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June 10

Functional powers

Beginning with the usual definition of f⁻¹ for a function f and

f^{n}:=\underbrace {f\circ f\circ \dots \circ f} _{n}

led me to the idea of functional roots (i.e. if g² = f, then g = f^1/2) and then to rational powers of functions (i.e. f^a/b = (f^1/b)^a). From here, I had several questions:

How could you extend the definition to real or complex powers of functions? Is it even possible, given the duplicity of things like f^1/2 (i.e. if f(x) = g(x) = x and h(x) = 1/x, then g(g(x)) = h(h(x)) = f(x), so g and h are both "f^1/2" for the right domain)? If it is possible, you could consider taking functions to the power of other functions, which is an interesting concept (to me, anyway).
How do you go about graphing or finding formulaic approximations for functions like rin = sin^1/2 (see here)? How was it determined that as n goes to infinity, sin^1/n goes to the sawtooth function?
Do these notions have any particular application?

Thanks in advance for taking the time with these loaded and naïve questions. —Anonymous Dissident^Talk 12:44, 10 June 2011 (UTC)[reply]

Your questions will take you into the area of functional equations. One complication you will encounter is that the functional square root of a function is usually far from unique - for example, the function

y(x)=x

has functional square roots

y_{(1/2)}={\frac {x+a}{ax-1}}

for any real (or complex) number a. Gandalf61 (talk) 14:13, 10 June 2011 (UTC)[reply]

The set of all "square roots" of the identity function on an arbitrary set

X

is in natural one-to-one correspondence with the set of all partitions of

X

into sets of size 1 or 2. --COVIZAPIBETEFOKY (talk) 15:00, 10 June 2011 (UTC)[reply]

You can define a generator of a function as follows. If:

$f(x)=\lim _{n\to \infty }\left[x+{\frac {g(x)}{n}}\right]^{\circ n}$

then we can consider g(x) to be a generator of f(x). Count Iblis (talk) 15:44, 10 June 2011 (UTC)[reply]

I get a one-to-one correspondence between the complement of a projective algebraic variety in the complex projective plane and the Möbius transformations that are functional square roots of g(z) = z. Let [a : b : c] be in CP², with a² − bc ≠ 0, and define

f(z):={\frac {az+b}{cz-a}}\,.

we see that ƒ has the property that (ƒ ∘ ƒ)(z) = z. — Fly by Night (talk) 11:32, 11 June 2011 (UTC)[reply]

If you're just interested in constructing the functional square root, then the method of Kneser (referenced in our article) seems to be worth studying. This paper constructs a functional square root of the exponential function, and the method seems like one could work it out without a great deal of specialized knowledge. If you're interested in looking at the whole semigroup (if there is one—which seems to me a little unlikely) of functional roots $\sin ^{\alpha }$ , then some basic papers on this subject appear to be Erdos and Jabotinsky (1960) [1] and Szekeres (1958) "Regular iteration of real and complex functions", Volume 100, Numbers 3-4, 203-258, DOI: 10.1007/BF02559539. It seems to be a theorem that there is no way to define non-integer iterates of the exponential function so that the semigroup property holds (maybe along with analyticity in the iteration parameter, it's not clear to me what the rules are). The following paper also seems to be worth looking at: Levy, (1928) "Fonctions à croissance régulière et itération d'ordre fractionnaire", Annali di Matematica Pura ed Applicata Volume 5, Number 1, 269-298, DOI: 10.1007/BF02415428. I haven't found any (reliable) papers that specifically address the sine function. There's this, which I find a little dubious. Sławomir Biały (talk) 13:55, 12 June 2011 (UTC)[reply]