Roots of sparse “quadratic-like” polynomial.
$begingroup$
So I know about this question and I've seen papers like this and this. But the former isn't exactly what I want and the latter two papers are too deep and I'm lazy and I wanna quick-and-easy answer and I'm sure you legitimate mathematicians are happy to supply this lazy electrical engineer with a quick-and-easy answer, right?
So, in general, an $N$th-order polynomial with real coefficients is:
$$ f(z) = sumlimits_{n=0}^{N} a_n , z^n qquad qquad text{where }Ninmathbb{Z}ge0, a_ninmathbb{R}, zinmathbb{C}$$
Now, consider this simple quadratic:
$$begin{align}
w^2 + bw + c &= w^2 (1 + bw^{-1} + cw^{-2})\
&= (w-r_1)(w-r_2) \
end{align}$$
again $b,cinmathbb{R}$ and $w,r_1,r_2inmathbb{C}$ . We know that
$$begin{align}
r_1 &= -tfrac{b}{2} + sqrt{left(tfrac{b}{2}right)^2 - c} \
r_2 &= -tfrac{b}{2} - sqrt{left(tfrac{b}{2}right)^2 - c} \
end{align}$$
if $b^2 ge 4c$ and
$$begin{align}
r_1 &= -tfrac{b}{2} + isqrt{c - left(tfrac{b}{2}right)^2} \
r_2 &= -tfrac{b}{2} - isqrt{c - left(tfrac{b}{2}right)^2} \
end{align}$$
if $b^2 < 4c$
In the latter complex-conjugate case, we know that
$$ |r_1| = |r_2| = c $$
which is both simple and handy.
Now, whether the roots are real or complex-conjugate, suppose we have chosen $b$ and $c$ so that both
$$begin{align}
|r_1| &< 1 \
|r_2| &< 1 \
end{align}$$
Fine, now let's return to the sparse polynomial $f(z)$. Suppose $N$ is even and all coefficients $a_n$ are zero except:
$$begin{align}
a_0 &= 1 \
a_{N/2} &= b \
a_N &= c \
end{align}$$
What are the roots of $f(z)$? Suppose $N$ is pretty big (and even), say $Napprox 1000$. If I can guarantee that $b$ and $c$ are chosen to insure that $|r_1|$ and $|r_2|$ are less than $1$, can I rely on all $N$ roots of $f(z)$ also being less than $1$?
I think that I can. I've been fiddling with the substitution of
$$ w = z^{N/2} $$
and I know about the Nth-roots-of-unity, but am I guaranteed that all of the roots of the $N$th-order polynomial have magnitude less than $1$?
polynomials factoring roots-of-unity
asked Dec 14 '18 at 1:17
robert bristow-johnsonrobert bristow-johnson
216117
$endgroup$
add a comment |
$begingroup$
So I know about this question and I've seen papers like this and this. But the former isn't exactly what I want and the latter two papers are too deep and I'm lazy and I wanna quick-and-easy answer and I'm sure you legitimate mathematicians are happy to supply this lazy electrical engineer with a quick-and-easy answer, right?
So, in general, an $N$th-order polynomial with real coefficients is:
$$ f(z) = sumlimits_{n=0}^{N} a_n , z^n qquad qquad text{where }Ninmathbb{Z}ge0, a_ninmathbb{R}, zinmathbb{C}$$
Now, consider this simple quadratic:
$$begin{align}
w^2 + bw + c &= w^2 (1 + bw^{-1} + cw^{-2})\
&= (w-r_1)(w-r_2) \
end{align}$$
again $b,cinmathbb{R}$ and $w,r_1,r_2inmathbb{C}$ . We know that
$$begin{align}
r_1 &= -tfrac{b}{2} + sqrt{left(tfrac{b}{2}right)^2 - c} \
r_2 &= -tfrac{b}{2} - sqrt{left(tfrac{b}{2}right)^2 - c} \
end{align}$$
if $b^2 ge 4c$ and
$$begin{align}
r_1 &= -tfrac{b}{2} + isqrt{c - left(tfrac{b}{2}right)^2} \
r_2 &= -tfrac{b}{2} - isqrt{c - left(tfrac{b}{2}right)^2} \
end{align}$$
if $b^2 < 4c$
In the latter complex-conjugate case, we know that
$$ |r_1| = |r_2| = c $$
which is both simple and handy.
Now, whether the roots are real or complex-conjugate, suppose we have chosen $b$ and $c$ so that both
$$begin{align}
|r_1| &< 1 \
|r_2| &< 1 \
end{align}$$
Fine, now let's return to the sparse polynomial $f(z)$. Suppose $N$ is even and all coefficients $a_n$ are zero except:
$$begin{align}
a_0 &= 1 \
a_{N/2} &= b \
a_N &= c \
end{align}$$
What are the roots of $f(z)$? Suppose $N$ is pretty big (and even), say $Napprox 1000$. If I can guarantee that $b$ and $c$ are chosen to insure that $|r_1|$ and $|r_2|$ are less than $1$, can I rely on all $N$ roots of $f(z)$ also being less than $1$?
I think that I can. I've been fiddling with the substitution of
$$ w = z^{N/2} $$
and I know about the Nth-roots-of-unity, but am I guaranteed that all of the roots of the $N$th-order polynomial have magnitude less than $1$?
polynomials factoring roots-of-unity
asked Dec 14 '18 at 1:17
robert bristow-johnsonrobert bristow-johnson
216117
$endgroup$
$begingroup$
I guess if $|w|<1$, we know that $big| z^{N/2} big| < 1$ and $|z|<1^{2/N}$ and $|z|<1$. So, although I dunno all $N$ roots of $f(z)$, I think I can say they are all inside the unit circle and that is the most important thing. (But I would like to know what the values of the roots are, anyway.)
$endgroup$
– robert bristow-johnson
Dec 14 '18 at 1:55
add a comment |
$begingroup$
So I know about this question and I've seen papers like this and this. But the former isn't exactly what I want and the latter two papers are too deep and I'm lazy and I wanna quick-and-easy answer and I'm sure you legitimate mathematicians are happy to supply this lazy electrical engineer with a quick-and-easy answer, right?
So, in general, an $N$th-order polynomial with real coefficients is:
$$ f(z) = sumlimits_{n=0}^{N} a_n , z^n qquad qquad text{where }Ninmathbb{Z}ge0, a_ninmathbb{R}, zinmathbb{C}$$
Now, consider this simple quadratic:
$$begin{align}
w^2 + bw + c &= w^2 (1 + bw^{-1} + cw^{-2})\
&= (w-r_1)(w-r_2) \
end{align}$$
again $b,cinmathbb{R}$ and $w,r_1,r_2inmathbb{C}$ . We know that
$$begin{align}
r_1 &= -tfrac{b}{2} + sqrt{left(tfrac{b}{2}right)^2 - c} \
r_2 &= -tfrac{b}{2} - sqrt{left(tfrac{b}{2}right)^2 - c} \
end{align}$$
if $b^2 ge 4c$ and
$$begin{align}
r_1 &= -tfrac{b}{2} + isqrt{c - left(tfrac{b}{2}right)^2} \
r_2 &= -tfrac{b}{2} - isqrt{c - left(tfrac{b}{2}right)^2} \
end{align}$$
if $b^2 < 4c$
In the latter complex-conjugate case, we know that
$$ |r_1| = |r_2| = c $$
which is both simple and handy.
Now, whether the roots are real or complex-conjugate, suppose we have chosen $b$ and $c$ so that both
$$begin{align}
|r_1| &< 1 \
|r_2| &< 1 \
end{align}$$
Fine, now let's return to the sparse polynomial $f(z)$. Suppose $N$ is even and all coefficients $a_n$ are zero except:
$$begin{align}
a_0 &= 1 \
a_{N/2} &= b \
a_N &= c \
end{align}$$
What are the roots of $f(z)$? Suppose $N$ is pretty big (and even), say $Napprox 1000$. If I can guarantee that $b$ and $c$ are chosen to insure that $|r_1|$ and $|r_2|$ are less than $1$, can I rely on all $N$ roots of $f(z)$ also being less than $1$?
I think that I can. I've been fiddling with the substitution of
$$ w = z^{N/2} $$
and I know about the Nth-roots-of-unity, but am I guaranteed that all of the roots of the $N$th-order polynomial have magnitude less than $1$?
polynomials factoring roots-of-unity
asked Dec 14 '18 at 1:17
robert bristow-johnsonrobert bristow-johnson
216117
$endgroup$
So I know about this question and I've seen papers like this and this. But the former isn't exactly what I want and the latter two papers are too deep and I'm lazy and I wanna quick-and-easy answer and I'm sure you legitimate mathematicians are happy to supply this lazy electrical engineer with a quick-and-easy answer, right?
So, in general, an $N$th-order polynomial with real coefficients is:
$$ f(z) = sumlimits_{n=0}^{N} a_n , z^n qquad qquad text{where }Ninmathbb{Z}ge0, a_ninmathbb{R}, zinmathbb{C}$$
Now, consider this simple quadratic:
$$begin{align}
w^2 + bw + c &= w^2 (1 + bw^{-1} + cw^{-2})\
&= (w-r_1)(w-r_2) \
end{align}$$
again $b,cinmathbb{R}$ and $w,r_1,r_2inmathbb{C}$ . We know that
$$begin{align}
r_1 &= -tfrac{b}{2} + sqrt{left(tfrac{b}{2}right)^2 - c} \
r_2 &= -tfrac{b}{2} - sqrt{left(tfrac{b}{2}right)^2 - c} \
end{align}$$
if $b^2 ge 4c$ and
$$begin{align}
r_1 &= -tfrac{b}{2} + isqrt{c - left(tfrac{b}{2}right)^2} \
r_2 &= -tfrac{b}{2} - isqrt{c - left(tfrac{b}{2}right)^2} \
end{align}$$
if $b^2 < 4c$
In the latter complex-conjugate case, we know that
$$ |r_1| = |r_2| = c $$
which is both simple and handy.
Now, whether the roots are real or complex-conjugate, suppose we have chosen $b$ and $c$ so that both
$$begin{align}
|r_1| &< 1 \
|r_2| &< 1 \
end{align}$$
Fine, now let's return to the sparse polynomial $f(z)$. Suppose $N$ is even and all coefficients $a_n$ are zero except:
$$begin{align}
a_0 &= 1 \
a_{N/2} &= b \
a_N &= c \
end{align}$$
What are the roots of $f(z)$? Suppose $N$ is pretty big (and even), say $Napprox 1000$. If I can guarantee that $b$ and $c$ are chosen to insure that $|r_1|$ and $|r_2|$ are less than $1$, can I rely on all $N$ roots of $f(z)$ also being less than $1$?
I think that I can. I've been fiddling with the substitution of
$$ w = z^{N/2} $$
and I know about the Nth-roots-of-unity, but am I guaranteed that all of the roots of the $N$th-order polynomial have magnitude less than $1$?
polynomials factoring roots-of-unity
polynomials factoring roots-of-unity
asked Dec 14 '18 at 1:17
robert bristow-johnsonrobert bristow-johnson
216117
asked Dec 14 '18 at 1:17
robert bristow-johnsonrobert bristow-johnson
216117
edited Dec 15 '18 at 0:38
robert bristow-johnson
asked Dec 14 '18 at 1:17
robert bristow-johnsonrobert bristow-johnson
216117
asked Dec 14 '18 at 1:17
robert bristow-johnsonrobert bristow-johnson
216117
asked Dec 14 '18 at 1:17
robert bristow-johnsonrobert bristow-johnson
216117
216117
$begingroup$
I guess if $|w|<1$, we know that $big| z^{N/2} big| < 1$ and $|z|<1^{2/N}$ and $|z|<1$. So, although I dunno all $N$ roots of $f(z)$, I think I can say they are all inside the unit circle and that is the most important thing. (But I would like to know what the values of the roots are, anyway.)
$endgroup$
– robert bristow-johnson
Dec 14 '18 at 1:55
add a comment |
$begingroup$
I guess if $|w|<1$, we know that $big| z^{N/2} big| < 1$ and $|z|<1^{2/N}$ and $|z|<1$. So, although I dunno all $N$ roots of $f(z)$, I think I can say they are all inside the unit circle and that is the most important thing. (But I would like to know what the values of the roots are, anyway.)
$endgroup$
– robert bristow-johnson
Dec 14 '18 at 1:55
$begingroup$
I guess if $|w|<1$, we know that $big| z^{N/2} big| < 1$ and $|z|<1^{2/N}$ and $|z|<1$. So, although I dunno all $N$ roots of $f(z)$, I think I can say they are all inside the unit circle and that is the most important thing. (But I would like to know what the values of the roots are, anyway.)
$endgroup$
– robert bristow-johnson
Dec 14 '18 at 1:55
$begingroup$
I guess if $|w|<1$, we know that $big| z^{N/2} big| < 1$ and $|z|<1^{2/N}$ and $|z|<1$. So, although I dunno all $N$ roots of $f(z)$, I think I can say they are all inside the unit circle and that is the most important thing. (But I would like to know what the values of the roots are, anyway.)
$endgroup$
– robert bristow-johnson
Dec 14 '18 at 1:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
So you have the equation:
$$cz^{2n}+bz^n+1 = 0$$,
where it's guaranteed that $z^n_{1,2}$ have modulus less than $1?$
If this is the case, then your intuition is correct because if:
$$|z^n| = |z|^n = |omega| <1,$$
it must necessarily follow that:
$$|z|<1$$
since $|z|$ is a positive real number.
answered Dec 15 '18 at 1:53
dezdichadodezdichado
6,3711929
$endgroup$
1
$begingroup$
or, at least a non-negative real number.
$endgroup$
– robert bristow-johnson
Dec 15 '18 at 1:56
add a comment |
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oldest
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1 Answer
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active
oldest
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oldest
votes
$begingroup$
So you have the equation:
$$cz^{2n}+bz^n+1 = 0$$,
where it's guaranteed that $z^n_{1,2}$ have modulus less than $1?$
If this is the case, then your intuition is correct because if:
$$|z^n| = |z|^n = |omega| <1,$$
it must necessarily follow that:
$$|z|<1$$
since $|z|$ is a positive real number.
answered Dec 15 '18 at 1:53
dezdichadodezdichado
6,3711929
$endgroup$
1
$begingroup$
or, at least a non-negative real number.
$endgroup$
– robert bristow-johnson
Dec 15 '18 at 1:56
add a comment |
$begingroup$
So you have the equation:
$$cz^{2n}+bz^n+1 = 0$$,
where it's guaranteed that $z^n_{1,2}$ have modulus less than $1?$
If this is the case, then your intuition is correct because if:
$$|z^n| = |z|^n = |omega| <1,$$
it must necessarily follow that:
$$|z|<1$$
since $|z|$ is a positive real number.
answered Dec 15 '18 at 1:53
dezdichadodezdichado
6,3711929
$endgroup$
1
$begingroup$
or, at least a non-negative real number.
$endgroup$
– robert bristow-johnson
Dec 15 '18 at 1:56
add a comment |
$begingroup$
So you have the equation:
$$cz^{2n}+bz^n+1 = 0$$,
where it's guaranteed that $z^n_{1,2}$ have modulus less than $1?$
If this is the case, then your intuition is correct because if:
$$|z^n| = |z|^n = |omega| <1,$$
it must necessarily follow that:
$$|z|<1$$
since $|z|$ is a positive real number.
answered Dec 15 '18 at 1:53
dezdichadodezdichado
6,3711929
$endgroup$
So you have the equation:
$$cz^{2n}+bz^n+1 = 0$$,
where it's guaranteed that $z^n_{1,2}$ have modulus less than $1?$
If this is the case, then your intuition is correct because if:
$$|z^n| = |z|^n = |omega| <1,$$
it must necessarily follow that:
$$|z|<1$$
since $|z|$ is a positive real number.
answered Dec 15 '18 at 1:53
dezdichadodezdichado
6,3711929
answered Dec 15 '18 at 1:53
dezdichadodezdichado
6,3711929
answered Dec 15 '18 at 1:53
dezdichadodezdichado
6,3711929
answered Dec 15 '18 at 1:53
dezdichadodezdichado
6,3711929
6,3711929
1
$begingroup$
or, at least a non-negative real number.
$endgroup$
– robert bristow-johnson
Dec 15 '18 at 1:56
add a comment |
1
$begingroup$
or, at least a non-negative real number.
$endgroup$
– robert bristow-johnson
Dec 15 '18 at 1:56
1
1
$begingroup$
or, at least a non-negative real number.
$endgroup$
– robert bristow-johnson
Dec 15 '18 at 1:56
$begingroup$
or, at least a non-negative real number.
$endgroup$
– robert bristow-johnson
Dec 15 '18 at 1:56
add a comment |
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$begingroup$
I guess if $|w|<1$, we know that $big| z^{N/2} big| < 1$ and $|z|<1^{2/N}$ and $|z|<1$. So, although I dunno all $N$ roots of $f(z)$, I think I can say they are all inside the unit circle and that is the most important thing. (But I would like to know what the values of the roots are, anyway.)
$endgroup$
– robert bristow-johnson
Dec 14 '18 at 1:55