continuous function

romano · Riemann Hypothesis Offline Joined: 10 Jun 2004 Posts: 304

Let be given a possitive integer n . Consider a continous function $f(x):[0;n]-->R$ satisfying $f(0)=f(n)$ . Prove that : there exist n couples of numbers $a_i , b_i (i=1,2,..,n)$ belonging to $[0;n]$ such that $b_i-a_i$ are possitive integers and $f(a_i) = f(b_i)$ for all $i=1,2,..,n$

**Posted:** Mon Jan 10, 2005 8:06 pm

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dickchimney · Poincare Conjecture Offline Joined: 06 Mar 2004 Posts: 218

Infact we can find infinitely many such couples
Let $c$ be a point in $(0,n)$ so that $f(c)$ is different than $f(0)$ and therefore $f(n)$
( If such $c$ does not exist then $f$ is constant and the conclusion holds trivially)
Apply the Intermediate value property, we find $a_1 \in (0,c)$ , $b_1\in (c,n)$ so that
$f(a_1) = f(b_1) = \frac {f(c) + f(0)}{2}$ .
Now consider $(a_1,b_1)$ alone and follow the same procedure above we have $a_2,b_2$ , then $a_3,b_3$ ....

**Posted:** Mon Jan 10, 2005 9:11 pm

Kent Merryfield

dickchimney, I don't see that you have shown that $b_1-a_1$ is an integer. In fact, with that integral-difference condition, there are in general only finitely many such pairs of ponts.

In fact, to be assured of having $n$ such pairs of points, we must include $(0,n)$ as one of the pairs.

**Posted:** Mon Jan 10, 2005 9:29 pm

grobber

If $(a_i,b_i)$ is such a couple, can we also take $(b_i,a_i)$ as being one?

**Posted:** Mon Jan 10, 2005 9:46 pm

dickchimney · Poincare Conjecture Offline Joined: 06 Mar 2004 Posts: 218

**Posted:** Mon Jan 10, 2005 10:49 pm

grobber

I do not think it's that easy, man. I think there can be some integers $m<n$ for which there's no $x$ s.t. $f(x)=f(x+m)$ . We can use your method to find such $x$ 's for the case $m|n$ with no problems, but it doesn't look good if we want to find $n$ pairs.. Confused

[Edit: I think I see what you mean now; I hadn't read it properly Smile

; sorry about that Blush

]

**Posted:** Mon Jan 10, 2005 10:58 pm

dickchimney · Poincare Conjecture Offline Joined: 06 Mar 2004 Posts: 218

**Posted:** Mon Jan 10, 2005 11:06 pm

dickchimney · Poincare Conjecture Offline Joined: 06 Mar 2004 Posts: 218

**Posted:** Mon Jan 10, 2005 11:08 pm

Kent Merryfield

Here's another approach.

For $k\in\{1,2,\dots\}$ , let $g_k(x)=f(x+k)-f(x).$ We note that $g_n(0)=0.$ We count that as one of our solutions. What we now need is for there to be a total of at least $n-1$ roots of the collection of equations $g_k(x)=0$ for $1\le k\le n-1.$

Consider $f$ to be continuous and periodic of period $n.$ (That's what $f(0)=f(n)$ is good for.) Then define each $g_k$ to be continous and periodic. Each $g_k$ has average value zero. Thus either $g_k$ is indentically zero (in which case we have infinitely many solutions) or it has at least one positive and at least one negative value in each period. Thus (by the Intermediate Value Theorem) for each $k,$ , $1\le k \le n-1,$ there are at least two roots of $g_k(x)=0$ in $[0,n).$

This apparently gives us $2(n-1)$ solutions, but can we count them all?

If $g_k(x)=0$ for $0\le x \le n-k,$ then this is a solution on our list. But if $n-k<x<n$ then we can rewrite this as $g_{n-k}(x-n+k)=0$ and it counts as a root of $g_{n-k}(x)=0.$ Thus each solution we found does count as one of our solutions, but we may be counting the same solution twice (but no more than twice). Allowing for the double counting, we still have $(n-1)$ solutions, plus the one solution of $g_n(x)=0$ which we handle separately.

**Posted:** Mon Jan 10, 2005 11:16 pm

dickchimney · Poincare Conjecture Offline Joined: 06 Mar 2004 Posts: 218

What Kent just proved is somewhat stronger than the desired result but I cannot state it clearly

**Posted:** Mon Jan 10, 2005 11:28 pm

romano · Riemann Hypothesis Offline Joined: 10 Jun 2004 Posts: 304

dear Kent Merryfield , why can you write $g_{n-k}(x-n+k)=0$ where $n-k\leq x\leq n$ ?

**Posted:** Tue Jan 11, 2005 7:36 am

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dickchimney · Poincare Conjecture Offline Joined: 06 Mar 2004 Posts: 218

**Posted:** Tue Jan 11, 2005 8:01 am

Kent Merryfield

Thanks for covering that, dickchimney. By the way, the double counting always happens. There is a one-to-one correspondence between a root of $g_k$ in the "wrong" part of the interval and a root of $g_{n-k}$ in the "right" part of the interval, and vice versa. The only way to have more than the specified minimum number of solutions is to have some $g_k$ have more than two roots.

It took me a while to understand your solution in #5, but now that I do, I see that it is quite elegant. You are inducting on $n$ , and your inductive hypothesis is that the whole theorem is true for $n-1.$

**Posted:** Tue Jan 11, 2005 10:26 am

dickchimney · Poincare Conjecture Offline Joined: 06 Mar 2004 Posts: 218

Thank u Kent! Wink

What is your opinion Romano ? Is it solved ?

**Posted:** Wed Jan 12, 2005 8:50 am

romano · Riemann Hypothesis Offline Joined: 10 Jun 2004 Posts: 304

Yeah , nice solutions , Kent , Dickchimney !

**Posted:** Thu Jan 13, 2005 1:46 am

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