An even function, $f(z)$, analytic near 0 can be written as another analytic function, $h(z^2)$
$begingroup$
If f is an even function, $f(z) = f(-z)$, and is analytic near 0, then there exists a function h, also analytic near 0, such that $f(z) = h(z^2)$
I suspect this statement is true because the analyticity of f near 0 allows us to write it as a taylor expansion around 0 and since it is even we have that
$$f(z) = sum a_n z^n = sum a_n (-z)^n$$
which implies that a_n does not depend on z and we can write
$h(z^2) = sum a_n z^{2n}$ but I am not sure where to go from here. Any hints are appreciated.
complex-analysis power-series taylor-expansion analyticity even-and-odd-functions
edited Dec 10 '18 at 22:14
José Carlos Santos
158k22126228
asked Dec 10 '18 at 21:48
Richard VillalobosRichard Villalobos
1687
$endgroup$
add a comment |
$begingroup$
If f is an even function, $f(z) = f(-z)$, and is analytic near 0, then there exists a function h, also analytic near 0, such that $f(z) = h(z^2)$
I suspect this statement is true because the analyticity of f near 0 allows us to write it as a taylor expansion around 0 and since it is even we have that
$$f(z) = sum a_n z^n = sum a_n (-z)^n$$
which implies that a_n does not depend on z and we can write
$h(z^2) = sum a_n z^{2n}$ but I am not sure where to go from here. Any hints are appreciated.
complex-analysis power-series taylor-expansion analyticity even-and-odd-functions
edited Dec 10 '18 at 22:14
José Carlos Santos
158k22126228
asked Dec 10 '18 at 21:48
Richard VillalobosRichard Villalobos
1687
$endgroup$
add a comment |
$begingroup$
If f is an even function, $f(z) = f(-z)$, and is analytic near 0, then there exists a function h, also analytic near 0, such that $f(z) = h(z^2)$
I suspect this statement is true because the analyticity of f near 0 allows us to write it as a taylor expansion around 0 and since it is even we have that
$$f(z) = sum a_n z^n = sum a_n (-z)^n$$
which implies that a_n does not depend on z and we can write
$h(z^2) = sum a_n z^{2n}$ but I am not sure where to go from here. Any hints are appreciated.
complex-analysis power-series taylor-expansion analyticity even-and-odd-functions
edited Dec 10 '18 at 22:14
José Carlos Santos
158k22126228
asked Dec 10 '18 at 21:48
Richard VillalobosRichard Villalobos
1687
$endgroup$
If f is an even function, $f(z) = f(-z)$, and is analytic near 0, then there exists a function h, also analytic near 0, such that $f(z) = h(z^2)$
I suspect this statement is true because the analyticity of f near 0 allows us to write it as a taylor expansion around 0 and since it is even we have that
$$f(z) = sum a_n z^n = sum a_n (-z)^n$$
which implies that a_n does not depend on z and we can write
$h(z^2) = sum a_n z^{2n}$ but I am not sure where to go from here. Any hints are appreciated.
complex-analysis power-series taylor-expansion analyticity even-and-odd-functions
complex-analysis power-series taylor-expansion analyticity even-and-odd-functions
edited Dec 10 '18 at 22:14
José Carlos Santos
158k22126228
asked Dec 10 '18 at 21:48
Richard VillalobosRichard Villalobos
1687
edited Dec 10 '18 at 22:14
José Carlos Santos
158k22126228
asked Dec 10 '18 at 21:48
Richard VillalobosRichard Villalobos
1687
edited Dec 10 '18 at 22:14
José Carlos Santos
158k22126228
edited Dec 10 '18 at 22:14
José Carlos Santos
158k22126228
edited Dec 10 '18 at 22:14
José Carlos Santos
158k22126228
158k22126228
asked Dec 10 '18 at 21:48
Richard VillalobosRichard Villalobos
1687
asked Dec 10 '18 at 21:48
Richard VillalobosRichard Villalobos
1687
asked Dec 10 '18 at 21:48
Richard VillalobosRichard Villalobos
1687
1687
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It follows from your equality$$sum_{n=0}^infty a_nz^n=sum_{n=0}^infty a_n(-z)^n$$than $(forall ninmathbb{Z}_+):a_n=(-1)^na_n$. Therefore, $a_n=0$ if $n$ is odd and so$$f(z)=sum_{n=0}^infty a_{2n}z^{2n}=sum_{n=0}^infty a_{2n}(z^2)^n.$$So, define$$h(z)=sum_{n=0}^infty a_{2n}z^n.$$
answered Dec 10 '18 at 21:54
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
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%2f3034534%2fan-even-function-fz-analytic-near-0-can-be-written-as-another-analytic-fun%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
$begingroup$
It follows from your equality$$sum_{n=0}^infty a_nz^n=sum_{n=0}^infty a_n(-z)^n$$than $(forall ninmathbb{Z}_+):a_n=(-1)^na_n$. Therefore, $a_n=0$ if $n$ is odd and so$$f(z)=sum_{n=0}^infty a_{2n}z^{2n}=sum_{n=0}^infty a_{2n}(z^2)^n.$$So, define$$h(z)=sum_{n=0}^infty a_{2n}z^n.$$
answered Dec 10 '18 at 21:54
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
add a comment |
$begingroup$
It follows from your equality$$sum_{n=0}^infty a_nz^n=sum_{n=0}^infty a_n(-z)^n$$than $(forall ninmathbb{Z}_+):a_n=(-1)^na_n$. Therefore, $a_n=0$ if $n$ is odd and so$$f(z)=sum_{n=0}^infty a_{2n}z^{2n}=sum_{n=0}^infty a_{2n}(z^2)^n.$$So, define$$h(z)=sum_{n=0}^infty a_{2n}z^n.$$
answered Dec 10 '18 at 21:54
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
add a comment |
$begingroup$
It follows from your equality$$sum_{n=0}^infty a_nz^n=sum_{n=0}^infty a_n(-z)^n$$than $(forall ninmathbb{Z}_+):a_n=(-1)^na_n$. Therefore, $a_n=0$ if $n$ is odd and so$$f(z)=sum_{n=0}^infty a_{2n}z^{2n}=sum_{n=0}^infty a_{2n}(z^2)^n.$$So, define$$h(z)=sum_{n=0}^infty a_{2n}z^n.$$
answered Dec 10 '18 at 21:54
José Carlos SantosJosé Carlos Santos
158k22126228
$endgroup$
It follows from your equality$$sum_{n=0}^infty a_nz^n=sum_{n=0}^infty a_n(-z)^n$$than $(forall ninmathbb{Z}_+):a_n=(-1)^na_n$. Therefore, $a_n=0$ if $n$ is odd and so$$f(z)=sum_{n=0}^infty a_{2n}z^{2n}=sum_{n=0}^infty a_{2n}(z^2)^n.$$So, define$$h(z)=sum_{n=0}^infty a_{2n}z^n.$$
answered Dec 10 '18 at 21:54
José Carlos SantosJosé Carlos Santos
158k22126228
answered Dec 10 '18 at 21:54
José Carlos SantosJosé Carlos Santos
158k22126228
answered Dec 10 '18 at 21:54
José Carlos SantosJosé Carlos Santos
158k22126228
answered Dec 10 '18 at 21:54
José Carlos SantosJosé Carlos Santos
158k22126228
158k22126228
add a comment |
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%2f3034534%2fan-even-function-fz-analytic-near-0-can-be-written-as-another-analytic-fun%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
