Compositions

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    Let f be a \mathbb{N} \rightarrow \mathbb{N} function defined by

    f(n) =\begin{Bmatrix}2n+1\ \ \ \ \ \mathrm{if}\ n>10, \\ 2n-2\ \ \ \ \ \ \mathrm{if} \ 0<n<10,

    How would I go about solving f \circ f?

    Any hints? I'm aware that this means f(f(n)) but I just don't know how to go about solving this.

    (I made this question up btw, the actual question is different but if I can understand what goes on here I can probably do the actual question)
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    (Original post by edothero)
    Let f be a \mathbb{N} \rightarrow \mathbb{N} function defined by

    f(n) =\begin{Bmatrix}2n+1\ \ \ \ \ \mathrm{if}\ n>10, \\ 2n-2\ \ \ \ \ \ \mathrm{if} \ 0<n<10,

    How would I go about solving f \circ f?

    Any hints? I'm aware that this means f(f(n)) but I just don't know how to go about solving this.

    (I made this question up btw, the actual question is different but if I can understand what goes on here I can probably do the actual question)
    The question you made up would need adusting since it's not a function (elements from the domain aren't mapped anywhere) and f \circ f wouldn't be defined for the full domain.

    Try this function instead with domain \mathbb{Z}.

    f(x) = \begin{cases}

x+2, \ x<-1 \\

x-1, \ x\geq -1

\end{cases}

    We'll start by trying different inputs e.g. a negative number like -10:

    Inputting -10 into f gives -8. Then input that into f again and we get -6.

    What's happening is we're getting f(f(x)) = f(x+2) = (x+2) + 2 = x+4.


    But what happens if we start with an input of -3. Putting that into f gives -1 and putting the result into f gives -2.

    What's happening here is f(f(x)) = f(x+2) = (x+2)-1 = x + 1.


    Try to think about all the different cases. Post all your ideas/working if you get stuck.
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    (Original post by notnek)
    The question you made up would need adusting since it's not a function (elements from the domain aren't mapped anywhere) and...
    So I've tried inputting various values and I've ended up with 4 cases:

    f(f(x)) = f(x+2) = (x+2)+2 = x + 4

    f(f(x)) = f(x+2) = (x+2)-1 = x + 1

    f(f(x)) = f(x+2) = (x-1)+2 = x + 1

    f(f(x)) = f(x+2) = (x-1)-1 = x - 2

    Though I'm not sure how the domain will be effected in each case;

    For example:

    f(-2) = -2+2 = 0
    f(0) = 0-1 = -1

    The input: -2<-1 but the output:  -1 \not< -1

    Quite confused right now :laugh: (I study Computer Science, not Maths)
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    (Original post by edothero)
    ...
    Essentially, you need to be worrying about the range of f as well.

    So, looking at the first part of the function (the x+2 part) you need to know when that goes into the first part, and when that goes into the second part.

    So for x+2 < -1 your f \circ f = (x+2) + 2 = x+4. This is your function. You only put x+2 into x+2 when both x+2 < -1 \iff x < -3 and when x < -1. So we fill in:

    f = x+4 when x<-3 (since that's when both conditions are fulfilled).

    This is because, for the first choice of which part of function to use, we look at what our x is, if it's less than -1, then we put it into x+2. We get an output from this, the output is x+2 (but remember that x < -1), we want to put this output into the function again. We need to look at what the output is. If the output is less than -1, then it goes into the first part of the function again. So if x+2 < -1 (which is the same as x<-3) then it goes into x+2 so you get (x+2) + 2 whenever x < -3.

    Now you repeat this process, when x &lt; -1, you get an output x+2 that you want to put into the function again. This time, you want to put it into the second part of the function. This means your ouput x+2 \geq -1 so x \geq -3 (along with x < -1) means that if you start with x satisfying those constraints, you put it into the first part of the function the first time around, get an output that tells you to put it into the second part of the function. So whenever -3 \leq  x &lt;-1 your function is (x+2) - 1 = x+1.

    Now we look at the second part of your function, we only use it when x \geq -1. When this happens, we get an output x-1. We want to see whether to put this output into the first part of the function or the second.

    Whenever x -1 &lt; -1 \iff x &lt; 0 happens, we put it into the second part of the function. So we get f(x) = (x-1) + 2 = x+1 whenever x &lt; 0 (along with x\geq -1) which combined together gives you -1 \leq x &lt; 0.

    See if you can figure this bit out yourself:
    Spoiler:
    Show
    You want to see when you put the second part of your function into the second part of your function. This means you have an x, it's x \geq -1. You get an output x-1. If this output is \geq -1, you compose it with the second bit of your function again.

    So you get f(x) = (x-1) - 1 = x-2 when both x \geq -1 and x-1 \geq -1 which combined together gives x \geq 0.

    (sorry for stealing your example notnek)

    Putting all of that together, our function is:
    Spoiler:
    Show

    \displaystyle

\begin{equation*}f(x) = \begin{cases} x+4, \, \, \, \,  x&lt;-3 \\ x+1,  \, \, \, \, -3 \leq x &lt; -1 \\ x+1,  \, \, \, \, -1 \leq x &lt; 0 \\ x-2, \, \, \, \, x \geq 0  \end{cases}\end{equation*}
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    (Original post by Zacken)
    Essentially, you need to be worrying about the range of f as well.

    So, looking at the first part...
    This is quite a good explanation! Thanks a ton. I was able to answer the original question; whether it's correct or not: time will tell :laugh:. Or you can be a bae and check it for me (its not assesed material).

    How is Cambridge treating you btw?

    Also thanks to notnek for the kick-start!
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    (Original post by edothero)
    This is quite a good explanation! Thanks a ton. I was able to answer the original question; whether it's correct or not: time will tell :laugh:. Or you can be a bae and check it for me (its not assesed material).

    Also thanks to notnek for the kick-start!
    Sure, I'm going to bed at 2:00, so if you get it up quickly, I'll have a look at it.
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    (Original post by Zacken)
    Sure, I'm going to bed at 2:00, so if you get it up quickly, I'll have a look at it.
    Can't seem to make the Latex work, if I end up posting it you can check it when you're free tomorrow.

    Have a good sleep!
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    (Original post by edothero)
    Can't seem to make the Latex work, if I end up posting it you can check it when you're free tomorrow.

    Have a good sleep!
    Just take a picture of it on paper or something or even write it out without latex, doesn't matter

    thanks, you too
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    (Original post by Zacken)
    Just take a picture of it on paper or something or even write it out without latex, doesn't matter

    thanks, you too
    Lmaooo well, lets just say I forgot to post it, but if you're still interested:

    Question:
    Spoiler:
    Show
    Let f be a \mathbb{N} \rightarrow \mathbb{N} function defined by:

    f(n) = n-1 when n &gt; 5
    f(n) = n+1 when 0 &lt; n \leq 5

    Find f \circ f
    Answer
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    f(n) = n-2 when n&gt;6
    f(n) = n when 5&lt;n \leq 6
    f(n) = n when 4&lt;n \leq 5
    f(n) = n+2 when -1\leq n \leq 4
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    (Original post by edothero)
    Lmaooo well, lets just say I forgot to post it, but if you're still interested:

    Question:
    Spoiler:
    Show
    Let f be a \mathbb{N} \rightarrow \mathbb{N} function defined by:

    f(n) = n-1 when n &gt; 5
    f(n) = n+1 when 0 &lt; n \leq 5

    Find f \circ f
    Answer
    Spoiler:
    Show
    f(n) = n-2 when n&gt;6
    f(n) = n when 5&lt;n \leq 6
    f(n) = n when 4&lt;n \leq 5
    f(n) = n+2 when -1\leq n \leq 4
    A few comments:

    1. Remember that you can collect the two middle functions together to get f(n) = n when 4 &lt; n \leqg 6

    2. You can immediately tell that the last part of your function is wrong, since it's mapping -1 to something, but -1 is not even in your domain! The problem is that you're forgetting to put together the conditions on n. n+1 goes into n+1 when TWO things are true. First, you need 0 &lt;n \leq 5 to hold so that you choose n+1 in the first place, you then get an output of n+1 (remembering that 0 &lt; n \leq 5) and then you put that back into n+1 if the output 0 &lt; n+1 \leq 5 holds.

    So f(n) = n+2 when you have (i) 0 &lt; n \leq 5 and (ii) -1 &lt; n \leq 4 which can be cleaned up as 0 &lt; n \leq 4.
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    (Original post by Zacken)
    A few comments:
    Thank you as always, I can imagine you're quite swamped with work at Cambridge so I appreciate taking time out to help :yy:
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    (Original post by edothero)
    Thank you as always, I can imagine you're quite swamped with work at Cambridge so I appreciate taking time out to help :yy:
    No problemo! And this is a good way of procrastinating on my four example sheets due in.
 
 
 
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