Mixing
Why does $sinleft(frac{pi}{z^2}right)$ not have a Laurent series at $z=0$?
$begingroup$
I tried to compute the Laurent series of $sinleft(frac{pi}{z^2}right)$ at $z=0$, then I realized something was off altogether and there simply doesn't exist a Laurent espansion centered at $z=0$ for that function (Mathematica says "no series expansion available"). I thought of a possible explanation, could you tell me if it makes sense?
If $C_{r,R}$ is the annulus of internal radius $r$ and external radius $R$ centered at $z=0$.
$$sin{frac{pi}{z^2}}=sum_{k=0}^{infty}{frac{(-1)^k pi^{2k+1}}{(2k+1)!}(z-0)^{-4k-2}},forall zin C_{r,R},space wherespace frac{1}{R}=lim_{krightarrowinfty}{supleft|frac{{(-1)^k pi^{2k+1}}}{(2k+1)!}right|^{1/k}=0}space andspace r=lim_{krightarrowinfty}{supleft|frac{{(-1)^-k pi^{-2k+1}}}{(-2k+1)!}right|^{1/k}=0}$$ $$iff$$ $$fspace holomorphicspace inspace C_{r,R}=C_{0,infty}space andspaceforall space closedspace pathspace gammain C_{0,infty},space c_{k}=frac{(-1)^kpi^{2k+1}}{(2k+1)!}=frac{1}{2pi i}oint_{gamma}{frac{sinleft({frac{pi}{z^2}}right)}{(z-0)^{k+1}}dz}$$
though if I compute the coefficients $c_k$ on a $gamma$ unitary circle centered at $z=0$: $frac{1}{2pi i}oint_{gamma}{frac{sinleft({frac{pi}{z^2}}right)}{(z-0)^{k+1}}dz}=frac{1}{2pi}int_{0}^{2pi}{frac{sinleft(pi e^{-2 i theta}right)}{e^{itheta k}}dtheta}=0neqfrac{(-1)^kpi^{2k+1}}{(2k+1)!}, forall kinmathbb{Z}$,
so the proposition on the left of the $iff$ can't be true.
Or said in another way, if the expansion exists it is unique, but then it would mean $f$ is zero everywhere which is obviously not true, so the expansion doesn't exist.
P.S. I'm sorry about poor formatting, if someone improves it I promise I'll look at the corrections and learn form them :)
complex-analysis laurent-series
asked Jan 19 at 20:27
EugenioDiPaolaEugenioDiPaola
515
$endgroup$
add a comment |
$begingroup$
I tried to compute the Laurent series of $sinleft(frac{pi}{z^2}right)$ at $z=0$, then I realized something was off altogether and there simply doesn't exist a Laurent espansion centered at $z=0$ for that function (Mathematica says "no series expansion available"). I thought of a possible explanation, could you tell me if it makes sense?
If $C_{r,R}$ is the annulus of internal radius $r$ and external radius $R$ centered at $z=0$.
$$sin{frac{pi}{z^2}}=sum_{k=0}^{infty}{frac{(-1)^k pi^{2k+1}}{(2k+1)!}(z-0)^{-4k-2}},forall zin C_{r,R},space wherespace frac{1}{R}=lim_{krightarrowinfty}{supleft|frac{{(-1)^k pi^{2k+1}}}{(2k+1)!}right|^{1/k}=0}space andspace r=lim_{krightarrowinfty}{supleft|frac{{(-1)^-k pi^{-2k+1}}}{(-2k+1)!}right|^{1/k}=0}$$ $$iff$$ $$fspace holomorphicspace inspace C_{r,R}=C_{0,infty}space andspaceforall space closedspace pathspace gammain C_{0,infty},space c_{k}=frac{(-1)^kpi^{2k+1}}{(2k+1)!}=frac{1}{2pi i}oint_{gamma}{frac{sinleft({frac{pi}{z^2}}right)}{(z-0)^{k+1}}dz}$$
though if I compute the coefficients $c_k$ on a $gamma$ unitary circle centered at $z=0$: $frac{1}{2pi i}oint_{gamma}{frac{sinleft({frac{pi}{z^2}}right)}{(z-0)^{k+1}}dz}=frac{1}{2pi}int_{0}^{2pi}{frac{sinleft(pi e^{-2 i theta}right)}{e^{itheta k}}dtheta}=0neqfrac{(-1)^kpi^{2k+1}}{(2k+1)!}, forall kinmathbb{Z}$,
so the proposition on the left of the $iff$ can't be true.
Or said in another way, if the expansion exists it is unique, but then it would mean $f$ is zero everywhere which is obviously not true, so the expansion doesn't exist.
P.S. I'm sorry about poor formatting, if someone improves it I promise I'll look at the corrections and learn form them :)
complex-analysis laurent-series
asked Jan 19 at 20:27
EugenioDiPaolaEugenioDiPaola
515
$endgroup$
1
$begingroup$
math.stackexchange.com/q/1564141
$endgroup$
– Did
Jan 19 at 20:44
add a comment |
$begingroup$
I tried to compute the Laurent series of $sinleft(frac{pi}{z^2}right)$ at $z=0$, then I realized something was off altogether and there simply doesn't exist a Laurent espansion centered at $z=0$ for that function (Mathematica says "no series expansion available"). I thought of a possible explanation, could you tell me if it makes sense?
If $C_{r,R}$ is the annulus of internal radius $r$ and external radius $R$ centered at $z=0$.
$$sin{frac{pi}{z^2}}=sum_{k=0}^{infty}{frac{(-1)^k pi^{2k+1}}{(2k+1)!}(z-0)^{-4k-2}},forall zin C_{r,R},space wherespace frac{1}{R}=lim_{krightarrowinfty}{supleft|frac{{(-1)^k pi^{2k+1}}}{(2k+1)!}right|^{1/k}=0}space andspace r=lim_{krightarrowinfty}{supleft|frac{{(-1)^-k pi^{-2k+1}}}{(-2k+1)!}right|^{1/k}=0}$$ $$iff$$ $$fspace holomorphicspace inspace C_{r,R}=C_{0,infty}space andspaceforall space closedspace pathspace gammain C_{0,infty},space c_{k}=frac{(-1)^kpi^{2k+1}}{(2k+1)!}=frac{1}{2pi i}oint_{gamma}{frac{sinleft({frac{pi}{z^2}}right)}{(z-0)^{k+1}}dz}$$
though if I compute the coefficients $c_k$ on a $gamma$ unitary circle centered at $z=0$: $frac{1}{2pi i}oint_{gamma}{frac{sinleft({frac{pi}{z^2}}right)}{(z-0)^{k+1}}dz}=frac{1}{2pi}int_{0}^{2pi}{frac{sinleft(pi e^{-2 i theta}right)}{e^{itheta k}}dtheta}=0neqfrac{(-1)^kpi^{2k+1}}{(2k+1)!}, forall kinmathbb{Z}$,
so the proposition on the left of the $iff$ can't be true.
Or said in another way, if the expansion exists it is unique, but then it would mean $f$ is zero everywhere which is obviously not true, so the expansion doesn't exist.
P.S. I'm sorry about poor formatting, if someone improves it I promise I'll look at the corrections and learn form them :)
complex-analysis laurent-series
asked Jan 19 at 20:27
EugenioDiPaolaEugenioDiPaola
515
$endgroup$
I tried to compute the Laurent series of $sinleft(frac{pi}{z^2}right)$ at $z=0$, then I realized something was off altogether and there simply doesn't exist a Laurent espansion centered at $z=0$ for that function (Mathematica says "no series expansion available"). I thought of a possible explanation, could you tell me if it makes sense?
If $C_{r,R}$ is the annulus of internal radius $r$ and external radius $R$ centered at $z=0$.
$$sin{frac{pi}{z^2}}=sum_{k=0}^{infty}{frac{(-1)^k pi^{2k+1}}{(2k+1)!}(z-0)^{-4k-2}},forall zin C_{r,R},space wherespace frac{1}{R}=lim_{krightarrowinfty}{supleft|frac{{(-1)^k pi^{2k+1}}}{(2k+1)!}right|^{1/k}=0}space andspace r=lim_{krightarrowinfty}{supleft|frac{{(-1)^-k pi^{-2k+1}}}{(-2k+1)!}right|^{1/k}=0}$$ $$iff$$ $$fspace holomorphicspace inspace C_{r,R}=C_{0,infty}space andspaceforall space closedspace pathspace gammain C_{0,infty},space c_{k}=frac{(-1)^kpi^{2k+1}}{(2k+1)!}=frac{1}{2pi i}oint_{gamma}{frac{sinleft({frac{pi}{z^2}}right)}{(z-0)^{k+1}}dz}$$
though if I compute the coefficients $c_k$ on a $gamma$ unitary circle centered at $z=0$: $frac{1}{2pi i}oint_{gamma}{frac{sinleft({frac{pi}{z^2}}right)}{(z-0)^{k+1}}dz}=frac{1}{2pi}int_{0}^{2pi}{frac{sinleft(pi e^{-2 i theta}right)}{e^{itheta k}}dtheta}=0neqfrac{(-1)^kpi^{2k+1}}{(2k+1)!}, forall kinmathbb{Z}$,
so the proposition on the left of the $iff$ can't be true.
Or said in another way, if the expansion exists it is unique, but then it would mean $f$ is zero everywhere which is obviously not true, so the expansion doesn't exist.
P.S. I'm sorry about poor formatting, if someone improves it I promise I'll look at the corrections and learn form them :)
complex-analysis laurent-series
complex-analysis laurent-series
asked Jan 19 at 20:27
EugenioDiPaolaEugenioDiPaola
515
asked Jan 19 at 20:27
EugenioDiPaolaEugenioDiPaola
515
edited Jan 19 at 20:35
EugenioDiPaola
asked Jan 19 at 20:27
EugenioDiPaolaEugenioDiPaola
515
asked Jan 19 at 20:27
EugenioDiPaolaEugenioDiPaola
515
asked Jan 19 at 20:27
EugenioDiPaolaEugenioDiPaola
515
515
1
$begingroup$
math.stackexchange.com/q/1564141
$endgroup$
– Did
Jan 19 at 20:44
add a comment |
1
$begingroup$
math.stackexchange.com/q/1564141
$endgroup$
– Did
Jan 19 at 20:44
1
1
$begingroup$
math.stackexchange.com/q/1564141
$endgroup$
– Did
Jan 19 at 20:44
$begingroup$
math.stackexchange.com/q/1564141
$endgroup$
– Did
Jan 19 at 20:44
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
It does have a Laurent series. Just render the usual Maclaurin series for $sin u$ and then put $u:=pi/z^2$. The negative powers are allowed, they need only to have integer exponents so that the terms are single-valued all around the central point (here, $z=0$). You discover, in fact, that there are infinitely many of such negative power terms. Look here for what that means.
answered Jan 19 at 20:36
Oscar LanziOscar Lanzi
12.9k12136
$endgroup$
$begingroup$
sure, I know what an essential singularity is, and that's what I first did (I was actually writing the Laurent series of $frac{sinleft(frac{pi}{z^2}right)}{(z-1)(z-2)^2}$), but then the fact that i was getting quite a mess and that Mathematica is not able to give a series expansion made me think that being able to write it is not enough to be sure the expansion actually exists.
$endgroup$
– EugenioDiPaola
Jan 19 at 21:18
add a comment |
$begingroup$
My math is limited compared to many here so from my limited understanding the Laurent Series by definition is about doubly infinite power series. Whilst z=0 it cannot expand beyond its initial value and therfore is itself. If z is inside the annulus of integration it will be valid but engender that integration is about area under the line of integration so that if the line has a zero value...
You should be able to follow it on logical from there.
answered Jan 20 at 16:09
RhodieRhodie
1077
$endgroup$
add a comment |
$begingroup$
Ok, found the mistakes, obviously I was wrong, $sinleft({frac{pi}{z^2}}right)$ does indeed have a Laurent expansion at $z=0$, namely $frac{pi}{z^2}-frac{pi^3}{z^6}+frac{pi^5}{z^10}-frac{pi^7}{z^14}+…$
First, the implications are slightly incorrect, then while computing $c_{n}=frac{1}{2pi i} oint_{gamma}{frac{sinleft({frac{pi}{z^2}}right)}{z^{k+1}}dz}$,
in order to check if I was getting the right coefficients I plugged in only positive $n$ in the integral on Wolphram Alpha obtaining always $0$ (and obviously for positive $n$ they are $0$), and so I assumed every coefficient was $0$.
For a generic circle of radius $R$ I have:
$$c_{n}=frac{1}{2pi i} oint_{gamma}{frac{sinleft({frac{pi}{z^2}}right)}{z^{k+1}}dz}
=frac{1}{2pi i} int_{0}^{2pi}{frac{sinleft({frac{pi}{R^2}e^{-i2theta}}right)}{R^n e^{itheta n}}dtheta}$$
Which gives the right coefficients for $n=-4k-2,kinmathbb{N}$ and $0$ for all other $n$.
My guess is that Wolphram Alpha says "series expansion not available" each time it cannot write the expansion as $somespace terms+O(something),space zrightarrow z_0$. For example, it says no series expansion available also for $sin{z}$ at $infty$.
answered Jan 23 at 14:01
EugenioDiPaolaEugenioDiPaola
515
$endgroup$
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%2f3079783%2fwhy-does-sin-left-frac-piz2-right-not-have-a-laurent-series-at-z-0%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It does have a Laurent series. Just render the usual Maclaurin series for $sin u$ and then put $u:=pi/z^2$. The negative powers are allowed, they need only to have integer exponents so that the terms are single-valued all around the central point (here, $z=0$). You discover, in fact, that there are infinitely many of such negative power terms. Look here for what that means.
answered Jan 19 at 20:36
Oscar LanziOscar Lanzi
12.9k12136
$endgroup$
$begingroup$
sure, I know what an essential singularity is, and that's what I first did (I was actually writing the Laurent series of $frac{sinleft(frac{pi}{z^2}right)}{(z-1)(z-2)^2}$), but then the fact that i was getting quite a mess and that Mathematica is not able to give a series expansion made me think that being able to write it is not enough to be sure the expansion actually exists.
$endgroup$
– EugenioDiPaola
Jan 19 at 21:18
add a comment |
$begingroup$
It does have a Laurent series. Just render the usual Maclaurin series for $sin u$ and then put $u:=pi/z^2$. The negative powers are allowed, they need only to have integer exponents so that the terms are single-valued all around the central point (here, $z=0$). You discover, in fact, that there are infinitely many of such negative power terms. Look here for what that means.
answered Jan 19 at 20:36
Oscar LanziOscar Lanzi
12.9k12136
$endgroup$
$begingroup$
sure, I know what an essential singularity is, and that's what I first did (I was actually writing the Laurent series of $frac{sinleft(frac{pi}{z^2}right)}{(z-1)(z-2)^2}$), but then the fact that i was getting quite a mess and that Mathematica is not able to give a series expansion made me think that being able to write it is not enough to be sure the expansion actually exists.
$endgroup$
– EugenioDiPaola
Jan 19 at 21:18
add a comment |
$begingroup$
It does have a Laurent series. Just render the usual Maclaurin series for $sin u$ and then put $u:=pi/z^2$. The negative powers are allowed, they need only to have integer exponents so that the terms are single-valued all around the central point (here, $z=0$). You discover, in fact, that there are infinitely many of such negative power terms. Look here for what that means.
answered Jan 19 at 20:36
Oscar LanziOscar Lanzi
12.9k12136
$endgroup$
It does have a Laurent series. Just render the usual Maclaurin series for $sin u$ and then put $u:=pi/z^2$. The negative powers are allowed, they need only to have integer exponents so that the terms are single-valued all around the central point (here, $z=0$). You discover, in fact, that there are infinitely many of such negative power terms. Look here for what that means.
answered Jan 19 at 20:36
Oscar LanziOscar Lanzi
12.9k12136
answered Jan 19 at 20:36
Oscar LanziOscar Lanzi
12.9k12136
answered Jan 19 at 20:36
Oscar LanziOscar Lanzi
12.9k12136
answered Jan 19 at 20:36
Oscar LanziOscar Lanzi
12.9k12136
12.9k12136
$begingroup$
sure, I know what an essential singularity is, and that's what I first did (I was actually writing the Laurent series of $frac{sinleft(frac{pi}{z^2}right)}{(z-1)(z-2)^2}$), but then the fact that i was getting quite a mess and that Mathematica is not able to give a series expansion made me think that being able to write it is not enough to be sure the expansion actually exists.
$endgroup$
– EugenioDiPaola
Jan 19 at 21:18
add a comment |
$begingroup$
sure, I know what an essential singularity is, and that's what I first did (I was actually writing the Laurent series of $frac{sinleft(frac{pi}{z^2}right)}{(z-1)(z-2)^2}$), but then the fact that i was getting quite a mess and that Mathematica is not able to give a series expansion made me think that being able to write it is not enough to be sure the expansion actually exists.
$endgroup$
– EugenioDiPaola
Jan 19 at 21:18
$begingroup$
sure, I know what an essential singularity is, and that's what I first did (I was actually writing the Laurent series of $frac{sinleft(frac{pi}{z^2}right)}{(z-1)(z-2)^2}$), but then the fact that i was getting quite a mess and that Mathematica is not able to give a series expansion made me think that being able to write it is not enough to be sure the expansion actually exists.
$endgroup$
– EugenioDiPaola
Jan 19 at 21:18
$begingroup$
sure, I know what an essential singularity is, and that's what I first did (I was actually writing the Laurent series of $frac{sinleft(frac{pi}{z^2}right)}{(z-1)(z-2)^2}$), but then the fact that i was getting quite a mess and that Mathematica is not able to give a series expansion made me think that being able to write it is not enough to be sure the expansion actually exists.
$endgroup$
– EugenioDiPaola
Jan 19 at 21:18
add a comment |
$begingroup$
My math is limited compared to many here so from my limited understanding the Laurent Series by definition is about doubly infinite power series. Whilst z=0 it cannot expand beyond its initial value and therfore is itself. If z is inside the annulus of integration it will be valid but engender that integration is about area under the line of integration so that if the line has a zero value...
You should be able to follow it on logical from there.
answered Jan 20 at 16:09
RhodieRhodie
1077
$endgroup$
add a comment |
$begingroup$
My math is limited compared to many here so from my limited understanding the Laurent Series by definition is about doubly infinite power series. Whilst z=0 it cannot expand beyond its initial value and therfore is itself. If z is inside the annulus of integration it will be valid but engender that integration is about area under the line of integration so that if the line has a zero value...
You should be able to follow it on logical from there.
answered Jan 20 at 16:09
RhodieRhodie
1077
$endgroup$
add a comment |
$begingroup$
My math is limited compared to many here so from my limited understanding the Laurent Series by definition is about doubly infinite power series. Whilst z=0 it cannot expand beyond its initial value and therfore is itself. If z is inside the annulus of integration it will be valid but engender that integration is about area under the line of integration so that if the line has a zero value...
You should be able to follow it on logical from there.
answered Jan 20 at 16:09
RhodieRhodie
1077
$endgroup$
My math is limited compared to many here so from my limited understanding the Laurent Series by definition is about doubly infinite power series. Whilst z=0 it cannot expand beyond its initial value and therfore is itself. If z is inside the annulus of integration it will be valid but engender that integration is about area under the line of integration so that if the line has a zero value...
You should be able to follow it on logical from there.
answered Jan 20 at 16:09
RhodieRhodie
1077
answered Jan 20 at 16:09
RhodieRhodie
1077
answered Jan 20 at 16:09
RhodieRhodie
1077
answered Jan 20 at 16:09
RhodieRhodie
1077
1077
add a comment |
add a comment |
$begingroup$
Ok, found the mistakes, obviously I was wrong, $sinleft({frac{pi}{z^2}}right)$ does indeed have a Laurent expansion at $z=0$, namely $frac{pi}{z^2}-frac{pi^3}{z^6}+frac{pi^5}{z^10}-frac{pi^7}{z^14}+…$
First, the implications are slightly incorrect, then while computing $c_{n}=frac{1}{2pi i} oint_{gamma}{frac{sinleft({frac{pi}{z^2}}right)}{z^{k+1}}dz}$,
in order to check if I was getting the right coefficients I plugged in only positive $n$ in the integral on Wolphram Alpha obtaining always $0$ (and obviously for positive $n$ they are $0$), and so I assumed every coefficient was $0$.
For a generic circle of radius $R$ I have:
$$c_{n}=frac{1}{2pi i} oint_{gamma}{frac{sinleft({frac{pi}{z^2}}right)}{z^{k+1}}dz}
=frac{1}{2pi i} int_{0}^{2pi}{frac{sinleft({frac{pi}{R^2}e^{-i2theta}}right)}{R^n e^{itheta n}}dtheta}$$
Which gives the right coefficients for $n=-4k-2,kinmathbb{N}$ and $0$ for all other $n$.
My guess is that Wolphram Alpha says "series expansion not available" each time it cannot write the expansion as $somespace terms+O(something),space zrightarrow z_0$. For example, it says no series expansion available also for $sin{z}$ at $infty$.
answered Jan 23 at 14:01
EugenioDiPaolaEugenioDiPaola
515
$endgroup$
add a comment |
$begingroup$
Ok, found the mistakes, obviously I was wrong, $sinleft({frac{pi}{z^2}}right)$ does indeed have a Laurent expansion at $z=0$, namely $frac{pi}{z^2}-frac{pi^3}{z^6}+frac{pi^5}{z^10}-frac{pi^7}{z^14}+…$
First, the implications are slightly incorrect, then while computing $c_{n}=frac{1}{2pi i} oint_{gamma}{frac{sinleft({frac{pi}{z^2}}right)}{z^{k+1}}dz}$,
in order to check if I was getting the right coefficients I plugged in only positive $n$ in the integral on Wolphram Alpha obtaining always $0$ (and obviously for positive $n$ they are $0$), and so I assumed every coefficient was $0$.
For a generic circle of radius $R$ I have:
$$c_{n}=frac{1}{2pi i} oint_{gamma}{frac{sinleft({frac{pi}{z^2}}right)}{z^{k+1}}dz}
=frac{1}{2pi i} int_{0}^{2pi}{frac{sinleft({frac{pi}{R^2}e^{-i2theta}}right)}{R^n e^{itheta n}}dtheta}$$
Which gives the right coefficients for $n=-4k-2,kinmathbb{N}$ and $0$ for all other $n$.
My guess is that Wolphram Alpha says "series expansion not available" each time it cannot write the expansion as $somespace terms+O(something),space zrightarrow z_0$. For example, it says no series expansion available also for $sin{z}$ at $infty$.
answered Jan 23 at 14:01
EugenioDiPaolaEugenioDiPaola
515
$endgroup$
add a comment |
$begingroup$
Ok, found the mistakes, obviously I was wrong, $sinleft({frac{pi}{z^2}}right)$ does indeed have a Laurent expansion at $z=0$, namely $frac{pi}{z^2}-frac{pi^3}{z^6}+frac{pi^5}{z^10}-frac{pi^7}{z^14}+…$
First, the implications are slightly incorrect, then while computing $c_{n}=frac{1}{2pi i} oint_{gamma}{frac{sinleft({frac{pi}{z^2}}right)}{z^{k+1}}dz}$,
in order to check if I was getting the right coefficients I plugged in only positive $n$ in the integral on Wolphram Alpha obtaining always $0$ (and obviously for positive $n$ they are $0$), and so I assumed every coefficient was $0$.
For a generic circle of radius $R$ I have:
$$c_{n}=frac{1}{2pi i} oint_{gamma}{frac{sinleft({frac{pi}{z^2}}right)}{z^{k+1}}dz}
=frac{1}{2pi i} int_{0}^{2pi}{frac{sinleft({frac{pi}{R^2}e^{-i2theta}}right)}{R^n e^{itheta n}}dtheta}$$
Which gives the right coefficients for $n=-4k-2,kinmathbb{N}$ and $0$ for all other $n$.
My guess is that Wolphram Alpha says "series expansion not available" each time it cannot write the expansion as $somespace terms+O(something),space zrightarrow z_0$. For example, it says no series expansion available also for $sin{z}$ at $infty$.
answered Jan 23 at 14:01
EugenioDiPaolaEugenioDiPaola
515
$endgroup$
Ok, found the mistakes, obviously I was wrong, $sinleft({frac{pi}{z^2}}right)$ does indeed have a Laurent expansion at $z=0$, namely $frac{pi}{z^2}-frac{pi^3}{z^6}+frac{pi^5}{z^10}-frac{pi^7}{z^14}+…$
First, the implications are slightly incorrect, then while computing $c_{n}=frac{1}{2pi i} oint_{gamma}{frac{sinleft({frac{pi}{z^2}}right)}{z^{k+1}}dz}$,
in order to check if I was getting the right coefficients I plugged in only positive $n$ in the integral on Wolphram Alpha obtaining always $0$ (and obviously for positive $n$ they are $0$), and so I assumed every coefficient was $0$.
For a generic circle of radius $R$ I have:
$$c_{n}=frac{1}{2pi i} oint_{gamma}{frac{sinleft({frac{pi}{z^2}}right)}{z^{k+1}}dz}
=frac{1}{2pi i} int_{0}^{2pi}{frac{sinleft({frac{pi}{R^2}e^{-i2theta}}right)}{R^n e^{itheta n}}dtheta}$$
Which gives the right coefficients for $n=-4k-2,kinmathbb{N}$ and $0$ for all other $n$.
My guess is that Wolphram Alpha says "series expansion not available" each time it cannot write the expansion as $somespace terms+O(something),space zrightarrow z_0$. For example, it says no series expansion available also for $sin{z}$ at $infty$.
answered Jan 23 at 14:01
EugenioDiPaolaEugenioDiPaola
515
edited Jan 23 at 16:12
answered Jan 23 at 14:01
EugenioDiPaolaEugenioDiPaola
515
answered Jan 23 at 14:01
EugenioDiPaolaEugenioDiPaola
515
answered Jan 23 at 14:01
EugenioDiPaolaEugenioDiPaola
515
515
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%2f3079783%2fwhy-does-sin-left-frac-piz2-right-not-have-a-laurent-series-at-z-0%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
1
$begingroup$
math.stackexchange.com/q/1564141
$endgroup$
– Did
Jan 19 at 20:44