The *triangular number* is given by

.

For example, .

• • • • • • • • • •

It turns out the inverse formula is (inventing some notation):

Thus, for example, .

Can this be generalised?

### Triangular Roots (2-Dimensional)

Given a , it would be good to have a formula to find . This could be called the *triangular root*, and denoted by .

There is an approximate formula for large . Therefore we have .

If we know that our is a triangular number, we also have that .

But what about an exact formula?

If we note that leads to the polynomial , then we can use the usual formula so solve this. If fact, it’s easier if we take the equation as . Thus

and we choose +, as we’re interested only in positive :

### Tetrahedral Roots (3-Dimensional)

Similarly, the *tetrahedral number* (the 3-dimensional equivalent of triangular number) is given by

.

For example, .

[The letter is a convenient choice, as it stands for **t**riangular, **t**etrahedral and (poly)**t**ope.]

Can the *tetrahedral root* be found? Can this be generalised?

Yes, to at least some extent it can.

We have the formulae for large : and

.

To attempt to find and exact formula, we can use the techniques of the algebraic solution of polynomials.

The corresponding polynomial for the tetrahedral case is .

Depress this with to . Substitute , and multiply through by to obtain , which is essentially quadratic (in ). Thus . Back-substituting after choosing + and the *real* cube (or cubic) root, we obtain the formula for the *tetrahedral root*:

where .

[We leave this in this form, as all the s need to be the same value, and mathematical notation doesn't easily cater for formulae that are DAGs (directed acyclic graphs) rather than trees.]

Thus, for example: where with . So, , so .

### Pentatope and other Hypertetrahedral Polytope Numbers

We define

.

The corresponding approximate formulae, for large are:

;

.

### Pentatopic Roots (4-Dimensional)

leads to

Depress this with :

which is just a quadratic in , because is just so happens that the depression has caused the term in addition to the term to vanish; thus this case is simpler than the 3-dimensional, tetrahedral, case.

(The result of this depression should not really surprise us, since the result is

and we know that

.)

Thus we have

.

So, choosing +, and back-substituting, we get the pentatopic root formula:

.

Example:

.

### 5 Dimensions?

Unfortunately, when the formula that arises in the 5-dimensional case is depressed by , we obtain . Although both the and terms have vanished, there is an inconvenient contant term. This may be soluble; I don’t know.

However, the 4-dimensional case makes me quite optimistic about the 6- and 8-dimensional cases…

### 6-Dimensional Hypertetrahedral Roots

leads to

Depress this with :

which is just a cubic in , so depress again with :

Substitute and multiply by :

which is now just a quadratic in , so

.

Thus, if we choose +, and back-substitute:

where

### 8-Dimensional Hypertetrahedral Roots

leads to

.

Depress with :

which is just a quartic in , so depress again with :

.

This is

with

, and .

Note that we know .

We try to factor this as a biquadratic:

.

Comparing coefficients, we thus have:

;

;

; and

.

Therefore

;

, so ;

, so .

We also have

;

;

so if we find , we will have ,

and we can then obtain k from and :

or .

Now,

, and

; so

.

Multiplying that by and rearranging, we have

.

This is just a cubic in , so we can depress this with :

.

Substitute , then multiply by :

where

, and

.

And, that is just a quadratic in .

So: .

Now, we just back-substitute to , solve the two quadratics, and back-substitute to . At various stages, there are multiple, complex roots, thus each must be tried.

The overall solution in this case is fairly complicated, but it does exist.

Tuesday, 6 September 2011 at 05:11 |

http://forums.delphiforums.com/figurate/messages/?msg=6.1