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Very interesting problem that puzzles me

by jdp
Tags: interesting, puzzles
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jdp
#1
Feb5-13, 08:36 PM
P: 3
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.
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Simon Bridge
#2
Feb5-13, 09:54 PM
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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
jfgobin
#3
Feb6-13, 12:25 PM
P: 90
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].

pwsnafu
#4
Feb6-13, 04:13 PM
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Very interesting problem that puzzles me

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.
jdp
#5
Feb6-13, 07:33 PM
P: 3
Yes, I clarified. Exactly like this. I just don't know where to start
Simon Bridge
#6
Feb6-13, 08:24 PM
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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))?
HallsofIvy
#7
Feb7-13, 09:48 AM
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Quote Quote by pwsnafu View Post
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.
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.


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