How can I prove that this function is periodic :












-2












$begingroup$


I know that a periodic function satisfies $F(x+a)=F(x)$




$y=leftlfloorfrac{sin x}xrightrfloor$




And the period of this fuction is $2pi$ But putting $f(x+2pi)$ I can't evaluate denominator inside the greatest integer function

So how can I prove this function periodic?






functional-analysis periodic-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why do you think the function has period $2pi$? That's the period of $sin x$, but that isn't your function.
    $endgroup$
    – Henrik
    Jan 11 at 20:14
















-2












$begingroup$


I know that a periodic function satisfies $F(x+a)=F(x)$




$y=leftlfloorfrac{sin x}xrightrfloor$




And the period of this fuction is $2pi$ But putting $f(x+2pi)$ I can't evaluate denominator inside the greatest integer function

So how can I prove this function periodic?






functional-analysis periodic-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why do you think the function has period $2pi$? That's the period of $sin x$, but that isn't your function.
    $endgroup$
    – Henrik
    Jan 11 at 20:14














-2












-2








-2


0



$begingroup$


I know that a periodic function satisfies $F(x+a)=F(x)$




$y=leftlfloorfrac{sin x}xrightrfloor$




And the period of this fuction is $2pi$ But putting $f(x+2pi)$ I can't evaluate denominator inside the greatest integer function

So how can I prove this function periodic?






functional-analysis periodic-functions







share|cite|improve this question











$endgroup$




I know that a periodic function satisfies $F(x+a)=F(x)$




$y=leftlfloorfrac{sin x}xrightrfloor$




And the period of this fuction is $2pi$ But putting $f(x+2pi)$ I can't evaluate denominator inside the greatest integer function

So how can I prove this function periodic?






functional-analysis periodic-functions




functional-analysis periodic-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 20:08









Henrik

6,04492030




6,04492030










asked Jan 5 at 8:17









Shubham GawriShubham Gawri

286




286












  • $begingroup$
    Why do you think the function has period $2pi$? That's the period of $sin x$, but that isn't your function.
    $endgroup$
    – Henrik
    Jan 11 at 20:14


















  • $begingroup$
    Why do you think the function has period $2pi$? That's the period of $sin x$, but that isn't your function.
    $endgroup$
    – Henrik
    Jan 11 at 20:14
















$begingroup$
Why do you think the function has period $2pi$? That's the period of $sin x$, but that isn't your function.
$endgroup$
– Henrik
Jan 11 at 20:14




$begingroup$
Why do you think the function has period $2pi$? That's the period of $sin x$, but that isn't your function.
$endgroup$
– Henrik
Jan 11 at 20:14










2 Answers
2






active

oldest

votes


















0












$begingroup$

For any single variable function to be periodic,



$$ f(x) = f(x+ lambda)$$



should be strictly satisfied. Just having (at least) two sequence of its roots in arithmetic progression is not enough.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ok that means that this function cannot be said periodic It just follows a pattern..
    $endgroup$
    – Shubham Gawri
    Jan 5 at 9:06








  • 1




    $begingroup$
    Yes, it cannot be said to be periodic. Even the roots of maxima/minima $tan x = x $ are not periodic.
    $endgroup$
    – Narasimham
    Jan 5 at 9:24



















3












$begingroup$

The result is wrong



First of all you have an issue to deal with at $0$.



Second $y$ is equal to zero on $I=(-pi, pi)setminus {0}$. And there is no other translated subset of $I$ in $mathbb R$ on which $y$ takes such value.






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    For any single variable function to be periodic,



    $$ f(x) = f(x+ lambda)$$



    should be strictly satisfied. Just having (at least) two sequence of its roots in arithmetic progression is not enough.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Ok that means that this function cannot be said periodic It just follows a pattern..
      $endgroup$
      – Shubham Gawri
      Jan 5 at 9:06








    • 1




      $begingroup$
      Yes, it cannot be said to be periodic. Even the roots of maxima/minima $tan x = x $ are not periodic.
      $endgroup$
      – Narasimham
      Jan 5 at 9:24
















    0












    $begingroup$

    For any single variable function to be periodic,



    $$ f(x) = f(x+ lambda)$$



    should be strictly satisfied. Just having (at least) two sequence of its roots in arithmetic progression is not enough.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Ok that means that this function cannot be said periodic It just follows a pattern..
      $endgroup$
      – Shubham Gawri
      Jan 5 at 9:06








    • 1




      $begingroup$
      Yes, it cannot be said to be periodic. Even the roots of maxima/minima $tan x = x $ are not periodic.
      $endgroup$
      – Narasimham
      Jan 5 at 9:24














    0












    0








    0





    $begingroup$

    For any single variable function to be periodic,



    $$ f(x) = f(x+ lambda)$$



    should be strictly satisfied. Just having (at least) two sequence of its roots in arithmetic progression is not enough.






    share|cite|improve this answer









    $endgroup$



    For any single variable function to be periodic,



    $$ f(x) = f(x+ lambda)$$



    should be strictly satisfied. Just having (at least) two sequence of its roots in arithmetic progression is not enough.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 5 at 8:44









    NarasimhamNarasimham

    21k62158




    21k62158












    • $begingroup$
      Ok that means that this function cannot be said periodic It just follows a pattern..
      $endgroup$
      – Shubham Gawri
      Jan 5 at 9:06








    • 1




      $begingroup$
      Yes, it cannot be said to be periodic. Even the roots of maxima/minima $tan x = x $ are not periodic.
      $endgroup$
      – Narasimham
      Jan 5 at 9:24


















    • $begingroup$
      Ok that means that this function cannot be said periodic It just follows a pattern..
      $endgroup$
      – Shubham Gawri
      Jan 5 at 9:06








    • 1




      $begingroup$
      Yes, it cannot be said to be periodic. Even the roots of maxima/minima $tan x = x $ are not periodic.
      $endgroup$
      – Narasimham
      Jan 5 at 9:24
















    $begingroup$
    Ok that means that this function cannot be said periodic It just follows a pattern..
    $endgroup$
    – Shubham Gawri
    Jan 5 at 9:06






    $begingroup$
    Ok that means that this function cannot be said periodic It just follows a pattern..
    $endgroup$
    – Shubham Gawri
    Jan 5 at 9:06






    1




    1




    $begingroup$
    Yes, it cannot be said to be periodic. Even the roots of maxima/minima $tan x = x $ are not periodic.
    $endgroup$
    – Narasimham
    Jan 5 at 9:24




    $begingroup$
    Yes, it cannot be said to be periodic. Even the roots of maxima/minima $tan x = x $ are not periodic.
    $endgroup$
    – Narasimham
    Jan 5 at 9:24











    3












    $begingroup$

    The result is wrong



    First of all you have an issue to deal with at $0$.



    Second $y$ is equal to zero on $I=(-pi, pi)setminus {0}$. And there is no other translated subset of $I$ in $mathbb R$ on which $y$ takes such value.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      The result is wrong



      First of all you have an issue to deal with at $0$.



      Second $y$ is equal to zero on $I=(-pi, pi)setminus {0}$. And there is no other translated subset of $I$ in $mathbb R$ on which $y$ takes such value.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        The result is wrong



        First of all you have an issue to deal with at $0$.



        Second $y$ is equal to zero on $I=(-pi, pi)setminus {0}$. And there is no other translated subset of $I$ in $mathbb R$ on which $y$ takes such value.






        share|cite|improve this answer











        $endgroup$



        The result is wrong



        First of all you have an issue to deal with at $0$.



        Second $y$ is equal to zero on $I=(-pi, pi)setminus {0}$. And there is no other translated subset of $I$ in $mathbb R$ on which $y$ takes such value.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 5 at 8:44

























        answered Jan 5 at 8:37









        mathcounterexamples.netmathcounterexamples.net

        27k22158




        27k22158






























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