DYNAMISM: THE FOURTH SEAL by GEORGE MARTIN WILLIAMS - HTML preview

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BOOK TWO:

RIGHTEOUSNESS!

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John 16:9-11

King James Version

8 And when he is come, he will reprove the world of sin, and of righteousness, and of judgment: 9 Of sin, because they believe not on me;

10 Of righteousness, because I go to my Father, and ye see me no more; 11 Of judgment, because the prince of this world is judged.

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'OF RIGHTEOUSNESS, BECAUSE I

GO TO MY FATHER, AND YE SEE

ME NO MORE.'

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Revelation 5:1-7

King James Version

5 And I saw in the right hand of him that sat on the throne a book written within and on the backside, sealed with seven seals.

2 And I saw a strong angel proclaiming with a loud voice, Who is worthy to open the book, and to loose the seals thereof?

3 And no man in heaven, nor in earth, neither under the earth, was able to open the book, neither to look thereon.

4 And I wept much, because no man was found worthy to open and to read the book, neither to look thereon.

5 And one of the elders saith unto me, Weep not: behold, the Lion of the tribe of Judah, the Root of David, hath prevailed to open the book, and to loose the seven seals thereof.

6 And I beheld, and, lo, in the midst of the throne and of the four beasts, and in the midst of the elders, stood a Lamb as it had been slain, having seven horns and seven eyes, which are the seven Spirits of God sent forth into all the earth.

7 And he came and took the book out of the right hand of him that sat upon the throne.

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THE LAMB WILL REVEAL THE TRUTH THAT COMES FROM GOD!

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The Millennium Problems: The Seven Greatest

Unsolved Mathematical Puzzles of Our Time

[Reviewed by David Roberts, on 02/7/2003]

The Millennium Problems: The Seven Greatest Unsolved Mathematical Puzzles of Our Time [Reviewed by David Roberts, on 02/7/2003] In May 2000, the Clay Mathematics Institute elevated seven long-standing open problems in mathematics to the status of "Millennium Prize Problems," endowing each with a million-dollar prize. The seven particular problems were chosen in part because of their difficulty, but even more so because of their central importance to modern mathematics. The problems and the corresponding general areas of mathematics are as follows.

1

The Riemann Hypothesis

Number Theory

2

Yang-Mills Existence and Mass Gap

Mathematical Physics

3

The P versus NP problem

Computer Science

4

Navier-Stokes Existence and Smoothness

Mathematical Physics

5

The Poincaré Conjecture

Topology

6

The Birch and Swinnerton-Dyer Conjecture

Number Theory

7

The Hodge Conjecture

Algebraic Geometry

In The Millennium Problems, Keith Devlin aims to communicate the essence of these seven problems to a broad readership. It is, of course, a very ambitious goal. The preface makes it clear what Devlin's ground rules are. First he assumes only "a good high school knowledge of mathematics." Second, he is writing "not for those who want to tackle one of the problems, but for readers — mathematician and non-mathematician alike — who are curious about the current state at the frontiers of humankind's oldest body of scientific knowledge." He is clear that the readership drives the level of the book, so that precise statements of the problems will not always be given. Rather the goal is "to provide the background to each problem, to describe how it arose, explain what makes it particularly difficult, and give... some sense of why mathematicians regard it as important."

After the short preface, the book has an interesting Chapter 0, and then one chapter for each problem in the above order.

These seven chapters are constructed similarly. Most have a long historical component, generally including biographical information about the person or persons after whom the conjecture is named. Each has substantial background mathematical information, with topics ranging from complex numbers in Chapter 1 and group theory in Chapter 2 to congruences in Chapter 6 and algebraic varieties in Chapter 7. Applications are emphasized when possible. A nice theme in Chapters 2 and 4 is that mathematicians are behind physicists and engineers and just trying to catch up. Each chapter concludes with a discussion of the millennium problem itself.

Chapter 5 illustrates how Devlin ties the various units of a chapter into a coherent narrative. It begins with four pages about the life and work of Henri Poincaré. It moves on to introduce "rubber sheet geometry" in terms of how subway maps and refrigerator wiring diagrams are not geometrically faithful to the physical objects they represent, but nonetheless clearly capture all relevant information. This unit is important as it will make readers feel that topology is natural, rather than weird. Chapter 5 next introduces the concepts of vertices, edges, faces and finally Euler characteristic in terms of the Königsberg bridge problem. It introduces non-orientable surfaces and makes the introduction of an ambient four-space seem natural, since it is necessary for an embedding of the Klein bottle. It topologically classifies closed surfaces first crudely in terms of their orientability, and then completely in terms of networks drawn upon them and the Euler characteristic of these networks. It gives a very attractive example of two seemingly linked rings that in fact can be pulled apart. This example shows the reader that not everything is geometrically obvious, and thus underscores the utility of algebraic invariants that can rigorously confirm that two objects are topologically different. It discusses how the ordinary two-sphere is characterized among all closed surfaces by having the property that any loop on it can be shrunk continuously to a point. Finally, by way of this two-dimensional analogy, it discusses the actual three-dimensional Poincaré conjecture.

The strain imposed by the challenge of communicating all seven millennium problems to a broad readership naturally shows at times. In the Navier-Stokes chapter, for example, the background mathematical information presented is calculus and specifically differentiation. Readers are instructed that "dy/dx" is to be read "dee-wye by dee-ex." Some seven pages later, the Navier-Stokes equations themselves are presented. They are four coupled non-linear partial differential equations in four independent variables. The exposition is gentle, but readers new to calculus will only understand at a superficial level. The strain is felt somewhat more in Chapter 6 and particularly so in Chapter 7. But these various strains are unavoidable, and I think in general Devlin has done a very good job giving general readers a feel for the seven millennium problems.

https://maa.org/press/maa-reviews/the-millennium-problems-the-seven-greatest-unsolved-mathematical-puzzles-of-our-time

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THE FOURTH SEAL: GRAPHING FOUR DIMENSIONAL

SPACE-TIME IMAGES

THIS CHAPTER WILL END WITH MY SOLUTION TO THE HODGE CONJECTURE. THE HODGE CONJECTURE IS

SIMILAR TO THE NAVIER-STOKES EQUATION IN THAT THEY ARE BOTH CALCULUS PROBLEMS.

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https://mathworld.wolfram.com/HodgeConjecture.html

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WHY IS IT DIFFICULT?

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https://www.redalyc.org/journal/5117/511766757039/html/

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'The Hodge conjecture[1] is difficult because complex submanifolds are very rigid objects (since they are defined by polynomials). In fact, it is very difficult to construct complex submanifolds, and there are very few of them. Proving that submanifolds exist without constructing them (a very helpful sort of indirect reasoning) has also been difficult.'

MY SOLUTION TO THE HODGE CONJECTURE WILL SIMPLY BE TO CONSTRUCT 'THE THING' IN THE ABOVE

PARAGRAPH, THAT THEY SAY IS DIFFICULT TO CONSTRUCT.

WHY IS A THEORY OF EVERYTHING IS IMPORTANT?

AT THE TIME OF THE WRITING OF THIS BOOK, WHICH IS THE YEAR 2025, MOST PEOPLE DON'T CARE ABOUT

'UNIFYING THE FIELDS', OR ABOUT 'GRAND UNIFIED THEORIES' BECAUSE THERE ARE MUCH MORE EXCITING

THINGS TO THINK ABOUT, SUCH AS ARTIFICIAL INTELLIGENCE OR QUANTUM COMPUTING. BUT WHAT IS THE

POINT OF UNDERSTANDING ANYTHING, IF NOT TO FIND A WAY TO USE IT IN SOME WAY THAT GIVES A BENEFIT.

THE IDEA THAT 'KNOWLEDGE IS POWER' IS A COMPLETE WASTE OF TIME IF MOST OF WHAT IS KNOWN IS

USELESS. THE ONLY 'KNOWLEDGE' THAT POTENTIALLY HAS 'POWER' ARE PRINCIPLES OF NATURE THAT CAN BE

APPLIED SUCCESSFULLY. IN HUMANITY'S FUTURE, A THOROUGH UNDERSTANDING OF THE FORCES OF NATURE

IS CRITICAL TO ALL THE FUTURE TECHNOLOGY THAT HUMANITY WILL POSSESS. TECHNOLOGY THAT WILL

ALLOW HUMAN BEINGS TO BECOME A SPACE FARING RACE THAT WILL COLONIZE NOT JUST OUR OWN SOLAR

SYSTEM, BUT THE WIDER GALAXY AND EVEN OTHER GALAXIES AS WELL.

IF YOU KNOW A THEORETICAL PHYSICIST; THEN YOU CAN ASK HIM THIS QUESTION ON MY BEHALF:

'IF RELATIVITY THEORY AND QUANTUM THEORY ARE EVENTUALLY UNIFIED AND ULTIMATELY PROVEN TO BE THE

TRUTH OF HOW THE UNIVERSE WORKS; WHAT KIND OF TECHNOLOGY DO YOU BELIEVE WILL NATURALLY

EMERGE FROM THIS UNIFIED FIELD THEORY, THAT WILL ULTIMATELY HELP ALL OF HUMANITY TO REACH THE

FAR DISTANT STARS?'

IN OTHER WORDS, 'HOW HELPFUL WILL SUCH A UNIFIED FIELD THEORY; WHICH UNIFIES THE STRONG FORCE, THE WEAK FORCE, ELECTROMAGNETISM AND GRAVITY; BE TO THE HUMAN RACE IN THE DEVELOPMENT OF

FUTURE TECHNOLOGY?

THE MOST INTIMIDATING PSYCHOLOGICAL PROBLEM

THE NAVIER-STOKES PROBLEM WAS TREATED AS A LEIBNIZIAN CALCULUS PROBLEM WHICH REQUIRED THE USE

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OF PARTIAL DERIVATIVES IN THREE SPATIAL DIMENSIONS. THIS MATHEMATICAL APPROACH TO WHAT IS

ESSENTIALLY A NEWTONIAN PHYSICS PROBLEM IS WHAT MADE IT IMPOSSIBLE TO SOLVE. THE PHYSICS

APPROACH TO CALCULUS USES THE TIME VARIABLE 'T' AS THE INDEPENDENT VARIABLE UPON WHICH THE

THREE SPATIAL VARIABLES OF 'X, Y, AND Z' ARE DEPENDANT VARIABLES. THIS ALONE IMMEDIATELY ELIMINATES

THE NEED FOR PARTIAL DERIVATIVES. THE FACT THAT MOMENTUM IS PROPORTIONAL TO VELOCITY IN THE

SAME WAY THAT FORCE IS PROPORTIONAL TO ACCELERATION IS ANOTHER PHYSICS PERSPECTIVE THAT WOULD

NOT BE OBVIOUS TO EVEN THE GREATEST OF MATHEMATICIANS. SINCE MASS IS A CONSTANT IN BOTH THE

MOMENTUM AND THE FORCE EQUATION, BY SIMPLY EQUATING MASS TO ONE 'M = 1' THE MASS VARIABLE IS

NEUTRALIZED AND MOMENTUM IMMEDIATELY BECOMES VELOCITY AND FORCE IMMEDIATELY BECOMES

ACCELERATION. IN NEWTONIAN PHYSICS, ACCELERATION IS THE DERIVATIVE OF VELOCITY. THE NAVIER STOKES

EQUATION REQUIRES THAT VISCOSITY IS THE SECOND DERIVATIVE OF THE VELOCITY FUNCION WHICH MEANS

THAT VISCOSITY IS THE FIRST DERIVATIVE OF THE ACCELERATION FUNCTION. ONCE THESE PHYSICAL

RELATIONSHIPS ARE UNDERSTOOD, THE REMAINING CALCULUS IS RELATIVELY STRAIGHTFORWARD.

THE SOLUTION TO THE NAVIER-STOKES EQUATION IS:

PRESSURE IN THE X-AXIAL DIRECTION = X'(T) - X(T) + X''(T) PRESSURE IN THE Y-AXIAL DIRECTION = Y'(T) - Y(T) + Y''(T) PRESSURE IN THE Z-AXIAL DIRECTION = Z'(T) - Z(T) + Z''(T) WHERE X(T), Y(T), Z(T) ARE THE VELOCITY FUNCTIONS IN THE THREE AXIAL DIRECTIONS.

WHERE X'(T), Y'(T), Z'(T) ARE THE ACCELERATION FUNCTIONS IN THE THREE AXIAL DIRECTIONS.

WHERE X''(T), Y''(T), Z''(T) ARE THE VISCOSITY FUNCTIONS IN THE THREE AXIAL DIRECTIONS.

AND X''(T) IS THE DERIVATIVE OF X'(T), WHICH IS ITSELF THE DERIVATIVE OF X(T).

AND Y''(T) IS THE DERIVATIVE OF Y'(T), WHICH IS ITSELF THE DERIVATIVE OF Y(T).

AND Z''(T) IS THE DERIVATIVE OF Z'(T), WHICH IS ITSELF THE DERIVATIVE OF Z(T).

NAVIER-STOKES IS A MILLENNIUM PROBLEM THAT HAS ELUDED SOLUTION FOR TWO HUNDRED YEARS. WHY DID

IT TAKE SO LONG BEFORE A SOLUTION WAS FOUND?

BECAUSE THE LONGER A MATH PROBLEM REMAINS UNSOLVED, THE MORE FAMOUS IT BECOMES, AND THE

GREATER ITS REPUTATION AS AN UNSOLVABLE PROBLEM BECOMES. THIS IS WHAT KEITH DEVLIN WROTE ABOUT

IT:

"...The resulting equations are known as the Navier-Stokes equations. Though these equations can be solved in the hypothetical two-dimensional case of an infinitely thin planar film of fluid, it is not known whether there is a solution in the (more realistic) three-dimensional case. Notice that the issue is not Do we know what the solution is? It's more basic than that. We don't even know whether there is a solution! "

THERE IS A CERTAIN MYSTIQUE THAT EVOLVES AND WHICH BECOMES GREATER AND GREATER AS TIME GOES BY.

THE REPUTATION OF THE NAVIER-STOKES PROBLEM IS THAT IT MAY NOT HAVE A SOLUTION BECAUSE IT HAS

BEEN TWO HUNDRED YEARS AND NO ONE HAS SOLVED IT. PSYCHOLOGICALLY SPEAKING, THIS IS AN EXTREMELY

INTIMIDATING THOUGHT. IF YOU ARE ONE OF THE BEST MATHEMATICIANS IN THE WORLD, AND YOU TALK TO

OTHER GREAT MATHEMATICIANS, AND ALL OF THEM DOUBT THAT THE PROBLEM EVEN HAS A SOLUTION, ARE

YOU GOING TO SPEND A YEAR OR TWO OF YOUR LIFE TRYING TO SOLVE IT?

THE ANSWER IS NO. THE OPPORTUNITY COST OF THE TIME THAT YOU SPEND TRYING TO SOLVE IT IS EVERY

OTHER PROBLEM THAT YOU COULD BE WORKING ON INSTEAD. THIS DIRECTLY LEADS TO A SITUATION WHERE

NONE OF THE BEST MATHEMATICIANS WILL EVEN MAKE A CASUAL ATTEMPT TO TRY AND SOLVE IT BECAUSE ALL

SUCH ATTEMPTS ARE CONSIDERED A WASTE OF TIME. THIS MEANS ONLY AMATEUR MATHEMATICIANS MAKE

ANY ATTEMPT TO SOLVE IT, AND IT THUS REMAINS UNSOLVED.

IF YOU HAVE FOLLOWED THIS LINE OF REASONING THUS FAR, THEN YOU MIGHT HAVE REALIZED THAT IF THIS

IS TRUE FOR ONE OF THE MILLENNIUM PROBLEMS, IT MIGHT BE TRUE FOR MORE THAN ONE OF THE

PROBLEMS.

BUT THE KNEE JERK REACTION OF MOST OF THE PROFESSIONAL MATHEMATICIANS THAT HAVE HEARD OF THE

NAVIER-STOKES PROBLEM AND THAT KNOW OF ITS REPUTATION AS BEING UNSOLVABLE, IS THAT IF ANYONE

DOES CLAIM TO HAVE A SOLUTION, ESPECIALLY AN AMATEUR MATHEMATICIAN LIKE MYSELF, THAT HE MUST BE

'DELUSIONAL'. THEY THEN TEND TO DISMISS THE SOLUTION AS BEING WRONG WITHOUT EVEN LOOKING AT IT.

THIS IS THE KIND OF INTELLECTUAL SNOBBERY THAT PERVADES THE ACADEMIC WORLD.

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Matthew 17:20

King James Version

20 And Jesus said unto them, Because of your unbelief: for verily I say unto you, If ye have faith as a grain of mustard seed, ye shall say unto this mountain, Remove hence to yonder place; and it shall remove; and nothing shall be impossible 6 of 37

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unto you.

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THUS THE ONLY WAY THAT ANY OF THE MILLENNIUM PROBLEMS CAN BE SOLVED IS BY SOMEONE WHO IS NOT

INTIMIDATED BY THEIR REPUTATION. IN OTHER WORDS, THE MAIN REQUIREMENT FOR SOLVING A MILLENNIUM

PROBLEM IS 'STRENGTH'. "BREAKING THE SEAL' WILL REQUIRE THE STRENGTH OF WILL TO PERSEVERE

THROUGH THICK AND THIN, THROUGH HIGHS AND LOW, THROUGH THE UPS AND THE DOWNS, THROUGH

SETBACK AFTER SETBACK, BELIEVING THAT FAILURE IS PART OF THE PROCESS, AND THAT IF ONE DOESN'T GIVE

UP, ONE WILL ULTIMATELY PREVAIL. IN SHORT, ONE HAS TO HAVE 'THE FAITH OF THE MUSTARD SEED'.

THE HODGE CONJECTURE.

THE HODGE CONJECTURE IS THE SINGLE MOST INTIMIDATING MATH PROBLEM, BY ALL ACCOUNTS, BECAUSE OF

THE INCOMPREHENSIBLE LANGUAGE THAT IT USES, AND BECAUSE; LIKE THE NAVIER-STOKES EQUATION, SO

MANY GREAT MATHEMATICIANS HAVE TRIED, AND FAILED, TO SOLVE IT. WHENEVER 'A BEAST' OF A PROBLEM

LIKE THIS ARISES WITHIN ANY SUBJECT AREA, IT TAKES ON 'A MYTHICAL', PERHAPS EVEN 'LEGENDARY', CHARACTERISTIC.

BUT IT IS A 'PAPER TIGER'!

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'During the nineteenth century, mathematicians took Descartes's approach a stem further. Instead of using algebra simply as a tool to help them reason about geometric objects--by writing down equations that determined those objects--they started with collections of algebraic equations and defined "geometric" objects to be the solutions to those equations.

Thus, instead of saying that the equation:

x^2 + y^2 = 4

provides an algebraic description of the circle with radius 2 and center the origin, they would simply study the object--

whatever it is--that arose from the equation.'

'THE MILLENNIUM PROBLEMS' BY KEITH DEVLIN [PAGE 217]

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'Actually, that's not exactly right. In defining an algebraic variety, mathematicians don't restrict themselves to a single algebraic equation; instead, you can start with any finite collection of equations. The variety then consists of all points that solve all the equations in the system. This makes the class of algebraic varieties richer than if you were only allowed to start with a single equation.'

'THE MILLENNIUM PROBLEMS' BY KEITH DEVLIN [PAGE 218]

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'Now, when you use calculus (as opposed to algebra) to define an object, the resulting object need not be in any sense

"geometric." The Hodge Conjecture says that the H-objects are exceptions to that last remark--or at least, almost so.

Though they may not themselves be geometric objects, they can be built up from geometric objects in a fairly straightforward (and calculus free) way. In the terminology of the conjecture, ... That is to say, any H-object can be built up from geometric objects in a purely algebraic way.

Thus, you can think of the Hodge Conjecture as saying, "Look, by using calculus on varieties, we've created a class of objects (the H-objects) that not only defy any hope of our being able to visualize them, we can't even describe them algebraically. However, these objects can be built up in an algebraic fashion from objects that can be described algebraically. So at least we still have a lifeline connecting us to firmer ground--a connection that we (i.e., the experts) might be able to use to carry the study of these objects further."

What the Hodge Conjecture does is provide the expert with some powerful mathematical structure that can be used to analyze H-objects. This is very typical of a lot of modern mathematics, where mathematicians are constantly looking for new structures on objects or links from one area or mathematics to another, so that they can adapt methods from one area for use in another.' 'THE MILLENNIUM PROBLEMS' BY KEITH DEVLIN [PAGE 219]

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'Before trying to convey some sense of the way the experts view the conjecture, let me make a few more remarks about its status in mathematics.

First, it has significant implications. A proof of the Hodge Conjecture would establish a fundamental link among the three disciplines of algebraic geometry, analysis, and topology.

Second, one case of the conjecture is known, but that case was established by the American mathematician Solomon Lefschetz in 1925, long before Hodge formulated the general conjecture. It's not much of an exaggeration to say that there has been essentially no progress on the problem since then.

Thus, to date, the Hodge Conjecture remains just that: a conjecture. Some would say it could be more accurately called a wild guess. But that has not prevented many mathematicians from trying to prove it or from investigating its 7 of 37

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consequences.' 'THE MILLENNIUM PROBLEMS' BY KEITH DEVLIN [PAGE 220]

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These equations are (and were then) familiar to physicists. They know them as Laplace's equations, and they play a major role in gravitational theory, electromagnetic theory, and fluid mechanics. (Moreover, closely related equations occur in the theories of heat flow, acoustics and the propagation of waves.) A solution of the Laplace equations is called a harmonic function. The discovery of this close relationship between the calculus of complex functions and the Laplace equations led to significant progress in mathematical physics, by providing a way to solve Laplace's equations in a variety of contexts.'

'THE MILLENNIUM PROBLEMS' BY KEITH DEVLIN [PAGE 225]

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COMPUTER AIDED NUMERICAL DATA ORDERING AND

RESTRUCTURING:

CANDOR

THE MILLENNIUM PROBLEMS WERE NOT ANSWERABLE WITHIN THE CONVENTIONAL MATHEMATICS THAT

EXISTED IN THE YEAR 2000; WHICH MEANS THAT NEW MATH AND NEW PHYSICS ARE REQUIRED TO SOLVE

THESE QUESTIONS. THE 'CANDOR' TOOLS THAT THE MATHEMATICS OF THE DYNAMISM MODEL ARE BASED ON, REPRESENT BOTH NEW MATH, AND NEW PHYSICS. THE FIRST OF THESE TOOLS IS CALLED 'FOUR-D', WHICH IS

THE PRIMARY TOOL FOR HANDLING FOUR DIMENSIONAL PROBLEMS; THE THREE DIMENSIONS OF 'SPACE'

COMBINED WITH THE FOURTH DIMENSION OF 'TIME'.

FOUR-D

THE 'FOUR-D' TOOL IS A TEMPORAL-SPATIAL GRAPHING METHOD WHICH PRODUCES POINTS THAT HAVE FOUR

COMPONENTS (X, Y, Z, T); AND WHICH THEN GRAPHS THESE POINT ON A THREE DIMENSIONAL CARTESIAN

PLANE.

FOR INSTANCE, A COMPLEX VALUED INPUT SUCH AS (X, Ti) OF ANY GIVEN POLYNOMIAL EQUATION, WOULD

RESULT IN A COMPLEX VALUE OUTPUT (Z, Yi). THE 'TRICK' TO GRAPHING THIS FUNCTION IS SIMPLY TO LET 'X =

T', SUCH THAT THE INPUT IS ACTUALLY (T, Ti) WHILE THE OUTPUT REMAINS (Z, Yi). THIS WOULD MEAN THAT

EVERY POINT TO BE GRAPHED WOULD TAKE THE FORM (X, Yi, Z, Ti). BUT NOTE THAT SINCE 'X = T', THE ACTUAL

POINT TO BE GRAPHED ONLY HAS THREE COMPONENTS, (X, Yi, Z).

THUS THE 'FOUR-D' TOOL TAKES ANY COMPLEX VALUED INPUT OF THE FORM (X, Ti) WHICH YIELDS THE

COMPLEX VALUED OUTPUT (Z, Yi) FROM ANY GIVEN POLYNOMIAL EQUATION, AND PLOTS THE POINTS: 1. (X(1), Yi(1), Z(1))

2. (X(2), Yi(2), Z(2))

3. (X(3), Yi(3), Z(3))

4. (X(4), Yi(4), Z(4))

5. (X(5), Yi(5), Z(5))

6. ... , ... , ...

7. ... , ... , ...

8. ... , ... , ...

9. (X(N- 2), Yi(N-2), Z(N-2))

10. (X(N-1), Yi(N-1), Z(N-1))

11. (X(N), Yi(N), Z(N))

WHICH IS SIMPLY:

(X(T), Yi(T), Z(T))

WHERE:

T = {1, 2, 3, 4, 5, . . ., N-2, N-1, N}

WHICH IS A NEW GRAPHING TECHNIQUE THAT USES THE THREE DIMENSIONAL CARTESIAN GRAPHING METHOD

TO DISPLAY FOUR DIMENSIONAL MATHEMATICAL FUNCTIONS.

THIS GRAPHING TECHNIQUE IS BASED ON ADDING A FOURTH AXIS, A 'TIME' AXIS TO THE NORMAL THREE

DIMENSIONAL CARTESIAN PLANE. THE FOURTH AXIS IS EQUAL TO THE LINE: X = Y = Z = T

T

X

Y

Z

.

.

.

.

-5

-5

-5

-5

-4

-4

-4

-4

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-3

-3

-3

-3

-2

-2

-2

-2

-1

-1

-1

-1

0

0

0

0

+1

+1

+1

+1

+2

+2

+2

+2

+3

+3

+3

+3

+4

+4

+4

+4

+5

+5

+5

+5

COMPLEX NUMBERS

THE MATHEMATICS OF THE HODGE CONJECTURE PRIMARILY DEALS WITH COMPLEX NUMBERS. THE MAIN

DIFFICULTY OF WORKING WITH COMPLEX NUMBERS IS THAT IT IS HARD TO VISUALIZE THE COMPLETE

FUNCTION, WITH BOTH THE COMPLEX VALUED INPUTS, AND THE COMPLEX VALUED OUTPUT ON THE SAME

GRAPH.

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https://www.varsitytutors.com/hotmath/hotmath_help/topics/operations-with-complex-numbers ___________________________________________________________________________________________________

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https://www.britannica.com/science/Laplaces-equation

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'In his address to the International Congress of Mathematicians in 1950, Hodge raised the possibility that for nonsingular projective complex algebraic varieties, that last property completely characterizes the algebraic cohomology classes. That is, every harmonic (p, p)-form is a rational combination of closed algebraic forms (loosely, could be built up in algebraic--

ie..., calculus-free--way).

Thus was the Hodge Conjecture born.

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But is the conjecture true? Nobody knows. At the moment there is no strong evidence to suggest that Hodge's intuition was correct.'

'THE MILLENNIUM PROBLEMS' BY KEITH DEVLIN [PAGE 228]

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THE REASON WHY I CALL THE HODGE CONJECTURE A 'PAPER TIGER' IS BECAUSE THE PROBLEM BASICALLY

AMOUNTS TO BEING ABLE TO ACCURATELY GRAPHICALLY PICTURE A 'COMPLEX VALUED FUNCTION' THAT HAS

BOTH A COMPLEX VALUED INPUT, AND A COMPLEX VALUED OUTPUT. THIS IS EXACTLY WHAT THE CANDOR

MATHEMATICAL TOOL; 'FOUR-D'; DOES. EVERYTHING ELSE IS SIMPLY 'ABSTRACTIONS, BUILT ON ABSTRACTIONS, BUILT ON ABSTRACTIONS', WHICH AMOUNTS TO MAKING A MOUNTAIN OUT OF A MOLEHILL.

BINOMIAL THEOREM

(a + ib)^n

WHERE 'n' IS A POSITIVE INTEGER.

THIS KIND OF EXPRESSION IS CALLED A BINOMIAL EXPRESSION. WHEN 'b' IS REPLACED BY 'ib', THE EXPRESSION

BECOMES A COMPLEX NUMBER, AND THE SECOND PARTIAL DERIVATIVE FOR 'u(a)', WITH RESPECT TO 'a', WILL

ALWAYS BE EQUAL AND OPPOSITE TO THE SECOND PARTIAL DERIVATIVE FOR 'v(b)' WITH RESPECT TO 'b'. THIS

MEANS THAT THE SECOND PARTIAL DERIVATIVE FOR THESE TYPES OF BINOMIAL EXPRESSIONS, WHEN

EXPANDED AS COMPLEX NUMBERS, IS ALWAYS EQUAL TO ZERO. THUS, THESE TYPES OF BINOMIAL

EXPRESSIONS, WHEN COMPLEX NUMBERS ARE USED AS THE INPUT, ALWAYS REPRESENT SOLUTIONS TO

LAPLACE'S EQUATION.

THE INPUT TO THE BINOMIAL EQUATION, IS 'X'; A COMPLEX VALUE: x = a + ib

NOTE THAT FOR ALL OF THE FOLLOWING, THE PARTIAL DERIVATIVE FOR 'X' IS: w(x) = w(a + ib)

w'(x) = 1

w''(x) = 0

u(a) = (a + ib)^2

u'(a) = 2(a + ib)

u''(a) = 2

v(b) = (a + ib)^2

v'(b) = 2i(a + ib)

v''(b) = -2

w''(x) + u''(a) + v''(b) = 0 + 2 − 2 = 0

u(a) = (a + ib)^3

u'(a) = 3(a + ib)^2

u''(a) = 6(a + ib)

v(b) = (a + ib)^3

v'(b) = 3i(a + ib)^2

v''(b) = -6(a + ib)

w''(x) + u''(a) + v''(b) = 0 + 6(a + ib) − 6(a + ib) = 0

u(a) = (a + ib)^4

u'(a) = 4(a + ib)^3

u''(a) = 12(a + ib)^2

v(b) = (a + ib)^4

v'(b) = 4i(a + ib)^3

v''(b) = -12(a + ib)^2

w''(x) + u''(a) + v''(b) = 0 + 12(a + ib)^2 − 12(a + ib)^2 = 0

u(a) = (a + ib)^5

u'(a) = 5(a + ib)^4

u''(a) = 20(a + ib)^3

v(b) = (a + ib)^5

v'(b) = 5i(a + ib)^4

v''(b) = -20(a + ib)^3

w''(x) + u''(a) + v''(b) = 0 + 20(a + ib)^3 − 20(a + ib)^3 = 0

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u(a) = (a + ib)^6

u'(a) = 6(a + ib)^5

u''(a) = 30(a + ib)^4

v(b) = (a + ib)^6

v'(b) = 6i(a + ib)^5

v''(b) = -30(a + ib)^4

w''(x) + u''(a) + v''(b) = 0 + 30(a + ib)^4 − 30(a + ib)^4 = 0

u(a) = (a + ib)^7

u'(a) = 7(a + ib)^6

u''(a) = 42(a + ib)^5

v(b) = (a + ib)^7

v'(b) = 7i(a + ib)^6

v''(b) = -42(a + ib)^5

w''(x) + u''(a) + v''(b) = 0 + 42(a + ib)^5 − 42(a + ib)^5 = 0

u(a) = (a + ib)^8

u'(a) = 8(a + ib)^7

u''(a) = 56(a + ib)^6

v(b) = (a + ib)^8

v'(b) = 8i(a + ib)^7

v''(b) = -56(a + ib)^6

w''(x) + u''(a) + v''(b) = 0 + 56(a + ib)^6 − 56(a + ib)^6 = 0

u(a) = (a + ib)^9

u'(a) = 9(a + ib)^8

u''(a) = 72(a + ib)^7

v(b) = (a + ib)^9

v'(b) = 9i(a + ib)^8

v''(b) = -72(a + ib)^7

w''(x) + u''(a) + v''(b) = 0 + 72(a + ib)^7 − 72(a + ib)^7 = 0

u(a) = (a + ib)^10

u'(a) = 10(a + ib)^9

u''(a) = 90(a + ib)^8

v(b) = (a + ib)^10

v'(b) = 10i(a + ib)^9

v''(b) = -90(a + ib)^8

w''(x) + u''(a) + v''(b) = 0 + 90(a + ib)^8 − 90(a + ib)^8 = 0

AS YOU CAN SEE, WHEN THE INPUT TO THE BINOMIAL EQUATION IS OF THE FORM 'x = a + ib', THEN EVERY

SECOND DERIVATIVE FOR ALL POSITIVE INTEGER 'n = {1, 2, 3, ..., -> infinity}' FOR THE PARTIAL DERIVATIVE, WILL

BE EQUAL TO ZERO.

THIS HOLDS TRUE EVEN WHEN 'X' IS CHANGED TO THE FORM 'x = a + ia', EVEN THOUGH YOU CANNOT TAKE THE

PARTIAL DERIVATIVE DIRECTLY FOR 'u(a)' AND 'v(b)', BECAUSE THERE IS NO 'b', SINCE IT HAS BEEN CHANGED TO

'a'. THIS IS A VERY IMPORTANT POINT TO UNDERSTAND. WHEN 'x = a + ib' IS THE INPUT TO ANY BINOMIAL, THAT

THE SECOND DERIVATIVE WILL ALWAYS BE EQUAL TO ZERO, USING PARTIAL DERIVATIVES; WHICH MAKES IT A SOLUTION TO LAPLACE'S EQUATION. THE SAME THING IS TRUE FOR 'x = a + ia' WHERE 'A' IS A SUBSTITUTE FOR

'B' IN THE EQUATION. WHEN USED AS AN INPUT TO ANY BINOMIAL EQUATION THE SECOND DERIVATIVE WILL

ALWAYS BE EQUAL TO ZERO, AND THIS MAKES IT A SOLUTION TO LAPLACE'S EQUATION; EVEN THOUGH YOU CAN

NOT USE PARTIAL DERIVATIVES.

(a + b)^1 =

a + b

(a+b)^2 =

a^2 + 2ab + b^2

(a+b)^3 =

a^3 + 3a^2b + 3ab^2 + b^3

(a+b)^4 =

a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4

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(a+b)^5 =

a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5

(a+b)^6 =

a^6 + 6a^5b + 15a^4b^2 + 20a^3b^3 + 15a^2b^4 + 6ab^5 + b^6

(a+b)^7 =

a^7 + 7a^6b + 21a^5b^2 + 35a^4b^3 + 35a^3b^4 + 21a^2b^5 + 7ab^6 + b^7

(a+b)^8 =

a^8 + 8a^7b + 28a^6b^2 + 56a^5b^3 + 70a^4b^4 + 56a^3b^5 + 28a^2b^6 + 8ab^7 + b^8

(a+b)^9 =

a^9 + 9a^8b + 36a^7b^2 + 84a^6b^3 + 126a^5b^4 + 126a^4b^5 + 84a^3b^6 + 36a^2b^7 + 9ab^8 + b^9

(a+b)^10 =

a^10 + 10a^9b + 45a^8b^2 + 120a^7b^3 + 210a^6b^4 + 252a^5b^5 + 210a^4b^6 + 120a^3b^7 + 45a^2b^8 +

10ab^9 + b^10

COMPLEX NUMBERS ARE CYCLICAL.

i^1 = i

i^2 = i * i = -1

i^3 = -1 * i = -i

i^4 = -i * i = -i^2 = 1

THE PATTERN ABOVE NOW REPEATS AS MULIPLICATION BY 'i' CONTINUES: i^5 = 1 * i = i

i^6 = i * i = -1

i^7 = -1 * i = -i

i^8 = -i * i = -i^2 = 1

LET 'i' BE AN IMAGINARY NUMBER: i^2 = -1.

LET (a + b)^n = (a + (ai))^n

N = 1: (x + xi)^n

(a + (ai))^1 =

a + (ai)

a + ai

(Y + Zi) = [a + ai] 1 = 2^0

N = 2: (x + xi)^n

(a + (ai))^2 =

a^2 + 2a(ai) + (ai)^2 =

a^2 + 2a^2i^2 + (ai)^2 =

0 + 2a^2i

(Y + Zi) = [0 + 2a^2i] 2 = 2^1

N = 3: (x + xi)^n

(a + (ai))^3 =

a^3 + 3a^2(ai) + 3a(ai)^2 + (ai)^3 =

a^3 + 3a^3i + 3a^3i^2 + a^3i^3 =

-2a^3 + 2a^3i

(Y + Zi) = [-2a^3 + 2a^3i] 2 = 2^1

N = 4: (x + xi)^n (a + (ai))^4 =

a^4 + 4a^3(ai) + 6a^2(ai)^2 + 4a(ai)^3 + (ai)^4 =

a^4 + 4a^4i + 6a^4i^2 + 4a^4i^3 + a^4i^4 =

-4a^4 + 0i

(Y + Zi) = [-4a^4 + 0i] 4 = 2^2

N = 5: (x + xi)^n

(a + (ai))^5 =

a^5 + 5a^4(ai) + 10a^3(ai)^2 + 10a^2(ai)^3 + 5a(ai)^4 + (ai)^5 =

a^5 + 5a^5i + 10a^5i^2 + 10a^5i^3 + 5a^5i^4 + a^5i^5 =

-4a^5 - 4a^5i

(Y + Zi) = [-4a^5 - 4a^5i] 4 = 2^2

N = 6: (x + xi)^n

(a + (ai))^6 =

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a^6 + 6a^5(ai) + 15a^4(ai)^2 + 20a^3(ai)^3 + 15a^2(ai)^4 + 6a(ai)^5 + (ai)^6 =

a^6 + 6a^6i + 15a^6i^2 + 20a^6i^3 + 15a^6i^4 + 6a^6i^5 + a^6i^6 =

0 - 8a^6i

(Y + Zi) = [0 - 8a^6i] 8 = 2^3

N = 7: (x + xi)^n

(a + (ai))^7 =

a^7 + 7a^6(ai) + 21a^5(ai)^2 + 35a^4(ai)^3 + 35a^3(ai)^4 + 21a^2(ai)^5 + 7a(ai)^6 + (ai)^7 =

a^7 + 7a^7i + 21a^7i^2 + 35a^7i^3 + 35a^7i^4 + 21a^7i^5 + 7a^7i^6 + a^7i^7 =

8a^7 - 8a^7i

(Y + Zi) = [8a^7 - 8a^7i] 8 = 2^3

N = 8: (x + xi)^n (a + (ai))^8 =

a^8 + 8a^7(ai) + 28a^6(ai)^2 + 56a^5(ai)^3 + 70a^4(ai)^4 + 56a^3(ai)^5 + 28a^2(ai)^6 + 8a(ai)^7 + (ai)^8 =

a^8 + 8a^8i + 28a^8i^2 + 56a^8i^3 + 70a^8i^4 + 56a^8i^5 + 28a^8i^6 + 8a^8i^7 + a^8i^8 =

16a^8 + 0i

(Y + Zi) = [16a^8 + 0i] 16 = 2^4

N = 9: (x + xi)^n

(a + (ai))^9 =

a^9 + 9a^8(ai) + 36a^7(ai)^2 + 84a^6(ai)^3 + 126a^5(ai)^4 + 126a^4(ai)^5 + 84a^3(ai)^6 + 36a^2(ai)^7 +

9a(ai)^8 + (ai)^9 =

a^9 + 9a^9i + 36a^9i^2 + 84a^9i^3 + 126a^9i^4 + 126a^9i^5 + 84a^9i^6 + 36a^9i^7 + 9a^9i^8 + a^9i^9 =

16a^9 + 16a^9i

(Y + Zi) = [16a^9 + 16a^9i] 16 = 2^4

N = 10: (x + xi)^n

(a + (ai))^10 =

a^10 + 10a^9(ai) + 45a^8(ai)^2 + 120a^7(ai)^3 + 210a^6(ai)^4 + 252a^5(ai)^5 + 210a^4(ai)^6 + 120a^3(ai)^7 +

45a^2(ai)^8 + 10a(ai)^9 + (ai)^10 =

a^10 + 10a^10i + 45a^10i^2 + 120a^10i^3 + 210a^10i^4 + 252a^10i^5 + 210a^10i^6 + 120a^10i^7 + 45a^10i^8

+ 10a^10i^9 + a^10i^10 =

0 + 32a^10i

(Y + Zi) = [0 + 32a^10i] 32 = 2^5

THUS THE FOLLOWING TABLE REPRESENTS NUMERICAL SOLUTIONS TO THE LAPLACE EQUATION: FOR

BINOMIAL EXPANSION FROM N = 1, TO N = 10.

N

X

Y

Z

EXP

.

.

.

.

.

1

a

2^0a

2^0ai

2^0

2

a

0

2^1a^2i

2^1

3

a

-2^1a^3

2^1a^3i

2^1

4

a

-2^2a^4

0

2^2

5

a

-2^2a^5

-2^2a^5i

2^2

6

a

0

-2^3a^6i

2^3

7

a

2^3a^7

-2^3a^7i

2^3

8

a

2^4a^8

0

2^4

9

a

2^4a^9

2^4a^9i

2^4

10

a

0

2^5a^10i

2^5

11

a

-2^5a^11

2^5a^11i

2^5

12

a

-2^6a^12

0

2^6

13

a

-2^6a^13

-2^6a^13i

2^6

14

a

0

-2^7a^14i

2^7

15

a

2^7a^15

-2^7a^15i

2^7

16

a

2^8a^16

0

2^8

17

a

2^8a^17

2^8a^17i

2^8

18

a

0

2^9a^18i

2^9

19

a

-2^9a^19

2^9a^19i

2^9

20

a

-2^10a^20

0

2^10

21

a

-2^10a^21

-2^10a^21i

2^10

22

a

0

-2^10a^22i

2^11

23

a

2^11a^23

-2^11a^23i

2^11

24

a

2^12a^24

0

2^12

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25

a

2^12a^25

2^12a^25i

2^12

26

a

0

2^13a^26i

2^13

27

a

-2^13a^27

2^13a^27i

2^13

28

a

-2^14a^28

0

2^14

29

a

-2^14a^29

-2^14a^29i

2^14

30

a

0

-2^15a^30i

2^15

31

a

2^15a^31

-2^15a^31i

2^15

32

a

2^16a^32

0

2^16

33

a

2^16a^33

2^16a^33i

2^16

34

a

0

2^17a^34i

2^17

35

a

-2^17a^35

2^17a^35i

2^17

36

a

-2^18a^36

0

2^18

37

a

-2^18a^37

-2^18a^37i

2^18

38

a

0

-2^19a^38i

2^19

39

a

2^19a^39

-2^19a^39i

2^19

40

a

2^20a^40

0

2^20

41

a

2^20a^41

2^20a^41i

2^20

42

a

0

2^21a^42i

2^21

43

a

-2^21a^43

2^21a^43i

2^21

44

a

-2^22a^44

0

2^22

45

a

-2^22a^45

-2^22a^45i

2^22

46

a

0

-2^23a^46i

2^23

47

a

2^23a^47

-2^23a^47i

2^23

48

a

2^24a^48

0

2^24

49

a

2^24a^49

2^24a^49i

2^24

50

a

0

2^25a^50i

2^25

51

a

-2^25a^51

2^25a^51i

2^25

52

a

-2^26a^52

0

2^26

53

a

-2^26a^53

-2^26a^53i

2^26

54

a

0

-2^27a^54i

2^27

55

a

2^27a^55

-2^27a^55i

2^27

56

a

2^28a^56

0

2^28

57

a

2^28a^57

2^28a^57i

2^28

58

a

0

2^29a^58i

2^29

59

a

-2^29a^59

2^29a^59i

2^29

60

a

-2^30a^60

0

2^30

61

a

-2^30a^61

-2^30a^61i

2^30

62

a

0

-2^30a^62i

2^31

63

a

2^31a^63

-2^31a^63i

2^31

64

a

2^32a^64

0

2^32

65

a

2^32a^65

2^32a^65i

2^32

66

a

0

2^33a^66i

2^33

67

a

-2^33a^67

2^33a^67i

2^33

68

a

-2^34a^68

0

2^34

69

a

-2^34a^69

-2^34a^69i

2^34

70

a

0

-2^35a^70i

2^35

71

a

2^35a^71

-2^35a^71i

2^35

72

a

2^36a^72

0

2^36

73

a

2^36a^73

2^36a^73i

2^36

74

a

0

2^37a^74i

2^37

75

a

-2^37a^75

2^37a^75i

2^37

76

a

-2^38a^76

0

2^38

77

a

-2^38a^77

-2^38a^77i

2^38

78

a

0

-2^39a^78i

2^39

79

a

2^39a^79

-2^39a^79i

2^39

80

a

2^40a^80

0

2^40

81

a

2^40a^81

2^40a^81i

2^40

82

a

0

2^41a^82i

2^41

83

a

-2^41a^83

2^41a^83i

2^41

84

a

-2^42a^84

0

2^42

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85

a

-2^42a^85

-2^42a^85i

2^42

86

a

0

-2^43a^86i

2^43

87

a

2^43a^87

-2^43a^87i

2^43

88

a

2^44a^88

0

2^44

89

a

2^44a^89

2^44a^89i

2^44

90

a

0

2^45a^90i

2^45

91

a

-2^45a^91

2^45a^91i

2^45

92

a

-2^46a^92

0

2^46

93

a

-2^46a^93

-2^46a^93i

2^46

94

a

0

-2^47a^94i

2^47

95

a

2^47a^95

-2^47a^95i

2^47

96

a

2^48a^96

0

2^48

97

a

2^48a^97

2^48a^97i

2^48

98

a

0

2^49a^98i

2^49

99

a

-2^49a^99

2^49a^99i

2^49

100

a

-2^50a^100

0

2^50

THE FOLLOWING PROGRAM IS WRITTEN FOR OPENSCAD AND PLOTS THE SOLUTIONS TO LAPLACE'S EQUATION

FOR TWENTY DIFFERENT EQUATIONS ON ONE THREE DIMENSIONAL GRAPH WITHOUT USING THE TECHNIQUES

OF CALCULUS.

THIS IS A PRACTICAL DEMONSTRATION OF THE CANDOR, FOUR-D, WHICH STANDS FOR 'FOUR DIMENSIONAL', TOOL AT WORK. NOTE THAT THE FOLLOWING PROGRAM CAN BE PLACED DIRECTLY IN ANY OPENSCAD EDITOR

AND IT WILL WORK AUTOMATICALLY.

___________________________________________________________________________________________________

// OPENSCAD PROGRAM FOR DRAWING LAPLACE EQUATION OBJECT.

$fn=25;

// EQUATION 01 X = a Y = 2^0a Z = 2^0

for (t=[-3:0.01:3]) color("Red"){

translate([t+1,t+1,t+1])

sphere(r=.1);

}

// EQUATION 02 X = a Y = 0 Z = 2^1a^2i

for (t=[-3:0.01:3]) color("Orange"){

translate([t+1,0,2*(t+1)^2])

sphere(r=.1);

}

// EQUATION 03 X = a Y = -2^1a^3 Z = 2^1a^3i for (t=[-3:0.01:3]) color("Yellow"){

translate([t+1,-2*(t+1)^3,2*(t+1)^3])

sphere(r=.1);

}

// EQUATION 04 X = a Y = -2^2a^4 Z = 0

for (t=[-3:0.01:3]) color("Green"){

translate([t+1,-4*(t+1)^4,0])

sphere(r=.1);

}

// EQUATION 05 X = a Y = -2^2a^5 Z = -2^2a^5i for (t=[-3:0.01:3]) color("Blue"){

translate([t+1,-4*(t+1)^5,-4*(t+1)^5])

sphere(r=.1);

}

// EQUATION 06 X = a Y = 0 Z = -2^3a^6i

for (t=[-3:0.01:3]) color("Purple"){

translate([t+1,0,-8*(t+1)^6])

sphere(r=.1);

}

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// EQUATION 07 X = a Y = 2^3a^7 Z = -2^3a^7i for (t=[-3:0.01:3]) color("Pink"){

translate([t+1,8*(t+1)^7,-8*(t+1)^7])

sphere(r=.1);

}

// EQUATION 08 X = a Y = 2^4a^8 Z = 0

for (t=[-3:0.01:3]) color("White"){

translate([t+1,16*(t+1)^8,0])

sphere(r=.1);

}

// EQUATION 09 X = a Y = 2^4a^9 Z = 2^4a^9i for (t=[-3:0.01:3]) color("Brown"){

translate([t+1,16*(t+1)^9,16*(t+1)^9])

sphere(r=.1);

}

// EQUATION 10 X = a Y = 0 Z = 2^5a^10i

for (t=[-3:0.01:3]) color("Black"){

translate([t+1,0,32*(t+1)^10])

sphere(r=.1);

}

// EQUATION 11 X = a Y = -2^5a^11 Z = 2^5a^11i for (t=[-3:0.01:3]) color("LightSalmon"){

translate([t+1,-32*(t+1)^11,32*(t+1)^11])

sphere(r=.1);

}

// EQUATION 12 X = a Y = -2^6a^12 Z = 0

for (t=[-3:0.01:3]) color("Tomato"){

translate([(t+1),-64*(t+1)^12,0])

sphere(r=.1);

}

// EQUATION 13 X = a Y = -2^6a^13 Z = -2^6a^13i for (t=[-3:0.01:3]) color("Gold"){

translate([(t+1),-64*(t+1)^13,-64*(t+1)^13])

sphere(r=.1);

}

// EQUATION 14 X = a Y = 0 Z = -2^7a^14i

for (t=[-3:0.01:3]) color("PaleGreen"){

translate([(t+1),0,-128*(t+1)^14])

sphere(r=.1);

}

// EQUATION 15 X = a Y = 2^7a^15 Z = -2^7a^15i for (t=[-3:0.01:3]) color("Aqua"){

translate([(t+1),128*(t+1)^15,-128*(t+1)^15])

sphere(r=.1);

}

// EQUATION 16 X = a Y = 2^8a^16 Z = 0

for (t=[-3:0.01:3]) color("Magenta"){

translate([(t+1),256*(t+1)^16,0])

sphere(r=.1);

}

// EQUATION 17 X = a Y = 2^8a^17 Z = 2^8a^17i for (t=[-3:0.01:3]) color("HotPink"){

translate([(t+1),256*(t+1)^17,256*(t+1)^17])

sphere(r=.1);

}

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// EQUATION 18 X = a Y = 0 Z = 2^9a^18i

for (t=[-3:0.01:3]) color("Beige"){

translate([(t+1),0,512*(t+1)^18])

sphere(r=.1);

}

// EQUATION 19 X = a Y = -2^9a^19 Z = 2^9a^19i for (t=[-3:0.01:3]) color("Chocolate"){

translate([(t+1),-512*(t+1)^19,512*(t+1)^19])

sphere(r=.1);

}

// EQUATION 20 X = a Y = -2^10a^20 Z = 0

for (t=[-3:0.01:3]) color("SlateGray"){

translate([(t+1),-1024*(t+1)^20,0])

sphere(r=.1);

}

OPENSCAD FOR LINUX OPERATING SYSTEMS

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NOTE THAT THE IN THE ABOVE, THE EXPRESSION THAT FOLLOWS THE 'TRANSLATE' COMMAND IS ALWAYS OF

THE FORM 'TRANSLATE([x(t), y(t), z(t)]). AS LONG AS THE HIGHEST EXPONENT OF 'x(t)' IS LESS THAN TWO, THEN

THE SECOND DERIVATIVE WILL BE EQUAL TO ZERO. IN THE ABOVE, 'x(t) = t + 1' EVERYTIME AND THUS MEETS

THIS REQUIREMENT. THE INPUT TO THE BINOMIAL EQUATION IS A COMPLEX VALUED INPUT WHICH ACTUALLY

TAKES THE FORM:

x(t) + ix(t) =

(t + 1) + i(t + 1)

THE OUTPUT IS ALSO A COMPLEX VALUED OUTPUT OF THE FORM: y(t) + iz(t) =

THUS EVERY POINT IN THE FOLLOWING 'GRAPH OBJECT' REPRESENTS A COMPLEX VALUED OUTPUT OF A COMPLEX VALUED INPUT WHICH IMPLICITLY CONTAINS THE INFORMATION FOR THE THREE DIMENSIONS OF

SPACE 'x', 'y' AND 'z', AND THE FOURTH DIMENSION OF TIME 't'. THEREFORE, THESE SOLUTIONS TO LAPLACE'S

EQUATION ARE CONTAINED IN ONE GRAPH OBJECT THAT WAS BUILT UP WITHOUT USING ANY OF THE

TECHNIQUES OF CALCULUS WHICH PROVES THAT HODGE'S INTUITION WAS CORRECT; AND THE FOLLOWING

IMAGES CLEARLY SHOW THAT THE HODGE CONJECTURE IS INDEED TRUE!

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THE 'LAPLACE EQUATION OBJECT'

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BREAKING THE FOURTH SEAL

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Revelation 6:7-8

King James Version

7 And when he had opened the fourth seal, I heard the voice of the fourth beast say, Come and see.

8 And I looked, and behold a pale horse: and his name that sat on him was Death, and Hell followed with him. And power was given unto them over the fourth part of the earth, to kill with sword, and with hunger, and with death, and with the beasts of the earth.

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Genesis 1:14-19

King James Version

14 And God said, Let there be lights in the firmament of the heaven to divide the day from the night; and let them be for signs, and for seasons, and for days, and years:

15 And let them be for lights in the firmament of the heaven to give light upon the earth: and it was so.

16 And God made two great lights; the greater light to rule the day, and the lesser light to rule the night: he made the stars also.

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17 And God set them in the firmament of the heaven to give light upon the earth, 18 And to rule over the day and over the night, and to divide the light from the darkness: and God saw that it was good.

19 And the evening and the morning were the fourth day.

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