Determination of the order of a pole












0














In the function $$f(z) =frac{sin(frac{pi}{2}(z+1))}{(z^2+2z+4)(z+1)^3}$$ the order of the pole in $z=-1$ is correctly two?
Or maybe it is an eliminable singularity?



I have a problem because often if I have an eliminable singularity I only have to Taylor expand the numerator of the fraction at the first term, in this situation, if I expand at the first term I have a singularity of order two and if expand at the second it disappears without becoming zero.






complex-analysis taylor-expansion singularity







share|cite|improve this question
























  • Yes, it is two. If you take the limit when $;zto-1;$ of the numerator with $;(z+1);$ in the denominator, the limit is finite , so the remaining exponent in the denominator of $;z+1;$ is the order of the pole at $;z=-1;$
    – DonAntonio
    Nov 27 '18 at 21:40












  • All you need to do is to plug the expansions $$sin u=uleft(1-frac{u^2}6+o(u^2)right)$$ and $$frac1{u^2+3}=frac13left(1-frac13u^2+o(u^2)right)$$ when $uto0$, in the fraction and, using the shorthand $w=z+1$, to deduce the correct expansion $$f(z)=frac{fracpi2w}{w^3}frac13left(1-frac{pi^2w^2}{24}+o(w^2)right)left(1-frac{w^2}3+o(w^3)right)$$ that is $$f(z)=fracpi{6w^2}left(1-cw^2+o(w^2)right)=fracpi{6w^2}-cfracpi6+o(1)$$ for some suitable $cne0$, indicating a pole of order $2$ and a residue $0$ at $z=-1$.
    – Did
    Nov 27 '18 at 21:50
















0














In the function $$f(z) =frac{sin(frac{pi}{2}(z+1))}{(z^2+2z+4)(z+1)^3}$$ the order of the pole in $z=-1$ is correctly two?
Or maybe it is an eliminable singularity?



I have a problem because often if I have an eliminable singularity I only have to Taylor expand the numerator of the fraction at the first term, in this situation, if I expand at the first term I have a singularity of order two and if expand at the second it disappears without becoming zero.






complex-analysis taylor-expansion singularity







share|cite|improve this question
























  • Yes, it is two. If you take the limit when $;zto-1;$ of the numerator with $;(z+1);$ in the denominator, the limit is finite , so the remaining exponent in the denominator of $;z+1;$ is the order of the pole at $;z=-1;$
    – DonAntonio
    Nov 27 '18 at 21:40












  • All you need to do is to plug the expansions $$sin u=uleft(1-frac{u^2}6+o(u^2)right)$$ and $$frac1{u^2+3}=frac13left(1-frac13u^2+o(u^2)right)$$ when $uto0$, in the fraction and, using the shorthand $w=z+1$, to deduce the correct expansion $$f(z)=frac{fracpi2w}{w^3}frac13left(1-frac{pi^2w^2}{24}+o(w^2)right)left(1-frac{w^2}3+o(w^3)right)$$ that is $$f(z)=fracpi{6w^2}left(1-cw^2+o(w^2)right)=fracpi{6w^2}-cfracpi6+o(1)$$ for some suitable $cne0$, indicating a pole of order $2$ and a residue $0$ at $z=-1$.
    – Did
    Nov 27 '18 at 21:50














0












0








0







In the function $$f(z) =frac{sin(frac{pi}{2}(z+1))}{(z^2+2z+4)(z+1)^3}$$ the order of the pole in $z=-1$ is correctly two?
Or maybe it is an eliminable singularity?



I have a problem because often if I have an eliminable singularity I only have to Taylor expand the numerator of the fraction at the first term, in this situation, if I expand at the first term I have a singularity of order two and if expand at the second it disappears without becoming zero.






complex-analysis taylor-expansion singularity







share|cite|improve this question















In the function $$f(z) =frac{sin(frac{pi}{2}(z+1))}{(z^2+2z+4)(z+1)^3}$$ the order of the pole in $z=-1$ is correctly two?
Or maybe it is an eliminable singularity?



I have a problem because often if I have an eliminable singularity I only have to Taylor expand the numerator of the fraction at the first term, in this situation, if I expand at the first term I have a singularity of order two and if expand at the second it disappears without becoming zero.






complex-analysis taylor-expansion singularity




complex-analysis taylor-expansion singularity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 21:51









Did

246k23220454




246k23220454










asked Nov 27 '18 at 21:35









pter26

307111




307111












  • Yes, it is two. If you take the limit when $;zto-1;$ of the numerator with $;(z+1);$ in the denominator, the limit is finite , so the remaining exponent in the denominator of $;z+1;$ is the order of the pole at $;z=-1;$
    – DonAntonio
    Nov 27 '18 at 21:40












  • All you need to do is to plug the expansions $$sin u=uleft(1-frac{u^2}6+o(u^2)right)$$ and $$frac1{u^2+3}=frac13left(1-frac13u^2+o(u^2)right)$$ when $uto0$, in the fraction and, using the shorthand $w=z+1$, to deduce the correct expansion $$f(z)=frac{fracpi2w}{w^3}frac13left(1-frac{pi^2w^2}{24}+o(w^2)right)left(1-frac{w^2}3+o(w^3)right)$$ that is $$f(z)=fracpi{6w^2}left(1-cw^2+o(w^2)right)=fracpi{6w^2}-cfracpi6+o(1)$$ for some suitable $cne0$, indicating a pole of order $2$ and a residue $0$ at $z=-1$.
    – Did
    Nov 27 '18 at 21:50


















  • Yes, it is two. If you take the limit when $;zto-1;$ of the numerator with $;(z+1);$ in the denominator, the limit is finite , so the remaining exponent in the denominator of $;z+1;$ is the order of the pole at $;z=-1;$
    – DonAntonio
    Nov 27 '18 at 21:40












  • All you need to do is to plug the expansions $$sin u=uleft(1-frac{u^2}6+o(u^2)right)$$ and $$frac1{u^2+3}=frac13left(1-frac13u^2+o(u^2)right)$$ when $uto0$, in the fraction and, using the shorthand $w=z+1$, to deduce the correct expansion $$f(z)=frac{fracpi2w}{w^3}frac13left(1-frac{pi^2w^2}{24}+o(w^2)right)left(1-frac{w^2}3+o(w^3)right)$$ that is $$f(z)=fracpi{6w^2}left(1-cw^2+o(w^2)right)=fracpi{6w^2}-cfracpi6+o(1)$$ for some suitable $cne0$, indicating a pole of order $2$ and a residue $0$ at $z=-1$.
    – Did
    Nov 27 '18 at 21:50
















Yes, it is two. If you take the limit when $;zto-1;$ of the numerator with $;(z+1);$ in the denominator, the limit is finite , so the remaining exponent in the denominator of $;z+1;$ is the order of the pole at $;z=-1;$
– DonAntonio
Nov 27 '18 at 21:40






Yes, it is two. If you take the limit when $;zto-1;$ of the numerator with $;(z+1);$ in the denominator, the limit is finite , so the remaining exponent in the denominator of $;z+1;$ is the order of the pole at $;z=-1;$
– DonAntonio
Nov 27 '18 at 21:40














All you need to do is to plug the expansions $$sin u=uleft(1-frac{u^2}6+o(u^2)right)$$ and $$frac1{u^2+3}=frac13left(1-frac13u^2+o(u^2)right)$$ when $uto0$, in the fraction and, using the shorthand $w=z+1$, to deduce the correct expansion $$f(z)=frac{fracpi2w}{w^3}frac13left(1-frac{pi^2w^2}{24}+o(w^2)right)left(1-frac{w^2}3+o(w^3)right)$$ that is $$f(z)=fracpi{6w^2}left(1-cw^2+o(w^2)right)=fracpi{6w^2}-cfracpi6+o(1)$$ for some suitable $cne0$, indicating a pole of order $2$ and a residue $0$ at $z=-1$.
– Did
Nov 27 '18 at 21:50




All you need to do is to plug the expansions $$sin u=uleft(1-frac{u^2}6+o(u^2)right)$$ and $$frac1{u^2+3}=frac13left(1-frac13u^2+o(u^2)right)$$ when $uto0$, in the fraction and, using the shorthand $w=z+1$, to deduce the correct expansion $$f(z)=frac{fracpi2w}{w^3}frac13left(1-frac{pi^2w^2}{24}+o(w^2)right)left(1-frac{w^2}3+o(w^3)right)$$ that is $$f(z)=fracpi{6w^2}left(1-cw^2+o(w^2)right)=fracpi{6w^2}-cfracpi6+o(1)$$ for some suitable $cne0$, indicating a pole of order $2$ and a residue $0$ at $z=-1$.
– Did
Nov 27 '18 at 21:50










1 Answer
1






active

oldest

votes


















1














We have



$$frac{sinfracpi2(z+1)}{left[3+(z+1)^2right](z+1)^3}=frac13cdotfrac1{1+left(frac{z+1}{sqrt3}right)^2}cdotfrac1{(z+1)^3}sum_{n=0}^inftyfrac{(-1)^npi^{2n+1}(z+1)^{2n+1}}{(2n+1)!}=$$



$$=frac13left(1-left(frac{z+1}{sqrt3}right)^2+left(frac{z+1}{sqrt3}right)^4-ldotsright)frac1{(z+1)^{color{red}2}}sum_{n=0}^inftyfrac{(-1)^npi^{2n+1}(z+1)^{color{red}{2n}}}{(2n+1)!}$$



and we can see clearly in the above Laurent expansion that the pole at $;z=-1;$ is of order two. Of course, the above is valid only in some neighbourhood of $;-1;$.






share|cite|improve this answer























  • Why is the first equality possible?
    – pter26
    Nov 27 '18 at 21:54










  • @pter26 That's merely algebra and the Taylor expansion of sine, which is true for any value of $;zinBbb C;$ .
    – DonAntonio
    Nov 27 '18 at 21:55










  • Thanks a lot! So can I tell the residue in $z=-1$ is one from this?
    – pter26
    Nov 27 '18 at 22:02












  • Well, I think it is zero since in the series the first non-constant element goes with $;(z+1)^2;$ , so " there is no coefficient of $;(z+1)^{-1};$" , meaning: its coefficient is zero
    – DonAntonio
    Nov 27 '18 at 22:15











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%2f3016333%2fdetermination-of-the-order-of-a-pole%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









1














We have



$$frac{sinfracpi2(z+1)}{left[3+(z+1)^2right](z+1)^3}=frac13cdotfrac1{1+left(frac{z+1}{sqrt3}right)^2}cdotfrac1{(z+1)^3}sum_{n=0}^inftyfrac{(-1)^npi^{2n+1}(z+1)^{2n+1}}{(2n+1)!}=$$



$$=frac13left(1-left(frac{z+1}{sqrt3}right)^2+left(frac{z+1}{sqrt3}right)^4-ldotsright)frac1{(z+1)^{color{red}2}}sum_{n=0}^inftyfrac{(-1)^npi^{2n+1}(z+1)^{color{red}{2n}}}{(2n+1)!}$$



and we can see clearly in the above Laurent expansion that the pole at $;z=-1;$ is of order two. Of course, the above is valid only in some neighbourhood of $;-1;$.






share|cite|improve this answer























  • Why is the first equality possible?
    – pter26
    Nov 27 '18 at 21:54










  • @pter26 That's merely algebra and the Taylor expansion of sine, which is true for any value of $;zinBbb C;$ .
    – DonAntonio
    Nov 27 '18 at 21:55










  • Thanks a lot! So can I tell the residue in $z=-1$ is one from this?
    – pter26
    Nov 27 '18 at 22:02












  • Well, I think it is zero since in the series the first non-constant element goes with $;(z+1)^2;$ , so " there is no coefficient of $;(z+1)^{-1};$" , meaning: its coefficient is zero
    – DonAntonio
    Nov 27 '18 at 22:15
















1














We have



$$frac{sinfracpi2(z+1)}{left[3+(z+1)^2right](z+1)^3}=frac13cdotfrac1{1+left(frac{z+1}{sqrt3}right)^2}cdotfrac1{(z+1)^3}sum_{n=0}^inftyfrac{(-1)^npi^{2n+1}(z+1)^{2n+1}}{(2n+1)!}=$$



$$=frac13left(1-left(frac{z+1}{sqrt3}right)^2+left(frac{z+1}{sqrt3}right)^4-ldotsright)frac1{(z+1)^{color{red}2}}sum_{n=0}^inftyfrac{(-1)^npi^{2n+1}(z+1)^{color{red}{2n}}}{(2n+1)!}$$



and we can see clearly in the above Laurent expansion that the pole at $;z=-1;$ is of order two. Of course, the above is valid only in some neighbourhood of $;-1;$.






share|cite|improve this answer























  • Why is the first equality possible?
    – pter26
    Nov 27 '18 at 21:54










  • @pter26 That's merely algebra and the Taylor expansion of sine, which is true for any value of $;zinBbb C;$ .
    – DonAntonio
    Nov 27 '18 at 21:55










  • Thanks a lot! So can I tell the residue in $z=-1$ is one from this?
    – pter26
    Nov 27 '18 at 22:02












  • Well, I think it is zero since in the series the first non-constant element goes with $;(z+1)^2;$ , so " there is no coefficient of $;(z+1)^{-1};$" , meaning: its coefficient is zero
    – DonAntonio
    Nov 27 '18 at 22:15














1












1








1






We have



$$frac{sinfracpi2(z+1)}{left[3+(z+1)^2right](z+1)^3}=frac13cdotfrac1{1+left(frac{z+1}{sqrt3}right)^2}cdotfrac1{(z+1)^3}sum_{n=0}^inftyfrac{(-1)^npi^{2n+1}(z+1)^{2n+1}}{(2n+1)!}=$$



$$=frac13left(1-left(frac{z+1}{sqrt3}right)^2+left(frac{z+1}{sqrt3}right)^4-ldotsright)frac1{(z+1)^{color{red}2}}sum_{n=0}^inftyfrac{(-1)^npi^{2n+1}(z+1)^{color{red}{2n}}}{(2n+1)!}$$



and we can see clearly in the above Laurent expansion that the pole at $;z=-1;$ is of order two. Of course, the above is valid only in some neighbourhood of $;-1;$.






share|cite|improve this answer














We have



$$frac{sinfracpi2(z+1)}{left[3+(z+1)^2right](z+1)^3}=frac13cdotfrac1{1+left(frac{z+1}{sqrt3}right)^2}cdotfrac1{(z+1)^3}sum_{n=0}^inftyfrac{(-1)^npi^{2n+1}(z+1)^{2n+1}}{(2n+1)!}=$$



$$=frac13left(1-left(frac{z+1}{sqrt3}right)^2+left(frac{z+1}{sqrt3}right)^4-ldotsright)frac1{(z+1)^{color{red}2}}sum_{n=0}^inftyfrac{(-1)^npi^{2n+1}(z+1)^{color{red}{2n}}}{(2n+1)!}$$



and we can see clearly in the above Laurent expansion that the pole at $;z=-1;$ is of order two. Of course, the above is valid only in some neighbourhood of $;-1;$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 27 '18 at 21:54

























answered Nov 27 '18 at 21:48









DonAntonio

177k1491225




177k1491225












  • Why is the first equality possible?
    – pter26
    Nov 27 '18 at 21:54










  • @pter26 That's merely algebra and the Taylor expansion of sine, which is true for any value of $;zinBbb C;$ .
    – DonAntonio
    Nov 27 '18 at 21:55










  • Thanks a lot! So can I tell the residue in $z=-1$ is one from this?
    – pter26
    Nov 27 '18 at 22:02












  • Well, I think it is zero since in the series the first non-constant element goes with $;(z+1)^2;$ , so " there is no coefficient of $;(z+1)^{-1};$" , meaning: its coefficient is zero
    – DonAntonio
    Nov 27 '18 at 22:15


















  • Why is the first equality possible?
    – pter26
    Nov 27 '18 at 21:54










  • @pter26 That's merely algebra and the Taylor expansion of sine, which is true for any value of $;zinBbb C;$ .
    – DonAntonio
    Nov 27 '18 at 21:55










  • Thanks a lot! So can I tell the residue in $z=-1$ is one from this?
    – pter26
    Nov 27 '18 at 22:02












  • Well, I think it is zero since in the series the first non-constant element goes with $;(z+1)^2;$ , so " there is no coefficient of $;(z+1)^{-1};$" , meaning: its coefficient is zero
    – DonAntonio
    Nov 27 '18 at 22:15
















Why is the first equality possible?
– pter26
Nov 27 '18 at 21:54




Why is the first equality possible?
– pter26
Nov 27 '18 at 21:54












@pter26 That's merely algebra and the Taylor expansion of sine, which is true for any value of $;zinBbb C;$ .
– DonAntonio
Nov 27 '18 at 21:55




@pter26 That's merely algebra and the Taylor expansion of sine, which is true for any value of $;zinBbb C;$ .
– DonAntonio
Nov 27 '18 at 21:55












Thanks a lot! So can I tell the residue in $z=-1$ is one from this?
– pter26
Nov 27 '18 at 22:02






Thanks a lot! So can I tell the residue in $z=-1$ is one from this?
– pter26
Nov 27 '18 at 22:02














Well, I think it is zero since in the series the first non-constant element goes with $;(z+1)^2;$ , so " there is no coefficient of $;(z+1)^{-1};$" , meaning: its coefficient is zero
– DonAntonio
Nov 27 '18 at 22:15




Well, I think it is zero since in the series the first non-constant element goes with $;(z+1)^2;$ , so " there is no coefficient of $;(z+1)^{-1};$" , meaning: its coefficient is zero
– DonAntonio
Nov 27 '18 at 22:15


















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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


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%2f3016333%2fdetermination-of-the-order-of-a-pole%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

Bundesstraße 106

Image
Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
Read more

Le Mesnil-Réaume

Image
Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
Read more

Ida-Boy-Ed-Garten

Image
Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
Read more