Is it possible to observe the residue?
0
$begingroup$
I was wondering if there is a way to recognise the value (or at least to say if it is non zero) just by observing the plot of a complex function? for instance, looking at (x,y,real(f)) or (x,y,im(f) ).
Do you have any idea ?
complex-analysis
asked Nov 28 '18 at 18:47
Marine GalantinMarine Galantin
705215
$endgroup$
add a comment |
0
$begingroup$
I was wondering if there is a way to recognise the value (or at least to say if it is non zero) just by observing the plot of a complex function? for instance, looking at (x,y,real(f)) or (x,y,im(f) ).
Do you have any idea ?
complex-analysis
asked Nov 28 '18 at 18:47
Marine GalantinMarine Galantin
705215
$endgroup$
add a comment |
0
0
0
$begingroup$
I was wondering if there is a way to recognise the value (or at least to say if it is non zero) just by observing the plot of a complex function? for instance, looking at (x,y,real(f)) or (x,y,im(f) ).
Do you have any idea ?
complex-analysis
asked Nov 28 '18 at 18:47
Marine GalantinMarine Galantin
705215
$endgroup$
I was wondering if there is a way to recognise the value (or at least to say if it is non zero) just by observing the plot of a complex function? for instance, looking at (x,y,real(f)) or (x,y,im(f) ).
Do you have any idea ?
complex-analysis
complex-analysis
asked Nov 28 '18 at 18:47
Marine GalantinMarine Galantin
705215
asked Nov 28 '18 at 18:47
Marine GalantinMarine Galantin
705215
asked Nov 28 '18 at 18:47
Marine GalantinMarine Galantin
705215
asked Nov 28 '18 at 18:47
Marine GalantinMarine Galantin
705215
asked Nov 28 '18 at 18:47
Marine GalantinMarine Galantin
705215
705215
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
No. Take $f(z)=frac1{z^2}$ and $g(z)=frac1{z^2}+frac1z$. Then $operatorname{res}_{z=0}bigl(f(z)bigr)=0$ and $operatorname{res}_{z=0}bigl(g(z)bigr)=1$. However, the graphs that you mentioned look the same with respect to both functions.
answered Nov 28 '18 at 18:59
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
$begingroup$
So is there a way to interpret the residue?
$endgroup$
– Marine Galantin
Nov 28 '18 at 19:00
2
$begingroup$
Sure there is. The residue at $a$ being non zere is equivalent to the assertion that $f$ has no primitive near $a$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:02
$begingroup$
Equivalent? Having a primitive Isn't equivalent to f being equal to the taylor expansion?
$endgroup$
– Marine Galantin
Nov 28 '18 at 19:16
$begingroup$
If you are talking about residues, you are talking about singularities. Therefore, there is no Taylor expansion. The function $frac1z$ has no Taylor expansion centered at $0$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:21
1
$begingroup$
But $frac1{z^2}$ has residue $0$ and it has a primitive: $-frac1z$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:28
|
show 2 more comments
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
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%2f3017534%2fis-it-possible-to-observe-the-residue%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
$begingroup$
No. Take $f(z)=frac1{z^2}$ and $g(z)=frac1{z^2}+frac1z$. Then $operatorname{res}_{z=0}bigl(f(z)bigr)=0$ and $operatorname{res}_{z=0}bigl(g(z)bigr)=1$. However, the graphs that you mentioned look the same with respect to both functions.
answered Nov 28 '18 at 18:59
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
$begingroup$
So is there a way to interpret the residue?
$endgroup$
– Marine Galantin
Nov 28 '18 at 19:00
2
$begingroup$
Sure there is. The residue at $a$ being non zere is equivalent to the assertion that $f$ has no primitive near $a$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:02
$begingroup$
Equivalent? Having a primitive Isn't equivalent to f being equal to the taylor expansion?
$endgroup$
– Marine Galantin
Nov 28 '18 at 19:16
$begingroup$
If you are talking about residues, you are talking about singularities. Therefore, there is no Taylor expansion. The function $frac1z$ has no Taylor expansion centered at $0$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:21
1
$begingroup$
But $frac1{z^2}$ has residue $0$ and it has a primitive: $-frac1z$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:28
|
show 2 more comments
1
$begingroup$
No. Take $f(z)=frac1{z^2}$ and $g(z)=frac1{z^2}+frac1z$. Then $operatorname{res}_{z=0}bigl(f(z)bigr)=0$ and $operatorname{res}_{z=0}bigl(g(z)bigr)=1$. However, the graphs that you mentioned look the same with respect to both functions.
answered Nov 28 '18 at 18:59
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
$begingroup$
So is there a way to interpret the residue?
$endgroup$
– Marine Galantin
Nov 28 '18 at 19:00
2
$begingroup$
Sure there is. The residue at $a$ being non zere is equivalent to the assertion that $f$ has no primitive near $a$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:02
$begingroup$
Equivalent? Having a primitive Isn't equivalent to f being equal to the taylor expansion?
$endgroup$
– Marine Galantin
Nov 28 '18 at 19:16
$begingroup$
If you are talking about residues, you are talking about singularities. Therefore, there is no Taylor expansion. The function $frac1z$ has no Taylor expansion centered at $0$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:21
1
$begingroup$
But $frac1{z^2}$ has residue $0$ and it has a primitive: $-frac1z$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:28
|
show 2 more comments
1
1
1
$begingroup$
No. Take $f(z)=frac1{z^2}$ and $g(z)=frac1{z^2}+frac1z$. Then $operatorname{res}_{z=0}bigl(f(z)bigr)=0$ and $operatorname{res}_{z=0}bigl(g(z)bigr)=1$. However, the graphs that you mentioned look the same with respect to both functions.
answered Nov 28 '18 at 18:59
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
No. Take $f(z)=frac1{z^2}$ and $g(z)=frac1{z^2}+frac1z$. Then $operatorname{res}_{z=0}bigl(f(z)bigr)=0$ and $operatorname{res}_{z=0}bigl(g(z)bigr)=1$. However, the graphs that you mentioned look the same with respect to both functions.
answered Nov 28 '18 at 18:59
José Carlos SantosJosé Carlos Santos
153k22123225
answered Nov 28 '18 at 18:59
José Carlos SantosJosé Carlos Santos
153k22123225
answered Nov 28 '18 at 18:59
José Carlos SantosJosé Carlos Santos
153k22123225
answered Nov 28 '18 at 18:59
José Carlos SantosJosé Carlos Santos
153k22123225
153k22123225
$begingroup$
So is there a way to interpret the residue?
$endgroup$
– Marine Galantin
Nov 28 '18 at 19:00
2
$begingroup$
Sure there is. The residue at $a$ being non zere is equivalent to the assertion that $f$ has no primitive near $a$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:02
$begingroup$
Equivalent? Having a primitive Isn't equivalent to f being equal to the taylor expansion?
$endgroup$
– Marine Galantin
Nov 28 '18 at 19:16
$begingroup$
If you are talking about residues, you are talking about singularities. Therefore, there is no Taylor expansion. The function $frac1z$ has no Taylor expansion centered at $0$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:21
1
$begingroup$
But $frac1{z^2}$ has residue $0$ and it has a primitive: $-frac1z$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:28
|
show 2 more comments
$begingroup$
So is there a way to interpret the residue?
$endgroup$
– Marine Galantin
Nov 28 '18 at 19:00
2
$begingroup$
Sure there is. The residue at $a$ being non zere is equivalent to the assertion that $f$ has no primitive near $a$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:02
$begingroup$
Equivalent? Having a primitive Isn't equivalent to f being equal to the taylor expansion?
$endgroup$
– Marine Galantin
Nov 28 '18 at 19:16
$begingroup$
If you are talking about residues, you are talking about singularities. Therefore, there is no Taylor expansion. The function $frac1z$ has no Taylor expansion centered at $0$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:21
1
$begingroup$
But $frac1{z^2}$ has residue $0$ and it has a primitive: $-frac1z$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:28
$begingroup$
So is there a way to interpret the residue?
$endgroup$
– Marine Galantin
Nov 28 '18 at 19:00
$begingroup$
So is there a way to interpret the residue?
$endgroup$
– Marine Galantin
Nov 28 '18 at 19:00
2
2
$begingroup$
Sure there is. The residue at $a$ being non zere is equivalent to the assertion that $f$ has no primitive near $a$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:02
$begingroup$
Sure there is. The residue at $a$ being non zere is equivalent to the assertion that $f$ has no primitive near $a$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:02
$begingroup$
Equivalent? Having a primitive Isn't equivalent to f being equal to the taylor expansion?
$endgroup$
– Marine Galantin
Nov 28 '18 at 19:16
$begingroup$
Equivalent? Having a primitive Isn't equivalent to f being equal to the taylor expansion?
$endgroup$
– Marine Galantin
Nov 28 '18 at 19:16
$begingroup$
If you are talking about residues, you are talking about singularities. Therefore, there is no Taylor expansion. The function $frac1z$ has no Taylor expansion centered at $0$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:21
$begingroup$
If you are talking about residues, you are talking about singularities. Therefore, there is no Taylor expansion. The function $frac1z$ has no Taylor expansion centered at $0$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:21
1
1
$begingroup$
But $frac1{z^2}$ has residue $0$ and it has a primitive: $-frac1z$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:28
$begingroup$
But $frac1{z^2}$ has residue $0$ and it has a primitive: $-frac1z$.
$endgroup$
– José Carlos Santos
Nov 28 '18 at 19:28
|
show 2 more comments
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
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%2f3017534%2fis-it-possible-to-observe-the-residue%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
