< Earlier Kibitzing · PAGE 4 OF 4 ·
|Jan-03-08|| ||MrMelad: I'm sorry to inspect but was that cinical? :)|
|Jan-04-08|| ||Open Defence: no not cynical at all.. I just wonder how i missed the material here before|
|Jan-04-08|| ||MrMelad: So what are your impressions, if I might ask?
|Jan-04-08|| ||Open Defence: well it requires careful reading and understanding to say so.. but i will say its interesting, some of it can be argued and challenged (as no doubt all works should be) but it is very interesting|
|Jan-05-08|| ||MrMelad: Well, thanks! Nice to hear that it interest you.. Btw, have you heard of Garret Lisie and his theory of everything?|
|Apr-12-08|| ||MrMelad: Hello all,
my new website:
one of the articles is what used to be parts 1+2+3 in my bio, the other article is new, re-written from what used to be sections 4+5.
|Dec-27-10|| ||zarg: Merry Christmas!
|Dec-27-10|| ||zarg: One of my favourites on youtube, and the part starting at 4:35 is just awesome. Love it!|
|Dec-27-10|| ||MrMelad: Thanks zarg :)
|May-18-11|| ||MrMelad: Gauss and Galois were probably the most prominent mathematicians to study the the multiplication table modulo n, they published about different symmetries in it, the table features in many mathematical fields since the early ages to modern mathematics, including linear algebra, group theory,discrete mathematics, topology, calculus, rational trigonometry, computer science, and more, and it also featured in many popular science fiction.|
However the multiplication table modulo n was never studied for its geometry of numbers. In fact, the geometry of numbers is the story of this post.
I have a series that my computer converges to a very close approximation of the square root of e minus one (0.648...). It is a conjecture because it can't be proven but only approximated with the thankful aid of the computer (to 0.0001 accuracy with my best algorithms) though the convergence itself is easily proven and the way in which this series is calculated offers a very deep route into the regularity of numbers in general and prime numbers in particular.
In fact, my entire work could have been “avoided” or "summed up" if a “simple” problem will be solved, if anyone can find the formula of the here by celebrated function S(n) which sums S(n,i) (to be defined). A simple formula for S(n) or S(n,i) doesn't exist at the moment and its existence will remove all obstacles from this project so here it is, the first ten members of S(n), in the name of intuition if you want give it your best shot:
S(1) = 0
S(2) = 0
S(3) = 0
S(4) = 2
S(5) = 14
S(6) = 28
S(7) = 70
S(8) = 108
S(9) = 205
Tip: S(n+1) > S(n) for any n>3.
Why is this S(n) so important and what is it in words? It is easy to say now in retrospect but was difficult to understand at first sight or measure. Nonetheless, calculating S(n) to low n requires little more than high school algebra, familiarity with the modulo operator and with Euclid's Elements (or rational trigonometry).
In one sentence, S(n) is the sum of the individual areas S(n,i) of the shapes that are created in the multiplication table modulo n when connecting its discrete points with straight lines where they share inner value i.
The little twist I bring into the multiplication table modulo n, my secret is that I play connect the dots. Like the diagrams of animals young children draw, I connect numbers on dots with straight lines and play space. My space is comprised of points which are given coordinates according to its relative position in the multiplication table and inner values according to the modular operator performed on a multiplication of the coordinates. The objects in my space are the shapes that created from connecting with straight lines members with the same inner value creating symmetry and geometrical patterns. A one is connected with ones, a two is connected with twos, a zero is connected with zeros and the lines are all straight.
For example, take the following picture for n=4, multiplication table modulo 4 From now on this will be our ongoing example. There is a reason 4 was chosen as a base for this example, and the reason is that 4 is not a prime number.
Multiplication table modulo 4:
0 0 0 0
0 1 2 3
0 2 0 2
0 3 2 1
First you have to forget about the zeros in the first row and column, they don't count in our game, they come from multiplying zero with all the numbers, we will always discuss just the inner part with out the border of zeros. The zeros are included on this one time occasion to explain why there are just 3 rows and columns in all the following but we are modulo 4.The zeroes are included just this once to emphasize they have to be chopped down. You can think about our example as a multiplication table modulo n starting with the number 1, where n-1 is the number of rows/columes defining the modulo.
To sum it up, when n=4 we are interested just in:
1 2 3
2 0 2
3 2 1
|May-18-11|| ||MrMelad: Now we play connect the dots. We make all the connections with straight lines – we connect every one to every other one, every zero to every other zero (Doesn't apply here, there is only one zero in the example) and every point with inner value i to any other point with inner value i. As simple as it might look, its important that you take a moment to do this on paper as accurately as possible with a symmetrical table and straight lines on a flat surface.|
The shape has a definite center and is symmetric towards its center (n/2,n/2), the element with inner value 0. in our example (4/2,4/2) = (2,2) = 2*2 mod 4 = 0.
You might want to try it with n=3, 5, 6 and so on. It will of course will get very complicated soon as the number of straight lines needed grow at a factorial rate with respect to n.
Now we are going to calculate S(4), a member of our desired S(n), so we will redo all that connecting of lines in just the example table, in a more precise attitude. This is the thorough treatment and from now on some things will get more abstract.
To make this post shorter and easier to read and also in order to keep all the information in text format, I have decided not to go deep into the proofs, but left just one proof at the end, to the fact that the conjectured sum I opened with actually converges. I did not want to attach files to avoid suspicion. Some of the missing proofs are provided at my website, I can of course provide any additional proofs for request.
First we give coordinates to all the points. The upper left point of the table with value 1 is always (1,1)=1*1 mod n = 1, and in general the coordinates are (x,y)=x*y mod n. They have inner value of i=x*y mod n.
We then define the multiplication table modulo n as a two dimensional Euclidean flat dotted space (lattice space), where the distance between two close points on the same row or column is 1. In other words Distance from (1,1) to (1,2) = 1. Any units of measurement is good for us for now, one meter, one centimeter, as long as the one is unity and minimal in our space and define a distance.
To illustrate, the length of the line that connects the ones in our example is the square root of 8 = 2.8284...
1 x x
x x x
x x 1
|May-18-11|| ||MrMelad: Stating that the minimal distance equals one is a definition in the manner that multiplication tables modulo n could be investigated with other geometric assumptions, for instance taking a different lattice space or assume a curvature to the table. Moreover, the multiplication table modulo n is used in a semi-unrelated way in the mathematics of both quantum mechanics and general relativity when it is sometimes used as a matrix usually along with matrix equations and operator analysis, Hamiltonians in particular. What is unrelated is that those theories doesn't include the fifth postulate of Euclid and they doesn't connect the dots, so to speak.|
The common aspect of this essay and those theories is that in all we are measuring space.
The multiplication tables modulo n are sometimes called Galois fields in Galois theory and Finite fields in linear algebra when n is a prime number or an exponent of an odd prime.
In contrast to the permutations and equations study which forms the basis to all the known methods to use the multiplication table modulo n, this is a more basic and abstract study of regularity in pure mathematics and the current status is that no one knows all the knowledge needed to complete the investigation, this work lie within the expertise of no one because it requires so many different mathematical approaches along the way, moreover, some of them hadn't been invented yet.
A positive thing that could be said about my research without discrediting anyone is that it includes and depends on all numbers for the modulo, not just prime numbers and exponents of prime numbers like it is usually the case. When it comes to multiplication tables modulo n, mathematicians do everything with it from symmetry and rings to Eigenvalues and Hamiltonians, defined only in exponents of prime numbers (due to Galois theory, and field definitions) and it never jumps from the numbers to the table itself and the geometrical shapes, except for the torus analogy in topology, and that analogy holds for multiplication tables modulo n also when n is not a prime number.
The prime numbers have a very important leading role in this project but it is not the limitation of the modulo.
Knowledge come in many faces and the consequence of all those important sciences using this multiplication table modulo n object was the reason that I chose to analyze it. This and the fact that it had to be introduced to me by one of my teachers Dr. Alex Eizenberg from the Jerusalem college for engineering. It wasn't in the syllabus btw. To illustrate what I think is relevant in today's mathematics, in the topological string theory they are in search of a geometry for space.
However this is not a theory in topology but in geometry and it is not a theory of the continuum but of the discrete.
Its kind of logically sad that quantum mechanics has to be given as an example for a continuum and not a discrete theory in comparison but nonetheless the focus in my research is on the natural and especially on the natural numbers. In this theory, the points of space are discrete themselves and not just the angular momentum and the particles. It goes to say that the limit (x->0) for f(x) = x is 1 when x is a distance approaching zero because x has to be a natural number bigger than zero but also the smallest of numbers.
|May-18-11|| ||MrMelad: Now we return to the MTM4 (Multiplication Table Modulo n=4) and the connecting of dots:|
x x x
x 0 x
x x x
We start with the zero element. It is a single point and we have nothing to do, no straight line to draw. Right now we are only interested in the areas that those shapes are “producing” and in this case our shape is just one point. A point has no area. But, an important thing could be said about this specific point, the zero in the 4th modulo is that it is the only zero, from all the zeros in all the other n<>4 that doesn't “produce” an area, except, and strongly except because it goes visa versa: only if n=4 or n is a prime number, the area of the shape of zero is zero. There are of course elementary proofs to all of this.
Since we “connected” the number 0 in a multiplication table modulo n=4 with all the other zeros and it didn't “produce” an area, We write its area down as S(4,0) = 0. This is our first S(n,i) the area of the shape of i=0 in n=4. Later on we will sum all the areas to produce S(4).
We move to i=1 and go on to start actually connecting some lines, one line to be exact:
1 x x
x x x
x x 1
This is that one straight line which we already encountered in the example on how to measure distance. A straight line, like a point also doesn't “produce” an area. We write this as S(4,1) = 0. We move to the next number:
x 2 x
2 x 2
x 2 x
Now we connect all the twos with a total of 6 lines to form a square with two diagonals. The total area of the square is 2. This is our first area. We write it down as S(4,2) = 2 and move to
x x 3
x x x
3 x x
S(4,3) = S(4,1) = 0.
We can now notice that the shape of the threes is exactly like the shape of the ones, only perpendicular in two dimensions. The ones and the threes are creating an X.
1 x 3
x x x
3 x 1
We make a note that Three and One are perpendicular in Four. A general proven statement is that the shapes of i and n-i are isomorphic and perpendicular in any base n and the equation S(n,i)=S(n,n-i) is true for the areas themselves for any n and 0<i<n. You might want to check it with larger N but the idea is very simple and easy to prove.
We are done with geometry and turn to the numbers to finally get to S(n) itself. S(4) = S(4,0) + S(4,1) + S(4,2) + S(4,3) = 0+0+2+0 = 2. S(n) is the summation of all the individual areas S(n,i) in the MTMn and in our example S(4)=2.
If you went deeper into n=5, n=6 and etc and noticed that S(n,i) is always a natural number give yourself a kodus, but if you are not familiar with Pick's theorem, this might be a mystery so go google it up. If you do know about Pick's Theorem give yourself another kodus, the thing is, almost no one is familiar with Pick's theorem, and that includes ALL the doctors to mathematics which I've asked. I know some people must know about it but I haven't run into one yet. I didn't know about it a long way into the project and I give my kodus to my uncle who told me and I was then able to prove that S(n,i) is a natural number.
|May-18-11|| ||MrMelad: I find Pick's theorem to be the single most important individual fact of flat dotted space (lattice space), on the same level as the Pythagoras theorem yet much simpler. Young children can understand it. Pick's theorem is a natural numbers theory of area-geometry where instead of multiplication of lengths we count points. Its a different way to look at areas since its units are not squared, all the elements in the Pick's theorem calculation are summed, and not multiplied, meaning, no ^2 for areas units. The amount of points P in a shape is always at the same rate of the area A in the two dimensional lattice space, P~A, and of the volume V in a three dimensional shape in lattice space is at the same rate of P as well, V~P. Pick's theorem is a very important aspect of a two dimensional lattice space because it can't be expanded into the third dimension, we can't count points to produce the exact volumes. Pick's theorem will come very handy to anyone trying to deal with the geometrical structure of the MTMn.|
At first I was just playing for fun drawing the shapes occasionally up to n=23 sometimes calculating S(n) until I turned to the computer but I conjectured the following conjecture before the project reached the computer era, so I will give S(n) first up to n=23 and then up to 999. A software and algorithms written in C# at my website.
I also wrote an advanced very different algorithm using python (Appendix 1) which enabled my computer to calculate up to n=3200 in less than a week and this algorithm is provided with the detailed results up to n=999 (Appendix 2). My website can serve as a reference if you want to get deeper into those ideas but the website doesn't portray a complete picture and is out dated (2007).
So for now we have S(4)=2. Notice that we had to make physical measurements to get a numerical value for S(n). It becomes exact only after the physical measurement analogous to the collapse of the wave function in quantum mechanics. The regularity of S(n) doesn't come with a formula but is possible to measure and is what yielded the most important result of this project, the conjecture.
It brings us unexpectedly to many other areas of mathematics with the extraction of one of the fundamental numbers, e, from the MTMn without previous assumptions or known formulas. It turns out to be another case of e taking part in the regularity of prime numbers at the infinite, only this time its in geometry and not just in numbers analysis.
The conjecture (with an accuracy of at least 10^-4):
1/S(4) + 1/S(5) + 1/S(6) + … + 1/S(infinity) = e^1/2 – 1
(real value) 0.648721270700
To give an intuitive perspective about the conjectured value, e^(1/2) - 1, it is the number created from a mix of inverses of all basic mathematical operations - logarithms, exponents, multiplication and addition. Take a calculator, start with 0, inverse lan it twice gives first 1 and then e, inverse square and it will give sqrt(e) and inverse addition using minus one to get the value.
Here is the sum of 1/S(n) up to n=23:
½ + 1/14 + 1/28 + 1/70 + 1/108 + 1/205 + 1/ 334 + 1/484 + 1/616 + 1/892 +1/1138 +1/1511 +1/1882 +1/2418 +1/2908 +1/3556 + 1/3962 + 1/4937 1/5876 +1/ 6926 = (up to 23) 0.647250533571794
This is the point where I conjectured. A 0.001 difference seemed plausible. With the computer aid I was able at first to get results first up to S(400):
1/S(4) + 1/S(5) + … + 1/S(400) = 0.6486208543
S(400) = 59953652
And then up to n=3200
1/S(4) + 1/S(5) + … + 1/S(3200) = 0.648624060519
S(3200) = 33309360592
My first algorithm was brute force oriented - it used pick's theorem itself, counting points on lines and inside shapes to accord with Pick's theorem and produce the area. It is a very low quality algorithm for an n and depends on the gcd.
|May-18-11|| ||MrMelad: To illustrate here is a problem: Given any two points with coordinates (a,b) and (c,d) where a,b,c,d are all natural numbers find the general solution to the number of points that lie exactly on the line that connects (a,b) and (c,d) in flat lattice space. |
1 x x x
x x x x
x x x x
x x x 1
(a,b) = (1,1) , (c,d) = (4,4) ,
In this example, the points on the line that connect (1,1) and (4,4) = 4 which are (1,1), (2,2), (3,3), (4,4).
The answer to the general problem is: Points = gcd(|c-a|,|d-b|) + 1.
In any case, the algorithm to calculate S(n) is available to download at my website.
Since the proof that S(n) converges without producing the actual value is a very interesting and rich topic, I decided to write it in this post, among things it gives the growth rate of S(n). This is a more complete version of the proof from the one at the website:
We know that the absolute maximum area each S(n,i) can “produce” is (n-1)^2 which is the size of the entire space of the MTMn. So the maximum Smax(n) = n*(n-1)^2 = n^3 – 2n^2 + n ~ n^3.
Smax(n,x) = n^2 >= S(n,x).
Smax(n) = n^3 >= S(n)
Since Smax is maximum, 1/Smax(n) produce the minimum sum - we first sum Smax(n,i) to create Smax(n) and then sum 1/Smax(n) to compare to the sum of 1/S(n). The sum 1/Smax(n) is minimal to 1/S(n) because a bigger value is at the denominator.
We have established the minimum for 1/Smax(n) ~ sum of 1/n^3 where n goes from 4 to infinity. This is of course the Zeta function and our minimum sum is Zeta(3) minus the first three elements i.e.: the sum of 1/n^3 from 4 to to infinity ~ 1.202 – 1.162 = 0.03 .
|May-18-11|| ||MrMelad: There is also a maximum for the sum of 1/S(n) and a minimum for S(n) itself which is Smin(n). It is comprised of the the first and last row and column of the table. For instance in the table we examined, the MTM4, the border was:|
1 2 3
2 x 2
3 2 1
All the MTMn share the same regularity of border by proof:
1 2 3 4
2 x x 3
3 x x 2
4 3 2 1
1 2 3 4 5
2 x x x 4
3 x x x 3
4 x x x 2
5 4 3 2 1
Smin(n) is calculated exactly like S(n) only on the border points, by drawing straight lines between identical inner value members and calculating the area that is created. We have simplified the problem to the summing of the areas of rectangles, analogous to the integral operation and old methods of proofs in ancient geometry.
Smin(n)<= S(n) because an area can not get any smaller when adding additional lines.
|May-18-11|| ||MrMelad: The minimum series could be calculated analytically from scratch like it is done in my website but luckily it comes with an easy formula, two actually:|
Remark: Smin(n) is the number of acute triangles made from the vertices of a regular (n-1) polygon.
for n>0 0, 0, 0, 2, 8, 20, 40, 70, 112 the formula is Smin(n) = (n-3)*(n-2)*(n-1) / 3
Or equivalently, we can define Smin(4) = 2, and then subsequently therecursion for n≥5 is:
Smin[n]=Smin[n-1] * (n-1)/(n-4)
We then take the sum of 1/Smin(n) from 4 to infinity = ¾ which is an established maximum. Smin(n,x) <= S(n,x) and 1/Smin(n) >= 1/S(n).
We have established 0.03<=sum of 1/S(n)<=0.75.
With a manual calculation of the first few members we get:
0.64<=sum of 1/S(n)<=0.75.
S(n) converges because the maximum sum of 1/Smin(n) converges and the proof is complete.
We proved that S(n) is in the desired range for the conjectured value and we also proved that the sum of n/Smin(n) converges as well, and as a result we have that the following sum of n/S(n):
4/S(4) + 5/S(5) + … n/S(n)
converges as well. Up to 400 the sum of n/S(n) ~ 2.99 with 10^-2 accuracy.
My article “On the geometry of natural numbers” which is on my website expands those issues in a very technical manner with almost no examples but with an axiomatic approach. It includes some complete proofs to what is stated here and goes much much further in to the realm of the third dimension and beyond. All we have discussed so far was just the two dimensional multiplication table, we can add another axis to create xyz geometrical flat frame of reference where (x,y,z) = x*y*z mod n and compute volumes created by surfaces. The concept and definition of a dimension is also explored in this article.
|May-18-11|| ||MrMelad: All the following are proven statements:
Any two dimensional multiplication table modulo n is symmetric towards the two dimensional center (n/2,n/2). Any three dimensional multiplication cube (3-table) modulo n is symmetric towards the three dimensional center (n/2,n/2,n/2) and in general, any X dimensional multiplication X-table modulo n is symmetric towards the X dimensional center (n/2,n/2,...).X is the number of the actual physical axis. Since its all made of easy calculated numbers we can create it for all dimensions where the old X-1 dimensional table becomes the first layer for the new X dimensional table maintaining self similarity. Its all proven of course and I can also prove thatin an D dimensional MTMn of a base n, all the (n-1) (D-1) dimensional layers are isomorphic to each other when n is a prime number.
An important remark about the the third dimensional MTMn:
Recently I managed to write a code to create graphical files to show the three dimensional shapes and was able to compute just a few volumes, all of which turned out to be rational numbers. I was also able to create three dimensional representations of the “inside” and “outside” of the shapes using a simple algorithm filtering vertexes if they are on the same plain or not, by that dividing the shape into two shapes the inner and outer shapes:
for any two vertexes:
if (x1=y1 or x2=y2 or x3=y)
you are in the inner structure of the 3 dimensional MTMn
you are in the outer surfaces structure, created from the surfaces of the two dimensional MTMn.
An important remark about the D dimensional MTMn:
A lie group is created by taking the continues symmetries of rotation of circles perpendicular between dimensions. The same analogues symmetry appear in the MTMn of general D dimensions, because due to the mod operation, the surface of the table is a map of the lie transformation. For instance, the two dimensional MTMn like the one we explored is analogous to a torus (like in this lecture http://www.youtube.com/watch?v=bGY3... at about 6:50) and a three dimensional is analogous to a rotating torus and etc. And the areas we are measuring are analogous to closed curves on the D-Torus space when the table is formed as one.
I am able to go on for ages about more things that I have to say about MTMn and modular arithmetic in general that I didn't include here or in website the mildest, all of it is proven.
The entire content of this post and website are just the representation of the MTMn in regards to the function f(n) = n, because f(n)=n is the function that creates the numbers on the axis that we are multiplying in the table, creating a multiplication table in our case – 1,2,3,etc * 1,2,3,etc while other continues functions not f(n)=n are sharing many similar symmetries as well, and the action of taking a double sum of areas of S(n) is analog to a partial double integral (another similarity to the wave function).
For example, an interesting and complicated idea is that due to Fermat's theorem, at any base n, if you take the exponent (n-1) on every inner value, all the inner values become 1, and the general S(n) becomes minimal, so it correspond to the function F(n)=i^n-1, where i is the inner value, x*y mod n in two dimensions, x*y*z mod n in 3 dimensions, etc...
Nonetheless I admit my defeat and I know that with all my best intentions I'm not an established expert in the required methods and I am unable to prove alone the main conjecture, people of different and more elaborated expertise should play a role especially number theory experts and it could and should interest physicists, especially quantum and relativity physicists. Combinatorics for instance is mentioned here on the fly and random numbers are created with the modulo operator in computer science. Getting those ideas to professional mathematicians and physicists, as I am not such by profession, will probably work miracles and my arguments might later seem pale and unimportant in comparison.
And it just might be the case that the infinite series converges to a different value then the conjectured value by a margin of 10^-4 and nothing could be said about the connection to the frequency of the prime numbers i.e. lnN as discovered by Gauss and is the subject of the Riemann hypothesis. But just remember the next fact: S(n,0) depends on the primilarity of n, S(n,0)=0 when n is a prime number and never otherwise except for n=4. A complete analytical solution to S(n,i) should include a private case solution for S(n,0), which combined with the proper action should be a very fast way to conclude with a proof that a number is prime.
S(n) is connected to the geometrical roots of modular arithmetic. The very first zero S(n,0), the first element of every S(n) is the regularity of the prime numbers.
|Aug-10-12|| ||MrMelad: a little update regarding those subjects which we discussed, |
If you take a a square of points (like our MTMn), and measure every point's distance to the center (n-1)/2,(n-1)/2, when n is the length of a side of the square and sum up all those distances squared you get
2 * (n ^2) * [(n ^ 2) – 1] / 12
which are the 4-dimensional pyramidal numbers and are also the number of independent elements in the Riemann tensor for general relativity.
This sequence is showing up at http://oeis.org/A002415
|Aug-13-12|| ||MrMelad: A little explanation, in regular graphs in calculus, the geometrical meaning of a solution t to the equation f(x,y) = t is the distance towards the direction of t of the point (x,y,t) from the point (x,y) that is on the plain created by the x and y axis.|
Close examination of the function g(x,y) = x*y mod n in natural numbers flat space on the same perpendicular axis of regular graphs led to the “popping up” of a new set of graphs and perpendicular axis originating from the center of the “square of solutions” (the MTMn) to the above function g(x,y). Those ideas were "summed up" in this project with S(n,i) and S(n)
That last piece of information that I posted shows how given a base n, the additional axis system, that is the center of the “square of solution points” on the regular axis, exists for every function f(x,y) = t and make it possible to alternate the geometrical meaning for the solution of the function f(x,y) = t to either
1. the distance of the solution point (x,y,t) to the (0,0,..n times..,0) point – which is what we call calculus
2. Given number of dimensions n, the distance of the solution point (x,y,t) to the center point of the “square of solutions” (n/2,n/2,...)
My article “On the philosophy of natural numbers” showed in a sense that the Lorentz transformation (special theory of relativity) is equivalent to the mod operator, and it chronologically preceded my analysis to the MTMn. Moreover, this was a result in physics and my analysis started on exactly this physical assumption, that the additional axis system which is used in the special theory of relativity (the other inertial system and its relation to c – the speed of light) is the mod operator.
Later I showed another physical result in the connection of those exact MTMn (and thus the mod operator) to torus geometry and membrane space which pretty much gives the MTMn geometrical basis for string theory
The general theory of relativity, an inherent of string theory, is a field theory and fields can be geometrically imagined using arrows on points in an axis system, and my last result is exactly about fields and looks like an attraction field if you draw the arrows - you take a square of points and measure the distance of every point to the center point of that square, sum those squared and you get the number of independent elements in the Riemann tensor. Notice that in this result we haven't defined the function, we measure general coordinates, so our result, as anticipated, doesn't tell us the structure of the elements of the Riemann tensor, but for this we have a model in which our number of independent elements match those of general relativity like a glove – the MTMn
|Aug-23-12|| ||MrMelad: In regular graphs and vector systems if one take a point (x,f(x)), draw a line from this point to the (0,0) origin, and measure the area between that line and the x axis, one finds out that the area is equal to A = x*f(x)/2. The result is a right sided triangle with projection on the x axis X=x and projection on the y axis Y=f(x) which area is A.|
If we assume another axis system originating from the center point of the "square of solutions" (n/2,n/2), then it has x' axis and y' axis in the same directions as x and y. If instead of drawing the vector from the point (x,f(x)) to the (0,0) origin we will draw it to the (n/2,n/2) origin' and measure the area between that line and the x' axis, one finds out that the area is equal to A' =[x*f(x) - n*(x+f(x))/2 + (n^2)/4] / 2. The result is a right sided triangle with projection on the x' axis X'=x-n/2 and projection on the y' axis Y'=f(x)-n/2 which area is A'.
the sum from x=1 to n of (x^2 - n/2) = (2n^3 + n)/3 which are the Octahedral numbers
|Apr-13-19|| ||MrMelad: Continued discussion from Gashimov Memorial (2019)|
<Pedro Fernandez: But now I want to refer to your work in geometry of numbers, in particular the multiplication table modulo n. I have read briefly your posts at respect and it seem some interesting due its novelty.>
It's hard to search the whole internet or tell what's going on in other people's minds but I've yet to find a single person that is charting those multiplication tables like me. I've always took comfort that if anything my research is at least creative. I'm glad you recognized that.
<BTW, I never heard about that Pick's theorem.>
Ha ha! That's not surprising that you haven't, few people have, but it's always been source of mystery to me, why is it that this theorem isn't more popular?
<Anyway, I did realize you found a discrete distribution S(n), so I wonder if you did a discrete Fourier analysis as I think it could be quite interesting.>
Yhea I have tried with my limited ability but couldn't find anything I thought was of value or that could advance me towards a formula. But I'd welcome any idea you have!
I did, however, expressed the "mod" operator with a Fourier series.
Here is that analysis pasted from my personal archive:
<The modulo operator has a Fourier transformation representation.>
This notation is derived from the Fourier expansion to the function for the fraction part of a number. It is used to derive a Fourier series expansion for x mod 1 for real x and then a general expansion for x mod n.
x mod n = n/2 - n/π * (∑_(k=1-∞)sin(2πxk)/k)
The ∑_(k=1-∞) is read as "sum of from k=1 to infinity". Without latex notation that's the best I can do here...
|Apr-13-19|| ||MrMelad: Anyways, my dear <Pedro Fernandez>, if this stuff interests you let me share with you some more points from my personal archive that I found interesting. I've halted the research sometimes in 2014 so those things might not be up to date or relevant. They might also not be true or even interesting. Apologies if that's the case! |
<A. The Riemann tensor (also written in my Aug-10-2012 post)>
My model illustrates very clearly the importance of the center point (n/2, n/2) to this geometrical construct. Every solution point p has an opposite solution point p' such that the line that connects them passes through this center point.
To understand how this center point with relation to “squares made of points that are solutions to equations” is connected to string theory, one must realize the following fact which I have discovered using an excel sheet and the online library of integer sequences.
In basic lattice space, if we take a square of n^2 points and measure every point's distance to the center point (n/2,n/2) and sum up all those distances squared we would get
(n^2) * [(n^2) – 1] / 6
which are exactly twice the number of distinct components of the Riemann curvature tensor.
<B. Torus topology and rotations of perpendicular dimensions>
To understand this connection one should consider multiplication cubes modulo n and multiplication D-tables modulo n where inner cells become voxels or n-cells instead of pixels and every inner value is equal to x*y*z*... mod n.
If we move along the x axis of the two dimensional multiplication table mod n we notice that after n steps in the right or the left direction we return to our origin point. So the x axis forms a closed loop, if we moved up and down in the table this was also true for the y axis which makes the multiplication table modulo n a torus and each row and column a closed loop (string).
If we extend this analogy, now we are taking about a D-table where each inner value is x*y*z*...*Xd mod n and do the same, ie connect with straight lines every same members, we can treat the resulting topology as a (D-1)-torus (and also construct inner torses from the S(n,i) objects which are positioned exactly 45 degrees to the torus I described but this is much too complicated subject to put forth in this post more than this intuitive hint, if you use scissors to cut those S(n,i) shapes, you can sometimes roll them to form a cylinder. The angle at which you should roll is 45 degrees to the angle needed to create the first cylinder from the entire square of points), and if we keep track of the border of our areas S(n,i) we discover that those closed polygons with deterministic area become closed continues loops around the surface of the D-torus knotting through its center without deterministic area (the area depends on the curvature).
I think that the same idea of symmetries of continues rotations of circles between dimensions is used in definitions for lie algebra.
<C. The Fourier transformation and the Lorentz transformation>
Another expansion for the mod function can be derived by dividing the unit circle to angles of multiples of pi/2n:
Let g(x) be the function:
g(x)=x mod n,when 0≤ πx/2n< π/2 or π≤ πx/2n< 3π/2
g(x)=n-(x mod n),when π/2≤ πx/2n< π or 3π/2≤ πx/2n< 2π
We can look at the Lorentz transformation as cos(a) in this layout:
v / c = cos(a)
sin(a) = 1 – (v^2/c^2)
where v is velocity and c is the speed of light
in which case x of g(x) becomes the x,y,z components of the velocity in 3 dimensions and we can use modular arithmetic velocities (angular momentum?) in respect to the speed of light and some base n. Interestingly enough, since speeds beyond the speed of light are prohibited, g(x) of real components of velocity will be confined to the positive quarter of the unit circle and will always be equal to x mod n (and not n-(x mod n)).
|Apr-13-19|| ||MrMelad: <D. A highly complicated intuitive connection to the holographic principle - part 1>|
To understand this connection one should remember Pick's theorem, which is used to calculate the area of polygons in lattice space by counting points on the boundary of the polygon and points inside the closed area “formed” (the “formed” used here has exactly the same meaning as the “formed” in the previous definition to the model) by this polygon.
There is a proof that such formula does not exist for the third dimensions, ie we can not count points to produce exact volumes. This fact hints to us that volumes are not fundamental to a physical world made of pixels/voxels/n-cells. If we assume that both areas and points are fundamental units that can not break even inside the event horizon of singularity, that might give intuitive (or even qualitative somehow) understanding of the two dimension quality of the event horizon of a singular object where all other dimensions break apart except the two dimensional lattice which can not break. This can also explain why black holes expand in volume when they consume energy – this is because the two dimensional lattice which is the surface of the event horizon can not break to smaller pieces (and become less thick) and thus have to expand outwards with new particles (points), not surprisingly in a formation symmetric with relation to its D dimensional center point.
If n is a prime number, there are no permutations and no “holes” inside the N-table, so in a D dimensional space, all cutting of a complete row (from the start of the table to its end), or complete surface, or complete volume, or complete N-area is isomorphic (only in geometrical shape, the numbers in each row and column, etc.. are different) to any other complete cutting of a similar dimensional row or surface, or etc... Moreover, every D dimensional cutting is self similar (has the same overall regularity) to a D-1 cutting and a D+1 cutting, which means it has infrastructure to interact and merge with less and more dimensional parts.
The following analogy might stretch this point even further
Consider making all connections of straight lines in a given D dimensional multiplication table mod n and considering that its only possible to be located on a straight line and on a discrete point and that we can “travel” only through the straight lines to other points directly in their path.
|Apr-13-19|| ||MrMelad: <D. A highly complicated intuitive connection to the holographic principle - part 2>|
Notation: The term CcutD(n) (C cut D) is used to describe a C dimensional complete cutting of a multiplication D-table mod n where each inner value is equal to x*y*z*...*Xd mod n
Consider the cutting CcutD(n) where C<D and D>=3 of different but closely located very large numbers n's.
If n is very large and the straight lines themselves (not to be confused with the size of the cell which might be much bigger) are very thin (maybe planck size), we would be able to visually see all the different CcutD - as 3 dimensional surfaces (even if C and D are larger than 3) of almost the same size and shape but with completely different attributes and symmetries.
Though they all have almost completely the same shape when looking at them from large distances, close inspection reveal major changes in most localities. Close and around the center of each object we could see a big almost perfect sphere (but not perfect if looking very closely), a polygon which extrapolation reaches exactly the limit of pi when n is infinite, where the lines density is maximum and the lines form an inpenetrateable spherical barrier. Exactly at the center point there will be absolutely nothing if n is odd. Its inner value is non existent if n is odd and is equal to ((n/2)^2 mod n) when n is even (Electric charge?).
Around the edges of this object things could only exist in discrete points in what would appear to be D dimensional orbits around a D dimensional center.
The vast majority of this object will consist of empty space (the lines are thin, the points are of no size and the inner values are invisible).
Those objects will consist of sub-objects that are larger dimensional S(n,i) that when going over i=1,2,...n in that order (time?) those objects will appear to be spinning around the D dimensional center. Due to the self similarity of multiplication tables mod n it would appear they are also spinning around their 3 dimensional center as well as the 2 dimensional and even 1 dimensional (each row is a loop = string) if n is a prime number. If n is not a prime number they will appear to spin around additional centers (n/x for instance if n is divisible by x).
S(n,i) is roughly asymptotically ~ (n/2)^2, and the amount of straight lines required for the object grow at factorial relation with respect to both the amount of solutions to the Diaphanous equations and D and C, and the total number of points initially needed for bigger n grow with respect to n^D.
The difference in our S(n) measurement between different n – which is the sum of the two dimensional orbits which are also the sum of the possible states (S(n,i) when i=1..n-1) - would be noticeable when measuring different n's (but the radius, n/2 would be almost identical from one object to another)
When n gets larger and we try to push this object there would be a lot more points and lines which would might cause it to be more difficult for us to push. We would call this thing resistance to acceleration and identify it with the mass.
To increase the n of an object we would need to insert into it more points - exactly (n-1)^D +1 new points, and we would probably need to insert them all at once (time period of 1/n). If n is large this amount of new points is quite large as well (E=mc^2?). decreasing n by 1 unit would cause n^(D-1) points to collapse to new D-1 dimensional relative formations and (n^D - n^(D-1)) points will roam free unless (n^(1/z)) is a natural number (where 1 < z < D is some natural number) in which case those points will arrange in a new formations of base z (Radioactive decay?).
A very elegant way of looking at the rest mass of any object is indeed exactly counting the number of space points (particles) the object is generally made of, or summing the number of possible states it can be at.
The rest mass remains depended on the original amount of space points and this quantity is conserved and could never change as long as n doesn't change.
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