$g^{(n)}(1)=frac{(f^{(n)}(1))^{a}}{(n!)^{a-1}}, a>0$; Prove that g can be extended to an entire function.












0












$begingroup$


Let $f(z)$ be entire and $g(z)$ be an analytic function in a neighborhood of $z=1$ which satisfies$$g^{(n)}(1)=frac{(f^{(n)}(1))^{a}}{(n!)^{a-1}}, a>0$$ Prove that g can be extended to an entire function.



My thought:



Since $f$ is entire, $f$ has a Taylor series expansion at $z=1$ as $$f(z)=f(1)+f'(1)z+frac{f''(1)}{2!}z^2+...$$
Where the coefficients are somehow related to $g$. But how can I possibly show that $g$ can be extended to an entire function? My guess is using Cauchy estimates, but since $f$ is entire, I'm not sure how I can bound the function itself.






complex-analysis







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let $f(z)$ be entire and $g(z)$ be an analytic function in a neighborhood of $z=1$ which satisfies$$g^{(n)}(1)=frac{(f^{(n)}(1))^{a}}{(n!)^{a-1}}, a>0$$ Prove that g can be extended to an entire function.



    My thought:



    Since $f$ is entire, $f$ has a Taylor series expansion at $z=1$ as $$f(z)=f(1)+f'(1)z+frac{f''(1)}{2!}z^2+...$$
    Where the coefficients are somehow related to $g$. But how can I possibly show that $g$ can be extended to an entire function? My guess is using Cauchy estimates, but since $f$ is entire, I'm not sure how I can bound the function itself.






    complex-analysis







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $f(z)$ be entire and $g(z)$ be an analytic function in a neighborhood of $z=1$ which satisfies$$g^{(n)}(1)=frac{(f^{(n)}(1))^{a}}{(n!)^{a-1}}, a>0$$ Prove that g can be extended to an entire function.



      My thought:



      Since $f$ is entire, $f$ has a Taylor series expansion at $z=1$ as $$f(z)=f(1)+f'(1)z+frac{f''(1)}{2!}z^2+...$$
      Where the coefficients are somehow related to $g$. But how can I possibly show that $g$ can be extended to an entire function? My guess is using Cauchy estimates, but since $f$ is entire, I'm not sure how I can bound the function itself.






      complex-analysis







      share|cite|improve this question









      $endgroup$




      Let $f(z)$ be entire and $g(z)$ be an analytic function in a neighborhood of $z=1$ which satisfies$$g^{(n)}(1)=frac{(f^{(n)}(1))^{a}}{(n!)^{a-1}}, a>0$$ Prove that g can be extended to an entire function.



      My thought:



      Since $f$ is entire, $f$ has a Taylor series expansion at $z=1$ as $$f(z)=f(1)+f'(1)z+frac{f''(1)}{2!}z^2+...$$
      Where the coefficients are somehow related to $g$. But how can I possibly show that $g$ can be extended to an entire function? My guess is using Cauchy estimates, but since $f$ is entire, I'm not sure how I can bound the function itself.






      complex-analysis




      complex-analysis






      share|cite|improve this question













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      asked Jan 16 at 17:22









      Ya GYa G

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      536211






















          1 Answer
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          $begingroup$

          Note that by the assumption that $f$ is entire, we have
          $$
          limsup_{nto infty}Big|frac{f^{(n)}(1)}{n!}Big|^{frac{1}{n}}=0.
          $$           This is because $f(z) = sum_{n=0}^infty frac{f^{(n)}(1)}{n!}(z-1)^n$ has a radius of convergence equal to $$R=left(limsup_{nto infty}Big|frac{f^{(n)}(1)}{n!}Big|^{frac{1}{n}}right)^{-1}=infty.$$ Now, note that the power series
          $$
          G(z)=sum_{n=0}^infty frac{g^{(n)}(1)}{n!}(z-1)^n=sum_{n=0}^infty left[frac{f^{(n)}(1)}{n!}right]^a(z-1)^n
          $$           has a radius of convergence equal to
          $$
          left(limsup_{nto infty}Big|frac{f^{(n)}(1)}{n!}Big|^{frac{a}{n}}right)^{-1}=0^{-1}=infty.
          $$           Thus $G$ defines an entire function. Suppose $g$ was initially defined on a connected open set $Omega supset {|z-1|<alpha}$. Since $G^{(n)}(1)=g^{(n)}(1)$ for all $nge 0$, it follows that $G$ and $g$ coincide on $|z-1|<alpha$. By the identity theorem, it follows that $g=G$ on $Omega$. This proves $G$ is an entire extension of $g$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Where did the power of $frac{1}{n}$ come from?
            $endgroup$
            – Ya G
            Jan 16 at 17:44










          • $begingroup$
            @YaG It comes from the root test of convergence. The root test: A series of complex numbers $sum_{j=1}^infty a_j$ converges if $limsup_j |a_j|^{1/j}<1$ and diverges if $limsup_j |a_j|^{1/j}>1$. You may be able to find it in en.wikipedia.org/wiki/Radius_of_convergence and related post about the root test.
            $endgroup$
            – Song
            Jan 16 at 17:48














          Your Answer





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          $begingroup$

          Note that by the assumption that $f$ is entire, we have
          $$
          limsup_{nto infty}Big|frac{f^{(n)}(1)}{n!}Big|^{frac{1}{n}}=0.
          $$           This is because $f(z) = sum_{n=0}^infty frac{f^{(n)}(1)}{n!}(z-1)^n$ has a radius of convergence equal to $$R=left(limsup_{nto infty}Big|frac{f^{(n)}(1)}{n!}Big|^{frac{1}{n}}right)^{-1}=infty.$$ Now, note that the power series
          $$
          G(z)=sum_{n=0}^infty frac{g^{(n)}(1)}{n!}(z-1)^n=sum_{n=0}^infty left[frac{f^{(n)}(1)}{n!}right]^a(z-1)^n
          $$           has a radius of convergence equal to
          $$
          left(limsup_{nto infty}Big|frac{f^{(n)}(1)}{n!}Big|^{frac{a}{n}}right)^{-1}=0^{-1}=infty.
          $$           Thus $G$ defines an entire function. Suppose $g$ was initially defined on a connected open set $Omega supset {|z-1|<alpha}$. Since $G^{(n)}(1)=g^{(n)}(1)$ for all $nge 0$, it follows that $G$ and $g$ coincide on $|z-1|<alpha$. By the identity theorem, it follows that $g=G$ on $Omega$. This proves $G$ is an entire extension of $g$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Where did the power of $frac{1}{n}$ come from?
            $endgroup$
            – Ya G
            Jan 16 at 17:44










          • $begingroup$
            @YaG It comes from the root test of convergence. The root test: A series of complex numbers $sum_{j=1}^infty a_j$ converges if $limsup_j |a_j|^{1/j}<1$ and diverges if $limsup_j |a_j|^{1/j}>1$. You may be able to find it in en.wikipedia.org/wiki/Radius_of_convergence and related post about the root test.
            $endgroup$
            – Song
            Jan 16 at 17:48


















          1












          $begingroup$

          Note that by the assumption that $f$ is entire, we have
          $$
          limsup_{nto infty}Big|frac{f^{(n)}(1)}{n!}Big|^{frac{1}{n}}=0.
          $$           This is because $f(z) = sum_{n=0}^infty frac{f^{(n)}(1)}{n!}(z-1)^n$ has a radius of convergence equal to $$R=left(limsup_{nto infty}Big|frac{f^{(n)}(1)}{n!}Big|^{frac{1}{n}}right)^{-1}=infty.$$ Now, note that the power series
          $$
          G(z)=sum_{n=0}^infty frac{g^{(n)}(1)}{n!}(z-1)^n=sum_{n=0}^infty left[frac{f^{(n)}(1)}{n!}right]^a(z-1)^n
          $$           has a radius of convergence equal to
          $$
          left(limsup_{nto infty}Big|frac{f^{(n)}(1)}{n!}Big|^{frac{a}{n}}right)^{-1}=0^{-1}=infty.
          $$           Thus $G$ defines an entire function. Suppose $g$ was initially defined on a connected open set $Omega supset {|z-1|<alpha}$. Since $G^{(n)}(1)=g^{(n)}(1)$ for all $nge 0$, it follows that $G$ and $g$ coincide on $|z-1|<alpha$. By the identity theorem, it follows that $g=G$ on $Omega$. This proves $G$ is an entire extension of $g$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Where did the power of $frac{1}{n}$ come from?
            $endgroup$
            – Ya G
            Jan 16 at 17:44










          • $begingroup$
            @YaG It comes from the root test of convergence. The root test: A series of complex numbers $sum_{j=1}^infty a_j$ converges if $limsup_j |a_j|^{1/j}<1$ and diverges if $limsup_j |a_j|^{1/j}>1$. You may be able to find it in en.wikipedia.org/wiki/Radius_of_convergence and related post about the root test.
            $endgroup$
            – Song
            Jan 16 at 17:48
















          1












          1








          1





          $begingroup$

          Note that by the assumption that $f$ is entire, we have
          $$
          limsup_{nto infty}Big|frac{f^{(n)}(1)}{n!}Big|^{frac{1}{n}}=0.
          $$           This is because $f(z) = sum_{n=0}^infty frac{f^{(n)}(1)}{n!}(z-1)^n$ has a radius of convergence equal to $$R=left(limsup_{nto infty}Big|frac{f^{(n)}(1)}{n!}Big|^{frac{1}{n}}right)^{-1}=infty.$$ Now, note that the power series
          $$
          G(z)=sum_{n=0}^infty frac{g^{(n)}(1)}{n!}(z-1)^n=sum_{n=0}^infty left[frac{f^{(n)}(1)}{n!}right]^a(z-1)^n
          $$           has a radius of convergence equal to
          $$
          left(limsup_{nto infty}Big|frac{f^{(n)}(1)}{n!}Big|^{frac{a}{n}}right)^{-1}=0^{-1}=infty.
          $$           Thus $G$ defines an entire function. Suppose $g$ was initially defined on a connected open set $Omega supset {|z-1|<alpha}$. Since $G^{(n)}(1)=g^{(n)}(1)$ for all $nge 0$, it follows that $G$ and $g$ coincide on $|z-1|<alpha$. By the identity theorem, it follows that $g=G$ on $Omega$. This proves $G$ is an entire extension of $g$.






          share|cite|improve this answer









          $endgroup$



          Note that by the assumption that $f$ is entire, we have
          $$
          limsup_{nto infty}Big|frac{f^{(n)}(1)}{n!}Big|^{frac{1}{n}}=0.
          $$           This is because $f(z) = sum_{n=0}^infty frac{f^{(n)}(1)}{n!}(z-1)^n$ has a radius of convergence equal to $$R=left(limsup_{nto infty}Big|frac{f^{(n)}(1)}{n!}Big|^{frac{1}{n}}right)^{-1}=infty.$$ Now, note that the power series
          $$
          G(z)=sum_{n=0}^infty frac{g^{(n)}(1)}{n!}(z-1)^n=sum_{n=0}^infty left[frac{f^{(n)}(1)}{n!}right]^a(z-1)^n
          $$           has a radius of convergence equal to
          $$
          left(limsup_{nto infty}Big|frac{f^{(n)}(1)}{n!}Big|^{frac{a}{n}}right)^{-1}=0^{-1}=infty.
          $$           Thus $G$ defines an entire function. Suppose $g$ was initially defined on a connected open set $Omega supset {|z-1|<alpha}$. Since $G^{(n)}(1)=g^{(n)}(1)$ for all $nge 0$, it follows that $G$ and $g$ coincide on $|z-1|<alpha$. By the identity theorem, it follows that $g=G$ on $Omega$. This proves $G$ is an entire extension of $g$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 16 at 17:35









          SongSong

          18.6k21651




          18.6k21651












          • $begingroup$
            Where did the power of $frac{1}{n}$ come from?
            $endgroup$
            – Ya G
            Jan 16 at 17:44










          • $begingroup$
            @YaG It comes from the root test of convergence. The root test: A series of complex numbers $sum_{j=1}^infty a_j$ converges if $limsup_j |a_j|^{1/j}<1$ and diverges if $limsup_j |a_j|^{1/j}>1$. You may be able to find it in en.wikipedia.org/wiki/Radius_of_convergence and related post about the root test.
            $endgroup$
            – Song
            Jan 16 at 17:48




















          • $begingroup$
            Where did the power of $frac{1}{n}$ come from?
            $endgroup$
            – Ya G
            Jan 16 at 17:44










          • $begingroup$
            @YaG It comes from the root test of convergence. The root test: A series of complex numbers $sum_{j=1}^infty a_j$ converges if $limsup_j |a_j|^{1/j}<1$ and diverges if $limsup_j |a_j|^{1/j}>1$. You may be able to find it in en.wikipedia.org/wiki/Radius_of_convergence and related post about the root test.
            $endgroup$
            – Song
            Jan 16 at 17:48


















          $begingroup$
          Where did the power of $frac{1}{n}$ come from?
          $endgroup$
          – Ya G
          Jan 16 at 17:44




          $begingroup$
          Where did the power of $frac{1}{n}$ come from?
          $endgroup$
          – Ya G
          Jan 16 at 17:44












          $begingroup$
          @YaG It comes from the root test of convergence. The root test: A series of complex numbers $sum_{j=1}^infty a_j$ converges if $limsup_j |a_j|^{1/j}<1$ and diverges if $limsup_j |a_j|^{1/j}>1$. You may be able to find it in en.wikipedia.org/wiki/Radius_of_convergence and related post about the root test.
          $endgroup$
          – Song
          Jan 16 at 17:48






          $begingroup$
          @YaG It comes from the root test of convergence. The root test: A series of complex numbers $sum_{j=1}^infty a_j$ converges if $limsup_j |a_j|^{1/j}<1$ and diverges if $limsup_j |a_j|^{1/j}>1$. You may be able to find it in en.wikipedia.org/wiki/Radius_of_convergence and related post about the root test.
          $endgroup$
          – Song
          Jan 16 at 17:48




















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