Alright, so we all know how single variable functions work with iteration:

If f(x) = x^2

Then f^2 (x) = (x^2)^2

This works really well because in with only one independent variable there is one input and one output. So it’s not so hard to understand something like Cos(x) being iterated 24 times.

With all this in mind, how does multivariate function iteration work.

f(x,y) = x+y

f^2 (x,y) = ? There are three combinations for n=2 (number of compositions or iterations) as far as I can tell:

= f( x, f(x,y) ) = x + (x+y) = 2x+y

= f( f(x,y), f(x,y) ) = (x+y) + (x+y)= 2x+2y

= f( f(x,y), y ) = (x+y) + y = x+2y

Unsure if placing y and x in for the first and second argument respectively would still achieve the goal I am looking for or not...

In any case, is it possible to generalize (or even notate at this point) the many combinations that arise as n increases? Has this been investigated before? If not, any ideas?


You're asking about composing a function with itself multiple times, which only makes sense if all elements in the range of the function are also in the domain. In the example $f(x, y)=x+y$ the domain is, I assume, $\mathbb{R}\times\mathbb{R}$ and the range is $\mathbb{R}$. Therefore one can't compose $f$ with itself, since outputs of $f$ are not valid inputs for $f$.

  • $\begingroup$ I see what you mean. In my way of switching the arguments around, is it possible to still find all of the combinations of arguments there could be (selecting from (x,y, f(x,y))? $\endgroup$ – Divide Me By Zero Oct 11 '19 at 22:24

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