In the specific case a = 4 , however, it turns out that by allowing more sophisticated mathematical functions one can get a complete formula: the result after any number of steps t can be written in any of the forms Sin[2 t ArcSin[ √ x ]] 2 (1 - Cos[2 t ArcCos[1 - 2 x]])/2 (1 - ChebyshevT[2 t , 1 - 2x])/2 where these follow from functional relations such as Sin[2x] 2  4 Sin[x] 2 (1 - Sin[x] 2 ) ChebyshevT[m n, x]  ChebyshevT[m, ChebyshevT[n, x]] For a = 2 it also turns out that there is a complete formula: (1 - (1 - 2 x) 2 t )/2 And the same is true for a = -2 : 1/2 - Cos[(1/3) ( π - (-2) t ( π - 3 ArcCos[1/2 - x]))] In all these examples t enters essentially only in a t . And if one assumes that this is a general feature then one can formally derive for any a the result 1/2 (1 - g[a t InverseFunction[g] [1 - 2x]]) where g is a function that satisfies the functional equation g[a x]  1 + (a/2) (g[x] 2 - 1) When a = 4 , g[x] is Cosh[Sqrt[2 x]] . … The series has an accumulation of poles on the circle Abs[a] 2  1 ; the coefficient of x m turns out to have denominator 2^(m - DigitCount[m, 2, 1]) Apply[Times, Table[Cyclotomic[s, a]^Floor[(m - 1)/s], {s, m - 1}]] For other iterated maps general formulas also seem rare.

One way to achieve this is to break the array into 2 n × 2 n blocks, then successively to fill in pixels in each block until the appropriate gray level is reached, as in the pictures below, in an order given for example by Nest[ Flatten2D[{{4 # + 0, 4 # + 2}, {4 # + 3, 4 # + 1}}] &, {{0}}, n] An alternative to this so-called ordered dither approach is the Floyd–Steinberg or error-diffusion method invented in 1976. … The method can be implemented using Module[{a = Flatten[data], r, s}, {r, s} = Dimensions[data]; Partition[Do[ a 〚 i + {1, s - 1, s, s + 1} 〛 += m (a 〚 i 〛 - If[a 〚 i 〛 < 1/2, 0, 1]), {i, r s - s - 1}]; Map[If[# < 1/2, 0, 1] &, a], s]] In its original version m = {7, 3, 5, 1}/16 , as in the first row of pictures below. But even with m = {1, 0, 1, 0}/2 the method generates fairly random patterns, as in the second row below.

Implementation [of finite automata for nested patterns] Given the rules for a substitution system in the form used on page 931 a finite automaton (as on page 957 ) which yields the color of each cell from the digit sequences of its position is Map[Flatten[MapIndexed[#2 - 1  Position[rules, #1  _] 〚 1, 1 〛 &, Last[#], {-1}]] &, rules] This works in any number of dimensions so long as each replacement yields a block of the same cuboidal form.

The underlying density of particles is approximately 1 per cell, or 1/6 the maximum possible—a density which more or less minimizes the viscosity of the fluid.

And in terms of differential geometry such Minkowski space can be specified by the metric DiagonalMatrix[{+1, -1, -1, -1}] (now taking c = 1 ). … To do this in general one starts by picking a vector e in a timelike direction, then normalizes it to be a unit vector so that e . g . e  -1 . … And the volume of such a cone then turns out to be s[d]t d + 1 (1 - t 2 (d + 1)(d RicciScalar + 2(d + 1)(RicciTensor . e . e))/((d + 2)(d + 3)) + … )/(d + 1)

In general s ≤ 2 r + 1 , and for a simple identity or shift rule, s = 1 . For k = 2 , r = 1 , it then turns out that all the reversible rules and their inverses have s = 1 . … For arbitrary k and r , it is not clear what the maximum s can be; the only bound rigorously established so far is s ≤ r + 1/2k 2 r + 1 (k 2 r - 1) .

Implementation [of 2D substitution systems] With the rule on page 187 given for example by {1  {{1, 0}, {1, 1}}, 0  {{0, 0}, {0, 0}}} the result of t steps in the evolution of a 2D substitution system from a initial condition such as {{1}} is given by SS2DEvolve[rule_, init_, t_] := Nest[Flatten2D[# /. rule] &, init, t] Flatten2D[list_] := Apply[Join, Map[MapThread[Join, #] &, list]]

Cylinder volumes In any d -dimensional space, the volume of a cylinder of length x and radius r whose direction is defined by a unit vector v turns out to be given by s[d - 1] r d - 1 x (1 - (d - 1)/(d + 1)(RicciScalar-RicciTensor . v . v) r 2 + …) Note that what determines the volume of the cylinder is curvature orthogonal to its direction—and this is what leads to the combination of Ricci scalar and tensor that appears.

The 3n+1 problem The system described here is similar to the so-called 3n+1 problem, in which one looks at the rule n  If[EvenQ[n], n/2, (3n + 1)/2] and asks whether for any initial value of n the system eventually evolves to 1 (and thereafter simply repeats the sequence 1, 2, 1, 2, ...). … (For negative initial n , the evolution appears always to reach -1, -5 or -17, and then repeat with periods 1, 3 or 11 respectively.) … Case (a) is the standard 3n+1 problem.

_Symbol  1 //. x_[y_]  1 + Max[x, y] .) With a list s of possible symbols, c[s, n] gives all possible expressions with LeafCount[expr]  n : c[s_, 1] = s; c[s_, n_] := Flatten[ Table[Outer[#1[#2] &, c[s, n - m], c[s, m]], {m, n - 1}]] There are a total of Binomial[2n - 2, n - 1] Length[s] n /n such expressions. When Length[s]  1 the expressions correspond to possible balanced sequences of opening and closing brackets (see page 989 ).