How do you define iterations of multivariable functions?

To be clear(example):

If $f: \mathbb R^2 \to \mathbb R$

How do you define

$f \circ f$, or $f \circ \cdots \circ f$?

I admit that this question sounds very odd, but I think I need to define or learn of this. (Why? I want to generalize this(Carleman matrix) to multivariable functions to solve this(Multivariable carleman matrix) or this(same but different sites) question!)

And I think this concept may be quite reasonable because there is a something like multiplication of matrices that have different dimensions.

My assumtion is that $f \circ f \cdots \circ f : \text{also } \mathbb R^2 \to \mathbb R$.

Any suggestions are appreciated.

  • $\begingroup$ I suppose you could define $\tilde f:\mathbb R^2\to\mathbb R^2$ by $\tilde f(x,y) = (f(x,y),f(x,y))$, then investigate $f_n := f\circ \tilde f\circ \tilde f\circ\dots \circ \tilde f$. $\endgroup$ Sep 11 '18 at 15:29
  • $\begingroup$ @MikeEarnest Worth trying! $\endgroup$ Sep 11 '18 at 15:36
  • 1
    $\begingroup$ There are definitely many possible versions... $F(x,y) =(0,f(x,y)); (f(x,y),f(x,y));$ or even $(f(x,y),f(y,x))$ (which is sort of a more ‘symmetric’ version of $f$). Which one you should consider depends highly on the property you want them to satisfy $\endgroup$ Sep 11 '18 at 15:59

The composition is undefined as $$f \circ f=\mathbb{R}^2\xrightarrow{f}\mathbb{R}\xrightarrow{?}\underline{?}.$$ However if you have a function $\mathbb{R}^2\xrightarrow{F}\mathbb{R}^2,$ then we can easily form the composition.

  • $\begingroup$ You're definitely right! But I know that too... I want to "define" it. Anyway, it seems that and your answer is saying that it has not been defined formally. $\endgroup$ Sep 11 '18 at 15:33
  • $\begingroup$ @KYHSGeekCode: How could we define something that cannot exist? The only way to get-rid of the problem is change the co-domain (naturally or artificially). $\endgroup$
    – Bumblebee
    Sep 11 '18 at 15:36

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