Is function composition associative?

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Is function composition associative?

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Answer 1 · 2015-07-24T10:41:45

Yes

Explanation:

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

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

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

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

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

Answer 2 · 2015-07-28T01:01:17

It is, if the following works:

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

That is, if:

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

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

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

Thus:

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

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

Then:

$(f \circ (g \circ h)) (x) = f (x^{6}) = 2 (x^{6}) = 2 x^{6}$ $(f@(g@h))(x) = f(x^6) = 2(x^6) = color(blue)(2x^6)$

$((f \circ g) \circ h) (x) = f (g (x^{3})) = f ({(x^{3})}^{2}) = f (x^{6}) = 2 x^{6}$ $((f@g)@h)(x) = f(g(x^3)) = f((x^3)^2) = f(x^6) = color(blue)(2x^6)$

Therefore they are associative.

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Is function composition associative?

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