Here is some nice function I've made that calculates a square root of a number:

k=>(f=>f(f))(f=>x=>x==(y=(x+k/x)/2)?x:f(f)(y))(1)

At first look this function is completely cryptic, however it produces exactly the same results for positive numbers as Math.sqrt() function. Let's take a closer look and find out how does it work.

To make our job easier lets do some formatting:

k =>
(f => f(f))

(f => x =>
x == (y = (x + k/x)/2)
? x
: f(f)(y)
)(1)

We've separated our code in two parts - the first part is an arrow function that we will dissect later, and the second part, where some arithmetic arrow function gets an argument 1. Let's check the second part because it looks simplier.