Fdtxjr

Can I board the first leg of the flight without having final country's visa?

Why don't programming languages automatically manage the synchronous/asynchronous problem?

Players Circumventing the limitations of Wish

"Eavesdropping" vs "Listen in on"

How to find image of a complex function with given constraints?

Film where the government was corrupt with aliens, people sent to kill aliens are given rigged visors not showing the right aliens

What difference does it make using sed with/without whitespaces?

Is French Guiana a (hard) EU border?

Spaces in which all closed sets are regular closed

Is it professional to write unrelated content in an almost-empty email?

Can this note be analyzed as a non-chord tone?

Yu-Gi-Oh cards in Python 3

Does higher Oxidation/ reduction potential translate to higher energy storage in battery?

(How) Could a medieval fantasy world survive a magic-induced "nuclear winter"?

Expressing the idea of having a very busy time

From jafe to El-Guest

What would be the main consequences for a country leaving the WTO?

Can someone explain this formula for calculating Manhattan distance?

Where do students learn to solve polynomial equations these days?

How to get the last not-null value in an ordered column of a huge table?

What flight has the highest ratio of timezone difference to flight time?

Aggressive Under-Indexing and no data for missing index

Redefining symbol midway through a document

Calculate the Mean mean of two numbers

Examples of $f:mathbbC rightarrow mathbbC$ entire function with $|f(z)| leq C|z|^2$ $forall z in mathbbC$

0

$begingroup$

Find the most general function $f: mathbbC rightarrow mathbbC$
such that $f$ is entire and $exists C > 0$ with $|f(z)| leq C|z|^2$
$forall z in mathbbC$.

I'm really not sure where to start with this problem.

complex-analysis entire-functions

share|cite|improve this question

edited Mar 19 at 20:01

Joseph Martin

688317

asked Mar 19 at 19:04

Amanda LococoAmanda Lococo

61

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have a look at this example and apply the same strategy.
  $endgroup$
  – rtybase
  Mar 19 at 20:33
  

  
  
  
  

add a comment | 

0

$begingroup$

Find the most general function $f: mathbbC rightarrow mathbbC$
such that $f$ is entire and $exists C > 0$ with $|f(z)| leq C|z|^2$
$forall z in mathbbC$.

I'm really not sure where to start with this problem.

complex-analysis entire-functions

share|cite|improve this question

edited Mar 19 at 20:01

Joseph Martin

688317

asked Mar 19 at 19:04

Amanda LococoAmanda Lococo

61

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have a look at this example and apply the same strategy.
  $endgroup$
  – rtybase
  Mar 19 at 20:33
  

  
  
  
  

add a comment | 

$begingroup$

Find the most general function $f: mathbbC rightarrow mathbbC$
such that $f$ is entire and $exists C > 0$ with $|f(z)| leq C|z|^2$
$forall z in mathbbC$.

I'm really not sure where to start with this problem.

complex-analysis entire-functions

share|cite|improve this question

edited Mar 19 at 20:01

Joseph Martin

688317

asked Mar 19 at 19:04

Amanda LococoAmanda Lococo

61

$endgroup$

share|cite|improve this question

edited Mar 19 at 20:01

Joseph Martin

688317

asked Mar 19 at 19:04

Amanda LococoAmanda Lococo

61

share|cite|improve this question

edited Mar 19 at 20:01

Joseph Martin

688317

asked Mar 19 at 19:04

Amanda LococoAmanda Lococo

61

edited Mar 19 at 20:01

Joseph Martin

688317

edited Mar 19 at 20:01

Joseph Martin

688317

asked Mar 19 at 19:04

Amanda LococoAmanda Lococo

61

asked Mar 19 at 19:04

Amanda LococoAmanda Lococo

61

3 Answers
3

active

oldest

votes

0

$begingroup$

Hint: Use the Cauchy integral formula on disks $D_R(0)$ and send $R to infty$ (this is fine since $f$ is entire). Doing so, you should be able to show that, above some order, the derivatives of $f$ vanish.

share|cite|improve this answer

answered Mar 19 at 19:13

Gary MoonGary Moon

92127

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We have not learned how to use the Cauchy integral formula on disks. Is there another way to do it?
  $endgroup$
  – Amanda Lococo
  Mar 19 at 19:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What tools do you have at your disposal?
  $endgroup$
  – Gary Moon
  Mar 19 at 19:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Cauchy integral test for complex functions in one dimension, Weierstrauss M-test, Lipschitz functions, uniform convergence, ratio test, domain of convergence, points of singularity. That's all we have covered so far in the course
  $endgroup$
  – Amanda Lococo
  Mar 19 at 19:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you talked about power series for holomorphic functions? If so, since $f$ is entire, it is everywhere equal to its power series centered at $z=0$. You could write $f(z) = sum_n=0^infty a_n z^n$ and go from there.
  $endgroup$
  – Gary Moon
  Mar 19 at 19:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We are talking about that next, but is not something we covered for this current exam, which could contain this question. The only other consideration is perhaps this practice exam was made with the intent we would have earned holomorphisms by now?
  $endgroup$
  – Amanda Lococo
  Mar 19 at 19:27
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

Hint: Observe that $f(z)/z^2$ is a bounded, entire function (can you see why it's holomorphic at $0$?)

share|cite|improve this answer

answered Mar 19 at 19:43

WojowuWojowu

19.3k23174

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see why it is holomorphic at 0, but still don't understand how to find the most general function that satisfies the equation. Can I have some more guidance?
  $endgroup$
  – Amanda Lococo
  Mar 19 at 20:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My intended solution proceeded by applying the Liouville's theorem to $f(z)/z^2$, but from your comments under the other answer it appears you have not covered that yet. Indeed, it seems that you have covered basically nothing regarding holomorphic functions, so I must say I have no idea how you could approach this question.
  $endgroup$
  – Wojowu
  Mar 19 at 20:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was mistaken underneath. We have been using the term analytic, not holomorphic, and I did not realize it was the same since it has never been introduced to me that way
  $endgroup$
  – Amanda Lococo
  Mar 19 at 20:50
  

  
  
  
  

add a comment | 

0

$begingroup$

As $f(z)$ is entire, it can be expressed as a Taylor's series $$f(z)=sum_n=0^infty a_nz^n$$therefore the most general form comes from $a_n=0$ for $n>2$ and $a_0=a_1=0$ so we have $$f(z)=a_2z^2$$

share|cite|improve this answer

answered Mar 19 at 23:53

Mostafa AyazMostafa Ayaz

18.3k31040

$endgroup$

add a comment | 

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

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3154467%2fexamples-of-f-mathbbc-rightarrow-mathbbc-entire-function-with-fz%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

0

$begingroup$

Hint: Use the Cauchy integral formula on disks $D_R(0)$ and send $R to infty$ (this is fine since $f$ is entire). Doing so, you should be able to show that, above some order, the derivatives of $f$ vanish.

share|cite|improve this answer

answered Mar 19 at 19:13

Gary MoonGary Moon

92127

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We have not learned how to use the Cauchy integral formula on disks. Is there another way to do it?
  $endgroup$
  – Amanda Lococo
  Mar 19 at 19:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What tools do you have at your disposal?
  $endgroup$
  – Gary Moon
  Mar 19 at 19:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Cauchy integral test for complex functions in one dimension, Weierstrauss M-test, Lipschitz functions, uniform convergence, ratio test, domain of convergence, points of singularity. That's all we have covered so far in the course
  $endgroup$
  – Amanda Lococo
  Mar 19 at 19:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you talked about power series for holomorphic functions? If so, since $f$ is entire, it is everywhere equal to its power series centered at $z=0$. You could write $f(z) = sum_n=0^infty a_n z^n$ and go from there.
  $endgroup$
  – Gary Moon
  Mar 19 at 19:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We are talking about that next, but is not something we covered for this current exam, which could contain this question. The only other consideration is perhaps this practice exam was made with the intent we would have earned holomorphisms by now?
  $endgroup$
  – Amanda Lococo
  Mar 19 at 19:27
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

Hint: Use the Cauchy integral formula on disks $D_R(0)$ and send $R to infty$ (this is fine since $f$ is entire). Doing so, you should be able to show that, above some order, the derivatives of $f$ vanish.

share|cite|improve this answer

answered Mar 19 at 19:13

Gary MoonGary Moon

92127

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We have not learned how to use the Cauchy integral formula on disks. Is there another way to do it?
  $endgroup$
  – Amanda Lococo
  Mar 19 at 19:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What tools do you have at your disposal?
  $endgroup$
  – Gary Moon
  Mar 19 at 19:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Cauchy integral test for complex functions in one dimension, Weierstrauss M-test, Lipschitz functions, uniform convergence, ratio test, domain of convergence, points of singularity. That's all we have covered so far in the course
  $endgroup$
  – Amanda Lococo
  Mar 19 at 19:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you talked about power series for holomorphic functions? If so, since $f$ is entire, it is everywhere equal to its power series centered at $z=0$. You could write $f(z) = sum_n=0^infty a_n z^n$ and go from there.
  $endgroup$
  – Gary Moon
  Mar 19 at 19:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We are talking about that next, but is not something we covered for this current exam, which could contain this question. The only other consideration is perhaps this practice exam was made with the intent we would have earned holomorphisms by now?
  $endgroup$
  – Amanda Lococo
  Mar 19 at 19:27
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Hint: Use the Cauchy integral formula on disks $D_R(0)$ and send $R to infty$ (this is fine since $f$ is entire). Doing so, you should be able to show that, above some order, the derivatives of $f$ vanish.

share|cite|improve this answer

answered Mar 19 at 19:13

Gary MoonGary Moon

92127

$endgroup$

share|cite|improve this answer

answered Mar 19 at 19:13

Gary MoonGary Moon

92127

answered Mar 19 at 19:13

Gary MoonGary Moon

92127

answered Mar 19 at 19:13

Gary MoonGary Moon

92127

0

$begingroup$

Hint: Observe that $f(z)/z^2$ is a bounded, entire function (can you see why it's holomorphic at $0$?)

share|cite|improve this answer

answered Mar 19 at 19:43

WojowuWojowu

19.3k23174

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see why it is holomorphic at 0, but still don't understand how to find the most general function that satisfies the equation. Can I have some more guidance?
  $endgroup$
  – Amanda Lococo
  Mar 19 at 20:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My intended solution proceeded by applying the Liouville's theorem to $f(z)/z^2$, but from your comments under the other answer it appears you have not covered that yet. Indeed, it seems that you have covered basically nothing regarding holomorphic functions, so I must say I have no idea how you could approach this question.
  $endgroup$
  – Wojowu
  Mar 19 at 20:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was mistaken underneath. We have been using the term analytic, not holomorphic, and I did not realize it was the same since it has never been introduced to me that way
  $endgroup$
  – Amanda Lococo
  Mar 19 at 20:50
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint: Observe that $f(z)/z^2$ is a bounded, entire function (can you see why it's holomorphic at $0$?)

share|cite|improve this answer

answered Mar 19 at 19:43

WojowuWojowu

19.3k23174

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see why it is holomorphic at 0, but still don't understand how to find the most general function that satisfies the equation. Can I have some more guidance?
  $endgroup$
  – Amanda Lococo
  Mar 19 at 20:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My intended solution proceeded by applying the Liouville's theorem to $f(z)/z^2$, but from your comments under the other answer it appears you have not covered that yet. Indeed, it seems that you have covered basically nothing regarding holomorphic functions, so I must say I have no idea how you could approach this question.
  $endgroup$
  – Wojowu
  Mar 19 at 20:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was mistaken underneath. We have been using the term analytic, not holomorphic, and I did not realize it was the same since it has never been introduced to me that way
  $endgroup$
  – Amanda Lococo
  Mar 19 at 20:50
  

  
  
  
  

add a comment | 

$begingroup$

Hint: Observe that $f(z)/z^2$ is a bounded, entire function (can you see why it's holomorphic at $0$?)

share|cite|improve this answer

answered Mar 19 at 19:43

WojowuWojowu

19.3k23174

$endgroup$

share|cite|improve this answer

answered Mar 19 at 19:43

WojowuWojowu

19.3k23174

answered Mar 19 at 19:43

WojowuWojowu

19.3k23174

answered Mar 19 at 19:43

WojowuWojowu

19.3k23174

0

$begingroup$

As $f(z)$ is entire, it can be expressed as a Taylor's series $$f(z)=sum_n=0^infty a_nz^n$$therefore the most general form comes from $a_n=0$ for $n>2$ and $a_0=a_1=0$ so we have $$f(z)=a_2z^2$$

share|cite|improve this answer

answered Mar 19 at 23:53

Mostafa AyazMostafa Ayaz

18.3k31040

$endgroup$

add a comment | 

0

$begingroup$

As $f(z)$ is entire, it can be expressed as a Taylor's series $$f(z)=sum_n=0^infty a_nz^n$$therefore the most general form comes from $a_n=0$ for $n>2$ and $a_0=a_1=0$ so we have $$f(z)=a_2z^2$$

share|cite|improve this answer

answered Mar 19 at 23:53

Mostafa AyazMostafa Ayaz

18.3k31040

$endgroup$

add a comment | 

$begingroup$

As $f(z)$ is entire, it can be expressed as a Taylor's series $$f(z)=sum_n=0^infty a_nz^n$$therefore the most general form comes from $a_n=0$ for $n>2$ and $a_0=a_1=0$ so we have $$f(z)=a_2z^2$$

share|cite|improve this answer

answered Mar 19 at 23:53

Mostafa AyazMostafa Ayaz

18.3k31040

$endgroup$

share|cite|improve this answer

answered Mar 19 at 23:53

Mostafa AyazMostafa Ayaz

18.3k31040

answered Mar 19 at 23:53

Mostafa AyazMostafa Ayaz

18.3k31040

answered Mar 19 at 23:53

Mostafa AyazMostafa Ayaz

18.3k31040