Max modulus Constrained Optimization
Clash Royale CLAN TAG#URR8PPP
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Let $z_1,z_2ldots,z_ninmathbbC$ such that
$$sum_i=1^nz_i^j=1quad,:forall j=1,2,ldots,n-1$$
and
$$sum_i=1^nz_i^n=qqquadtextrm with |q|<1.$$
I am interested in finding a bound on the maximum modulus
$$max_1leq ileq n$$
Specifically, if conditions can be imposed on $|q|$ such that these bounds are less that 1 i.e. all $z_i$ lie within the unit circle.
inequality optimization complex-numbers nonlinear-optimization
asked Aug 7 at 14:54
RTJ
3,5762621
add a comment |Â
up vote
2
down vote
favorite
Let $z_1,z_2ldots,z_ninmathbbC$ such that
$$sum_i=1^nz_i^j=1quad,:forall j=1,2,ldots,n-1$$
and
$$sum_i=1^nz_i^n=qqquadtextrm with |q|<1.$$
I am interested in finding a bound on the maximum modulus
$$max_1leq ileq n$$
Specifically, if conditions can be imposed on $|q|$ such that these bounds are less that 1 i.e. all $z_i$ lie within the unit circle.
inequality optimization complex-numbers nonlinear-optimization
asked Aug 7 at 14:54
RTJ
3,5762621
You're asking for $displaystyleminleftz_k$. I would try KKT (after squaring the objective function, though it promises to be somewhat difficult...).
– metamorphy
Aug 7 at 15:34
@metamorphy Thank you for your comment. I am looking for something more elegant and simple. Note that the points $z_1,cdots,z_n$ are solutions of the equation $z^n-z^n-1+(1-q)=0$. Thus, Schur-Cohn theorem can be used to find necessary and sufficient condition so that all $z_i$ are within the unit circle but I do not want to calculate all these lengthy determinants. I am just searching for a sufficient condition on this.
– RTJ
Aug 7 at 15:46
1
@RTJ: in order to find the number of roots of $z^n-z^n-1+(1-q)$ inside the unit circle you just need a numerical approximation of $$frac12pi ioint_zfracn z^n-1-(n-1)z^n-2z^n-z^n-1+(1-q),dz$$
– Jack D'Aurizio♦
Aug 7 at 19:06
and in a geometric framework, given the relation between the topological degree of a curve and the number of roots of a polynomial in a the circle, you just need to find sufficient conditions on $q$ such that the origin lies inside the inner loop of the parametric curve $cos(ntheta)-cos((n-1)theta)+1-q,sin(ntheta)-sin((n-1)theta)$. $1-fracCnleq q< 1$ for some positive constant $C$ should suffice.
– Jack D'Aurizio♦
Aug 7 at 19:11
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $z_1,z_2ldots,z_ninmathbbC$ such that
$$sum_i=1^nz_i^j=1quad,:forall j=1,2,ldots,n-1$$
and
$$sum_i=1^nz_i^n=qqquadtextrm with |q|<1.$$
I am interested in finding a bound on the maximum modulus
$$max_1leq ileq n$$
Specifically, if conditions can be imposed on $|q|$ such that these bounds are less that 1 i.e. all $z_i$ lie within the unit circle.
inequality optimization complex-numbers nonlinear-optimization
asked Aug 7 at 14:54
RTJ
3,5762621
Let $z_1,z_2ldots,z_ninmathbbC$ such that
$$sum_i=1^nz_i^j=1quad,:forall j=1,2,ldots,n-1$$
and
$$sum_i=1^nz_i^n=qqquadtextrm with |q|<1.$$
I am interested in finding a bound on the maximum modulus
$$max_1leq ileq n$$
Specifically, if conditions can be imposed on $|q|$ such that these bounds are less that 1 i.e. all $z_i$ lie within the unit circle.
inequality optimization complex-numbers nonlinear-optimization
asked Aug 7 at 14:54
RTJ
3,5762621
asked Aug 7 at 14:54
RTJ
3,5762621
asked Aug 7 at 14:54
RTJ
3,5762621
asked Aug 7 at 14:54
RTJ
3,5762621
3,5762621
You're asking for $displaystyleminleftz_k$. I would try KKT (after squaring the objective function, though it promises to be somewhat difficult...).
– metamorphy
Aug 7 at 15:34
@metamorphy Thank you for your comment. I am looking for something more elegant and simple. Note that the points $z_1,cdots,z_n$ are solutions of the equation $z^n-z^n-1+(1-q)=0$. Thus, Schur-Cohn theorem can be used to find necessary and sufficient condition so that all $z_i$ are within the unit circle but I do not want to calculate all these lengthy determinants. I am just searching for a sufficient condition on this.
– RTJ
Aug 7 at 15:46
1
@RTJ: in order to find the number of roots of $z^n-z^n-1+(1-q)$ inside the unit circle you just need a numerical approximation of $$frac12pi ioint_zfracn z^n-1-(n-1)z^n-2z^n-z^n-1+(1-q),dz$$
– Jack D'Aurizio♦
Aug 7 at 19:06
and in a geometric framework, given the relation between the topological degree of a curve and the number of roots of a polynomial in a the circle, you just need to find sufficient conditions on $q$ such that the origin lies inside the inner loop of the parametric curve $cos(ntheta)-cos((n-1)theta)+1-q,sin(ntheta)-sin((n-1)theta)$. $1-fracCnleq q< 1$ for some positive constant $C$ should suffice.
– Jack D'Aurizio♦
Aug 7 at 19:11
add a comment |Â
You're asking for $displaystyleminleftz_k$. I would try KKT (after squaring the objective function, though it promises to be somewhat difficult...).
– metamorphy
Aug 7 at 15:34
@metamorphy Thank you for your comment. I am looking for something more elegant and simple. Note that the points $z_1,cdots,z_n$ are solutions of the equation $z^n-z^n-1+(1-q)=0$. Thus, Schur-Cohn theorem can be used to find necessary and sufficient condition so that all $z_i$ are within the unit circle but I do not want to calculate all these lengthy determinants. I am just searching for a sufficient condition on this.
– RTJ
Aug 7 at 15:46
1
@RTJ: in order to find the number of roots of $z^n-z^n-1+(1-q)$ inside the unit circle you just need a numerical approximation of $$frac12pi ioint_zfracn z^n-1-(n-1)z^n-2z^n-z^n-1+(1-q),dz$$
– Jack D'Aurizio♦
Aug 7 at 19:06
and in a geometric framework, given the relation between the topological degree of a curve and the number of roots of a polynomial in a the circle, you just need to find sufficient conditions on $q$ such that the origin lies inside the inner loop of the parametric curve $cos(ntheta)-cos((n-1)theta)+1-q,sin(ntheta)-sin((n-1)theta)$. $1-fracCnleq q< 1$ for some positive constant $C$ should suffice.
– Jack D'Aurizio♦
Aug 7 at 19:11
You're asking for $displaystyleminleftz_k$. I would try KKT (after squaring the objective function, though it promises to be somewhat difficult...).
– metamorphy
Aug 7 at 15:34
You're asking for $displaystyleminleftz_k$. I would try KKT (after squaring the objective function, though it promises to be somewhat difficult...).
– metamorphy
Aug 7 at 15:34
@metamorphy Thank you for your comment. I am looking for something more elegant and simple. Note that the points $z_1,cdots,z_n$ are solutions of the equation $z^n-z^n-1+(1-q)=0$. Thus, Schur-Cohn theorem can be used to find necessary and sufficient condition so that all $z_i$ are within the unit circle but I do not want to calculate all these lengthy determinants. I am just searching for a sufficient condition on this.
– RTJ
Aug 7 at 15:46
@metamorphy Thank you for your comment. I am looking for something more elegant and simple. Note that the points $z_1,cdots,z_n$ are solutions of the equation $z^n-z^n-1+(1-q)=0$. Thus, Schur-Cohn theorem can be used to find necessary and sufficient condition so that all $z_i$ are within the unit circle but I do not want to calculate all these lengthy determinants. I am just searching for a sufficient condition on this.
– RTJ
Aug 7 at 15:46
1
1
@RTJ: in order to find the number of roots of $z^n-z^n-1+(1-q)$ inside the unit circle you just need a numerical approximation of $$frac12pi ioint_zfracn z^n-1-(n-1)z^n-2z^n-z^n-1+(1-q),dz$$
– Jack D'Aurizio♦
Aug 7 at 19:06
@RTJ: in order to find the number of roots of $z^n-z^n-1+(1-q)$ inside the unit circle you just need a numerical approximation of $$frac12pi ioint_zfracn z^n-1-(n-1)z^n-2z^n-z^n-1+(1-q),dz$$
– Jack D'Aurizio♦
Aug 7 at 19:06
and in a geometric framework, given the relation between the topological degree of a curve and the number of roots of a polynomial in a the circle, you just need to find sufficient conditions on $q$ such that the origin lies inside the inner loop of the parametric curve $cos(ntheta)-cos((n-1)theta)+1-q,sin(ntheta)-sin((n-1)theta)$. $1-fracCnleq q< 1$ for some positive constant $C$ should suffice.
– Jack D'Aurizio♦
Aug 7 at 19:11
and in a geometric framework, given the relation between the topological degree of a curve and the number of roots of a polynomial in a the circle, you just need to find sufficient conditions on $q$ such that the origin lies inside the inner loop of the parametric curve $cos(ntheta)-cos((n-1)theta)+1-q,sin(ntheta)-sin((n-1)theta)$. $1-fracCnleq q< 1$ for some positive constant $C$ should suffice.
– Jack D'Aurizio♦
Aug 7 at 19:11
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Reformulating, we want some conditions on $q$ such that all the roots of the polynomial $z^n-z^n-1+(1-q)$ lie inside the unit circle. This is pretty much the same exercise appearing at page 127 of my notes, serving as an introduction to the original proof of the Fundamental Theorem of Algebra provided by Gauss. Let $tau=1-q$. Given the relation between the topological degree of curves and the number of roots of polynomials inside disks, it is enough to understand which values of $tau$ ensure that the origin belongs to the inner loop of the parametric curve
$$gamma(s)=left(cos(ns)-cos((n-1)s)+tau,sin(ns)-sin((n-1)s)right).$$
For instance, here it is our curve for $n=7$ and $tau=0$:
The upper part of the inner loop is given by values of $s$ between $0$ and $fracpi2n+1$. If we want the origin to be enclosed by the inner loop, we need that $0<tau<2sinleft(fracpi4n+2right)$.
If we allow complex $tau$s,
$$ left|tau-sinleft(tfracpi4n+2right)right|leq tfrac12,sinleft(tfracpi4n+2right)$$
is a sufficient condition for $x^n-x^n-1+tau$ to have all its zeroes inside the unit disk.
answered Aug 7 at 19:35
Jack D'Aurizio♦
270k31266630
Great answer, thank you Jack!
– RTJ
Aug 7 at 22:46
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Reformulating, we want some conditions on $q$ such that all the roots of the polynomial $z^n-z^n-1+(1-q)$ lie inside the unit circle. This is pretty much the same exercise appearing at page 127 of my notes, serving as an introduction to the original proof of the Fundamental Theorem of Algebra provided by Gauss. Let $tau=1-q$. Given the relation between the topological degree of curves and the number of roots of polynomials inside disks, it is enough to understand which values of $tau$ ensure that the origin belongs to the inner loop of the parametric curve
$$gamma(s)=left(cos(ns)-cos((n-1)s)+tau,sin(ns)-sin((n-1)s)right).$$
For instance, here it is our curve for $n=7$ and $tau=0$:
The upper part of the inner loop is given by values of $s$ between $0$ and $fracpi2n+1$. If we want the origin to be enclosed by the inner loop, we need that $0<tau<2sinleft(fracpi4n+2right)$.
If we allow complex $tau$s,
$$ left|tau-sinleft(tfracpi4n+2right)right|leq tfrac12,sinleft(tfracpi4n+2right)$$
is a sufficient condition for $x^n-x^n-1+tau$ to have all its zeroes inside the unit disk.
answered Aug 7 at 19:35
Jack D'Aurizio♦
270k31266630
Great answer, thank you Jack!
– RTJ
Aug 7 at 22:46
add a comment |Â
up vote
3
down vote
accepted
Reformulating, we want some conditions on $q$ such that all the roots of the polynomial $z^n-z^n-1+(1-q)$ lie inside the unit circle. This is pretty much the same exercise appearing at page 127 of my notes, serving as an introduction to the original proof of the Fundamental Theorem of Algebra provided by Gauss. Let $tau=1-q$. Given the relation between the topological degree of curves and the number of roots of polynomials inside disks, it is enough to understand which values of $tau$ ensure that the origin belongs to the inner loop of the parametric curve
$$gamma(s)=left(cos(ns)-cos((n-1)s)+tau,sin(ns)-sin((n-1)s)right).$$
For instance, here it is our curve for $n=7$ and $tau=0$:
The upper part of the inner loop is given by values of $s$ between $0$ and $fracpi2n+1$. If we want the origin to be enclosed by the inner loop, we need that $0<tau<2sinleft(fracpi4n+2right)$.
If we allow complex $tau$s,
$$ left|tau-sinleft(tfracpi4n+2right)right|leq tfrac12,sinleft(tfracpi4n+2right)$$
is a sufficient condition for $x^n-x^n-1+tau$ to have all its zeroes inside the unit disk.
answered Aug 7 at 19:35
Jack D'Aurizio♦
270k31266630
Great answer, thank you Jack!
– RTJ
Aug 7 at 22:46
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Reformulating, we want some conditions on $q$ such that all the roots of the polynomial $z^n-z^n-1+(1-q)$ lie inside the unit circle. This is pretty much the same exercise appearing at page 127 of my notes, serving as an introduction to the original proof of the Fundamental Theorem of Algebra provided by Gauss. Let $tau=1-q$. Given the relation between the topological degree of curves and the number of roots of polynomials inside disks, it is enough to understand which values of $tau$ ensure that the origin belongs to the inner loop of the parametric curve
$$gamma(s)=left(cos(ns)-cos((n-1)s)+tau,sin(ns)-sin((n-1)s)right).$$
For instance, here it is our curve for $n=7$ and $tau=0$:
The upper part of the inner loop is given by values of $s$ between $0$ and $fracpi2n+1$. If we want the origin to be enclosed by the inner loop, we need that $0<tau<2sinleft(fracpi4n+2right)$.
If we allow complex $tau$s,
$$ left|tau-sinleft(tfracpi4n+2right)right|leq tfrac12,sinleft(tfracpi4n+2right)$$
is a sufficient condition for $x^n-x^n-1+tau$ to have all its zeroes inside the unit disk.
answered Aug 7 at 19:35
Jack D'Aurizio♦
270k31266630
Reformulating, we want some conditions on $q$ such that all the roots of the polynomial $z^n-z^n-1+(1-q)$ lie inside the unit circle. This is pretty much the same exercise appearing at page 127 of my notes, serving as an introduction to the original proof of the Fundamental Theorem of Algebra provided by Gauss. Let $tau=1-q$. Given the relation between the topological degree of curves and the number of roots of polynomials inside disks, it is enough to understand which values of $tau$ ensure that the origin belongs to the inner loop of the parametric curve
$$gamma(s)=left(cos(ns)-cos((n-1)s)+tau,sin(ns)-sin((n-1)s)right).$$
For instance, here it is our curve for $n=7$ and $tau=0$:
The upper part of the inner loop is given by values of $s$ between $0$ and $fracpi2n+1$. If we want the origin to be enclosed by the inner loop, we need that $0<tau<2sinleft(fracpi4n+2right)$.
If we allow complex $tau$s,
$$ left|tau-sinleft(tfracpi4n+2right)right|leq tfrac12,sinleft(tfracpi4n+2right)$$
is a sufficient condition for $x^n-x^n-1+tau$ to have all its zeroes inside the unit disk.
answered Aug 7 at 19:35
Jack D'Aurizio♦
270k31266630
edited Aug 7 at 19:44
answered Aug 7 at 19:35
Jack D'Aurizio♦
270k31266630
answered Aug 7 at 19:35
Jack D'Aurizio♦
270k31266630
answered Aug 7 at 19:35
Jack D'Aurizio♦
270k31266630
270k31266630
Great answer, thank you Jack!
– RTJ
Aug 7 at 22:46
add a comment |Â
Great answer, thank you Jack!
– RTJ
Aug 7 at 22:46
Great answer, thank you Jack!
– RTJ
Aug 7 at 22:46
Great answer, thank you Jack!
– RTJ
Aug 7 at 22:46
add a comment |Â
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You're asking for $displaystyleminleftz_k$. I would try KKT (after squaring the objective function, though it promises to be somewhat difficult...).
– metamorphy
Aug 7 at 15:34
@metamorphy Thank you for your comment. I am looking for something more elegant and simple. Note that the points $z_1,cdots,z_n$ are solutions of the equation $z^n-z^n-1+(1-q)=0$. Thus, Schur-Cohn theorem can be used to find necessary and sufficient condition so that all $z_i$ are within the unit circle but I do not want to calculate all these lengthy determinants. I am just searching for a sufficient condition on this.
– RTJ
Aug 7 at 15:46
1
@RTJ: in order to find the number of roots of $z^n-z^n-1+(1-q)$ inside the unit circle you just need a numerical approximation of $$frac12pi ioint_zfracn z^n-1-(n-1)z^n-2z^n-z^n-1+(1-q),dz$$
– Jack D'Aurizio♦
Aug 7 at 19:06
and in a geometric framework, given the relation between the topological degree of a curve and the number of roots of a polynomial in a the circle, you just need to find sufficient conditions on $q$ such that the origin lies inside the inner loop of the parametric curve $cos(ntheta)-cos((n-1)theta)+1-q,sin(ntheta)-sin((n-1)theta)$. $1-fracCnleq q< 1$ for some positive constant $C$ should suffice.
– Jack D'Aurizio♦
Aug 7 at 19:11