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Very interesting problem that puzzles me

  1. Feb 5, 2013 #1

    jdp

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    One of my friends showed me this problem and I've been thinking about it all day. It's something like if f(2x) is equal to f(f(y)) and that f(2y) + 1 is f(2y +1) and f(0) is 0, what is f(n). It seems very maclaurin esque to me but... anyway... first post and pretty nice problem. Would love help figuring this beast out.
     
  2. jcsd
  3. Feb 5, 2013 #2

    Simon Bridge

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    Number them:
    1. f(2x)=f(f(y))
    2. f(2y)+1=f(2y+1)
    3. f(0)=0

    guessing: x,y,n are integers?
    are x and y specific numbers or is just any arbitrary number?
    i.e. does this also hold: ##f(2x+1)=f(2x)+1## ?

    If so then I'd start from the observations:

    #1 tells you f(f(n))=f(some even number).
    #2 (with #3) tells you that if n is odd, then f(n)=f(n-1)+1
     
  4. Feb 6, 2013 #3
    That looks weird. Are you sure there isn't something missing?

    By #3, [itex]f(0)=0[/itex], using [itex]y=0[/itex], by #2, [itex]f(1)=1[/itex]. Using [itex]2y=1[/itex] and #2 again, [itex]f(2)=2[/itex], and so forth.

    So at least for [itex]x \in \mathbb{N}[/itex], [itex]f(x)=x[/itex].
     
  5. Feb 6, 2013 #4

    pwsnafu

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    I think OP's problem is the following:

    Find ##f : \mathbb{N} \to \mathbb{N}## satisfying
    1. ##f(0) = 0##,
    2. For each ##k \in \mathbb{N}## we have ##f(2k) = f(f(k))##,
    3. For each ##k \in \mathbb{N}## we have ##f(2k+1) = f(2k) + 1##.

    At the very least it prevents the trivial solution.

    Note: I'm including 0 in the naturals for this problem.
     
  6. Feb 6, 2013 #5

    jdp

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    Yes, I clarified. Exactly like this. I just don't know where to start
     
  7. Feb 6, 2013 #6

    Simon Bridge

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    Have you had a play with the relations, looking at different numbers?

    Using pwsnafu's formulation in post #4:

    If ##f^{-1}\big ( f(k)\big )=k## is the inverse function in action,
    then, from #2:
    ##f^{-1}\big ( f(2k)\big ) = f^{-1}\big ( f(f(k)\big ) \\ \Rightarrow f(k)=2k##​
    But that does not fit the other clues.

    Notice: 2k is always even, 2k+1 is always odd.
    ... so the clues are telling you what happens in the case of odd or even n.

    This is how you go about exploring mathematical relations - you end up making a list of things that the relations are telling you.
    It's important that you get used to this process - after a bit you spot a pattern that looks likely and you set out to prove it.

    eg.
    What happens when you apply the formula to k=0?
    What is f(f(2k))? f(f(2k+1))?
     
  8. Feb 7, 2013 #7

    HallsofIvy

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    Start by doing the obvious calculations. 1 is odd: 1= 2(0)+ 1 so f(1)= f(2(0)+ 1)= f(2(0))+ 1= f(0)+ 1= 1. 2 is even: 2= 2(1) so f(2)= f(f(1))= f(1)= 1. 3 is odd: 3= 2(1)+ 1 so f(3)= f(2(1)+ 1)= f(2(1))+ 1= f(2)+ 1= 1+ 1= 2. 4 is even: 4= 2(2) so f(4)= f(2(2))= f(f(2))= f(1)= 1. 5 is odd: 5= 2(2)+ 1 so f(5)= f(2(2)+ 1)= f(2(2))+ 1= f(4)+ 1= 1+ 1= 2.

    You should be able to see a very easy pattern for f(n) when n is even and for f(n) when n is odd. And then use "proof by induction" to show that you are correct.
     
    Last edited: Feb 7, 2013
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