$f$ 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







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














$f(z)$ 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

votes


















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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062782%2ff-is-analytic-in-the-unit-disk-such-that-f1-n-leq-e-sqrtn-prove-t%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062782%2ff-is-analytic-in-the-unit-disk-such-that-f1-n-leq-e-sqrtn-prove-t%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Mario Kart Wii

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
Read more

What does “Dominus providebit” mean?

Image
5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
Read more

Antonio Litta Visconti Arese

Image
Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
Read more