Wikipedia:Reference desk/Archives/Mathematics/2016 June 14

= June 14 =

f(e^x) = e^f(x)
Functions for which this is true include the Half-exponential function since f(f(x))=e^x imples that f(f(f(x)))) = f(e^x) = e^f(x). It appears that this can also be done with the "one third" exponential function, and I *think* any rational iterate of the Exponential function. What I'm unclear on is whether than can be expanded beyond the rationals to the reals and whether by definition, such a function has to be an iterate of e^x.Naraht (talk) 14:21, 14 June 2016 (UTC)
 * If one requires of a fractional or real iterate of a function that it adheres to the identities f1(x) = f(x) and fa+b(x) = fa(fb(x)) for an interval of x and the interesting exponents, then it seems trivial to show that any iterate f of ex will satisfy f(ex) = ef(x). The converse, however is not true: not every function that satisfies the latter identity will be an iterate of ex though; a counterexample is the identity function f(x) = x.  It might help to clarify exactly what you are asking.  The question of existence of an iterate for a specific exponents is thorny, I guess, and probably constraints (such as continuity in both the argument and in the exponent, monotonicity in the argument, with domain and range conditions) might make the problem more tractable.  —Quondum 16:42, 14 June 2016 (UTC)
 * The Identity function is the zero-th interate of any function. If fa = gb then f and g are fractional interates of each other f is the b/a-th interate, but I'm not sure how that gets extended to the reals given the setup. So I guess the second half of the question more or less walks away from the specific for e^x, but rather the following for who functions f and g if fg = gf for the entire domain, *must* there exist numbers a and b such that fa = gb? if not, is it true if g(x) = e^x? Note if f is the identity then b = 0 and it works.Naraht (talk) 17:52, 14 June 2016 (UTC)
 * I seem to have inverted my ordering – I was thinking of functions such that an iterate is ex. So, yes, I must concede that.
 * This does not feel like it is adequately constrained. If if fg = gf, and we impose the condition that the functions be invertible, then inherently there exists numbers a and b such that this is true, even under the simplest definitions of iterate.  For example, a = 1 and b = −1, or both zero.  Extending iterates to the reals seems to be a more complex subject, but feels like it might be reasonable under tight constraints.  —Quondum 20:04, 14 June 2016 (UTC)
 * Not sure about ex, but if we're talking about general invertable functions on the reals (implicitly assuming continuity as well) then consider f(x) = sinx + x and g(x) = x + 2π. It's easy to verify that f(g(x))=g(f(x))=sinx + x + 2π. But f leaves the interval (0, π) invariant so all of its iterates must as well, while the iterates of g applied to (0, π) give disjoint sets. So fa = gb implies b=0 and fa is the identity. In addition, f(x) > x for x in (0, π), so fa(x) > x if a>0 and fa(x) < x if a<0, so the only a for which fa is the identity is a=0. This gives fa = gb implies a=b=0 which I assume you want to exclude since f0 = g0 is true whether f and g commute or not. The function ex makes things more complicated since you're now dealing with a semigroup structure instead of a group structure. But the fact that ex has no fixed points might help. --RDBury (talk) 01:03, 15 June 2016 (UTC)
 * "f(x) > x for x in (0, π), so fa(x) > x if a>0 and fa(x) < x if a<0" – doesn't this make some unstated (and not necessarily valid) assumptions on the fractional iterate? For example, we have not explicitly excluded the half-iterate f1/2(x) = −x of the identity function f(x) = x. I'd expect we'd need to formalize the idea of a "principal iterate" first.  —Quondum 02:41, 15 June 2016 (UTC)
 * Actually I wasn't talking about about fractional iterates but about the question of whether fg = gf implies fa = gb for some a and b; I was assuming that a and b are integers though. But it doesn't matter; if a and b are fractions, say a=m/n and b=p/q where m, n, p, q are integers, then we have fm/n=gp/q. If you iterate this nq times you get fmq=gpn and you're back to the integer case to get mq=pn=0. The denominators n and q can't be 0 so m and p are 0 and the conclusion is the same. --RDBury (talk) 03:05, 15 June 2016 (UTC)
 * Is the following reframing of the original question accurate? (rational case) Given the monoid of real functions under composition, does a pair of elements commuting imply that there exists an element such that each of the members of the pair is an integer power (repeated composition) of it?  The answer to this is clearly 'no' (take one function being the zero function).  So does the question need to be rephrased/constrained?  —Quondum 15:18, 15 June 2016 (UTC)
 * If you restrict to continuous, one-one, onto functions (i.e. the homeomorphism group of the reals), the answer is no. But this group is big and complex and messy so examples can be tricky to come up with and awkward to prove. So perhaps it would be better to start with a much smaller and better understood group such as the group of affine functions f(x) = ax+b, a≠0. If f(x) = 2x and g(x) = 3x then f and g clearly commute but if they have non-zero powers in common then you end up with 2a = 3b for non-zero integers and this isn't possible. On the other hand fractional iterates make more sense in this context and you could start talking about irrational iterates. So you could make a case that flog23=g. The perspective from which I'm looking at it is that we're talking about permutation groups acting on the reals. For permutation groups in general there's no reason to expect that elements that commute have powers in common unless you're talking about a torsion group in which case any two elements have powers in common. The fact that we're requiring that the group acts on the reals is simply making examples difficult to find.
 * Btw, the case f(x) = ex in the original question is not as bad as I thought. Let g(x) be any continuous function from [0, 1] to [0, 1] so that g(0)=0 and g(1)=1. The g can be continued to a continuous function on R which commutes with f similar to the construction in the half-exponential function article. Namely
 * $$ h(x) =

\begin{cases} g(x) & \mbox{if } x \in [0,1], \\ \exp (g( \ln (x))) & \mbox{if } x \in (1,\infty), \\ \ln (g( \exp (x))) & \mbox{if } x \in (-\infty,0). \\ \end{cases} $$
 * If you then take g so g(x)>x on (0, 1) then the proof given above shows that no (positive integer) power of g can equal a power of f. --RDBury (talk) 20:14, 15 June 2016 (UTC)