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ganeshie8

  • one year ago

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

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  1. welshfella
    • one year ago
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    is there a difference?

  2. ganeshie8
    • one year ago
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    It turns out the compositions are associative, but what exactly is the difference between them ?

  3. zzr0ck3r
    • one year ago
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    Hint: Function composition is associative

  4. ikram002p
    • one year ago
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    there is no difference only be careful about range and domain

  5. ganeshie8
    • one year ago
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    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
    • one year ago
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    order might affect on that only.

  7. ganeshie8
    • one year ago
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    Hmm okay, domain and codomain are valid in both cases ``` h : X -> Y g : Y -> Z f : Z -> W ```

  8. freckles
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    how does that work ? say \(f(x) = x\) \(g(x) = 2x\) \(h(x)=3x\)

  13. ganeshie8
    • one year ago
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    \((f\circ g)\circ h(2) = ?\) what does it mean to compose \(f\) and \(g\) first, here ?

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

  15. ganeshie8
    • one year ago
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    I think I get it! thanks a lot for breaking it down !

  16. freckles
    • one year ago
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    I kind of like to leave the x off sometimes

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

  18. ganeshie8
    • one year ago
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    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
    • one year ago
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    \(f\) 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
    • one year ago
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    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
    • one year ago
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    ur doing abstract now ?

  22. freckles
    • one year ago
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    what?!? you really are doing abstract algebra?

  23. zzr0ck3r
    • one year ago
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    \(f:\mathbb{R}\rightarrow \mathbb{R}, f(x) = 3x+2\) is a function

  24. freckles
    • one year ago
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    I thought you already graduated and stuff. You know everything. :p

  25. zzr0ck3r
    • one year ago
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    \(f:\underset{x\mapsto 3x+2}{\mathbb{R}\rightarrow \mathbb{R}}\)

  26. zzr0ck3r
    • one year ago
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    I really like that last one, but it takes so long in latex

  27. ikram002p
    • one year ago
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    hmmm yeah using this notation is better

  28. zzr0ck3r
    • one year ago
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    I would not call this abstract algebra. Everything in math is a set. Almost everything is a function.

  29. ganeshie8
    • one year ago
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    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
    • one year ago
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    nah some times u are perfect in math gane.

  31. ganeshie8
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    and in this situation we could say \(f(\{1,2,4\})=\{2,4,5\}\).

  35. zzr0ck3r
    • one year ago
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    some people would want \(f[\{1,2,4\}]=\{2,4,5\}\)

  36. ikram002p
    • one year ago
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    i thought this is also correcto \(f (\{(1,2),(2,4),(4,5)\} )\)

  37. anonymous
    • one year ago
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    \((f\circ g)(x)=f(g(x))\\((f\circ g)\circ h)(x)=(f\circ g)(h(x))=f(g(h(x)))\\(f\circ(g\circ h))(x)=f((g\circ h)(x))=f(g(h(x)))\)

  38. anonymous
    • one year ago
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    this is the 'difference' in order, @ganeshie8

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