2 X + K Define F(X) by ------ ; then 2 X N N 2 2 (X + sqrt(K)) + (X - sqrt(K)) F(F(F(...(X)))) = sqrt(k) --------------------------------- N N 2 2 (X + sqrt(K)) - (X - sqrt(K)) which = sqrt(K) (coth 2 (arccoth X/sqrt(K))).

Known solutions are:

- Various linear things.
- X to different powers, sometimes multiplied by roots of 1.
- P and Q are each another polynomial R composed with itself different numbers of times.
- Solutions arising out of the flow of X^2 - 2, as follows:
suppose X = Y + 1/Y, then Y^N + Y^-N can be written as a polynomial in X.

For example, P = the expression for squares = X^2 - 2 (N = 2) and Q = the expression for cubes = X^3 - 3 X (N = 3)

- Replace X by Y-A, then add A to the original constants in both P and Q. For example, P = X^2 and Q = X^3, then P = 1 + (Y-1)^2 = Y^2 - 2 Y + 2 and Q = 1 + (Y-1)^3, then P(Q) = 1 + (Y-1)^6 = Q(P). Similarly, replacing X with A Y + B works.
- There are no more through degrees 3 and 4 (checked with Mathlab); but are there any more at all?

(n + ...101010) XOR (...101010) (reversible).

FLOW sub ZERO = X FLOW sub ONE = F(X) FLOW sub TWO = F(F(X)) etc.NOTE (Gosper): If we remove the restriction that F has a power series, the functions that satisfy an equation of the form F(F(X)) = sin X can be put into one-to-one correspondence with the set of all functions.

PROBLEM: Is the flow interpolation unique? If it is not, what extra conditions are necessary to make it unique for natural cases like X^N ?

Set aside a table TAB(J), 0 <= J <= log 2 (Largest possible period). Let C = the number of time F has been applied, initially 0. Compare each new value of F for equality with those table entries which contain old values of F. These will be the first S entries, where S is the number of times C can be right shifted before becoming 0. No match means F hasn't been looping very long, so increment C and store this latest value of F into TAB(J), where J is the number of trailing zero bits in the binary of C. (The first 16 values of J are: 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, ...; Eric Jensen calls this the RULER function.) A match with entry E means the loop length is 1 more than the low E+2 bits of C - 2^(E+1).

- 1 the zero loop, 0 -> 0
- 2 a positive loop, 4 -> 2 -> 1 -> 4
- 3 three negative loops
(equivalent to the 3N-1 problem with positive N)
- -2 -> -1 -> -2
- -5 -> -7 -> -10 -> -5
- -17 -> -25 -> -37 -> -55 -> -82 -> -41 -> -61 -> -91 -> -136 -> -68 -> -34 -> -17

What was the drawing?

Answer: It is a cousin to both the dragon and snowflake curves (and arose as a bug in a spacefilling curve). It can be constructed as follows. Start with a line segment. Replace it with the two legs of the isosceles right triangle of which it is the hypotenuse. Repeat this for the two new segments, always bulging outward in the same direction. We now have four segments forming half a square, with the middle two segments collinear. Replacing these four segments with eight and then sixteen, we find the middle two segments superimposed. As the process continues, the curve crosses itself more and more often, eventually taking on the shape of a wildly curly letter C which forms the envelope of a myriad of epicyclic octagons.

A faster way to approach the same limiting curve is to substitute the curve itself for each of its 2^2^n segments, starting with a 90 degree "<".

Yet another way to construct it is to iteratively connect opposite ends of two copies at a 90 degree angle. (The archeologist did this with his double exposure.) If we reduce the scale by sqrt(2) each time, the distance between the endpoints stays the same. If the initial line segment is red and there is some other blue shape elsewhere in the picture, the iteration will simultaneously proliferate and shrink the blue shapes, until they are all piled up along the red "C". Thus, no matter what you start with, you eventually get something that looks like the "C" curve.

There are other pictures besides the C curve which are preserved by this process, but they are of infinite size. You can get them by starting with anything and running the iteration backwards as well as forwards, superimposing all the results. A backward step consists of rotating the two copies in directions opposite those in the forward step and stretching by sqrt(2) instead of shrinking. David silver has sketched an arrangement of mirrors which might do this to a real scene.

Figure 8. Two orders of the "C" curve.