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

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is associative, that is

##([email protected]([email protected]))(x) = (([email protected])@h)(x)##

There is no discord in the outcome, though the steps may be developed differently.

##([email protected]([email protected]))(x) = f(g(h(x))) = (([email protected])@h)(x)##

For issue, suppose:

##f(x) = x^2##

##g(x) = 1/x##

##h(x) = x + 1##

Then:

##([email protected])(x) = g(h(x)) = 1/(x+1)##

##([email protected]([email protected]))(x) = f(([email protected])(x)) = f(1/(x+1)) = 1/(x+1)^2##

##([email protected])(x) = f(g(x)) = 1/x^2##

##(([email protected])@h)(x) = ([email protected])(h(x)) = ([email protected])(x+1) = 1/(x+1)^2##


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