## ganeshie8 one year ago whats the difference between the function compositions $$(f\circ g)\circ h$$ and $$f \circ (g\circ h)$$

1. welshfella

is there a difference?

2. ganeshie8

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

3. zzr0ck3r

Hint: Function composition is associative

4. ikram002p

there is no difference only be careful about range and domain

5. 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..

6. ikram002p

order might affect on that only.

7. ganeshie8

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

8. freckles

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

9. 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 ?

10. 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}

11. 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

12. ganeshie8

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

13. ganeshie8

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

14. freckles

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

15. ganeshie8

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

16. freckles

I kind of like to leave the x off sometimes

17. 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

18. 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!

19. 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 :)

20. ganeshie8

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

21. ikram002p

ur doing abstract now ?

22. freckles

what?!? you really are doing abstract algebra?

23. zzr0ck3r

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

24. freckles

25. zzr0ck3r

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

26. zzr0ck3r

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

27. ikram002p

hmmm yeah using this notation is better

28. zzr0ck3r

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

29. 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.

30. ikram002p

nah some times u are perfect in math gane.

31. 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 $$3x+2$$

32. zzr0ck3r

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

33. zzr0ck3r

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

34. zzr0ck3r

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

35. zzr0ck3r

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

36. ikram002p

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

37. anonymous

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

38. anonymous

this is the 'difference' in order, @ganeshie8