Mixing
Constructing explicit connected path for coefficients of monic polynomials with roots lying in the open left half plane?
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
This is a bijection $pi$ between $mathbb C^n$ and the set of monic polynomials of degree $n$ defined by
beginalign*
pi(alpha_n-1, alpha_n-2, dots, alpha_0) = t^n + alpha_n-1 t^n-1 + dots + alpha_0.
endalign*
Let
beginalign*
E = zeta in mathbb C^n: textAll roots of pi(zeta) text have negative real parts.
endalign*
By Vieta's formulas, we know $E$ is connected. Indeed, if we let $Delta = z in mathbb C: textRe(z) < 0$. Then $E$ is the image of $underbraceDelta times Delta times dots times Delta_n text times$ under a continuous function. I am wondering whether it is easy to construct some explicit path. That is, suppose we have $a = (a_n-1, dots, a_0), b= (b_n-1, dots, b_0) in E$, can we construct a continuous path $gamma: [0,1] to E$ with $gamma(0) = a$ and $gamma(1)=b$ explicitly?
linear-algebra general-topology polynomials connectedness path-connected
asked Jul 16 at 17:58
user9527
925525
add a comment |Â
up vote
0
down vote
favorite
This is a bijection $pi$ between $mathbb C^n$ and the set of monic polynomials of degree $n$ defined by
beginalign*
pi(alpha_n-1, alpha_n-2, dots, alpha_0) = t^n + alpha_n-1 t^n-1 + dots + alpha_0.
endalign*
Let
beginalign*
E = zeta in mathbb C^n: textAll roots of pi(zeta) text have negative real parts.
endalign*
By Vieta's formulas, we know $E$ is connected. Indeed, if we let $Delta = z in mathbb C: textRe(z) < 0$. Then $E$ is the image of $underbraceDelta times Delta times dots times Delta_n text times$ under a continuous function. I am wondering whether it is easy to construct some explicit path. That is, suppose we have $a = (a_n-1, dots, a_0), b= (b_n-1, dots, b_0) in E$, can we construct a continuous path $gamma: [0,1] to E$ with $gamma(0) = a$ and $gamma(1)=b$ explicitly?
linear-algebra general-topology polynomials connectedness path-connected
asked Jul 16 at 17:58
user9527
925525
It is irritating to begin your question with "Suppose we have a monic polynomial ...". I suggest to do it like that: There is bijection $pi$ between $mathbbC^n$ and the set of monic polynomials of degree $n$ defined by $pi(alpha_n-1,...,alpha_0) = ...$. Then define $E = zeta in mathbbC^n mid textAll roots of pi(zeta) text have a negative real part $. Can you make explicit which continuous function has image $E$?
– Paul Frost
Jul 16 at 23:13
@PaulFrost: Thanks for your editing suggestion. The function is elementary symmetric functions of the roots given by Vieta's formulas.
– user9527
Jul 17 at 4:33
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This is a bijection $pi$ between $mathbb C^n$ and the set of monic polynomials of degree $n$ defined by
beginalign*
pi(alpha_n-1, alpha_n-2, dots, alpha_0) = t^n + alpha_n-1 t^n-1 + dots + alpha_0.
endalign*
Let
beginalign*
E = zeta in mathbb C^n: textAll roots of pi(zeta) text have negative real parts.
endalign*
By Vieta's formulas, we know $E$ is connected. Indeed, if we let $Delta = z in mathbb C: textRe(z) < 0$. Then $E$ is the image of $underbraceDelta times Delta times dots times Delta_n text times$ under a continuous function. I am wondering whether it is easy to construct some explicit path. That is, suppose we have $a = (a_n-1, dots, a_0), b= (b_n-1, dots, b_0) in E$, can we construct a continuous path $gamma: [0,1] to E$ with $gamma(0) = a$ and $gamma(1)=b$ explicitly?
linear-algebra general-topology polynomials connectedness path-connected
asked Jul 16 at 17:58
user9527
925525
This is a bijection $pi$ between $mathbb C^n$ and the set of monic polynomials of degree $n$ defined by
beginalign*
pi(alpha_n-1, alpha_n-2, dots, alpha_0) = t^n + alpha_n-1 t^n-1 + dots + alpha_0.
endalign*
Let
beginalign*
E = zeta in mathbb C^n: textAll roots of pi(zeta) text have negative real parts.
endalign*
By Vieta's formulas, we know $E$ is connected. Indeed, if we let $Delta = z in mathbb C: textRe(z) < 0$. Then $E$ is the image of $underbraceDelta times Delta times dots times Delta_n text times$ under a continuous function. I am wondering whether it is easy to construct some explicit path. That is, suppose we have $a = (a_n-1, dots, a_0), b= (b_n-1, dots, b_0) in E$, can we construct a continuous path $gamma: [0,1] to E$ with $gamma(0) = a$ and $gamma(1)=b$ explicitly?
linear-algebra general-topology polynomials connectedness path-connected
asked Jul 16 at 17:58
user9527
925525
edited Jul 17 at 4:59
asked Jul 16 at 17:58
user9527
925525
asked Jul 16 at 17:58
user9527
925525
asked Jul 16 at 17:58
user9527
925525
925525
It is irritating to begin your question with "Suppose we have a monic polynomial ...". I suggest to do it like that: There is bijection $pi$ between $mathbbC^n$ and the set of monic polynomials of degree $n$ defined by $pi(alpha_n-1,...,alpha_0) = ...$. Then define $E = zeta in mathbbC^n mid textAll roots of pi(zeta) text have a negative real part $. Can you make explicit which continuous function has image $E$?
– Paul Frost
Jul 16 at 23:13
@PaulFrost: Thanks for your editing suggestion. The function is elementary symmetric functions of the roots given by Vieta's formulas.
– user9527
Jul 17 at 4:33
add a comment |Â
It is irritating to begin your question with "Suppose we have a monic polynomial ...". I suggest to do it like that: There is bijection $pi$ between $mathbbC^n$ and the set of monic polynomials of degree $n$ defined by $pi(alpha_n-1,...,alpha_0) = ...$. Then define $E = zeta in mathbbC^n mid textAll roots of pi(zeta) text have a negative real part $. Can you make explicit which continuous function has image $E$?
– Paul Frost
Jul 16 at 23:13
@PaulFrost: Thanks for your editing suggestion. The function is elementary symmetric functions of the roots given by Vieta's formulas.
– user9527
Jul 17 at 4:33
It is irritating to begin your question with "Suppose we have a monic polynomial ...". I suggest to do it like that: There is bijection $pi$ between $mathbbC^n$ and the set of monic polynomials of degree $n$ defined by $pi(alpha_n-1,...,alpha_0) = ...$. Then define $E = zeta in mathbbC^n mid textAll roots of pi(zeta) text have a negative real part $. Can you make explicit which continuous function has image $E$?
– Paul Frost
Jul 16 at 23:13
It is irritating to begin your question with "Suppose we have a monic polynomial ...". I suggest to do it like that: There is bijection $pi$ between $mathbbC^n$ and the set of monic polynomials of degree $n$ defined by $pi(alpha_n-1,...,alpha_0) = ...$. Then define $E = zeta in mathbbC^n mid textAll roots of pi(zeta) text have a negative real part $. Can you make explicit which continuous function has image $E$?
– Paul Frost
Jul 16 at 23:13
@PaulFrost: Thanks for your editing suggestion. The function is elementary symmetric functions of the roots given by Vieta's formulas.
– user9527
Jul 17 at 4:33
@PaulFrost: Thanks for your editing suggestion. The function is elementary symmetric functions of the roots given by Vieta's formulas.
– user9527
Jul 17 at 4:33
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
If $e_0, e_1, ... ,e_n$ denote the elementary symmetric polynomials in $n$ variables, we get a continuous map
$$sigma : mathbbC^n to mathbbC^n, sigma(zeta) = (-e_1(zeta), e_2(zeta),...,(-1)^n e_n(zeta)) .$$
You may interpret $sigma(zeta)$ as the coefficients of a monic polynomial $p_zeta$ of degree $n$ having as roots the coordinates of $zeta$. In fact, $p_zeta = pi(sigma(zeta))$ where $pi$ was introduced in the question.
We have $E = sigma(Delta^n)$. This means that the set $E$ could alternatively be defined without reference to monic polynomials and their roots. Let $s : Delta^n to E$ denote the restriction.
Now it depends on what you understand by "explicitly construct some path from $a$ to $b$". You find $a', b' in Delta^n$ such that $s(a') = a, s(b') = b$. Define $u : [0,1] to Delta^n, u(t) = (1-t)a' + tb'$. This is a path connecting $a'$ and $b'$, therefore $s circ u$ is a path connecting $a$ and $b$. However, it is not obvious how to find $a',b'$. There is no general method to express $a',b'$ as a function of $a,b$ and in that sense $a',b'$ cannot be made explicit (if you could, you would have a solution formula for polynomials of arbitrary degree $n$ which in fact only exists for $n le 4$).
Let us finally consider the map $sigma$. For each $eta in mathbbC^n$ the inverse image $sigma^-1(eta)$ consists of all $zeta = (z_1,....z_n)$ such that the $ z_1,....z_n $ is the set of all roots of $pi(eta)$. Hence $sigma$ is surjective. Moreover, the symmetric group $S_n$ of $n$ elements operates by permutation of coordinates on $mathbbC^n$; the fibres $sigma^-1(eta)$ agree with the orbits of this operation. Therefore $sigma$ induces a bijection $sigma': mathbbC^n/S_n to mathbbC^n$.
We show that $sigma$ is a closed map. It is known that $max1, lvert a_n-1 rvert, ... , lvert a_0 rvert $ is an upper bound for the abolute values of the roots of $z^n + a_n-1z^n+-1 + ... + a_1z + a_0$. This implies that $sigma^-1(B)$ is bounded if $B$ is bounded. Now let $A subset mathbbC^n$ be closed and $(eta_m)$ be a sequence in $sigma(A)$ converging to some $eta in mathbbC^n$. Then $Y = eta_m mid m in mathbbN $ is bounded, hence $Z = sigma^-1(Y)$ is bounded. Choose $zeta_m in A$ such that $sigma(zeta_m) = eta_n$. Since $zeta_m in Z$, we see that $(zeta_m)$ is a bounded sequence and has a convergent subsequence with limit $zeta$. Since $A$ is closed, $zeta in A$. By continuity of $sigma$ we conclude $eta in sigma(A)$.
This implies that $sigma$ is an identification map. Therefore, if we give $mathbbC^n/S_n$ the quotient topology induced by the canonical quotient function $mathbbC^n to mathbbC^n/S_n$, we see that $sigma'$ is a homeomorphism.
answered Jul 17 at 9:51
Paul Frost
3,703420
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If $e_0, e_1, ... ,e_n$ denote the elementary symmetric polynomials in $n$ variables, we get a continuous map
$$sigma : mathbbC^n to mathbbC^n, sigma(zeta) = (-e_1(zeta), e_2(zeta),...,(-1)^n e_n(zeta)) .$$
You may interpret $sigma(zeta)$ as the coefficients of a monic polynomial $p_zeta$ of degree $n$ having as roots the coordinates of $zeta$. In fact, $p_zeta = pi(sigma(zeta))$ where $pi$ was introduced in the question.
We have $E = sigma(Delta^n)$. This means that the set $E$ could alternatively be defined without reference to monic polynomials and their roots. Let $s : Delta^n to E$ denote the restriction.
Now it depends on what you understand by "explicitly construct some path from $a$ to $b$". You find $a', b' in Delta^n$ such that $s(a') = a, s(b') = b$. Define $u : [0,1] to Delta^n, u(t) = (1-t)a' + tb'$. This is a path connecting $a'$ and $b'$, therefore $s circ u$ is a path connecting $a$ and $b$. However, it is not obvious how to find $a',b'$. There is no general method to express $a',b'$ as a function of $a,b$ and in that sense $a',b'$ cannot be made explicit (if you could, you would have a solution formula for polynomials of arbitrary degree $n$ which in fact only exists for $n le 4$).
Let us finally consider the map $sigma$. For each $eta in mathbbC^n$ the inverse image $sigma^-1(eta)$ consists of all $zeta = (z_1,....z_n)$ such that the $ z_1,....z_n $ is the set of all roots of $pi(eta)$. Hence $sigma$ is surjective. Moreover, the symmetric group $S_n$ of $n$ elements operates by permutation of coordinates on $mathbbC^n$; the fibres $sigma^-1(eta)$ agree with the orbits of this operation. Therefore $sigma$ induces a bijection $sigma': mathbbC^n/S_n to mathbbC^n$.
We show that $sigma$ is a closed map. It is known that $max1, lvert a_n-1 rvert, ... , lvert a_0 rvert $ is an upper bound for the abolute values of the roots of $z^n + a_n-1z^n+-1 + ... + a_1z + a_0$. This implies that $sigma^-1(B)$ is bounded if $B$ is bounded. Now let $A subset mathbbC^n$ be closed and $(eta_m)$ be a sequence in $sigma(A)$ converging to some $eta in mathbbC^n$. Then $Y = eta_m mid m in mathbbN $ is bounded, hence $Z = sigma^-1(Y)$ is bounded. Choose $zeta_m in A$ such that $sigma(zeta_m) = eta_n$. Since $zeta_m in Z$, we see that $(zeta_m)$ is a bounded sequence and has a convergent subsequence with limit $zeta$. Since $A$ is closed, $zeta in A$. By continuity of $sigma$ we conclude $eta in sigma(A)$.
This implies that $sigma$ is an identification map. Therefore, if we give $mathbbC^n/S_n$ the quotient topology induced by the canonical quotient function $mathbbC^n to mathbbC^n/S_n$, we see that $sigma'$ is a homeomorphism.
answered Jul 17 at 9:51
Paul Frost
3,703420
add a comment |Â
up vote
1
down vote
accepted
If $e_0, e_1, ... ,e_n$ denote the elementary symmetric polynomials in $n$ variables, we get a continuous map
$$sigma : mathbbC^n to mathbbC^n, sigma(zeta) = (-e_1(zeta), e_2(zeta),...,(-1)^n e_n(zeta)) .$$
You may interpret $sigma(zeta)$ as the coefficients of a monic polynomial $p_zeta$ of degree $n$ having as roots the coordinates of $zeta$. In fact, $p_zeta = pi(sigma(zeta))$ where $pi$ was introduced in the question.
We have $E = sigma(Delta^n)$. This means that the set $E$ could alternatively be defined without reference to monic polynomials and their roots. Let $s : Delta^n to E$ denote the restriction.
Now it depends on what you understand by "explicitly construct some path from $a$ to $b$". You find $a', b' in Delta^n$ such that $s(a') = a, s(b') = b$. Define $u : [0,1] to Delta^n, u(t) = (1-t)a' + tb'$. This is a path connecting $a'$ and $b'$, therefore $s circ u$ is a path connecting $a$ and $b$. However, it is not obvious how to find $a',b'$. There is no general method to express $a',b'$ as a function of $a,b$ and in that sense $a',b'$ cannot be made explicit (if you could, you would have a solution formula for polynomials of arbitrary degree $n$ which in fact only exists for $n le 4$).
Let us finally consider the map $sigma$. For each $eta in mathbbC^n$ the inverse image $sigma^-1(eta)$ consists of all $zeta = (z_1,....z_n)$ such that the $ z_1,....z_n $ is the set of all roots of $pi(eta)$. Hence $sigma$ is surjective. Moreover, the symmetric group $S_n$ of $n$ elements operates by permutation of coordinates on $mathbbC^n$; the fibres $sigma^-1(eta)$ agree with the orbits of this operation. Therefore $sigma$ induces a bijection $sigma': mathbbC^n/S_n to mathbbC^n$.
We show that $sigma$ is a closed map. It is known that $max1, lvert a_n-1 rvert, ... , lvert a_0 rvert $ is an upper bound for the abolute values of the roots of $z^n + a_n-1z^n+-1 + ... + a_1z + a_0$. This implies that $sigma^-1(B)$ is bounded if $B$ is bounded. Now let $A subset mathbbC^n$ be closed and $(eta_m)$ be a sequence in $sigma(A)$ converging to some $eta in mathbbC^n$. Then $Y = eta_m mid m in mathbbN $ is bounded, hence $Z = sigma^-1(Y)$ is bounded. Choose $zeta_m in A$ such that $sigma(zeta_m) = eta_n$. Since $zeta_m in Z$, we see that $(zeta_m)$ is a bounded sequence and has a convergent subsequence with limit $zeta$. Since $A$ is closed, $zeta in A$. By continuity of $sigma$ we conclude $eta in sigma(A)$.
This implies that $sigma$ is an identification map. Therefore, if we give $mathbbC^n/S_n$ the quotient topology induced by the canonical quotient function $mathbbC^n to mathbbC^n/S_n$, we see that $sigma'$ is a homeomorphism.
answered Jul 17 at 9:51
Paul Frost
3,703420
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If $e_0, e_1, ... ,e_n$ denote the elementary symmetric polynomials in $n$ variables, we get a continuous map
$$sigma : mathbbC^n to mathbbC^n, sigma(zeta) = (-e_1(zeta), e_2(zeta),...,(-1)^n e_n(zeta)) .$$
You may interpret $sigma(zeta)$ as the coefficients of a monic polynomial $p_zeta$ of degree $n$ having as roots the coordinates of $zeta$. In fact, $p_zeta = pi(sigma(zeta))$ where $pi$ was introduced in the question.
We have $E = sigma(Delta^n)$. This means that the set $E$ could alternatively be defined without reference to monic polynomials and their roots. Let $s : Delta^n to E$ denote the restriction.
Now it depends on what you understand by "explicitly construct some path from $a$ to $b$". You find $a', b' in Delta^n$ such that $s(a') = a, s(b') = b$. Define $u : [0,1] to Delta^n, u(t) = (1-t)a' + tb'$. This is a path connecting $a'$ and $b'$, therefore $s circ u$ is a path connecting $a$ and $b$. However, it is not obvious how to find $a',b'$. There is no general method to express $a',b'$ as a function of $a,b$ and in that sense $a',b'$ cannot be made explicit (if you could, you would have a solution formula for polynomials of arbitrary degree $n$ which in fact only exists for $n le 4$).
Let us finally consider the map $sigma$. For each $eta in mathbbC^n$ the inverse image $sigma^-1(eta)$ consists of all $zeta = (z_1,....z_n)$ such that the $ z_1,....z_n $ is the set of all roots of $pi(eta)$. Hence $sigma$ is surjective. Moreover, the symmetric group $S_n$ of $n$ elements operates by permutation of coordinates on $mathbbC^n$; the fibres $sigma^-1(eta)$ agree with the orbits of this operation. Therefore $sigma$ induces a bijection $sigma': mathbbC^n/S_n to mathbbC^n$.
We show that $sigma$ is a closed map. It is known that $max1, lvert a_n-1 rvert, ... , lvert a_0 rvert $ is an upper bound for the abolute values of the roots of $z^n + a_n-1z^n+-1 + ... + a_1z + a_0$. This implies that $sigma^-1(B)$ is bounded if $B$ is bounded. Now let $A subset mathbbC^n$ be closed and $(eta_m)$ be a sequence in $sigma(A)$ converging to some $eta in mathbbC^n$. Then $Y = eta_m mid m in mathbbN $ is bounded, hence $Z = sigma^-1(Y)$ is bounded. Choose $zeta_m in A$ such that $sigma(zeta_m) = eta_n$. Since $zeta_m in Z$, we see that $(zeta_m)$ is a bounded sequence and has a convergent subsequence with limit $zeta$. Since $A$ is closed, $zeta in A$. By continuity of $sigma$ we conclude $eta in sigma(A)$.
This implies that $sigma$ is an identification map. Therefore, if we give $mathbbC^n/S_n$ the quotient topology induced by the canonical quotient function $mathbbC^n to mathbbC^n/S_n$, we see that $sigma'$ is a homeomorphism.
answered Jul 17 at 9:51
Paul Frost
3,703420
If $e_0, e_1, ... ,e_n$ denote the elementary symmetric polynomials in $n$ variables, we get a continuous map
$$sigma : mathbbC^n to mathbbC^n, sigma(zeta) = (-e_1(zeta), e_2(zeta),...,(-1)^n e_n(zeta)) .$$
You may interpret $sigma(zeta)$ as the coefficients of a monic polynomial $p_zeta$ of degree $n$ having as roots the coordinates of $zeta$. In fact, $p_zeta = pi(sigma(zeta))$ where $pi$ was introduced in the question.
We have $E = sigma(Delta^n)$. This means that the set $E$ could alternatively be defined without reference to monic polynomials and their roots. Let $s : Delta^n to E$ denote the restriction.
Now it depends on what you understand by "explicitly construct some path from $a$ to $b$". You find $a', b' in Delta^n$ such that $s(a') = a, s(b') = b$. Define $u : [0,1] to Delta^n, u(t) = (1-t)a' + tb'$. This is a path connecting $a'$ and $b'$, therefore $s circ u$ is a path connecting $a$ and $b$. However, it is not obvious how to find $a',b'$. There is no general method to express $a',b'$ as a function of $a,b$ and in that sense $a',b'$ cannot be made explicit (if you could, you would have a solution formula for polynomials of arbitrary degree $n$ which in fact only exists for $n le 4$).
Let us finally consider the map $sigma$. For each $eta in mathbbC^n$ the inverse image $sigma^-1(eta)$ consists of all $zeta = (z_1,....z_n)$ such that the $ z_1,....z_n $ is the set of all roots of $pi(eta)$. Hence $sigma$ is surjective. Moreover, the symmetric group $S_n$ of $n$ elements operates by permutation of coordinates on $mathbbC^n$; the fibres $sigma^-1(eta)$ agree with the orbits of this operation. Therefore $sigma$ induces a bijection $sigma': mathbbC^n/S_n to mathbbC^n$.
We show that $sigma$ is a closed map. It is known that $max1, lvert a_n-1 rvert, ... , lvert a_0 rvert $ is an upper bound for the abolute values of the roots of $z^n + a_n-1z^n+-1 + ... + a_1z + a_0$. This implies that $sigma^-1(B)$ is bounded if $B$ is bounded. Now let $A subset mathbbC^n$ be closed and $(eta_m)$ be a sequence in $sigma(A)$ converging to some $eta in mathbbC^n$. Then $Y = eta_m mid m in mathbbN $ is bounded, hence $Z = sigma^-1(Y)$ is bounded. Choose $zeta_m in A$ such that $sigma(zeta_m) = eta_n$. Since $zeta_m in Z$, we see that $(zeta_m)$ is a bounded sequence and has a convergent subsequence with limit $zeta$. Since $A$ is closed, $zeta in A$. By continuity of $sigma$ we conclude $eta in sigma(A)$.
This implies that $sigma$ is an identification map. Therefore, if we give $mathbbC^n/S_n$ the quotient topology induced by the canonical quotient function $mathbbC^n to mathbbC^n/S_n$, we see that $sigma'$ is a homeomorphism.
answered Jul 17 at 9:51
Paul Frost
3,703420
edited Jul 17 at 15:48
answered Jul 17 at 9:51
Paul Frost
3,703420
answered Jul 17 at 9:51
Paul Frost
3,703420
answered Jul 17 at 9:51
Paul Frost
3,703420
3,703420
add a comment |Â
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2853674%2fconstructing-explicit-connected-path-for-coefficients-of-monic-polynomials-with%23new-answer', 'question_page');
);
Post as a guest
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
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
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
It is irritating to begin your question with "Suppose we have a monic polynomial ...". I suggest to do it like that: There is bijection $pi$ between $mathbbC^n$ and the set of monic polynomials of degree $n$ defined by $pi(alpha_n-1,...,alpha_0) = ...$. Then define $E = zeta in mathbbC^n mid textAll roots of pi(zeta) text have a negative real part $. Can you make explicit which continuous function has image $E$?
– Paul Frost
Jul 16 at 23:13
@PaulFrost: Thanks for your editing suggestion. The function is elementary symmetric functions of the roots given by Vieta's formulas.
– user9527
Jul 17 at 4:33