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If $f$ represents the function $f(x)$ and $g$ represents the functions $g(x)$, is $(f \circ g)\circ f$ essentially $f(g(f(x)))$?

I know that:

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

however I'm not sure if the brackets in my equations make a difference to this new function.

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    $\begingroup$ short answer: yes! $\endgroup$
    – Yanko
    Feb 28, 2018 at 22:19
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    $\begingroup$ Function composition is associative, so $\,(f \circ g) \circ f = f \circ (g \circ f) = f \circ g \circ f\,$. $\endgroup$
    – dxiv
    Feb 28, 2018 at 22:19

2 Answers 2

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The brackets make no difference: the composition of functions is completely associative, meaning that $(f\circ g)\circ h=f\circ(g\circ h)$.

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Yes it is. Furthermore, $(f \circ g) \circ f = f \circ g \circ f = f \circ (g \circ f)$ by associativity.

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