I have two equations, first says the sum of x and y equals a constant or invariant D,
$$x+y=D$$
Second formula writes the multiplication of those variables x and y in the form of D,
$$xy=(D/2)^2$$
When you sum both we obtain:
$$x+y+xy=D+(D/2)^2$$
After this, as we graphic this result we see that we want the first equation to have more "power" reason why it's multiplied by a leverage χ obtaining:
$$χ(x+y)+xy=χD+(D/2)^2$$
Now we want to make χ dimensionless and not depending on the amount in x and y.(so that x=4, y=6 work the same as x=40, y=60, only the relation is relevant) So for example if we set D=10, χ=1 we would obtain the same graphic result as if we do D=100, χ=1. (Because the relation in % is the same for x,y) In order to this instead of multypling the first equation by χ we do it by χD, obtaining:
$$χD(x+y)+xy=χD^2+(D/2)^2$$
Which actually works as intended, my question is, why does multypling by D make χ dimensionless?
I understand the objective of making a parameter dimensionless but I do not understand how to obtain that result. Why multiply by D? Is it trial error or how is the deduction made?
All this I from the StableSwap invariant construction https://classic.curve.fi/files/stableswap-paper.pdf
The first equation $x+y=D$ states that $D$ has the same dimension as $x$ and $y$, and the second equation $xy=D^2/4$ is conforming to to this. In the sum $x+y+xy$ there is a term of dimension $D$ plus a term of dimension $D^2$, so it does not make sense if these were variables in physics. To get the same units in both terms, the simplest way to get a meaningful number with a well-defined unit is to multiply the $x+y$ (which has already dimension $D$) by another $D$ so it has the same $D^2$ dimension as the second.