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 triangular, tetrahedral and (poly)tope.]
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:
[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
The corresponding approximate formulae, for large are:
Pentatopic Roots (4-Dimensional)
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:
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
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:
8-Dimensional Hypertetrahedral Roots
Depress with :
which is just a quartic in , so depress again with :
, and .
Note that we know .
We try to factor this as a biquadratic:
Comparing coefficients, we thus have:
, so ;
, so .
We also have
so if we find , we will have ,
and we can then obtain k from and :
Multiplying that by and rearranging, we have
This is just a cubic in , so we can depress this with :
Substitute , then multiply by :
And, that is just a quadratic in .
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.