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Q: Math question ( Answered ,   13 Comments ) Question
 Subject: Math question Category: Science > Math Asked by: antthrush-ga List Price: \$2.00 Posted: 12 Jul 2005 22:31 PDT Expires: 11 Aug 2005 22:31 PDT Question ID: 542904
 ```Someone asked me the following question. The person who told it to him said there was an easy answer, but neither of us can figure it out! The question is What function satisfies f(f(x)) = x^2 + 2``` Subject: Re: Math question Answered By: mathtalk-ga on 15 Jul 2005 21:54 PDT Rated: ```Hi, antthrush-ga: Let g(x) = x^2 + 2. This is a continuous function on the real numbers, it has "even" symmetry, g(-x) = g(x), and it is strictly increasing for nonnegative x. We can show how to define/construct all possible solutions to: f(f(x)) = g(x) which share these characteristics. The construction is pretty easy, but in a sense it uses mathematical induction. First off let's consider what f(0) might be. Since g(0) = 2, it's not hard to convince oneself that for f to be an even function and strictly increasing on the nonnegative real numbers, we will need f(0) > 0. Let's call f(0) = a. Note that f(a) = f(f(0)) = g(0) = 2, and since f is strictly increasing: 0 < a implies a = f(0) < f(a) = 2 In fact we have an increasing sequence of real numbers: x_0 = 0, x_1 = a, x_2 = 2, ... , x_k = f(x_{k-1}) with the property (since f(f(x)) = g(x)) that: x_{k+2} = g(x_k) = (x_k)^2 + 2 Clearly the repeated applications of g(x) guarantee that the x_k's will increase without limit (tend to positive infinity), so that every nonnegative real number belongs to exactly one interval [x_k, x_{k+1}). The good news is that we can define f(x) to be any strictly increasing function at all that maps [0,a] onto [a,2], and then the rest of the definition of f(x) will automatically follow, giving a solution to f(f(x)) = g(x). Since the definition of f(x) on [0,a] = [x_0,x_1] determines its definition for all nonnegative real numbers (and thus, because of its evenness, on all real numbers), you will hopefully not be shocked to learn that this definition is "piecemeal" in the sense of the following: / | (f?�(x))^2 + 2 for x in [a,2] = [x_1,x_2] f(x) = < | g?(f(g??(x))) for x in [x_{2n},x_{2n+2}] \ A few words may clarify the definition. Since f is chosen to be continuous and strictly increasing in mapping [0,a] to [a,2] = [x_1,x_2], it has a continuous (and monotone increasing) inverse f?�. That first formula is then a matter of "one step backwards, two steps forward" as we apply g to the result of f?�, so for x in [a,2], we move backwards once to [0,a], then ahead twice (of f, once of g) into [2, a^2 + 2] = [x_2,x_3]. Then f:[x_{2k,x_{2k+2}] --> [x_{2k+1},x_{2k+2}] is defined by pulling x back to [0,2] with k applications of g?�, applying f there and advancing again with k applications of g. The continuity of this patchwork f depends on its continuity on each of the pieces [x_k,x_{k+1}], together with the consistency of f at the boundaries between the pieces. But f agrees at the boundary between [0,a] and [a,2], because on the lower subinterval we have chosen function f so that f(a) = 2, and on the upper subinterval, taking f?�(a) = 0 and then g(0) = 2 gives the same answer. The consistency of f at all other boundaries (and thus f's continuity) is a corollary of this agreement. We give a couple of examples to illustrate that f can be defined flexibly on the first subinterval [0,a], but we simplify the notation from hereout by taking a = 1. Any value for a between 0 and 2 is feasible, just more effort to express. Example 1. ---------- Let f:[0,1] --> [1,2] be defined by f(x) = x + 1. Then f:[1,2] --> [2,3] is defined by f(x) = x^2 - 2x + 3, _____ f:[2,3] --> [3,6] is defined by f(x) = x + 2?x - 2 + 1, _____ and f:[3,6] --> [6,11] by f(x) = x^2 - 4(x+1)?x - 2 + 6x - 5. Example 2. ---------- Let f:[0,1] --> [1,2] be defined by f(x) = x^2 + 1. Then f:[1,2] --> [2,3] is defined by f(x) = x + 1, f:[2,3] --> [3,6] is defined by f(x) = x^2 - 2x + 3, _____ and f:[3,6] --> [6,11] by f(x) = x + 2?x - 2 + 1. The general formulation simplifies in both cases in that the formulas on any particular interval [x_k,x_{k+1}] can be expressed "by radicals" because it is in fact a composition of the polynomial functions f and g and the easily expressed inverse g?�: _____ g?�(x) = ?x - 2 Final Remarks ============= The two examples above share a strange affinity. What is it and why does it happen? Neither of the examples above give differentiability of f at the boundary "knots" x_k. What "knot", and what would the "smoothest" possible solution be, characterized in terms of a definition f on [0,a]? regards, mathtalk-ga``` Request for Answer Clarification by antthrush-ga on 16 Jul 2005 13:20 PDT ```Hi Mathtalk, Your solution is a good start, but not terribly satisfying. It's not clear to me that the function f(x) has to be strictly increasing. for example, f(x) = 1/(x+1) is decreasing, but f(f(x)) is increasing. I think there should be an analytic solution. Perhaps in can be constructed your way. But perhaps not. It might be possible to construct an analytic function by taking some limit of your functions, and values of a. But I can't even construct a solution with a continuous first derivative. If you have any other thoughts, about whether there should be a smooth solution, or even a way to solve this numerically, I would be happy to hear them. --Antthrush``` Clarification of Answer by mathtalk-ga on 16 Jul 2005 23:17 PDT ```Antthrush-ga wrote: > Hi Mathtalk, > Your solution is a good start, but not terribly satisfying. Sorry about that, but I appreciate your giving me the chance to extend and clarify my remarks. > It's not clear to me that the function f(x) has to be strictly > increasing. for example, f(x) = 1/(x+1) is decreasing, but f(f(x)) > is increasing. My goal was to characterize (using "easy" math) _all_ solutions which share with g(x) = x^2 + 2 the properties: - continuous on the real numbers - even symmetry - strictly increasing for nonnegative x I therefore omitted any proof of this last property, though I'll be happy to discuss its status with respect to the possibility of more general solutions. It is certainly possible to obtain more general solutions, for example by removing the requirement of continuity or even that the function be defined at x=0. Since the latter prevents the functional equation f(f(x)) = g(x) from being even defined at x = 0, in my opinion such solutions are "inferior" to ones which are continuous on all real numbers. As far as f(x) being strictly increasing on the nonnegative real numbers goes, let's start with the observation that g(x) is 1-1 on the nonnegative real numbers, which follows from the existence of its inverse g?� on [2,+oo). Then f(x) must also be 1-1 on the nonnegative real numbers, since f(a) = f(b) would imply g(a) = g(b), and hence a = b. Consider what it means for a continuous function like f to be 1-1 on an interval [a,b]. It means (by the Intermediate Value Theorem) that f is strictly monotone, whether increasing or decreasing, on [a,b]. By extension this is so (f strictly monotone) on all nonnegative real numbers Of course as your example f(x) = 1/(x+1) shows, the composition of a strictly decreasing function with itself does produce a strictly increasing result. However in this case we can prove that f(x) is strictly increasing on the nonnegative real numbers, given the other assumptions of continuity and evenness. There are two cases to consider. Either f(0) = a > 2 or a < 2. Since f(a) = g(0) = 2 and f(2) = g(a) = a^2 + 2 > 2, we have: A) if a > 2, then f is strictly decreasing on [2,a] & thus on [0,+oo) B) if a < 2, then f is strictly increasing on [a,2] & thus on [0,+oo) But if A) and f were strictly decreasing on [0,+oo), then f would be bounded above on the nonnegative real numbers by f(0) = a. Invoking the evenness of f would be bound f above by a for all real numbers. Hence g would be bounded above by a. But since g(x) = x^2 + 2 is not bounded above, that would be a contradiction. While we're at it, we may as well complete the demonstration that: f(0) = a > 0 Clearly a = 0 cannot be true, since that gives this contradiction: 0 = a = f(0) = f(a) = 2 Suppose, also for sake of contradiction, that a < 0. Then since: f(2) = f(g(0)) = g(f(0)) = a^2 + 2 > 2 f changes signs on [0,2], and hence has root f(b) = 0 for some b in [0,2] by the Intermediate Value Theorem. But then g(b) = f(f(b)) = f(0) = a < 0, which contradicts that g(b) = b^2 + 2 is not less than 2. > I think there should be an analytic solution. Perhaps [it] can be > constructed your way. But perhaps not. It might be possible to > construct an analytic function by taking some limit of your functions, > and values of a. But I can't even construct a solution with a > continuous first derivative. Requiring the solution to be analytic is not likely to be enough to guarantee uniqueness. For a discussion (using substantial mathematical machinery, but with some really good diagrams, such as unfortunately I cannot provide directly in this text-based medium) of the issue of which solution of the functional equation is "best", see this link to a PDF download of a research paper by Resch, Stenger, and Waldvogel that appeared in 2000: [ Functional Equations related to the Iteration of Functions] (Aequationes Math. 60, 2000, 25-37) http://www.sam.math.ethz.ch/~waldvoge/Papers/functequ.html They develop a general theory, give an analytical solution (Taylor series) for one specific problem, and discuss "the question of selecting distinguished solutions from an continuum of possible solutions." The computation of an analytic solution is best approached through a change of domain and range. Note that if h(x) = 1/f(1/x), then h satisfies a functional equation: x^2 h(h(x)) = 1/f(1/h(x)) = 1/f(f(1/x)) = 1/(x^-2 + 2) = -------- 2x^2 + 1 Basically we've made a similarity transformation that centers attention at the origin being a fixed point for h(x), where previously the equivalent fixed point for f(x) would have to be "at infinity". In short one way to pick a "best" solution f(x) is to specify the one with the most regular behavior at infinity. Before pursuing further the topic of analytic solutions, let's construct by the elementary method outlined above "a solution with a continuous first derivative". Leave f(0) = a as a free parameter, 0 < a < 2, and define f:[0,a] --> [a,2]: f(x) = ((2 - a)/a^2) * x^2 + a Then f:[a,2] --> [2,a^2 + 2] has this first-degree polynomial formula: f(x) = (a^2)(x - a)/(2 - a) + 2 The first derivative of f(x) as we approach x = a from below is: LIMIT (2(2 - a)x)/(a^2) = 2(2 - a)/a x -> a- while the first derivative of f(x) approaching x = a from above is constantly: (a^2)/(2 - a) Naturally we will choose free parameter a to make these derivatives agree: 2(2 - a)/a = (a^2)/(2 - a) a^3 = 2(a - 2)^2 a^3 - 2a^2 + 8a - 8 = 0 and Cardano's formula gives an explicit value for the root of interest: a = (2/3) + c - 20/(9c) where c = ( 4*SQRT(69)/9 + (44/27) )^(1/3) and numerically a = 1.139680581996106... > If you have any other thoughts, about whether there should be a smooth > solution, or even a way to solve this numerically, I would be happy to > hear them. Yes, there should be a smooth solution in the sense of a series expansion "around infinity". An expository paper on the general topic of "fractional iterates" of a general quadratic is: R.E. Rice, B. Schweizer & A. Sklar, When is f(f(z)) = az^2 + bz + c for all complex z? Amer. Math. Monthly 87 (1980), 252-263 The demonstrate that if (b - 1)^2 - 4ac =< 1, then for every integer r > 1 there exists solutions such that f^r = g, ie. chain composing r copies of f gives g. Note that in your case a = 1, b = 0, and c = 2, and that the inequality: (b - 1)^2 - 4ac = 1 - 8 =< 1 is certainly satisfied. To clarify any doubts about whether the definition I've outlined can produce explicit values for a solution, I propose that you pick: - some value a between 0 and 2 - some strictly increasing function f:[0,a] --> [a,2] - some argument x Provided the function f and its inverse are reasonably explicit, I will then evaluate the function f at your chosen argument x. regards, mathtalk-ga``` Clarification of Answer by mathtalk-ga on 17 Jul 2005 23:16 PDT ```I'd like to comment on some of the interesting Comments posted to this thread. reinedd-ga made a suggestion that the problem statement is perhaps flawed, and should be f(f(x)) = x^(2+2) = x^4, which as hfshaw-ga points out would have a simple solution f(x) = x^2. Curiously enough there is also a fairly compact solution for the modified problem: f(f(x)) = x^2 - 2 This polynomial g(x) = x^2 - 2 is related to the Chebyshev polynomial P_2 by a simple "normalization". One often finds the definition: P_n(cos(t)) = cos(n*t) for this family of orthogonal polynomials. If instead one defines: F_n(2*cos(t/2)) = 2*cos(n*t/2) then F_2(x) = 2*(2*(x/2)^2 - 1) = x^2 - 2 = g(x). But the family of Chebshev polynomials has the property: P_m(P_n(x)) = P_mn(x) which is "similarly" shared by the family F_n. So one can formally assert that a solution to f(f(x)) = x^2 - 2 is given by: _ _ _ F_?2(x) where F_?2(2*cos(t/2)) = 2*cos(?2 * t/2) In other words: _ _ F_?2(x) = 2*cos(?2 * arccos(x/2)) For the solution of this modified problem I am indebted to David G. Cantor's March 1985 post (to an ARPAnet list?), available in Google Groups in a slightly redacted form here: [PROLOG Digest V3 #11] http://groups-beta.google.com/group/net.lang.prolog/msg/746e409319986d69 "The solution is essentially contained in the article by Michael Restivo in the March 20 issue of Prolog Digest." One is tempted to conceive that the original problem statement might have been in this easier form. Even so I would like to present a full solution to the problem as it has been assigned to us by antthrush-ga. regards, mathtalk-ga``` Request for Answer Clarification by antthrush-ga on 18 Jul 2005 11:31 PDT ```Now you're talking. This solution is much better. In fact, I think it gives the solution to my original problem. If we take f(x) = a Cos[ sqrt(2) ArcCos( x/a ) ] then f(f(x)) = a Cos[ sqrt(2) ArcCos( cos (sqrt(2) ArcCos(x/a) ] = a Cos[2 ArcCos(x/a) ] = 2 x^2/a - a thus we take a = -2 and x-> I x. so the solution is f(x) = -2 Cos [ sqrt(2) ArcCos( - i x/a) ] I don't even think there's an ambiguity about the definition of ArcCos here, since arccos( i x) = pi/2 + i arcsinh(x), and arcsinh is well defined. thanks for your help!``` Clarification of Answer by mathtalk-ga on 18 Jul 2005 21:01 PDT ```Hi, antthrush-ga: The solution of f(f(x)) = g(x) = x^2 - 2 does allow equally the solution of: -g(-x) = -(x^2) + 2 but I do not think we can change x |--> ix afterwards within the context of the problem. Notice that the formal definition: _ f(2 cos(t/2)) = 2 cos(?2 * t/2) only provides values for f(x) on [-2,2]. However the hyperbolic cosine satisfies the same "double angle" identity as cosine: cosh(2t) = 2 cosh^2(t) - 1 and so the definition may be extended outside of [-2,2] by simply replacing the circular function by its hyperbolic counterpart. To apply this technique to solving g(x) = x^2 + 2 would require that we have a function satisfying (say): h(2t) = 2 (h(t))^2 + 1 Unfortunately the closest that elementary transcendental functions can bring us to such an identity is probably this one that mixes two different functions: cosh(2t) = 2 sinh^2(t) + 1 Moreover, if we take t = 0 in the functional equation above for h, we get: 2(h(0))^2 - h(0) + 1 = 0 _ whose only solutions are complex: h(0) = (1 � i?7)/4. This, together with your impulse to substitute x |--> ix, points to expanding the search for solutions from real valued to complex valued functions. regards, mathtalk-ga``` Clarification of Answer by mathtalk-ga on 28 Jul 2005 22:53 PDT ```There are some loose threads I'd like to tie up in connection with this Question, having to do with the existence of solutions to: f(f(x)) = g(x) where g(x) = x^2 + 2, or more generally some other quadratic. The theme that I'd like to pick up is the importance of specifying a domain for f (and hence for g) in posing such problems. The paper by Rice, Schweizer, and Sklar which we referenced above has as its first theorem the answer for every quadratic g(x) to existence of functions of a complex variable: Thm. No f:C -> C satisfies the above equation for all x in C. Their proof uses a simple combinatorial approach that assumes no analyticity or even continuity of the proposed solutions. The complete proof would deal with a few special cases, but we can explain how the proof applies to g(x) = x^2 + 2 directly. The fixed points of g(x) = x^2 + 2 are the two roots of this: g(x) = x x^2 - x + 2 = 0 x = (1 � i SQRT(7))/2 We are interested in showing that g(x) has one and only one pair of points a,b such that: g(a) = b, g(b) = a, a not equal to b These points are in fact the two roots of the quartic equation: g(g(x)) = x (x^2 + 2)^2 + 2 = x that are _not_ fixed points of g, ie. that are different from (1 � i SQRT(7))/2. We may refer to the set {a,b} as a two-cycle (or orbit of length 2) as g transposes them. Next we consider what values f takes on this set. Since f(a) = f(g(b)) and f(b) = f(g(a)), it follows from the commutativity of f and g that: g(g(f(a))) = f(g(g(a))) = f(a) so f(a) is a root of g(g(x)) = x and likewise so is f(b). But g(a) = b, g(b) = a implies neither f(a) nor f(b) can be fixed points of g, since if g(f(a)) = f(a), then applying f to both sides: f(g(f(a))) = f(f(a)) g(g(a)) = g(a) a = b (contradiction) Since f(a), and similarly f(b), is not a fixed point of g, {f(a),f(b)} is by implication the set {a,b}. We could not have f(a) = a, since this would imply g(a) = a, so instead it must be that f(a) = b. But this implies f(b) = a, and hence g(a) = f(f(a)) = a. By construction a is known not to be a fixed point of g. Contradiction. Thus there is no functional square root of g(x) = x^2 + 2 define on the (entire) complex domain, so it makes sense to look more carefully at solutions on the real domain. We quoted earlier from a theorem which Rice, Schweizer, and Sklar give (without proof) at the end of their cited paper. They give a result that establishes that if g(x) = ax^2 + bx + c has real coefficients and d = (b - 1)^2 - 4ac, then there is a functional square root over the real domain if and only if d <= 1. The negative result for d > 1 can be established by the same logic that guided our treatment of the complex domain. We will illustrate the argument for the apparently "nice" case: g(x) = x^2 - 2 presented earlier with a "formal solution": f(2 cos(t/2)) = 2 cos(SQRT(2)*t/2) A bit of algebra shows that the two fixed points of g(x): g(x) = x are x = 2,-1. Furthermore the fixed points of g(g(x)) that are not fixed points of g(x) are real roots of this quadratic: x^2 + x - 1 = 0 -1 � SQRT(5) x = -------------- 2 As argued before (or as one can directly calculate), this pair of values x = a,b has the property that g(a)=b and g(b)=a. They thus form a two-cycle under the action of g, and it follows that no functional square root of g can be satisfactorily defined on {a,b}. But what is wrong with the definition of "formal solution" f as presented above? The first point to make is that f is not well- defined; that is the equation above that supposedly defines f does not assign unique values for f(x) given a definite value of x. In brief the problem is that cos(t/2) has period 4pi, but unless k is an integer, cos(k * t/2) will not have period 4pi. The "result" 2 cos(SQRT(2)*t/2) will therefore depend on which "representative" t is chosen to make x = 2 cos(t/2). Efforts to repair the definition are doomed to failure on the interval [-2,2], which is where the cosine function would seem to apply above, essentially because this "region" of the real numbers contains the two-cycle derived above. On the other hand using the hyperbolic cosine to assign values for x >= 2 and/or x <= -2 does work (because 2 cosh(t/2) is not periodic on the reals, or rather has a period of imaginary modulus). The more explicit construction given by David G. Cantor that we cited earlier is also limited to values outside [-2,2] since his formulation uses terms like SQRT(x^2 - 4). This echoing note on the importance of the domain in posing a question about the existence of f(x) s.t. f(f(x)) = g(x) is a good place to stop. regards, mathtalk-ga```
 antthrush-ga rated this answer: ```Great job researching and explaining the answer. But I still think there's an analytic solution...``` Subject: Re: Math question From: reinedd-ga on 13 Jul 2005 06:55 PDT
 `could the function be f(f(x)) = x^(2+2)`
 Subject: Re: Math question From: antthrush-ga on 13 Jul 2005 11:22 PDT
 ```Do you mean f(f(x))=x^4? The question is what is f(x), not what is f(f(x)).```
 Subject: Re: Math question From: biophysicist-ga on 14 Jul 2005 05:22 PDT
 ```I thought about it for a little while, and I don't see any obvious answer. I also don't know a general procedure for solving such problems. Could your friend have been mistaken when he said it was easy? f(x) can't be simply a polynomial in x. If f were quadratic, f(f(x)) would involve x^4. If f were linear, f(f(x)) would also be linear. So f would have to be something more complicated. If you do find out the answer, please post it here so I'll know what it is!```
 Subject: Re: Math question From: juanitin-ga on 14 Jul 2005 12:42 PDT
 ```I have a rather straightforward solution but I don't know whether it is the one you are interested in. We can define the following function separately in two regions: f(x)= x+j*sqrt(2) if Im(x)=0; f(x)=x*(x^*) if Im(x) is not zero where x^* means complex conjugate and Im(x) means imaginary part of x and j*j=1. Thus, if x is a real variable, we have that f(x)=x+i*sqrt(2), and f(f(x))=x^2+2. Of course, I have assumed that f is a function of complex variable, while x is real. If somebody has any information about how to attack these problems in a more general form please let me know! I have found the problem appealing.```
 Subject: Re: Math question From: hfshaw-ga on 14 Jul 2005 13:39 PDT
 ```I think what reinedd was asking is whether the original questioner might have misread or misunderstood the problem. Reinedd was asking whether the problem involved finding f(x) when f(f(x)) = x^(2+2) rather than (x^2)+ 2. The former has a straightforward solution (i.e., f(x) = x^2). The latter problem is (despite what juanitin wrote) is not what most would consider "straightforward"!```
 Subject: Re: Math question From: antthrush-ga on 14 Jul 2005 14:35 PDT
 ```juanitin's answer is cute, but somewhat unsatisfying. especially since it doesn't work if x is not real. i don't know if the function is supposed to be real or not, but i think there's probably a better solution.```
 Subject: Re: Math question From: juanitin-ga on 15 Jul 2005 07:29 PDT
 ```I agree with you, the solution I posted is somewhat "ugly". I have another solution, although probably there is another more elegant (I'm afraid this one is "ugly" too!). But this time x is real and the function also,i. e., f:R->R. The function is the following: f(x)=lim(t->0) (1+d(e^(t*x^2))/dt) This is, f(x) is the limit when t->0 of 1 plus the derivative with respect to t of e^(t*x^2) The calculation gives: f(x)=1+x^2 f(f(x))=2+x^2 If I have time I would try something easier like a fraction of two polynomials, or any other more often used function, but in any case the function given above is a solution.```
 Subject: Re: Math question From: juanitin-ga on 15 Jul 2005 09:26 PDT
 ```By the way, I agree with biophysicist. It doesn' seem that a polynomial will work. PD: when I say I would try something easier, I mean that I will try to find a function which can be expressed in a simplier and nicer form. But to me it's more difficult to find such a solution than the ones I have posted since from the first attempts I made with "easy" functions I got nothing. Thanks for the question, antthrush!```
 Subject: Re: Math question From: antthrush-ga on 15 Jul 2005 09:46 PDT
 ```I don't understand juanitin's solution. Isn't f(x) = lim 1 + d/dt exp(t x^2) = 1+x^2 in which case f(f(x))= 2 + 2 x^2 +x^4? A finite polynomial probably won't work. But an infinite series might. We can write f(x) = sum a_n x^2n, then f(f(x)) is a series. matching to x^2 + 2 gives a countable series of equations which can be solved, in principle. then maybe the series can be summed giving a closed form answer. but i haven't had any luck with this approach.```
 Subject: Re: Math question From: juanitin-ga on 16 Jul 2005 15:01 PDT
 ```Very good point, mathtalk! But it seems that antthrush (which was completely right when he said that my posted solution for real functions was wrong, I made a silly miscalculation!) looks for a smoother function. In any case I have found your posted solution a bright answer, although it doesn't cover all the casuistics as antthrush pointed out. I will think more on that (and I will be more careful!)```
 Subject: Re: Math question From: mathtalk-ga on 20 Jul 2005 08:32 PDT
 ```The word "analytic" may suggest something different to interested readers than what it means in higher mathematics. An analytic function is one with a derivative in the sense of a function of a complex variable. This turns out to be a much stronger notion than having a derivative in the sense of a function of a real variable. Indeed being analytic implies having derivatives of all orders and a Taylor series expansion in the neighborhood of any point where the function is analytic. Often the phrase "analytic function" needs to be qualified as to where, in the complex plane, the function is analytic. Functions that are analytic everywhere in the complex plane are then called "entire" functions. Functions that are real-valued on the real numbers and have convergent power series expansions in a neighborhood of each real argument are called "real analytic". This is the sense of my asserting that there is an analytic solution, not ncessarily a "neat, compactly defined" one, but a convergent power series, and further that requiring the solution to be analytic will not make the answer unique. A nice collection of references to the literature, but unfortunately few articles available on-line, is here: [Iterative Roots and Fractional Iteration] (references collected by Lars Kindermann) http://reglos.de/lars/ffx.html regards, mathtalk-ga```
 Subject: Re: Math question From: dyscogitator-ga on 12 Sep 2005 07:47 PDT
 ```This message thread was recently brought to my attention. I believe that in all likelihood I am the "person who told it to him," as referenced in the original question. Except that when I first posed it, I thought that I specified "for all Real x." . . . smooth . . . analytic . . . continuous . . . Put all your maths away. You're killing me! Let's think very simply about this, and, at the same time, let's generalize: Let f(#n)(x) denote n iterations of the function f(x). [e.g. f(#5)(x) = (f(f(f(f(f(x)))))]. Consider problems of the shape f(#n)(x) = g(x) in which the domain of f(#n)(x) and the range of g(x) are both Real, where f(x) is to be determined. Please consider: / | x+(n-1)i for REAL x f(x) = < g(x-i) for REAL (x-i) | x-i for everything else \ Does this not satisfy?```
 Subject: Re: Math question From: mathtalk-ga on 12 Sep 2005 09:23 PDT
 ```dyscogitator-ga asked: "Does this not satisfy?" The challenge of this problem lies in finding a way to satisfy the criterion on the entire domain of f. The domain of the nth iterate f(#n) of f, to use dyscognitator-ga's notation, is in fact the same as the domain of f. It would therefore not be accurate to claim that "the domain of f(#n)(x) and the range of g(x) are both Real." There is no difficulty in finding any number of solutions where: f(f(x)) = x^2 + 2 is satisfied on only a part of the domain of f. There are piecwise polynomial real-valued functions f which are continuous and satisfies the equation for all real numbers. Continuity can certainly be improved to differentiability, and possibly to real analyticity. Notice that the approach of defining a complex-valued function on R by adding an imaginary constant to shift the argument away from the reals, to a line parallel in the complex plane, then defining the "solution" by assigning values there, was offered by juanitin-ga on July 14, 2005 (see above). Such a solution bears no essential relationship to the prescribed real-valued function x^2 + 2. This technique can supply any desired function as a "self-composition" on the reals as a subdomain of the complex numbers, a fact dyscognitator-ga sees as "generalization" but which the concensus of the Commenters and the Customer made out to be "unsatisfying". Note that antthrush-ga (July 14, 2005) wrote "juanitin's answer is cute, but somewhat unsatisfying. especially since it doesn't work if x is not real." regards, mathtalk-ga``` 