(f o g) o f composition function

Question

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.

Function composition is associative, so $,(f \circ g) \circ f = f \circ (g \circ f) = f \circ g \circ f,$. — dxiv, Feb 28 '18 at 22:19

score 1 · Accepted Answer · answered Feb 28 '18 at 22:20

1

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

answered Feb 28 '18 at 22:20

Álvaro Serrano

score 1 · Answer 2 · answered Feb 28 '18 at 22:20

1

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

answered Feb 28 '18 at 22:20

ArsenBerk

2 Answers2