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Math Help - Function Question

  1. #1
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    Function Question

    Thanks in advance!
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  2. #2
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    How much of this have you done?
    Have you done any of it?
    If you have, what pattern do you see?
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  3. #3
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    Quote Originally Posted by Plato View Post
    How much of this have you done?
    Have you done any of it?
    If you have, what pattern do you see?
    For part a, for f2(x) i get f(f(x)), so what would f(f(n)) be? Without knowing n, how would I calculate this? maybe i'm just confused about what to provide for an answer.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by shawn View Post
    For part a, for f2(x) i get f(f(x)), so what would f(f(n)) be? Without knowing n, how would I calculate this? maybe i'm just confused about what to provide for an answer.
    Well,
    f(n) = \left \{ \begin{array}{rr}n + 3,\text{ if n is even} \\ n - 1,\text{ if n is odd} \end{array} \right .

    So to calculate f(f(n)),
    If n is even this is f(f(n)) = f(n + 3) = (n + 3) - 1 = n + 2
    since n + 3 is odd.

    If n is odd then this is f(f(n)) = f(n - 1) = (n - 1) + 3 = n + 2
    since n - 1 is even.

    So f^2(n) = n + 2 for all n.

    You do the rest.

    -Dan
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  5. #5
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    Quote Originally Posted by topsquark View Post
    Well,
    f(n) = \left \{ \begin{array}{rr}n + 3,\text{ if n is even} \\ n - 1,\text{ if n is odd} \end{array} \right .

    So to calculate f(f(n)),
    If n is even this is f(f(n)) = f(n + 3) = (n + 3) - 1 = n + 2
    since n + 3 is odd.

    If n is odd then this is f(f(n)) = f(n - 1) = (n - 1) + 3 = n + 2
    since n - 1 is even.

    So f^2(n) = n + 2 for all n.

    You do the rest.

    -Dan
    Ah okay got it..thanks.

    So, f^2(n) = n + 2 when n is odd
    and f^2(n) = n + 2 when n is even

    and f^3(n) = n + 5 when n is odd
    and f^3(n) = n + 1 when n is even

    and f^4(n) = n + 6 when n is odd
    and f^4(n) = n + 4 when n is even

    I don't see a pattern for a simple formula?
    Last edited by shawn; April 7th 2008 at 05:03 PM.
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  6. #6
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    Inductive Proof

    I'm doing part c now,
    Basis is easy when k = 3.

    Inductive step

    Let x >=3 be an integer.
    Our IH is (from the formula's I found for part b)
    f^x (n) = n + x when x is even, and
    f^x (n) = n + x + 2(-1)^n when x is odd

    We want to prove that
    f^x+1(n)=n+x+1 when x+1 is even, and
    f^x+1(n)=n+x+1+2(-1)^n when x+1 is odd

    Is the above correct?

    So we have two cases from here:

    Case 1: when x+1 is even, x is odd
    Now, f^x+1(n) = f(f^x(n)) = f(n +x + 2(-1)^n)

    Now, I think we have two more sub cases but I'm not sure?

    Case 2: when x+1 is odd, x is even
    Now, f^x+1(n) = f(f^x(n)) = f(n + x)
    Now, I think we have two more sub cases but I'm not sure?


    Can someone help me finish the sub cases?

    Thanks!
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