Functional square root - Misplaced Pages

(Redirected from Half-iterate) Function that, applied twice, gives another function Not to be confused with Root of a function.

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In mathematics, a functional square root (sometimes called a half iterate) is a square root of a function with respect to the operation of function composition. In other words, a functional square root of a function g is a function f satisfying f(f(x)) = g(x) for all x.

Notation

Notations expressing that f is a functional square root of g are f = g and f = g_1/2, or rather f = g (see Iterated function#Fractional_iterates_and_flows,_and_negative_iterates), although this leaves the usual ambiguity with taking the function to that power in the multiplicative sense, just as f ² = f ∘ f can be misinterpreted as x ↦ f(x)².

History

The functional square root of the exponential function (now known as a half-exponential function) was studied by Hellmuth Kneser in 1950.
The solutions of f(f(x)) = x over $\mathbb {R}$ (the involutions of the real numbers) were first studied by Charles Babbage in 1815, and this equation is called Babbage's functional equation. A particular solution is f(x) = (b − x)/(1 + cx) for bc ≠ −1. Babbage noted that for any given solution f, its functional conjugate Ψ∘ f ∘ Ψ by an arbitrary invertible function Ψ is also a solution. In other words, the group of all invertible functions on the real line acts on the subset consisting of solutions to Babbage's functional equation by conjugation.

Solutions

A systematic procedure to produce arbitrary functional n-roots (including arbitrary real, negative, and infinitesimal n) of functions $g:\mathbb {C} \rightarrow \mathbb {C}$ relies on the solutions of Schröder's equation. Infinitely many trivial solutions exist when the domain of a root function f is allowed to be sufficiently larger than that of g.

Examples

f(x) = 2x is a functional square root of g(x) = 8x.
A functional square root of the nth Chebyshev polynomial, $g(x)=T_{n}(x)$ , is $f(x)=\cos {({\sqrt {n}}\arccos(x))}$ , which in general is not a polynomial.
$f(x)=x/({\sqrt {2}}+x(1-{\sqrt {2}}))$ is a functional square root of $g(x)=x/(2-x)$ .