Functional Equation ```Name: Unknown Status: N/A Age: N/A Location: N/A Country: N/A Date: Around 1993 ``` Question: Is there any way to approximate the solution of f(x) given f(f(x))=g(x)? i.e., if f(f(x))=x^4, then f(X)=x^2? I would like any method for approximating the solution to f(f(x))=e^x. Replies: Hey, this is a really interesting problem! First off, there is no unique solution to the problem as stated. For example, even for the identity under composition, g(x)=x, ANY f(x) that just transposes x values will work. For example, -x, or 1/x will do just as well as f(x) = x. f(x) = x is unique, however, under the requirement that f be monotonic increasing (which only makes sense because g is monotonic increasing). However, even monotonicity is not enough to determine f uniquely if g increases too fast, and for g(x) = exp(x). I suspect that you need to require all derivatives to be positive, or maybe something even stronger (any clues out there?). This non uniqueness is a little surprising, since there are "obvious" solutions in many cases. For example, for g(x) = x^p, f(x) = x^sqrt(p) is the obvious solution. For g(x) = exp(exp(x)), f(x) = exp(x) is the obvious solution. So, is there an obvious solution for g(x) = exp(x)? There must be in some sense, and maybe the condition on derivatives is the most sensible. But, I can construct a solution in an essentially arbitrary manner. Note that the solution must (for large x) rise faster than any power, and yet slower than any exponential of a (positive) power of x. So this f(x) is somewhere between power-law and exponential growth for large x. For x close to 0, exp(x) is approximately 1 + x. For g(x) = 1+x, the (obvious) solution is f(x) = 1/2 + x. So for x in the range 0 to 1/2, define f(x) = 1/2 + x. This then defines f(x) for all x! To see that this is so, consider f(0) = 1/2. Then f(f(0)) = exp(0) = 1 is satisfied because f(1/2) = 1. Then f(1) = f(f(1/2)) = sqrt(e) is immediately defined, and so is f(y) in the interval: y in [1/2, 1] => f(y) = f(f(y - 1/2)) = exp(y - 1/2). Then: y in [1, sqrt(e)] => f(y) = f(f(log(y) + 1/2)) = y*sqrt(e), so f(sqrt(e)) = e. Then: y in [sqrt(e), e] => f(y) = f(f(y/sqrt(e))) = exp(y/sqrt(e)) so f(e) = e^sqrt(e). Etc. The result is two alternating sequences of intervals in which f is defined by (ever more complicated) combinations of elementary functions. In the first set of intervals (successive exponentiation of [0, 1/2]), f(x) follows the sequence: x + 1/2, sqrt(e) x, x^(sqrt(e)), exp(log(x)^sqrt(e)), or in general (defining exp^(n)(x) to be exp(exp(...(exp(x))...)) with n exps, and log^(n)(x). Similarly, f(x) = exp^(n)((log^(n)(x))^sqrt(e)). This effectively bounds the function of interest from below by powers and faster rising functions. In the second set of intervals (successive exponentiation of [1/2,1]) f(x) follows the sequence: e^(x - 1/2), e^(x/sqrt(e)),e^(x^1/sqrt(e)), e^(e^(log(x)^1/sqrt(e))), or in general exp^(n+1)((log^(n)(x))^1/sqrt(e)), effectively bounding the function of interest from above by exponentials of powers and more slowly rising functions. (Note that these general expressions work for negative n, realizing that exp^(-n)(x) = log^(n)(x)). i.e., I would guess that for large x, exp^(n)((log^(n)(x))^sqrt(e)) < f(x) ``` if (x < 0) { neg = 1; nxr = -1 shiftedx = exp(x); /* Between 0 and 1 */} else { neg =0; nxr = 0 shiftedx = x; while (shiftedx > 1) {shiftedx = log(shiftedx); nxr++:} if (shiftedx < 0.5) { fshift = shiftedx + 0.5;} else {fshift = exp(shiftedx - 0.5);} if (neg == 1) {res = log(fshift);} else {res = fshift; for (i=0; i < nxr; i++) {res = exp(res); ``` I looked this up, and it turns out that the idea of "fractional iteration" is quite an old one, and has been solved somewhat satisfactorily. I could not find any explicit solution for g(x) = exp(x), however, so I will comment on that later. The general method of solution (when it can be done) is to try to find a solution to "Abel's equation": phi(g(x)) = phi(x) + 1. Then the fractional iterate f(x) = g^k(x) (k = 1/2 in our example, but it could be anything) is given by: f(x) = phi^-1 ( phi(x) + k). The solution given above for g(x) = exp(x) corresponds to taking phi(x) = 1 + x in the interval x = 0 to 1, and applying Abel's equation to find phi(x) everywhere else. This turns out to be an excellent approximation to the true (infinitely differentiable) phi(x). In fact, we can satisfy n derivatives of Abel's equation at x = 0 by expressing phi as a n+2'th order polynomial between 0 and 1, and this appears to converge to a power series: phi(x) approx = 1 + 0.9159456 x + 0.2493525 x^2 -0.11046 x^3 + ... Unfortunately this power series seems to have only a finite radius of convergence, but it does include the endpoints 0 and 1. The straight line 1+x is very close to this function. Arthur Smith Click here to return to the Mathematics Archives

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