Mixing
Zeros of an affine combination of two coprime polynomials
$begingroup$
Suppose $f, g in mathbb R[x]$ are two coprime monic polynomials with
begin{align*}
f(x) = x^n + a_{n-1} x^{n-1} + dots + a_0 \
g(x) = x^m + b_{m-1} x^{m-1} + dots + b_0,
end{align*}
and $n > m$.
Let $h(x,t) = f(x) - t g(x)$ where $t in mathbb R$. I read in a book without explanation that: If $t in [0, infty)$ and as $t to infty$, there are $n-m$ zeros of $h$ that would be asymptotically close to $n-m$ rays passing through $z = -frac{a_{n-1}-b_{m-1}}{n-m}$. As $t to infty$, the $n-m$ zeros are arbitrarily close to
begin{align*}
z = -frac{a_{n-1} - b_{m-1} }{n-m} + t^{1/(n-m)}omega_j,
end{align*}
where $omega_j$'s are the $(n-m)^{text{th}}$ roots of unity. How do we formally prove this argument?
I can only $n-m$ zeros going to infinity by Rouche's Theorem. We can compare $tg$ and $f-tg$ on a sphere that encloses all the zeros of $g$. But have not idea on where the asymptotes come from.
abstract-algebra complex-analysis polynomials
edited Jan 29 at 9:05
Eric Wofsey
191k14216349
asked Jan 28 at 23:58
user1101010user1101010
9011830
$endgroup$
add a comment |
$begingroup$
Suppose $f, g in mathbb R[x]$ are two coprime monic polynomials with
begin{align*}
f(x) = x^n + a_{n-1} x^{n-1} + dots + a_0 \
g(x) = x^m + b_{m-1} x^{m-1} + dots + b_0,
end{align*}
and $n > m$.
Let $h(x,t) = f(x) - t g(x)$ where $t in mathbb R$. I read in a book without explanation that: If $t in [0, infty)$ and as $t to infty$, there are $n-m$ zeros of $h$ that would be asymptotically close to $n-m$ rays passing through $z = -frac{a_{n-1}-b_{m-1}}{n-m}$. As $t to infty$, the $n-m$ zeros are arbitrarily close to
begin{align*}
z = -frac{a_{n-1} - b_{m-1} }{n-m} + t^{1/(n-m)}omega_j,
end{align*}
where $omega_j$'s are the $(n-m)^{text{th}}$ roots of unity. How do we formally prove this argument?
I can only $n-m$ zeros going to infinity by Rouche's Theorem. We can compare $tg$ and $f-tg$ on a sphere that encloses all the zeros of $g$. But have not idea on where the asymptotes come from.
abstract-algebra complex-analysis polynomials
edited Jan 29 at 9:05
Eric Wofsey
191k14216349
asked Jan 28 at 23:58
user1101010user1101010
9011830
$endgroup$
$begingroup$
What is $K$? And have you tried yourself?
$endgroup$
– Servaes
Jan 29 at 0:01
add a comment |
$begingroup$
Suppose $f, g in mathbb R[x]$ are two coprime monic polynomials with
begin{align*}
f(x) = x^n + a_{n-1} x^{n-1} + dots + a_0 \
g(x) = x^m + b_{m-1} x^{m-1} + dots + b_0,
end{align*}
and $n > m$.
Let $h(x,t) = f(x) - t g(x)$ where $t in mathbb R$. I read in a book without explanation that: If $t in [0, infty)$ and as $t to infty$, there are $n-m$ zeros of $h$ that would be asymptotically close to $n-m$ rays passing through $z = -frac{a_{n-1}-b_{m-1}}{n-m}$. As $t to infty$, the $n-m$ zeros are arbitrarily close to
begin{align*}
z = -frac{a_{n-1} - b_{m-1} }{n-m} + t^{1/(n-m)}omega_j,
end{align*}
where $omega_j$'s are the $(n-m)^{text{th}}$ roots of unity. How do we formally prove this argument?
I can only $n-m$ zeros going to infinity by Rouche's Theorem. We can compare $tg$ and $f-tg$ on a sphere that encloses all the zeros of $g$. But have not idea on where the asymptotes come from.
abstract-algebra complex-analysis polynomials
edited Jan 29 at 9:05
Eric Wofsey
191k14216349
asked Jan 28 at 23:58
user1101010user1101010
9011830
$endgroup$
Suppose $f, g in mathbb R[x]$ are two coprime monic polynomials with
begin{align*}
f(x) = x^n + a_{n-1} x^{n-1} + dots + a_0 \
g(x) = x^m + b_{m-1} x^{m-1} + dots + b_0,
end{align*}
and $n > m$.
Let $h(x,t) = f(x) - t g(x)$ where $t in mathbb R$. I read in a book without explanation that: If $t in [0, infty)$ and as $t to infty$, there are $n-m$ zeros of $h$ that would be asymptotically close to $n-m$ rays passing through $z = -frac{a_{n-1}-b_{m-1}}{n-m}$. As $t to infty$, the $n-m$ zeros are arbitrarily close to
begin{align*}
z = -frac{a_{n-1} - b_{m-1} }{n-m} + t^{1/(n-m)}omega_j,
end{align*}
where $omega_j$'s are the $(n-m)^{text{th}}$ roots of unity. How do we formally prove this argument?
I can only $n-m$ zeros going to infinity by Rouche's Theorem. We can compare $tg$ and $f-tg$ on a sphere that encloses all the zeros of $g$. But have not idea on where the asymptotes come from.
abstract-algebra complex-analysis polynomials
abstract-algebra complex-analysis polynomials
edited Jan 29 at 9:05
Eric Wofsey
191k14216349
asked Jan 28 at 23:58
user1101010user1101010
9011830
edited Jan 29 at 9:05
Eric Wofsey
191k14216349
asked Jan 28 at 23:58
user1101010user1101010
9011830
edited Jan 29 at 9:05
Eric Wofsey
191k14216349
edited Jan 29 at 9:05
Eric Wofsey
191k14216349
edited Jan 29 at 9:05
Eric Wofsey
191k14216349
191k14216349
asked Jan 28 at 23:58
user1101010user1101010
9011830
asked Jan 28 at 23:58
user1101010user1101010
9011830
asked Jan 28 at 23:58
user1101010user1101010
9011830
9011830
$begingroup$
What is $K$? And have you tried yourself?
$endgroup$
– Servaes
Jan 29 at 0:01
add a comment |
$begingroup$
What is $K$? And have you tried yourself?
$endgroup$
– Servaes
Jan 29 at 0:01
$begingroup$
What is $K$? And have you tried yourself?
$endgroup$
– Servaes
Jan 29 at 0:01
$begingroup$
What is $K$? And have you tried yourself?
$endgroup$
– Servaes
Jan 29 at 0:01
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note first that if we substitute $x+c$ for $x$ in our polynomials, $a_{n-1}$ increases by $nc$ and $b_{m-1}$ increases by $mc$ and so $-frac{a_{n-1}-b_{m-1}}{n-m}$ decreases by $c$ and the truth of the statement does not change. In particular, taking $c=-b_{m-1}/m$, we may assume $b_{m-1}=0$.
Now let $d=-frac{a_{n-1}}{n-m}$, and let $f_0(x)=f(x)-x^m(x-d)^{n-m}$ and let $g_0(x)=g(x)-x^m$. Note that $deg f_0leq n-2$ and $deg g_0leq m-2$. We have $$h(x,t)=x^m((x-d)^{n-m}-t)+f_0(x)-tg_0(x).$$ Let $r=d+t^{1/(n-m)}omega$ where $omega$ is an $(n-m)$th root of unity and observe that $r$ is a root of $H(x,t)=x^m((x-d)^{n-m}-t)$. The idea is now to show that $H(x,t)$ dominates $f_0(x)-tg_0(x)$ as $x$ goes around a loop near $r$, so that by Rouche's theorem $h(x,t)$ will have to also have a root near $r$.
Fix $epsilon>0$. We see that if $|x-r|=epsilon$ then $|x^m|$ is within a constant factor of $t^{m/(n-m)}$ for $t$ sufficiently large. Also, $$frac{(x-d)^{n-m}-t}{x-r}=frac{(t^{1/(n-m)}omega+(x-r))^{n-m}-(t^{1/(n-m)}omega)^{n-m}}{x-r}$$ is equal to the derivative of $zmapsto z^{n-m}$ at some point within $epsilon$ of $t^{1/(n-m)}omega$ by the mean value theorem. It follows that $|(x-d)^{n-m}-t|$ is within a constant factor of $t^{(n-m-1)/(n-m)}$ for $t$ sufficiently large. We thus see that for $|x-r|<epsilon$, $|H(x,t)|$ is bounded below by a constant multiple of $t^{(n-1)/(n-m)}$ for $t$ sufficiently large.
On the other hand, for $|x-r|=epsilon$, $|f_0(x)|$ is bounded above by a constant multiple of $t^{(n-2)/(n-m)}$ and $|g_0(x)|$ is bounded above by a constant multiple of $t^{(m-2)/(n-m)}$. It follows that $|f_0(x)-tg_0(x)|$ is bounded above by a constant multiple of $t^{(n-2)/(n-m)}$.
Thus, we see that if $t$ is sufficiently large, then $|H(x,t)|>|f_0(x)-tg_0(x)|$ for all $x$ such that $|x-r|=epsilon$. So, by Rouche's theorem, since $H(x,t)$ has a root at $x=r$, we conclude that $h(x,t)=H(x,t)+f_0(x)-tg_0(x)$ must also have a root in the disk $|x-r|leqepsilon$.
answered Jan 29 at 8:49
Eric WofseyEric Wofsey
191k14216349
$endgroup$
add a comment |
$begingroup$
To avoid the roots set $t=s^{n-m}$. Now scale the problem by setting $x=su$. Then
$$
s^{-n}h(su, s^{n-m})=(u^n+s^{-1}a_{n-1}u^{n-1}+...+s^{-n}a_0)-(u^m+s^{-1}b_{m-1}u^{m-1}+...+s^{-m}b_0).$$
For $stoinftyiff ttoinfty$ the remaining polynomial $u^n-u^m$ has an $m$- fold root at $u=0$ and the other roots at the $(m-n)$th unit roots. For $sapproxinfty$ very large or $s^{-1}$ very small, the perturbation due to the next terms in
$$
0=u^m(u^{n-m}-1)+s^{-1}u^{m-1}(a_{n-1}u^{n-m}-b_{m-1})+O(s^{-2}u^{m-2})
$$
gives $m$ very small roots of size $O(s^{-1})$ close to the roots of $0=g(su)$. From each of the unperturbed unit roots $u=ω^k$, $ω^{n-m}=1$, the perturbation will result in a root $u=ω^k(1+s^{-1}v)+O(s^{-2})$, where $v$ solves
$$
0=ω^{mk}(n-m)v+ω^{(m-1)k}(a_{n-1}-b_{m-1})implies v=-ω^{-k}frac{a_{n-1}-b_{m-1}}{n-m}
$$
so that in total
$$
u=ω^k-s^{-1}frac{a_{n-1}-b_{m-1}}{n-m}+O(s^{-2})~~text{ or }~~ x=sω^k-frac{a_{n-1}-b_{m-1}}{n-m}+O(s^{-1})
$$
As an example I plotted the root locations for polynomials with $n=9,m=4$, $a_8=2$, $b_3=-3$, left for the $u$ parametrization, right for the $x$ parametrization. $s$ varies from $1$ to $50$ ($s^{-1}$ in equidistant steps). In light gray are the unit root locations resp. the asymptotic rays.
answered Jan 31 at 21:00
LutzLLutzL
60.1k42057
$endgroup$
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%2f3091568%2fzeros-of-an-affine-combination-of-two-coprime-polynomials%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note first that if we substitute $x+c$ for $x$ in our polynomials, $a_{n-1}$ increases by $nc$ and $b_{m-1}$ increases by $mc$ and so $-frac{a_{n-1}-b_{m-1}}{n-m}$ decreases by $c$ and the truth of the statement does not change. In particular, taking $c=-b_{m-1}/m$, we may assume $b_{m-1}=0$.
Now let $d=-frac{a_{n-1}}{n-m}$, and let $f_0(x)=f(x)-x^m(x-d)^{n-m}$ and let $g_0(x)=g(x)-x^m$. Note that $deg f_0leq n-2$ and $deg g_0leq m-2$. We have $$h(x,t)=x^m((x-d)^{n-m}-t)+f_0(x)-tg_0(x).$$ Let $r=d+t^{1/(n-m)}omega$ where $omega$ is an $(n-m)$th root of unity and observe that $r$ is a root of $H(x,t)=x^m((x-d)^{n-m}-t)$. The idea is now to show that $H(x,t)$ dominates $f_0(x)-tg_0(x)$ as $x$ goes around a loop near $r$, so that by Rouche's theorem $h(x,t)$ will have to also have a root near $r$.
Fix $epsilon>0$. We see that if $|x-r|=epsilon$ then $|x^m|$ is within a constant factor of $t^{m/(n-m)}$ for $t$ sufficiently large. Also, $$frac{(x-d)^{n-m}-t}{x-r}=frac{(t^{1/(n-m)}omega+(x-r))^{n-m}-(t^{1/(n-m)}omega)^{n-m}}{x-r}$$ is equal to the derivative of $zmapsto z^{n-m}$ at some point within $epsilon$ of $t^{1/(n-m)}omega$ by the mean value theorem. It follows that $|(x-d)^{n-m}-t|$ is within a constant factor of $t^{(n-m-1)/(n-m)}$ for $t$ sufficiently large. We thus see that for $|x-r|<epsilon$, $|H(x,t)|$ is bounded below by a constant multiple of $t^{(n-1)/(n-m)}$ for $t$ sufficiently large.
On the other hand, for $|x-r|=epsilon$, $|f_0(x)|$ is bounded above by a constant multiple of $t^{(n-2)/(n-m)}$ and $|g_0(x)|$ is bounded above by a constant multiple of $t^{(m-2)/(n-m)}$. It follows that $|f_0(x)-tg_0(x)|$ is bounded above by a constant multiple of $t^{(n-2)/(n-m)}$.
Thus, we see that if $t$ is sufficiently large, then $|H(x,t)|>|f_0(x)-tg_0(x)|$ for all $x$ such that $|x-r|=epsilon$. So, by Rouche's theorem, since $H(x,t)$ has a root at $x=r$, we conclude that $h(x,t)=H(x,t)+f_0(x)-tg_0(x)$ must also have a root in the disk $|x-r|leqepsilon$.
answered Jan 29 at 8:49
Eric WofseyEric Wofsey
191k14216349
$endgroup$
add a comment |
$begingroup$
Note first that if we substitute $x+c$ for $x$ in our polynomials, $a_{n-1}$ increases by $nc$ and $b_{m-1}$ increases by $mc$ and so $-frac{a_{n-1}-b_{m-1}}{n-m}$ decreases by $c$ and the truth of the statement does not change. In particular, taking $c=-b_{m-1}/m$, we may assume $b_{m-1}=0$.
Now let $d=-frac{a_{n-1}}{n-m}$, and let $f_0(x)=f(x)-x^m(x-d)^{n-m}$ and let $g_0(x)=g(x)-x^m$. Note that $deg f_0leq n-2$ and $deg g_0leq m-2$. We have $$h(x,t)=x^m((x-d)^{n-m}-t)+f_0(x)-tg_0(x).$$ Let $r=d+t^{1/(n-m)}omega$ where $omega$ is an $(n-m)$th root of unity and observe that $r$ is a root of $H(x,t)=x^m((x-d)^{n-m}-t)$. The idea is now to show that $H(x,t)$ dominates $f_0(x)-tg_0(x)$ as $x$ goes around a loop near $r$, so that by Rouche's theorem $h(x,t)$ will have to also have a root near $r$.
Fix $epsilon>0$. We see that if $|x-r|=epsilon$ then $|x^m|$ is within a constant factor of $t^{m/(n-m)}$ for $t$ sufficiently large. Also, $$frac{(x-d)^{n-m}-t}{x-r}=frac{(t^{1/(n-m)}omega+(x-r))^{n-m}-(t^{1/(n-m)}omega)^{n-m}}{x-r}$$ is equal to the derivative of $zmapsto z^{n-m}$ at some point within $epsilon$ of $t^{1/(n-m)}omega$ by the mean value theorem. It follows that $|(x-d)^{n-m}-t|$ is within a constant factor of $t^{(n-m-1)/(n-m)}$ for $t$ sufficiently large. We thus see that for $|x-r|<epsilon$, $|H(x,t)|$ is bounded below by a constant multiple of $t^{(n-1)/(n-m)}$ for $t$ sufficiently large.
On the other hand, for $|x-r|=epsilon$, $|f_0(x)|$ is bounded above by a constant multiple of $t^{(n-2)/(n-m)}$ and $|g_0(x)|$ is bounded above by a constant multiple of $t^{(m-2)/(n-m)}$. It follows that $|f_0(x)-tg_0(x)|$ is bounded above by a constant multiple of $t^{(n-2)/(n-m)}$.
Thus, we see that if $t$ is sufficiently large, then $|H(x,t)|>|f_0(x)-tg_0(x)|$ for all $x$ such that $|x-r|=epsilon$. So, by Rouche's theorem, since $H(x,t)$ has a root at $x=r$, we conclude that $h(x,t)=H(x,t)+f_0(x)-tg_0(x)$ must also have a root in the disk $|x-r|leqepsilon$.
answered Jan 29 at 8:49
Eric WofseyEric Wofsey
191k14216349
$endgroup$
add a comment |
$begingroup$
Note first that if we substitute $x+c$ for $x$ in our polynomials, $a_{n-1}$ increases by $nc$ and $b_{m-1}$ increases by $mc$ and so $-frac{a_{n-1}-b_{m-1}}{n-m}$ decreases by $c$ and the truth of the statement does not change. In particular, taking $c=-b_{m-1}/m$, we may assume $b_{m-1}=0$.
Now let $d=-frac{a_{n-1}}{n-m}$, and let $f_0(x)=f(x)-x^m(x-d)^{n-m}$ and let $g_0(x)=g(x)-x^m$. Note that $deg f_0leq n-2$ and $deg g_0leq m-2$. We have $$h(x,t)=x^m((x-d)^{n-m}-t)+f_0(x)-tg_0(x).$$ Let $r=d+t^{1/(n-m)}omega$ where $omega$ is an $(n-m)$th root of unity and observe that $r$ is a root of $H(x,t)=x^m((x-d)^{n-m}-t)$. The idea is now to show that $H(x,t)$ dominates $f_0(x)-tg_0(x)$ as $x$ goes around a loop near $r$, so that by Rouche's theorem $h(x,t)$ will have to also have a root near $r$.
Fix $epsilon>0$. We see that if $|x-r|=epsilon$ then $|x^m|$ is within a constant factor of $t^{m/(n-m)}$ for $t$ sufficiently large. Also, $$frac{(x-d)^{n-m}-t}{x-r}=frac{(t^{1/(n-m)}omega+(x-r))^{n-m}-(t^{1/(n-m)}omega)^{n-m}}{x-r}$$ is equal to the derivative of $zmapsto z^{n-m}$ at some point within $epsilon$ of $t^{1/(n-m)}omega$ by the mean value theorem. It follows that $|(x-d)^{n-m}-t|$ is within a constant factor of $t^{(n-m-1)/(n-m)}$ for $t$ sufficiently large. We thus see that for $|x-r|<epsilon$, $|H(x,t)|$ is bounded below by a constant multiple of $t^{(n-1)/(n-m)}$ for $t$ sufficiently large.
On the other hand, for $|x-r|=epsilon$, $|f_0(x)|$ is bounded above by a constant multiple of $t^{(n-2)/(n-m)}$ and $|g_0(x)|$ is bounded above by a constant multiple of $t^{(m-2)/(n-m)}$. It follows that $|f_0(x)-tg_0(x)|$ is bounded above by a constant multiple of $t^{(n-2)/(n-m)}$.
Thus, we see that if $t$ is sufficiently large, then $|H(x,t)|>|f_0(x)-tg_0(x)|$ for all $x$ such that $|x-r|=epsilon$. So, by Rouche's theorem, since $H(x,t)$ has a root at $x=r$, we conclude that $h(x,t)=H(x,t)+f_0(x)-tg_0(x)$ must also have a root in the disk $|x-r|leqepsilon$.
answered Jan 29 at 8:49
Eric WofseyEric Wofsey
191k14216349
$endgroup$
Note first that if we substitute $x+c$ for $x$ in our polynomials, $a_{n-1}$ increases by $nc$ and $b_{m-1}$ increases by $mc$ and so $-frac{a_{n-1}-b_{m-1}}{n-m}$ decreases by $c$ and the truth of the statement does not change. In particular, taking $c=-b_{m-1}/m$, we may assume $b_{m-1}=0$.
Now let $d=-frac{a_{n-1}}{n-m}$, and let $f_0(x)=f(x)-x^m(x-d)^{n-m}$ and let $g_0(x)=g(x)-x^m$. Note that $deg f_0leq n-2$ and $deg g_0leq m-2$. We have $$h(x,t)=x^m((x-d)^{n-m}-t)+f_0(x)-tg_0(x).$$ Let $r=d+t^{1/(n-m)}omega$ where $omega$ is an $(n-m)$th root of unity and observe that $r$ is a root of $H(x,t)=x^m((x-d)^{n-m}-t)$. The idea is now to show that $H(x,t)$ dominates $f_0(x)-tg_0(x)$ as $x$ goes around a loop near $r$, so that by Rouche's theorem $h(x,t)$ will have to also have a root near $r$.
Fix $epsilon>0$. We see that if $|x-r|=epsilon$ then $|x^m|$ is within a constant factor of $t^{m/(n-m)}$ for $t$ sufficiently large. Also, $$frac{(x-d)^{n-m}-t}{x-r}=frac{(t^{1/(n-m)}omega+(x-r))^{n-m}-(t^{1/(n-m)}omega)^{n-m}}{x-r}$$ is equal to the derivative of $zmapsto z^{n-m}$ at some point within $epsilon$ of $t^{1/(n-m)}omega$ by the mean value theorem. It follows that $|(x-d)^{n-m}-t|$ is within a constant factor of $t^{(n-m-1)/(n-m)}$ for $t$ sufficiently large. We thus see that for $|x-r|<epsilon$, $|H(x,t)|$ is bounded below by a constant multiple of $t^{(n-1)/(n-m)}$ for $t$ sufficiently large.
On the other hand, for $|x-r|=epsilon$, $|f_0(x)|$ is bounded above by a constant multiple of $t^{(n-2)/(n-m)}$ and $|g_0(x)|$ is bounded above by a constant multiple of $t^{(m-2)/(n-m)}$. It follows that $|f_0(x)-tg_0(x)|$ is bounded above by a constant multiple of $t^{(n-2)/(n-m)}$.
Thus, we see that if $t$ is sufficiently large, then $|H(x,t)|>|f_0(x)-tg_0(x)|$ for all $x$ such that $|x-r|=epsilon$. So, by Rouche's theorem, since $H(x,t)$ has a root at $x=r$, we conclude that $h(x,t)=H(x,t)+f_0(x)-tg_0(x)$ must also have a root in the disk $|x-r|leqepsilon$.
answered Jan 29 at 8:49
Eric WofseyEric Wofsey
191k14216349
edited Jan 29 at 9:03
answered Jan 29 at 8:49
Eric WofseyEric Wofsey
191k14216349
answered Jan 29 at 8:49
Eric WofseyEric Wofsey
191k14216349
answered Jan 29 at 8:49
Eric WofseyEric Wofsey
191k14216349
191k14216349
add a comment |
add a comment |
$begingroup$
To avoid the roots set $t=s^{n-m}$. Now scale the problem by setting $x=su$. Then
$$
s^{-n}h(su, s^{n-m})=(u^n+s^{-1}a_{n-1}u^{n-1}+...+s^{-n}a_0)-(u^m+s^{-1}b_{m-1}u^{m-1}+...+s^{-m}b_0).$$
For $stoinftyiff ttoinfty$ the remaining polynomial $u^n-u^m$ has an $m$- fold root at $u=0$ and the other roots at the $(m-n)$th unit roots. For $sapproxinfty$ very large or $s^{-1}$ very small, the perturbation due to the next terms in
$$
0=u^m(u^{n-m}-1)+s^{-1}u^{m-1}(a_{n-1}u^{n-m}-b_{m-1})+O(s^{-2}u^{m-2})
$$
gives $m$ very small roots of size $O(s^{-1})$ close to the roots of $0=g(su)$. From each of the unperturbed unit roots $u=ω^k$, $ω^{n-m}=1$, the perturbation will result in a root $u=ω^k(1+s^{-1}v)+O(s^{-2})$, where $v$ solves
$$
0=ω^{mk}(n-m)v+ω^{(m-1)k}(a_{n-1}-b_{m-1})implies v=-ω^{-k}frac{a_{n-1}-b_{m-1}}{n-m}
$$
so that in total
$$
u=ω^k-s^{-1}frac{a_{n-1}-b_{m-1}}{n-m}+O(s^{-2})~~text{ or }~~ x=sω^k-frac{a_{n-1}-b_{m-1}}{n-m}+O(s^{-1})
$$
As an example I plotted the root locations for polynomials with $n=9,m=4$, $a_8=2$, $b_3=-3$, left for the $u$ parametrization, right for the $x$ parametrization. $s$ varies from $1$ to $50$ ($s^{-1}$ in equidistant steps). In light gray are the unit root locations resp. the asymptotic rays.
answered Jan 31 at 21:00
LutzLLutzL
60.1k42057
$endgroup$
add a comment |
$begingroup$
To avoid the roots set $t=s^{n-m}$. Now scale the problem by setting $x=su$. Then
$$
s^{-n}h(su, s^{n-m})=(u^n+s^{-1}a_{n-1}u^{n-1}+...+s^{-n}a_0)-(u^m+s^{-1}b_{m-1}u^{m-1}+...+s^{-m}b_0).$$
For $stoinftyiff ttoinfty$ the remaining polynomial $u^n-u^m$ has an $m$- fold root at $u=0$ and the other roots at the $(m-n)$th unit roots. For $sapproxinfty$ very large or $s^{-1}$ very small, the perturbation due to the next terms in
$$
0=u^m(u^{n-m}-1)+s^{-1}u^{m-1}(a_{n-1}u^{n-m}-b_{m-1})+O(s^{-2}u^{m-2})
$$
gives $m$ very small roots of size $O(s^{-1})$ close to the roots of $0=g(su)$. From each of the unperturbed unit roots $u=ω^k$, $ω^{n-m}=1$, the perturbation will result in a root $u=ω^k(1+s^{-1}v)+O(s^{-2})$, where $v$ solves
$$
0=ω^{mk}(n-m)v+ω^{(m-1)k}(a_{n-1}-b_{m-1})implies v=-ω^{-k}frac{a_{n-1}-b_{m-1}}{n-m}
$$
so that in total
$$
u=ω^k-s^{-1}frac{a_{n-1}-b_{m-1}}{n-m}+O(s^{-2})~~text{ or }~~ x=sω^k-frac{a_{n-1}-b_{m-1}}{n-m}+O(s^{-1})
$$
As an example I plotted the root locations for polynomials with $n=9,m=4$, $a_8=2$, $b_3=-3$, left for the $u$ parametrization, right for the $x$ parametrization. $s$ varies from $1$ to $50$ ($s^{-1}$ in equidistant steps). In light gray are the unit root locations resp. the asymptotic rays.
answered Jan 31 at 21:00
LutzLLutzL
60.1k42057
$endgroup$
add a comment |
$begingroup$
To avoid the roots set $t=s^{n-m}$. Now scale the problem by setting $x=su$. Then
$$
s^{-n}h(su, s^{n-m})=(u^n+s^{-1}a_{n-1}u^{n-1}+...+s^{-n}a_0)-(u^m+s^{-1}b_{m-1}u^{m-1}+...+s^{-m}b_0).$$
For $stoinftyiff ttoinfty$ the remaining polynomial $u^n-u^m$ has an $m$- fold root at $u=0$ and the other roots at the $(m-n)$th unit roots. For $sapproxinfty$ very large or $s^{-1}$ very small, the perturbation due to the next terms in
$$
0=u^m(u^{n-m}-1)+s^{-1}u^{m-1}(a_{n-1}u^{n-m}-b_{m-1})+O(s^{-2}u^{m-2})
$$
gives $m$ very small roots of size $O(s^{-1})$ close to the roots of $0=g(su)$. From each of the unperturbed unit roots $u=ω^k$, $ω^{n-m}=1$, the perturbation will result in a root $u=ω^k(1+s^{-1}v)+O(s^{-2})$, where $v$ solves
$$
0=ω^{mk}(n-m)v+ω^{(m-1)k}(a_{n-1}-b_{m-1})implies v=-ω^{-k}frac{a_{n-1}-b_{m-1}}{n-m}
$$
so that in total
$$
u=ω^k-s^{-1}frac{a_{n-1}-b_{m-1}}{n-m}+O(s^{-2})~~text{ or }~~ x=sω^k-frac{a_{n-1}-b_{m-1}}{n-m}+O(s^{-1})
$$
As an example I plotted the root locations for polynomials with $n=9,m=4$, $a_8=2$, $b_3=-3$, left for the $u$ parametrization, right for the $x$ parametrization. $s$ varies from $1$ to $50$ ($s^{-1}$ in equidistant steps). In light gray are the unit root locations resp. the asymptotic rays.
answered Jan 31 at 21:00
LutzLLutzL
60.1k42057
$endgroup$
To avoid the roots set $t=s^{n-m}$. Now scale the problem by setting $x=su$. Then
$$
s^{-n}h(su, s^{n-m})=(u^n+s^{-1}a_{n-1}u^{n-1}+...+s^{-n}a_0)-(u^m+s^{-1}b_{m-1}u^{m-1}+...+s^{-m}b_0).$$
For $stoinftyiff ttoinfty$ the remaining polynomial $u^n-u^m$ has an $m$- fold root at $u=0$ and the other roots at the $(m-n)$th unit roots. For $sapproxinfty$ very large or $s^{-1}$ very small, the perturbation due to the next terms in
$$
0=u^m(u^{n-m}-1)+s^{-1}u^{m-1}(a_{n-1}u^{n-m}-b_{m-1})+O(s^{-2}u^{m-2})
$$
gives $m$ very small roots of size $O(s^{-1})$ close to the roots of $0=g(su)$. From each of the unperturbed unit roots $u=ω^k$, $ω^{n-m}=1$, the perturbation will result in a root $u=ω^k(1+s^{-1}v)+O(s^{-2})$, where $v$ solves
$$
0=ω^{mk}(n-m)v+ω^{(m-1)k}(a_{n-1}-b_{m-1})implies v=-ω^{-k}frac{a_{n-1}-b_{m-1}}{n-m}
$$
so that in total
$$
u=ω^k-s^{-1}frac{a_{n-1}-b_{m-1}}{n-m}+O(s^{-2})~~text{ or }~~ x=sω^k-frac{a_{n-1}-b_{m-1}}{n-m}+O(s^{-1})
$$
As an example I plotted the root locations for polynomials with $n=9,m=4$, $a_8=2$, $b_3=-3$, left for the $u$ parametrization, right for the $x$ parametrization. $s$ varies from $1$ to $50$ ($s^{-1}$ in equidistant steps). In light gray are the unit root locations resp. the asymptotic rays.
answered Jan 31 at 21:00
LutzLLutzL
60.1k42057
answered Jan 31 at 21:00
LutzLLutzL
60.1k42057
answered Jan 31 at 21:00
LutzLLutzL
60.1k42057
answered Jan 31 at 21:00
LutzLLutzL
60.1k42057
60.1k42057
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%2f3091568%2fzeros-of-an-affine-combination-of-two-coprime-polynomials%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
$begingroup$
What is $K$? And have you tried yourself?
$endgroup$
– Servaes
Jan 29 at 0:01