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Discrete Math

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Find g ° f

f(x) = {(1, 2), (2,1), (3,1), (4, 4)}
g(x) = {(1, 2), (2,4), (3, 1), (4, 3)}

  • Discrete Math - ,

    Can you check if there is a typo?
    As it is, f(x) is not onto, i.e. f(2)=f(3)=1.


    To find g°f, complete the following table:
    x u=f(x) g(u) = g ° f(x)
    1 2 4 = g°f(x)
    2 1 2
    3 1 2
    4 4 3

    Therefore
    g°f : {(1,4),(2,2) ....}

    As supplementary information, read up
    http://en.wikipedia.org/wiki/Function_composition

  • Discrete Math - ,

    Well, I am suppose to find the composition of functions from a figure. If you don't mind I uploaded a photo of it on photobucket. Can you take a look, and offer any suggestions for the first problem, so that I can get an idea? Since this forum will not let me post a link I can give directions on how to find the photo. First, go to search bar and type: flutegirl516. Then, this message will appear: Are you looking for the Photobucket user flutegirl516? Click on this. Then you will see the photo album. There is only one picture.

    Also, click "View as slide show" to make it larger. Tell me if you can view it okay. Thanks for any helpful replies :)

  • Discrete Math - ,

    I read the the link you provided previous to posting this discussion, but I was still confused.

  • Discrete Math - ,

    Sorry, I am not able to find the link according to your directions, probably we have different search bars.

    You can try posting the link that following the prefix

    http://media.photobucket.com/

    and I can append it myself.

  • Discrete Math - ,

    I'm using the search bar at the very top right-hand corner, then I enter: flutegirl516. It will say no matches found, but it will say: Are you looking for the Photobucket user flutegirl516?

    You can also try using the search bar drop down menu and clicking on "Users," and then enter flutegirl516. The album should pop up. Try it, hopefully you will be able to view it.

  • Discrete Math - ,

    Yes, indeed, I found it when I used Photobucket's search bar instead of the browser's search bar.

    For reference, here's the link:
    http://s1193.photobucket.com/albums/aa348/flutegirl516/?action=view¤t=Screenshot2011-02-15at61144PM.png

    You have correctly represented the two functions, so there is no typo.

    In case it is not clear how composition works, you only have to imagine that the range of f(x), i.e. the right part of f(x), is merged with the domain of g(x), or the left part of g(x).

    You can then follow the arrows from the domain of f(x) to the codomain of g(x). For example, f(1)=2, g(2)=4, which is exactly what is in the table I posted earlier, namely
    g°f=g(f(x))=g(f(1))=g(2)=4

    It is easier than you can imagine!

  • Discrete Math - ,

    ooOo I think I get it now. . .

  • Discrete Math - ,

    Great!

  • Discrete Math - ,

    Ok so

    g ° f = {(1, 4), (2, 2), (3, 2), (4,3)}

    There are two 1s in the range of f(x) though (1,2) and (3,1). . .Does that mean anything?

    Is this correct:

    Does g ° h = {(1, 2), (2?, 2?), (3, 2), (4, 1)}? For this one there was no 2 in the range of h.

    Also, for h^2 = h ° h, what am I suppose to square? Domain (x) or Range (y).

  • Discrete Math - ,

    "There are two 1s in the range of f(x) though (1,2) **I meant to say (2,1)** and (3,1). . .Does that mean anything?" Anyway, scratch this statement of the previous post. . .

  • Discrete Math - ,

    Here are the functions from the figures:

    • f(x) = {(1, 2), (2,1), (3,1), (4, 4)}
    • g(x) = {(1, 2), (2,4), (3, 1), (4, 3)}
    • h(x) = {(1, 1), (2, 3), (3, 1), (4, 3)}

  • Discrete Math - ,

    Yes, the (2,1) and (3,1) mean something.

    You will notice that f(x) does not map to the number 3. This makes f(x) not surjective (or "onto"). If it were, all the elements in the codomain {1,2,3,4} would be mapped by f(x).

    As another example, g(x) is bijective, because it is one-to-one and onto (or injective and surjective).

    For g°h, you just have to create a similar table as in f°g:
    x h(x) g°h=g(h(x))
    1 h(1)=1 g(h(1))=g(1)=2
    2 h(2)=3 g(h(2))=g(3)=1
    3 h(3)=1 g(h(3))=g(1)=2
    4 h(4)=4 g(h(4))=g(4)=3

    Note: It is OK that there is no element 2 in the range of h, just like there is no element 3 in the range of f.

    You can give it a try for h°h in a similar way.

  • Discrete Math - ,

    Some reading about surjection:

    http://en.wikipedia.org/wiki/Surjective_function

  • Discrete Math - ,

    So is this correct?

    g ° f = {(1, 4), (2, 2), (3, 2), (4,3)}

  • Discrete Math - ,

    There is something not connecting in my thought process. I'm still confused with g ° h does it = {(1,2), (3,1), (1,2), (4,3)}? I feel way off. . .I am doing something terribly wrong. IDK

  • Discrete Math - ,

    Correct!

    WOuld you like to try g°h again?
    This one was not correct:
    "Does g ° h = {(1, 2), (2?, 2?), (3, 2), (4, 1)}? For this one there was no 2 in the range of h. "
    Hint: (g°h)(2)=g(h(2))=g(3)=1

  • Discrete Math - ,

    "Correct!" applied to g ° f = {(1, 4), (2, 2), (3, 2), (4,3)}
    not g°h, sorry.

    FOR G°h:
    To do g°h, you can draw it graphically (at least for now), with the domain of h on the left, the range of h in the middle, and the range of g on the right.

    Between the left and the middle, draw arrows according to h(x), and between the middle and the right, draw arrows according to g(x).

    Start from any value x0 on the left, travel to the right through the middle according to the arrows, where you arrive is (g°h)(x0), or g(h(x0)).

    Your previous attempt was almost correct, just (2,2) was not.

  • Discrete Math - ,

    Is this correct?

    g ° h ={(1,2), (3,1), (1,2), (4,3)}

  • Discrete Math - ,

    g ° h ={(1,2), (3,1), (1,2), (4,3)}
    is correct.
    Hope you get it now, and ready to try h²!

  • Discrete Math-correction - ,

    g ° h ={(1,2), (3,1), (1,2), (4,3)}
    is not correct. I went too fast.

    (g°h)(1)=g(h(1))=g(1)=2
    (g°h)(2)=g(h(2))=g(3)=1
    (g°h)(3)=g(h(3))=g(1)=2
    (g°h)(4)=g(h(4))=g(3)=1

    So g°h : {(1,2),(2,1),(3,2),(4,1)}


    Hope you're ready to try h²!

  • Discrete Math - ,

    Lol. . .But really Thank You! You are really helping me to understand. I know you are probably annoyed by my silly questions, but I am really starting to get a better understanding.

    Ok so would h ° h = {(1,1), (2,1), (3,1), (4,1)}? But the h² is throwing me off.

  • Discrete Math - ,

    (h°h)(1)=h(h(1))=h(1)=1
    (h°h)(2)=h(h(2))=h(3)=1
    (h°h)(3)=h(h(3))=h(1)=1
    (h°h)(4)=h(h(4))=h(3)=1

    Great,you got the correct answers!

    (Even though they appear a little strange).

  • Discrete Math - ,

    They appear a little strange. . .how so? The h² does not effect the answer at all?

    OK so last one f ° g ° h. . .I will attempt now, and check back and see if I am on the right track. . .

  • Discrete Math - ,

    Good!

  • Discrete Math - ,

    Oh it looks strange b/c they all end in zero. Just noticed. . .

    Is this correct?

    f ° g ° h = {(1,1), (2,2), (3,1), (4,2)

  • Discrete Math - ,

    Hey, I know I said the previous would be the last one but can you check this one too. . .

    Let f: A→B be a function from A to B. f = {(w, 1), (x, 2), (y, 3), (z, 2)}. Find f^-1.

    Answer: f^-1 = {(1, w), (2, x), (3, y), (2, z)}

  • Discrete Math - ,

    Perfect!

    I forgot to mention that the expression f°g°h is associative, which means that you can evaluate it as f°(g°h) or (f°g)°h. This is why the parentheses have been left out.

    If you want extra practice, try both and verify the associativity.

  • Discrete Math - ,

    The function A->B is surjective, but not injective. The domain has more elements than codomain.

    The inverse relation B->A is not a function, since f-1(x) is undefined when x=2, namely f-1(2)=x or z?

    Under these circumstances, the inverse does not exist.

  • Discrete Math - ,

    How about this?

    Let f: A→B be a function from A to B. f = {(w, 1), (x, 2), (y, 3), (z, 2)}. Find f^-1.

    Answer: f^-1 = {(1, w), (2, x), (3, y), (2, z)}

    So would this one not have an inverse?

  • Discrete Math - ,

    For further reading, see:
    http://en.wikipedia.org/wiki/Inverse_function

  • Discrete Math - ,

    Oh okay we posted at the same time. . .You were such a big help! I really understand this stuff a lot better. Thank you a thousands times!

  • Discrete Math - ,

    Also thank you for the time you took out to help me :)

  • Discrete Math - ,

    The basic definition of a function is f(x) must have a unique value for a given x.

    A-> satisfies this requirement, but since it is not one-to-one (injective), it maps both x and z to 2, which does not stop f(x) from being a function.

    However, if you draw the diagram, you will see that the inverse function, if it exists, will have f-1 pointing to both x and z, which makes f-1 not a function. Therefore under these circumstances, f-1 is not defined.

  • Discrete Math :) - ,

    Keep up the good work, and post if you have other questions.

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