Is function composition associative?

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Answer 1 · 2015-07-24T10:41:45

Yes

Given composable functions $f$ , $g$ and $h$

$(f@(g@h))(x)$

$= f((g@h)(x)) = f(g(h(x))) = (f@g)(h(x))$

$= ((f@g)@h)(x)$

So $f@(g@h) = (f@g)@h$

Answer 2 · 2015-07-28T01:01:17

It is, if the following works:

$(f@(g@h))(x) = ((f@g)@h)(x)$

That is, if:

$f(x) = "something"$
$g(h(x)) = (g@h)(x) = "something else"$
$f(g(x)) = (f@g)(x) = "something else again"$
$h(x) = "something else yet again"$

...and you can use these together to satisfy the first expression, then they are associative. Let:

$f(x) = 2x$
$g(x) = x^2$
$h(x) = x^3$

Thus:

$g(h(x)) = g(x^3) = (x^3)^2 = x^6$

$f(g(x)) = g(x^2) = 2(x^2) = 2x^2$

Then:

$(f@(g@h))(x) = f(x^6) = 2(x^6) = color(blue)(2x^6)$

$((f@g)@h)(x) = f(g(x^3)) = f((x^3)^2) = f(x^6) = color(blue)(2x^6)$

Therefore they are associative.