The Geometry of Irregular Forms: Part 3

(Introduction to this series is at this link.)

Let's return to my experiment with the calculator's square root key. Although I didn't know it at the time, what I had stumbled upon was an iterative algorithm. That's fancy words for a process that's kind of like hitting a nail with a hammer. Every time you hit the nail, you drive it into the two-by-four a little bit. That's one iteration of the process. Then you take the output of that iteration (a nail that's part-way into the board), and you do the same thing to it. You do this over and over, until a terminating condition is met -- in this case, when the nail is all the way into the board.

When you start an iterative algorithm, you need an initial condition. From my square root experiment, 0.0000001 and 99999999 are examples of initial conditions. If we take the variable x to be the number on the calculator's display, then the initial condition is x₀.

For the first iteration, we plug x₀ into the process, and it yields the next x, or x₁. And the second iteration of the process yields x₂, and so on. In general:

the next x = do-something-to( the current x ), or
x_n+1 = do-something-to( x_n)

In my square root experiment, the "do-something-to" is the square root function. In the hammer-and-nail example, the "do-something-to" is striking the nail with the hammer. Using the x_n notation, the sequence from Part 1 looks like:

x₀ = 0.0000001
x₁ = 0.0003162
x₂ = 0.0177828
x₃ = 0.1333521
x₄ = 0.3651741
x₅ = 0.6042964
x₆ = 0.7773650
x₇ = 0.8816831
x₈ = 0.9389798
... 15 more ...
x₂₄ = 0.9999990
x₂₅ = 0.9999995
x₂₆ = 0.9999998
x₂₇ = 0.9999999
x₂₈ = 0.9999999
x₂₉ = 1

After hitting the square root key 29 times, I arrived at a terminating condition -- the calculator read 1, and the value was no longer changing -- and so I stopped.