Exploring the Unit Cube in [tex]\ R^{n}[/itex]

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In summary: How about this- suppose the origin is not at the center of the "unit cube". Suppose, for instance, that the origin is at the edge of the "unit cube". In that case, the distance between any two vertices is not 1, but the distance between the center of the cube and the origin. You could then take the square root of the distance between the center of the cube and the origin, and that would be the average distance of the vertices to the origin.
  • #1
shwin
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I'm having trouble visualizing [tex]\ R^{4}[/itex](a domain of reals in four dimensions).

1. Describe a procedure in given 3 vectors, finds a fourth vector perpendicular to those three. Explain why we can use it in analogous fashion to the normal vector to a plane in [tex]\ R^{3}[/itex].

Here, I'm thinking taking the normal vector of each vector and then adding the three normals would be sufficient, but I am not sure how this is analogous to the normal vector to a plane in the xyz system.

2. How many vertices does the unit cube have in [tex]\ R^{n}[/itex] have? What is the furthest distance from the origin that one can be on the unit cube in [tex]\ R^{n}[/itex]? What is the average distance of the vertices to the origin?

First part I have [tex]\ 2^{n}[/itex], second part a square root of [ a summation from i = 1 to i = n of [tex]\ n^{2}[/itex]]

But the last part stumps me...average distance? and writing a formula for this is a bit confusing too.
 
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  • #2
for the first you should look at the gram-schmidt proces:

http://en.wikipedia.org/wiki/Gram-Schmidt_process

for the second, you are right about the [tex]2^n[/tex]. In general the distance from a vertice to the origin is given by

[tex] \frac{\sqrt{1^2+1^2+\dots+1^2}}{2} = \frac{\sqrt{n}}{2} [/tex]

so the average must be

[tex] lim_{n\rightarrow \infty}\frac{1}{n}\sum_i^{n} \frac{\sqrt{i}}{2} = lim_{n\rightarrow \infty} \frac{1}{n}[\frac{\sqrt{1}+\sqrt{2}+\dots+\sqrt{n}}{2}] = lim_{n\rightarrow \infty} \frac{\sqrt{1}}{2n}+\frac{\sqrt{2}}{2n}+\dots +\frac{\sqrt{n}}{2n} = \infty [/tex]

edited from:

[tex] lim_{n\rightarrow \infty}\frac{1}{n}\sum_i^{n} \frac{\sqrt{i}}{2} = lim_{n\rightarrow \infty} \frac{1}{n}[\frac{\sqrt{1}+\sqrt{2}+\dots+\sqrt{n}}{2}] = lim_{n\rightarrow \infty} \frac{\sqrt{1}}{2n}+\frac{\sqrt{2}}{2n}+\dots +\frac{\sqrt{n}}{2n} = 0[/tex]

if the average is over dimension.
 
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  • #3
ups, don't think the limit converge
 
  • #4
then I'm not sure what average you should calculate
 
  • #5
shwin said:
I'm having trouble visualizing [tex]\ R^{4}[/itex](a domain of reals in four dimensions).

1. Describe a procedure in given 3 vectors, finds a fourth vector perpendicular to those three. Explain why we can use it in analogous fashion to the normal vector to a plane in [tex]\ R^{3}[/itex].

Here, I'm thinking taking the normal vector of each vector and then adding the three normals would be sufficient, but I am not sure how this is analogous to the normal vector to a plane in the xyz system.
There is no such thing as "the" normal vector to a vector- any vector in the 3 dimensional "plane" normal to that vector. And even in 3 dimensions, adding normal vectors to two given vectors will not in general give you a vector normal to both. In 3 dimensions you have to take the cross product of the two given vectors. Can you thknk of a procedure analagous to that?

2. How many vertices does the unit cube have in [tex]\ R^{n}[/itex] have? What is the furthest distance from the origin that one can be on the unit cube in [tex]\ R^{n}[/itex]? What is the average distance of the vertices to the origin?

First part I have [tex]\ 2^{n}[/itex], second part a square root of [ a summation from i = 1 to i = n of [tex]\ n^{2}[/itex]]

But the last part stumps me...average distance? and writing a formula for this is a bit confusing too.
Assuming by "unit cube" they mean an n-dimensional "cube" with center at the origin and each side of length 1, then you have, NOT a " summation from i = 1 to i = n of n^2" but only a summation from 1 to n of 1 (the square of the side length= 1)- and that is just n. The diagonal distance between opposite vertices is [itex]\sqrt{n}[/itex] and if the origin is at the center of the "cube" the distance you want is [itex]\sqrt{n}/2[/itex] (I see that mrandersdk has already given that). Now I guess you could divide that by n and sum to get an average but that does not converge!
 

Related to Exploring the Unit Cube in [tex]\ R^{n}[/itex]

1. What is a unit cube in [tex]\ R^{n}[/itex]?

A unit cube in [tex]\ R^{n}[/itex] is a geometric shape with n dimensions, where all sides have a length of 1 unit. It is a fundamental object in mathematics and is often used as a building block for more complex shapes and structures.

2. How many dimensions does a unit cube in [tex]\ R^{n}[/itex] have?

A unit cube in [tex]\ R^{n}[/itex] has n dimensions, as it is defined as a shape with the same number of dimensions as the space it exists in. For example, a unit cube in 3-dimensional space (or [tex]\ R^{3}[/itex]) has 3 dimensions.

3. What is the volume of a unit cube in [tex]\ R^{n}[/itex]?

The volume of a unit cube in [tex]\ R^{n}[/itex] is always equal to 1. This is because the unit cube has a length of 1 unit on all sides, so the volume is simply 1 x 1 x 1 = 1.

4. How can the unit cube be explored in [tex]\ R^{n}[/itex]?

The unit cube in [tex]\ R^{n}[/itex] can be explored through various geometric operations and transformations, such as rotations, translations, and reflections. It can also be divided into smaller cubes or combined with other shapes to create more complex figures.

5. What are the practical applications of exploring the unit cube in [tex]\ R^{n}[/itex]?

Exploring the unit cube in [tex]\ R^{n}[/itex] has practical applications in fields such as architecture, engineering, and computer graphics. It also helps in understanding higher-dimensional spaces and their properties, which have implications in physics and other scientific fields.

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