Sphericality wrote:JMBR wrote: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.

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`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:

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` 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

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` o ^ w-axis`

/ \ | (the t means, triangle)

t---o ----->x,y,z-axes

for 5D, this'd be very similar:

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` 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:

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` 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)

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` 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