n dimensional simplex vertex to centerpoint length ?

[Sphericality]

I am trying to figure out the limit of the ratio of the edge of an n dimensional simplex to its vertex to center length.
I think another way to say this is Im trying to find the distance of the vertex to center point of the infinite dimensional regular simplex of edge length one.

Am I correct the series begins like this:
Line segment distance to centerpoint from end points = .5
Equilateral Triangle distance to centerpoint = 2/3 √3 (not sure if this is how to write 2/3 of the square root of 3?)
Tetrahedron distance to centerpoint = √6/6 (is this right? Ive adapted it from Wikipedia but it seems too small)

Please excuse my lack of mathematical training.
Would someone who actually knows how this works enlighten me as to the logic of the progression and the limit it is heading towards ~

[student5]

have you considered asking wolframalpha?

note that the distance to the centerpoint is nothing but the circumradius of the polytope (since it's regular, all points lie on an n-sphere) and the radius of this n-sphere would be the answer to your question

so, I filled in circumradius n-simplex, and it gave:
sqrt(n/(2n+2))

which should be your answer

[Sphericality]

Thanks Student5 -
that not only answered my question
and pointed out the obvious circumradius correlation
it also turned me on to Wolframalpha !

[JMBR]

You can calculate the limit of this to infinity

lim√(x/(2x+2))=lim√(x/(2(x+1)))=1/√2 lim√(x/(x+1))=1/√2 lim√(x/(x(1+1/x)))=1/√2 lim√(1/(1+1/x))=1/√2 * 1/1=1/√2

The formula for the n-cross is constant, with this value (1/√2). Does that mean that they would have the same size?

And the formula for the n-cube is √x/2, which diverges at infinity. Should this be possible?

[Klitzing]

you also might like to have a look here.
--- rk

[Marek14]

You can calculate the limit of this to infinity

lim√(x/(2x+2))=lim√(x/(2(x+1)))=1/√2 lim√(x/(x+1))=1/√2 lim√(x/(x(1+1/x)))=1/√2 lim√(1/(1+1/x))=1/√2 * 1/1=1/√2

The formula for the n-cross is constant, with this value (1/√2). Does that mean that they would have the same size?

And the formula for the n-cube is √x/2, which diverges at infinity. Should this be possible?

Sure, the bigger the dimension of the cube, the larger its diagonal. It has no upper limit.

[Sphericality]

You can calculate the limit of this to infinity

lim√(x/(2x+2))=lim√(x/(2(x+1)))=1/√2 lim√(x/(x+1))=1/√2 lim√(x/(x(1+1/x)))=1/√2 lim√(1/(1+1/x))=1/√2 * 1/1=1/√2

Thats definitely the other part of what Im after, but Ive done some research and Im afraid Im still not able to understand what limit it is approaching at infinity. Can you please translate it into English for a non mathematician such as myself ?

I am very interested to know what the approximate maximum distance is from the vertex to the centre of a high dimensional simplex. Intuitively I thought it might be approaching 1:1 with the edge length but it seems from the formula to be much less than that?
(I have trouble imagining how the 5 vertexes of the pentachoron relate to the w x y z axis that divides 4 space into 16 quadrants.)

Very interesting observation about the cross and the measure and I understand that much better as its sort of obvious on the coordinate system how those distances will increase.

[student5]

You can calculate the limit of this to infinity

lim√(x/(2x+2))=lim√(x/(2(x+1)))=1/√2 lim√(x/(x+1))=1/√2 lim√(x/(x(1+1/x)))=1/√2 lim√(1/(1+1/x))=1/√2 * 1/1=1/√2

Thats definitely the other part of what Im after, but Ive done some research and Im afraid Im still not able to understand what limit it is approaching at infinity. Can you please translate it into English for a non mathematician such as myself ?

well, then the question is: how much of a non-mathematician are you?
do you, for example get the steps from
lim √(x/(2x+2)=1/√2 * lim√x/(x+1)?

from then on the intuitive explanation could be (so not mathematically rigorous)
take a look at x/(x+1):
lim x/(x+1) basically says, what does x/(x+1) get close to, when x becomes really really large?
well, for a very, very large number of x, you could say that x is almost equal to x+1 (take, 100000000000 the difference with 100000000001 is so small that, well 100000000000/100000000001 is practically 1)

filling this in, you'd get, 1/√2 * √1 = 1/√2

I am very interested to know what the approximate maximum distance is from the vertex to the centre of a high dimensional simplex. Intuitively I thought it might be approaching 1:1 with the edge length but it seems from the formula to be much less than that?
(I have trouble imagining how the 5 vertexes of the pentachoron relate to the w x y z axis that divides 4 space into 16 quadrants.)

Very interesting observation about the cross and the measure and I understand that much better as its sort of obvious on the coordinate system how those distances will increase.

one thing about visualizing 4D-polytopes: don't try it's really difficult, and more often puzzles you, than makes things clearer.

what does work well, is try to imagine what a 2D person would think about 3D space (or 1D about 2D, 0D about 1D...)
take the notion of simplex, which is basically a n-1 simplex, with a point added to it.

Code: Select all

o (0D) (point) (center = point)

o-o (1D) (line) (center = halfway between)

  o---o 2D (triangle)
   \ /     (center = √2/5)
    o


now for a 2D person, making a tetrahedon (triangular pyramid) would be as much as saying: "put a point in the middle of this triangle, with length 1 to all these points" which is of course utterly impossible in 2D!
you'd need to go to 3D, by adding a z-axis (which a 2D-person can not imagine)
a 2D person could also tell, that the point needs to be in the middle of the triangle, but it should be at some extra distance above it. the question is: how high?
this solely depends on the circumradius of the n-1 simplex. if it were 1, the point would be in the same plane (it would not be in the extra dimension), and if it were 0, the distance would be 1.
so now, I'll rotate the triangle so it's perpendicular to our viewpoint and becomes a line, and draw a point above it:

Code: Select all

    o    ^ z-axis
   / \   |      (the x means, two points behind each other, or a line)
  x---o  ----->x,y-axes


now, to go to 4D, we'd need to rotate this pyramid, so the x,y,z-axes are on the horizontal line, and the w-axis is vertical

Code: Select all

    o    ^ w-axis
   / \   |      (the t means, triangle)
  t---o  ----->x,y,z-axes


for 5D, this'd be very similar:

Code: Select all

    o    ^ w2-axis
   / \   |      (the p means, triangular pyramid)
  p---o  ----->x,y,z,w1-axes


etc for 6+D
I hope you'll agree that the circumradius will increase, as the dimension increases, so the horizontal distance gets bigger and bigger.
the question is, where does this stop?
take a very large simplex:

Code: Select all

    o    ^ w_n-axis
   / \   |      (the s means, n-2 simplex)
  s---o  ----->x,y,z,w1.....w_n-1-axes


we know, that for a very large simplex, the circumradius almost doesn't change, so now the question is, at which height does this occur?
the circumradius won't change anymore, if the added height is equal to the circumadius of the n-1 simplex.
and that just happens to be when this height is 1/√2 (and the circumradius the same)

Code: Select all

    o   
   /|\       
  s---o
let's zoom in on one of the two perpendicular triangles:
o   both the height, and the bottom (which is the circumradius) are 1/√2
|\   the hypothenuse then is, according to pythagoras: √ (1/√2)^2+ (1/√2)^2 = √1/2+1/2 = √1 = 1
| \
---o


so if the circumference of a simplex were to be 1/√2, it wouldn't change anymore

[Klitzing]

That you can understand the simplex as the alternation of the measure polytope (cube) only is special for 3D. Actually it doesn't remains true neither in 2D nor 4D nor anywhere beyond that dimensionality.

But you can consider the simplex as the vertex figure either of the measure polytope (cube, tesseract, etc.) or of the dodecahedron. Only that the edge size then is scaled up according to the secant of the according polygon (squares: factor = q = sqrt(2) = 1.4.14..., pentagons: factor = f = (1+sqrt(5))/2 = 1.618...). While the series of H_n = {5,3,...,3} becomes hyperbolical after 4D, the series C_n = {4,3,...,3} remains in spherical geometry (i.e. embeddable as compact subset in euclidean space of according dimension) always.

That is, you could describe the vertices of a simplex always as the points (-1/2,1/2,1/2,1/2,...), (1/2,-1/2,1/2,1/2,...), (1/2,1/2,-1/2,1/2,...), ... All those then are in the according hyperplane x₁ + x₂ + x₃ + ... = (D-2)/2 of the vertex figure. As you can see. the simplex edges then will have size sqrt(2).

Alternatively you could consider the simplices as a facet, i.e. (D-1)-element of the dual crosspolytope (octahedron, 16-cell, ...). Then you'd have the description of the vertex points according to (1/sqrt(2),0,0,0,...), (0,1/sqrt(2),0,0,...), (0,0,1/sqrt(2),0,...), ... Again the crosspolytope (just as the measure polytope before) here is designed to be of unit edge size. And thus its facet, the considered simplex, too has unit edge size. And, once more, the description is taken within one dimension plus, i.e. the vertices all fall into a single hyperplane, this time defined by x₁ + x₂ + x₃ + ... = 1/sqrt(2).

Within either description, the circumcenter point obviously is the point (c,c,c,c,...) of the according hyperplane. In the last description e.g. this happens to evaluate to c=1/(D*sqrt(2)).

Now you can see what happens when D runs to infinity: c obviously decreases to zero. And thus, in this limit, the circumradius is just the same as the single non-zero coordinate: 1/sqrt(2).

--- rk

[Sphericality]

Thanks for your elucidations.
I will chew on all this for quite some time before swallowing.