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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 $$f(0)=0,f(1)=1,f(2)=0$$

tarit goswami
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math110
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2 Answers2

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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 $4$.

amsmath
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  • Why must $x$ be so large? I can't tell... – Jason Kim Aug 29 '18 at 13:46
  • What is it you don't understand in my reasoning? – amsmath Aug 29 '18 at 15:52
  • $N=1+2+3...+2011$ – Jason Kim Aug 29 '18 at 21:21
  • @JasonKim How do you think I can understand what you mean? Try to show a bit of empathy. That's also very important when writing proofs in mathematics. You don't write them to show others that you have understood the problem, but for making others understand. So, when you write something mathematic, think twice about whether other people can understand your stuff. What I have written above and what you stated in your last comment is the definition of a number, nothing more. – amsmath Aug 29 '18 at 21:41
  • I don't understand why we need to prove for just all numbers $N$ greater than $1.$ Also, I suggest writing a few more terms to make the reader understand how it went from all $1$'s to $2011.$ – Jason Kim Aug 30 '18 at 00:12
  • @JasonKim "I don't understand why we need to prove for just all numbers N greater than 1" Prove what? That $x > 1$ cannot be a fixed point? – amsmath Aug 30 '18 at 00:41
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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 $\boxed{\text{Q.E.D.}}.$

Jason Kim
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