Fractional Function Iteration

Let \(f\colon X \to X\) be an endofunction on a set \(X\) so that it is possible to chain \(f\) with itself. We have \(f^0 = \mathsf{id}\), \(f^1 = f\), \(f^2 = f \circ f\), \(f^3 = f \circ f \circ f\), etc. all as endofunctions \(X \to X\) again. In general, \(f^0 = \mathsf{id}\) and \(f^{k+1} = f \circ f^{k}\) for all \(k \in \omega\). But what about \(f^{\frac{1}{2}}\) -- or any other fractional exponent for that matter? We want to require that \(f^a \circ f^b = f^{a + b}\) holds for all \(a, b \in \mathbb{Q}_{\geq 0}\). This property clearly is satisfied for the natural numbers case. But now we attempt to solve this system of functional equations for all fractional exponents. In general, such functional equations can be very hard: for instance just have a look at the massive amount of work published about partial differential equations, which are just a small subset. Most students already struggle with ordinary differential equations (I do so myself!), which may be seen as the most convenient setting. In general we cannot even expect to find a solution, since \(f\) may have weird incontinuities. So let's enrich \(X\) with a topology, and restrict ourselves to the continuous endofunctions \(f\colon \mathcal{X} \to \mathcal{X}\) instead. In particular, we may hope to find solutions for polynomial functions, since this class is very well behaved.

With the problem setting out of the way, let's start at hunting a solution. As a first example, let \(f(x) = a x\) as a function \(\mathbb{R} \to \mathbb{R}\) and some \(a \in \mathbb{R}\). We easily find the solution that \(f^r(x) = a^r x\) since \((f^p \circ f^q)(x) = a^p a^q x = a^{p+q} x = f^{p+q}(x)\) holds for all \(x \in \mathbb{R}\). Let us consider \(f(x) = x^2\) next. We have \(f^{\frac{1}{2}}(x) = x^{\sqrt{2}}\) because \(f^{\frac{1}{2}}(f^{\frac{1}{2}}(x)) = (x^{\sqrt{2}})^{\sqrt{2}} = x^{\sqrt{2} \cdot \sqrt{2}} = x^2 = f(x)\) holds for all \(x \in \mathbb{R}\). We can expand this nicely to any monomial function \(f(x) = x^b\) and any fractional exponent \(r\), having \(f^r(x) = x^{(b^r)}\). We basically have exploited the \(b\)-homogeneity of the function \(f\).

However, the problem already becomes harder if we mix both the coefficient and exponent, and consider functions of the form \(f(x) = a x^2\). The reason for this is that \(a\) appears within the exponent of the outer function application. Thus, the naive attempt \(f^{\frac{1}{2}}(x) = \sqrt{a} x^{\sqrt{2}}\) already fails: [f^{\frac{1}{2}}(f^{\frac{1}{2}}(x)) = \sqrt{a} (\sqrt{a} x^{\sqrt{2}})^{\sqrt{2}} = \sqrt{a} \sqrt{a}^\sqrt{2} x^2 \neq a x^2 = f(x)] for any \(x \neq 0\). From this we can read off that we need to find a coefficient \(a_{\frac{1}{2}} \in \mathbb{R}\) that solves \(c \cdot c^{\sqrt{2}} = a\) as an equation of \(c\). Obviously such a number exists,^[Indeed, \(a_{\frac{1}{2}} = a^{\frac{1}{1 + \sqrt{2}}}\) is the unique solution.] and thus we are able to find \(f^{\frac{1}{2}} = a_{\frac{1}{2}} x^{\sqrt{2}}\) with the desired property. But does this still work for other exponents? Let us try \(\frac{1}{4}\) next, since then we may immediately want to check that \(f^{\frac{1}{2}} = f^{\frac{1}{4}} \circ f^{\frac{1}{4}}\) holds. With an analogous attempt, we define \(a_{\frac{1}{4}}\) as the (unique) solution to \(c^{2^{0/4}} \cdot c^{2^{1/4}} \cdot c^{2^{2/4}} \cdot c^{2^{3/4}} = a\) as an equation of \(c\).^[I hope you catch the pattern from the prior equation.] We compute \((f^{\frac{1}{4}} \circ f^{\frac{1}{4}})(x) = a_{\frac{1}{4}} \big(a_{\frac{1}{4}} x^{2^{\frac{1}{4}}}\big)^{2^{\frac{1}{4}}} = a_{\frac{1}{4}} a_{\frac{1}{4}}^{2^{\frac{1}{4}}} x^{\frac{1}{2}}\) which is equal to \(f^{\frac{1}{2}}(x)\) for all \(x \in \mathbb{R}\) if and only if \(c \coloneqq a_{\frac{1}{4}} a_{\frac{1}{4}}^{2^{\frac{1}{4}}}\) is equal to \(a_{\frac{1}{2}}\).^[Since the equation that defines \(a_{\frac{1}{2}}\) has a unique solution!] Indeed, checking the defining equation for \(a_{\frac{1}{2}}\) we have that \(c \cdot c^{2^{\frac{1}{2}}} = a_{\frac{1}{4}} a_{\frac{1}{4}}^{2^{\frac{1}{4}}} a_{\frac{1}{4}}^{2^{\frac{2}{4}}} a_{\frac{1}{4}}^{2^{\frac{3}{4}}} = a\) by our definition of \(a_{\frac{1}{4}}\). Note that we should actually check all permutations of these functions, since function composition \(\circ\) is not a commutative operation! Gladly, here this boils down to commutativity of multiplication. We expect this pattern to generalize for other exponents \(f(x) = a x^b\) and all other fractional functional exponents \(f^r\).

Now sums don't play so nicely with exponents. By the distributive law, we have covered already functions like \(f(x) = a_1 x + a_2 x\) by the simple case \(f(x) = (a_1 + a_2) x\). What about \(f(x) = x + c\)? This is still easy and we get \(f^r(x) = x + rc\). Next, let us take a look at \(f(x) = x^2 + x\) and try to find \(f^{\frac{1}{2}}\). This is not so easy anymore, and indeed, I struggle to solve it.^[I just wrote all this up on a Sunday's afternoon!] Since the monomial case with coefficients was already quite involved, I deem it doomed to find a solution for arbitrary polynomials. I am actually quite astonished to have gotten this far already! In the future I plan to look more into linear functions \(f\) but for general vector spaces instead of just \(\mathbb{R}\), since this is the case I would need to solve for my original problem. There, I expect to encounter the exponential map for matrices from Lie-theory.

As an addendum, we have (by definition) \(f(x^\ast) = x^\ast\) if \(x^\ast\) is a fixed point of \(f\). One expects that \(f^r(x^\ast) = x^\ast\) but in general this need not be the case -- not even if \(f^r\) varies continuously with \(r\). But conversely, if \(x^\ast\) is a fixed point of \(f^{\frac{1}{k}}\) then it must also be one for \(f\): By induction, we get \(f(x^\ast) = (f^{\frac{1}{k}})^k(x^\ast) = x^\ast ~.\) Note that this may not hold for arbitrary fractional exponents \(f^{\frac{n}{m}}\). Here, by the same argument, we can only guarantee that \(x^\ast\) is a fixed point of \(f^{\frac{\operatorname{lcm}(n,m)}{m}} = f^k\) where potentially \(k \geq 2\), which may have additional fixed points to that from \(f\).

After some discussion with a few friends, Linus found a paper by R. E. Rice that addresses where I left off and references more general work about fractional iterations by R. Isaacs. However, I found it fun to attempt the problem myself, and haven't read any of this yet.