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A nice equation with the floor function

  1. Jun 18, 2016 #1
    • Moved from a technical forum, so homework template missing
    Hi all!
    I should solve the following equation involving the ceil or floor function:
    [tex]
    x \cdot \lceil x \cdot \lceil x \rceil \rceil = 82
    [/tex]
    I have found just one solution 41/9, but I have no idea about how finding the other solutions.
    Thanks you in advance!
     
  2. jcsd
  3. Jun 18, 2016 #2

    jbriggs444

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    What convinces you that there are other solutions?
     
  4. Jun 18, 2016 #3

    mfb

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    Can the expression stay the same or decrease if you increase x?
    Can the expression stay the same or increase if you decrease x?
     
  5. Jun 18, 2016 #4
    The problem says: 'Find all possibile solutions'. I find this solution numerically.
    The expression must be the same.
     
  6. Jun 18, 2016 #5

    mfb

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    Right, so can this be possible with a different x?
    That does not tell you how many solutions there are.
     
  7. Jun 18, 2016 #6
    I propose this solution, check it out!
    We must find solution on the positive real axis.
    Let be [itex]x= n + \delta[/itex], where [itex]n[/itex] is the integer part and [itex]\delta[/itex] is the fractional part with [itex]0 \leq \delta < 1[/itex].
    Given that [itex]x \lfloor x \lfloor x \rfloor \rfloor=82[/itex] and [itex]x \lfloor x \lfloor x \rfloor \rfloor \leq x^3[/itex] , we get [itex] x \geq 82^{\frac{1}{3}} \approx 4.344[/itex]. Therefore [itex]n \geq 4[/itex].
    If 5 is not solution the equation, then every number greater than 5 are not solutions. Therefore [itex]n=4[/itex].
    With the previous statements, the equation becames:
    [itex](4+\delta)\cdot \lfloor 16+4\delta \rfloor=82 [/itex]
    This equation is equivalent to the following equations:
    [itex]0 \leq \delta < 1/4[/itex] and [itex](4+\delta)\cdot(16+0)=82[/itex]
    [itex]1/4 \leq \delta < 2/4[/itex] and [itex](4+\delta)\cdot(16+1)=82[/itex]
    [itex]2/4 \leq \delta < 3/4[/itex] and [itex](4+\delta)\cdot(16+2)=82[/itex]
    [itex]3/4 \leq \delta < 4/4[/itex] and [itex](4+\delta)\cdot(16+3)=82[/itex]
    Only the third equation admits a solution that is [itex]\delta=5/9[/itex].
     
    Last edited: Jun 18, 2016
  8. Jun 18, 2016 #7

    fresh_42

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    What I don't understand here is why there is a solution at all?
    Obviously x has to be between 4 and 5, excluding 4 and 5. So the inner ceiling is 5 and the outer at least 21. But four times twenty-one is already 84.
     
  9. Jun 18, 2016 #8

    mfb

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    You write the ceil function and then derive things for the floor function? For the floor function that works.
    See the floor/ceil mixup.
     
  10. Jun 18, 2016 #9

    fresh_42

    Staff: Mentor

  11. Jun 18, 2016 #10
    With [itex] \lceil x \rceil [/itex] I want to denote the floor function of [itex] x [/itex].
     
  12. Jun 18, 2016 #11

    SammyS

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    That's how to denote the ceiling function.

    Use lfloor and rfloor for the floor function. ##\ \lfloor x \rfloor##
     
  13. Jun 18, 2016 #12

    Ray Vickson

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    No. The function ##f(x) = x \lfloor x \lfloor x \rfloor \rfloor## is strictly increasing in ##x##, sometimes linearly and sometimes through a jump discontinuity.
    To see this, let ##x = n + r## where ##n \geq 0 ## is an integer and ##0 \leq r < 1##. Then the inner floor function is ##\lfloor n + r \rfloor = n##, so
    [tex] f(n+r) = (n+r) \lfloor (n+r)n \rfloor [/tex]
    The function ## g(n,r) = \lfloor (n+r)n \rfloor## is positve and nondecreasing in ##n## and ##r##, so ##f(n+r) = (n+r) g(n,r)## is strictly increasing.

    That means that ##f(x) = 82## has at most one root.
     
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