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Given a function $f(x)$ whose image is a subset of its domain, we can define $$ f^n(x) = \underbrace{f(f(f(\dots f(x) \dots )))}_{n \text{ times}} $$

This makes sense when $n$ is a nonnegative integer.

Can we extend this definition to continuous values of $n$? Such as $f^{\frac{1}{2}}(x)$?

Alex Van de Kleut
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  • Maybe, $f^{1/2}$ is defined so that $f^{1/2}\circ f^{1/2}=f$. However, this definition is not good, because in general $f^{1/2}$ is not the only such function (why?). – azif00 Oct 18 '19 at 03:01
  • It turns out it's called a fractional iterated function! – Alex Van de Kleut Oct 18 '19 at 03:02
  • Consult the posts here with the tag (fractional-iteration) then tell us why this one should not be closed as a duplicate. – GEdgar Oct 18 '19 at 07:42

2 Answers2

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Let $f(x)=x$. You would want $f^{1/2}$ defined so that $$\left(f^{1/2}\circ f^{1/2}\right)(x)=f(x)=x$$

Here are five options, for starters:

What would be the basis to choose one of these? I could understand a preference for the first option here. But with general $f$, if this is evidence that there are many possibilities for $f^{1/2}$, you might want to have a definition that somehow picks one canonically.

Or maybe you just let $f^{1/2}$ stand for the class of all functions that meet the condition $f^{1/2}\circ f^{1/2}=f$.

2'5 9'2
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You see here $n$ is a whole number. If you take it as 0.5 then problems will arise. For example is $n=2$ then that means we input function into itself two times but what $n=0.5$ would mean. Would it mean that we only input half of f or the first term of f or the last term.This becomes ridiculous when n goes out of range

Who am I
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