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January 26th, 2013, 01:14 PM  #1 
Senior Member Joined: Jan 2013 Posts: 209 Thanks: 3  How to calculate number of dimensions using only dot product
The surface of a sphere has slightly less than 2 dimensions because you can't draw 2 exactly perpendicular lines from any point. If you could, it would be like a flat piece of paper, but if you try to crush that paper onto any small part of a sphere, it gets a little wrinkle. That wrinkle is the small Amount Of Dimension less than 2, the difference in dimensions between the paper and sphere surface. There can exist no 2 parallel lines on a sphere surface, regardless of how small an epsilon between them, because they always intersect somewhere. It is wrong to integrate the surface of a sphere using dx and dy or any 2 variables. Instead we can define the surface of a circle, sphere, or any dimension of hypersphere, using almost any set of points on its surface and the Amount Of Dimension (aod) equation. The curve of spacetime, for example, includes the amount of curve you would expect from embedding a hypersphere onto the surface of another hypersphere, which you could do this way by defining the vectors as on that surface. Start with the equation of the surface of a hypersphere of radius 1: a^2 + b^2 + c^2 + ... = 1 Then add Amount Of Dimension, starting at 1, for each dimension, and its the same equation: aod(a)*a^2 + aod(b)*b^2 + aod(c)*c^2 + ... = 1 Now divide one of the dimensions in half and have 2 halfdimensions, and its the same equation: (aod(a)/2)*a^2 + (aod(d)/2)*d^2 + aod(b)*b^2 + aod(c)*c^2 + ... = 1 The surprising thing is the dimensions can be any length 1 vectors and do not have be perpendicular to eachother. If you create n random vectors in d dimensional space, each length 1, and run the following algorithm, then the sum of all Amount Of Dimensions quickly converges to slightly less than or equal to d. It gets more accurate as n gets larger, even though you're adding more randomness. Repeat approximately log number of times for each vector all at once: Set each aod(self) to: aod(self)/sum(aod(eachVector)*dotProductBetweenSelfAndThatVe ctor^2) The dot product between any length 1 vector and itself is 1. The others range 1 to 1. Then, sum of all aod(eachVector) converges to d, even though the original number of dimensions d was erased and only existed in the dotproducts between the vectors which are known to be length 1. The Amount Of Dimension equation and its converging algorithm form allow us to efficiently calculate sparse variable dimensional manifolds using sets of points, vectors on the surface of a hypersphere of length 1. The main difficulty in using such sets of points is the Triangle Inequality, which says that the lengths of any 2 sides of a triangle must sum to more than the other side. In my experience, vectors/points that violate triangle inequality tend to shrink to zero Amount Of Dimension, but it needs to be explored. 
February 7th, 2013, 10:08 AM  #2  
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: How to calculate number of dimensions using only dot pro Quote:
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You may have been thinking that "The surface of a sphere has slightly less than 2 dimensions" because you straight lines (or vectors) do not lie in the sphere. However, such vectors, for general differentiable manifolds lie in the tangent plane at each point, not in the manifold itelf. Quote:
Last edited by skipjack; September 22nd, 2014 at 10:37 PM.  
February 15th, 2013, 02:42 PM  #3  
Senior Member Joined: Jan 2013 Posts: 209 Thanks: 3  Re: How to calculate number of dimensions using only dot pro Quote:
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March 8th, 2013, 01:43 PM  #4 
Senior Member Joined: Jan 2013 Posts: 209 Thanks: 3  Re: How to calculate number of dimensions using only dot pro
IMPORTANT EDIT: Instead of "20 vectors of length .05 each, then they are .05 amount of dimension", all vectors are always length 1, but it still finds their Amount Of Dimension. Also, I've been researching how this fits with density fields, like a bell curve. My theory is dot product is some function of the amount of overlap between each pairs of bell curve, so there should be a way to interpolate very accurately in a sparse network of points modelled as high dimensional bell curves. This would be useful for painting colors (or other variables) in high dimensions instead of pictures just being 2 or volumes 3 dimensional, so any kind of flowing information or art can be represented in practically small sparse information. The core math operator of this is the Lambda Tensor, at least in theory... "Lambda Tensor  a minimalist general computing math operator" viewtopic.php?f=20&t=38864 
March 9th, 2013, 12:33 PM  #5 
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: How to calculate number of dimensions using only dot pro
You still haven't answered the questions I asked. Please define your terms. What do you mean by "amount of dimension"? And what definition of "dimension" are you using?

March 20th, 2013, 10:53 AM  #6 
Senior Member Joined: Jan 2013 Posts: 209 Thanks: 3  Re: How to calculate number of dimensions using only dot pro
degrees of freedom is not always an integer amount

March 20th, 2013, 06:22 PM  #7 
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: How to calculate number of dimensions using only dot pro
So you cannot define the words you are using?

March 21st, 2013, 11:36 AM  #8 
Senior Member Joined: Jan 2013 Posts: 209 Thanks: 3  Re: How to calculate number of dimensions using only dot pro
Dimension is a Degree Of Freedom in which a vector can have a position independent of other Dimensions. If 2 Dimensions are Entangled, for example required to have equal position in each (x=y), then there is 1 less Degree Of Freedom and Dimension. If this Entanglement restricts them less than that, the number of degrees of freedom is between 1 and 2. For example, NAND is a constraint on 3 dimensions, normally defined as binary values instead of scalars, that z=not(and(x,y)), so in the 8 possible extremes of each of x, y, and z being true or false, only 4 are allowed (001, 011, 101, and 110). In that case, NAND reduces 3 dimensions to 2 because 4 (2^2) of 8 (2^3) possibilities remain. 

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