2
$\begingroup$

Find the complete solution set of the equation $2|x-1|=\frac{[x]([x]-1)(2[x]-1)}{6}+[x]^2\{x\}$

This is a part of a longer question and I am stuck at this step. I'm not able to figure out how to solve this equation. It would be great if I could get some hint here.

[]-greatest integer function
{}-fractional part function

  • 0
    `{}-fractional part function` Where does that figure into the question? – 2017-02-24
  • 0
    @dxiv oh wait these braces didn't show in mathjax...made the edit – 2017-02-24
  • 0
    I think you can try to estimate a bound for $[x]$. For example, it's not hard to prove (by induction, calculus, etc.) for $n\geq 3, $n\in \mathbb{N}$, $n(n-1)(2n-1)/6>2(n-1)$, that means you cannot have $[x]>=3$. Similarly you can find a lower bound for $[x]$. After that you'll reduce to finite possible value of $[x]$ and the problem can be solved case by case. – 2017-02-24
  • 0
    Can someone help me removing the above comment? The x button overlap with the hyperlink of related problem and I cannot seem to click it. – 2017-02-24
  • 0
    Try to estimate $[x]$. For example, prove for $n\geq 3$, $n\in \mathbb{N}$, we have $2(n-1)$[x]\geq 3$. Similarly try to find a lower bound of $[x]$. After that you reduce the possible value of $[x]$ to finite. – 2017-02-24
  • 0
    @frank000 can you elaborate on this...maybe as an answer – 2017-02-24
  • 0
    @Osheen Sachdev, I'll do it later if no one do it before me. It's late in my place now. – 2017-02-24
  • 0
    @frank000 I think you would have to – 2017-02-25

1 Answers 1

0

For simplicity I'll use $a=[x]$, $b=\{x\}$, $x=a+b$. The equation becomes $$2|a+b-1|=a(a-1)(2a-1)/6+a^2b$$.

First consider the case $a\geq 1$, then $x\geq 1$ We can estimate $$RHS-LHS\geq a(a-1)(2a-1)/6-2(a+b-1)\geq a(a-1)(2a-1)/6-2a=\\ \frac{a(2a^2-3a-11)}{6}$$

Last formula is strictly positive when $a\geq 4$ (Routine calculus or induction exercise). So in this case all possible values of $a$ are $1, 2, 3$.

Similarly when $a\leq 0$, $$RHS-LHS=a(a-1)(2a-1)/6+a^2b-2(1-a-b)\leq a(a-1)(2a-1)/6+a^2+2a=\frac{a(2a^2+3a+13)}{6}$$. Again use calculus or induction to prove last expression is strictly negative when $a\leq -1$. So in this case all possible value of $a$ is $0$.

Now plug in $a=0,1,2,3$ and solve $b$ for each case (Don't forget $0\leq b<1$). $x=a+b$ and if I compute it correctly $x=1$ or $5/2$