$f$ is analytic in the unit disk such that $|f(1/n)|leq e^{-sqrt{n}}$. Prove that $f$ Is identically zero.












1














$f(z)$ is analytic in the unit disk such that $|f(1/n)|leq e^{-sqrt{n}}$ for $n=2,3,...$ Prove that $f$ Is identically equal to zero.



My Attempt:



Consider the Taylor expansion $$f(z)=f(0)+f'(0)+frac{f''(0)}{2!}+frac{f'''(0)}{3!}+...$$By Cauchy estimate, for some $r=frac{1}{n}$, we have $$frac{|f^{(k)}|}{k!}leqfrac{e^{-sqrt{n}}}{(1/n)^k}$$



As $nrightarrow infty$, for each $k$, we have $frac{|f^{(k)}|}{k!}=0$. Then, Taylor expansion suggests that $f(z)$ is identically zero.



I feel like I'm missing a lot of pieces. Any help or input is appreciated.






complex-analysis analytic-functions







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  • You're not applying Cauchy's estimate correctly. The RHS of Cauchy's estimate involves a $sup_{|z| = r} |f(z)|$ term, not just $|f(r)|$.
    – mathworker21
    2 days ago
















1














$f(z)$ is analytic in the unit disk such that $|f(1/n)|leq e^{-sqrt{n}}$ for $n=2,3,...$ Prove that $f$ Is identically equal to zero.



My Attempt:



Consider the Taylor expansion $$f(z)=f(0)+f'(0)+frac{f''(0)}{2!}+frac{f'''(0)}{3!}+...$$By Cauchy estimate, for some $r=frac{1}{n}$, we have $$frac{|f^{(k)}|}{k!}leqfrac{e^{-sqrt{n}}}{(1/n)^k}$$



As $nrightarrow infty$, for each $k$, we have $frac{|f^{(k)}|}{k!}=0$. Then, Taylor expansion suggests that $f(z)$ is identically zero.



I feel like I'm missing a lot of pieces. Any help or input is appreciated.






complex-analysis analytic-functions







share|cite|improve this question
























  • You're not applying Cauchy's estimate correctly. The RHS of Cauchy's estimate involves a $sup_{|z| = r} |f(z)|$ term, not just $|f(r)|$.
    – mathworker21
    2 days ago














1












1








1







$f(z)$ is analytic in the unit disk such that $|f(1/n)|leq e^{-sqrt{n}}$ for $n=2,3,...$ Prove that $f$ Is identically equal to zero.



My Attempt:



Consider the Taylor expansion $$f(z)=f(0)+f'(0)+frac{f''(0)}{2!}+frac{f'''(0)}{3!}+...$$By Cauchy estimate, for some $r=frac{1}{n}$, we have $$frac{|f^{(k)}|}{k!}leqfrac{e^{-sqrt{n}}}{(1/n)^k}$$



As $nrightarrow infty$, for each $k$, we have $frac{|f^{(k)}|}{k!}=0$. Then, Taylor expansion suggests that $f(z)$ is identically zero.



I feel like I'm missing a lot of pieces. Any help or input is appreciated.






complex-analysis analytic-functions







share|cite|improve this question















$f(z)$ is analytic in the unit disk such that $|f(1/n)|leq e^{-sqrt{n}}$ for $n=2,3,...$ Prove that $f$ Is identically equal to zero.



My Attempt:



Consider the Taylor expansion $$f(z)=f(0)+f'(0)+frac{f''(0)}{2!}+frac{f'''(0)}{3!}+...$$By Cauchy estimate, for some $r=frac{1}{n}$, we have $$frac{|f^{(k)}|}{k!}leqfrac{e^{-sqrt{n}}}{(1/n)^k}$$



As $nrightarrow infty$, for each $k$, we have $frac{|f^{(k)}|}{k!}=0$. Then, Taylor expansion suggests that $f(z)$ is identically zero.



I feel like I'm missing a lot of pieces. Any help or input is appreciated.






complex-analysis analytic-functions




complex-analysis analytic-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









José Carlos Santos

152k22123226




152k22123226










asked 2 days ago









Ya GYa G

47629




47629












  • You're not applying Cauchy's estimate correctly. The RHS of Cauchy's estimate involves a $sup_{|z| = r} |f(z)|$ term, not just $|f(r)|$.
    – mathworker21
    2 days ago


















  • You're not applying Cauchy's estimate correctly. The RHS of Cauchy's estimate involves a $sup_{|z| = r} |f(z)|$ term, not just $|f(r)|$.
    – mathworker21
    2 days ago
















You're not applying Cauchy's estimate correctly. The RHS of Cauchy's estimate involves a $sup_{|z| = r} |f(z)|$ term, not just $|f(r)|$.
– mathworker21
2 days ago




You're not applying Cauchy's estimate correctly. The RHS of Cauchy's estimate involves a $sup_{|z| = r} |f(z)|$ term, not just $|f(r)|$.
– mathworker21
2 days ago










1 Answer
1






active

oldest

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3














I don't know what you mean when you write “for some $r=frac1n$”.



It follows from the squeeze theorem that $lim_{ntoinfty}fleft(frac1nright)=0$. So, unless $f$ is the null function, the Taylor series of $f$ about $0$ will be of the type$$f(z)=a_kz^k+a_{k+1}z^{k+1}+cdots$$with $kinmathbb N$ and $a_kneq0$. So $left(leftlvert fleft(frac1nright)rightrvertright)_{ninmathbb N}$ behaves as $frac{lvert a_krvert}{n^k}$. But $left(e^{-sqrt n}right)_{ninmathbb N}$ goes to $0$ much faster than that:$$lim_{ntoinfty}frac{frac{lvert a_krvert}{n^k}}{e^{-sqrt n}}=lvert a_krvertlim_{ntoinfty}frac{e^{sqrt n}}{n^k}=infty.$$






share|cite|improve this answer























  • I'm just curious about the last line. What's the purpose of that in the proof? How does showing the ratio diverges show that $f$ is identically zero?
    – Ya G
    2 days ago










  • I assumed that $f$ is not the null function. Then I reached a contradiction (that final limit shouldn't be $infty$). Therefore, $f$ is the null function.
    – José Carlos Santos
    2 days ago










  • @JoséCarlosSantos Nobody is saying $f$ behaves as $e^{-sqrt{n}}$, just that $f$ decays quicker than it. The inequalities implicit in your proof work in your favor though, so all is fine.
    – mathworker21
    2 days ago










  • @mathworker21 You are right. I've edited my answer. I hope that everything is correct now.
    – José Carlos Santos
    2 days ago










  • @JoséCarlosSantos I think you mean "with $k in mathbb{N}$ and $a_k not = 0$"
    – mathworker21
    2 days ago











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

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3














I don't know what you mean when you write “for some $r=frac1n$”.



It follows from the squeeze theorem that $lim_{ntoinfty}fleft(frac1nright)=0$. So, unless $f$ is the null function, the Taylor series of $f$ about $0$ will be of the type$$f(z)=a_kz^k+a_{k+1}z^{k+1}+cdots$$with $kinmathbb N$ and $a_kneq0$. So $left(leftlvert fleft(frac1nright)rightrvertright)_{ninmathbb N}$ behaves as $frac{lvert a_krvert}{n^k}$. But $left(e^{-sqrt n}right)_{ninmathbb N}$ goes to $0$ much faster than that:$$lim_{ntoinfty}frac{frac{lvert a_krvert}{n^k}}{e^{-sqrt n}}=lvert a_krvertlim_{ntoinfty}frac{e^{sqrt n}}{n^k}=infty.$$






share|cite|improve this answer























  • I'm just curious about the last line. What's the purpose of that in the proof? How does showing the ratio diverges show that $f$ is identically zero?
    – Ya G
    2 days ago










  • I assumed that $f$ is not the null function. Then I reached a contradiction (that final limit shouldn't be $infty$). Therefore, $f$ is the null function.
    – José Carlos Santos
    2 days ago










  • @JoséCarlosSantos Nobody is saying $f$ behaves as $e^{-sqrt{n}}$, just that $f$ decays quicker than it. The inequalities implicit in your proof work in your favor though, so all is fine.
    – mathworker21
    2 days ago










  • @mathworker21 You are right. I've edited my answer. I hope that everything is correct now.
    – José Carlos Santos
    2 days ago










  • @JoséCarlosSantos I think you mean "with $k in mathbb{N}$ and $a_k not = 0$"
    – mathworker21
    2 days ago
















3














I don't know what you mean when you write “for some $r=frac1n$”.



It follows from the squeeze theorem that $lim_{ntoinfty}fleft(frac1nright)=0$. So, unless $f$ is the null function, the Taylor series of $f$ about $0$ will be of the type$$f(z)=a_kz^k+a_{k+1}z^{k+1}+cdots$$with $kinmathbb N$ and $a_kneq0$. So $left(leftlvert fleft(frac1nright)rightrvertright)_{ninmathbb N}$ behaves as $frac{lvert a_krvert}{n^k}$. But $left(e^{-sqrt n}right)_{ninmathbb N}$ goes to $0$ much faster than that:$$lim_{ntoinfty}frac{frac{lvert a_krvert}{n^k}}{e^{-sqrt n}}=lvert a_krvertlim_{ntoinfty}frac{e^{sqrt n}}{n^k}=infty.$$






share|cite|improve this answer























  • I'm just curious about the last line. What's the purpose of that in the proof? How does showing the ratio diverges show that $f$ is identically zero?
    – Ya G
    2 days ago










  • I assumed that $f$ is not the null function. Then I reached a contradiction (that final limit shouldn't be $infty$). Therefore, $f$ is the null function.
    – José Carlos Santos
    2 days ago










  • @JoséCarlosSantos Nobody is saying $f$ behaves as $e^{-sqrt{n}}$, just that $f$ decays quicker than it. The inequalities implicit in your proof work in your favor though, so all is fine.
    – mathworker21
    2 days ago










  • @mathworker21 You are right. I've edited my answer. I hope that everything is correct now.
    – José Carlos Santos
    2 days ago










  • @JoséCarlosSantos I think you mean "with $k in mathbb{N}$ and $a_k not = 0$"
    – mathworker21
    2 days ago














3












3








3






I don't know what you mean when you write “for some $r=frac1n$”.



It follows from the squeeze theorem that $lim_{ntoinfty}fleft(frac1nright)=0$. So, unless $f$ is the null function, the Taylor series of $f$ about $0$ will be of the type$$f(z)=a_kz^k+a_{k+1}z^{k+1}+cdots$$with $kinmathbb N$ and $a_kneq0$. So $left(leftlvert fleft(frac1nright)rightrvertright)_{ninmathbb N}$ behaves as $frac{lvert a_krvert}{n^k}$. But $left(e^{-sqrt n}right)_{ninmathbb N}$ goes to $0$ much faster than that:$$lim_{ntoinfty}frac{frac{lvert a_krvert}{n^k}}{e^{-sqrt n}}=lvert a_krvertlim_{ntoinfty}frac{e^{sqrt n}}{n^k}=infty.$$






share|cite|improve this answer














I don't know what you mean when you write “for some $r=frac1n$”.



It follows from the squeeze theorem that $lim_{ntoinfty}fleft(frac1nright)=0$. So, unless $f$ is the null function, the Taylor series of $f$ about $0$ will be of the type$$f(z)=a_kz^k+a_{k+1}z^{k+1}+cdots$$with $kinmathbb N$ and $a_kneq0$. So $left(leftlvert fleft(frac1nright)rightrvertright)_{ninmathbb N}$ behaves as $frac{lvert a_krvert}{n^k}$. But $left(e^{-sqrt n}right)_{ninmathbb N}$ goes to $0$ much faster than that:$$lim_{ntoinfty}frac{frac{lvert a_krvert}{n^k}}{e^{-sqrt n}}=lvert a_krvertlim_{ntoinfty}frac{e^{sqrt n}}{n^k}=infty.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered 2 days ago









José Carlos SantosJosé Carlos Santos

152k22123226




152k22123226












  • I'm just curious about the last line. What's the purpose of that in the proof? How does showing the ratio diverges show that $f$ is identically zero?
    – Ya G
    2 days ago










  • I assumed that $f$ is not the null function. Then I reached a contradiction (that final limit shouldn't be $infty$). Therefore, $f$ is the null function.
    – José Carlos Santos
    2 days ago










  • @JoséCarlosSantos Nobody is saying $f$ behaves as $e^{-sqrt{n}}$, just that $f$ decays quicker than it. The inequalities implicit in your proof work in your favor though, so all is fine.
    – mathworker21
    2 days ago










  • @mathworker21 You are right. I've edited my answer. I hope that everything is correct now.
    – José Carlos Santos
    2 days ago










  • @JoséCarlosSantos I think you mean "with $k in mathbb{N}$ and $a_k not = 0$"
    – mathworker21
    2 days ago


















  • I'm just curious about the last line. What's the purpose of that in the proof? How does showing the ratio diverges show that $f$ is identically zero?
    – Ya G
    2 days ago










  • I assumed that $f$ is not the null function. Then I reached a contradiction (that final limit shouldn't be $infty$). Therefore, $f$ is the null function.
    – José Carlos Santos
    2 days ago










  • @JoséCarlosSantos Nobody is saying $f$ behaves as $e^{-sqrt{n}}$, just that $f$ decays quicker than it. The inequalities implicit in your proof work in your favor though, so all is fine.
    – mathworker21
    2 days ago










  • @mathworker21 You are right. I've edited my answer. I hope that everything is correct now.
    – José Carlos Santos
    2 days ago










  • @JoséCarlosSantos I think you mean "with $k in mathbb{N}$ and $a_k not = 0$"
    – mathworker21
    2 days ago
















I'm just curious about the last line. What's the purpose of that in the proof? How does showing the ratio diverges show that $f$ is identically zero?
– Ya G
2 days ago




I'm just curious about the last line. What's the purpose of that in the proof? How does showing the ratio diverges show that $f$ is identically zero?
– Ya G
2 days ago












I assumed that $f$ is not the null function. Then I reached a contradiction (that final limit shouldn't be $infty$). Therefore, $f$ is the null function.
– José Carlos Santos
2 days ago




I assumed that $f$ is not the null function. Then I reached a contradiction (that final limit shouldn't be $infty$). Therefore, $f$ is the null function.
– José Carlos Santos
2 days ago












@JoséCarlosSantos Nobody is saying $f$ behaves as $e^{-sqrt{n}}$, just that $f$ decays quicker than it. The inequalities implicit in your proof work in your favor though, so all is fine.
– mathworker21
2 days ago




@JoséCarlosSantos Nobody is saying $f$ behaves as $e^{-sqrt{n}}$, just that $f$ decays quicker than it. The inequalities implicit in your proof work in your favor though, so all is fine.
– mathworker21
2 days ago












@mathworker21 You are right. I've edited my answer. I hope that everything is correct now.
– José Carlos Santos
2 days ago




@mathworker21 You are right. I've edited my answer. I hope that everything is correct now.
– José Carlos Santos
2 days ago












@JoséCarlosSantos I think you mean "with $k in mathbb{N}$ and $a_k not = 0$"
– mathworker21
2 days ago




@JoséCarlosSantos I think you mean "with $k in mathbb{N}$ and $a_k not = 0$"
– mathworker21
2 days ago


















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