Doctordick 42,287 Posted November 7, 2006 Author Report Share Posted November 7, 2006 Hi turtle, Do to the extended lack of action on the “philosophy of science” forum, I reviewed the “Philosophy of Science” thread. I noted a minor misunderstanding in your perception of the purpose of my proof. There is no real planed application of the proof, I hold that the proof itself is the profound consequence. Read the last two paragraphs of the original post again:Thus it is seen that absolutely any three dimensional universe consisting of n+1 points for n greater than four can be seen as an n dimensional equilateral polyhedron with unit edges projected on a three dimensional space. That any means absolutely any configuration of points conceivable. Talk about "emergent" phenomena, this picture is totally open ended. Any collection of points can be so represented! Consider the republican convention at noon of the second day (together with every object and every person in the rest of the world; and all the planets; and all the galaxies ...) where the collection of the positions of all the fundamental particles in the universe is no more than a projection of some n dimensional equilateral polyhedron of unit size on a three dimensional space. Talk about emergent phenomena! On top of that, if nothing in the universe can move instantaneously from one position to another, it follows that the future (another distribution of that collection of positions of all the fundamental particles in the universe) is no more than another orientation of that n dimensional polyhedron and the evolution of the universe in every detail must correspond to continuous rotation of that figure. Think about that view of a rather simple geometric construct and the complex phenomena which is directly emergent from the fundamental perspective.Certainly anyone’s picture of the current state of the universe (including all the people, all the letters, all the books, all the movies,…, all of everything you can conceive of) can be seen as an extremely large collection of points distributed in a three dimensional space (most of the points would be simple electrons, protons and neutrons). Let’s say a number of points amounting to something around ten to the fiftieth power would certainly suffice; that volume of information produces no problems at all in the abstract. Thus, an excellent explanation of the universe is, it IS “a rotating n dimensional equilateral polyhedron with unit edges projected on a three dimensional space”. The evolution of that universe can be nothing more than the defined rotation of that polyhedron. It is not too difficult to show that simple conservation of momentum in each and every component of that defined rotation yields exactly the common laws of physics for the apparent relationships between collections of those projected points. In fact, that is exactly how I happened to stumble upon the proof. Have fun -- Dick Quote Link to post Share on other sites

Turtle 389,602 Posted December 1, 2006 Report Share Posted December 1, 2006 Hi turtle, Do to the extended lack of action on the “philosophy of science” forum, I reviewed the “Philosophy of Science” thread. I noted a minor misunderstanding in your perception of the purpose of my proof. There is no real planed application of the proof, I hold that the proof itself is the profound consequence. Read the last two paragraphs of the original post again:Certainly anyone’s picture of the current state of the universe (including all the people, all the letters, all the books, all the movies,…, all of everything you can conceive of) can be seen as an extremely large collection of points distributed in a three dimensional space (most of the points would be simple electrons, protons and newtrons). Let’s say a number of points amounting to something around ten to the fiftieth power would certainly suffice; that volume of information produces no problems at all in the abstract. Thus, an excellent explanation of the universe is, it IS “a rotating n dimensional equilateral polyhedron with unit edges projected on a three dimensional space”. The evolution of that universe can be nothing more than the defined rotation of that polyhedron. It is not too difficult to show that simple conservation of momentum in each and every component of that defined rotation yields exactly the common laws of physics for the apparent relationships between collections of those projected points. In fact, that is exactly how I happened to stumble upon the proof. Have fun -- Dick Hi Doc, My lack of response is no indication of my lack of interest, rather of everyday distractions coupled with trying to form up an adequate perspective of all we have discussed. While you say there is no planned application, do you suspect there may be an emergent application? That is to say if I came to understand everything you have explained, might I (or someone else) discover an application? We talked somewhere about the improvement in probabilities under your description, and I started to wonder if this improvement is in reference to 'we' as individuals during a lifetime, or 'we' as a species still extant? As I reread your last post here I also came to wonder if the rotation of the n-dimensional equilateral polyhedron repeats? I hope I haven't drifted too far afield again and that I have formed some applicable questions.Best Cheerful Regards,Turtle :) Quote Link to post Share on other sites

Doctordick 42,287 Posted December 2, 2006 Author Report Share Posted December 2, 2006 Hi Turtle, Actually, I was somewhat surprised by your lack of response but I can well understand the burden of everyday distractions.While you say there is no planned application, do you suspect there may be an emergent application? That is to say if I came to understand everything you have explained, might I (or someone else) discover an application?Oh, I think that is not only possible but probably quite probable. The central issue of it all is the fundamental foundation of thought itself. (I am, at the moment, composing an essay on how my stuff impacts the fundamentals of philosophy as presented by the argument between Kant and Hume, though I suspect few will follow my thoughts). The issue central to this thread is the realization that a three dimensional projection of an n-dimensional polyhedron can appear to be exactly what is seen by us in our supposed three dimensional universe. I doubt there are many people who can comprehend the true complexity of a projection of an n-dimensional polyhedron. As I have said somewhere else, I discovered that proof when I carried my solution technique to higher dimensions. If one begins with my original fundamental equation (deduced in "A Universal Analytical Model of Explanation Itself"), there exists an attack on that equation which will yield an approximate solution for the probabilistic behavior of any single element taken to be the only unknown phenomena in the universe. That approximation turns out to be exactly Schroedinger's equation in one dimension. That result can be simply expanded into a three dimensional version via the simple idea of regarding the collection of elements to be represented by three independent sets. Relativistic effects arise because of the need for an additional dimension (tau). Taken together these things encourage one to look at higher dimensional representations of that equation. If one takes all elements to be independent of one another (the data thus being represented by one point in an n dimensional universe), the resulting fundamental equation takes the following rather simple form: [math] \left( \nabla^2 + g^2® \right) \vec{\Psi} = -K\vec{\Psi}[/math] where [math]\nabla^2 = \sum_{i=1}^n \frac{\partial^2}{\partial x^2}[/math] which is the n dimensional generalization of the Dalembertian operator. This equation is quite straight forward to solve and is, in fact exactly the quantum mechanical solution for a rotating n-dimensional equilateral polyhedron. That implies that the universe may be represented by a rotating n-dimensional equilateral polyhedron; thus arose my interest in proving that fact. As I reread your last post here I also came to wonder if the rotation of the n-dimensional equilateral polyhedron repeats? I hope I haven't drifted too far afield again and that I have formed some applicable questions.Clearly, the rotations must be quantized solutions thus the total solution of the universe can be expressed as a set of n quantum numbers. It follows that, if sufficient time is allowed, it will certainly repeat itself ; however, if time is to be defined in terms of the most rapidly changing orientation, one must wait until the slowest has completed one cycle and all other rotations are in the same phase as when the example started (if all quantum numbers happened to be prime, the relative time would be on the order of the product of those n prime numbers). It pretty well follows that a repetition is quite a rare event and is seriously dependent upon the particular quantum state of the system (wobbles in n-dimensions can be quite complex). We talked somewhere about the improvement in probabilities under your description, and I started to wonder if this improvement is in reference to 'we' as individuals during a lifetime, or 'we' as a species still extant?I am sorry but I don't recall the reference. But needless to say, if the probabilities of your expectations do not obey quantum mechanics, I would venture they are going to be found erroneous no matter what the subject. Have fun -- Dick Quote Link to post Share on other sites

Bombadil 14,583 Posted June 7, 2008 Report Share Posted June 7, 2008 There is an interesting corollary to the above proof. Notice that the rotation specified in the final paragraph changes only the components of the collection of vertices along the x axis and the nth axis. All other components of that collection of vertices remain exactly as they were. Since the order used to establish the coordinates of our polyhedron is immaterial to the resultant construct, the nth axis can be a line through the center of the polyhedron and any point except the first and second (which essentially establish the x axis under our current perspective). It follows that for any such n dimensional polyhedron for n greater than three (any x projection universe containing more than four points) there always exists n-2 axes orthogonal to both the x and y axes. These n-2 axes may be established in any orientation of interest so long as they are orthogonal to each other and the x,y plane. Thus, by construction, for any point (excepting the first and the second which establish the x axis) there exists an orientation of these n-2 axes such that one will be parallel to the line between that point and the center of the polyhedron. Any rotation in the plane of that axis and the y axis will do nothing but scale the y components of all the points and move that point through the collection, making no change whatsoever in the projection on the x axis. How I’m understanding this is that for the case of projecting a 3 dimensional object in a three dimensional plane (or a 2 dimensional object in a 2 dimensional plane) or projecting any size tetrahedron on a plane with more then one dimension, that you are rotating it in a plane that is perpendicular to all planes except one plane which has one point in it that is on the tetrahedron (would this be a plane with one more dimension then the dimension of the tetrahedron)? Another thing that seems possible is that you are using some of the points of the tetrahedron as the axis’ and not counting them as points in the plane that the tetrahedron is being projected on. Now, would I also be correct in understanding that you are defining the unit vector separately for each dimension of the plan that you are projecting the points onto? Is there anything that would keep us from generalizing this so that we could project an N dimensional tetrahedron onto a plane made up of any number of dimensions? I’m really not sure of the full relevance of this proof. It is an interesting proof and what it seems to do is show that any set of points can be seen as a set of wave equations and a scaling factor or at least a set of sine functions that describe the position of an N dimensional tetrahedron projected on a plane. In your last post in this topic you seem to suggest a connection between this proof and the fundamental equation and possibly relativistic effects but you don’t seem to say much about how it comes about. One of the links there leads to a page where you had been discussing the fundamental equation. I haven’t had a chance to take a good look at it but it looks like you didn’t get much past where you have already gotten to in the thread “What can we know of reality?” so it might not be worth trying to go into the details that you go into there but rather just wait until we get there. Quote Link to post Share on other sites

Doctordick 42,287 Posted June 10, 2008 Author Report Share Posted June 10, 2008 Hi Bombadil! You have surprised me. You have obviously taken the trouble to examine my proof; something no one else seems to have done; however, you don't seem to have much understanding of analytical geometry. How I’m understanding this is that for the case of projecting a 3 dimensional object in a three dimensional plane (or a 2 dimensional object in a 2 dimensional plane) or projecting any size tetrahedron on a plane with more then one dimension, that you are rotating it in a plane that is perpendicular to all planes except one plane which has one point in it that is on the tetrahedron (would this be a plane with one more dimension then the dimension of the tetrahedron)?Size and dimensionality are entirely different concepts. Size has to do with length measurements and dimensionality has to do with the geometry used to display the characteristics of an object. I do not know if you understand the original proof or not, That proof has to do with a distribution of points on a line (a line is a one dimensional entity: i.e., only one variable is necessary to specify a position on a line).The assertion which will be proved is that every conceivable distribution of points on the line segment is achievable by a specifying a particular rotational orientation of the polyhedron relative to the line segment of interest.The “projection” of a vertex of that polyhedron on a line can be seen as the shadow of that vertex caused by a light shining in a direction perpendicular to the line. The analytical display must always be done in a space having a dimensionality equal to or greater than the polyhedron under consideration; an idea which seems to be absent from your comments. “... projecting a 3 dimensional object in a three dimensional plane ...”. Projection is always onto something, not in a thing and a plane is a two dimensional object: i.e., it requires two variables to specify a position on a plane. The object being projected is an “n” dimensional equilateral polyhedron. The term “n dimensional” means that it requires “n” numbers to specify a position of a point in the space being used to display the object. Equilateral means all the edges of the object under examination have the same length (that is a reference to the size of the object, not the dimensionality) and polyhedron means that it has multiple faces but a face is a plane so the there can be no faces until n exceeds two: i.e., the common meaning of the term “polyhedron” really isn't applicable until we get into at least three dimensions. To review, the case “n=0” is a zero dimensional object and the number of points necessary to describe a zero dimensional object is “one”: i.e., it is simply a point. The projection of a point on a line is “one point”. We can draw a picture of that: draw our line and one point not on that line and show how a light shining perpendicular to that line can produce a shadow of that point on the line. The point, plus the line on which the projection is to be made, define a plane and the projecting light becomes a line in that plane. The case “n=1” is a one dimensional object and the number of points required to define a one dimensional object is “two”: i.e., the object is a line and the end points of that line define it. The projection of the endpoints of that line on a different line consists of two points. The distance between the two points is equal to the length of the original line if and only if both lines lie in the same plane; however, if the universe is the projection, there is no way (within that universe) of defining length other than as the distance between the two projected points. Since there is nothing else to which one can compare that distance, that length is pretty meaningless. We cannot actually draw a picture of this circumstance because all pictures lie in a plane and the construct we have described requires a three dimensional space (the two lines need not lie in the same plane); however, if we require the lines to lie in the same plane (in the same two dimensional space) we can draw the circumstance. The case “n=2” is a two dimensional object. At this point the adjectives minimal and equilateral have meaning. Minimal means it is the simplest two dimensional object possible (that would be a triangle) and equilateral means all three legs of that triangle have the same length. If we require the line on which those three points defining that triangle are projected also lies in the same plane as the triangle, we can again draw a picture of this circumstance. The case “n=3” is a three dimensional object. Life has now become complex as, since all pictures lie in a plane, we cannot draw this object. The best we can do is to draw a picture of the projection of this object on a plane. We are, at this point, quite lucky because our brains have come to the conclusion that the universe we live in is describable as a three dimensional space. In fact our subconscious is actually quite agile at reconstructing three dimensional mental image of objects from two dimensional projections. The process works both ways, we are also quite agile at creating the correct projections of three dimensional objects onto a two dimensional space. (What I am saying is that, not only can we interpret what is being represented in a common picture but we can in fact draw such pictures.) So I have “drawn a picture” (a two dimensional representation) of the case n=3. That is the only case where the minimal equilateral object is a tetrahedron (a common object of classical geometry). The case n=4, for which the object has five vertexes, ten edges and five tetrahedral boundaries between the inside and outside of the object, is essentially beyond the comprehension of our subconscious mind. We can only reconstruct its properties via logic together with mathematical reconstruction of its elements and that is the very subject of “analytical geometry” Summarizing: n=0 The object is a point. There is no such thing as “inside and outside”n=1 The object is a line. It has two boundaries between “inside the line” and “outside the line” and those boundaries are called end points of the line.n=2 The object is a triangle. It has three boundaries between “inside the triangle” and “outside the triangle” and those boundaries are called edges which consist of lines.n=3 The object is a tetrahedron. It has four boundaries between “inside the tetrahedron” and “outside the tetrahedron” and those boundaries are called faces which consist of triangles. Each face has three edges which yield a total of twelve edges; however, all edges are shared with one of the other edges thus there are only six actual edges.n=4 The object is ??? as far as I know no one has ever named such a thing. I do know it has five boundaries between its “inside” and its “outside” and that these boundaries are also unnamed (though I have heard them called “solids”) and they consist of tetrahedrons. Each “solid” has four “faces” which yield a total of twenty faces; however, all faces are shared with one of the other solids thus there are only ten actual faces. They also posses ten edges and five vertexes. Though I can work out their characteristics analytically, I cannot picture such a thing; it is purely beyond my comprehension. In my life I have met many people who insist they can visualize such objects but I really do not believe them. It seems to me that, if they can indeed visualize these objects, they should be able to count the number of embedded subparts without analytical analysis and I have met no one who can clearly “count” these parts. My whole reason for this discourse was to point out the necessity to understand analytical geometry before trying to comprehend my proof. Is there anything that would keep us from generalizing this so that we could project an N dimensional tetrahedron onto a plane made up of any number of dimensions?Considering the confusion expressed in this statement, I will presume what you mean is, “Is there anything that would keep us from generalizing this so that we could consider an n dimensional minimal equilateral polyhedron projected upon an n dimensional space”. That would serve us no purpose as it is no more than the minimal equilateral polyhedron itself.I’m really not sure of the full relevance of this proof. It is an interesting proof and what it seems to do is show that any set of points can be seen as a set of wave equations and a scaling factor or at least a set of sine functions that describe the position of an N dimensional tetrahedron projected on a plane.You have apparently misconstrued the proof entirely. What is proved is that absolutely any collection of points in a three dimensional universe (i.e., the universe we actually see) can be seen to be no more than the projection of a minimal equilateral polyhedron projected upon a three dimensional space. A rather simple minded TOE so to speak.In your last post in this topic you seem to suggest a connection between this proof and the fundamental equation and possibly relativistic effects but you don’t seem to say much about how it comes about. One of the links there leads to a page where you had been discussing the fundamental equation. I haven’t had a chance to take a good look at it but it looks like you didn’t get much past where you have already gotten to in the thread “What can we know of reality?” so it might not be worth trying to go into the details that you go into there but rather just wait until we get there.Yeh, I think it is best to put this stuff off for a while. Try to understand the fundamental proof first and then we can go on to the corollary as the corollary proof depends very strongly on issues within the original proof. Meanwhile, if you want to understand the central issue of special relativity, you might look at post #96. Have fun -- Dick Quote Link to post Share on other sites

ughaibu 30,835 Posted June 11, 2008 Report Share Posted June 11, 2008 n=4 The object is ??? as far as I know no one has ever named such a thingIt's called a pentachoron or 4-simplex. Quote Link to post Share on other sites

Doctordick 42,287 Posted June 11, 2008 Author Report Share Posted June 11, 2008 It's called a pentachoron or 4-simplex.Thank you -- Dick Quote Link to post Share on other sites

Bombadil 14,583 Posted June 18, 2008 Report Share Posted June 18, 2008 Hi Bombadil!The “projection” of a vertex of that polyhedron on a line can be seen as the shadow of that vertex caused by a light shining in a direction perpendicular to the line. The analytical display must always be done in a space having a dimensionality equal to or greater than the polyhedron under consideration; an idea which seems to be absent from your comments. Wouldn’t this also be equivalent to finding the dot product of the point we wish to project on the line with a unit vector along the axis we chose to project on. While this seems that it will project the points onto a line it also seems that this is not going to give us the same result as what you have arrived at in that this will only give it in the same length measurement as the coordinate system that the tetrahedron is in. While we could then divide this result by the length of the unit vector in this coordinate system this seems like it is not staying to the idea of treating the projection like its own universe simply because it defines the origin as the location of the center of the tetrahedron. So is there a place in which the origin of such a coordinate system would be naturally placed or is distance all that can be defined. It seems to me that distance is all that can be defined. The case n=4, for which the object has five vertexes, ten edges and five tetrahedral boundaries between the inside and outside of the object, is essentially beyond the comprehension of our subconscious mind. We can only reconstruct its properties via logic together with mathematical reconstruction of its elements and that is the very subject of “analytical geometry” Am I correct in understanding that a boundary of an n dimensional tetrahedron is a plane of dimension n-1 that passes through n-1 of the points on the tetrahedron? Also what is the minimum information necessary to uniquely define an n dimensional tetrahedron or to uniquely define its projection on a line? Quote Link to post Share on other sites

Doctordick 42,287 Posted June 18, 2008 Author Report Share Posted June 18, 2008 Wouldn’t this also be equivalent to finding the dot product of the point we wish to project on the line with a unit vector along the axis we chose to project on.The concept of a “dot” product applies only to vector operations. The “dot product of the point” is a meaningless phrase. Now what you may have in mind is the fact that the dot product between the projection line (what I referred to as a beam of light providing a shadow of the point) and the axis we chose to project on must vanish (those lines must be orthogonal).While this seems that it will project the points onto a line it also seems that this is not going to give us the same result as what you have arrived at in that this will only give it in the same length measurement as the coordinate system that the tetrahedron is in.You are missing the very important point that distances “in a universe” must be defined via some mechanism using elements which are part of that universe. That is, in order to define “a distance” in your universe, you need to have some structure within that universe to provide a basis for the definition. If your universe consists of one point, distance is an indefinable concept. If you have a universe consisting of two points, you certainly can define “a distance” as the separation between those two points but such a thing has little meaning as, having used those points to define distance, you are left with nothing to measure. Only when your universe contains three or more points can you define a usable “distance”. You could, for example, define your standard unit of measure to be the distance between two specific points. With that definition, you could specify the position of the third point (on the projection line) in terms of that defined distance and the positions of the given two points.It seems to me that distance is all that can be definedWith two points, you can define a unit distance on the projection axis to be the distance between those two points. Now, having defined a unit distance, you can define your origin to be half way between the two points. If you have three points, you can define a unit distance on the projection axis to be the distance between a specific pair. Then you can use that distance as a measure and specify the distance between the third point and each of the distance defining points. Define the origin as the point where the average distance to the three points is zero. What I am trying to point out is that, if we have two or more points, we can define a distance and an origin. Am I correct in understanding that a boundary of an n dimensional tetrahedron is a plane of dimension n-1 that passes through n-1 of the points on the tetrahedron? A tetrahedron is a three dimensional object: i.e., an n dimensional tetrahedron is not a defined object. That is why I referred to the object of interest as a minimal equilateral n dimensional polyhedron. My specialty is nuclear structure theory (and that is as of forty years ago) and not n-dimensional analytical geometry thus my knowledge of geometric terms is limited. According to ughaibu a four dimensional instance is called a pentachoron or 4-simplex (ughaibu, I think you meant to say “a pentachoron”). If that is the case, then it seems that the object of interest in this proof would be called an “n-simplex”. I would be happy if we were to adopt that convention. If we accept that convention, the boundary of an n-simplex is an (n-1)-simplex. Unless you are defining a “plane of dimension n-1” to be an (n-1)-simplex”, your understanding would be flawed: i.e., a plane is ordinarily considered to be a two dimensional object. I don’t think that is the real problem here. I suspect the real problem is one of vocabulary. We need reference terms the meanings of which we can agree on.Also what is the minimum information necessary to uniquely define an [n-simplex] or to uniquely define its projection on a line?The minimum information necessary to uniquely define an n-simplex is the value of n. The minimum information necessary to define the projection on a line is the orientation of the line relative to the orientation of the n-simplex. Yon need to understand n-dimensional analytical geometry. I don’t know where to send you or what books to recommend because everything I know I learned from math teachers and not from books. Have fun -- Dick. Quote Link to post Share on other sites

ughaibu 30,835 Posted June 18, 2008 Report Share Posted June 18, 2008 ughaibu, I think you meant to say “a pentachoron”Yes, I think so too:It's called a pentachoron or 4-simplex.Do you mean it's better to say "a 2-sphere" rather than "a circle"? Quote Link to post Share on other sites

Doctordick 42,287 Posted June 18, 2008 Author Report Share Posted June 18, 2008 Yes, I think so too:Do you mean it's better to say "a 2-sphere" rather than "a circle"?That is not the issue. When it comes to a proof which is valid for all n, giving an English name to every possibility of interest is ridiculous. English is very small basis within which to talk about any realistic structure of interest. It is much better to use a name such as n-simplex when you realize n can be on the order of [imath]10^{80}[/imath]. (Arthur Eddington's best estimate of the number of protons in the universe was [imath]1.575\;x\;10^{79}[/imath]). It should be quite clear to you that the complexities of an explanation of the universe in mathematical terms can easily exceed anything which could ever be explained in English prose. That is the issue being discussed here. Have fun -- Dick Quote Link to post Share on other sites

ughaibu 30,835 Posted June 19, 2008 Report Share Posted June 19, 2008 When it comes to a proof which is valid for all n, giving an English name to every possibility of interest is ridiculousBut, you were listing english names:n=0 The object is a point. n=1 The object is a line. n=2 The object is a triangle. n=3 The object is a tetrahedron. n=4 The object is ??? as far as I know no one has ever named such a thing.and I'm not presenting a proof. Quote Link to post Share on other sites

Doctordick 42,287 Posted June 19, 2008 Author Report Share Posted June 19, 2008 Do you mean it's better to say "a 2-sphere" rather than "a circle"?Use whatever terminology you like. All that is important is that the two parties agree as to the meanings. Have fun -- Dick Quote Link to post Share on other sites

Bombadil 14,583 Posted June 25, 2008 Report Share Posted June 25, 2008 If we accept that convention, the boundary of an n-simplex is an (n-1)-simplex. Unless you are defining a “plane of dimension n-1” to be an (n-1)-simplex”, your understanding would be flawed: i.e., a plane is ordinarily considered to be a two dimensional object. I don’t think that is the real problem here. I suspect the real problem is one of vocabulary. We need reference terms the meanings of which we can agree on. This is not quite what I was thinking of although this is what I think it would turn into in the case of an n-simplex. What I was thinking of is a hyperplane of dimension n-1 that passes through the maximum possible number of points (which can be only (n-1) points) of course only the area of this hyperplane that is inside of the points is of interest in this case a (n-1)-simplex. The result will be an n dimensional equilateral polyhedron with unit edge which will project to exactly the same distribution of points obtained from the previous n-1 dimensional polyhedron with one additional point at the center of our reference line segment. If our n dimensional polyhedron is rotated on an axis perpendicular to both the reference line segment and the nth axis just added, the only effect on the original distribution will be to adjust the scale of every point via the scale factor [math]costheta[/math]. That is, the new x_i is obtained by [math]x_{1i New} = x_{1iOld}cos\theta + sqrt{\frac{n}{2(n+1)}}sin\theta [/math], So we add a new axis and move the n-simplex off of the current axis we then position the new point along the negative axis at the location [imath]x_{n+1}=-\sqrt{\frac{n+1}{2(n+2)}}[/imath] to form a (n+1)-simplex while the reminder of the new (n+1)-simplex is on the positive side of the axis that we just added. Then a rotation perpendicular to the axis that was just moved off of and perpendicular to all sides that made up the previous n-simplex will have only the effect of scaling the projection of the (n) points that are on one side of the axis while the projection of the remaining point will move along the axis form plus to minus infinity? Quote Link to post Share on other sites

Doctordick 42,287 Posted July 5, 2008 Author Report Share Posted July 5, 2008 This is not quite what I was thinking of although this is what I think it would turn into in the case of an n-simplex.Let us please settle upon a vocabulary which we agree upon so that confusion can be removed from this conversation. What I was thinking of is a hyperplane of dimension n-1 that passes through the maximum possible number of points (which can be only (n-1) points) of course only the area of this hyperplane that is inside of the points is of interest in this case a (n-1)-simplex.For example, you are now talking about the “area” of a hyperplane without defining what you mean by area. This just isn't conducive to a logical discussion.So we add a new axis and move the n-simplex off of the current axis we then position the new point along the negative axis at the location Again your description of the situation uses confused references. Exactly what axis are you referring to as the “current axis”? You can not talk about manipulations with regard to undefined references. The correct description of what I am doing is expressed in my original post where we start with a simple entity constructed of n points displayed in an n-1 dimensional space (our initial (n-1)-simplex) and create an n dimensional space by adding a new axis orthogonal to every axis in the original (n-1)-dimensional space (using the exact same point for the “center” of our new coordinate system as we used for the center of the initial (n-1)-simplex).Thus we can move the lower order entity up from the center of our new coordinate axis and add a new point on the new axis below the center. In this case, the coordinates of previous polyhedron (as displayed in the n Euclidean space) remain exactly what they were for the n-1 established coordinates and are all shifted by the same distance from zero along the new axis.To paraphrase your comment, “we add a new axis and move the (n-1)-simplex [up along that new axis] and then position [a single new point on that same axis] at the location”.[math]x_n=-\sqrt{\frac{n}{2(n+1)}}[/math] Note that every point in the original (n-1)-simplex (which was originally at zero of the new axis) has been moved up a distance of [math]x_n=\frac{1}{\sqrt{2n(n+1)}}[/math] so that the distance from the new point to every point in the original (n-1)-simplex is exactly equal to unity. (View these points in the plane formed by the new axis and the axis to the earlier point of interest noting that the radius to the point is always given by [imath]r_n=\sqrt{r_n^2+r_{n-1}^2}[/imath].)The final part of the proof (if it is true for an n-1 dimensional figure, it is true for an n dimensional figure) requires a little thought:I quote my equations in order not to avoid confusion. You appear to be talking about change from an n-dimensional figure to an n+1 dimensional figure; in which case your expression is correct:[math]x_{n+1}=-\sqrt{\frac{n+1}{2(n+2)}}.[/math] ...to form a (n+1)-simplex while the reminder of the new (n+1)-simplex is on the positive side of the axis that we just added. Then a rotation perpendicular to the axis that was just moved off of and perpendicular to all sides that made up the previous n-simplex will have only the effect of scaling the projection of the (n) points that are on one side of the axis while the projection of the remaining point will move along the axis form plus to minus infinity?Your expression of a rotation being “perpendicular to the axis that was just moved off of and perpendicular to all sides that made up the previous n-simplex” is particularly confusing. Rotation is defined by the plane of transformation. A plane is defined by two lines. In this case, the plane of transformation is defined by that line upon which the projection has been made and the new axis which was just created and the expression for the change in the projection upon that lime which the projection has been made is expressly given by [math]x_{1_i \; New} = x_{1_i \; Old} cos\theta + \sqrt{\frac{n}{2(n+1)}} sin\theta [/math], where [imath]\theta[/imath] is the angle of that rotation from the construction orientation. If the unit of measurement within the universe described by that projection of points is to be obtained by some procedure from that distribution of points, then the added point may be moved to any spot within the universe described by those points. Plus and minus infinity are included by allowing the angle of rotation to be ninety degrees as, when the rotation reaches ninety degrees, the entire distribution of the previous collection of points collapses to a projection of zero and any defined unit of measure defined within that original projection goes to zero when compared to the distance of the new point from the origin. This implies the distance of the new point from the origin is infinite compared to the standard of measure already established. Have fun -- Dick Quote Link to post Share on other sites

Bombadil 14,583 Posted July 10, 2008 Report Share Posted July 10, 2008 Let us please settle upon a vocabulary which we agree upon so that confusion can be removed from this conversation. Yes, this seem like it is a good idea, so I will define some of the terms that seem to me to be of use. I suspect that part of the problem is that not knowing much about analytical geometry I’m thinking more along the lines of liner algebra. And while what I know about analytical geometry suggests to me that these are related it also seems that there are important differences some of which I am certainly overlooking. I think that the use of the term n-simplex is perhaps the best term to refer to a minimal equilateral of n dimensions I don’t think either of us have any objections to using the term in this way. For example, you are now talking about the “area” of a hyperplane without defining what you mean by area. This just isn't conducive to a logical discussion. Before I define a hyperplane I think it will simplify things to define a subspace as the set of points that satisfy the equations [imath]\sum_{i=1}^{n}a_{1i} x_i =0[/imath] [imath]\sum_{i=1}^{n}a_{2i} x_i =0[/imath]………[imath]\sum_{i=1}^{n}a_{ni} x_i =0[/imath] where [imath]a_{ki}[/imath] are constants and [imath]x_i[/imath] are the coordinates of the points that satisfy the equations (note that the solution [math]x_i=0[/math] need not be the only solution).I will define a hyperplane or the area of a hyperplane as the set of points satisfying the equation [math]x=x_0+y[/math] in which y is the set of points in a subspace, [math]x_0[/math] is any vector and x is then the set of point making up the hyperplane. The dimension of a hyperplane is then the number of linearly independent vectors that are in the corresponding subspace. I hope that these definitions are sufficient. If you have any problems with them or think that they are not sufficient just say so and I can try and revise them some. I suspect that you have encountered these ideas before although perhaps not recently and/or perhaps not by the same name. quote my equations in order not to avoid confusion. You appear to be talking about change from an n-dimensional figure to an n+1 dimensional figure; in which case your expression is correct: [math]x_{n+1}=-\sqrt{\frac{n+1}{2(n+2)}}.[/math] Yes this is correct. This is probably due to thinking of induction as proving for n then for (n+1). This seems completely equivalent to first proving for (n-1) then for n. It is probably best if we continue using this method, as it is what you do in the first post. Your expression of a rotation being “perpendicular to the axis that was just moved off of and perpendicular to all sides that made up the previous n-simplex” is particularly confusing. Rotation is defined by the plane of transformation. A plane is defined by two lines. In this case, the plane of transformation is defined by that line upon which the projection has been made and the new axis which was just created and the expression for the change in the projection upon that lime which the projection has been made is expressly given by [math]x_{1_i ; New} = x_{1_i ; Old} cos\theta + \sqrt{\frac{n}{2(n+1)}} sintheta,[/math] where theta is the angle of that rotation from the construction orientation. This is perhaps due to how I am thinking of a rotation and perhaps my wording could be better. How I understand it, you are rotating all of the points by rotation of a plane that they are all in ( a plane being 2 dimensional), while I am thinking of a rotation about a line in the case of rotation of a plane the line would be perpendicular to the plane that is being rotated. If this is the case I don’t see a difference in the end result for the case of a 3 dimensional space in which a rotation of any three points on a 3-simplex is being performed assuming that one point is not being rotated and three of the points are in the plane that is being rotated, although if the plane is all that is being rotated in a space of greater then three dimensions and an n-simplex with n>3 is being rotated then there would seem to me to be a change in the result. Quote Link to post Share on other sites

Doctordick 42,287 Posted July 11, 2008 Author Report Share Posted July 11, 2008 I hope that these definitions are sufficient.They are fine but they don't answer my original question. I have no idea as to why you should want this “hyperplane” as part of our discussion.Yes this is correct. This is probably due to thinking of induction as proving for n then for (n+1). This seems completely equivalent to first proving for (n-1) then for n. It is probably best if we continue using this method, as it is what you do in the first post.I am not sure here but I definitely get the impression that you seem to miss the point of algebraic substitution. The relationship expressed in[math]x_{n+1}=-\sqrt{\frac{n+1}{2(n+2)}}[/math] is absolutely identical to the relationship expressed in [math]x_n=-\sqrt{\frac{n}{2(n+1)}}[/math] or, for that matter, in[math]x_{n+537}=-\sqrt{\frac{n+537}{2(n+538)}}[/math]. But it is at the next step where we get seriously out of line.How I understand it, you are rotating all of the points by rotation of a plane that they are all in ( a plane being 2 dimensional), while I am thinking of a rotation about a line in the case of rotation of a plane the line would be perpendicular to the plane that is being rotated.Your concept of rotation is totally dominated by your intuitive understanding of three dimensional relationships. As I said to you long ago, the human mind has no experience with four dimensional relationships so we must be very careful. We cannot use our intuition at all and must depend upon analytical geometry, not mental pictures. In analytical geometry, all constructs are handled by expressing the coordinates of all relevant points as specific collections of numbers and then talking about the algebraic relationships between these sets of numbers. In analytical geometry there exist some very specific well defined operations which transform one set of points into a second set. One of these operations is “rotation”. This “rotation” is identical to the common concept of rotation in a three dimensional space. Rotation in three dimensions is associated with “an axis” not because such a relationship is required but rather because rotation is an operation defined in a two dimensional space and, in order to define rotation in a three dimensional space, one must eliminate one of the dimensions. Defining an axis of rotation is no more than a convenient way of reducing a three dimensional space to two dimensions. In higher dimensional spaces, the concept of "an axis of rotation" simply fails establish what rotation is intended because there are many planes orthogonal to any given axis. In an n-dimensional space one needs to explicitly define exactly what plane is to be used to express the rotation. In order to do that, one needs to specify two lines as two lines define a plane. In addition, we need to specify the direction of the rotation (which axis, as seen on the object, leads and towards which axis that point goes in the rotation). The common practice in analytical geometry is to specify two orthogonal axes because that definition makes defining the rotation quite simple. Under such a defined rotation, for any point in any structure, only the coordinates specified by those two axes change. All other coordinates stay exactly the same as they were prior to the rotation. For the two axes of interest, which, for the simplicity I will call [imath]\hat{a}[/imath] and [imath]\hat{b}[/imath] (rotating from [imath]\hat{a}[/imath] towards [imath]\hat{b}[/imath]) the new coordinates for the ith point in the structure defined in the geometry are given by:[math]a_i' = a_i cos \theta -b_i sin \theta[/math] and [math]b_i' = a_i sin \theta +b_i cos \theta[/math] where [imath]\theta[/imath] is the angle of rotation. If this is the case I don’t see a difference in the end result for the case of a 3 dimensional space in which a rotation of any three points on a 3-simplex is being performed assuming that one point is not being rotated and three of the points are in the plane that is being rotated, although if the plane is all that is being rotated in a space of greater then three dimensions and an n-simplex with n>3 is being rotated then there would seem to me to be a change in the result.”There would seem to me to be a change in the result!” only because your intuition does not serve you for dimensions greater than 3. By the way, "the plane" is not "all that is being rotated"; the entire object is being rotated. The plane merely defines the orientation of the rotation being specified. Have fun -- Dick Quote Link to post Share on other sites

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