```|Portada|Blog|Space|
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```                        Square Root of a Function.

1. Introduction.

In math it's common to study when a function has inverse, denoting it with
f^-1, since f^-1 ∘ f = id.  Following the same syntax it makes sense to
define f^n = f ∘ ... ∘ f {n times}.

We then introduce √f = g iff g∘g = f, and in this work we try to derive
the square root of a functions for a few simple cases, for which we use
PostScript as it's a concatenative language.

Also we should note that there's as well multiple possible values for
the square root of a given function, some cases being more useful than
others.  Such an example of a trivial useless solution could be checking if
a special private singleton is located at the top of the stack, if it's not
then we push it, otherwise we pop it and apply f.  Here we will strive for

2. Basic examples for Real arithmetic.

Single operations:

* √{n add} -> {m add} where m = n/2,
as: x m add m add  ->  x <2m> add  ->  x n add

* √{n sub} -> {m sub} where m = n/2.

* √{n mul} -> {m mul} where m = √n.
being careful with negative values of n, as PostScript doesn't provide
complex math support by default (though they could be represented as
arrays of two numbers).

* √{n sub} -> {m sub} where m = n/2.
* √{n div} -> {m div} where m = √n.

In the following case we observe how √(f∘g) = √f∘√g does always not hold.

* √{a mul b add} -> {c mul d add} where c=√a, d=b/(1+√a).
as: x c mul d add c mul d add  ->  <c(cx+d)+d = c²x+cd+d = ax+b>
a=c²  ->  c=√a,
b=cd+d  ->  b=(c+1)d  ->  d = b/(c+1) = b/(1+√a)

* √{a mul b add} -> {d add c mul} where c=√a, d=b/(a+√a)
as x d add c mul d add c mul  ->  <c(c(x+d)+d) = c(cx+cd+d) = c²x+c²d+cd>
b = c²d+cd = (c²+c)d  ->  d = b/(a+√a).

* √{b add a mul} -> {c mul d add} where c=√a, d=ab/(1+√a)
* √{b add a mul} -> {d add c mul} where c=√a, d=ab/(a+√a)
since √{b add a mul} -> √{a mul b' add} with b'=ab, as a(x+b)=ax+ab.

3. Integer arithmetic.

There are two cases to consider here: idiv and mod.  For √{n idiv} we know
that if n is the square of an integer we can apply the same derivation as
in real algebra.  The case where n=2 is trickier as {2 idiv} is in essence
the same as {-1 bitshift} or in other words dropping the least significant
bit, then √{2 idiv} would mean shifting right half a bit.

TODO:
* Think about the half-bit case.```
```document.title="wiki: misc/funcsqr"
```