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











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3047143%2ffloor-equation-lfloor-3x-x2-rfloor-lfloor-x2-1-2-rfloor%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    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











    $endgroup$


















      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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3047143%2ffloor-equation-lfloor-3x-x2-rfloor-lfloor-x2-1-2-rfloor%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Probability when a professor distributes a quiz and homework assignment to a class of n students.

            Image
            0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
            Read more

            Aardman Animations

            Image
            Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
            Read more

            Are they similar matrix

            Image
            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
            Read more