Hypertetrahedral Polytopic Roots

The $n^{\rm th}$ triangular number is given by

$t = \bigtriangleup_2(n) = \bigtriangleup(n) = \dfrac{n(n+1)}{2} = \dbinom{n+1}{2}$.

For example, $\bigtriangleup(4)=10$.

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

$n = \sqrt[\bigtriangleup]{t} = \dfrac{\sqrt{8t+1}-1}{2}$

Thus, for example, $\sqrt[\bigtriangleup]{10} = \dfrac{\sqrt{80+1}-1}{2} = 4$.

Can this be generalised?

Triangular Roots (2-Dimensional)

Given a $t \text{ where } \bigtriangleup(n) = t$, it would be good to have a formula to find $n$. This $n$ could be called the triangular root, and denoted by $\sqrt[\bigtriangleup]{t}$.

There is an approximate formula $\bigtriangleup(n) \approx \dfrac{(n+\frac{1}{2})^2}{2}$ for large $n$. Therefore we have $\sqrt[\bigtriangleup]{t} \approx \sqrt{2t}-\frac{1}{2}$.

If we know that our $t$ is a triangular number, we also have that $\sqrt[\bigtriangleup]{t} = \lfloor \sqrt{2t} \rfloor$.

But what about an exact formula?

If we note that $\bigtriangleup(n) = t$ leads to the polynomial $0 = \bigtriangleup(n) - t$, then we can use the usual formula so solve this. If fact, it’s easier if we take the equation as $0 = 2 (\bigtriangleup(n) - t) = n^2+n-2t$. Thus

$n = \dfrac{\pm \sqrt{8t+1}-1}{2}$ and we choose +, as we’re interested only in positive $n$:

$\sqrt[\bigtriangleup]{t} = \dfrac{\sqrt{8t+1}-1}{2}$

Tetrahedral Roots (3-Dimensional)

Similarly, the $n^{\rm th}$ tetrahedral number (the 3-dimensional equivalent of triangular number) is given by

$t = \bigtriangleup_3(n) = \dfrac{n(n+1)(n+2)}{3!} = \dbinom{n+2}{3}$.

For example, $\bigtriangleup_3(6)=56$.

[The letter $t$ 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 $n$: $\bigtriangleup_3(n) \approx \dfrac{(n+1)^3}{6}$ and
$\sqrt[\bigtriangleup_3]{t} \approx \sqrt[3]{6t} - 1$.

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 $0 = 6 (\bigtriangleup_3(n) - t) = n^3+3n^2+2n-6t$.

Depress this with $n = m - 1$ to $0 = m^3 - m - 6t$. Substitute $m = r+\dfrac{1}{3r}$, and multiply through by $r^3$ to obtain $0 = r^6 - 6t r^3 + \frac{1}{27}$, which is essentially quadratic (in $r^3$). Thus $r = \sqrt[3]{3t \pm \sqrt{9t^2-\frac{1}{27}}}$. Back-substituting after choosing + and the real cube (or cubic) root, we obtain the formula for the tetrahedral root:

$\sqrt[\bigtriangleup_3]{t} = \dfrac{3r^2-3r+1}{3r}$

where $r = \sqrt[3]{3t + \sqrt{9t^2-\frac{1}{27}}}$.

[We leave this in this form, as all the $r$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: $n = \sqrt[\bigtriangleup_3]{56} = \dfrac{3r^2-3r+1}{3r}$ where $r = \sqrt[3]{3t + \sqrt{9t^2-\frac{1}{27}}}$ with $t=56$. So, $r = \sqrt[3]{3\times 56 + \sqrt{9\times 56^2-\frac{1}{27}}} \approx 6.952$, so $n = 6$.

Pentatope and other Hypertetrahedral Polytope Numbers

We define

$t = \bigtriangleup_d(n) = \dbinom{n+d-1}{d}$.

The corresponding approximate formulae, for large $n$ are:

$\bigtriangleup_d(n) \approx \dfrac{\left( n+\dfrac{d-1}{2} \right)^d}{d!}$;

$\sqrt[\bigtriangleup_d]{t} \approx \sqrt[d]{d! t} - \dfrac{d-1}{2}$.

Pentatopic Roots (4-Dimensional)

$t = \bigtriangleup_4(n) = \dbinom{n+3}{4} = \frac{1}{24}n^4 + \frac{1}{4}n^3 + \frac{11}{24}n^2 + \frac{1}{4}n$
$0 = n^4 + 6n^3 + 11n^2 + 6n - 24t$

Depress this with $n=m-\frac{3}{2}$:
$0 = m^4 - \frac{5}{2}m^2 + (\frac{9}{16} - 24t)$
which is just a quadratic in $m^2$, because is just so happens that the depression has caused the $m$ term in addition to the $m^3$ 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
$0 = (n - \frac{3}{2})(n - \frac{1}{2})(n + \frac{1}{2})(n + \frac{3}{2})$
and we know that
$(n-c)(n+c) = n^2 - c^2$.)

Thus we have
$m^2 = \dfrac{(\frac{5}{2} \pm \sqrt{(\frac{5}{2})^2 - 4(\frac{9}{16} - 24t)})}{2} = \frac{5}{4} \pm \sqrt{24t + 1}$.

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

$\sqrt[\bigtriangleup_4]{t} = n = \sqrt{\frac{5}{4} + \sqrt{24t + 1}} - \frac{3}{2}$.

Example:
$\bigtriangleup_4(5) = 70$
$\sqrt[\bigtriangleup_4]{70} = \sqrt{\frac{5}{4} + \sqrt{24 \times 70 + 1}} - \frac{3}{2} = \sqrt{\frac{5}{4} + \sqrt{1681}} - \frac{3}{2} = \sqrt{\frac{169}{4}} - \frac{3}{2} = \frac{13}{2} - \frac{3}{2} = 5$.

5 Dimensions?

Unfortunately, when the formula that arises in the 5-dimensional case $0 = n^5 + 10n^4 + 35n^3 + 50n^2 + 24n - 120t$ is depressed by $n=m-2$, we obtain $0 = m^5 - 5m^3 + 4m - 120t$. Although both the $m^4$ and $m^2$ 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

$t = \bigtriangleup_6(n) = \dbinom{n+5}{6}$
$0 = n^6 + 15n^5 + 85n^4 + 225n^3 + 274n^2 + 120n - 720t$

Depress this with $n=m-\frac{5}{2}$:
$0 = m^6 - \frac{35}{4}m^4 + \frac{259}{16}m^2 - (\frac{225}{64} + 720t)$
which is just a cubic in $m^2$, so depress again with $m^2 = k + \frac{35}{12}$:
$0 = k^3 - \frac{28}{3}k - (\frac{160}{27} + 720t)$
Substitute $k = h + \frac{28}{9h}$ and multiply by $h^3$:
$0 = h^6 - (720t+\frac{160}{27})h^3 + \frac{21952}{739}$
which is now just a quadratic in $h^3$, so
$h^3 = \dfrac{720t + \frac{160}{27} \pm \sqrt{(720t + \frac{160}{27})^2 - 4 \times \frac{21952}{729}}}{2}$
$= 360t + \frac{80}{27} \pm \sqrt{\frac{64}{3}(6075t^2 + 100t - 1)}$.

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

$\sqrt[\bigtriangleup_6]{t} = n = \sqrt{(h+\frac{28}{9h} + \frac{35}{12}} - \frac{5}{2}$ where

$h = \sqrt[3]{360t + \frac{80}{27} + \sqrt{\frac{64}{3}(6075t^2 + 100t - 1)}}$

8-Dimensional Hypertetrahedral Roots

$t = \bigtriangleup_8(n) = \dbinom{n+7}{8}$
$0 = n^8 + 28 n^7 + 322 n^6 + 1960 n^5 + 6769 n^4 + 13132 n^3 + 13068 n^2 + 5040 n - 40320 t$.

Depress with $n=m-\frac{7}{2}$:

$0 = m^8 - 21m^6 + \frac{987}{8}m^4 + \frac{3229}{16}m^2 + (\frac{11025}{256} - 40320t)$
which is just a quartic in $m^2$, so depress again with $m^2 = k + \frac{21}{4}$:
$0 = k^4 - 42 k^2 - 64 k + (105 -40320t)$.

This is
$0 = k^4 + c k^2 + b k + a$
with
$c = -42$, $b = -64$ and $a = 105 - 40320 t$.

Note that we know $\langle c, b, a \rangle$.

We try to factor this as a biquadratic:
$0 = k^4 + c k^2 +b k + a$
$= (k^2 + p k + s)(k^2 + q k + r)$
$0 = k^4 + (p+q) k^3 + (pq+r+s) k^2 + (pr+qs) k + rs$.

Comparing coefficients, we thus have:
$0 = p+q$;
$c = pq+r+s$;
$b = pr+qs$; and
$a = rs$.

Therefore
$q = -p$;
$c = r+s-p^2$, so $r+s = c+p^2$;
$b = p(r-s)$, so $r-s = b/p$.

We also have
$s = \frac{1}{2}(c+p^2-b/p)$;
$r = \frac{1}{2}(c+p^2+b/p)$;
so if we find $p$, we will have $\langle p, q, r, s \rangle$,
and we can then obtain k from $0 = (k^2 + p k + s)$ and $0 = (k^2 + q k + r)$:

$k = \dfrac{-p \pm \sqrt{p^2 - 4s}}{2}$ or $\dfrac{-q \pm \sqrt{q^2 - 4r}}{2}$.

Now,
$(c+p^2)^2 = (r+s)^2 = r^2 + s^2 + 2rs$, and
$(b/p)^2 = (r-s)^2 = r^2 + s^2 - 2rs$; so
$4a = 4rs = (c+p^2)^2 - (b/p)^2$.

Multiplying that by $p^2$ and rearranging, we have

$0 = p^6 + 2c p^4 + (c^2 - 4a) p^2 - b^2$
$= p^6 - 84p^4 + 1344(120t+1)p^2 - 4096$.

This is just a cubic in $p^2$, so we can depress this with $p^2 = h+28$:

$0 = h^3 + 1008(160t-1)h + 1152(3920t-9)$.

Substitute $h = g - \dfrac{336(160t-1)}{g}$, then multiply by $g^3$:

$0 = g^6 + v g^3 + w$ where
$v = 1152(3290t - 9)$, and
$w = - (336(160t - 1))^3$.

And, that is just a quadratic in $g^3$.

So: $g = \sqrt[3]{\dfrac{-v \pm \sqrt{v^2 - 4w}}{2}}$.

Now, we just back-substitute to $p$, solve the two quadratics, and back-substitute to $n$. 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.

3 Responses to “Hypertetrahedral Polytopic Roots”

1. Randolph Says:

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

2. Heritier Says:

I like this article. I just want to let you know that there is no need to make individual formula for each level of the simplexes. I have developed a general formula which works for all of them and gives the exact roots. By the way, I am an undergraduate studying mathematics in London.

• Rob Says:

Thanks for your comment. I’d be interested in seeing the details. I’d be especially interested to know whether the (general) formulae are soluble by radicals.

If you have any notes on-line, could you provide a link?

Thanks.

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