Find number of roots of a polynomial inside $ |z| < 2 $
$begingroup$
The question asks me to find the number of roots of $ z^6 - 5z^4 + 8z - 1 $ in the domain $ |z| < 2 $. This looks like a Rouché's theorem question, however $ |z^6 + 8z - 1| leq 81 $ while $ |-5z^4| = 80 $. How do I solve this question?
complex-analysis
asked Jan 6 at 9:29
calmcalm
1387
$endgroup$
add a comment |
$begingroup$
The question asks me to find the number of roots of $ z^6 - 5z^4 + 8z - 1 $ in the domain $ |z| < 2 $. This looks like a Rouché's theorem question, however $ |z^6 + 8z - 1| leq 81 $ while $ |-5z^4| = 80 $. How do I solve this question?
complex-analysis
asked Jan 6 at 9:29
calmcalm
1387
$endgroup$
$begingroup$
A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
$endgroup$
– LutzL
Jan 6 at 11:59
$begingroup$
So maybe this is not Rouché's theorem?
$endgroup$
– calm
Jan 6 at 12:02
add a comment |
$begingroup$
The question asks me to find the number of roots of $ z^6 - 5z^4 + 8z - 1 $ in the domain $ |z| < 2 $. This looks like a Rouché's theorem question, however $ |z^6 + 8z - 1| leq 81 $ while $ |-5z^4| = 80 $. How do I solve this question?
complex-analysis
asked Jan 6 at 9:29
calmcalm
1387
$endgroup$
The question asks me to find the number of roots of $ z^6 - 5z^4 + 8z - 1 $ in the domain $ |z| < 2 $. This looks like a Rouché's theorem question, however $ |z^6 + 8z - 1| leq 81 $ while $ |-5z^4| = 80 $. How do I solve this question?
complex-analysis
complex-analysis
asked Jan 6 at 9:29
calmcalm
1387
asked Jan 6 at 9:29
calmcalm
1387
edited Jan 6 at 12:01
calm
asked Jan 6 at 9:29
calmcalm
1387
asked Jan 6 at 9:29
calmcalm
1387
asked Jan 6 at 9:29
calmcalm
1387
1387
$begingroup$
A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
$endgroup$
– LutzL
Jan 6 at 11:59
$begingroup$
So maybe this is not Rouché's theorem?
$endgroup$
– calm
Jan 6 at 12:02
add a comment |
$begingroup$
A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
$endgroup$
– LutzL
Jan 6 at 11:59
$begingroup$
So maybe this is not Rouché's theorem?
$endgroup$
– calm
Jan 6 at 12:02
$begingroup$
A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
$endgroup$
– LutzL
Jan 6 at 11:59
$begingroup$
A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
$endgroup$
– LutzL
Jan 6 at 11:59
$begingroup$
So maybe this is not Rouché's theorem?
$endgroup$
– calm
Jan 6 at 12:02
$begingroup$
So maybe this is not Rouché's theorem?
$endgroup$
– calm
Jan 6 at 12:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
plot of the function values on the circle of radius $2$
Graphically, I find that with $g(z)=-4z^4+8z-1$ the inequality $$|f(z)|+|g(z)|ge 2+|f(z)-g(z)|tag{*}label{ineq:rouche}$$ holds for $|z|=2$. $g$ has all its 4 roots inside $|z|<2$. This allows to apply a version of Rouché where $f$ and $g$ have the same number of roots if $|f|+|g|>|f-g|$. Of course, only after closing the hole in the argumentation by strictly proving the claimed inequality eqref{ineq:rouche}.
The roots of $g$ are actually quite close to the roots of $f$, as the plot of the roots of $tf(z)+(1-t)g(z)=tz^6-(4+t)z^4+8z-1$ shows:
answered Jan 6 at 13:38
LutzLLutzL
59.3k42057
$endgroup$
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%2f3063643%2ffind-number-of-roots-of-a-polynomial-inside-z-2%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
$begingroup$
A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
plot of the function values on the circle of radius $2$
Graphically, I find that with $g(z)=-4z^4+8z-1$ the inequality $$|f(z)|+|g(z)|ge 2+|f(z)-g(z)|tag{*}label{ineq:rouche}$$ holds for $|z|=2$. $g$ has all its 4 roots inside $|z|<2$. This allows to apply a version of Rouché where $f$ and $g$ have the same number of roots if $|f|+|g|>|f-g|$. Of course, only after closing the hole in the argumentation by strictly proving the claimed inequality eqref{ineq:rouche}.
The roots of $g$ are actually quite close to the roots of $f$, as the plot of the roots of $tf(z)+(1-t)g(z)=tz^6-(4+t)z^4+8z-1$ shows:
answered Jan 6 at 13:38
LutzLLutzL
59.3k42057
$endgroup$
add a comment |
$begingroup$
A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
plot of the function values on the circle of radius $2$
Graphically, I find that with $g(z)=-4z^4+8z-1$ the inequality $$|f(z)|+|g(z)|ge 2+|f(z)-g(z)|tag{*}label{ineq:rouche}$$ holds for $|z|=2$. $g$ has all its 4 roots inside $|z|<2$. This allows to apply a version of Rouché where $f$ and $g$ have the same number of roots if $|f|+|g|>|f-g|$. Of course, only after closing the hole in the argumentation by strictly proving the claimed inequality eqref{ineq:rouche}.
The roots of $g$ are actually quite close to the roots of $f$, as the plot of the roots of $tf(z)+(1-t)g(z)=tz^6-(4+t)z^4+8z-1$ shows:
answered Jan 6 at 13:38
LutzLLutzL
59.3k42057
$endgroup$
add a comment |
$begingroup$
A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
plot of the function values on the circle of radius $2$
Graphically, I find that with $g(z)=-4z^4+8z-1$ the inequality $$|f(z)|+|g(z)|ge 2+|f(z)-g(z)|tag{*}label{ineq:rouche}$$ holds for $|z|=2$. $g$ has all its 4 roots inside $|z|<2$. This allows to apply a version of Rouché where $f$ and $g$ have the same number of roots if $|f|+|g|>|f-g|$. Of course, only after closing the hole in the argumentation by strictly proving the claimed inequality eqref{ineq:rouche}.
The roots of $g$ are actually quite close to the roots of $f$, as the plot of the roots of $tf(z)+(1-t)g(z)=tz^6-(4+t)z^4+8z-1$ shows:
answered Jan 6 at 13:38
LutzLLutzL
59.3k42057
$endgroup$
A naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
plot of the function values on the circle of radius $2$
Graphically, I find that with $g(z)=-4z^4+8z-1$ the inequality $$|f(z)|+|g(z)|ge 2+|f(z)-g(z)|tag{*}label{ineq:rouche}$$ holds for $|z|=2$. $g$ has all its 4 roots inside $|z|<2$. This allows to apply a version of Rouché where $f$ and $g$ have the same number of roots if $|f|+|g|>|f-g|$. Of course, only after closing the hole in the argumentation by strictly proving the claimed inequality eqref{ineq:rouche}.
The roots of $g$ are actually quite close to the roots of $f$, as the plot of the roots of $tf(z)+(1-t)g(z)=tz^6-(4+t)z^4+8z-1$ shows:
answered Jan 6 at 13:38
LutzLLutzL
59.3k42057
answered Jan 6 at 13:38
LutzLLutzL
59.3k42057
answered Jan 6 at 13:38
LutzLLutzL
59.3k42057
answered Jan 6 at 13:38
LutzLLutzL
59.3k42057
59.3k42057
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%2f3063643%2ffind-number-of-roots-of-a-polynomial-inside-z-2%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 naive Newton-polygon-type analysis finds 2 roots close to $z=pmsqrt5$, 3 roots at about $z=q^ksqrt[3]{frac85}$ and one root close to $z=frac18$, thus $4$ roots inside $|z|<2$. Numerics moves the root guessed at $sqrt5$ to $z=2.02332..$, making the application of Rouché for the circle $|z|=2$ non-trivial.
$endgroup$
– LutzL
Jan 6 at 11:59
$begingroup$
So maybe this is not Rouché's theorem?
$endgroup$
– calm
Jan 6 at 12:02