If $f$ is entire such that $|f(z)|leq C|z|^{5/2}$, $f$ is polynomial of degree two.












0












$begingroup$


$f$ is entire and there exists $C>0$ and $M>0$ such that $|f(z)|leq C|z|^{5/2}$ for all $zinmathbb{C}$ where $|z|>M$. Prove that $f$ is polynomial of degree two.



I don't have a clear idea but given an entire function and we can bound it $dfrac{|f(z)|}{|z|^{5/2}}leq C$, and showing that it has removable singularity at $z=0$, using Riemann's theorem, I can show that $dfrac{f(z)}{z^{5/2}}$ is a constant by Liouville's theorem. However, I'm not sure what I should do to show that it is intact a polynomial.






complex-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    No! $f(z)/z^{5/2}$ is not a holomorphic function in $Bbb Csetminus{0}$.
    $endgroup$
    – David C. Ullrich
    Dec 21 '18 at 18:09






  • 2




    $begingroup$
    Cauchy's Estimate
    $endgroup$
    – Story123
    Dec 21 '18 at 18:54
















0












$begingroup$


$f$ is entire and there exists $C>0$ and $M>0$ such that $|f(z)|leq C|z|^{5/2}$ for all $zinmathbb{C}$ where $|z|>M$. Prove that $f$ is polynomial of degree two.



I don't have a clear idea but given an entire function and we can bound it $dfrac{|f(z)|}{|z|^{5/2}}leq C$, and showing that it has removable singularity at $z=0$, using Riemann's theorem, I can show that $dfrac{f(z)}{z^{5/2}}$ is a constant by Liouville's theorem. However, I'm not sure what I should do to show that it is intact a polynomial.






complex-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    No! $f(z)/z^{5/2}$ is not a holomorphic function in $Bbb Csetminus{0}$.
    $endgroup$
    – David C. Ullrich
    Dec 21 '18 at 18:09






  • 2




    $begingroup$
    Cauchy's Estimate
    $endgroup$
    – Story123
    Dec 21 '18 at 18:54














0












0








0


1



$begingroup$


$f$ is entire and there exists $C>0$ and $M>0$ such that $|f(z)|leq C|z|^{5/2}$ for all $zinmathbb{C}$ where $|z|>M$. Prove that $f$ is polynomial of degree two.



I don't have a clear idea but given an entire function and we can bound it $dfrac{|f(z)|}{|z|^{5/2}}leq C$, and showing that it has removable singularity at $z=0$, using Riemann's theorem, I can show that $dfrac{f(z)}{z^{5/2}}$ is a constant by Liouville's theorem. However, I'm not sure what I should do to show that it is intact a polynomial.






complex-analysis







share|cite|improve this question











$endgroup$




$f$ is entire and there exists $C>0$ and $M>0$ such that $|f(z)|leq C|z|^{5/2}$ for all $zinmathbb{C}$ where $|z|>M$. Prove that $f$ is polynomial of degree two.



I don't have a clear idea but given an entire function and we can bound it $dfrac{|f(z)|}{|z|^{5/2}}leq C$, and showing that it has removable singularity at $z=0$, using Riemann's theorem, I can show that $dfrac{f(z)}{z^{5/2}}$ is a constant by Liouville's theorem. However, I'm not sure what I should do to show that it is intact a polynomial.






complex-analysis




complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 21 '18 at 18:10







Ya G

















asked Dec 21 '18 at 17:55









Ya GYa G

536211




536211








  • 1




    $begingroup$
    No! $f(z)/z^{5/2}$ is not a holomorphic function in $Bbb Csetminus{0}$.
    $endgroup$
    – David C. Ullrich
    Dec 21 '18 at 18:09






  • 2




    $begingroup$
    Cauchy's Estimate
    $endgroup$
    – Story123
    Dec 21 '18 at 18:54














  • 1




    $begingroup$
    No! $f(z)/z^{5/2}$ is not a holomorphic function in $Bbb Csetminus{0}$.
    $endgroup$
    – David C. Ullrich
    Dec 21 '18 at 18:09






  • 2




    $begingroup$
    Cauchy's Estimate
    $endgroup$
    – Story123
    Dec 21 '18 at 18:54








1




1




$begingroup$
No! $f(z)/z^{5/2}$ is not a holomorphic function in $Bbb Csetminus{0}$.
$endgroup$
– David C. Ullrich
Dec 21 '18 at 18:09




$begingroup$
No! $f(z)/z^{5/2}$ is not a holomorphic function in $Bbb Csetminus{0}$.
$endgroup$
– David C. Ullrich
Dec 21 '18 at 18:09




2




2




$begingroup$
Cauchy's Estimate
$endgroup$
– Story123
Dec 21 '18 at 18:54




$begingroup$
Cauchy's Estimate
$endgroup$
– Story123
Dec 21 '18 at 18:54










4 Answers
4






active

oldest

votes


















2












$begingroup$

Almost certainly, the author of this exercise expected you to use Cauchy's estimates:
$$frac{|f^{(n)}(a)|}{n!}lefrac{sup_t{|f(a+re^{it}|)}}{r^n}$$
for $r>0$. Here, then
$$frac{|f^{(n)}(0)|}{n!}lefrac{Cr^{5/2}}{r^n}$$
for $r>R$. If $nge3$ and we let $rtoinfty$ we get $f^{(n)}(0)=0$.
In the power series $f(z)=sum a_nz^n$ then $a_3=a_4=cdots=0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Two questions. Why are we considering $a+re^{it}$? What is $a$ here? And, how do we get the supremum $Mr^{5/2}$?
    $endgroup$
    – Ya G
    Dec 21 '18 at 20:24










  • $begingroup$
    @YaG That's three questions! $a+re^it$ is the circle, centre $a$ radius $r$. $a=0$ in your problem. The supremum comes from your condition on $f$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 22 '18 at 5:56



















7












$begingroup$

Let $f(z) = sum_{n=0}^infty a_n z^n$, and let $g(z) = sum_{k=0}^infty a_{k+3} z^k$, so that
$$
[f(z) - a_0 - a_1 z - a_2 z^2] = z^3 g(z).
$$
The function $g$ is entire and, by assumption, $lim_{|z| to +infty} g(z) = 0$:
$$
|g(z)| = frac{|f(z) - a_0 - a_1 z - a_2 z^2|}{|z|^{5/2}} cdot frac{1}{|z|^{1/2}}
leq left(C + frac{|a_0| + |a_1|, |z| + |a_2|, |z|^2}{|z|^{5/2}}right)frac{1}{|z|^{1/2}} to 0.
$$
Hence, by Liouville's theorem, we must have $g = 0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Would you please explain a little bit more? This does make sense by itself but how did you come up with the $g(z)$ other than the fact that to make a relation to $f(z)$ as you set it up. How is $|z|^{5/2}$ and $C$ used this this case?
    $endgroup$
    – Ya G
    Dec 21 '18 at 18:19






  • 1




    $begingroup$
    Added a line in the proof.
    $endgroup$
    – Rigel
    Dec 21 '18 at 18:24



















1












$begingroup$

It may be interesting to note this follows from basic facts about Fourier series (Parseval) with more or less no complex analysis. Say $f(z)=sum c_nz^n$. Then $$C^2r^5gefrac1{2pi}int_0^{2pi}|f(re^{it})|^2,dt=sum_j|c_j|^2r^{2j}ge|c_n|^2r^{2n};$$hence $c_n=0$ for $nge3$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    A lot of complex analysis follows from basic facts about Fourier series.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 18:55



















0












$begingroup$

Your approach can also be made to work, but you need to show that $h(z)=frac{f(z)}{z^3}$ has a removable singularity at $z=0$.



This can be done as follows:
Let $g(z)=frac{f(z)}{z^2}$. Then, as $lim_{z to 0}g(z)=0$, $g$ has a removable singularity at $z=0$, and if we remove it we have $g(0)=0$.



Now since, after removing the singularity, $g$ is entire and $g(0)=0$, then $h(z)=frac{g(z)}{z}$ also has a removable singularity at $0$.



Then $h$ becomes an entire function and
$$lim_{z to infty} h(z)=0$$
From here it is easy to conclude that $h$ is constant.






share|cite|improve this answer









$endgroup$













    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%2f3048745%2fif-f-is-entire-such-that-fz-leq-cz5-2-f-is-polynomial-of-degree%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Almost certainly, the author of this exercise expected you to use Cauchy's estimates:
    $$frac{|f^{(n)}(a)|}{n!}lefrac{sup_t{|f(a+re^{it}|)}}{r^n}$$
    for $r>0$. Here, then
    $$frac{|f^{(n)}(0)|}{n!}lefrac{Cr^{5/2}}{r^n}$$
    for $r>R$. If $nge3$ and we let $rtoinfty$ we get $f^{(n)}(0)=0$.
    In the power series $f(z)=sum a_nz^n$ then $a_3=a_4=cdots=0$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Two questions. Why are we considering $a+re^{it}$? What is $a$ here? And, how do we get the supremum $Mr^{5/2}$?
      $endgroup$
      – Ya G
      Dec 21 '18 at 20:24










    • $begingroup$
      @YaG That's three questions! $a+re^it$ is the circle, centre $a$ radius $r$. $a=0$ in your problem. The supremum comes from your condition on $f$.
      $endgroup$
      – Lord Shark the Unknown
      Dec 22 '18 at 5:56
















    2












    $begingroup$

    Almost certainly, the author of this exercise expected you to use Cauchy's estimates:
    $$frac{|f^{(n)}(a)|}{n!}lefrac{sup_t{|f(a+re^{it}|)}}{r^n}$$
    for $r>0$. Here, then
    $$frac{|f^{(n)}(0)|}{n!}lefrac{Cr^{5/2}}{r^n}$$
    for $r>R$. If $nge3$ and we let $rtoinfty$ we get $f^{(n)}(0)=0$.
    In the power series $f(z)=sum a_nz^n$ then $a_3=a_4=cdots=0$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Two questions. Why are we considering $a+re^{it}$? What is $a$ here? And, how do we get the supremum $Mr^{5/2}$?
      $endgroup$
      – Ya G
      Dec 21 '18 at 20:24










    • $begingroup$
      @YaG That's three questions! $a+re^it$ is the circle, centre $a$ radius $r$. $a=0$ in your problem. The supremum comes from your condition on $f$.
      $endgroup$
      – Lord Shark the Unknown
      Dec 22 '18 at 5:56














    2












    2








    2





    $begingroup$

    Almost certainly, the author of this exercise expected you to use Cauchy's estimates:
    $$frac{|f^{(n)}(a)|}{n!}lefrac{sup_t{|f(a+re^{it}|)}}{r^n}$$
    for $r>0$. Here, then
    $$frac{|f^{(n)}(0)|}{n!}lefrac{Cr^{5/2}}{r^n}$$
    for $r>R$. If $nge3$ and we let $rtoinfty$ we get $f^{(n)}(0)=0$.
    In the power series $f(z)=sum a_nz^n$ then $a_3=a_4=cdots=0$.






    share|cite|improve this answer











    $endgroup$



    Almost certainly, the author of this exercise expected you to use Cauchy's estimates:
    $$frac{|f^{(n)}(a)|}{n!}lefrac{sup_t{|f(a+re^{it}|)}}{r^n}$$
    for $r>0$. Here, then
    $$frac{|f^{(n)}(0)|}{n!}lefrac{Cr^{5/2}}{r^n}$$
    for $r>R$. If $nge3$ and we let $rtoinfty$ we get $f^{(n)}(0)=0$.
    In the power series $f(z)=sum a_nz^n$ then $a_3=a_4=cdots=0$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 22 '18 at 5:57

























    answered Dec 21 '18 at 19:03









    Lord Shark the UnknownLord Shark the Unknown

    107k1162135




    107k1162135












    • $begingroup$
      Two questions. Why are we considering $a+re^{it}$? What is $a$ here? And, how do we get the supremum $Mr^{5/2}$?
      $endgroup$
      – Ya G
      Dec 21 '18 at 20:24










    • $begingroup$
      @YaG That's three questions! $a+re^it$ is the circle, centre $a$ radius $r$. $a=0$ in your problem. The supremum comes from your condition on $f$.
      $endgroup$
      – Lord Shark the Unknown
      Dec 22 '18 at 5:56


















    • $begingroup$
      Two questions. Why are we considering $a+re^{it}$? What is $a$ here? And, how do we get the supremum $Mr^{5/2}$?
      $endgroup$
      – Ya G
      Dec 21 '18 at 20:24










    • $begingroup$
      @YaG That's three questions! $a+re^it$ is the circle, centre $a$ radius $r$. $a=0$ in your problem. The supremum comes from your condition on $f$.
      $endgroup$
      – Lord Shark the Unknown
      Dec 22 '18 at 5:56
















    $begingroup$
    Two questions. Why are we considering $a+re^{it}$? What is $a$ here? And, how do we get the supremum $Mr^{5/2}$?
    $endgroup$
    – Ya G
    Dec 21 '18 at 20:24




    $begingroup$
    Two questions. Why are we considering $a+re^{it}$? What is $a$ here? And, how do we get the supremum $Mr^{5/2}$?
    $endgroup$
    – Ya G
    Dec 21 '18 at 20:24












    $begingroup$
    @YaG That's three questions! $a+re^it$ is the circle, centre $a$ radius $r$. $a=0$ in your problem. The supremum comes from your condition on $f$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 22 '18 at 5:56




    $begingroup$
    @YaG That's three questions! $a+re^it$ is the circle, centre $a$ radius $r$. $a=0$ in your problem. The supremum comes from your condition on $f$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 22 '18 at 5:56











    7












    $begingroup$

    Let $f(z) = sum_{n=0}^infty a_n z^n$, and let $g(z) = sum_{k=0}^infty a_{k+3} z^k$, so that
    $$
    [f(z) - a_0 - a_1 z - a_2 z^2] = z^3 g(z).
    $$
    The function $g$ is entire and, by assumption, $lim_{|z| to +infty} g(z) = 0$:
    $$
    |g(z)| = frac{|f(z) - a_0 - a_1 z - a_2 z^2|}{|z|^{5/2}} cdot frac{1}{|z|^{1/2}}
    leq left(C + frac{|a_0| + |a_1|, |z| + |a_2|, |z|^2}{|z|^{5/2}}right)frac{1}{|z|^{1/2}} to 0.
    $$
    Hence, by Liouville's theorem, we must have $g = 0$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Would you please explain a little bit more? This does make sense by itself but how did you come up with the $g(z)$ other than the fact that to make a relation to $f(z)$ as you set it up. How is $|z|^{5/2}$ and $C$ used this this case?
      $endgroup$
      – Ya G
      Dec 21 '18 at 18:19






    • 1




      $begingroup$
      Added a line in the proof.
      $endgroup$
      – Rigel
      Dec 21 '18 at 18:24
















    7












    $begingroup$

    Let $f(z) = sum_{n=0}^infty a_n z^n$, and let $g(z) = sum_{k=0}^infty a_{k+3} z^k$, so that
    $$
    [f(z) - a_0 - a_1 z - a_2 z^2] = z^3 g(z).
    $$
    The function $g$ is entire and, by assumption, $lim_{|z| to +infty} g(z) = 0$:
    $$
    |g(z)| = frac{|f(z) - a_0 - a_1 z - a_2 z^2|}{|z|^{5/2}} cdot frac{1}{|z|^{1/2}}
    leq left(C + frac{|a_0| + |a_1|, |z| + |a_2|, |z|^2}{|z|^{5/2}}right)frac{1}{|z|^{1/2}} to 0.
    $$
    Hence, by Liouville's theorem, we must have $g = 0$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Would you please explain a little bit more? This does make sense by itself but how did you come up with the $g(z)$ other than the fact that to make a relation to $f(z)$ as you set it up. How is $|z|^{5/2}$ and $C$ used this this case?
      $endgroup$
      – Ya G
      Dec 21 '18 at 18:19






    • 1




      $begingroup$
      Added a line in the proof.
      $endgroup$
      – Rigel
      Dec 21 '18 at 18:24














    7












    7








    7





    $begingroup$

    Let $f(z) = sum_{n=0}^infty a_n z^n$, and let $g(z) = sum_{k=0}^infty a_{k+3} z^k$, so that
    $$
    [f(z) - a_0 - a_1 z - a_2 z^2] = z^3 g(z).
    $$
    The function $g$ is entire and, by assumption, $lim_{|z| to +infty} g(z) = 0$:
    $$
    |g(z)| = frac{|f(z) - a_0 - a_1 z - a_2 z^2|}{|z|^{5/2}} cdot frac{1}{|z|^{1/2}}
    leq left(C + frac{|a_0| + |a_1|, |z| + |a_2|, |z|^2}{|z|^{5/2}}right)frac{1}{|z|^{1/2}} to 0.
    $$
    Hence, by Liouville's theorem, we must have $g = 0$.






    share|cite|improve this answer











    $endgroup$



    Let $f(z) = sum_{n=0}^infty a_n z^n$, and let $g(z) = sum_{k=0}^infty a_{k+3} z^k$, so that
    $$
    [f(z) - a_0 - a_1 z - a_2 z^2] = z^3 g(z).
    $$
    The function $g$ is entire and, by assumption, $lim_{|z| to +infty} g(z) = 0$:
    $$
    |g(z)| = frac{|f(z) - a_0 - a_1 z - a_2 z^2|}{|z|^{5/2}} cdot frac{1}{|z|^{1/2}}
    leq left(C + frac{|a_0| + |a_1|, |z| + |a_2|, |z|^2}{|z|^{5/2}}right)frac{1}{|z|^{1/2}} to 0.
    $$
    Hence, by Liouville's theorem, we must have $g = 0$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 21 '18 at 18:24

























    answered Dec 21 '18 at 18:03









    RigelRigel

    11.4k11320




    11.4k11320












    • $begingroup$
      Would you please explain a little bit more? This does make sense by itself but how did you come up with the $g(z)$ other than the fact that to make a relation to $f(z)$ as you set it up. How is $|z|^{5/2}$ and $C$ used this this case?
      $endgroup$
      – Ya G
      Dec 21 '18 at 18:19






    • 1




      $begingroup$
      Added a line in the proof.
      $endgroup$
      – Rigel
      Dec 21 '18 at 18:24


















    • $begingroup$
      Would you please explain a little bit more? This does make sense by itself but how did you come up with the $g(z)$ other than the fact that to make a relation to $f(z)$ as you set it up. How is $|z|^{5/2}$ and $C$ used this this case?
      $endgroup$
      – Ya G
      Dec 21 '18 at 18:19






    • 1




      $begingroup$
      Added a line in the proof.
      $endgroup$
      – Rigel
      Dec 21 '18 at 18:24
















    $begingroup$
    Would you please explain a little bit more? This does make sense by itself but how did you come up with the $g(z)$ other than the fact that to make a relation to $f(z)$ as you set it up. How is $|z|^{5/2}$ and $C$ used this this case?
    $endgroup$
    – Ya G
    Dec 21 '18 at 18:19




    $begingroup$
    Would you please explain a little bit more? This does make sense by itself but how did you come up with the $g(z)$ other than the fact that to make a relation to $f(z)$ as you set it up. How is $|z|^{5/2}$ and $C$ used this this case?
    $endgroup$
    – Ya G
    Dec 21 '18 at 18:19




    1




    1




    $begingroup$
    Added a line in the proof.
    $endgroup$
    – Rigel
    Dec 21 '18 at 18:24




    $begingroup$
    Added a line in the proof.
    $endgroup$
    – Rigel
    Dec 21 '18 at 18:24











    1












    $begingroup$

    It may be interesting to note this follows from basic facts about Fourier series (Parseval) with more or less no complex analysis. Say $f(z)=sum c_nz^n$. Then $$C^2r^5gefrac1{2pi}int_0^{2pi}|f(re^{it})|^2,dt=sum_j|c_j|^2r^{2j}ge|c_n|^2r^{2n};$$hence $c_n=0$ for $nge3$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      A lot of complex analysis follows from basic facts about Fourier series.
      $endgroup$
      – Lord Shark the Unknown
      Dec 21 '18 at 18:55
















    1












    $begingroup$

    It may be interesting to note this follows from basic facts about Fourier series (Parseval) with more or less no complex analysis. Say $f(z)=sum c_nz^n$. Then $$C^2r^5gefrac1{2pi}int_0^{2pi}|f(re^{it})|^2,dt=sum_j|c_j|^2r^{2j}ge|c_n|^2r^{2n};$$hence $c_n=0$ for $nge3$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      A lot of complex analysis follows from basic facts about Fourier series.
      $endgroup$
      – Lord Shark the Unknown
      Dec 21 '18 at 18:55














    1












    1








    1





    $begingroup$

    It may be interesting to note this follows from basic facts about Fourier series (Parseval) with more or less no complex analysis. Say $f(z)=sum c_nz^n$. Then $$C^2r^5gefrac1{2pi}int_0^{2pi}|f(re^{it})|^2,dt=sum_j|c_j|^2r^{2j}ge|c_n|^2r^{2n};$$hence $c_n=0$ for $nge3$.






    share|cite|improve this answer









    $endgroup$



    It may be interesting to note this follows from basic facts about Fourier series (Parseval) with more or less no complex analysis. Say $f(z)=sum c_nz^n$. Then $$C^2r^5gefrac1{2pi}int_0^{2pi}|f(re^{it})|^2,dt=sum_j|c_j|^2r^{2j}ge|c_n|^2r^{2n};$$hence $c_n=0$ for $nge3$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 21 '18 at 18:16









    David C. UllrichDavid C. Ullrich

    61.6k43995




    61.6k43995












    • $begingroup$
      A lot of complex analysis follows from basic facts about Fourier series.
      $endgroup$
      – Lord Shark the Unknown
      Dec 21 '18 at 18:55


















    • $begingroup$
      A lot of complex analysis follows from basic facts about Fourier series.
      $endgroup$
      – Lord Shark the Unknown
      Dec 21 '18 at 18:55
















    $begingroup$
    A lot of complex analysis follows from basic facts about Fourier series.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 18:55




    $begingroup$
    A lot of complex analysis follows from basic facts about Fourier series.
    $endgroup$
    – Lord Shark the Unknown
    Dec 21 '18 at 18:55











    0












    $begingroup$

    Your approach can also be made to work, but you need to show that $h(z)=frac{f(z)}{z^3}$ has a removable singularity at $z=0$.



    This can be done as follows:
    Let $g(z)=frac{f(z)}{z^2}$. Then, as $lim_{z to 0}g(z)=0$, $g$ has a removable singularity at $z=0$, and if we remove it we have $g(0)=0$.



    Now since, after removing the singularity, $g$ is entire and $g(0)=0$, then $h(z)=frac{g(z)}{z}$ also has a removable singularity at $0$.



    Then $h$ becomes an entire function and
    $$lim_{z to infty} h(z)=0$$
    From here it is easy to conclude that $h$ is constant.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Your approach can also be made to work, but you need to show that $h(z)=frac{f(z)}{z^3}$ has a removable singularity at $z=0$.



      This can be done as follows:
      Let $g(z)=frac{f(z)}{z^2}$. Then, as $lim_{z to 0}g(z)=0$, $g$ has a removable singularity at $z=0$, and if we remove it we have $g(0)=0$.



      Now since, after removing the singularity, $g$ is entire and $g(0)=0$, then $h(z)=frac{g(z)}{z}$ also has a removable singularity at $0$.



      Then $h$ becomes an entire function and
      $$lim_{z to infty} h(z)=0$$
      From here it is easy to conclude that $h$ is constant.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Your approach can also be made to work, but you need to show that $h(z)=frac{f(z)}{z^3}$ has a removable singularity at $z=0$.



        This can be done as follows:
        Let $g(z)=frac{f(z)}{z^2}$. Then, as $lim_{z to 0}g(z)=0$, $g$ has a removable singularity at $z=0$, and if we remove it we have $g(0)=0$.



        Now since, after removing the singularity, $g$ is entire and $g(0)=0$, then $h(z)=frac{g(z)}{z}$ also has a removable singularity at $0$.



        Then $h$ becomes an entire function and
        $$lim_{z to infty} h(z)=0$$
        From here it is easy to conclude that $h$ is constant.






        share|cite|improve this answer









        $endgroup$



        Your approach can also be made to work, but you need to show that $h(z)=frac{f(z)}{z^3}$ has a removable singularity at $z=0$.



        This can be done as follows:
        Let $g(z)=frac{f(z)}{z^2}$. Then, as $lim_{z to 0}g(z)=0$, $g$ has a removable singularity at $z=0$, and if we remove it we have $g(0)=0$.



        Now since, after removing the singularity, $g$ is entire and $g(0)=0$, then $h(z)=frac{g(z)}{z}$ also has a removable singularity at $0$.



        Then $h$ becomes an entire function and
        $$lim_{z to infty} h(z)=0$$
        From here it is easy to conclude that $h$ is constant.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 21 '18 at 19:14









        N. S.N. S.

        105k7114210




        105k7114210






























            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3048745%2fif-f-is-entire-such-that-fz-leq-cz5-2-f-is-polynomial-of-degree%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

            Bundesstraße 106

            Image
            Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück ...
            Read more

            Verónica Boquete

            Image
            Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl ...
            Read more

            Ida-Boy-Ed-Garten

            Image
            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
            Read more