Advanced Flat Earth Theory

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sandokhan

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Re: Advanced Flat Earth Theory
« Reply #360 on: March 24, 2017, 07:38:23 AM »
THE RETURN OF KING ARTHUR



In the hour of Britain's greatest need, King Arthur will return to rescue his people.

In the new radical chronology of history, King Arthur (Akhenaten), the founder of the British Empire, lived some 250 years ago.

After leaving Egypt, passing through modern day Palestine, Akhenaten set for a voyage finally reaching the British Isles. King Arthur is even described as having arrived in North America, the final destination being Avalon.

The Isle of Avallonis is reported as being far away, to reach this island would require a long journey by sea.

And King Arthur is not the only historical figure awaiting a mysterious return, in a time of great peril: Apollo (Horus) is set to return one day, from Hiperborea. There is also a third character who is destined to come back at the end of the fifth age.

In order to understand how such a thing would be possible, we need to study subquark biochirality and its relationship to the human aura.

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1488624#msg1488624 (biochirality and terrestrial gravity)

A very high supply of laevorotatory subquarks in the air, much higher than usual, would constitute a very effective barrier against the effects related to the dextrorotatory biochirality of the human aura, caused by terrestrial gravity.

http://creation.com/origin-of-life-the-chirality-problem

http://creation.com/god-left-handed

http://creationbc.org/index.php/right-handed-amino-acids-can-they-smack-down-the-evolutionists-chirality-problem/

https://web.archive.org/web/20140921043113/https://creationresearch.org/members-only/crsq/50/50_2/CRSQ%20Fall%202013%20lo%20res%20bookmarked%20for%20web.pdf

http://www.evolutionnews.org/2012/05/homochirality_i059531.html

http://www.creationismonline.com/YEC/The_Origin_Of_Life.pdf






« Last Edit: December 28, 2018, 09:51:42 AM by sandokhan »

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sandokhan

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Re: Advanced Flat Earth Theory
« Reply #361 on: March 24, 2017, 08:48:47 AM »
LARGE SEMIPRIME FACTORIZATION - RSA CYPHER I

Presently, nobody has able to provide significant methods for pushing integer factoring toward P . Then again, factoring is probably the hardest problem in analytical number theory.

In my opinion, the factorization of large semiprimes must be as easy as multiplying two integers: the algorithm should be proportional to the number of digits of the semiprime, and not an extremely difficult search using very sophisticated procedures.


For a 200 digit number (semiprime), the required computational time (1990) for the methods then used in integer factorization will take 4 x 1015 years.

For a 300 digit number, we would need 5 x 1021 years

For a 500 digit number, the figure would rise to 4.2 x 1032 years.


"RSA is a public key encryption algorithm which uses two different keys for encryption and decryption.

Procedure :

1.Select two very large prime numbers. (>200 digits)
2. Private key (decryption key) is calculated using p,q which are kept secret
3. Public key (encryption key) calculated as n=p*q , n is made public

Security of RSA relies on the fact that there is no Polynomial time algorithm for factorizing and integers into prime factors. Even if the adversary knows the value of n, it is computationally infeasible to factorize n to retrieve values of p and q. Therefore, the security of internet transactions is intact unless there is any polynomial time factorization algorithm."

We have seen that the sacred cubit fractal is the hidden template behind the distribution of the zeros of Riemann's zeta function.

Is there a relationship between the sacred cubit and semiprime factorization?


Background information

b1, a1 and c1 are the three sides of a right triangle

b12 + a12 = c12

b1 = d1 x d2 (divisors of b1)

a1 = (d12 - d22)/2

c1 = (d12 +d22/2


If b1 is prime, then b12 + a22 = c22 (where c2 = (b12 +1)/2 )


Modern geometry/trigonometry tells us that Pythagoras' theorem is the only known relationship relating the three sides of a right triangle, in a single equation.


But there is another equation, involving of course the sacred cubit, relating the three sides of right triangle:

b12sc + a12sc =~ [(b1 + a1 + c1)/2]2sc + ...


Since a1 + c1 = d12, with a reasonable estimate for a1, we can obtain a very good approximation for d1.


The Fibonacci numbers are actually sacred cubit numbers.

1,618034 = 4sc2 (1sc = 0.636009827, in this case)

Then Fn = 1/(8sc2 -1) x 22n x sc2n


Another formula:

b11/sc + a11/sc=~ c11/sc + ...


The first formula proves to be enough to completely solve the large integer factorization involving a semiprime having 10 or less digits. The right side of the equation is an asymptotic expansion, I was able to obtain the main term; of course, adding more terms of this expansion (a very difficult endeavor), would mean we can factorize semiprimes which have more than 10 digits, the accuracy depending on the number of terms in the expansion.

Of course, to attempt to solve the large semiprime factorization problem beyond the case where b1 has more than 10-20 digits, would mean we need a more precise algorithm, involving sacred cubits.

« Last Edit: March 24, 2017, 09:02:57 AM by sandokhan »

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sandokhan

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Re: Advanced Flat Earth Theory
« Reply #362 on: March 24, 2017, 10:57:12 AM »
LARGE SEMIPRIME FACTORIZATION - RSA CYPHER II

"Because both the system's privacy and the security of digital money depend on encryption, a breakthrough in mathematics or computer science that defeats the cryptographic system could be a disaster. The obvious mathematical breakthrough would be the development of an easy way to factor large prime numbers"
(Bill Gates 1995)


http://www.muppetlabs.com/~breadbox/txt/rsa.html

http://web.archive.org/web/20130904000227/http://csis.bits-pilani.ac.in/faculty/murali/netsec-09/seminar/refs/atharvasrep.pdf

http://empslocal.ex.ac.uk/people/staff/mrwatkin/zeta/tutorial.htm#q16

http://mathworld.wolfram.com/PrimeNumber.html



List of Fibonacci numbers (Fn) (sacred cubit sequences):

http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibtable.html


b1 has less than 10 digits

How to obtain a reasonable estimate for a1


b1 = 8141 x 131071 = 1073602561

1073602561 = 286572 + 252378911 = 28657 x 46368 - 255165215 

F23 = 28657

F24 = 46368

If b1<a1, then the a1 term will be of the form F242 - ..., F24 x F25 - ..., F25 x F26 - ..., or F252 - ...

if b1>a1, then a1 will equal F23 x F24 - ..., F222 - ... , that is, only 4-6 possible choices.


In order to get a very good estimate for d1, we will use the first remainder (and a few subsequent remainders if needed, more explanation below) obtained from the b1 for each of the above choices .

For the a1 =  F25 x F26 - ... choice, using a10 = 255165215, and substituting in the first formula, we get:

d1 = 132578.957, an excellent approximation.

Actually, a1 = 8556257280 = 750252 + 2927506655 = 75025 x 121393 - 551252545


b1 = 65537 x 131071 = 8590000127 = 750252 + 2961249502 = 75025 x 121393 - 517509698

F25 = 75025

Using the same reasoning and the same formula, we get a first estimate for d1, d1 = 130095.707


It is only by using the power of the sacred cubit that we can actually get these estimates, impossible to obtain otherwise by any other method, without resorting to sophisticated factoring methods.


The sacred cubit hidden pattern of the natural number system can be used to obtain as much information as possible out of the b1 semiprime.

b1 = 821 x 941 = 772561

772561 = 610 x 987 + 170491 = 9872 - 201608

170491 = 3772 + 28362 = 377 x 610 - 59479

F15 = 610
F16 = 987

We use each and every remainder obtained by dividing b1 by Fibonacci numbers, in a similar sequence: each subsequent remainder expressed as in the classic division formula (a = qd + r, where q and d are Fibonacci numbers, while r is the remainder to be used in the next division process)

201608 = 377 x 610 - 28362 = 3772 + 59479

F14 = 377

28362 = 144 x 233 - 5190 = 1442 + 7899

59479 = 2332 + 5190 = 233 x 377 - 28362

7899 = 892 - 22 = 89 x 55 + 3004

5190 = 89 x 55 + 295 = 892 - 2731


3004 = 552 - 21 = 55 x 34 + 1134

2731 = 552 - 294 = 55 x 34 + 861


1134 = 342 - 22 = 34 x 21 + 420

861 = 34 x 21 + 147 = 342 - 295

420 = 212 - 20 = 21 x 13 + 147

294 = 21 x 13 + 21 = 212 - 147 ; 147 + 21 = 168

147 = 21 x 8 - 21 = 13 x 8 + 43

43 = 8 x 5 + 3 = 82 - 21

21 = 5 x 3 + 6 = 52 - 4


Interestingly, we can immediately obtain a first approximation for d1, d1 = 918; by summing the remainders of b1 in their corresponding order, for a 3 digit d1 divisor. Several such sums can be obtained (where d1 can be assumed to have 3, 4, 5 digits) and one of them will actually represent a nice estimate of d1.


The crucial observation is that we can actually get the remainders of the a1 term either by noticing that 6 and 4 (remainders obtained by dividing 21 by F5 and F4) can be used to initiate the a1 sequence of remainders starting from the bottom up, or by using a very interesting shortcut involving b1sc, where this can be applied.

Actually, a1 = 105720 = 3772 - 36409 = 377 x 233 + 17879

Using the same scheme as above for the a1 term (same division by Fibonacci numbers algorithm as was utilized for the b1 term) we finally get:

40 = 82 - 24 = 8 x 3 + 16

16 = 52 - 9 = 32 + 7

9 = 3 x 5 - 6 = 2 x 3 - 3


65 = 82 + 1 = 8 x 13 - 39

39 = 8 x 5 - 1 = 52 + 14

14 = 3 x 5 -1 = 2 x 5 + 4


Knowing that 6 and 4 are the remainders of a1, we can see that from the possible choices we eventually get (11, 19, 9, and 14) only 9 and 14 will make any sense, given the fact that the remainders at each stage of the calculation have to be expressed as in the classic division formula (a = qd + r, where q and d are Fibonacci numbers, while r is the remainder).


One of the remainders of a1 will be 2857.


3004 - 2857 = 147



772561sc = 5530


5530 - 5063 = 2 x 233

5530 - 2857 = 89 x 30

(5063, another a1 remainder)

That is, there is a certain symmetry and relationship between b1sc and some of the a1 remainders.



Another example.

b1 = 1000009

For 1000009 = 3413 x 293, we get a first estimate of 3486, and by summing the remainders of b1 (576230 + 204130 + 62001 + 25840 + 5104 + 2817 + 947 ...) we get an estimate of 3400, which is amazing, because we only use the remainders from b1 and very simple approximations.


For 1000009, b1sc = 6515.72

9368 - 6515.72 = 610 x 4.66 = 987 x 2.88  (4.66 = 2 x 2.33 , and 2.88 = 2 x 1.44, both 233 and 144 are Fibonacci numbers)

9368 is one of the a1 remainders

Another a1 remainder is 3448

6515.72 - 3448 =~ 552 = 233 x 13



Thus, the factorization of semiprimes is related to the sacred cubit, and I believe the above algorithm is a start in studying further this new approach to solving this problem, based on the power of the sacred cubit.

For moderately large b1 such as:

231 - 1 = 2147483647

261 - 1 = 2.3059 x 1018

b1 = (231 - 1) x (261 - 1) = 4.951760152 x 1027

a computer which can handle the full/entire number of digits could be used to verify the algorithm proposed above, and to see if the same relationship exists between the sequence of remainders obtained for both b1 and a1.

In fact, with an a1 trial function 4 x 1035, we get an estimate for d1 = 2.353 x 1018.

Of course for the exact answer, we would need to verify the correctness of the above algorithm, involving the sequence of remainders obtained upon division by the corresponding Fibonacci numbers.




« Last Edit: March 25, 2017, 12:13:47 AM by sandokhan »

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sandokhan

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Re: Advanced Flat Earth Theory
« Reply #363 on: March 24, 2017, 11:49:30 AM »
ORIGIN OF URANIUM PARADOX

Uranium exists as high-grade ore at 200,000 ppm (as common as tin or zinc).

http://www.world-nuclear.org/information-library/nuclear-fuel-cycle/uranium-resources/supply-of-uranium.aspx

"Along with most people, I hadn't understood until today how prevalent Uranium is in the Earth's crust. Discovering that made me suspicious once more, because it doesn't make any sense, given the current theory of element production.

Elements are said to be produced by fusion in stars. But most stars don't fuse past element number 2, Helium. None fuse past iron. Since Uranium is element number 92, it could only be produced by the very largest stars in collapse. It would spread out across the galaxy when they went supernova. But given how diffuse the galaxy is, you wouldn't expect planets to contain so much Uranium. I will be told that the galaxy is very old, so we have Uranium left over from eons of big stars going nova. Yes, but the half-life of Uranium is about 4.5 billion years, which is about half the lifespan of a star like the Sun.

So while the Sun is alive, ¾ of the existing Uranium will break down. So you see, Uranium doesn't persist to be recycled through several star-cycles. We can't get that sort of buildup over time. Plus, since the Earth is said to be exactly that old, it would have originally had twice as much Uranium as now, doubling our problem."


"We will now ASSUME that the clouds formed themselves into what evolutionists call proto-stars, or first-generation stars.

STARS EXPLODE AND SUPERNOVAS

PRODUCE HEAVY ELEMENTS

The problem—The Big Bang only produced hydrogen and helium. Somehow, the 90 heavier (post-helium) elements had to be made. The theorists had to figure out a way to account for their existence.

The theory—The first stars, which were formed, were so-called 'first-generation stars' (also called 'population III stars'). They contained only lighter elements (hydrogen and helium). Then all of these stars repeatedly exploded. Billions upon billions of stars kept exploding, for billions of years. Gradually, these explosions are said to have produced all our heavier elements.

This concept is as wild as those preceding it.

1 - Another imaginative necessity. Like all the other aspects of this theory, this one is included in order to somehow get the heavier (post-helium) elements into the universe. The evolutionists admit that the Big Bang would only have produced hydrogen and helium.

2 - The nuclear gaps at mass 5 and 8 make it impossible for hydrogen or helium to change itself into any of the heavier elements. This is an extremely important point, and is called the 'helium mass 4 gap' (that is, there is a gap immediately after helium 4). Therefore exploding stars could not produce the heavier elements. (Some scientists speculate that a little might be produced, but even that would not be enough to supply all the heavier elements now in our universe.) Among nuclides that can actually be formed, gaps exists at mass 5 and 8. Neither hydrogen nor helium can jump the gap at mass 5. This first gap is caused by the fact that neither a proton nor a neutron can be attached to a helium nucleus of mass 4. Because of this gap, the only element that hydrogen can normally change into is helium. Even if it spanned this gap, it would be stopped again at mass 8. Hydrogen bomb explosions produce deuterum (hydrogen 2), which, in turn, forms helium 4. In theory, the hydrogen bomb chain reaction of nuclear changes could continue changing into ever heavier elements until it reached uranium;—but the process is stopped at the gap at mass 5. If it were not for that gap, our sun would be radiating uranium toward us!

'In the sequence of atomic weight numbers 5 and 8 are vacant. That is, there is no stable atom of mass 5 or mass 8 . . The question then is: How can the build-up of elements by neutron capture get by these gaps? The process could not go beyond helium 4 and even if it spanned this gap it would be stopped again at mass 8. This basic objection to Gamow’s theory is a great disappointment in view of the promise and philosophical attractiveness of the idea.'—*William A. Fowler, California Institute of Technology, quoted in Creation Science, p. 90.

Clarification: If you will look at any standard table of the elements, you will find that the atomic weight of hydrogen is 1.008. (Deuterum is a form of hydrogen with a weight of 2.016.) Next comes helium (4.003), followed by lithium (6.939), beryllium (9.012), boron (10.811), etc. Gaps in atomic weight exist at mass 5 and 8.

But cannot hydrogen explosions cross those gaps? No. Nuclear fision (a nuclear bomb or reactor) splits (unevenly halves) uranium into barium and technetium. Nuclear fusion (a hydrogen bomb) combines (doubles) hydrogen into deuterum (helium 2), which then doubles into helium 4—and stops there. So a hydrogen explosion (even in a star) does not go across the mass 5 gap.

We will now ASSUME that hydrogen and helium explosions could go across the gaps at mass 5 and 8:

3 - There has not been enough theoretical time to produce all the needed heavier elements that now exist. We know from spectrographs that heavier elements are found all over the universe. The first stars are said to have formed about 250 million years after the initial Big Bang explosion. (No one ever dates the Big Bang over 20 billion years ago, and the date has recently been lowered to 15 billions years ago.) At some lengthy time after the gas coalesced into 'first-generation' stars, most of them are theorized to have exploded and then, 250 million years later, reformed into 'second-generation' stars. These are said to have exploded into 'third-generation' stars. Our sun is supposed to be a second- or third-generation star.

4 - There are no population III stars (also called first-generation stars) in the sky. According to the theory, there should be 'population III' stars, containing only hydrogen and helium, many of which exploded and made 'population II' (second-generation stars), but there are only population I and II stars (*Isaac Asimov, Asimov’s New Guide to Science, 1984, pp. 35-36).

5 - Random explosions do not produce intricate orbits. The theory requires that countless billions of stars exploded. How could haphazard explosions result in the marvelously intricate circlings that we find in the orbits of suns, stars, binary stars, galaxies, and star clusters? Within each galactic system, hundreds of billions of stars are involved in these interrelated orbits. Were these careful balancings not maintained, the planets would fall into the stars, and the stars would fall into their galactic centers—or they would fly apart! Over half of all the stars in the sky are in binary systems, with two or more stars circling one another. How could such astonishing patterns be the result of explosions? Because there are no 'first generation' ('Population I') stars, the Big Bang theory requires that every star exploded at least one or two times. But random explosions never produce orbits.

6 - There are not enough supernova explosions to produce the needed heavier elements. There are 81 stable elements and 90 natural elements. Each one has unusual properties and intricate orbits. When a star explodes, it is called a nova. When a large star explodes, it becomes extremely bright for a few weeks or months and is called a supernova. It is said that only the explosions of supernovas could produce much of the needed heavier elements, yet there have been relatively few such explosions.

7 - Throughout all recorded history, there have been relatively few supernova explosions. If the explosions occurred in the past, they should be occurring now. Research astronomers tell us that one or two supernova explosions are seen every century, and only 16 have exploded in our galaxy in the past 2,000 years. Past civilizations carefully recorded each one. The Chinese observed one, in A.D. 185, and another in A.D. 1006. The one in 1054 produced the Crab nebula, and was visible in broad daylight for weeks. It was recorded both in Europe and the Far East. Johannes Kepler wrote a book about the next one, in 1604. The next bright one was 1918 in Aquila, and the latest in the Veil Nebula in the Large Magellanic Cloud on February 24, 1987.

'Supernovae are quite different . . and astronomers are eager to study their spectra in detail. The main difficulty is their rarity. About 1 per 650 years is the average for any one galaxy . . The 1885 supernova of Andromeda was the closest to us in the last 350 years.'—*Isaac Asimov, New Guide to Science (1984), p. 48.

8 - Why did the stellar explosions mysteriously stop? The theory required that all the stars exploded, often. The observable facts are that, throughout recorded history, stars only rarely explode. In order to explain this, evolutionists postulate that 5 billion years ago, the explosions suddenly stopped. Very convenient. When the theory was formulated in the 1940s, through telescopes astronomers could see stars whose light left them 5 billion light-years ago. But today, we can see stars that are 15 billion light-years away. Why are we not seeing massive numbers of stellar explosions far out in space? The stars are doing just fine; it is the theory which is wrong.

9 - The most distant stars, which are said to date nearly to the time of the Big Bang explosion, are not exploding,—and yet they contain heavier elements. We can now see out in space to nearly the beginning of the Big Bang time. Because of the Hubble telescope, we can now see almost as far out in space as the beginning of the evolutionists’ theoretical time. But, as with nearby stars, the farthest ones have heavier elements (are 'second-generation'), and they are not exploding any more frequently than are the nearby ones.

10 - Supernovas do not throw off enough matter to make additional stars. There are not many stellar explosions and most of them are small-star (nova) explosions. Yet novas cast off very little matter. A small-star explosion only loses a hundred-thousandth of its matter; a supernova explosion loses about 10 percent; yet even that amount is not sufficient to produce all the heavier elements found in the planets, interstellar gas, and stars. So supernovas—Gamow’s fuel source for nearly all the elements in the universe—occur far too infrequently and produce far too small an amount of heavy elements—to produce the vast amount that exists in the universe.

11 - Only hydrogen and helium have been found in the outflowing gas from supernova explosions. The theory requires lots of supernova explosions in order to produce heavy elements. But there are not enough supernovas,—and research indicates that they do not produce heavy elements! All that was needed was to turn a spectroscope toward an exploded supernova and analyze the elements in the outflowing gas from the former star. *K. Davidson did that in 1982, and found that the Crab nebula (resulting from an A.D. 1054 supernova) only has hydrogen and helium. This means that, regardless of the temperature of the explosion, the helium mass 4 gap was never bridged. (It had been theorized that a supernova would generate temperatures high enough to bridge the gap. But the gap at mass 4 and 8 prevented it from occurring.)

12 - An explosion of a star would not produce another star. It has been theorized that supernova explosions would cause nearby gas to compress and form itself into new stars. But if a star exploded, it would only shoot outward and any gas encountered would be pushed along with it."

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sandokhan

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Re: Advanced Flat Earth Theory
« Reply #364 on: March 24, 2017, 11:52:54 AM »
ORIGIN OF URANIUM PARADOX II

Helium Flash Paradox

"The fusion of hydrogen to helium by either the PP chain or the CNO cycle requires temperatures of the order of 10,000,000 K or higher, since only at those temperatures will there be enough hydrogen ions in the plasma with high enough velocities to tunnel through the Coulomb barrier at sufficient rates.


The Mass-5 and Mass-8 Bottlenecks

The helium that is produced as the "ash" in this thermonuclear "burning" cannot undergo fusion reactions at these temperatures or even substantially above because of a basic fact of nuclear physics in our Universe: there are no stable isotopes (of any element) having atomic masses 5 or 8. This means that the two most likely initial steps for the fusion of helium-4 (the next most abundant isotope in stars after hydrogen-1) involve combining the He-4 with H-1 to form a mass-5 isotope, or combining two He-4 nuclei to form a mass-8 isotope. But both are unstable, and so immediately fly apart before they can undergo any further reactions. This produces a bottleneck to further fusion at mass 5 and at mass 8.


High Temperatures and Helium Fusion

Only at extremely high temperatures, of order 100 million K, can this bottleneck be circumvented by a highly improbable reaction. At those temperatures, the fusion of two He-4 nuclei forms highly unstable Beryllium-8 at a fast enough rate that there is always a very small equilibrium concentration of Be-8 at any one instant.

The situation is somewhat like running water through a sieve. Normally the sieve holds no water because it drains out as fast as it is added. However, if the flow of water into the sieve is made fast enough, a small equilibrium amount of water will be in the sieve at any instant because even the sieve cannot empty the water fast enough to keep up with the incoming water.

This small concentration of Be-8 can begin to undergo reactions with other He-4 nuclei to produce an excited state of the mass-12 isotope of Carbon. This excited state is unstable, but a few of these excited Carbon nuclei emit a gamma-ray quickly enough to become stable before they disintegrate. This extremely improbable sequence is called the triple-alpha process because the net effect is to combine 3 alpha particles (that is, 3 He-4 nuclei) to form a C-12 nucleus."


And, of course, this scenario is based on the following assumption: gravity compresses the core where, at high temperature and pressure, nuclear fusion occurs.

But there is no such thing as attractive gravity: the Biefeld-Brown effect, the Lamoreaux effect, the experiments carried out by Dr. N. Kozyrev, Dr. Bruce DePalma, defy the notion of attractive gravity, as do the quotes attributed to Newton himself, in letters to Bentley, Halley and Oldenburg.

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sandokhan

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Re: Advanced Flat Earth Theory
« Reply #365 on: March 25, 2017, 12:11:21 AM »
QUANTUM RIEMANN'S ZETA FUNCTION XIII

Some of the best mathematicians in the world are beginning to realize that all of the zeros of Riemann's zeta function are actually interconnected/related to each other.

http://www2.le.ac.uk/departments/mathematics/research/research-reports-2/reports-2012/MA12_03Matiyasevich.pdf

Y. Matiyasevich provided the negative solution to Hilbert's tenth problem:

http://www.math.le.ac.uk/people/ag153/homepage/Matiyasevich.htm

http://www.cis.upenn.edu/~jean/old511/html/Martin-Davis-Hilberts-10th.pdf

Now, the algorithm in the paper makes use of very large numbers, but it proves that each zero of Riemann's zeta function depends/is related to the previous set of zeros.


The Riemann-Siegel formula describes the end product, that is, the values of the zeros of the zeta function, but does not explain why they are located exactly at those precise points on the 1/2 line:

http://www.dtc.umn.edu/~odlyzko/zeta_tables/zeros1

     14.134725142
     21.022039639
     25.010857580
     30.424876126
     32.935061588
     37.586178159
     40.918719012
     43.327073281
     48.005150881
     49.773832478
     52.970321478
     56.446247697
     59.347044003
     60.831778525
     65.112544048
     67.079810529
     69.546401711
     72.067157674
     75.704690699
     77.144840069
     79.337375020
     82.910380854
     84.735492981
     87.425274613
     88.809111208
     92.491899271
     94.651344041
     95.870634228
     98.831194218
...

    211.690862595
    213.347919360
    214.547044783
    216.169538508
    219.067596349
    220.714918839
    221.430705555
    224.007000255
    224.983324670
    227.421444280
    229.337413306
    231.250188700
    231.987235253
    233.693404179
    236.524229666
    237.769820481
    239.555477573


The previous twelve messages on this subject prove that the five elements law applied to the sacred cubit distance of 0.636621... will create a sacred cubit fractal, whose values will coincide with values of the zeros of the zeta function.

The zeros of Riemann's zeta function are generated by the subdivision of the sacred cubit distance according to the five elements law.

In turn, these zeros describe completely the distribution of the prime numbers.


The relationship between power tower series and the zeta function:

http://file.scirp.org/pdf/APM_2016042615075938.pdf


While the fact that if Riemann's hypothesis is true could lead in finding a polynomial time algorithm for integer factorization, such a computational breakthrough still has to be discovered on its own. The fact that semiprime factorization is directly related to the sacred cubit, in my opinion, would provide the basis to find an elegant solution (see the two messages posted on this page, on this subject).

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sandokhan

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Re: Advanced Flat Earth Theory
« Reply #366 on: March 25, 2017, 03:01:57 AM »
SAGNAC EFFECT VIII


The Sagnac effect is far larger than the effect forecast by relativity theory.

STR has no possible function in explaining the Sagnac effect.

The Sagnac effect is a non-relativistic effect.

COMPARISON OF THE SAGNAC EFFECT WITH SPECIAL RELATIVITY, starts on page 7, calculations/formulas on page 8