Is $sqrt{sin x}$ periodic?












8












$begingroup$


$sin^2(x)$ has period $pi$ but it seems to me $sqrt{sin x}$ is not periodic since inside square root has to be positive and when it is negative, it is not defined.



Does it creates problem for periodicity? Can we say square root of the periodic function need to be periodic? Thanks for your help.



enter image description here



This is graph of the $sqrt{sin x}$ above.










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$endgroup$








  • 7




    $begingroup$
    Well, it's periodic where it is defined. That's something!
    $endgroup$
    – lulu
    Mar 30 at 13:21










  • $begingroup$
    You could ask the same question of $tan(x)$, because it's not defined at odd multiples of $pi/2$.
    $endgroup$
    – Barry Cipra
    Mar 31 at 13:43
















8












$begingroup$


$sin^2(x)$ has period $pi$ but it seems to me $sqrt{sin x}$ is not periodic since inside square root has to be positive and when it is negative, it is not defined.



Does it creates problem for periodicity? Can we say square root of the periodic function need to be periodic? Thanks for your help.



enter image description here



This is graph of the $sqrt{sin x}$ above.










share|cite|improve this question











$endgroup$








  • 7




    $begingroup$
    Well, it's periodic where it is defined. That's something!
    $endgroup$
    – lulu
    Mar 30 at 13:21










  • $begingroup$
    You could ask the same question of $tan(x)$, because it's not defined at odd multiples of $pi/2$.
    $endgroup$
    – Barry Cipra
    Mar 31 at 13:43














8












8








8


1



$begingroup$


$sin^2(x)$ has period $pi$ but it seems to me $sqrt{sin x}$ is not periodic since inside square root has to be positive and when it is negative, it is not defined.



Does it creates problem for periodicity? Can we say square root of the periodic function need to be periodic? Thanks for your help.



enter image description here



This is graph of the $sqrt{sin x}$ above.










share|cite|improve this question











$endgroup$




$sin^2(x)$ has period $pi$ but it seems to me $sqrt{sin x}$ is not periodic since inside square root has to be positive and when it is negative, it is not defined.



Does it creates problem for periodicity? Can we say square root of the periodic function need to be periodic? Thanks for your help.



enter image description here



This is graph of the $sqrt{sin x}$ above.







calculus functions radicals periodic-functions






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share|cite|improve this question













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share|cite|improve this question








edited Mar 30 at 17:18









user21820

40.1k544162




40.1k544162










asked Mar 30 at 13:20









izaagizaag

421210




421210








  • 7




    $begingroup$
    Well, it's periodic where it is defined. That's something!
    $endgroup$
    – lulu
    Mar 30 at 13:21










  • $begingroup$
    You could ask the same question of $tan(x)$, because it's not defined at odd multiples of $pi/2$.
    $endgroup$
    – Barry Cipra
    Mar 31 at 13:43














  • 7




    $begingroup$
    Well, it's periodic where it is defined. That's something!
    $endgroup$
    – lulu
    Mar 30 at 13:21










  • $begingroup$
    You could ask the same question of $tan(x)$, because it's not defined at odd multiples of $pi/2$.
    $endgroup$
    – Barry Cipra
    Mar 31 at 13:43








7




7




$begingroup$
Well, it's periodic where it is defined. That's something!
$endgroup$
– lulu
Mar 30 at 13:21




$begingroup$
Well, it's periodic where it is defined. That's something!
$endgroup$
– lulu
Mar 30 at 13:21












$begingroup$
You could ask the same question of $tan(x)$, because it's not defined at odd multiples of $pi/2$.
$endgroup$
– Barry Cipra
Mar 31 at 13:43




$begingroup$
You could ask the same question of $tan(x)$, because it's not defined at odd multiples of $pi/2$.
$endgroup$
– Barry Cipra
Mar 31 at 13:43










4 Answers
4






active

oldest

votes


















10












$begingroup$

The function $sqrt{sin x}$ is periodic. If you want (but don't do it in public) think of "undefined" as a real number. Then if $sin x$ is negative, you have $$sqrt{sin x} = mbox{ undefined } =sqrt{sin(x+2pi)}.$$



Or you can extend to the complex numbers and you'll have periodicity everywhere.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Now we just need to choose how to define $sqrt{z}$ in general, but that's doable.
    $endgroup$
    – J.G.
    Mar 31 at 13:21



















8












$begingroup$

$f$ is periodic, if there is a positive real number $p $ (the period) such that for every $x$ from the domain of $f$ the value of $x$ is the same as value of $(x+p)$ - more formal and more exactly:



$$f text{ is periodic }iffexists p>0: forall x in D(f): x+p in D(f) land f(x) = f(x+p)$$



For the function $g = sqrt f$ of a periodic function $f$ obviously $$x in D(g) iff x + k in D(g)$$



(because $x in D(g) iff f(x) ge 0 iff f(x+k)=f(x) ge 0 $)



and



$$f(x) = f(x+p),$$



so the square root of a periodic function is a periodic one, too.






share|cite|improve this answer











$endgroup$





















    4












    $begingroup$

    If a function $f$ is periodic, then there exists $kne0$ such that $f(x)=f(x+k)$ for all $xinmathbb{R}$. So if $g$ is its square root, we have
    $$g(x)=sqrt{f(x)}=sqrt{f(x+k)}=g(x+k).$$
    And, hence, $g$ is also periodic. BUT note that the square root is only defined in the intervals in which $f$ is nonnegative. Out of them, $g$ is not periodic because it is not even defined.






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      The periodicity holds if we can show $f(x+T) = f(x)$. We can see if the periodicity holds if we do
      $$
      begin{align}
      f(x+T) &= sqrt{sin(x+T)} \
      &= sqrt{sin(x)cos(T)+cos(t)sin(T)} tag{1}
      end{align}
      $$

      We need to search for the smallest value of $T$ in Eq.(1) that leads to $f(x)$. Indeed, if $T=2pi$ sec, then Eq.(1) becomes
      $$
      begin{align}
      f(x+T)
      &= sqrt{sin(x)cos(T)+cos(t)sin(T)} \
      &= sqrt{sin(x)(1)+cos(t)(0)} \
      &= sqrt{sin(x)} \
      &= f(x) tag{2}
      end{align}
      $$

      Eq.(2) proves the periodicity of the function with a fundamental period $T_o=2pi$ sec. We can plot the function and check if it indeed repeats itself after $T=2pi=6.28$ sec.



      enter image description here






      share|cite|improve this answer









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        4 Answers
        4






        active

        oldest

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        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        10












        $begingroup$

        The function $sqrt{sin x}$ is periodic. If you want (but don't do it in public) think of "undefined" as a real number. Then if $sin x$ is negative, you have $$sqrt{sin x} = mbox{ undefined } =sqrt{sin(x+2pi)}.$$



        Or you can extend to the complex numbers and you'll have periodicity everywhere.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Now we just need to choose how to define $sqrt{z}$ in general, but that's doable.
          $endgroup$
          – J.G.
          Mar 31 at 13:21
















        10












        $begingroup$

        The function $sqrt{sin x}$ is periodic. If you want (but don't do it in public) think of "undefined" as a real number. Then if $sin x$ is negative, you have $$sqrt{sin x} = mbox{ undefined } =sqrt{sin(x+2pi)}.$$



        Or you can extend to the complex numbers and you'll have periodicity everywhere.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          Now we just need to choose how to define $sqrt{z}$ in general, but that's doable.
          $endgroup$
          – J.G.
          Mar 31 at 13:21














        10












        10








        10





        $begingroup$

        The function $sqrt{sin x}$ is periodic. If you want (but don't do it in public) think of "undefined" as a real number. Then if $sin x$ is negative, you have $$sqrt{sin x} = mbox{ undefined } =sqrt{sin(x+2pi)}.$$



        Or you can extend to the complex numbers and you'll have periodicity everywhere.






        share|cite|improve this answer









        $endgroup$



        The function $sqrt{sin x}$ is periodic. If you want (but don't do it in public) think of "undefined" as a real number. Then if $sin x$ is negative, you have $$sqrt{sin x} = mbox{ undefined } =sqrt{sin(x+2pi)}.$$



        Or you can extend to the complex numbers and you'll have periodicity everywhere.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 30 at 13:46









        B. GoddardB. Goddard

        20.2k21543




        20.2k21543












        • $begingroup$
          Now we just need to choose how to define $sqrt{z}$ in general, but that's doable.
          $endgroup$
          – J.G.
          Mar 31 at 13:21


















        • $begingroup$
          Now we just need to choose how to define $sqrt{z}$ in general, but that's doable.
          $endgroup$
          – J.G.
          Mar 31 at 13:21
















        $begingroup$
        Now we just need to choose how to define $sqrt{z}$ in general, but that's doable.
        $endgroup$
        – J.G.
        Mar 31 at 13:21




        $begingroup$
        Now we just need to choose how to define $sqrt{z}$ in general, but that's doable.
        $endgroup$
        – J.G.
        Mar 31 at 13:21











        8












        $begingroup$

        $f$ is periodic, if there is a positive real number $p $ (the period) such that for every $x$ from the domain of $f$ the value of $x$ is the same as value of $(x+p)$ - more formal and more exactly:



        $$f text{ is periodic }iffexists p>0: forall x in D(f): x+p in D(f) land f(x) = f(x+p)$$



        For the function $g = sqrt f$ of a periodic function $f$ obviously $$x in D(g) iff x + k in D(g)$$



        (because $x in D(g) iff f(x) ge 0 iff f(x+k)=f(x) ge 0 $)



        and



        $$f(x) = f(x+p),$$



        so the square root of a periodic function is a periodic one, too.






        share|cite|improve this answer











        $endgroup$


















          8












          $begingroup$

          $f$ is periodic, if there is a positive real number $p $ (the period) such that for every $x$ from the domain of $f$ the value of $x$ is the same as value of $(x+p)$ - more formal and more exactly:



          $$f text{ is periodic }iffexists p>0: forall x in D(f): x+p in D(f) land f(x) = f(x+p)$$



          For the function $g = sqrt f$ of a periodic function $f$ obviously $$x in D(g) iff x + k in D(g)$$



          (because $x in D(g) iff f(x) ge 0 iff f(x+k)=f(x) ge 0 $)



          and



          $$f(x) = f(x+p),$$



          so the square root of a periodic function is a periodic one, too.






          share|cite|improve this answer











          $endgroup$
















            8












            8








            8





            $begingroup$

            $f$ is periodic, if there is a positive real number $p $ (the period) such that for every $x$ from the domain of $f$ the value of $x$ is the same as value of $(x+p)$ - more formal and more exactly:



            $$f text{ is periodic }iffexists p>0: forall x in D(f): x+p in D(f) land f(x) = f(x+p)$$



            For the function $g = sqrt f$ of a periodic function $f$ obviously $$x in D(g) iff x + k in D(g)$$



            (because $x in D(g) iff f(x) ge 0 iff f(x+k)=f(x) ge 0 $)



            and



            $$f(x) = f(x+p),$$



            so the square root of a periodic function is a periodic one, too.






            share|cite|improve this answer











            $endgroup$



            $f$ is periodic, if there is a positive real number $p $ (the period) such that for every $x$ from the domain of $f$ the value of $x$ is the same as value of $(x+p)$ - more formal and more exactly:



            $$f text{ is periodic }iffexists p>0: forall x in D(f): x+p in D(f) land f(x) = f(x+p)$$



            For the function $g = sqrt f$ of a periodic function $f$ obviously $$x in D(g) iff x + k in D(g)$$



            (because $x in D(g) iff f(x) ge 0 iff f(x+k)=f(x) ge 0 $)



            and



            $$f(x) = f(x+p),$$



            so the square root of a periodic function is a periodic one, too.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Mar 30 at 16:55

























            answered Mar 30 at 13:54









            MarianDMarianD

            2,2611618




            2,2611618























                4












                $begingroup$

                If a function $f$ is periodic, then there exists $kne0$ such that $f(x)=f(x+k)$ for all $xinmathbb{R}$. So if $g$ is its square root, we have
                $$g(x)=sqrt{f(x)}=sqrt{f(x+k)}=g(x+k).$$
                And, hence, $g$ is also periodic. BUT note that the square root is only defined in the intervals in which $f$ is nonnegative. Out of them, $g$ is not periodic because it is not even defined.






                share|cite|improve this answer











                $endgroup$


















                  4












                  $begingroup$

                  If a function $f$ is periodic, then there exists $kne0$ such that $f(x)=f(x+k)$ for all $xinmathbb{R}$. So if $g$ is its square root, we have
                  $$g(x)=sqrt{f(x)}=sqrt{f(x+k)}=g(x+k).$$
                  And, hence, $g$ is also periodic. BUT note that the square root is only defined in the intervals in which $f$ is nonnegative. Out of them, $g$ is not periodic because it is not even defined.






                  share|cite|improve this answer











                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    If a function $f$ is periodic, then there exists $kne0$ such that $f(x)=f(x+k)$ for all $xinmathbb{R}$. So if $g$ is its square root, we have
                    $$g(x)=sqrt{f(x)}=sqrt{f(x+k)}=g(x+k).$$
                    And, hence, $g$ is also periodic. BUT note that the square root is only defined in the intervals in which $f$ is nonnegative. Out of them, $g$ is not periodic because it is not even defined.






                    share|cite|improve this answer











                    $endgroup$



                    If a function $f$ is periodic, then there exists $kne0$ such that $f(x)=f(x+k)$ for all $xinmathbb{R}$. So if $g$ is its square root, we have
                    $$g(x)=sqrt{f(x)}=sqrt{f(x+k)}=g(x+k).$$
                    And, hence, $g$ is also periodic. BUT note that the square root is only defined in the intervals in which $f$ is nonnegative. Out of them, $g$ is not periodic because it is not even defined.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Mar 31 at 13:09

























                    answered Mar 30 at 13:54









                    AugSBAugSB

                    3,46421734




                    3,46421734























                        1












                        $begingroup$

                        The periodicity holds if we can show $f(x+T) = f(x)$. We can see if the periodicity holds if we do
                        $$
                        begin{align}
                        f(x+T) &= sqrt{sin(x+T)} \
                        &= sqrt{sin(x)cos(T)+cos(t)sin(T)} tag{1}
                        end{align}
                        $$

                        We need to search for the smallest value of $T$ in Eq.(1) that leads to $f(x)$. Indeed, if $T=2pi$ sec, then Eq.(1) becomes
                        $$
                        begin{align}
                        f(x+T)
                        &= sqrt{sin(x)cos(T)+cos(t)sin(T)} \
                        &= sqrt{sin(x)(1)+cos(t)(0)} \
                        &= sqrt{sin(x)} \
                        &= f(x) tag{2}
                        end{align}
                        $$

                        Eq.(2) proves the periodicity of the function with a fundamental period $T_o=2pi$ sec. We can plot the function and check if it indeed repeats itself after $T=2pi=6.28$ sec.



                        enter image description here






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          The periodicity holds if we can show $f(x+T) = f(x)$. We can see if the periodicity holds if we do
                          $$
                          begin{align}
                          f(x+T) &= sqrt{sin(x+T)} \
                          &= sqrt{sin(x)cos(T)+cos(t)sin(T)} tag{1}
                          end{align}
                          $$

                          We need to search for the smallest value of $T$ in Eq.(1) that leads to $f(x)$. Indeed, if $T=2pi$ sec, then Eq.(1) becomes
                          $$
                          begin{align}
                          f(x+T)
                          &= sqrt{sin(x)cos(T)+cos(t)sin(T)} \
                          &= sqrt{sin(x)(1)+cos(t)(0)} \
                          &= sqrt{sin(x)} \
                          &= f(x) tag{2}
                          end{align}
                          $$

                          Eq.(2) proves the periodicity of the function with a fundamental period $T_o=2pi$ sec. We can plot the function and check if it indeed repeats itself after $T=2pi=6.28$ sec.



                          enter image description here






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            The periodicity holds if we can show $f(x+T) = f(x)$. We can see if the periodicity holds if we do
                            $$
                            begin{align}
                            f(x+T) &= sqrt{sin(x+T)} \
                            &= sqrt{sin(x)cos(T)+cos(t)sin(T)} tag{1}
                            end{align}
                            $$

                            We need to search for the smallest value of $T$ in Eq.(1) that leads to $f(x)$. Indeed, if $T=2pi$ sec, then Eq.(1) becomes
                            $$
                            begin{align}
                            f(x+T)
                            &= sqrt{sin(x)cos(T)+cos(t)sin(T)} \
                            &= sqrt{sin(x)(1)+cos(t)(0)} \
                            &= sqrt{sin(x)} \
                            &= f(x) tag{2}
                            end{align}
                            $$

                            Eq.(2) proves the periodicity of the function with a fundamental period $T_o=2pi$ sec. We can plot the function and check if it indeed repeats itself after $T=2pi=6.28$ sec.



                            enter image description here






                            share|cite|improve this answer









                            $endgroup$



                            The periodicity holds if we can show $f(x+T) = f(x)$. We can see if the periodicity holds if we do
                            $$
                            begin{align}
                            f(x+T) &= sqrt{sin(x+T)} \
                            &= sqrt{sin(x)cos(T)+cos(t)sin(T)} tag{1}
                            end{align}
                            $$

                            We need to search for the smallest value of $T$ in Eq.(1) that leads to $f(x)$. Indeed, if $T=2pi$ sec, then Eq.(1) becomes
                            $$
                            begin{align}
                            f(x+T)
                            &= sqrt{sin(x)cos(T)+cos(t)sin(T)} \
                            &= sqrt{sin(x)(1)+cos(t)(0)} \
                            &= sqrt{sin(x)} \
                            &= f(x) tag{2}
                            end{align}
                            $$

                            Eq.(2) proves the periodicity of the function with a fundamental period $T_o=2pi$ sec. We can plot the function and check if it indeed repeats itself after $T=2pi=6.28$ sec.



                            enter image description here







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Apr 4 at 5:15









                            CroCoCroCo

                            283221




                            283221






























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