Falling and Rising Roots

In my post on hypertetrahedral polytopic roots, I gave a few formulae for functions such as the triangular root.

Here I give some very similar formulae for some falling and rising roots.

The notations for rising and falling factorials or powers are, respectively:

x^{\overline{n}} = \dfrac{(x+n-1)!}{(x-1)!} = x(x+1)(x+2)\cdots(x+n-1)

x^{\underline{n}} =\dfrac{x!}{(x-n+1)!} = x(x-1)(x-2)\cdots(x-n+1)

where

n \in \mathbb{N}

For example:

x^{\overline{2}} = \dfrac{(x+1)!}{(x-1)!} = x(x+1) = x^2 + x

x^{\underline{2}} =\dfrac{x!}{(x-1)!} = x(x-1) = x^2 - x

If t = x^{\overline{2}} = x^2 + x then 0 = x^2 + x - t, and so (inventing some notation):

\sqrt[\overline{2}]{t} = \dfrac{-1 \pm \sqrt{1 + 4t}}{2} = - \frac{1}{2} \pm \sqrt{\frac{1}{4} + t}.

Similarly

\sqrt[\underline{2}]{t} = (+) \frac{1}{2} \pm \sqrt{\frac{1}{4} + t}.

Recall that

\sqrt[\bigtriangleup_{2}]{t} =\sqrt[\bigtriangleup]{t} = \dfrac{(\pm)\sqrt{8t+1}-1}{2}.

It is easy to see that these are related: this is because

2 \bigtriangleup(x) = 2 \bigtriangleup_2(x) = x^{\overline{2}}

Likewise,

n! \bigtriangleup_n(x) = x^{\overline{n}}

hence

\sqrt[\overline{n}]{t} = \sqrt[\bigtriangleup_{n}]{\dfrac{t}{n!}}.

We also have that

x^{\underline{n}} = (x-n+1)^{\overline{n}}

hence

\sqrt[\underline{n}]{t} = \sqrt[\overline{n}]{t}+n-1.

From these together with my earlier post, we therefore have, for example:

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

\sqrt[\underline{3}]{t} = \dfrac{3r^2+3r+1}{3r}

where

r = \sqrt[3]{\dfrac{t}{2} + \sqrt{\left( \dfrac{t}{2} \right)^2 - \dfrac{1}{27}}}

and:

\sqrt[\overline{4}]{t} = \sqrt{\dfrac{5}{4} + \sqrt{t + 1}} - \dfrac{3}{2}

\sqrt[\underline{4}]{t} = \sqrt{\dfrac{5}{4} + \sqrt{t + 1}} + \dfrac{3}{2}

Similarly the other formulae given in that earlier post give rise easily to a few further formulae.

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