Find complex residue and Laurent series expansion












0












$begingroup$


$$ w = sin(z) * sin(frac{1}{:z}) $$



special point is $$ z_0 = 0 $$
$$ lim _{zto 0}left(sinleft(zright)cdot :sin:left(frac{1}{z}right)right) $$ isn't exist, next I have to decompose both functions in Laurent series.
$$sin z=z-frac{z^3}{3!}+frac{z^5text{}}{5!}-... $$
$$ sinfrac{1}{z}=frac{1}{z}-frac{1}{z^33!}+frac{1}{z^55!} $$
after, multiplication series I have to find $$ c^{-1} $$ addend in Laurent series, and coefficients with it will be residue?
but
$$ 1+frac{1}{3!cdot 3!}+frac{1}{5!cdot 5!} $$






complex-analysis taylor-expansion residue-calculus laurent-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    $LaTeX$ advice: use sin for sine function.
    $endgroup$
    – Chinnapparaj R
    Dec 17 '18 at 15:22










  • $begingroup$
    thanks for hint
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 15:41
















0












$begingroup$


$$ w = sin(z) * sin(frac{1}{:z}) $$



special point is $$ z_0 = 0 $$
$$ lim _{zto 0}left(sinleft(zright)cdot :sin:left(frac{1}{z}right)right) $$ isn't exist, next I have to decompose both functions in Laurent series.
$$sin z=z-frac{z^3}{3!}+frac{z^5text{}}{5!}-... $$
$$ sinfrac{1}{z}=frac{1}{z}-frac{1}{z^33!}+frac{1}{z^55!} $$
after, multiplication series I have to find $$ c^{-1} $$ addend in Laurent series, and coefficients with it will be residue?
but
$$ 1+frac{1}{3!cdot 3!}+frac{1}{5!cdot 5!} $$






complex-analysis taylor-expansion residue-calculus laurent-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    $LaTeX$ advice: use sin for sine function.
    $endgroup$
    – Chinnapparaj R
    Dec 17 '18 at 15:22










  • $begingroup$
    thanks for hint
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 15:41














0












0








0





$begingroup$


$$ w = sin(z) * sin(frac{1}{:z}) $$



special point is $$ z_0 = 0 $$
$$ lim _{zto 0}left(sinleft(zright)cdot :sin:left(frac{1}{z}right)right) $$ isn't exist, next I have to decompose both functions in Laurent series.
$$sin z=z-frac{z^3}{3!}+frac{z^5text{}}{5!}-... $$
$$ sinfrac{1}{z}=frac{1}{z}-frac{1}{z^33!}+frac{1}{z^55!} $$
after, multiplication series I have to find $$ c^{-1} $$ addend in Laurent series, and coefficients with it will be residue?
but
$$ 1+frac{1}{3!cdot 3!}+frac{1}{5!cdot 5!} $$






complex-analysis taylor-expansion residue-calculus laurent-series







share|cite|improve this question











$endgroup$




$$ w = sin(z) * sin(frac{1}{:z}) $$



special point is $$ z_0 = 0 $$
$$ lim _{zto 0}left(sinleft(zright)cdot :sin:left(frac{1}{z}right)right) $$ isn't exist, next I have to decompose both functions in Laurent series.
$$sin z=z-frac{z^3}{3!}+frac{z^5text{}}{5!}-... $$
$$ sinfrac{1}{z}=frac{1}{z}-frac{1}{z^33!}+frac{1}{z^55!} $$
after, multiplication series I have to find $$ c^{-1} $$ addend in Laurent series, and coefficients with it will be residue?
but
$$ 1+frac{1}{3!cdot 3!}+frac{1}{5!cdot 5!} $$






complex-analysis taylor-expansion residue-calculus laurent-series




complex-analysis taylor-expansion residue-calculus laurent-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 15:21









Chinnapparaj R

5,4872928




5,4872928










asked Dec 17 '18 at 14:48









Dmitry SokolovDmitry Sokolov

526




526












  • $begingroup$
    $LaTeX$ advice: use sin for sine function.
    $endgroup$
    – Chinnapparaj R
    Dec 17 '18 at 15:22










  • $begingroup$
    thanks for hint
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 15:41


















  • $begingroup$
    $LaTeX$ advice: use sin for sine function.
    $endgroup$
    – Chinnapparaj R
    Dec 17 '18 at 15:22










  • $begingroup$
    thanks for hint
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 15:41
















$begingroup$
$LaTeX$ advice: use sin for sine function.
$endgroup$
– Chinnapparaj R
Dec 17 '18 at 15:22




$begingroup$
$LaTeX$ advice: use sin for sine function.
$endgroup$
– Chinnapparaj R
Dec 17 '18 at 15:22












$begingroup$
thanks for hint
$endgroup$
– Dmitry Sokolov
Dec 17 '18 at 15:41




$begingroup$
thanks for hint
$endgroup$
– Dmitry Sokolov
Dec 17 '18 at 15:41










1 Answer
1






active

oldest

votes


















4












$begingroup$

The residue is $0$ since two odd terms $z^m$ and $z^{-k}$ will always make an even term when multiplied: $z^mz^{-k} = z^{m-k}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, Should I find residue at infinity?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 16:10










  • $begingroup$
    *Must I have to find?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 16:29










  • $begingroup$
    @DmitrySokolov Residue at infinity is the residue of $-frac{1}{z^2} f(frac{1}{z})$ at $0$. Now, we have $f(frac{1}{z}) = f(z)$ and the even term $frac{1}{z^2}$ won't change the parity of a term it multiplies, so again we have zero residue at infinity.
    $endgroup$
    – ploosu2
    Dec 17 '18 at 18:13












  • $begingroup$
    I mean:Is Infinity isolated singular point?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 18:22












  • $begingroup$
    @DmitrySokolov Yes it is. Notice also that both $0$ and infinity are essential singularities.
    $endgroup$
    – ploosu2
    Dec 17 '18 at 21:04











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%2f3044039%2ffind-complex-residue-and-laurent-series-expansion%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









4












$begingroup$

The residue is $0$ since two odd terms $z^m$ and $z^{-k}$ will always make an even term when multiplied: $z^mz^{-k} = z^{m-k}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, Should I find residue at infinity?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 16:10










  • $begingroup$
    *Must I have to find?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 16:29










  • $begingroup$
    @DmitrySokolov Residue at infinity is the residue of $-frac{1}{z^2} f(frac{1}{z})$ at $0$. Now, we have $f(frac{1}{z}) = f(z)$ and the even term $frac{1}{z^2}$ won't change the parity of a term it multiplies, so again we have zero residue at infinity.
    $endgroup$
    – ploosu2
    Dec 17 '18 at 18:13












  • $begingroup$
    I mean:Is Infinity isolated singular point?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 18:22












  • $begingroup$
    @DmitrySokolov Yes it is. Notice also that both $0$ and infinity are essential singularities.
    $endgroup$
    – ploosu2
    Dec 17 '18 at 21:04
















4












$begingroup$

The residue is $0$ since two odd terms $z^m$ and $z^{-k}$ will always make an even term when multiplied: $z^mz^{-k} = z^{m-k}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, Should I find residue at infinity?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 16:10










  • $begingroup$
    *Must I have to find?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 16:29










  • $begingroup$
    @DmitrySokolov Residue at infinity is the residue of $-frac{1}{z^2} f(frac{1}{z})$ at $0$. Now, we have $f(frac{1}{z}) = f(z)$ and the even term $frac{1}{z^2}$ won't change the parity of a term it multiplies, so again we have zero residue at infinity.
    $endgroup$
    – ploosu2
    Dec 17 '18 at 18:13












  • $begingroup$
    I mean:Is Infinity isolated singular point?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 18:22












  • $begingroup$
    @DmitrySokolov Yes it is. Notice also that both $0$ and infinity are essential singularities.
    $endgroup$
    – ploosu2
    Dec 17 '18 at 21:04














4












4








4





$begingroup$

The residue is $0$ since two odd terms $z^m$ and $z^{-k}$ will always make an even term when multiplied: $z^mz^{-k} = z^{m-k}$.






share|cite|improve this answer









$endgroup$



The residue is $0$ since two odd terms $z^m$ and $z^{-k}$ will always make an even term when multiplied: $z^mz^{-k} = z^{m-k}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 17 '18 at 15:13









ploosu2ploosu2

4,6451024




4,6451024












  • $begingroup$
    Thanks, Should I find residue at infinity?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 16:10










  • $begingroup$
    *Must I have to find?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 16:29










  • $begingroup$
    @DmitrySokolov Residue at infinity is the residue of $-frac{1}{z^2} f(frac{1}{z})$ at $0$. Now, we have $f(frac{1}{z}) = f(z)$ and the even term $frac{1}{z^2}$ won't change the parity of a term it multiplies, so again we have zero residue at infinity.
    $endgroup$
    – ploosu2
    Dec 17 '18 at 18:13












  • $begingroup$
    I mean:Is Infinity isolated singular point?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 18:22












  • $begingroup$
    @DmitrySokolov Yes it is. Notice also that both $0$ and infinity are essential singularities.
    $endgroup$
    – ploosu2
    Dec 17 '18 at 21:04


















  • $begingroup$
    Thanks, Should I find residue at infinity?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 16:10










  • $begingroup$
    *Must I have to find?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 16:29










  • $begingroup$
    @DmitrySokolov Residue at infinity is the residue of $-frac{1}{z^2} f(frac{1}{z})$ at $0$. Now, we have $f(frac{1}{z}) = f(z)$ and the even term $frac{1}{z^2}$ won't change the parity of a term it multiplies, so again we have zero residue at infinity.
    $endgroup$
    – ploosu2
    Dec 17 '18 at 18:13












  • $begingroup$
    I mean:Is Infinity isolated singular point?
    $endgroup$
    – Dmitry Sokolov
    Dec 17 '18 at 18:22












  • $begingroup$
    @DmitrySokolov Yes it is. Notice also that both $0$ and infinity are essential singularities.
    $endgroup$
    – ploosu2
    Dec 17 '18 at 21:04
















$begingroup$
Thanks, Should I find residue at infinity?
$endgroup$
– Dmitry Sokolov
Dec 17 '18 at 16:10




$begingroup$
Thanks, Should I find residue at infinity?
$endgroup$
– Dmitry Sokolov
Dec 17 '18 at 16:10












$begingroup$
*Must I have to find?
$endgroup$
– Dmitry Sokolov
Dec 17 '18 at 16:29




$begingroup$
*Must I have to find?
$endgroup$
– Dmitry Sokolov
Dec 17 '18 at 16:29












$begingroup$
@DmitrySokolov Residue at infinity is the residue of $-frac{1}{z^2} f(frac{1}{z})$ at $0$. Now, we have $f(frac{1}{z}) = f(z)$ and the even term $frac{1}{z^2}$ won't change the parity of a term it multiplies, so again we have zero residue at infinity.
$endgroup$
– ploosu2
Dec 17 '18 at 18:13






$begingroup$
@DmitrySokolov Residue at infinity is the residue of $-frac{1}{z^2} f(frac{1}{z})$ at $0$. Now, we have $f(frac{1}{z}) = f(z)$ and the even term $frac{1}{z^2}$ won't change the parity of a term it multiplies, so again we have zero residue at infinity.
$endgroup$
– ploosu2
Dec 17 '18 at 18:13














$begingroup$
I mean:Is Infinity isolated singular point?
$endgroup$
– Dmitry Sokolov
Dec 17 '18 at 18:22






$begingroup$
I mean:Is Infinity isolated singular point?
$endgroup$
– Dmitry Sokolov
Dec 17 '18 at 18:22














$begingroup$
@DmitrySokolov Yes it is. Notice also that both $0$ and infinity are essential singularities.
$endgroup$
– ploosu2
Dec 17 '18 at 21:04




$begingroup$
@DmitrySokolov Yes it is. Notice also that both $0$ and infinity are essential singularities.
$endgroup$
– ploosu2
Dec 17 '18 at 21:04


















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%2f3044039%2ffind-complex-residue-and-laurent-series-expansion%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

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
Read more

build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

Image
0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
Read more