Is the set of periodic functions from $mathbb R$ to $mathbb R$a subspace of $mathbb R^{mathbb R}$?












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A function $f:mathbb R→mathbb R$ is called periodic if there exists a positive number $p$ such that $f(x)=f(x+p)$ for all $xinmathbb R$. Is the set of periodic functions from $mathbb{R}$ to $mathbb{R}$ a subspace of $mathbb{R}^mathbb{R}$? Explain.



The question has been asked here



I still don't quite understand why the ratio of two periods should not be irrational.



For example if $beta/alpha=r/s$ and $r/s = sqrt{2}$, the following equation still works?
$$h(x+sbeta)=f(x+ralpha)+g(x+sbeta)=f(x)+g(x)=h(x)$$
$$h(x+beta)=f(x+sqrt{2}alpha)+g(x+beta)=f(x)+g(x)=h(x)$$






linear-algebra proof-explanation







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    0












    $begingroup$


    A function $f:mathbb R→mathbb R$ is called periodic if there exists a positive number $p$ such that $f(x)=f(x+p)$ for all $xinmathbb R$. Is the set of periodic functions from $mathbb{R}$ to $mathbb{R}$ a subspace of $mathbb{R}^mathbb{R}$? Explain.



    The question has been asked here



    I still don't quite understand why the ratio of two periods should not be irrational.



    For example if $beta/alpha=r/s$ and $r/s = sqrt{2}$, the following equation still works?
    $$h(x+sbeta)=f(x+ralpha)+g(x+sbeta)=f(x)+g(x)=h(x)$$
    $$h(x+beta)=f(x+sqrt{2}alpha)+g(x+beta)=f(x)+g(x)=h(x)$$






    linear-algebra proof-explanation







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      A function $f:mathbb R→mathbb R$ is called periodic if there exists a positive number $p$ such that $f(x)=f(x+p)$ for all $xinmathbb R$. Is the set of periodic functions from $mathbb{R}$ to $mathbb{R}$ a subspace of $mathbb{R}^mathbb{R}$? Explain.



      The question has been asked here



      I still don't quite understand why the ratio of two periods should not be irrational.



      For example if $beta/alpha=r/s$ and $r/s = sqrt{2}$, the following equation still works?
      $$h(x+sbeta)=f(x+ralpha)+g(x+sbeta)=f(x)+g(x)=h(x)$$
      $$h(x+beta)=f(x+sqrt{2}alpha)+g(x+beta)=f(x)+g(x)=h(x)$$






      linear-algebra proof-explanation







      share|cite|improve this question











      $endgroup$




      A function $f:mathbb R→mathbb R$ is called periodic if there exists a positive number $p$ such that $f(x)=f(x+p)$ for all $xinmathbb R$. Is the set of periodic functions from $mathbb{R}$ to $mathbb{R}$ a subspace of $mathbb{R}^mathbb{R}$? Explain.



      The question has been asked here



      I still don't quite understand why the ratio of two periods should not be irrational.



      For example if $beta/alpha=r/s$ and $r/s = sqrt{2}$, the following equation still works?
      $$h(x+sbeta)=f(x+ralpha)+g(x+sbeta)=f(x)+g(x)=h(x)$$
      $$h(x+beta)=f(x+sqrt{2}alpha)+g(x+beta)=f(x)+g(x)=h(x)$$






      linear-algebra proof-explanation




      linear-algebra proof-explanation






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      share|cite|improve this question













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      share|cite|improve this question








      edited Dec 25 '18 at 0:41









      Chris Custer

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      14.3k3827










      asked Dec 25 '18 at 0:16









      JOHN JOHN

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      4349






















          2 Answers
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          $begingroup$

          If $m(x)$ is periodic with $p$, then it is also periodic with integral multiples of $p$, because $m(x)=m(x+p)=m(x+p+p)...$, but we can't say about non-integral multiples of $p$.



          Note that you can only write $f(x+ralpha)=f(x),g(x+sbeta)=g(x)$ in general if $r,s$ are integers. If $r/s$ is irrational, at-least one of $r,s$ is non-integral, and the equality doesn't work in general. As an example, take $f(x)=sin(sqrt2 x),g(x)=sin(sqrt3x)$.






          share|cite|improve this answer









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          • 1




            $begingroup$
            An even better example might be $f(x)=cos(sqrt2 x)$ and $g(x)=cos(sqrt3 x)$ -- where there is a particularly direct argument that $f+g$ cannot be periodic with any period, namely that $f(x)+g(x)=2$ has exactly one solution, $x=0$.
            $endgroup$
            – Henning Makholm
            Dec 25 '18 at 3:03



















          1












          $begingroup$

          $f$ has period $p$ if $f(x+p)=f(x)$ for all $x$. (Equivalently, $f(x+np)=f(x)$ for all $x$ and for all integers $n$). If $r$ is a real number and $f(x+rp)=f(x)$ for all $x$ we cannot say $f$ is periodic. In the quoted proof $r$ and $s$ are integers and that makes $frac r s$ rational.






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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

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            active

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            active

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            2












            $begingroup$

            If $m(x)$ is periodic with $p$, then it is also periodic with integral multiples of $p$, because $m(x)=m(x+p)=m(x+p+p)...$, but we can't say about non-integral multiples of $p$.



            Note that you can only write $f(x+ralpha)=f(x),g(x+sbeta)=g(x)$ in general if $r,s$ are integers. If $r/s$ is irrational, at-least one of $r,s$ is non-integral, and the equality doesn't work in general. As an example, take $f(x)=sin(sqrt2 x),g(x)=sin(sqrt3x)$.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              An even better example might be $f(x)=cos(sqrt2 x)$ and $g(x)=cos(sqrt3 x)$ -- where there is a particularly direct argument that $f+g$ cannot be periodic with any period, namely that $f(x)+g(x)=2$ has exactly one solution, $x=0$.
              $endgroup$
              – Henning Makholm
              Dec 25 '18 at 3:03
















            2












            $begingroup$

            If $m(x)$ is periodic with $p$, then it is also periodic with integral multiples of $p$, because $m(x)=m(x+p)=m(x+p+p)...$, but we can't say about non-integral multiples of $p$.



            Note that you can only write $f(x+ralpha)=f(x),g(x+sbeta)=g(x)$ in general if $r,s$ are integers. If $r/s$ is irrational, at-least one of $r,s$ is non-integral, and the equality doesn't work in general. As an example, take $f(x)=sin(sqrt2 x),g(x)=sin(sqrt3x)$.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              An even better example might be $f(x)=cos(sqrt2 x)$ and $g(x)=cos(sqrt3 x)$ -- where there is a particularly direct argument that $f+g$ cannot be periodic with any period, namely that $f(x)+g(x)=2$ has exactly one solution, $x=0$.
              $endgroup$
              – Henning Makholm
              Dec 25 '18 at 3:03














            2












            2








            2





            $begingroup$

            If $m(x)$ is periodic with $p$, then it is also periodic with integral multiples of $p$, because $m(x)=m(x+p)=m(x+p+p)...$, but we can't say about non-integral multiples of $p$.



            Note that you can only write $f(x+ralpha)=f(x),g(x+sbeta)=g(x)$ in general if $r,s$ are integers. If $r/s$ is irrational, at-least one of $r,s$ is non-integral, and the equality doesn't work in general. As an example, take $f(x)=sin(sqrt2 x),g(x)=sin(sqrt3x)$.






            share|cite|improve this answer









            $endgroup$



            If $m(x)$ is periodic with $p$, then it is also periodic with integral multiples of $p$, because $m(x)=m(x+p)=m(x+p+p)...$, but we can't say about non-integral multiples of $p$.



            Note that you can only write $f(x+ralpha)=f(x),g(x+sbeta)=g(x)$ in general if $r,s$ are integers. If $r/s$ is irrational, at-least one of $r,s$ is non-integral, and the equality doesn't work in general. As an example, take $f(x)=sin(sqrt2 x),g(x)=sin(sqrt3x)$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 25 '18 at 0:35









            Shubham JohriShubham Johri

            5,515818




            5,515818








            • 1




              $begingroup$
              An even better example might be $f(x)=cos(sqrt2 x)$ and $g(x)=cos(sqrt3 x)$ -- where there is a particularly direct argument that $f+g$ cannot be periodic with any period, namely that $f(x)+g(x)=2$ has exactly one solution, $x=0$.
              $endgroup$
              – Henning Makholm
              Dec 25 '18 at 3:03














            • 1




              $begingroup$
              An even better example might be $f(x)=cos(sqrt2 x)$ and $g(x)=cos(sqrt3 x)$ -- where there is a particularly direct argument that $f+g$ cannot be periodic with any period, namely that $f(x)+g(x)=2$ has exactly one solution, $x=0$.
              $endgroup$
              – Henning Makholm
              Dec 25 '18 at 3:03








            1




            1




            $begingroup$
            An even better example might be $f(x)=cos(sqrt2 x)$ and $g(x)=cos(sqrt3 x)$ -- where there is a particularly direct argument that $f+g$ cannot be periodic with any period, namely that $f(x)+g(x)=2$ has exactly one solution, $x=0$.
            $endgroup$
            – Henning Makholm
            Dec 25 '18 at 3:03




            $begingroup$
            An even better example might be $f(x)=cos(sqrt2 x)$ and $g(x)=cos(sqrt3 x)$ -- where there is a particularly direct argument that $f+g$ cannot be periodic with any period, namely that $f(x)+g(x)=2$ has exactly one solution, $x=0$.
            $endgroup$
            – Henning Makholm
            Dec 25 '18 at 3:03











            1












            $begingroup$

            $f$ has period $p$ if $f(x+p)=f(x)$ for all $x$. (Equivalently, $f(x+np)=f(x)$ for all $x$ and for all integers $n$). If $r$ is a real number and $f(x+rp)=f(x)$ for all $x$ we cannot say $f$ is periodic. In the quoted proof $r$ and $s$ are integers and that makes $frac r s$ rational.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              $f$ has period $p$ if $f(x+p)=f(x)$ for all $x$. (Equivalently, $f(x+np)=f(x)$ for all $x$ and for all integers $n$). If $r$ is a real number and $f(x+rp)=f(x)$ for all $x$ we cannot say $f$ is periodic. In the quoted proof $r$ and $s$ are integers and that makes $frac r s$ rational.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                $f$ has period $p$ if $f(x+p)=f(x)$ for all $x$. (Equivalently, $f(x+np)=f(x)$ for all $x$ and for all integers $n$). If $r$ is a real number and $f(x+rp)=f(x)$ for all $x$ we cannot say $f$ is periodic. In the quoted proof $r$ and $s$ are integers and that makes $frac r s$ rational.






                share|cite|improve this answer









                $endgroup$



                $f$ has period $p$ if $f(x+p)=f(x)$ for all $x$. (Equivalently, $f(x+np)=f(x)$ for all $x$ and for all integers $n$). If $r$ is a real number and $f(x+rp)=f(x)$ for all $x$ we cannot say $f$ is periodic. In the quoted proof $r$ and $s$ are integers and that makes $frac r s$ rational.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 25 '18 at 0:20









                Kavi Rama MurthyKavi Rama Murthy

                72.6k53170




                72.6k53170






























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