Can You Prove the Volume of an N-Dimensional Trianglehedron?

[T@P]

if you take two unit orthogonal vectors in R2, the triangle they form has area 1/2. take 3 in R3, the volume is 1/6. i claim (and half proved) that the voluarea of an n dimensional bunch of orthognal vectors would give you 1/n!. can anyone prove it fully?

(go easy on the linear algebra, I am infantile when it comes to that)

 

[NateTG]

I take it you mean the n-volume of the convex hull formed by the origin and the usual unit vectors in $\Re^n$ under the usual metric.

Let's instead of looking at just the unit length version, look at $v_n(r)$ the 'trianglehedron' with legs of length $r$.

Now, let's prove by induction that
$$v_n(r)=\frac{1}{n!} r^n$$
Then it's easy to see that
$$v_1(r)=\frac{1}{n!} r^1$$
And that for the n+1 case, we can integrate by slicing to get:
$$v_{n+1}(r)=\int_0^rv_n(x)dx$$
but from induction we have:
$$v_n(x)=\frac{1}{n!}x^n$$
so
$$v_{n+1}(r)=\int_0^r\frac{1}{n!}x^n dx$$
which readily works out to
$$v_{n+1}(r)=\frac{1}{n!} \times \frac{1}{n+1} r^{n+1}=\frac{1}{(n+1)!}r^{n+1}$$

Sepecifically,
$$v_n(1)=\frac{1}{n!}$$

 

[R0man]

The concept of a "N-dimensional trianglehedron" is not a commonly used term in mathematics. However, based on the description provided, it seems to refer to a geometric shape formed by a set of orthogonal vectors in N-dimensional space.

The statement that the volume or area of this shape is related to the factorial of the dimension (n!) is an interesting observation. It is true that in a two-dimensional space, the area of a triangle formed by two orthogonal unit vectors is 1/2, and in a three-dimensional space, the volume of the shape formed by three orthogonal unit vectors is 1/6. This pattern can be extended to higher dimensions as well.

To prove this, we can use the concept of determinants in linear algebra. The determinant of a square matrix is a scalar value that represents the "size" of the matrix. In two dimensions, the determinant of a matrix is equal to the area of the parallelogram formed by the two column vectors of the matrix. In three dimensions, the determinant of a matrix is equal to the volume of the parallelepiped formed by the three column vectors of the matrix.

Now, in N-dimensional space, the volume of the shape formed by N orthogonal unit vectors can be calculated using the determinant of a N x N matrix, where each column represents one of the unit vectors. The determinant of this matrix will be equal to the volume of the N-dimensional shape, which is 1/n! times the volume of a unit cube (where n is the dimension).

To prove this fully, we can use mathematical induction. The base case is n=2, where the determinant of a 2 x 2 matrix is equal to the area of the shape formed by two orthogonal unit vectors, which is 1/2. For the inductive step, assume that this statement is true for n=k. Then, for n=k+1, the determinant of a (k+1) x (k+1) matrix will be equal to 1/(k+1)! times the volume of a unit cube, which is 1/(k+1)! times 1/k!. This simplifies to 1/(k+1)!, proving the statement for n=k+1.

In conclusion, the volume or area of a N-dimensional shape formed by orthogonal unit vectors is indeed related to the factorial of the dimension (n!). This can be proven using the concept of determinants and mathematical induction.