If $Re(f)$ is a polynomial then $f$ is a polynomial, where$f$ is entire
$begingroup$
If $f$ is entire, and $Re(f)$ is a polynomial in x,y I am trying to show that f is a polynomial in z.
We can write $f(z) = u(z) + iv(z)$. where $Re(f) = u(z)$
I have solved problems similar to this one before, except usually the real part is bounded which allows me to use Liouville's theorem, here since there is no such condition I am unsure of how to proceed.
I also know that since v is the harmonic conjugate of u, it is uniquely determined up to addition of a constant, I believe this fact is helpful but not sure how to begin with this.
complex-analysis analytic-functions entire-functions
asked Dec 10 '18 at 23:41
Richard VillalobosRichard Villalobos
1687
$endgroup$
add a comment |
$begingroup$
If $f$ is entire, and $Re(f)$ is a polynomial in x,y I am trying to show that f is a polynomial in z.
We can write $f(z) = u(z) + iv(z)$. where $Re(f) = u(z)$
I have solved problems similar to this one before, except usually the real part is bounded which allows me to use Liouville's theorem, here since there is no such condition I am unsure of how to proceed.
I also know that since v is the harmonic conjugate of u, it is uniquely determined up to addition of a constant, I believe this fact is helpful but not sure how to begin with this.
complex-analysis analytic-functions entire-functions
asked Dec 10 '18 at 23:41
Richard VillalobosRichard Villalobos
1687
$endgroup$
add a comment |
$begingroup$
If $f$ is entire, and $Re(f)$ is a polynomial in x,y I am trying to show that f is a polynomial in z.
We can write $f(z) = u(z) + iv(z)$. where $Re(f) = u(z)$
I have solved problems similar to this one before, except usually the real part is bounded which allows me to use Liouville's theorem, here since there is no such condition I am unsure of how to proceed.
I also know that since v is the harmonic conjugate of u, it is uniquely determined up to addition of a constant, I believe this fact is helpful but not sure how to begin with this.
complex-analysis analytic-functions entire-functions
asked Dec 10 '18 at 23:41
Richard VillalobosRichard Villalobos
1687
$endgroup$
If $f$ is entire, and $Re(f)$ is a polynomial in x,y I am trying to show that f is a polynomial in z.
We can write $f(z) = u(z) + iv(z)$. where $Re(f) = u(z)$
I have solved problems similar to this one before, except usually the real part is bounded which allows me to use Liouville's theorem, here since there is no such condition I am unsure of how to proceed.
I also know that since v is the harmonic conjugate of u, it is uniquely determined up to addition of a constant, I believe this fact is helpful but not sure how to begin with this.
complex-analysis analytic-functions entire-functions
complex-analysis analytic-functions entire-functions
asked Dec 10 '18 at 23:41
Richard VillalobosRichard Villalobos
1687
asked Dec 10 '18 at 23:41
Richard VillalobosRichard Villalobos
1687
asked Dec 10 '18 at 23:41
Richard VillalobosRichard Villalobos
1687
asked Dec 10 '18 at 23:41
Richard VillalobosRichard Villalobos
1687
asked Dec 10 '18 at 23:41
Richard VillalobosRichard Villalobos
1687
1687
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$f$ can be written as power series $f(z) = sum_{n=0}^infty a_n z^n$ with radius of convergence = $infty$. If $f$ is not a polynomial, then neccessarily infinitely many $a_n ne 0$. In that case $Re(f)$ is not a polynomial.
answered Dec 10 '18 at 23:52
Paul FrostPaul Frost
10.3k3933
$endgroup$
1
$begingroup$
"In that case Re(f) is not a polynomial." This maybe so, but why?
$endgroup$
– Did
Dec 11 '18 at 0:07
2
$begingroup$
Write $a_n = r_n + i s_n$, $r_n, s_n in mathbb{R}$. If $a_n ne 0$, then $r_n ne 0$ or $s_n ne 0$. But then $Re(f)$ contains a term $r_n x^n$ and a term $-n s_n x^{n-1}y$. Therefore, if infinitely many $a_n ne 0$, then $Re(f)$ is obviously not a polynomial.
$endgroup$
– Paul Frost
Dec 11 '18 at 0:16
add a comment |
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
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3034658%2fif-ref-is-a-polynomial-then-f-is-a-polynomial-wheref-is-entire%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
$begingroup$
$f$ can be written as power series $f(z) = sum_{n=0}^infty a_n z^n$ with radius of convergence = $infty$. If $f$ is not a polynomial, then neccessarily infinitely many $a_n ne 0$. In that case $Re(f)$ is not a polynomial.
answered Dec 10 '18 at 23:52
Paul FrostPaul Frost
10.3k3933
$endgroup$
1
$begingroup$
"In that case Re(f) is not a polynomial." This maybe so, but why?
$endgroup$
– Did
Dec 11 '18 at 0:07
2
$begingroup$
Write $a_n = r_n + i s_n$, $r_n, s_n in mathbb{R}$. If $a_n ne 0$, then $r_n ne 0$ or $s_n ne 0$. But then $Re(f)$ contains a term $r_n x^n$ and a term $-n s_n x^{n-1}y$. Therefore, if infinitely many $a_n ne 0$, then $Re(f)$ is obviously not a polynomial.
$endgroup$
– Paul Frost
Dec 11 '18 at 0:16
add a comment |
$begingroup$
$f$ can be written as power series $f(z) = sum_{n=0}^infty a_n z^n$ with radius of convergence = $infty$. If $f$ is not a polynomial, then neccessarily infinitely many $a_n ne 0$. In that case $Re(f)$ is not a polynomial.
answered Dec 10 '18 at 23:52
Paul FrostPaul Frost
10.3k3933
$endgroup$
1
$begingroup$
"In that case Re(f) is not a polynomial." This maybe so, but why?
$endgroup$
– Did
Dec 11 '18 at 0:07
2
$begingroup$
Write $a_n = r_n + i s_n$, $r_n, s_n in mathbb{R}$. If $a_n ne 0$, then $r_n ne 0$ or $s_n ne 0$. But then $Re(f)$ contains a term $r_n x^n$ and a term $-n s_n x^{n-1}y$. Therefore, if infinitely many $a_n ne 0$, then $Re(f)$ is obviously not a polynomial.
$endgroup$
– Paul Frost
Dec 11 '18 at 0:16
add a comment |
$begingroup$
$f$ can be written as power series $f(z) = sum_{n=0}^infty a_n z^n$ with radius of convergence = $infty$. If $f$ is not a polynomial, then neccessarily infinitely many $a_n ne 0$. In that case $Re(f)$ is not a polynomial.
answered Dec 10 '18 at 23:52
Paul FrostPaul Frost
10.3k3933
$endgroup$
$f$ can be written as power series $f(z) = sum_{n=0}^infty a_n z^n$ with radius of convergence = $infty$. If $f$ is not a polynomial, then neccessarily infinitely many $a_n ne 0$. In that case $Re(f)$ is not a polynomial.
answered Dec 10 '18 at 23:52
Paul FrostPaul Frost
10.3k3933
answered Dec 10 '18 at 23:52
Paul FrostPaul Frost
10.3k3933
answered Dec 10 '18 at 23:52
Paul FrostPaul Frost
10.3k3933
answered Dec 10 '18 at 23:52
Paul FrostPaul Frost
10.3k3933
10.3k3933
1
$begingroup$
"In that case Re(f) is not a polynomial." This maybe so, but why?
$endgroup$
– Did
Dec 11 '18 at 0:07
2
$begingroup$
Write $a_n = r_n + i s_n$, $r_n, s_n in mathbb{R}$. If $a_n ne 0$, then $r_n ne 0$ or $s_n ne 0$. But then $Re(f)$ contains a term $r_n x^n$ and a term $-n s_n x^{n-1}y$. Therefore, if infinitely many $a_n ne 0$, then $Re(f)$ is obviously not a polynomial.
$endgroup$
– Paul Frost
Dec 11 '18 at 0:16
add a comment |
1
$begingroup$
"In that case Re(f) is not a polynomial." This maybe so, but why?
$endgroup$
– Did
Dec 11 '18 at 0:07
2
$begingroup$
Write $a_n = r_n + i s_n$, $r_n, s_n in mathbb{R}$. If $a_n ne 0$, then $r_n ne 0$ or $s_n ne 0$. But then $Re(f)$ contains a term $r_n x^n$ and a term $-n s_n x^{n-1}y$. Therefore, if infinitely many $a_n ne 0$, then $Re(f)$ is obviously not a polynomial.
$endgroup$
– Paul Frost
Dec 11 '18 at 0:16
1
1
$begingroup$
"In that case Re(f) is not a polynomial." This maybe so, but why?
$endgroup$
– Did
Dec 11 '18 at 0:07
$begingroup$
"In that case Re(f) is not a polynomial." This maybe so, but why?
$endgroup$
– Did
Dec 11 '18 at 0:07
2
2
$begingroup$
Write $a_n = r_n + i s_n$, $r_n, s_n in mathbb{R}$. If $a_n ne 0$, then $r_n ne 0$ or $s_n ne 0$. But then $Re(f)$ contains a term $r_n x^n$ and a term $-n s_n x^{n-1}y$. Therefore, if infinitely many $a_n ne 0$, then $Re(f)$ is obviously not a polynomial.
$endgroup$
– Paul Frost
Dec 11 '18 at 0:16
$begingroup$
Write $a_n = r_n + i s_n$, $r_n, s_n in mathbb{R}$. If $a_n ne 0$, then $r_n ne 0$ or $s_n ne 0$. But then $Re(f)$ contains a term $r_n x^n$ and a term $-n s_n x^{n-1}y$. Therefore, if infinitely many $a_n ne 0$, then $Re(f)$ is obviously not a polynomial.
$endgroup$
– Paul Frost
Dec 11 '18 at 0:16
add a comment |
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.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3034658%2fif-ref-is-a-polynomial-then-f-is-a-polynomial-wheref-is-entire%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
