now the tractrix is always a laggard and so we know for sure that the > circle at > 10^600 is the same area as the circle at 10^601 through 10^603 since > there > are zeroes and so the circle cannot build anymore area during that > ride from > 10^601 through 10^603. However the tractrix can add more area during > the ride > from 10^601 through 10^603. > > So let us say that the ending last six digits of the tractrix at > 10^600 is this ending: > --404011 whereas the circle at 10^600 is exactly --405132. So at > 10^600 the tractrix > laggs behind the circle by the amount of area of 405132 subtract > 404011 which is 1121 amount > of area. > > So the question is, since the tractrix has to wander down three > exponents of 10^601, 10^602, > and 10^603, can it pick up enough area of that asymptotic curve to add > up to 1121 amount > of area? I believe so, and that it does in fact pick up the area so as > to equal the circle area > for the first time. And I believe the exact number for the x distance > of the tractrix that picks up > that area is the number 9.999..x10^603. > > Now I believe calculus can do this task because we can segment the > calculus. We can segment > that portion of the tractrix from 10^601 through 10^603. And we can > segment other portions of the > tractrix and in the end piece together all the segments for the final > area. So I believe someone with > a supercomputer and good program can actually do the above task. Find > what specific number that the > tractrix area exactly equals the circle area. Is it the number > 9.999..x10^603 the maximum perfect cube > divisible by 120 in the interval of 10^603 to 10^604. > > Now there maybe an analysis trick also that is supporting evidence > that the number where the areas are > equal is 9.999..x10^603 do to the fact that both pi and this border > number are both divisible by 120 and one is a perfect cube. The > perfect cube is important because when we need exactly 1/2 the volume > for the pseudosphere versus the sphere of 4/3 pi r^3 so the unit > sphere volume involves a 4/3 factor and is handled > by the 120 divisibility. >
Let me describe the second method proof that I started a few days ago. Just in case Calculus cannot prove that the pseudosphere exactly matches the area of sphere with pi truncated at 10^-603 and where infinity starts exactly at 9.999..x10^603.
What we do in the second method is create what is going to be the smallest bit for 2nd dimension and 3rd dimension. The number 1/9.99..x10^603 is the smallest nonzero positive number. It is close to 10^-604 but larger by a small amount.
So that the smallest bit in this second method is for area that of 1.1..x10^-604 x(10^-604) or area approx 1.1x10^-1208.
Now we look at the unit circle with pi ending in digits --32000 and for radius 1 the unit circle area would be that pi. It is evenly divisible by 4 since we want only the first quadrant. And we divide it by 4 and is approx 0.785..
Now we ask, how many bits of area of 1.1x10^-1208 are contained in the first quadrant unit circle. and so divide that into 0.785..
Now we find the complement set, the set that is of area of the unit circle but not of the unit tractrix.
So that of that 0.785.. unit circle area is the complement set of circle with tractrix. So how many of these tiny rectangles of 1.1x10^-1208 area are contained in the complement set?
Now we do construction. For we take the number of those complement rectangles and build out the rest of the arm of the tractrix.
If my hunch is correct then those complement rectangles will reach a maximum distance of 9.99..x10^603 distance and run out of these tiny rectangles.
That is the essence of the second method, provided if the Calculus, modern day Calculus is unable to do the proof.
The second method will depend on the "thickness of the arm" as it reaches each exponent and leaves that exponent to construct the tractrix arm.
I am optimistic of this method because a trick of analysis would apply that can prove the entire conjecture that 9.99..x10^603 is where the distance makes the areas exactly equal.
A similar type of argument would also prove the volume of unit pseudosphere equals exactly 1/2 of the unit sphere volume at that distance.