whats the difference between the function compositions \((f\circ g)\circ h\) and \(f \circ (g\circ h)\)

- ganeshie8

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- welshfella

is there a difference?

- ganeshie8

It turns out the compositions are associative, but what exactly is the difference between them ?

- zzr0ck3r

Hint: Function composition is associative

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## More answers

- ikram002p

there is no difference only be careful about range and domain

- ganeshie8

i mean, in terms of execution order... what difference that parenthesis ordering make ?
idk im having a hard time moving forward with my review of abstract algebra because of this..

- ikram002p

order might affect on that only.

- ganeshie8

Hmm okay, domain and codomain are valid in both cases
```
h : X -> Y
g : Y -> Z
f : Z -> W
```

- freckles

hey so you do you mean like what would \[f \circ (g \circ h) \text{ verus what would it look like to find } (f \circ g) \circ h\]
\[f=x^2 \\ g=x-1 \\ h= \sqrt{x} \\ f \circ g =(x-1)^2 \\ g \circ h=\sqrt{x}-1 \\ ( f \circ g) \circ h=(\sqrt{x}-1)^2 \\ f \circ (g \circ h)=(\sqrt{x}-1)^2\]?

- ganeshie8

I see compositions don't commute in general : \(f\circ g \ne g\circ f\)
But I am studying associativity as of now.
for example, we knw that addition over real numbers is associative :
```
(a + b) + c = a + (b + c)
```
here we agree that "a+b" is executed first on left hand side.
is there a similar thing for compositions ?

- ikram002p

take this example
h : {b,1,2,3}->{1 }
g : {1,2,3}->{4,5,a,c}
f : {.4,1,2,3}->{a,b,c}

- freckles

yes you can find f composed with g first then plug in h
or you can find g composed with h then plug that result into f

- ganeshie8

how does that work ?
say
\(f(x) = x\)
\(g(x) = 2x\)
\(h(x)=3x\)

- ganeshie8

\((f\circ g)\circ h(2) = ?\)
what does it mean to compose \(f\) and \(g\) first, here ?

- freckles

\[(f \circ g)(x)=2x \\ (g \circ h)(x)=2(3x) \\ \text{ now we will replace } x \text{ in first one with } h(x) \\ (f \circ g)(h(x))=2(3x) \\ \text{ now we will replace the } x \text{ in } f(x)=x \text{ with } (g \circ h)(x) \\ \text{ so we have } \\ f((g \circ h)(x))=2(3x)\]

- ganeshie8

I think I get it! thanks a lot for breaking it down !

- freckles

I kind of like to leave the x off sometimes

- freckles

\[(f \circ g) \circ h \\ \text{ where } f=x^2 \\ g=x-1 \\ h=\sqrt{x} \\ f \circ g =(x-1)^2 \\ \text{ then the input is } h \\ (f \circ g) (h)=(\sqrt{x}-1)^2\]
it seems less confusing to me for some reason without writing the x

- ganeshie8

I see... guess I never played with these much, its kinda fun reviewing/learning these now :)
Was going through this proof earlier when I got stuck on compositions :
It can be shown that every square matrix represents a composition, therefore the matrix multiplication is also associative!

- zzr0ck3r

\(f\) is the function
\(f(x) = 3x+2\) is the rule
\(f(x)\) does not really make sense, but we abuse the hell out of the notation :)

- ganeshie8

f : x -> 3x+2
is more proper i think, hope ill be able to appreciate these fine details after finishing abstract algebra xD

- ikram002p

ur doing abstract now ?

- freckles

what?!? you really are doing abstract algebra?

- zzr0ck3r

\(f:\mathbb{R}\rightarrow \mathbb{R}, f(x) = 3x+2\) is a function

- freckles

I thought you already graduated and stuff. You know everything. :p

- zzr0ck3r

\(f:\underset{x\mapsto 3x+2}{\mathbb{R}\rightarrow \mathbb{R}}\)

- zzr0ck3r

I really like that last one, but it takes so long in latex

- ikram002p

hmmm yeah using this notation is better

- zzr0ck3r

I would not call this abstract algebra.
Everything in math is a set.
Almost everything is a function.

- ganeshie8

lol i suck at math, i am an engineer, what do you think haha! I think abstract algebra is more like pure mathematics , never had to touch this before.

- ikram002p

nah some times u are perfect in math gane.

- ganeshie8

Ahh nice, is it correct to interpret that as :
\(f\) is a relation between the sets \(\mathbb R\) and \(\mathbb{R}\)
such that \(x\) maps to \(3x+2\)

- zzr0ck3r

I would say that \(f\) is the name of the function, the name of the rule that gives the relation is \(f(x) = 2x\).

- zzr0ck3r

so we could say \(f= \{(1,2),(2,4),(4,5)\}\) but \(f(x)= \{(1,2),(2,4),(4,5)\}\) makes no sense.

- zzr0ck3r

and in this situation we could say \(f(\{1,2,4\})=\{2,4,5\}\).

- zzr0ck3r

some people would want \(f[\{1,2,4\}]=\{2,4,5\}\)

- ikram002p

i thought this is also correcto
\(f (\{(1,2),(2,4),(4,5)\} )\)

- anonymous

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

- anonymous

this is the 'difference' in order, @ganeshie8

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