Mixing
zeros of convex combination of monic real polynomials with degree $2$
$begingroup$
Suppose $f(x) = x^2 + a_1 x + a_0 in mathbb R[x]$ and $g(x) = x^2 + b_1 x + b_0 in mathbb R[x]$ with zeroes in the open unit disk of $mathbb C$. Let $h_t(x) = (1-t) f + t g$ where $t in [0,1]$. Are zeros of $h_t(x)$ inside the open unit disk of $mathbb C$?
In Fell's "On the zeros of convex combinations of polynomials", Theorem $4$ suggests that if $f$ and $g$ have both zeros in the closed unit disk, then $h_t$ has zeros on the closed unit disk (apply $n=2$ there).
I am interested in the open disk case. $f, g$ are simple enough. I tried to show this directly by formula for quadratic equations. But the formulas become too complicated.
polynomials roots
asked Feb 1 at 9:02
MyCindy2012MyCindy2012
12911
$endgroup$
add a comment |
$begingroup$
Suppose $f(x) = x^2 + a_1 x + a_0 in mathbb R[x]$ and $g(x) = x^2 + b_1 x + b_0 in mathbb R[x]$ with zeroes in the open unit disk of $mathbb C$. Let $h_t(x) = (1-t) f + t g$ where $t in [0,1]$. Are zeros of $h_t(x)$ inside the open unit disk of $mathbb C$?
In Fell's "On the zeros of convex combinations of polynomials", Theorem $4$ suggests that if $f$ and $g$ have both zeros in the closed unit disk, then $h_t$ has zeros on the closed unit disk (apply $n=2$ there).
I am interested in the open disk case. $f, g$ are simple enough. I tried to show this directly by formula for quadratic equations. But the formulas become too complicated.
polynomials roots
asked Feb 1 at 9:02
MyCindy2012MyCindy2012
12911
$endgroup$
add a comment |
$begingroup$
Suppose $f(x) = x^2 + a_1 x + a_0 in mathbb R[x]$ and $g(x) = x^2 + b_1 x + b_0 in mathbb R[x]$ with zeroes in the open unit disk of $mathbb C$. Let $h_t(x) = (1-t) f + t g$ where $t in [0,1]$. Are zeros of $h_t(x)$ inside the open unit disk of $mathbb C$?
In Fell's "On the zeros of convex combinations of polynomials", Theorem $4$ suggests that if $f$ and $g$ have both zeros in the closed unit disk, then $h_t$ has zeros on the closed unit disk (apply $n=2$ there).
I am interested in the open disk case. $f, g$ are simple enough. I tried to show this directly by formula for quadratic equations. But the formulas become too complicated.
polynomials roots
asked Feb 1 at 9:02
MyCindy2012MyCindy2012
12911
$endgroup$
Suppose $f(x) = x^2 + a_1 x + a_0 in mathbb R[x]$ and $g(x) = x^2 + b_1 x + b_0 in mathbb R[x]$ with zeroes in the open unit disk of $mathbb C$. Let $h_t(x) = (1-t) f + t g$ where $t in [0,1]$. Are zeros of $h_t(x)$ inside the open unit disk of $mathbb C$?
In Fell's "On the zeros of convex combinations of polynomials", Theorem $4$ suggests that if $f$ and $g$ have both zeros in the closed unit disk, then $h_t$ has zeros on the closed unit disk (apply $n=2$ there).
I am interested in the open disk case. $f, g$ are simple enough. I tried to show this directly by formula for quadratic equations. But the formulas become too complicated.
polynomials roots
polynomials roots
asked Feb 1 at 9:02
MyCindy2012MyCindy2012
12911
asked Feb 1 at 9:02
MyCindy2012MyCindy2012
12911
edited Feb 1 at 9:11
MyCindy2012
asked Feb 1 at 9:02
MyCindy2012MyCindy2012
12911
asked Feb 1 at 9:02
MyCindy2012MyCindy2012
12911
asked Feb 1 at 9:02
MyCindy2012MyCindy2012
12911
12911
add a comment |
add a comment |
1 Answer
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If h has complex roots, they are conjugate so if one is on the unit circle, both are and then the free term of h is 1 (being the product of the roots) and that is not possible from the fact that $a_0, b_0$ are both strictly less than 1 in absolute value by your assumption about f, g having roots inside the open disc and said free term is a convex combination of those two. If we assume the roots of h are real and at least one on the circle, then it is -1 or 1 and the other some number c strictly less than 1 in absolute value (product must be strictly less than 1 in absolute value). But now $f(1), f(-1), g(1), g(-1)$ are all positive (if say f has complex roots, then f is positive everywhere on the reals, while if f has real roots they are between -1 and 1 strictly so f is positive at -1 and 1 also, same for g) and then $h(-1), h(1)$ are positive as convex combination of positive numbers, so again contradiction.
answered Feb 1 at 16:56
ConradConrad
1,42245
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1 Answer
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1 Answer
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$begingroup$
If h has complex roots, they are conjugate so if one is on the unit circle, both are and then the free term of h is 1 (being the product of the roots) and that is not possible from the fact that $a_0, b_0$ are both strictly less than 1 in absolute value by your assumption about f, g having roots inside the open disc and said free term is a convex combination of those two. If we assume the roots of h are real and at least one on the circle, then it is -1 or 1 and the other some number c strictly less than 1 in absolute value (product must be strictly less than 1 in absolute value). But now $f(1), f(-1), g(1), g(-1)$ are all positive (if say f has complex roots, then f is positive everywhere on the reals, while if f has real roots they are between -1 and 1 strictly so f is positive at -1 and 1 also, same for g) and then $h(-1), h(1)$ are positive as convex combination of positive numbers, so again contradiction.
answered Feb 1 at 16:56
ConradConrad
1,42245
$endgroup$
add a comment |
$begingroup$
If h has complex roots, they are conjugate so if one is on the unit circle, both are and then the free term of h is 1 (being the product of the roots) and that is not possible from the fact that $a_0, b_0$ are both strictly less than 1 in absolute value by your assumption about f, g having roots inside the open disc and said free term is a convex combination of those two. If we assume the roots of h are real and at least one on the circle, then it is -1 or 1 and the other some number c strictly less than 1 in absolute value (product must be strictly less than 1 in absolute value). But now $f(1), f(-1), g(1), g(-1)$ are all positive (if say f has complex roots, then f is positive everywhere on the reals, while if f has real roots they are between -1 and 1 strictly so f is positive at -1 and 1 also, same for g) and then $h(-1), h(1)$ are positive as convex combination of positive numbers, so again contradiction.
answered Feb 1 at 16:56
ConradConrad
1,42245
$endgroup$
add a comment |
$begingroup$
If h has complex roots, they are conjugate so if one is on the unit circle, both are and then the free term of h is 1 (being the product of the roots) and that is not possible from the fact that $a_0, b_0$ are both strictly less than 1 in absolute value by your assumption about f, g having roots inside the open disc and said free term is a convex combination of those two. If we assume the roots of h are real and at least one on the circle, then it is -1 or 1 and the other some number c strictly less than 1 in absolute value (product must be strictly less than 1 in absolute value). But now $f(1), f(-1), g(1), g(-1)$ are all positive (if say f has complex roots, then f is positive everywhere on the reals, while if f has real roots they are between -1 and 1 strictly so f is positive at -1 and 1 also, same for g) and then $h(-1), h(1)$ are positive as convex combination of positive numbers, so again contradiction.
answered Feb 1 at 16:56
ConradConrad
1,42245
$endgroup$
If h has complex roots, they are conjugate so if one is on the unit circle, both are and then the free term of h is 1 (being the product of the roots) and that is not possible from the fact that $a_0, b_0$ are both strictly less than 1 in absolute value by your assumption about f, g having roots inside the open disc and said free term is a convex combination of those two. If we assume the roots of h are real and at least one on the circle, then it is -1 or 1 and the other some number c strictly less than 1 in absolute value (product must be strictly less than 1 in absolute value). But now $f(1), f(-1), g(1), g(-1)$ are all positive (if say f has complex roots, then f is positive everywhere on the reals, while if f has real roots they are between -1 and 1 strictly so f is positive at -1 and 1 also, same for g) and then $h(-1), h(1)$ are positive as convex combination of positive numbers, so again contradiction.
answered Feb 1 at 16:56
ConradConrad
1,42245
edited Feb 1 at 17:04
answered Feb 1 at 16:56
ConradConrad
1,42245
answered Feb 1 at 16:56
ConradConrad
1,42245
answered Feb 1 at 16:56
ConradConrad
1,42245
1,42245
add a comment |
add a comment |
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