# The Reveal, Part 2

Here are the functions uses in this Alien Math scheme:

D(x) = tanh(x) or the hyberbolic tangent.

$tanh(a+b) = \frac{tanh(a)+tanh(b)}{1+tanh(a) tanh(b)}$

$Z(a,b) = \frac{a+b}{1+ab}$

$C(x) = atanh(x) = \frac{1}{2} log(\frac{1+a}{1-a})$

Z(D(a), D(b)) then is tanh(a + b). So that is why C(Z(D(a), D(b))) is a + b

This is all pretty familiar stuff. You can the expression in Z if you take Einstein’s formula for relativistic velocity addition, and set the units so that c = 1.

In the wikipedia articles on Special Relativity, they mention something called rapidity which is the arctanh of the velocity. All of the relativistic velocity stuff can be turned into normal addition by using this arctanh.

W(a, b) was then the expression for getting the original velocity given the combined velocity and one of the originals.

$W(a,b) = \frac{b^2 - a*b}{a b^2 - b}$

If a = 0, then W(a,b) is -b which is a handy way to get negatives.

The next step was to see if something akin to relativistic multiplication could be done. If you calculate Z(a,a), Z(a,Z(a,a)) and so on, eventually you discover a pattern of odd polynomial terms in the numerator and even terms in the denominator. These are the terms of the binomial expansion of $(a+1)^b$

An example of what this looks like would be:

$Z(a, Z(a,a)) = \frac{a^3+3 a}{3 a^2 + 1}$
$Z(a, Z(a, Z(a,a))) = \frac{4 a^3+4 a}{a^4 + 6 a^2+1}$

I studied these even and odd sequences and learned that you can represent them using two binomials

Even terms: $(a+1)^b + (1-a)^b$
Odd terms: $(a+1)^b - (1-a)^b$

Extending the exponents to all values, then, we get the analogue of multiplication in this system:

$K(a,b) = \frac{(a+1)^b - (1-a)^b}{(a+1)^b + (1-a)^b}$

So D and C are the hyberbolic tangent and its inverse, which can be used to exploit the tanh’s angle addition identity. The same addition identify can be extended to K to prove that C(K(D(a), b)) = a b.

And you need a separate value for 1 for two reasons. If aliens scaled their numbers so that 1 was C, it would make sense for the numeral system to rely on smaller numbers. Also 1 can’t be produced in the system, and the only reciprocal of 1 is 1 itself, so you need a special number E for 1.

Since K is already non-associative, I don’t think any extension to an analogue of exponentiation can be made. K is already an exponential function.