Where does a Laurent Series converge?
$begingroup$
From the Cauchy integral formula, I get
$f(z)=int_{C} frac{f(zeta)}{zeta -z}dzeta$
manipulating, =$int_{C} f(zeta)(frac{1}{zeta -p -(z-p)})$=$int_{C} f(zeta)(frac{1}{(zeta -p)(1 -frac{(z-p)}{zeta -p}})$=$int_{C} f(zeta) sum_{n=0}^{infty} (frac{z-p}{zeta -p})^n$ which is valid for $|frac{z-p}{zeta -p}|<1$.
Geometrically, I interpret this as the points closer to $z$ than the boundary of C
Or, I can choose to do a different power series.
$f(z)=int_{C} frac{f(zeta)}{zeta -z}dzeta$ = manipulating differently, =$int_{C} f(zeta)(frac{1}{zeta -p -(z-p)})$=$int_{C} f(zeta)(frac{1}{-(z -p)(1 -frac{(zeta-p)}{z -p}})$=$-int_{C} f(zeta) sum_{n=0}^{infty} (frac{zeta-p}{z -p})^n$ which is valid for $|frac{zeta-p}{z -p}|<1$.
Which I am then interpreting as the outside region of the circle
My question is, as you may have guessed, Where does the Laurent series given by
$f(z)=sum_{n=1}^{infty} a_n(z-p)^n +sum_{n=1}^{infty} frac{b_n}{(z-p)^n}$ converge? I do not understand where the annulus of convergence can be.
Thanks in advance.
complex-analysis
asked Nov 27 '18 at 4:07
JungleshrimpJungleshrimp
19311
$endgroup$
add a comment |
$begingroup$
From the Cauchy integral formula, I get
$f(z)=int_{C} frac{f(zeta)}{zeta -z}dzeta$
manipulating, =$int_{C} f(zeta)(frac{1}{zeta -p -(z-p)})$=$int_{C} f(zeta)(frac{1}{(zeta -p)(1 -frac{(z-p)}{zeta -p}})$=$int_{C} f(zeta) sum_{n=0}^{infty} (frac{z-p}{zeta -p})^n$ which is valid for $|frac{z-p}{zeta -p}|<1$.
Geometrically, I interpret this as the points closer to $z$ than the boundary of C
Or, I can choose to do a different power series.
$f(z)=int_{C} frac{f(zeta)}{zeta -z}dzeta$ = manipulating differently, =$int_{C} f(zeta)(frac{1}{zeta -p -(z-p)})$=$int_{C} f(zeta)(frac{1}{-(z -p)(1 -frac{(zeta-p)}{z -p}})$=$-int_{C} f(zeta) sum_{n=0}^{infty} (frac{zeta-p}{z -p})^n$ which is valid for $|frac{zeta-p}{z -p}|<1$.
Which I am then interpreting as the outside region of the circle
My question is, as you may have guessed, Where does the Laurent series given by
$f(z)=sum_{n=1}^{infty} a_n(z-p)^n +sum_{n=1}^{infty} frac{b_n}{(z-p)^n}$ converge? I do not understand where the annulus of convergence can be.
Thanks in advance.
complex-analysis
asked Nov 27 '18 at 4:07
JungleshrimpJungleshrimp
19311
$endgroup$
$begingroup$
a Laurent series converges between the point where it is constructed around (in your case $p$) and up to the closer singularity to the function that it represent. If the function doesn't have any singularity then the Laurent series is just a Taylor series with radius of convergence infinity
$endgroup$
– Masacroso
Nov 27 '18 at 4:17
$begingroup$
@Masacroso Here yes but in general the domain of convergence can be any anulus, eg. $sum_n (1/2)^{|n|} z^n$ converges for $|z| < 1/2, |1/z| < 1/2 = |z| in (1/2,2)$. The (open) annulus where $sum_n a_n z^n$ converges is $|z|in (1/r,R)$ where $R,r$ are the radius of convergence of $sum_{n ge 0} a_n z^n,sum_{n > 0} a_{-n} z^n$. There is a singularity on $|z| = R, |z| =1/r$
$endgroup$
– reuns
Nov 27 '18 at 4:25
add a comment |
$begingroup$
From the Cauchy integral formula, I get
$f(z)=int_{C} frac{f(zeta)}{zeta -z}dzeta$
manipulating, =$int_{C} f(zeta)(frac{1}{zeta -p -(z-p)})$=$int_{C} f(zeta)(frac{1}{(zeta -p)(1 -frac{(z-p)}{zeta -p}})$=$int_{C} f(zeta) sum_{n=0}^{infty} (frac{z-p}{zeta -p})^n$ which is valid for $|frac{z-p}{zeta -p}|<1$.
Geometrically, I interpret this as the points closer to $z$ than the boundary of C
Or, I can choose to do a different power series.
$f(z)=int_{C} frac{f(zeta)}{zeta -z}dzeta$ = manipulating differently, =$int_{C} f(zeta)(frac{1}{zeta -p -(z-p)})$=$int_{C} f(zeta)(frac{1}{-(z -p)(1 -frac{(zeta-p)}{z -p}})$=$-int_{C} f(zeta) sum_{n=0}^{infty} (frac{zeta-p}{z -p})^n$ which is valid for $|frac{zeta-p}{z -p}|<1$.
Which I am then interpreting as the outside region of the circle
My question is, as you may have guessed, Where does the Laurent series given by
$f(z)=sum_{n=1}^{infty} a_n(z-p)^n +sum_{n=1}^{infty} frac{b_n}{(z-p)^n}$ converge? I do not understand where the annulus of convergence can be.
Thanks in advance.
complex-analysis
asked Nov 27 '18 at 4:07
JungleshrimpJungleshrimp
19311
$endgroup$
From the Cauchy integral formula, I get
$f(z)=int_{C} frac{f(zeta)}{zeta -z}dzeta$
manipulating, =$int_{C} f(zeta)(frac{1}{zeta -p -(z-p)})$=$int_{C} f(zeta)(frac{1}{(zeta -p)(1 -frac{(z-p)}{zeta -p}})$=$int_{C} f(zeta) sum_{n=0}^{infty} (frac{z-p}{zeta -p})^n$ which is valid for $|frac{z-p}{zeta -p}|<1$.
Geometrically, I interpret this as the points closer to $z$ than the boundary of C
Or, I can choose to do a different power series.
$f(z)=int_{C} frac{f(zeta)}{zeta -z}dzeta$ = manipulating differently, =$int_{C} f(zeta)(frac{1}{zeta -p -(z-p)})$=$int_{C} f(zeta)(frac{1}{-(z -p)(1 -frac{(zeta-p)}{z -p}})$=$-int_{C} f(zeta) sum_{n=0}^{infty} (frac{zeta-p}{z -p})^n$ which is valid for $|frac{zeta-p}{z -p}|<1$.
Which I am then interpreting as the outside region of the circle
My question is, as you may have guessed, Where does the Laurent series given by
$f(z)=sum_{n=1}^{infty} a_n(z-p)^n +sum_{n=1}^{infty} frac{b_n}{(z-p)^n}$ converge? I do not understand where the annulus of convergence can be.
Thanks in advance.
complex-analysis
complex-analysis
asked Nov 27 '18 at 4:07
JungleshrimpJungleshrimp
19311
asked Nov 27 '18 at 4:07
JungleshrimpJungleshrimp
19311
asked Nov 27 '18 at 4:07
JungleshrimpJungleshrimp
19311
asked Nov 27 '18 at 4:07
JungleshrimpJungleshrimp
19311
asked Nov 27 '18 at 4:07
JungleshrimpJungleshrimp
19311
19311
$begingroup$
a Laurent series converges between the point where it is constructed around (in your case $p$) and up to the closer singularity to the function that it represent. If the function doesn't have any singularity then the Laurent series is just a Taylor series with radius of convergence infinity
$endgroup$
– Masacroso
Nov 27 '18 at 4:17
$begingroup$
@Masacroso Here yes but in general the domain of convergence can be any anulus, eg. $sum_n (1/2)^{|n|} z^n$ converges for $|z| < 1/2, |1/z| < 1/2 = |z| in (1/2,2)$. The (open) annulus where $sum_n a_n z^n$ converges is $|z|in (1/r,R)$ where $R,r$ are the radius of convergence of $sum_{n ge 0} a_n z^n,sum_{n > 0} a_{-n} z^n$. There is a singularity on $|z| = R, |z| =1/r$
$endgroup$
– reuns
Nov 27 '18 at 4:25
add a comment |
$begingroup$
a Laurent series converges between the point where it is constructed around (in your case $p$) and up to the closer singularity to the function that it represent. If the function doesn't have any singularity then the Laurent series is just a Taylor series with radius of convergence infinity
$endgroup$
– Masacroso
Nov 27 '18 at 4:17
$begingroup$
@Masacroso Here yes but in general the domain of convergence can be any anulus, eg. $sum_n (1/2)^{|n|} z^n$ converges for $|z| < 1/2, |1/z| < 1/2 = |z| in (1/2,2)$. The (open) annulus where $sum_n a_n z^n$ converges is $|z|in (1/r,R)$ where $R,r$ are the radius of convergence of $sum_{n ge 0} a_n z^n,sum_{n > 0} a_{-n} z^n$. There is a singularity on $|z| = R, |z| =1/r$
$endgroup$
– reuns
Nov 27 '18 at 4:25
$begingroup$
a Laurent series converges between the point where it is constructed around (in your case $p$) and up to the closer singularity to the function that it represent. If the function doesn't have any singularity then the Laurent series is just a Taylor series with radius of convergence infinity
$endgroup$
– Masacroso
Nov 27 '18 at 4:17
$begingroup$
a Laurent series converges between the point where it is constructed around (in your case $p$) and up to the closer singularity to the function that it represent. If the function doesn't have any singularity then the Laurent series is just a Taylor series with radius of convergence infinity
$endgroup$
– Masacroso
Nov 27 '18 at 4:17
$begingroup$
@Masacroso Here yes but in general the domain of convergence can be any anulus, eg. $sum_n (1/2)^{|n|} z^n$ converges for $|z| < 1/2, |1/z| < 1/2 = |z| in (1/2,2)$. The (open) annulus where $sum_n a_n z^n$ converges is $|z|in (1/r,R)$ where $R,r$ are the radius of convergence of $sum_{n ge 0} a_n z^n,sum_{n > 0} a_{-n} z^n$. There is a singularity on $|z| = R, |z| =1/r$
$endgroup$
– reuns
Nov 27 '18 at 4:25
$begingroup$
@Masacroso Here yes but in general the domain of convergence can be any anulus, eg. $sum_n (1/2)^{|n|} z^n$ converges for $|z| < 1/2, |1/z| < 1/2 = |z| in (1/2,2)$. The (open) annulus where $sum_n a_n z^n$ converges is $|z|in (1/r,R)$ where $R,r$ are the radius of convergence of $sum_{n ge 0} a_n z^n,sum_{n > 0} a_{-n} z^n$. There is a singularity on $|z| = R, |z| =1/r$
$endgroup$
– reuns
Nov 27 '18 at 4:25
add a comment |
0
active
oldest
votes
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%2f3015323%2fwhere-does-a-laurent-series-converge%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3015323%2fwhere-does-a-laurent-series-converge%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
$begingroup$
a Laurent series converges between the point where it is constructed around (in your case $p$) and up to the closer singularity to the function that it represent. If the function doesn't have any singularity then the Laurent series is just a Taylor series with radius of convergence infinity
$endgroup$
– Masacroso
Nov 27 '18 at 4:17
$begingroup$
@Masacroso Here yes but in general the domain of convergence can be any anulus, eg. $sum_n (1/2)^{|n|} z^n$ converges for $|z| < 1/2, |1/z| < 1/2 = |z| in (1/2,2)$. The (open) annulus where $sum_n a_n z^n$ converges is $|z|in (1/r,R)$ where $R,r$ are the radius of convergence of $sum_{n ge 0} a_n z^n,sum_{n > 0} a_{-n} z^n$. There is a singularity on $|z| = R, |z| =1/r$
$endgroup$
– reuns
Nov 27 '18 at 4:25