Vtyjkyuk

Let $xin mathbbR$, 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
$$f(0)=0,f(1)=1,f(2)=0$$

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 $zge 0$. 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 $f$:

$$
beginmatrix
k & |k-a|\
2011 & boxed2010\
boxed2010 & 0 \
2009 & 2009\
2008 & 1\
2007 & boxed2006\
boxed2006 & 0\
vdots & vdots\
& boxed2\
boxed2 & 0\
1 & 1
endmatrix
$$
The boxed values are the ones that are even but not divisible by $4$.

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 $boxedtextQ.E.D..$