Question

What is the difference between (fo(goh)(x) and ((fog)oh)(x)?

Answer 1

Function composition is associative, that is

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

There is no difference in the result, though the steps may be expressed differently.

Explanation:

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

For example, suppose:

`f(x) = x^2`

`g(x) = 1/x`

`h(x) = x + 1`

Then:

`(g@h)(x) = g(h(x)) = 1/(x+1)`

`(f@(g@h))(x) = f((g@h)(x)) = f(1/(x+1)) = 1/(x+1)^2`

`(f@g)(x) = f(g(x)) = 1/x^2`

`((f@g)@h)(x) = (f@g)(h(x)) = (f@g)(x+1) = 1/(x+1)^2`