If $x\in \mathbb{R}$ such $|1-|2-|3-|4-\cdots-|2011-x|\cdots||||=x$

Question

Let $x\in \mathbb{R}$, and such $$|1-|2-|3-|4-\cdots-|2011-x|\cdots||||=x$$ find the maximum of the $x$

I have try $f(x)=|1-|2-|3-|4-\cdots-|2011-x|\cdots||||$,I have find $$f(0)=0,f(1)=1,f(2)=0$$

Answer 1

Let us show that $x=1$ is the largest fixed point of $f(x) = |1-|2-|\ldots-|2011-x|\ldots|||$. First of all, it can be easily seen that it is indeed a fixed point (see the table below).

Now, let $N := 1+\ldots+2011$. We definitely have $f(N) = 0$ and $f(N+z) = z$ for $z\ge 0$. So, $f$ cannot have a fixed point larger than $N$.

Assume that $x > 1$ is a fixed point of $f$. Then $1 < x = |1-|2-|\ldots|||$. That means that the non-negative number $|2-|\ldots||$ has distance larger than $1$ to $1$. That is, $|2-|\ldots|| > 2$. Hence, similarly, $|3-|\ldots|| > 4 = 1 + 1 + 2$. Therefore, $|4-|\ldots|| > 7 = 1 + 1 + 2 + 3$ and so on. In the end, $|2011-x| > 1 + 1 +\ldots + 2010$. So $x > 1+N$, which is a contradiction.

Here is why $x=1$ is a fixed point of $f$:

$$ \begin{matrix} k & |k-a|\\ 2011 & \boxed{2010}\\ \boxed{2010} & 0 \\ 2009 & 2009\\ 2008 & 1\\ 2007 & \boxed{2006}\\ \boxed{2006} & 0\\ \vdots & \vdots\\ & \boxed{2}\\ \boxed{2} & 0\\ 1 & 1 \end{matrix} $$ The boxed values are the ones that are even but not divisible by $4$.

Answer 2

We claim that $1$ is the highest value for which $f_{2011}(x)=x.$

Let $f_a(x)$ represent $|1-|2-|3-|4-...-|a-x|...||||.$

We look for a pattern to calculate $f_{2011}(1).$

Take note that for $c>4,$ we have $f_{c}(1)=f_{c-1}(|c-1|)=f_{c-1}(|c-1|)=f_{c-2}(|(c-1)-(c-1)|)=f_{c-2}(0)=f_{c-3}(|c-3|)=f_{c-3}(c-3)=f_{c-4}((c-4)-(c-3))=f_{c-4}(|-1|)=f_{c-4}(1).$ Applying this repeatedly gives $f_{2011}(1)=f_{2007}(1)=f_{2003}(1)=...=f_{3}(1)=|3-|2-|1-1|||=|3-|2-0||=|3-2|=1.$ Therefore, $f_{2011}(x)=x$ is satisfied for $x=1.$

Now we will prove that that is the maximum possible value of $x$ which can work.

We will evaluate $f_{2011}(2).$ We let $d$ be a positive integer such that $g<4.$ We will start to evaluate $f_{a}(2)$. Calculating, we have $f_{a}(2)=f_{a-1}(|a-2|)=f_{a-1}(a-2)=f_{a-2}(|(a-1)-(a-2)|)=f_{a-2}(1).$ Applying this, we get $f_{2011}(2)=f_{2009}(1)=f_{1}(1)=|1-1|=0.$

We know that the slope of the absolute value is $-1$ or $1$ since there is just one variable $x,$ and the absolute value would change its coefficient, and nothing else.

Since we know that the function is continuous, we have that from $f_{2011}(1)$ to $f_{2011}(2)$ that the slope is $-1.$ We know that from there, even if it it has a slope of $1$ $f_{2011}(2)$ to infinity, it would still be an parallel line that never intersects. That means for every real number greater than $2,$ since $y=x$ can't intersect with $f_{2011}(x),$ $x=1$ is the greatest solution $\boxed{\text{Q.E.D.}}.$