Prove that $sin^2(x)/x=x-lfloor xrfloor$ has infinite solutions












0












$begingroup$


I am trying to prove that
$sin^2(x)/x=x-lfloor xrfloor$ has infinite solutions.



I know that $sin^2(x) $is between [0,1] and $|x|>|sin(x)|$ therefore $x>sin^2x$
so the left side of the equation is between [0,1].
for right side of the equation I know that it between [0,1) but I don`t know how to continue from these info in order to show that it has infinite solutions.



Also, I am thinking the Intermediate value theorem might help to prove it, however I don`t know how to show more than one solution with this theorem.



Any suggestions?






calculus







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$endgroup$












  • $begingroup$
    The intermediate value theorem is the right idea. Note that the left-hand side is $0$ at multiple of $pi$ and the right-hand side is $0$ at the integers.
    $endgroup$
    – saulspatz
    Dec 7 '18 at 14:47










  • $begingroup$
    What you probably want to do is to show that you have at least one solution in an interval $(kpi,(k+1)pi)$
    $endgroup$
    – Andrei
    Dec 7 '18 at 14:47










  • $begingroup$
    What is $[x]$ ?
    $endgroup$
    – Rhys Hughes
    Dec 7 '18 at 14:50






  • 2




    $begingroup$
    @JohnD The intermediate value theorem guarantees you at least one solution, not only one.
    $endgroup$
    – Federico
    Dec 7 '18 at 15:02






  • 1




    $begingroup$
    @AlexVong for [1.75] it returns 1.
    $endgroup$
    – John D
    Dec 7 '18 at 15:25
















0












$begingroup$


I am trying to prove that
$sin^2(x)/x=x-lfloor xrfloor$ has infinite solutions.



I know that $sin^2(x) $is between [0,1] and $|x|>|sin(x)|$ therefore $x>sin^2x$
so the left side of the equation is between [0,1].
for right side of the equation I know that it between [0,1) but I don`t know how to continue from these info in order to show that it has infinite solutions.



Also, I am thinking the Intermediate value theorem might help to prove it, however I don`t know how to show more than one solution with this theorem.



Any suggestions?






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    The intermediate value theorem is the right idea. Note that the left-hand side is $0$ at multiple of $pi$ and the right-hand side is $0$ at the integers.
    $endgroup$
    – saulspatz
    Dec 7 '18 at 14:47










  • $begingroup$
    What you probably want to do is to show that you have at least one solution in an interval $(kpi,(k+1)pi)$
    $endgroup$
    – Andrei
    Dec 7 '18 at 14:47










  • $begingroup$
    What is $[x]$ ?
    $endgroup$
    – Rhys Hughes
    Dec 7 '18 at 14:50






  • 2




    $begingroup$
    @JohnD The intermediate value theorem guarantees you at least one solution, not only one.
    $endgroup$
    – Federico
    Dec 7 '18 at 15:02






  • 1




    $begingroup$
    @AlexVong for [1.75] it returns 1.
    $endgroup$
    – John D
    Dec 7 '18 at 15:25














0












0








0


0



$begingroup$


I am trying to prove that
$sin^2(x)/x=x-lfloor xrfloor$ has infinite solutions.



I know that $sin^2(x) $is between [0,1] and $|x|>|sin(x)|$ therefore $x>sin^2x$
so the left side of the equation is between [0,1].
for right side of the equation I know that it between [0,1) but I don`t know how to continue from these info in order to show that it has infinite solutions.



Also, I am thinking the Intermediate value theorem might help to prove it, however I don`t know how to show more than one solution with this theorem.



Any suggestions?






calculus







share|cite|improve this question











$endgroup$




I am trying to prove that
$sin^2(x)/x=x-lfloor xrfloor$ has infinite solutions.



I know that $sin^2(x) $is between [0,1] and $|x|>|sin(x)|$ therefore $x>sin^2x$
so the left side of the equation is between [0,1].
for right side of the equation I know that it between [0,1) but I don`t know how to continue from these info in order to show that it has infinite solutions.



Also, I am thinking the Intermediate value theorem might help to prove it, however I don`t know how to show more than one solution with this theorem.



Any suggestions?






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 14:53









Andrei

11.8k21026




11.8k21026










asked Dec 7 '18 at 14:41









John DJohn D

366




366












  • $begingroup$
    The intermediate value theorem is the right idea. Note that the left-hand side is $0$ at multiple of $pi$ and the right-hand side is $0$ at the integers.
    $endgroup$
    – saulspatz
    Dec 7 '18 at 14:47










  • $begingroup$
    What you probably want to do is to show that you have at least one solution in an interval $(kpi,(k+1)pi)$
    $endgroup$
    – Andrei
    Dec 7 '18 at 14:47










  • $begingroup$
    What is $[x]$ ?
    $endgroup$
    – Rhys Hughes
    Dec 7 '18 at 14:50






  • 2




    $begingroup$
    @JohnD The intermediate value theorem guarantees you at least one solution, not only one.
    $endgroup$
    – Federico
    Dec 7 '18 at 15:02






  • 1




    $begingroup$
    @AlexVong for [1.75] it returns 1.
    $endgroup$
    – John D
    Dec 7 '18 at 15:25


















  • $begingroup$
    The intermediate value theorem is the right idea. Note that the left-hand side is $0$ at multiple of $pi$ and the right-hand side is $0$ at the integers.
    $endgroup$
    – saulspatz
    Dec 7 '18 at 14:47










  • $begingroup$
    What you probably want to do is to show that you have at least one solution in an interval $(kpi,(k+1)pi)$
    $endgroup$
    – Andrei
    Dec 7 '18 at 14:47










  • $begingroup$
    What is $[x]$ ?
    $endgroup$
    – Rhys Hughes
    Dec 7 '18 at 14:50






  • 2




    $begingroup$
    @JohnD The intermediate value theorem guarantees you at least one solution, not only one.
    $endgroup$
    – Federico
    Dec 7 '18 at 15:02






  • 1




    $begingroup$
    @AlexVong for [1.75] it returns 1.
    $endgroup$
    – John D
    Dec 7 '18 at 15:25
















$begingroup$
The intermediate value theorem is the right idea. Note that the left-hand side is $0$ at multiple of $pi$ and the right-hand side is $0$ at the integers.
$endgroup$
– saulspatz
Dec 7 '18 at 14:47




$begingroup$
The intermediate value theorem is the right idea. Note that the left-hand side is $0$ at multiple of $pi$ and the right-hand side is $0$ at the integers.
$endgroup$
– saulspatz
Dec 7 '18 at 14:47












$begingroup$
What you probably want to do is to show that you have at least one solution in an interval $(kpi,(k+1)pi)$
$endgroup$
– Andrei
Dec 7 '18 at 14:47




$begingroup$
What you probably want to do is to show that you have at least one solution in an interval $(kpi,(k+1)pi)$
$endgroup$
– Andrei
Dec 7 '18 at 14:47












$begingroup$
What is $[x]$ ?
$endgroup$
– Rhys Hughes
Dec 7 '18 at 14:50




$begingroup$
What is $[x]$ ?
$endgroup$
– Rhys Hughes
Dec 7 '18 at 14:50




2




2




$begingroup$
@JohnD The intermediate value theorem guarantees you at least one solution, not only one.
$endgroup$
– Federico
Dec 7 '18 at 15:02




$begingroup$
@JohnD The intermediate value theorem guarantees you at least one solution, not only one.
$endgroup$
– Federico
Dec 7 '18 at 15:02




1




1




$begingroup$
@AlexVong for [1.75] it returns 1.
$endgroup$
– John D
Dec 7 '18 at 15:25




$begingroup$
@AlexVong for [1.75] it returns 1.
$endgroup$
– John D
Dec 7 '18 at 15:25










2 Answers
2






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0












$begingroup$

For positive integers $n$



On every interval $(n, n+1]$



$frac{sin^2 x}{x} - x + lfloor x rfloor$ is continuous.



$lim_limits{xto n^+} frac{sin^2 x}{x} - x + lfloor x rfloor > 0$



and



$frac{sin^2 (n+1)}{n} - n + lfloor n rfloor < 0$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    It is not restrictive to look for solutions $>1$.



    Consider
    $$
    f(x)=frac{sin^2x}{x}-(x-lfloor xrfloor)
    $$
    Then, for integer $n>1$,
    $$
    f(n)=frac{sin^2n}{n}>0
    $$
    and
    $$
    lim_{xto (n+1)^-}f(x)=frac{sin^2n}{n}-1<0
    $$






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

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      0












      $begingroup$

      For positive integers $n$



      On every interval $(n, n+1]$



      $frac{sin^2 x}{x} - x + lfloor x rfloor$ is continuous.



      $lim_limits{xto n^+} frac{sin^2 x}{x} - x + lfloor x rfloor > 0$



      and



      $frac{sin^2 (n+1)}{n} - n + lfloor n rfloor < 0$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        For positive integers $n$



        On every interval $(n, n+1]$



        $frac{sin^2 x}{x} - x + lfloor x rfloor$ is continuous.



        $lim_limits{xto n^+} frac{sin^2 x}{x} - x + lfloor x rfloor > 0$



        and



        $frac{sin^2 (n+1)}{n} - n + lfloor n rfloor < 0$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          For positive integers $n$



          On every interval $(n, n+1]$



          $frac{sin^2 x}{x} - x + lfloor x rfloor$ is continuous.



          $lim_limits{xto n^+} frac{sin^2 x}{x} - x + lfloor x rfloor > 0$



          and



          $frac{sin^2 (n+1)}{n} - n + lfloor n rfloor < 0$






          share|cite|improve this answer









          $endgroup$



          For positive integers $n$



          On every interval $(n, n+1]$



          $frac{sin^2 x}{x} - x + lfloor x rfloor$ is continuous.



          $lim_limits{xto n^+} frac{sin^2 x}{x} - x + lfloor x rfloor > 0$



          and



          $frac{sin^2 (n+1)}{n} - n + lfloor n rfloor < 0$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 7 '18 at 17:11









          Doug MDoug M

          44.6k31854




          44.6k31854























              0












              $begingroup$

              It is not restrictive to look for solutions $>1$.



              Consider
              $$
              f(x)=frac{sin^2x}{x}-(x-lfloor xrfloor)
              $$
              Then, for integer $n>1$,
              $$
              f(n)=frac{sin^2n}{n}>0
              $$
              and
              $$
              lim_{xto (n+1)^-}f(x)=frac{sin^2n}{n}-1<0
              $$






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                It is not restrictive to look for solutions $>1$.



                Consider
                $$
                f(x)=frac{sin^2x}{x}-(x-lfloor xrfloor)
                $$
                Then, for integer $n>1$,
                $$
                f(n)=frac{sin^2n}{n}>0
                $$
                and
                $$
                lim_{xto (n+1)^-}f(x)=frac{sin^2n}{n}-1<0
                $$






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  It is not restrictive to look for solutions $>1$.



                  Consider
                  $$
                  f(x)=frac{sin^2x}{x}-(x-lfloor xrfloor)
                  $$
                  Then, for integer $n>1$,
                  $$
                  f(n)=frac{sin^2n}{n}>0
                  $$
                  and
                  $$
                  lim_{xto (n+1)^-}f(x)=frac{sin^2n}{n}-1<0
                  $$






                  share|cite|improve this answer











                  $endgroup$



                  It is not restrictive to look for solutions $>1$.



                  Consider
                  $$
                  f(x)=frac{sin^2x}{x}-(x-lfloor xrfloor)
                  $$
                  Then, for integer $n>1$,
                  $$
                  f(n)=frac{sin^2n}{n}>0
                  $$
                  and
                  $$
                  lim_{xto (n+1)^-}f(x)=frac{sin^2n}{n}-1<0
                  $$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 7 '18 at 16:42

























                  answered Dec 7 '18 at 15:49









                  egregegreg

                  181k1485202




                  181k1485202






























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