So I've been using this website for some insight into the weird world of the higher dimension. And ran into some really weird stuff. Go figure.
Anyways, more math please.

That website is gold. I don't have anything particular to discuss/post about right now, but I'll enjoy reading it and hopefully find something. Thanks!

My stats teacher recommended it to me today when I started asking her about the nature of a 4-sphere. We were able to follow it up til the point where it started explaining how to find content and surface area, but my ADD was acting up pretty bad by then, so wasn't getting much out of it anyways...
anyways, yeah, she said that was the best math website in the world, and I'm inclined to agree with her.

By a 4-sphere, do you mean a 3-dimensional sphere lying in 4-space, or a 4-dimensional sphere lying in 5-space? (A normal sphere lying in 3-space is a 2D manifold.)

Hey, this site uses the term "glome" for a 3-sphere...I'd never heard that one before. Always good to have real mathematical terms for objects in higher dimensions; it lessens confusion.

I thought I was referring to a 4-dimensional sphere lying in 4-space, but I guess it's in 5-space...

A "sphere" is defined as the set of points equidistant from a given point, or center. So if you have a sphere in 4-space,, that's a 3-dimensional sphere, because every point on the sphere can be expressed with three coordinates. Similarly, a typical sphere is 2-dimensional, even though it lies in 3-space. A "ball" is the interior of an n-sphere, which is (n+1)-dimensional.

These topics always make my head hurt. I still don't get the n-sphere stuff.

math scares me.

I love higher-dimensional maths. I need to find a book I read in college that was similar to Flatland, but had even more of a fantasy milleu to it. In it, there was a beast that devoured its prey by surrounding it and smothering it slowly. The author of the account, witness to this happening to a 2D guy, saves the 2D guy from the beast and, in so doing, exposes the 2D guy to sights beyond eyes and sounds beyond ears.

The article at Wikipedia provides a pretty good introduction. As I've said before, Wikipedia is actually a surprisingly good website for detailed and easy to understand mathematical information. The fundamental definition of an n-sphere is the set of all points in (n+1)-dimensional space that are a given distance from a given point. In symbols,

where r is the radius and C is a center point with (n+1) coordinates. Think of a circle (1-sphere) and regular sphere (2-sphere); the n-spheres are just generalizations of those forms to higher dimensions.

For Sam, since you said you had trouble finding the "content" and "surface area": the important formula to know is , where represents the n-dimensional content of the n-sphere itself (i.e. the circumference of a circle or the surface area of a sphere). The gamma function is a generalization of the factorial to the entire real number line and is defined by . From that formula, you can easily derive a formula for , the (n+1)-dimensional content of the interior of the n-sphere, known as the (n+1)-dimensional ball, using calculus.

I think this is the problem we're having.

Yeah, that is inconvenient. I've become a firm convert to the topologists' convention, although I used to find it confusing.

Ok, I understand that a topologist sees an object as a collection of an infinite number of points combined into a solid surface (and the common explanation that a topologist can't tell the difference between a coffee mug and a donut) but what about geometers?

I'm really not sure why topologists and geometers refer to them differently; not far enough into the field of mathematics yet to be familiar with such subtleties.

Ok, so I was attempting to get back on a regular sleep cycle last night, and sat and thought about this sometime between 12 and 1 am.....and I'm pretty sure I have this thing figured out, by graphing rough approximations of a series of points.

I figured the easiest way to visualize a 4-sphere (or 3-sphere to a topologist) is as a unit sphere that grows and shrinks as it travels along the 4th dimension, such that if you connect the circumferences of all the possible spheres, you end up tracing the surface of another sphere occupying (I think only) the 2nd, 3rd, and 4th dimension. Then, you can extrapolate to the 5th dimension. If you graph all the points (x,y,z,w,v) where (0,0,0,w,v) on the w-v plane, you trace a unit circle. Finally, you can extrapolate this to the 6th dimension, and all the points (0,0,0,w,v,u) form a sphere, and you can start this process over going through the 7th, 8th, and 9th dimensions, then 10th, 11th, 12th, etc.

Once you can wrap your mind around the 4-, 5-, and 6-spheres (using geometers' notation), any further dimensions can be extrapolated from there. And it's effing awesome.

Excellent way to think about it. That is indeed how a glome (3-sphere) would appear to us: a series of regular spheres that gradually got bigger and then smaller again.

Jeez, I'm still having trouble picturing it....
this is why I don't plan on taking higher math classes.

try thinking about it while high. I work out most of this stuff while half-asleep, which feels very similar to what people describe being high feels like.

not really, I don't condone the use of illegal substances.......but it might be worth a try...

I'll get right on that....

ok, now that the hypersphere has been conquered, does anyone have an similarly simple way of visualizing a hypercube?

oh, hey, I just got this. I just had a topological epiphany.

The best way to intuitively understand higher-dimensional geometry is always to think about lower-dimensional analogues. You can't visualize a tesseract (4D hypercube), because your brain is three-dimensional, but you can visualize its cross-sections falling through our space. So what would those cross-sections look like? Well, consider the analogue one dimension lower: a cube falling through a flat plane (for example, Edwin Abbott's Flatland, if you've read that book).

There are three ways for a cube to fall through a plane:
1. Face-first: The cube starts out sitting flat on one of its square faces, with 4 vertices touching the plane. In this case, the cross-section is simply a square that appears for a while. It then disappears as the opposite square face, and the other 4 vertices, pass through the plane.
2. Edge-first: The cube starts out balanced on one of its edges. The cross-section starts out as an edge (2 vertices touching the plane) and becomes a gradually expanding rectangle. When the plane passes through the middle of the cube, 4 more vertices touch, and the cross-section is a rectangle with sides in the ratio 1:sqrt(2). The rectangle then becomes smaller again and diminishes to an edge, when the last 2 vertices pass through.
3. Vertex-first: The cube starts out balanced on a point (1 vertex touching). The three faces meeting at that vertex then begin to pass through the plane, creating a cross-section that is a regular triangle. When the next 3 vertices touch the plane, the edges of the triangle become truncated, so that the cross-section is a hexagon---and in the exact middle of the cube, it becomes a perfect regular hexagon. The process then reverses itself, and the cross-section dwindles to a regular triangle (where 3 more vertices touch) and then finally to a point (where the last vertex touches).

So, what are the 3-dimensional cross-sections of a tesseract falling through our 3-dimensional space? Well, there are four possible ways for a tesseract to fall through:
1. Cell-first: One of the tesseract's eight cubical cells (and thereby 8 of its vertices) first enters the space. The cross-section then is simply a cube that remains for a while. It disappears when the opposite cubical cell, and the other 8 vertices, pass through space. In other words, this is exactly like a cube falling through a plane face-first, but with a dimension added.
2. Face-first: This is analogous to a cube falling through a plane edge-first, but again with a dimension added. Thus, the cross-section starts out as a square (4 vertices touching) and gradually expands to a square prism with sides in the ratio 1:1:sqrt(2), with 8 more vertices touching. Then it diminishes back to a square, when the last 4 vertices touch.
3. Edge-first: Once again, we take the lower-dimensional analogy---a cube falling through a plane vertex-first---and add a dimension. So the cross-section starts out as an edge (2 vertices), expands to a regular triangular prism (6 more vertices), truncates to a hexagonal prism, and then reverses itself as before.

Now we've seen three of the cases, but what about the fourth and most interesting case, the one with no direct lower-dimensional analogy as described: a tesseract falling through our space vertex-first? Well, let's look at a pattern in the way the vertices touch as a unit cube of each dimension falls vertex-first through a Euclidean space of one dimension lower:

1D (line segment through point): 1, 1
2D (square through line): 1, 2, 1
3D (cube through plane): 1, 3, 3, 1

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
...Seen this pattern before? Pascal's triangle! So, assuming that a hypercube falling vertex-first through 3D space follows the vertex pattern 1, 4, 6, 4, 1, what 3-dimensional shapes would that create? Obviously, the "1" on each side is just a point. The only polyhedron with 4 vertices is the tetrahedron, and since the tesseract is symmetrical in every dimension, its cross-section a quarter of the way through its fall will be a perfect regular tetrahedron. Halfway through, it's a perfect regular octahedron, the polyhedron with 6 vertices. The process then reverses itself. Can you figure out what the cross-section would look like three-eighths of the way through---that is, right in between the tetrahedron and the octahedron? How about the cross-sections of other regular polychora falling through space?

I'm gonna have to go back and read that ^ later, mind's too tired for it at the moment.
However, does anyone have any knowledge of the formulas used by Lockheed Martin's Skunk Works program to develop stealth jets, or at least a fundamental knowledge of the principals behind the formulas?

This thread is amazing. I wish I knew more math, so I could understand some of this stuff, but AK_WDB, your description was superb. Like, I understand what you're saying completely. Using that same system, that is why we would see a 4-sphere as a growing and subsequently shrinking sphere?

I read that part in a stoner voice... It fit too well.

yes, that is exactly why

I'm reviving this thread with a couple of fairly simple, interesting math problems my brother and I were discussing today:

1. a. Prove that for any value n>0, lim_x->inf(ln(x)/xⁿ) = 0.
b. Find the maximum value of n for which there exists some value x such that ln(x) = xⁿ.

2. Prove that if the lengths of all sides of a right triangle are integers, then its inradius is also an integer.

For part a, can you just use induction?

No; n is any real number, not just integers.

L'Hopital's Rule?

lim x->inf (1/x)/(nx^n-1)=lim x->inf 1/(nx^n-1)=0

Right. That's the easy part really, just to see the point of doing part b.

Yeah, I figured.

For 2,

Let a and b equal the legs and c equal the hypotenuse.
Let p=the perimeter, r=the inradius, and k=the area.

k=a*b/2=r*p/2
r=a*b/p
=a*b/(a+b+c)
Multiplying by (a+b-c)/(a+b-c)
r=a*b*(a+b-c)/((a+ b )^2-c^2)
=a*b*(a+b-c)/(2*a*b+a^+b^2-c^2)
a^2+b^2-c^2=0, so,
r=(a+b-c)/2
I won't go into details but either a,b, and c are all even or only one of a, b, and c is even. Therefore, r must be an integer.

Now I have a cool problem.

A class is arranged as 5 rows of 5 desks with a student at each desk. Each student is asked to move to a neighboring desk to the back or front, left or right. Show how the students can do this or show that it is impossible.

Scratch that.

Well done, Negotiator. It seems to me that your problem is the same as asking if it's possible to draw a closed path through all 25 desks; is that correct?

yeah, pretty sure that is what it comes down to. forgot the proof, but it's impossible, has to do with the fact that it's an odd number in both dimensions; I run into the same problem when I try to plow or harvest all the squares on my farmville farm in a closed path to maximize efficiency ....

not quite.

Yeah, not quite, you can have as many closed paths of length greater than 1 as you want as long as every desk is covered exactly once.

True; good point. In fact concentric squares were what I thought of first, but of course that doesn't work with an odd number of desks. Still don't know how to prove it's impossible though...

OK, Negotiator:

Let D = {(x,y)|x,y in {0,1,2,3,4}}. Suppose that a particle starts at the point (a,b) in D and traces out a closed, oriented path consisting only of line segments of length 1 between two points of D. For every line segment on which the particle's position increases by 1 in the x (y) direction, there must be a corresponding line segment on which the position decreases by 1 in the x (y) direction, since the particle's net movement in each direction must be 0 for it to end up at (a,b). Therefore, the perimeter of the closed path must be an even integer. Since the path consists of line segments between two points of D, its perimeter is equal to the number of points in D it contains. So, because any such closed path contains an even number of points and D contains 25 points, an odd number, there exists no set of such paths for which each point in D is on exactly one path.

Now, nobody's done part b of my first problem yet.

I don't know about part B, as it is late and I'm exhausted and full, but I've been having fun finding proofs for .999999999999... = 1. I have 3 so far: 1/9 + 8/9 expressed as decimals and then fractions, the sum of the series 9/10^n from 1 to infiniti, and
x = .9999999...
10x = 9.999999...
10x - x = 9x = 9.99999... - .9999999... = 9
9 = 9x
x = 1


anyone got any others?

since nobody else seems to have wanted to play....

let XYZ be a pythagorean right triangle with legs x, y; hypotenuse z; incenter I and inradius r.

the area of XYZ is equivalent to the sums of the areas of constructed triangles XIY, XIZ and YIZ and thus

0.5xy = 0.5rx + 0.5ry + 0.5rz and therefore

r = xy / (x + y + z).

---

without loss of generality (from euclid), let x = m^2 - n^2, y = 2mn and thus z = m^2 + n^2 with m > n > 0 and m, n existing in N.

thus r = [(m^2 - n^2)(2mn)] / (m^2 - n^2 + 2mn + m^2 + n^2)
= [2mn(m+n)(m-n)] / [2m(m+n)]
= m(m-n)

which is a positive integer, as desired.

that is all

Negotiator already proved that one, although by a different procedure. The problem I'm referring to is:



Really? I thought only some Pythagorean triples were of this form, not all.

it was such a cute solution, too.

and the euclid thing should be true (i think... feel free to verify) if i insert a constant, which im too lazy to do.

Do you want a specific value for n?
Or is it like (ln(lnx))/(lnx)?

Yes, read the question carefully; it's asking for a specific value of n.

Hint: It's what you think it is.

11/30?

I get 1/e... I'm not too sure.