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












0














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
























  • 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.
    – saulspatz
    Dec 7 at 14:47










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










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






  • 2




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






  • 1




    @AlexVong for [1.75] it returns 1.
    – John D
    Dec 7 at 15:25
















0














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
























  • 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.
    – saulspatz
    Dec 7 at 14:47










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










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






  • 2




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






  • 1




    @AlexVong for [1.75] it returns 1.
    – John D
    Dec 7 at 15:25














0












0








0


0





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















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 at 14:53









Andrei

10.9k21025




10.9k21025










asked Dec 7 at 14:41









John D

103




103












  • 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.
    – saulspatz
    Dec 7 at 14:47










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










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






  • 2




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






  • 1




    @AlexVong for [1.75] it returns 1.
    – John D
    Dec 7 at 15:25


















  • 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.
    – saulspatz
    Dec 7 at 14:47










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










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






  • 2




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






  • 1




    @AlexVong for [1.75] it returns 1.
    – John D
    Dec 7 at 15:25
















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.
– saulspatz
Dec 7 at 14:47




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.
– saulspatz
Dec 7 at 14:47












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




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












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




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




2




2




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




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




1




1




@AlexVong for [1.75] it returns 1.
– John D
Dec 7 at 15:25




@AlexVong for [1.75] it returns 1.
– John D
Dec 7 at 15:25










2 Answers
2






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0














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































    0














    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





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

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      active

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      active

      oldest

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      0














      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




























        0














        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


























          0












          0








          0






          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














          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 at 16:42

























          answered Dec 7 at 15:49









          egreg

          177k1484200




          177k1484200























              0














              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


























                0














                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
























                  0












                  0








                  0






                  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












                  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 at 17:11









                  Doug M

                  43.9k31854




                  43.9k31854






























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