Floor equation $lfloor 3x-x^2 rfloor = lfloor x^2 + 1/2 rfloor$












1












$begingroup$



Solve the equation:



$$left lfloor 3x-x^2 right rfloor = left lfloor x^2 + 1/2 right rfloor$$




In the solution it writes




We notice that $x^{2}+frac{1}{2}> 0$ therfore $left lfloor x^2 - 1/2 right rfloor geq 0$ . From there $left lfloor 3x-x^2 right rfloor = n geq 0$.




So far all of this I understand but than it writes:




But $3x-x^2 leq frac{9}{4}$ and $left lfloor 3x-x^2 right rfloor< 3$.




I dont understand this last part how did they get $3x-x^2 leq frac{9}{4}$ ?






floor-function







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    I fail to see step 1. Say $x=0$ then $x^2+1/2$ has floor zero so $x^2-1/2$ has floor $-1$ which isn't $ge 0.$
    $endgroup$
    – coffeemath
    Dec 20 '18 at 4:15










  • $begingroup$
    For $3x-x^2le frac 94$, note that $-(x-frac32)^2$ is the underlying quadratic...
    $endgroup$
    – abiessu
    Dec 20 '18 at 4:18










  • $begingroup$
    I think the key observation is that $x^2+frac{1}{2}$ and $3x-x^2$ are a convex and a concave parabola.
    $endgroup$
    – Mr.Robot
    Dec 20 '18 at 4:23
















1












$begingroup$



Solve the equation:



$$left lfloor 3x-x^2 right rfloor = left lfloor x^2 + 1/2 right rfloor$$




In the solution it writes




We notice that $x^{2}+frac{1}{2}> 0$ therfore $left lfloor x^2 - 1/2 right rfloor geq 0$ . From there $left lfloor 3x-x^2 right rfloor = n geq 0$.




So far all of this I understand but than it writes:




But $3x-x^2 leq frac{9}{4}$ and $left lfloor 3x-x^2 right rfloor< 3$.




I dont understand this last part how did they get $3x-x^2 leq frac{9}{4}$ ?






floor-function







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    I fail to see step 1. Say $x=0$ then $x^2+1/2$ has floor zero so $x^2-1/2$ has floor $-1$ which isn't $ge 0.$
    $endgroup$
    – coffeemath
    Dec 20 '18 at 4:15










  • $begingroup$
    For $3x-x^2le frac 94$, note that $-(x-frac32)^2$ is the underlying quadratic...
    $endgroup$
    – abiessu
    Dec 20 '18 at 4:18










  • $begingroup$
    I think the key observation is that $x^2+frac{1}{2}$ and $3x-x^2$ are a convex and a concave parabola.
    $endgroup$
    – Mr.Robot
    Dec 20 '18 at 4:23














1












1








1


1



$begingroup$



Solve the equation:



$$left lfloor 3x-x^2 right rfloor = left lfloor x^2 + 1/2 right rfloor$$




In the solution it writes




We notice that $x^{2}+frac{1}{2}> 0$ therfore $left lfloor x^2 - 1/2 right rfloor geq 0$ . From there $left lfloor 3x-x^2 right rfloor = n geq 0$.




So far all of this I understand but than it writes:




But $3x-x^2 leq frac{9}{4}$ and $left lfloor 3x-x^2 right rfloor< 3$.




I dont understand this last part how did they get $3x-x^2 leq frac{9}{4}$ ?






floor-function







share|cite|improve this question











$endgroup$





Solve the equation:



$$left lfloor 3x-x^2 right rfloor = left lfloor x^2 + 1/2 right rfloor$$




In the solution it writes




We notice that $x^{2}+frac{1}{2}> 0$ therfore $left lfloor x^2 - 1/2 right rfloor geq 0$ . From there $left lfloor 3x-x^2 right rfloor = n geq 0$.




So far all of this I understand but than it writes:




But $3x-x^2 leq frac{9}{4}$ and $left lfloor 3x-x^2 right rfloor< 3$.




I dont understand this last part how did they get $3x-x^2 leq frac{9}{4}$ ?






floor-function




floor-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 4:33









Kemono Chen

3,1891844




3,1891844










asked Dec 20 '18 at 4:04









Ivan LjiljaIvan Ljilja

205




205








  • 4




    $begingroup$
    I fail to see step 1. Say $x=0$ then $x^2+1/2$ has floor zero so $x^2-1/2$ has floor $-1$ which isn't $ge 0.$
    $endgroup$
    – coffeemath
    Dec 20 '18 at 4:15










  • $begingroup$
    For $3x-x^2le frac 94$, note that $-(x-frac32)^2$ is the underlying quadratic...
    $endgroup$
    – abiessu
    Dec 20 '18 at 4:18










  • $begingroup$
    I think the key observation is that $x^2+frac{1}{2}$ and $3x-x^2$ are a convex and a concave parabola.
    $endgroup$
    – Mr.Robot
    Dec 20 '18 at 4:23














  • 4




    $begingroup$
    I fail to see step 1. Say $x=0$ then $x^2+1/2$ has floor zero so $x^2-1/2$ has floor $-1$ which isn't $ge 0.$
    $endgroup$
    – coffeemath
    Dec 20 '18 at 4:15










  • $begingroup$
    For $3x-x^2le frac 94$, note that $-(x-frac32)^2$ is the underlying quadratic...
    $endgroup$
    – abiessu
    Dec 20 '18 at 4:18










  • $begingroup$
    I think the key observation is that $x^2+frac{1}{2}$ and $3x-x^2$ are a convex and a concave parabola.
    $endgroup$
    – Mr.Robot
    Dec 20 '18 at 4:23








4




4




$begingroup$
I fail to see step 1. Say $x=0$ then $x^2+1/2$ has floor zero so $x^2-1/2$ has floor $-1$ which isn't $ge 0.$
$endgroup$
– coffeemath
Dec 20 '18 at 4:15




$begingroup$
I fail to see step 1. Say $x=0$ then $x^2+1/2$ has floor zero so $x^2-1/2$ has floor $-1$ which isn't $ge 0.$
$endgroup$
– coffeemath
Dec 20 '18 at 4:15












$begingroup$
For $3x-x^2le frac 94$, note that $-(x-frac32)^2$ is the underlying quadratic...
$endgroup$
– abiessu
Dec 20 '18 at 4:18




$begingroup$
For $3x-x^2le frac 94$, note that $-(x-frac32)^2$ is the underlying quadratic...
$endgroup$
– abiessu
Dec 20 '18 at 4:18












$begingroup$
I think the key observation is that $x^2+frac{1}{2}$ and $3x-x^2$ are a convex and a concave parabola.
$endgroup$
– Mr.Robot
Dec 20 '18 at 4:23




$begingroup$
I think the key observation is that $x^2+frac{1}{2}$ and $3x-x^2$ are a convex and a concave parabola.
$endgroup$
– Mr.Robot
Dec 20 '18 at 4:23










1 Answer
1






active

oldest

votes


















4












$begingroup$

If we put $-x^2+3x$ into translated form we get:



$$-(x^2-3x)=-left[left(x-frac 3 2right)^2-left(frac32right)^2right]=-left(x-frac 3 2right)^2+frac94$$



We can now see that this is $xmapsto x^2$ that has been dilated by a factor of -1 (graphically: reflected about the $x$ axis) and translated $frac 3 2$ units right and $frac 9 4$ units up. (In simple terms, the graph is open downwards and the vertex is at $left(frac 3 2, frac 9 4right)$.) Hence, for all $x$, $-x^2+3xle frac 9 4$.



Edit



I think there's a typo in the first part of your presented solution (see coffeemath's comment): it should be:




... therefore $leftlfloor x^2+frac 1 2rightrfloorge0$. From there...




In any case, I think the solution is directing you to to identify potential cases to investigate: this equation can only hold for floors of 0, 1 and 2.



For example, investigating the floor 0 case:



$$begin{align}
leftlfloor x^2+frac 1 2rightrfloor = 0&Rightarrow 0le x^2+frac 1 2<1\
x^2+frac 1 2ge 0&Rightarrow xinmathbb R\
x^2+frac 1 2 < 1&Rightarrow
xinleft]-frac{sqrt 2}{2},frac{sqrt2}{2}right[\\
leftlfloor-x^2+3xrightrfloor=0&Rightarrow0le-x^2+3x<1\
-x^2+3xge0&Rightarrow xinleft[0, 3right]\
-x^2+3x<1&Rightarrow xinleft]-infty,frac{3-sqrt 5}{2}right[;;bigcup;;left]frac{3+sqrt5}{2},inftyright[
end{align}$$



Taking the intersection of all these sets gives us part of the solution to the original problem:



$$xinleft[0, frac{3-sqrt 5}{2}right[$$



Similarly, investigating the floor 1 case gives us that the equation holds for $xinleft[frac{sqrt 2}{2}, 1right[$, and the floor 2 case gives us that the equation holds for $xinleft[frac{sqrt6}{2},frac{sqrt{10}}{2}right[$.



The solution set to the problem is thus:



$$left[0, frac{3-sqrt 5}{2}right[;;bigcup;;left[frac{sqrt 2}{2}, 1right[;;bigcup;;left[frac{sqrt6}{2},frac{sqrt{10}}{2}right[$$






share|cite|improve this answer











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    1 Answer
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    1 Answer
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    active

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    4












    $begingroup$

    If we put $-x^2+3x$ into translated form we get:



    $$-(x^2-3x)=-left[left(x-frac 3 2right)^2-left(frac32right)^2right]=-left(x-frac 3 2right)^2+frac94$$



    We can now see that this is $xmapsto x^2$ that has been dilated by a factor of -1 (graphically: reflected about the $x$ axis) and translated $frac 3 2$ units right and $frac 9 4$ units up. (In simple terms, the graph is open downwards and the vertex is at $left(frac 3 2, frac 9 4right)$.) Hence, for all $x$, $-x^2+3xle frac 9 4$.



    Edit



    I think there's a typo in the first part of your presented solution (see coffeemath's comment): it should be:




    ... therefore $leftlfloor x^2+frac 1 2rightrfloorge0$. From there...




    In any case, I think the solution is directing you to to identify potential cases to investigate: this equation can only hold for floors of 0, 1 and 2.



    For example, investigating the floor 0 case:



    $$begin{align}
    leftlfloor x^2+frac 1 2rightrfloor = 0&Rightarrow 0le x^2+frac 1 2<1\
    x^2+frac 1 2ge 0&Rightarrow xinmathbb R\
    x^2+frac 1 2 < 1&Rightarrow
    xinleft]-frac{sqrt 2}{2},frac{sqrt2}{2}right[\\
    leftlfloor-x^2+3xrightrfloor=0&Rightarrow0le-x^2+3x<1\
    -x^2+3xge0&Rightarrow xinleft[0, 3right]\
    -x^2+3x<1&Rightarrow xinleft]-infty,frac{3-sqrt 5}{2}right[;;bigcup;;left]frac{3+sqrt5}{2},inftyright[
    end{align}$$



    Taking the intersection of all these sets gives us part of the solution to the original problem:



    $$xinleft[0, frac{3-sqrt 5}{2}right[$$



    Similarly, investigating the floor 1 case gives us that the equation holds for $xinleft[frac{sqrt 2}{2}, 1right[$, and the floor 2 case gives us that the equation holds for $xinleft[frac{sqrt6}{2},frac{sqrt{10}}{2}right[$.



    The solution set to the problem is thus:



    $$left[0, frac{3-sqrt 5}{2}right[;;bigcup;;left[frac{sqrt 2}{2}, 1right[;;bigcup;;left[frac{sqrt6}{2},frac{sqrt{10}}{2}right[$$






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      If we put $-x^2+3x$ into translated form we get:



      $$-(x^2-3x)=-left[left(x-frac 3 2right)^2-left(frac32right)^2right]=-left(x-frac 3 2right)^2+frac94$$



      We can now see that this is $xmapsto x^2$ that has been dilated by a factor of -1 (graphically: reflected about the $x$ axis) and translated $frac 3 2$ units right and $frac 9 4$ units up. (In simple terms, the graph is open downwards and the vertex is at $left(frac 3 2, frac 9 4right)$.) Hence, for all $x$, $-x^2+3xle frac 9 4$.



      Edit



      I think there's a typo in the first part of your presented solution (see coffeemath's comment): it should be:




      ... therefore $leftlfloor x^2+frac 1 2rightrfloorge0$. From there...




      In any case, I think the solution is directing you to to identify potential cases to investigate: this equation can only hold for floors of 0, 1 and 2.



      For example, investigating the floor 0 case:



      $$begin{align}
      leftlfloor x^2+frac 1 2rightrfloor = 0&Rightarrow 0le x^2+frac 1 2<1\
      x^2+frac 1 2ge 0&Rightarrow xinmathbb R\
      x^2+frac 1 2 < 1&Rightarrow
      xinleft]-frac{sqrt 2}{2},frac{sqrt2}{2}right[\\
      leftlfloor-x^2+3xrightrfloor=0&Rightarrow0le-x^2+3x<1\
      -x^2+3xge0&Rightarrow xinleft[0, 3right]\
      -x^2+3x<1&Rightarrow xinleft]-infty,frac{3-sqrt 5}{2}right[;;bigcup;;left]frac{3+sqrt5}{2},inftyright[
      end{align}$$



      Taking the intersection of all these sets gives us part of the solution to the original problem:



      $$xinleft[0, frac{3-sqrt 5}{2}right[$$



      Similarly, investigating the floor 1 case gives us that the equation holds for $xinleft[frac{sqrt 2}{2}, 1right[$, and the floor 2 case gives us that the equation holds for $xinleft[frac{sqrt6}{2},frac{sqrt{10}}{2}right[$.



      The solution set to the problem is thus:



      $$left[0, frac{3-sqrt 5}{2}right[;;bigcup;;left[frac{sqrt 2}{2}, 1right[;;bigcup;;left[frac{sqrt6}{2},frac{sqrt{10}}{2}right[$$






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        If we put $-x^2+3x$ into translated form we get:



        $$-(x^2-3x)=-left[left(x-frac 3 2right)^2-left(frac32right)^2right]=-left(x-frac 3 2right)^2+frac94$$



        We can now see that this is $xmapsto x^2$ that has been dilated by a factor of -1 (graphically: reflected about the $x$ axis) and translated $frac 3 2$ units right and $frac 9 4$ units up. (In simple terms, the graph is open downwards and the vertex is at $left(frac 3 2, frac 9 4right)$.) Hence, for all $x$, $-x^2+3xle frac 9 4$.



        Edit



        I think there's a typo in the first part of your presented solution (see coffeemath's comment): it should be:




        ... therefore $leftlfloor x^2+frac 1 2rightrfloorge0$. From there...




        In any case, I think the solution is directing you to to identify potential cases to investigate: this equation can only hold for floors of 0, 1 and 2.



        For example, investigating the floor 0 case:



        $$begin{align}
        leftlfloor x^2+frac 1 2rightrfloor = 0&Rightarrow 0le x^2+frac 1 2<1\
        x^2+frac 1 2ge 0&Rightarrow xinmathbb R\
        x^2+frac 1 2 < 1&Rightarrow
        xinleft]-frac{sqrt 2}{2},frac{sqrt2}{2}right[\\
        leftlfloor-x^2+3xrightrfloor=0&Rightarrow0le-x^2+3x<1\
        -x^2+3xge0&Rightarrow xinleft[0, 3right]\
        -x^2+3x<1&Rightarrow xinleft]-infty,frac{3-sqrt 5}{2}right[;;bigcup;;left]frac{3+sqrt5}{2},inftyright[
        end{align}$$



        Taking the intersection of all these sets gives us part of the solution to the original problem:



        $$xinleft[0, frac{3-sqrt 5}{2}right[$$



        Similarly, investigating the floor 1 case gives us that the equation holds for $xinleft[frac{sqrt 2}{2}, 1right[$, and the floor 2 case gives us that the equation holds for $xinleft[frac{sqrt6}{2},frac{sqrt{10}}{2}right[$.



        The solution set to the problem is thus:



        $$left[0, frac{3-sqrt 5}{2}right[;;bigcup;;left[frac{sqrt 2}{2}, 1right[;;bigcup;;left[frac{sqrt6}{2},frac{sqrt{10}}{2}right[$$






        share|cite|improve this answer











        $endgroup$



        If we put $-x^2+3x$ into translated form we get:



        $$-(x^2-3x)=-left[left(x-frac 3 2right)^2-left(frac32right)^2right]=-left(x-frac 3 2right)^2+frac94$$



        We can now see that this is $xmapsto x^2$ that has been dilated by a factor of -1 (graphically: reflected about the $x$ axis) and translated $frac 3 2$ units right and $frac 9 4$ units up. (In simple terms, the graph is open downwards and the vertex is at $left(frac 3 2, frac 9 4right)$.) Hence, for all $x$, $-x^2+3xle frac 9 4$.



        Edit



        I think there's a typo in the first part of your presented solution (see coffeemath's comment): it should be:




        ... therefore $leftlfloor x^2+frac 1 2rightrfloorge0$. From there...




        In any case, I think the solution is directing you to to identify potential cases to investigate: this equation can only hold for floors of 0, 1 and 2.



        For example, investigating the floor 0 case:



        $$begin{align}
        leftlfloor x^2+frac 1 2rightrfloor = 0&Rightarrow 0le x^2+frac 1 2<1\
        x^2+frac 1 2ge 0&Rightarrow xinmathbb R\
        x^2+frac 1 2 < 1&Rightarrow
        xinleft]-frac{sqrt 2}{2},frac{sqrt2}{2}right[\\
        leftlfloor-x^2+3xrightrfloor=0&Rightarrow0le-x^2+3x<1\
        -x^2+3xge0&Rightarrow xinleft[0, 3right]\
        -x^2+3x<1&Rightarrow xinleft]-infty,frac{3-sqrt 5}{2}right[;;bigcup;;left]frac{3+sqrt5}{2},inftyright[
        end{align}$$



        Taking the intersection of all these sets gives us part of the solution to the original problem:



        $$xinleft[0, frac{3-sqrt 5}{2}right[$$



        Similarly, investigating the floor 1 case gives us that the equation holds for $xinleft[frac{sqrt 2}{2}, 1right[$, and the floor 2 case gives us that the equation holds for $xinleft[frac{sqrt6}{2},frac{sqrt{10}}{2}right[$.



        The solution set to the problem is thus:



        $$left[0, frac{3-sqrt 5}{2}right[;;bigcup;;left[frac{sqrt 2}{2}, 1right[;;bigcup;;left[frac{sqrt6}{2},frac{sqrt{10}}{2}right[$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 16 at 2:23

























        answered Dec 20 '18 at 4:16









        Richard AmblerRichard Ambler

        1,308515




        1,308515






























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