Existance of an analytic function on unit disc












2












$begingroup$


Is there an analytic function $f:B_1(0)to B_1(0)$ such that $f(0)=1/2$ and $f^{prime}(0)=3/4$? If it exists, is it unique?



The answer to the first part of the question is affirmative. We can use Scharz Pick lemma to find a Mobius transformation $f$ satisfying the condition. But is the function unique? I got stuck here. Please help.






complex-analysis holomorphic-functions mobius-transformation







share|cite|improve this question









$endgroup$












  • $begingroup$
    here $B_1(0)$ is the open ball or radius 1 centered at zero, or there is something different?
    $endgroup$
    – Masacroso
    Nov 14 '18 at 9:26






  • 1




    $begingroup$
    @Anupam You may be interested in knowing that 3/4 is the largest possible value of $|f'(0)|$. You can look for "An extremal problem" in Rudin's RCA (chapter on MMP).
    $endgroup$
    – Kavi Rama Murthy
    Nov 14 '18 at 12:19
















2












$begingroup$


Is there an analytic function $f:B_1(0)to B_1(0)$ such that $f(0)=1/2$ and $f^{prime}(0)=3/4$? If it exists, is it unique?



The answer to the first part of the question is affirmative. We can use Scharz Pick lemma to find a Mobius transformation $f$ satisfying the condition. But is the function unique? I got stuck here. Please help.






complex-analysis holomorphic-functions mobius-transformation







share|cite|improve this question









$endgroup$












  • $begingroup$
    here $B_1(0)$ is the open ball or radius 1 centered at zero, or there is something different?
    $endgroup$
    – Masacroso
    Nov 14 '18 at 9:26






  • 1




    $begingroup$
    @Anupam You may be interested in knowing that 3/4 is the largest possible value of $|f'(0)|$. You can look for "An extremal problem" in Rudin's RCA (chapter on MMP).
    $endgroup$
    – Kavi Rama Murthy
    Nov 14 '18 at 12:19














2












2








2





$begingroup$


Is there an analytic function $f:B_1(0)to B_1(0)$ such that $f(0)=1/2$ and $f^{prime}(0)=3/4$? If it exists, is it unique?



The answer to the first part of the question is affirmative. We can use Scharz Pick lemma to find a Mobius transformation $f$ satisfying the condition. But is the function unique? I got stuck here. Please help.






complex-analysis holomorphic-functions mobius-transformation







share|cite|improve this question









$endgroup$




Is there an analytic function $f:B_1(0)to B_1(0)$ such that $f(0)=1/2$ and $f^{prime}(0)=3/4$? If it exists, is it unique?



The answer to the first part of the question is affirmative. We can use Scharz Pick lemma to find a Mobius transformation $f$ satisfying the condition. But is the function unique? I got stuck here. Please help.






complex-analysis holomorphic-functions mobius-transformation




complex-analysis holomorphic-functions mobius-transformation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 14 '18 at 8:54









AnupamAnupam

2,4731824




2,4731824












  • $begingroup$
    here $B_1(0)$ is the open ball or radius 1 centered at zero, or there is something different?
    $endgroup$
    – Masacroso
    Nov 14 '18 at 9:26






  • 1




    $begingroup$
    @Anupam You may be interested in knowing that 3/4 is the largest possible value of $|f'(0)|$. You can look for "An extremal problem" in Rudin's RCA (chapter on MMP).
    $endgroup$
    – Kavi Rama Murthy
    Nov 14 '18 at 12:19


















  • $begingroup$
    here $B_1(0)$ is the open ball or radius 1 centered at zero, or there is something different?
    $endgroup$
    – Masacroso
    Nov 14 '18 at 9:26






  • 1




    $begingroup$
    @Anupam You may be interested in knowing that 3/4 is the largest possible value of $|f'(0)|$. You can look for "An extremal problem" in Rudin's RCA (chapter on MMP).
    $endgroup$
    – Kavi Rama Murthy
    Nov 14 '18 at 12:19
















$begingroup$
here $B_1(0)$ is the open ball or radius 1 centered at zero, or there is something different?
$endgroup$
– Masacroso
Nov 14 '18 at 9:26




$begingroup$
here $B_1(0)$ is the open ball or radius 1 centered at zero, or there is something different?
$endgroup$
– Masacroso
Nov 14 '18 at 9:26




1




1




$begingroup$
@Anupam You may be interested in knowing that 3/4 is the largest possible value of $|f'(0)|$. You can look for "An extremal problem" in Rudin's RCA (chapter on MMP).
$endgroup$
– Kavi Rama Murthy
Nov 14 '18 at 12:19




$begingroup$
@Anupam You may be interested in knowing that 3/4 is the largest possible value of $|f'(0)|$. You can look for "An extremal problem" in Rudin's RCA (chapter on MMP).
$endgroup$
– Kavi Rama Murthy
Nov 14 '18 at 12:19










3 Answers
3






active

oldest

votes


















3












$begingroup$

$f(z)=frac {2z+1} {z+2}$ is one function with these properties. If $g$ is another such function define $h(z)=phi (g(z))$ where $phi (z)=frac {z-frac 1 2} {1-frac 1 2 z}$. You can verify that $h$ maps $B(0,1)$ into itself and vanishes at $0$. A simple calculation shows $h'(0)=1$. By Schwarz Lemma we get $h(z)=z$ for all $z$ from which we get $g(z)=f(z)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @ Kavi Rama Murthty: That's so nice. Thanks. Can we say anything on existence and uniqueness of an analytic function from open unit disc to open unit disc fatisfying $f(0)=1/2, f^{prime}(0)=3/8?$. Here the problem is that equality in Schawrz-Pick lemma does not hold.
    $endgroup$
    – Anupam
    Nov 14 '18 at 16:26












  • $begingroup$
    @Anupam $frac {3z} 8 +frac 1 2$ and $frac 1 2 e^{frac 3 4 z}$ are two solutions in this case.
    $endgroup$
    – Kavi Rama Murthy
    Nov 14 '18 at 23:11












  • $begingroup$
    @ Kavi Rama Murthy: Is there any standard way to find the solutions?
    $endgroup$
    – Anupam
    Nov 14 '18 at 23:26










  • $begingroup$
    @ Kavi Rama Murthy: Will $1/2e^{frac{3}{4}z}$ keep images from open unit disc to open unit disc?
    $endgroup$
    – Anupam
    Nov 15 '18 at 1:57












  • $begingroup$
    @Anupam I am sorry, it does not. My example is wrong. There is no general method for finding such functions.
    $endgroup$
    – Kavi Rama Murthy
    Nov 15 '18 at 5:37



















2












$begingroup$

By Schwarz-Pick Lemma, for all $|z|<1$,
$$|f'(z)|leq frac{1-|f(z)|^2}{1-|z|^2}.$$
Note that if equality holds at some point then
$f$ must be an analytic automorphism of the unit disc.



Since for $z=0$ the above equality holds then
$$f(z)=e^{itheta}frac{z-a}{1-bar{a}z}$$
with $|a|<1$ and $thetainmathbb{R}$.
Now
$$
begin{cases}f(0)=-e^{itheta}a=1/2\
f'(0)=e^{itheta}(1-|a|^2)=3/4
end{cases}
Leftrightarrow
begin{cases}a=-1/2\
e^{itheta}=1
end{cases}
$$
and we may conclude that $f$ exists and it is unique:
$$f(z)=frac{2z+1}{z+2}.$$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    The Mobius tfm $zmapsto frac{1+2z}{2+z}$ does the job. If you want $f$ to be a bijection, then the answer is yes because the only transforms that map the disk into itself bijectively are $zmapsto frac{z-a}{1-overline{a}z}e^{itheta}$, and $a,theta$ are determined by the conditions. There is something special about the choice $f'(0)=frac{3}{4}$, because otherwise you could choose $zmapsto frac{z-a}{1-overline{a}z}rho$ with $|rho|<1$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Kavi's answer was better than this
      $endgroup$
      – Richard Martin
      Nov 14 '18 at 12:58











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    $f(z)=frac {2z+1} {z+2}$ is one function with these properties. If $g$ is another such function define $h(z)=phi (g(z))$ where $phi (z)=frac {z-frac 1 2} {1-frac 1 2 z}$. You can verify that $h$ maps $B(0,1)$ into itself and vanishes at $0$. A simple calculation shows $h'(0)=1$. By Schwarz Lemma we get $h(z)=z$ for all $z$ from which we get $g(z)=f(z)$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      @ Kavi Rama Murthty: That's so nice. Thanks. Can we say anything on existence and uniqueness of an analytic function from open unit disc to open unit disc fatisfying $f(0)=1/2, f^{prime}(0)=3/8?$. Here the problem is that equality in Schawrz-Pick lemma does not hold.
      $endgroup$
      – Anupam
      Nov 14 '18 at 16:26












    • $begingroup$
      @Anupam $frac {3z} 8 +frac 1 2$ and $frac 1 2 e^{frac 3 4 z}$ are two solutions in this case.
      $endgroup$
      – Kavi Rama Murthy
      Nov 14 '18 at 23:11












    • $begingroup$
      @ Kavi Rama Murthy: Is there any standard way to find the solutions?
      $endgroup$
      – Anupam
      Nov 14 '18 at 23:26










    • $begingroup$
      @ Kavi Rama Murthy: Will $1/2e^{frac{3}{4}z}$ keep images from open unit disc to open unit disc?
      $endgroup$
      – Anupam
      Nov 15 '18 at 1:57












    • $begingroup$
      @Anupam I am sorry, it does not. My example is wrong. There is no general method for finding such functions.
      $endgroup$
      – Kavi Rama Murthy
      Nov 15 '18 at 5:37
















    3












    $begingroup$

    $f(z)=frac {2z+1} {z+2}$ is one function with these properties. If $g$ is another such function define $h(z)=phi (g(z))$ where $phi (z)=frac {z-frac 1 2} {1-frac 1 2 z}$. You can verify that $h$ maps $B(0,1)$ into itself and vanishes at $0$. A simple calculation shows $h'(0)=1$. By Schwarz Lemma we get $h(z)=z$ for all $z$ from which we get $g(z)=f(z)$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      @ Kavi Rama Murthty: That's so nice. Thanks. Can we say anything on existence and uniqueness of an analytic function from open unit disc to open unit disc fatisfying $f(0)=1/2, f^{prime}(0)=3/8?$. Here the problem is that equality in Schawrz-Pick lemma does not hold.
      $endgroup$
      – Anupam
      Nov 14 '18 at 16:26












    • $begingroup$
      @Anupam $frac {3z} 8 +frac 1 2$ and $frac 1 2 e^{frac 3 4 z}$ are two solutions in this case.
      $endgroup$
      – Kavi Rama Murthy
      Nov 14 '18 at 23:11












    • $begingroup$
      @ Kavi Rama Murthy: Is there any standard way to find the solutions?
      $endgroup$
      – Anupam
      Nov 14 '18 at 23:26










    • $begingroup$
      @ Kavi Rama Murthy: Will $1/2e^{frac{3}{4}z}$ keep images from open unit disc to open unit disc?
      $endgroup$
      – Anupam
      Nov 15 '18 at 1:57












    • $begingroup$
      @Anupam I am sorry, it does not. My example is wrong. There is no general method for finding such functions.
      $endgroup$
      – Kavi Rama Murthy
      Nov 15 '18 at 5:37














    3












    3








    3





    $begingroup$

    $f(z)=frac {2z+1} {z+2}$ is one function with these properties. If $g$ is another such function define $h(z)=phi (g(z))$ where $phi (z)=frac {z-frac 1 2} {1-frac 1 2 z}$. You can verify that $h$ maps $B(0,1)$ into itself and vanishes at $0$. A simple calculation shows $h'(0)=1$. By Schwarz Lemma we get $h(z)=z$ for all $z$ from which we get $g(z)=f(z)$.






    share|cite|improve this answer









    $endgroup$



    $f(z)=frac {2z+1} {z+2}$ is one function with these properties. If $g$ is another such function define $h(z)=phi (g(z))$ where $phi (z)=frac {z-frac 1 2} {1-frac 1 2 z}$. You can verify that $h$ maps $B(0,1)$ into itself and vanishes at $0$. A simple calculation shows $h'(0)=1$. By Schwarz Lemma we get $h(z)=z$ for all $z$ from which we get $g(z)=f(z)$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 14 '18 at 9:27









    Kavi Rama MurthyKavi Rama Murthy

    60.5k42161




    60.5k42161












    • $begingroup$
      @ Kavi Rama Murthty: That's so nice. Thanks. Can we say anything on existence and uniqueness of an analytic function from open unit disc to open unit disc fatisfying $f(0)=1/2, f^{prime}(0)=3/8?$. Here the problem is that equality in Schawrz-Pick lemma does not hold.
      $endgroup$
      – Anupam
      Nov 14 '18 at 16:26












    • $begingroup$
      @Anupam $frac {3z} 8 +frac 1 2$ and $frac 1 2 e^{frac 3 4 z}$ are two solutions in this case.
      $endgroup$
      – Kavi Rama Murthy
      Nov 14 '18 at 23:11












    • $begingroup$
      @ Kavi Rama Murthy: Is there any standard way to find the solutions?
      $endgroup$
      – Anupam
      Nov 14 '18 at 23:26










    • $begingroup$
      @ Kavi Rama Murthy: Will $1/2e^{frac{3}{4}z}$ keep images from open unit disc to open unit disc?
      $endgroup$
      – Anupam
      Nov 15 '18 at 1:57












    • $begingroup$
      @Anupam I am sorry, it does not. My example is wrong. There is no general method for finding such functions.
      $endgroup$
      – Kavi Rama Murthy
      Nov 15 '18 at 5:37


















    • $begingroup$
      @ Kavi Rama Murthty: That's so nice. Thanks. Can we say anything on existence and uniqueness of an analytic function from open unit disc to open unit disc fatisfying $f(0)=1/2, f^{prime}(0)=3/8?$. Here the problem is that equality in Schawrz-Pick lemma does not hold.
      $endgroup$
      – Anupam
      Nov 14 '18 at 16:26












    • $begingroup$
      @Anupam $frac {3z} 8 +frac 1 2$ and $frac 1 2 e^{frac 3 4 z}$ are two solutions in this case.
      $endgroup$
      – Kavi Rama Murthy
      Nov 14 '18 at 23:11












    • $begingroup$
      @ Kavi Rama Murthy: Is there any standard way to find the solutions?
      $endgroup$
      – Anupam
      Nov 14 '18 at 23:26










    • $begingroup$
      @ Kavi Rama Murthy: Will $1/2e^{frac{3}{4}z}$ keep images from open unit disc to open unit disc?
      $endgroup$
      – Anupam
      Nov 15 '18 at 1:57












    • $begingroup$
      @Anupam I am sorry, it does not. My example is wrong. There is no general method for finding such functions.
      $endgroup$
      – Kavi Rama Murthy
      Nov 15 '18 at 5:37
















    $begingroup$
    @ Kavi Rama Murthty: That's so nice. Thanks. Can we say anything on existence and uniqueness of an analytic function from open unit disc to open unit disc fatisfying $f(0)=1/2, f^{prime}(0)=3/8?$. Here the problem is that equality in Schawrz-Pick lemma does not hold.
    $endgroup$
    – Anupam
    Nov 14 '18 at 16:26






    $begingroup$
    @ Kavi Rama Murthty: That's so nice. Thanks. Can we say anything on existence and uniqueness of an analytic function from open unit disc to open unit disc fatisfying $f(0)=1/2, f^{prime}(0)=3/8?$. Here the problem is that equality in Schawrz-Pick lemma does not hold.
    $endgroup$
    – Anupam
    Nov 14 '18 at 16:26














    $begingroup$
    @Anupam $frac {3z} 8 +frac 1 2$ and $frac 1 2 e^{frac 3 4 z}$ are two solutions in this case.
    $endgroup$
    – Kavi Rama Murthy
    Nov 14 '18 at 23:11






    $begingroup$
    @Anupam $frac {3z} 8 +frac 1 2$ and $frac 1 2 e^{frac 3 4 z}$ are two solutions in this case.
    $endgroup$
    – Kavi Rama Murthy
    Nov 14 '18 at 23:11














    $begingroup$
    @ Kavi Rama Murthy: Is there any standard way to find the solutions?
    $endgroup$
    – Anupam
    Nov 14 '18 at 23:26




    $begingroup$
    @ Kavi Rama Murthy: Is there any standard way to find the solutions?
    $endgroup$
    – Anupam
    Nov 14 '18 at 23:26












    $begingroup$
    @ Kavi Rama Murthy: Will $1/2e^{frac{3}{4}z}$ keep images from open unit disc to open unit disc?
    $endgroup$
    – Anupam
    Nov 15 '18 at 1:57






    $begingroup$
    @ Kavi Rama Murthy: Will $1/2e^{frac{3}{4}z}$ keep images from open unit disc to open unit disc?
    $endgroup$
    – Anupam
    Nov 15 '18 at 1:57














    $begingroup$
    @Anupam I am sorry, it does not. My example is wrong. There is no general method for finding such functions.
    $endgroup$
    – Kavi Rama Murthy
    Nov 15 '18 at 5:37




    $begingroup$
    @Anupam I am sorry, it does not. My example is wrong. There is no general method for finding such functions.
    $endgroup$
    – Kavi Rama Murthy
    Nov 15 '18 at 5:37











    2












    $begingroup$

    By Schwarz-Pick Lemma, for all $|z|<1$,
    $$|f'(z)|leq frac{1-|f(z)|^2}{1-|z|^2}.$$
    Note that if equality holds at some point then
    $f$ must be an analytic automorphism of the unit disc    .



    Since for $z=0$ the above equality holds then
    $$f(z)=e^{itheta}frac{z-a}{1-bar{a}z}$$
    with $|a|<1$ and $thetainmathbb{R}$.
    Now
    $$
    begin{cases}f(0)=-e^{itheta}a=1/2\
    f'(0)=e^{itheta}(1-|a|^2)=3/4
    end{cases}
    Leftrightarrow
    begin{cases}a=-1/2\
    e^{itheta}=1
    end{cases}
    $$
    and we may conclude that $f$ exists and it is unique:
    $$f(z)=frac{2z+1}{z+2}.$$






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      By Schwarz-Pick Lemma, for all $|z|<1$,
      $$|f'(z)|leq frac{1-|f(z)|^2}{1-|z|^2}.$$
      Note that if equality holds at some point then
      $f$ must be an analytic automorphism of the unit disc      .



      Since for $z=0$ the above equality holds then
      $$f(z)=e^{itheta}frac{z-a}{1-bar{a}z}$$
      with $|a|<1$ and $thetainmathbb{R}$.
      Now
      $$
      begin{cases}f(0)=-e^{itheta}a=1/2\
      f'(0)=e^{itheta}(1-|a|^2)=3/4
      end{cases}
      Leftrightarrow
      begin{cases}a=-1/2\
      e^{itheta}=1
      end{cases}
      $$
      and we may conclude that $f$ exists and it is unique:
      $$f(z)=frac{2z+1}{z+2}.$$






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        By Schwarz-Pick Lemma, for all $|z|<1$,
        $$|f'(z)|leq frac{1-|f(z)|^2}{1-|z|^2}.$$
        Note that if equality holds at some point then
        $f$ must be an analytic automorphism of the unit disc        .



        Since for $z=0$ the above equality holds then
        $$f(z)=e^{itheta}frac{z-a}{1-bar{a}z}$$
        with $|a|<1$ and $thetainmathbb{R}$.
        Now
        $$
        begin{cases}f(0)=-e^{itheta}a=1/2\
        f'(0)=e^{itheta}(1-|a|^2)=3/4
        end{cases}
        Leftrightarrow
        begin{cases}a=-1/2\
        e^{itheta}=1
        end{cases}
        $$
        and we may conclude that $f$ exists and it is unique:
        $$f(z)=frac{2z+1}{z+2}.$$






        share|cite|improve this answer











        $endgroup$



        By Schwarz-Pick Lemma, for all $|z|<1$,
        $$|f'(z)|leq frac{1-|f(z)|^2}{1-|z|^2}.$$
        Note that if equality holds at some point then
        $f$ must be an analytic automorphism of the unit disc        .



        Since for $z=0$ the above equality holds then
        $$f(z)=e^{itheta}frac{z-a}{1-bar{a}z}$$
        with $|a|<1$ and $thetainmathbb{R}$.
        Now
        $$
        begin{cases}f(0)=-e^{itheta}a=1/2\
        f'(0)=e^{itheta}(1-|a|^2)=3/4
        end{cases}
        Leftrightarrow
        begin{cases}a=-1/2\
        e^{itheta}=1
        end{cases}
        $$
        and we may conclude that $f$ exists and it is unique:
        $$f(z)=frac{2z+1}{z+2}.$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 14 '18 at 11:57

























        answered Nov 14 '18 at 9:15









        Robert ZRobert Z

        97.7k1066137




        97.7k1066137























            1












            $begingroup$

            The Mobius tfm $zmapsto frac{1+2z}{2+z}$ does the job. If you want $f$ to be a bijection, then the answer is yes because the only transforms that map the disk into itself bijectively are $zmapsto frac{z-a}{1-overline{a}z}e^{itheta}$, and $a,theta$ are determined by the conditions. There is something special about the choice $f'(0)=frac{3}{4}$, because otherwise you could choose $zmapsto frac{z-a}{1-overline{a}z}rho$ with $|rho|<1$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Kavi's answer was better than this
              $endgroup$
              – Richard Martin
              Nov 14 '18 at 12:58
















            1












            $begingroup$

            The Mobius tfm $zmapsto frac{1+2z}{2+z}$ does the job. If you want $f$ to be a bijection, then the answer is yes because the only transforms that map the disk into itself bijectively are $zmapsto frac{z-a}{1-overline{a}z}e^{itheta}$, and $a,theta$ are determined by the conditions. There is something special about the choice $f'(0)=frac{3}{4}$, because otherwise you could choose $zmapsto frac{z-a}{1-overline{a}z}rho$ with $|rho|<1$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Kavi's answer was better than this
              $endgroup$
              – Richard Martin
              Nov 14 '18 at 12:58














            1












            1








            1





            $begingroup$

            The Mobius tfm $zmapsto frac{1+2z}{2+z}$ does the job. If you want $f$ to be a bijection, then the answer is yes because the only transforms that map the disk into itself bijectively are $zmapsto frac{z-a}{1-overline{a}z}e^{itheta}$, and $a,theta$ are determined by the conditions. There is something special about the choice $f'(0)=frac{3}{4}$, because otherwise you could choose $zmapsto frac{z-a}{1-overline{a}z}rho$ with $|rho|<1$.






            share|cite|improve this answer











            $endgroup$



            The Mobius tfm $zmapsto frac{1+2z}{2+z}$ does the job. If you want $f$ to be a bijection, then the answer is yes because the only transforms that map the disk into itself bijectively are $zmapsto frac{z-a}{1-overline{a}z}e^{itheta}$, and $a,theta$ are determined by the conditions. There is something special about the choice $f'(0)=frac{3}{4}$, because otherwise you could choose $zmapsto frac{z-a}{1-overline{a}z}rho$ with $|rho|<1$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 14 '18 at 9:32

























            answered Nov 14 '18 at 9:24









            Richard MartinRichard Martin

            1,61318




            1,61318












            • $begingroup$
              Kavi's answer was better than this
              $endgroup$
              – Richard Martin
              Nov 14 '18 at 12:58


















            • $begingroup$
              Kavi's answer was better than this
              $endgroup$
              – Richard Martin
              Nov 14 '18 at 12:58
















            $begingroup$
            Kavi's answer was better than this
            $endgroup$
            – Richard Martin
            Nov 14 '18 at 12:58




            $begingroup$
            Kavi's answer was better than this
            $endgroup$
            – Richard Martin
            Nov 14 '18 at 12:58


















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