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# How to solve the equation f(f(x))=x^2+1 ?

Posted 12 years ago
 In order to contextualize the question, let us first examine some examples.  Function Nest is useful to generate the multiple compositions of a given function: NestList[f, x, 5]{x, f[x], f[f[x]], f[f[f[x]]], f[f[f[f[x]]]], f[f[f[f[f[x]]]]]}For instance, if  f (x) = 2 x + 1 Simplify[NestList[2 # + 1 &, x, 5]]{x, 1 + 2 x, 3 + 4 x, 7 + 8 x, 15 + 16 x, 31 + 32 x}Given the equation f (f (x)) = a x + b, with a > 0 its solution can be readily verified  to be:  Simplify[Sqrt[a] x + b/(1 + Sqrt[a]) /. x -> Sqrt[a] x + b/(1 + Sqrt[a])]b + a xor alternatively : Simplify[Nest[Sqrt[a] # + b/(1 + Sqrt[a]) &, x, 2], a > 0]b + a xTo obtain the solution to f (f (f (x))) = a x + b we use Reduce Reduce[Nest[c # + d &, x, 3] == a x + b, {x}]and verify the answer FullSimplify[ Nest[(Power[a, (3)^-1] # + b/(     1 + Power[a, (3)^-1] + Power[a^2, (3)^-1])) &, x, 3], a > 0]b + a xNow let us consider quadratic equations. Consider solving the equation f (f (x)) = x^2. Martin Gardner solved it a while ago but do not look it up!.  It would be worthwhile the time you spend in trying to solve it.After you have solved it, consider now the equation f (f (x)) = x^2 + 1. How can we solve it? Do we need to consider the complex domain, that is f (f (z)) = z^2 + 1? Do we have to resort to numerical methods, at least to have an idea of what it looks like? If so, I would still be interested.
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Posted 11 years ago
 It seems like any strictly increasing function on the interval 0<=x<=f[0] (say after a choice of f[0]) with the restriction that f[0]<1 and f[f[0]]==1, gives rise to a continuous solution for positive numbers which can be defined by iteration of {x,y} -> {y,x^2+1}  e.g.,ListLinePlot[Flatten[NestList[{#[[2]], #[[1]]^2 + 1} & /@ # &, Table[{x, 1/2 + x}, {x, 0, 1/2, .1}], 4], 1]]starts with f[0]=1/2 and f=x+1/2Looking at this picture it is evident that the derivative goes through some jumps.  I suspect it is possible to show that there are no solutions which are differentiable everywhere because the way the derivative changes betweens iterations depends on x and the sided derivatives will not always agree.
Posted 12 years ago
 How to solve the equation f(f(x))=x^2+1 ?
Posted 12 years ago
 Hi Jaime,This article shows that, strictly speaking, the equation f[ f[ z ] ] == a z^2 + b z + c has no solution for f defined for all complex numbers:http://yaroslavvb.com/papers/rice-when.pdfHowever it is possible to find solutions defined over a smaller domain. For example, using the properties of the Chebyshev polynomials, a solution of f[ f[ x] ] == x^2 -2 is f[ x ] = 2 Cos[?2 ArcCos[ x / 2 ] ]. I'm afraid Gardner's solution suffers from the same problem...Jose.
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