IFS with Memory

Constant Cycles

The simplest repeated sequence is constant, just repeat the same number forever. For example, 11111... = 1^infinity.

Starting with (1/2, 1/2), applying T₁(x, y) = (x/2, y/2) repeatedly produces the sequence of points

T1(1/2, 1/2) = (1/4, 1/4),

T1(1/4, 1/4) = (1/8, 1/8),

T1(1/8, 1/8) = (1/16, 1/16),

...

converging to the lower left corner of the unit square, as pictured below.

Because it is gotten by applying T₁ infinitely many times, the address of this point is 11111... = 1^∞.

Recalling

T1(x, y) = (x/2, y/2) =	((x+0)/2, (y+0)/2),	the midpoint of (x, y) and (0, 0)
T2(x, y) = (x/2, y/2) + (1/2, 0) =	((x+1)/2, (y+0)/2),	the midpoint of (x, y) and (1, 0)
T3(x, y) = (x/2, y/2) + (0, 1/2) =	((x+0)/2, (y+1)/2),	the midpoint of (x, y) and (0, 1)
T4(x, y) = (x/2, y/2) + (1/2, 1/2) =	((x+1)/2, (y+1)/2),	the midpoint of (x, y) and (1, 1)

we see the cycles 2^∞, 3^∞, and 4^∞ generate sequences of points that converge to (1, 0), (0, 1), and (1,1), respectively.

3∞	4∞
1∞	2∞

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