Notations expressing that f is a functional square root of g are f = g_[1/2] and f = g_1/2^{[citation needed][dubious – discuss]}, or rather f = g ^1/2 (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.

• The functional square root of the exponential function (now known as a half-exponential function) was studied by Hellmuth Kneser in 1950, later providing the basis for extending tetration to non-integer heights in 2017.^[1]
• The solutions of f(f(x)) = x over ${\displaystyle \mathbb {R} }$  (the involutions of the real numbers) were first studied by Charles Babbage in 1815, and this equation is called Babbage's functional equation.^[2] 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⁻¹∘ 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.

A systematic procedure to produce arbitrary functional n-roots (including arbitrary real, negative, and infinitesimal n) of functions ${\displaystyle g:\mathbb {C} \rightarrow \mathbb {C} }$  relies on the solutions of Schröder's equation.^[3][4][5] Infinitely many trivial solutions exist when the domain of a root function f is allowed to be sufficiently larger than that of g.

• f(x) = 2x² is a functional square root of g(x) = 8x⁴.
• A functional square root of the nth Chebyshev polynomial, ${\displaystyle g(x)=T_{n}(x)}$ , is ${\displaystyle f(x)=\cos {({\sqrt {n}}\arccos(x))}}$ , which in general is not a polynomial.
• ${\displaystyle f(x)=x/({\sqrt {2}}+x(1-{\sqrt {2}}))}$  is a functional square root of ${\displaystyle g(x)=x/(2-x)}$ .

sin_[2](x) = sin(sin(x)) [red curve]

sin_[1](x) = sin(x) = rin(rin(x)) [blue curve]

sin_[⁠1/2⁠](x) = rin(x) = qin(qin(x)) [orange curve], although this is not unique, the opposite - rin being a solution of sin = rin ∘ rin, too.

sin_[⁠1/4⁠](x) = qin(x) [black curve above the orange curve]

sin_[–1](x) = arcsin(x) [dashed curve]

Using this extension, sin_[⁠1/2⁠](1) can be shown to be approximately equal to 0.90871.^[6]

(See.^[7] For the notation, see [1] Archived 2022-12-05 at the Wayback Machine.)