An even function, $f(z)$, analytic near 0 can be written as another analytic function, $h(z^2)$












1












$begingroup$


If f is an even function, $f(z) = f(-z)$, and is analytic near 0, then there exists a function h, also analytic near 0, such that $f(z) = h(z^2)$



I suspect this statement is true because the analyticity of f near 0 allows us to write it as a taylor expansion around 0 and since it is even we have that
$$f(z) = sum a_n z^n = sum a_n (-z)^n$$



which implies that a_n does not depend on z and we can write



$h(z^2) = sum a_n z^{2n}$ but I am not sure where to go from here. Any hints are appreciated.






complex-analysis power-series taylor-expansion analyticity even-and-odd-functions







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    1












    $begingroup$


    If f is an even function, $f(z) = f(-z)$, and is analytic near 0, then there exists a function h, also analytic near 0, such that $f(z) = h(z^2)$



    I suspect this statement is true because the analyticity of f near 0 allows us to write it as a taylor expansion around 0 and since it is even we have that
    $$f(z) = sum a_n z^n = sum a_n (-z)^n$$



    which implies that a_n does not depend on z and we can write



    $h(z^2) = sum a_n z^{2n}$ but I am not sure where to go from here. Any hints are appreciated.






    complex-analysis power-series taylor-expansion analyticity even-and-odd-functions







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      If f is an even function, $f(z) = f(-z)$, and is analytic near 0, then there exists a function h, also analytic near 0, such that $f(z) = h(z^2)$



      I suspect this statement is true because the analyticity of f near 0 allows us to write it as a taylor expansion around 0 and since it is even we have that
      $$f(z) = sum a_n z^n = sum a_n (-z)^n$$



      which implies that a_n does not depend on z and we can write



      $h(z^2) = sum a_n z^{2n}$ but I am not sure where to go from here. Any hints are appreciated.






      complex-analysis power-series taylor-expansion analyticity even-and-odd-functions







      share|cite|improve this question











      $endgroup$




      If f is an even function, $f(z) = f(-z)$, and is analytic near 0, then there exists a function h, also analytic near 0, such that $f(z) = h(z^2)$



      I suspect this statement is true because the analyticity of f near 0 allows us to write it as a taylor expansion around 0 and since it is even we have that
      $$f(z) = sum a_n z^n = sum a_n (-z)^n$$



      which implies that a_n does not depend on z and we can write



      $h(z^2) = sum a_n z^{2n}$ but I am not sure where to go from here. Any hints are appreciated.






      complex-analysis power-series taylor-expansion analyticity even-and-odd-functions




      complex-analysis power-series taylor-expansion analyticity even-and-odd-functions






      share|cite|improve this question















      share|cite|improve this question













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      edited Dec 10 '18 at 22:14









      José Carlos Santos

      158k22126228




      158k22126228










      asked Dec 10 '18 at 21:48









      Richard VillalobosRichard Villalobos

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          It follows from your equality$$sum_{n=0}^infty a_nz^n=sum_{n=0}^infty a_n(-z)^n$$than $(forall ninmathbb{Z}_+):a_n=(-1)^na_n$. Therefore, $a_n=0$ if $n$ is odd and so$$f(z)=sum_{n=0}^infty a_{2n}z^{2n}=sum_{n=0}^infty a_{2n}(z^2)^n.$$So, define$$h(z)=sum_{n=0}^infty a_{2n}z^n.$$






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            $begingroup$

            It follows from your equality$$sum_{n=0}^infty a_nz^n=sum_{n=0}^infty a_n(-z)^n$$than $(forall ninmathbb{Z}_+):a_n=(-1)^na_n$. Therefore, $a_n=0$ if $n$ is odd and so$$f(z)=sum_{n=0}^infty a_{2n}z^{2n}=sum_{n=0}^infty a_{2n}(z^2)^n.$$So, define$$h(z)=sum_{n=0}^infty a_{2n}z^n.$$






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              It follows from your equality$$sum_{n=0}^infty a_nz^n=sum_{n=0}^infty a_n(-z)^n$$than $(forall ninmathbb{Z}_+):a_n=(-1)^na_n$. Therefore, $a_n=0$ if $n$ is odd and so$$f(z)=sum_{n=0}^infty a_{2n}z^{2n}=sum_{n=0}^infty a_{2n}(z^2)^n.$$So, define$$h(z)=sum_{n=0}^infty a_{2n}z^n.$$






              share|cite|improve this answer









              $endgroup$
















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                3





                $begingroup$

                It follows from your equality$$sum_{n=0}^infty a_nz^n=sum_{n=0}^infty a_n(-z)^n$$than $(forall ninmathbb{Z}_+):a_n=(-1)^na_n$. Therefore, $a_n=0$ if $n$ is odd and so$$f(z)=sum_{n=0}^infty a_{2n}z^{2n}=sum_{n=0}^infty a_{2n}(z^2)^n.$$So, define$$h(z)=sum_{n=0}^infty a_{2n}z^n.$$






                share|cite|improve this answer









                $endgroup$



                It follows from your equality$$sum_{n=0}^infty a_nz^n=sum_{n=0}^infty a_n(-z)^n$$than $(forall ninmathbb{Z}_+):a_n=(-1)^na_n$. Therefore, $a_n=0$ if $n$ is odd and so$$f(z)=sum_{n=0}^infty a_{2n}z^{2n}=sum_{n=0}^infty a_{2n}(z^2)^n.$$So, define$$h(z)=sum_{n=0}^infty a_{2n}z^n.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 10 '18 at 21:54









                José Carlos SantosJosé Carlos Santos

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