Let $f:\mathbb{N} \rightarrow \mathbb{N}$. When does there exist a $g: \mathbb{N}\rightarrow \mathbb{N}$ such that $f(n)=g(g(n))$ for all $n$?. I don't think its always possible, for example $f(n)=n+1$, doesn't seem likely that $g$ exists. The problem could be easier if we say $f$ is injective. Then $g$ must be injective. I didn't find this problem anywhere so I don't have a reference, I just thought of it. Im hoping dealing with $\mathbb{N}$ will give a nice answer. The problem has a nice answer for invertible matrices.
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4See also this post. – Dietrich Burde Jun 07 '22 at 13:05
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4Does this answer your question? Characterising functions $f$ that can be written as $f = g \circ g$? – Clyde Kertzer Feb 16 '23 at 02:47
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Specifically for the function $f(n)=n+1$, it is easy to show there is no such $g$.
First, recall the observation (made initially by user GMC2 in a now-deleted post) that $f=g\circ g$ implies $f\circ g = g\circ g\circ g = g\circ f$, meaning $f$ commutes with $g$.
Unfolding this for our particular function $f(n)=n+1$, we get $g(n+1) = g(n)+1$. By induction, this means $g(n) = g(0)+n$. Thus, this function is completely determined by $g(0)$, and this is the general form of any function commuting with $f$.
Next, we know that $f=g\circ g$. Compute $(g\circ g)(n) = g(g(n)) = g(g(0)+n) = g(0) + g(0) + n = 2g(0) + n$. We want this to be equal to $f(n)=n+1$, so $2g(0)+n=n+1 \Longrightarrow 2g(0)=1$, which doesn't have solutions in natural numbers.
So, there is no square root of the function $f$.
lisyarus
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