Identity theorem for a holomorphic funtion defined near zero












1














I have to show, whether there is a holomorphic funtion $f$ defined in an open neighborhood of zero, such that:
$$ fleft(frac{1}{n}right)=(-1)^n frac{1}{n^3}$$ for all positive integer $n$.



My idea was to apply the identity theorem for holomorphic funtions.
How can I do that?
Maybe I must consider the subsequences $ frac{1}{2k}, frac{1}{2k+1}$. Can somebody help me?






sequences-and-series complex-analysis holomorphic-functions entire-functions







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  • "f in zero"? What do you mean by that?
    – Lord Shark the Unknown
    Dec 9 at 17:59






  • 1




    I don't know for sure what you want. This is my guess. Please change it if it is not what you wanted.
    – Batominovski
    Dec 9 at 23:49












  • Thank you, Your guess was right:)
    – Steven33
    Dec 10 at 7:30
















1














I have to show, whether there is a holomorphic funtion $f$ defined in an open neighborhood of zero, such that:
$$ fleft(frac{1}{n}right)=(-1)^n frac{1}{n^3}$$ for all positive integer $n$.



My idea was to apply the identity theorem for holomorphic funtions.
How can I do that?
Maybe I must consider the subsequences $ frac{1}{2k}, frac{1}{2k+1}$. Can somebody help me?






sequences-and-series complex-analysis holomorphic-functions entire-functions







share|cite|improve this question
























  • "f in zero"? What do you mean by that?
    – Lord Shark the Unknown
    Dec 9 at 17:59






  • 1




    I don't know for sure what you want. This is my guess. Please change it if it is not what you wanted.
    – Batominovski
    Dec 9 at 23:49












  • Thank you, Your guess was right:)
    – Steven33
    Dec 10 at 7:30














1












1








1







I have to show, whether there is a holomorphic funtion $f$ defined in an open neighborhood of zero, such that:
$$ fleft(frac{1}{n}right)=(-1)^n frac{1}{n^3}$$ for all positive integer $n$.



My idea was to apply the identity theorem for holomorphic funtions.
How can I do that?
Maybe I must consider the subsequences $ frac{1}{2k}, frac{1}{2k+1}$. Can somebody help me?






sequences-and-series complex-analysis holomorphic-functions entire-functions







share|cite|improve this question















I have to show, whether there is a holomorphic funtion $f$ defined in an open neighborhood of zero, such that:
$$ fleft(frac{1}{n}right)=(-1)^n frac{1}{n^3}$$ for all positive integer $n$.



My idea was to apply the identity theorem for holomorphic funtions.
How can I do that?
Maybe I must consider the subsequences $ frac{1}{2k}, frac{1}{2k+1}$. Can somebody help me?






sequences-and-series complex-analysis holomorphic-functions entire-functions




sequences-and-series complex-analysis holomorphic-functions entire-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 at 23:47









Batominovski

33.7k33292




33.7k33292










asked Dec 9 at 17:57









Steven33

144




144












  • "f in zero"? What do you mean by that?
    – Lord Shark the Unknown
    Dec 9 at 17:59






  • 1




    I don't know for sure what you want. This is my guess. Please change it if it is not what you wanted.
    – Batominovski
    Dec 9 at 23:49












  • Thank you, Your guess was right:)
    – Steven33
    Dec 10 at 7:30


















  • "f in zero"? What do you mean by that?
    – Lord Shark the Unknown
    Dec 9 at 17:59






  • 1




    I don't know for sure what you want. This is my guess. Please change it if it is not what you wanted.
    – Batominovski
    Dec 9 at 23:49












  • Thank you, Your guess was right:)
    – Steven33
    Dec 10 at 7:30
















"f in zero"? What do you mean by that?
– Lord Shark the Unknown
Dec 9 at 17:59




"f in zero"? What do you mean by that?
– Lord Shark the Unknown
Dec 9 at 17:59




1




1




I don't know for sure what you want. This is my guess. Please change it if it is not what you wanted.
– Batominovski
Dec 9 at 23:49






I don't know for sure what you want. This is my guess. Please change it if it is not what you wanted.
– Batominovski
Dec 9 at 23:49














Thank you, Your guess was right:)
– Steven33
Dec 10 at 7:30




Thank you, Your guess was right:)
– Steven33
Dec 10 at 7:30










1 Answer
1






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oldest

votes


















1














While I don't know clearly what your question is, I assume you want to find all entire function $f$ such that $f(1/n)=(-1)^n/n^3$ for all positive integers $n$. I want to point out that $f$ does not exist, and your idea of considering $1/(2k)$ and $1/(2k+1)$ is a good idea.



Note that $f(z)=z^3$ for all $z$ of the form $1/(2k)$ where $k$ is a positive integer. Since the set of $1/(2k)$ has an accumulation point (namely, $0$) in $Bbb{C}$, so $f(z)=z^3$ must hold for all $zin Bbb{C}$, but then the condition says that $fbig(1/(2k+1)big)$ is $-1/(2k+1)^3$, not $1/(2k+1)^3$. This is a contradiction.






share|cite|improve this answer





















  • Thank you for your answer:) I dont understand how you get $ f(z)=z^3$ When you consider $1/2k$, then you get $ f(1/2k)= 1/(2k)^3 = frac{1}{8 z^3 } $?
    – Steven33
    Dec 9 at 18:53








  • 1




    If $z=1/(2k)$, then $1/(2k)^3=z^3$.
    – Snookie
    Dec 9 at 21:34










  • Ok, you put $z=frac{1}{2k} Rightarrow g(z)=z^3 $, so $f(z)=g(z) forall z=1/2k Rightarrow f=g$ , but $ f(1/2k+1)= .-1/(2k+1)^3 ne 1/(2k+1)^3 =g(1/2k+1) $
    – Steven33
    Dec 10 at 7:26












  • Ok I think I understand that now:)
    – Steven33
    Dec 10 at 7:29








  • 1




    I think something is wrong with your last example. If $f(1/n)=frac{1}{n^2-1}$, then $$f(1/n)=frac{(1/n)^2}{1-(1/n)^2}.$$ Therefore, if $z=1/n$, then $f(z)=frac{1}{1-z^2}$.
    – Snookie
    Dec 10 at 11:55











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1














While I don't know clearly what your question is, I assume you want to find all entire function $f$ such that $f(1/n)=(-1)^n/n^3$ for all positive integers $n$. I want to point out that $f$ does not exist, and your idea of considering $1/(2k)$ and $1/(2k+1)$ is a good idea.



Note that $f(z)=z^3$ for all $z$ of the form $1/(2k)$ where $k$ is a positive integer. Since the set of $1/(2k)$ has an accumulation point (namely, $0$) in $Bbb{C}$, so $f(z)=z^3$ must hold for all $zin Bbb{C}$, but then the condition says that $fbig(1/(2k+1)big)$ is $-1/(2k+1)^3$, not $1/(2k+1)^3$. This is a contradiction.






share|cite|improve this answer





















  • Thank you for your answer:) I dont understand how you get $ f(z)=z^3$ When you consider $1/2k$, then you get $ f(1/2k)= 1/(2k)^3 = frac{1}{8 z^3 } $?
    – Steven33
    Dec 9 at 18:53








  • 1




    If $z=1/(2k)$, then $1/(2k)^3=z^3$.
    – Snookie
    Dec 9 at 21:34










  • Ok, you put $z=frac{1}{2k} Rightarrow g(z)=z^3 $, so $f(z)=g(z) forall z=1/2k Rightarrow f=g$ , but $ f(1/2k+1)= .-1/(2k+1)^3 ne 1/(2k+1)^3 =g(1/2k+1) $
    – Steven33
    Dec 10 at 7:26












  • Ok I think I understand that now:)
    – Steven33
    Dec 10 at 7:29








  • 1




    I think something is wrong with your last example. If $f(1/n)=frac{1}{n^2-1}$, then $$f(1/n)=frac{(1/n)^2}{1-(1/n)^2}.$$ Therefore, if $z=1/n$, then $f(z)=frac{1}{1-z^2}$.
    – Snookie
    Dec 10 at 11:55
















1














While I don't know clearly what your question is, I assume you want to find all entire function $f$ such that $f(1/n)=(-1)^n/n^3$ for all positive integers $n$. I want to point out that $f$ does not exist, and your idea of considering $1/(2k)$ and $1/(2k+1)$ is a good idea.



Note that $f(z)=z^3$ for all $z$ of the form $1/(2k)$ where $k$ is a positive integer. Since the set of $1/(2k)$ has an accumulation point (namely, $0$) in $Bbb{C}$, so $f(z)=z^3$ must hold for all $zin Bbb{C}$, but then the condition says that $fbig(1/(2k+1)big)$ is $-1/(2k+1)^3$, not $1/(2k+1)^3$. This is a contradiction.






share|cite|improve this answer





















  • Thank you for your answer:) I dont understand how you get $ f(z)=z^3$ When you consider $1/2k$, then you get $ f(1/2k)= 1/(2k)^3 = frac{1}{8 z^3 } $?
    – Steven33
    Dec 9 at 18:53








  • 1




    If $z=1/(2k)$, then $1/(2k)^3=z^3$.
    – Snookie
    Dec 9 at 21:34










  • Ok, you put $z=frac{1}{2k} Rightarrow g(z)=z^3 $, so $f(z)=g(z) forall z=1/2k Rightarrow f=g$ , but $ f(1/2k+1)= .-1/(2k+1)^3 ne 1/(2k+1)^3 =g(1/2k+1) $
    – Steven33
    Dec 10 at 7:26












  • Ok I think I understand that now:)
    – Steven33
    Dec 10 at 7:29








  • 1




    I think something is wrong with your last example. If $f(1/n)=frac{1}{n^2-1}$, then $$f(1/n)=frac{(1/n)^2}{1-(1/n)^2}.$$ Therefore, if $z=1/n$, then $f(z)=frac{1}{1-z^2}$.
    – Snookie
    Dec 10 at 11:55














1












1








1






While I don't know clearly what your question is, I assume you want to find all entire function $f$ such that $f(1/n)=(-1)^n/n^3$ for all positive integers $n$. I want to point out that $f$ does not exist, and your idea of considering $1/(2k)$ and $1/(2k+1)$ is a good idea.



Note that $f(z)=z^3$ for all $z$ of the form $1/(2k)$ where $k$ is a positive integer. Since the set of $1/(2k)$ has an accumulation point (namely, $0$) in $Bbb{C}$, so $f(z)=z^3$ must hold for all $zin Bbb{C}$, but then the condition says that $fbig(1/(2k+1)big)$ is $-1/(2k+1)^3$, not $1/(2k+1)^3$. This is a contradiction.






share|cite|improve this answer












While I don't know clearly what your question is, I assume you want to find all entire function $f$ such that $f(1/n)=(-1)^n/n^3$ for all positive integers $n$. I want to point out that $f$ does not exist, and your idea of considering $1/(2k)$ and $1/(2k+1)$ is a good idea.



Note that $f(z)=z^3$ for all $z$ of the form $1/(2k)$ where $k$ is a positive integer. Since the set of $1/(2k)$ has an accumulation point (namely, $0$) in $Bbb{C}$, so $f(z)=z^3$ must hold for all $zin Bbb{C}$, but then the condition says that $fbig(1/(2k+1)big)$ is $-1/(2k+1)^3$, not $1/(2k+1)^3$. This is a contradiction.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 at 18:37









Snookie

1,30017




1,30017












  • Thank you for your answer:) I dont understand how you get $ f(z)=z^3$ When you consider $1/2k$, then you get $ f(1/2k)= 1/(2k)^3 = frac{1}{8 z^3 } $?
    – Steven33
    Dec 9 at 18:53








  • 1




    If $z=1/(2k)$, then $1/(2k)^3=z^3$.
    – Snookie
    Dec 9 at 21:34










  • Ok, you put $z=frac{1}{2k} Rightarrow g(z)=z^3 $, so $f(z)=g(z) forall z=1/2k Rightarrow f=g$ , but $ f(1/2k+1)= .-1/(2k+1)^3 ne 1/(2k+1)^3 =g(1/2k+1) $
    – Steven33
    Dec 10 at 7:26












  • Ok I think I understand that now:)
    – Steven33
    Dec 10 at 7:29








  • 1




    I think something is wrong with your last example. If $f(1/n)=frac{1}{n^2-1}$, then $$f(1/n)=frac{(1/n)^2}{1-(1/n)^2}.$$ Therefore, if $z=1/n$, then $f(z)=frac{1}{1-z^2}$.
    – Snookie
    Dec 10 at 11:55


















  • Thank you for your answer:) I dont understand how you get $ f(z)=z^3$ When you consider $1/2k$, then you get $ f(1/2k)= 1/(2k)^3 = frac{1}{8 z^3 } $?
    – Steven33
    Dec 9 at 18:53








  • 1




    If $z=1/(2k)$, then $1/(2k)^3=z^3$.
    – Snookie
    Dec 9 at 21:34










  • Ok, you put $z=frac{1}{2k} Rightarrow g(z)=z^3 $, so $f(z)=g(z) forall z=1/2k Rightarrow f=g$ , but $ f(1/2k+1)= .-1/(2k+1)^3 ne 1/(2k+1)^3 =g(1/2k+1) $
    – Steven33
    Dec 10 at 7:26












  • Ok I think I understand that now:)
    – Steven33
    Dec 10 at 7:29








  • 1




    I think something is wrong with your last example. If $f(1/n)=frac{1}{n^2-1}$, then $$f(1/n)=frac{(1/n)^2}{1-(1/n)^2}.$$ Therefore, if $z=1/n$, then $f(z)=frac{1}{1-z^2}$.
    – Snookie
    Dec 10 at 11:55
















Thank you for your answer:) I dont understand how you get $ f(z)=z^3$ When you consider $1/2k$, then you get $ f(1/2k)= 1/(2k)^3 = frac{1}{8 z^3 } $?
– Steven33
Dec 9 at 18:53






Thank you for your answer:) I dont understand how you get $ f(z)=z^3$ When you consider $1/2k$, then you get $ f(1/2k)= 1/(2k)^3 = frac{1}{8 z^3 } $?
– Steven33
Dec 9 at 18:53






1




1




If $z=1/(2k)$, then $1/(2k)^3=z^3$.
– Snookie
Dec 9 at 21:34




If $z=1/(2k)$, then $1/(2k)^3=z^3$.
– Snookie
Dec 9 at 21:34












Ok, you put $z=frac{1}{2k} Rightarrow g(z)=z^3 $, so $f(z)=g(z) forall z=1/2k Rightarrow f=g$ , but $ f(1/2k+1)= .-1/(2k+1)^3 ne 1/(2k+1)^3 =g(1/2k+1) $
– Steven33
Dec 10 at 7:26






Ok, you put $z=frac{1}{2k} Rightarrow g(z)=z^3 $, so $f(z)=g(z) forall z=1/2k Rightarrow f=g$ , but $ f(1/2k+1)= .-1/(2k+1)^3 ne 1/(2k+1)^3 =g(1/2k+1) $
– Steven33
Dec 10 at 7:26














Ok I think I understand that now:)
– Steven33
Dec 10 at 7:29






Ok I think I understand that now:)
– Steven33
Dec 10 at 7:29






1




1




I think something is wrong with your last example. If $f(1/n)=frac{1}{n^2-1}$, then $$f(1/n)=frac{(1/n)^2}{1-(1/n)^2}.$$ Therefore, if $z=1/n$, then $f(z)=frac{1}{1-z^2}$.
– Snookie
Dec 10 at 11:55




I think something is wrong with your last example. If $f(1/n)=frac{1}{n^2-1}$, then $$f(1/n)=frac{(1/n)^2}{1-(1/n)^2}.$$ Therefore, if $z=1/n$, then $f(z)=frac{1}{1-z^2}$.
– Snookie
Dec 10 at 11:55


















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