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

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welshfella
 one year ago
Best ResponseYou've already chosen the best response.1is there a difference?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1It turns out the compositions are associative, but what exactly is the difference between them ?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Hint: Function composition is associative

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1there is no difference only be careful about range and domain

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1i 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
 one year ago
Best ResponseYou've already chosen the best response.1order might affect on that only.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Hmm okay, domain and codomain are valid in both cases ``` h : X > Y g : Y > Z f : Z > W ```

freckles
 one year ago
Best ResponseYou've already chosen the best response.1hey so you do you mean like what would \[f \circ (g \circ h) \text{ verus what would it look like to find } (f \circ g) \circ h\] \[f=x^2 \\ g=x1 \\ h= \sqrt{x} \\ f \circ g =(x1)^2 \\ g \circ h=\sqrt{x}1 \\ ( f \circ g) \circ h=(\sqrt{x}1)^2 \\ f \circ (g \circ h)=(\sqrt{x}1)^2\]?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.1take 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
 one year ago
Best ResponseYou've already chosen the best response.1yes 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
 one year ago
Best ResponseYou've already chosen the best response.1how does that work ? say \(f(x) = x\) \(g(x) = 2x\) \(h(x)=3x\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\((f\circ g)\circ h(2) = ?\) what does it mean to compose \(f\) and \(g\) first, here ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[(f \circ g)(x)=2x \\ (g \circ h)(x)=2(3x) \\ \text{ now we will replace } x \text{ in first one with } h(x) \\ (f \circ g)(h(x))=2(3x) \\ \text{ now we will replace the } x \text{ in } f(x)=x \text{ with } (g \circ h)(x) \\ \text{ so we have } \\ f((g \circ h)(x))=2(3x)\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I think I get it! thanks a lot for breaking it down !

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I kind of like to leave the x off sometimes

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[(f \circ g) \circ h \\ \text{ where } f=x^2 \\ g=x1 \\ h=\sqrt{x} \\ f \circ g =(x1)^2 \\ \text{ then the input is } h \\ (f \circ g) (h)=(\sqrt{x}1)^2\] it seems less confusing to me for some reason without writing the x

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.2\(f\) 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
 one year ago
Best ResponseYou've already chosen the best response.1f : x > 3x+2 is more proper i think, hope ill be able to appreciate these fine details after finishing abstract algebra xD

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1ur doing abstract now ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1what?!? you really are doing abstract algebra?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2\(f:\mathbb{R}\rightarrow \mathbb{R}, f(x) = 3x+2\) is a function

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I thought you already graduated and stuff. You know everything. :p

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2\(f:\underset{x\mapsto 3x+2}{\mathbb{R}\rightarrow \mathbb{R}}\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2I really like that last one, but it takes so long in latex

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1hmmm yeah using this notation is better

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2I would not call this abstract algebra. Everything in math is a set. Almost everything is a function.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1lol 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
 one year ago
Best ResponseYou've already chosen the best response.1nah some times u are perfect in math gane.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Ahh 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
 one year ago
Best ResponseYou've already chosen the best response.2I would say that \(f\) is the name of the function, the name of the rule that gives the relation is \(f(x) = 2x\).

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2so we could say \(f= \{(1,2),(2,4),(4,5)\}\) but \(f(x)= \{(1,2),(2,4),(4,5)\}\) makes no sense.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2and in this situation we could say \(f(\{1,2,4\})=\{2,4,5\}\).

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2some people would want \(f[\{1,2,4\}]=\{2,4,5\}\)

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.1i thought this is also correcto \(f (\{(1,2),(2,4),(4,5)\} )\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\((f\circ g)(x)=f(g(x))\\((f\circ g)\circ h)(x)=(f\circ g)(h(x))=f(g(h(x)))\\(f\circ(g\circ h))(x)=f((g\circ h)(x))=f(g(h(x)))\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is the 'difference' in order, @ganeshie8
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