Mixing
Can $frac{xp(x)-ap(a)}{x-a}$ have an inside root if $p(x)$ does not?
$begingroup$
Let $p(x)$ be a polynomial that has all of its roots outside of the unit circle and let $a$ be some real number that is not a root of $p(x)$. Can the polynomial
$$ frac{xp(x)-ap(a)}{x-a}$$
have a root inside the unit circle?
polynomials roots
asked Jan 16 at 14:14
mzpmzp
1,24721234
$endgroup$
|
show 7 more comments
$begingroup$
Let $p(x)$ be a polynomial that has all of its roots outside of the unit circle and let $a$ be some real number that is not a root of $p(x)$. Can the polynomial
$$ frac{xp(x)-ap(a)}{x-a}$$
have a root inside the unit circle?
polynomials roots
asked Jan 16 at 14:14
mzpmzp
1,24721234
$endgroup$
1
$begingroup$
A transformation that somewhat simplifies the issue : writing the numerator under the form $(x-a)p(x)+a (p(x)-p(a))$ gives a quotient of the form : $p(x)+a frac{p(x)-p(a)}{x-a}=p(x)+a(p'(a)+frac12(x-a)p''(a)+...)$ (the last expression coming from Taylor's expansion of $p$ at point $a$).
$endgroup$
– Jean Marie
Jan 16 at 16:21
1
$begingroup$
@JeanMarie I follow what you have written so far, but I don't see how it helps to figure out whether or not there is an inside root. Is it obvious?
$endgroup$
– mzp
Jan 16 at 16:33
1
$begingroup$
Had you known complex function theory, I would have advise you to try to use the so-called "Rouché's theorem" en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem because it has the form p(x)+q(x)$, as my transformation evidences it,
$endgroup$
– Jean Marie
Jan 16 at 16:34
1
$begingroup$
@JeanMarie I see. Thank you! Then the problem becomes that of showing that $|p(x)|>|a(p'(a)+frac{1}{2}(x-a)p''(a)+cdots)|$ for $x$ inside the unit circle, right?
$endgroup$
– mzp
Jan 16 at 16:39
1
$begingroup$
I don't say I have a solution : I just indicated this transformation as a way to simplify your issue. But I think it can be a path towards the proof...
$endgroup$
– Jean Marie
Jan 16 at 16:42
|
show 7 more comments
$begingroup$
Let $p(x)$ be a polynomial that has all of its roots outside of the unit circle and let $a$ be some real number that is not a root of $p(x)$. Can the polynomial
$$ frac{xp(x)-ap(a)}{x-a}$$
have a root inside the unit circle?
polynomials roots
asked Jan 16 at 14:14
mzpmzp
1,24721234
$endgroup$
Let $p(x)$ be a polynomial that has all of its roots outside of the unit circle and let $a$ be some real number that is not a root of $p(x)$. Can the polynomial
$$ frac{xp(x)-ap(a)}{x-a}$$
have a root inside the unit circle?
polynomials roots
polynomials roots
asked Jan 16 at 14:14
mzpmzp
1,24721234
asked Jan 16 at 14:14
mzpmzp
1,24721234
edited Jan 16 at 14:22
mzp
asked Jan 16 at 14:14
mzpmzp
1,24721234
asked Jan 16 at 14:14
mzpmzp
1,24721234
asked Jan 16 at 14:14
mzpmzp
1,24721234
1,24721234
1
$begingroup$
A transformation that somewhat simplifies the issue : writing the numerator under the form $(x-a)p(x)+a (p(x)-p(a))$ gives a quotient of the form : $p(x)+a frac{p(x)-p(a)}{x-a}=p(x)+a(p'(a)+frac12(x-a)p''(a)+...)$ (the last expression coming from Taylor's expansion of $p$ at point $a$).
$endgroup$
– Jean Marie
Jan 16 at 16:21
1
$begingroup$
@JeanMarie I follow what you have written so far, but I don't see how it helps to figure out whether or not there is an inside root. Is it obvious?
$endgroup$
– mzp
Jan 16 at 16:33
1
$begingroup$
Had you known complex function theory, I would have advise you to try to use the so-called "Rouché's theorem" en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem because it has the form p(x)+q(x)$, as my transformation evidences it,
$endgroup$
– Jean Marie
Jan 16 at 16:34
1
$begingroup$
@JeanMarie I see. Thank you! Then the problem becomes that of showing that $|p(x)|>|a(p'(a)+frac{1}{2}(x-a)p''(a)+cdots)|$ for $x$ inside the unit circle, right?
$endgroup$
– mzp
Jan 16 at 16:39
1
$begingroup$
I don't say I have a solution : I just indicated this transformation as a way to simplify your issue. But I think it can be a path towards the proof...
$endgroup$
– Jean Marie
Jan 16 at 16:42
|
show 7 more comments
1
$begingroup$
A transformation that somewhat simplifies the issue : writing the numerator under the form $(x-a)p(x)+a (p(x)-p(a))$ gives a quotient of the form : $p(x)+a frac{p(x)-p(a)}{x-a}=p(x)+a(p'(a)+frac12(x-a)p''(a)+...)$ (the last expression coming from Taylor's expansion of $p$ at point $a$).
$endgroup$
– Jean Marie
Jan 16 at 16:21
1
$begingroup$
@JeanMarie I follow what you have written so far, but I don't see how it helps to figure out whether or not there is an inside root. Is it obvious?
$endgroup$
– mzp
Jan 16 at 16:33
1
$begingroup$
Had you known complex function theory, I would have advise you to try to use the so-called "Rouché's theorem" en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem because it has the form p(x)+q(x)$, as my transformation evidences it,
$endgroup$
– Jean Marie
Jan 16 at 16:34
1
$begingroup$
@JeanMarie I see. Thank you! Then the problem becomes that of showing that $|p(x)|>|a(p'(a)+frac{1}{2}(x-a)p''(a)+cdots)|$ for $x$ inside the unit circle, right?
$endgroup$
– mzp
Jan 16 at 16:39
1
$begingroup$
I don't say I have a solution : I just indicated this transformation as a way to simplify your issue. But I think it can be a path towards the proof...
$endgroup$
– Jean Marie
Jan 16 at 16:42
1
1
$begingroup$
A transformation that somewhat simplifies the issue : writing the numerator under the form $(x-a)p(x)+a (p(x)-p(a))$ gives a quotient of the form : $p(x)+a frac{p(x)-p(a)}{x-a}=p(x)+a(p'(a)+frac12(x-a)p''(a)+...)$ (the last expression coming from Taylor's expansion of $p$ at point $a$).
$endgroup$
– Jean Marie
Jan 16 at 16:21
$begingroup$
A transformation that somewhat simplifies the issue : writing the numerator under the form $(x-a)p(x)+a (p(x)-p(a))$ gives a quotient of the form : $p(x)+a frac{p(x)-p(a)}{x-a}=p(x)+a(p'(a)+frac12(x-a)p''(a)+...)$ (the last expression coming from Taylor's expansion of $p$ at point $a$).
$endgroup$
– Jean Marie
Jan 16 at 16:21
1
1
$begingroup$
@JeanMarie I follow what you have written so far, but I don't see how it helps to figure out whether or not there is an inside root. Is it obvious?
$endgroup$
– mzp
Jan 16 at 16:33
$begingroup$
@JeanMarie I follow what you have written so far, but I don't see how it helps to figure out whether or not there is an inside root. Is it obvious?
$endgroup$
– mzp
Jan 16 at 16:33
1
1
$begingroup$
Had you known complex function theory, I would have advise you to try to use the so-called "Rouché's theorem" en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem because it has the form p(x)+q(x)$, as my transformation evidences it,
$endgroup$
– Jean Marie
Jan 16 at 16:34
$begingroup$
Had you known complex function theory, I would have advise you to try to use the so-called "Rouché's theorem" en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem because it has the form p(x)+q(x)$, as my transformation evidences it,
$endgroup$
– Jean Marie
Jan 16 at 16:34
1
1
$begingroup$
@JeanMarie I see. Thank you! Then the problem becomes that of showing that $|p(x)|>|a(p'(a)+frac{1}{2}(x-a)p''(a)+cdots)|$ for $x$ inside the unit circle, right?
$endgroup$
– mzp
Jan 16 at 16:39
$begingroup$
@JeanMarie I see. Thank you! Then the problem becomes that of showing that $|p(x)|>|a(p'(a)+frac{1}{2}(x-a)p''(a)+cdots)|$ for $x$ inside the unit circle, right?
$endgroup$
– mzp
Jan 16 at 16:39
1
1
$begingroup$
I don't say I have a solution : I just indicated this transformation as a way to simplify your issue. But I think it can be a path towards the proof...
$endgroup$
– Jean Marie
Jan 16 at 16:42
$begingroup$
I don't say I have a solution : I just indicated this transformation as a way to simplify your issue. But I think it can be a path towards the proof...
$endgroup$
– Jean Marie
Jan 16 at 16:42
|
show 7 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Let :
$$Q_a(z)=frac{zP(z)-aP(a)}{z-a}=P(z)+aleft(frac{P(z)-P(a)}{z-a}right)tag{1}$$
Let $mathbb{U}$ denote the unit disk.
Let $P(z)=z+b$, then $Q_a(z)=z+a+b$.
It suffices then to take $a$ and $b$ in such a way that :
$$|b|>1 text{and} |a+b|<1$$
knowing that this can be done in many different ways !
(see a similar answer, though less general, in Can $frac{zR(z)-aR(a)}{z-a}$ have an inside root if $R$ does not? ).
We could find similar counter-examples for higher degree polynomials. For example here is one for a quadratic polynomial and $a=1$ : take $P(z)=z^2-2$, whose roots are $z=pmsqrt{2}$, thus outside $mathbb{U}$, whereas, according to (1) :
$$Q_a(z)=z^2-2-2left(frac{z^2-2+1}{z-1}right)=z^2+z-1$$
has one of its roots $z_1:=frac{-1+sqrt{5}}{2}$ inside $mathbb{U}$, the other one $z_1:=frac{-1+sqrt{5}}{2}$ being outside $mathbb{U}$ (have you recognized the golden ratio ?).
answered Jan 19 at 14:50
Jean MarieJean Marie
30.2k42051
$endgroup$
$begingroup$
Thank you for this, I think it is a useful example. However, I don't think it counts as a counter-example since $a$ is assumed to be real.
$endgroup$
– mzp
Jan 21 at 14:43
$begingroup$
I failed to notice this restriction...
$endgroup$
– Jean Marie
Jan 21 at 22:03
$begingroup$
Can p have complex coefficients ?
$endgroup$
– Jean Marie
Jan 21 at 23:01
1
$begingroup$
I should have specified this in the question, sorry. I meant to have $p$ be real valued.
$endgroup$
– mzp
Jan 21 at 23:04
1
$begingroup$
Thank you very much! I really appreciate your help!
$endgroup$
– mzp
Jan 22 at 9:14
|
show 1 more comment
Your Answer
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Let :
$$Q_a(z)=frac{zP(z)-aP(a)}{z-a}=P(z)+aleft(frac{P(z)-P(a)}{z-a}right)tag{1}$$
Let $mathbb{U}$ denote the unit disk.
Let $P(z)=z+b$, then $Q_a(z)=z+a+b$.
It suffices then to take $a$ and $b$ in such a way that :
$$|b|>1 text{and} |a+b|<1$$
knowing that this can be done in many different ways !
(see a similar answer, though less general, in Can $frac{zR(z)-aR(a)}{z-a}$ have an inside root if $R$ does not? ).
We could find similar counter-examples for higher degree polynomials. For example here is one for a quadratic polynomial and $a=1$ : take $P(z)=z^2-2$, whose roots are $z=pmsqrt{2}$, thus outside $mathbb{U}$, whereas, according to (1) :
$$Q_a(z)=z^2-2-2left(frac{z^2-2+1}{z-1}right)=z^2+z-1$$
has one of its roots $z_1:=frac{-1+sqrt{5}}{2}$ inside $mathbb{U}$, the other one $z_1:=frac{-1+sqrt{5}}{2}$ being outside $mathbb{U}$ (have you recognized the golden ratio ?).
answered Jan 19 at 14:50
Jean MarieJean Marie
30.2k42051
$endgroup$
$begingroup$
Thank you for this, I think it is a useful example. However, I don't think it counts as a counter-example since $a$ is assumed to be real.
$endgroup$
– mzp
Jan 21 at 14:43
$begingroup$
I failed to notice this restriction...
$endgroup$
– Jean Marie
Jan 21 at 22:03
$begingroup$
Can p have complex coefficients ?
$endgroup$
– Jean Marie
Jan 21 at 23:01
1
$begingroup$
I should have specified this in the question, sorry. I meant to have $p$ be real valued.
$endgroup$
– mzp
Jan 21 at 23:04
1
$begingroup$
Thank you very much! I really appreciate your help!
$endgroup$
– mzp
Jan 22 at 9:14
|
show 1 more comment
$begingroup$
Let :
$$Q_a(z)=frac{zP(z)-aP(a)}{z-a}=P(z)+aleft(frac{P(z)-P(a)}{z-a}right)tag{1}$$
Let $mathbb{U}$ denote the unit disk.
Let $P(z)=z+b$, then $Q_a(z)=z+a+b$.
It suffices then to take $a$ and $b$ in such a way that :
$$|b|>1 text{and} |a+b|<1$$
knowing that this can be done in many different ways !
(see a similar answer, though less general, in Can $frac{zR(z)-aR(a)}{z-a}$ have an inside root if $R$ does not? ).
We could find similar counter-examples for higher degree polynomials. For example here is one for a quadratic polynomial and $a=1$ : take $P(z)=z^2-2$, whose roots are $z=pmsqrt{2}$, thus outside $mathbb{U}$, whereas, according to (1) :
$$Q_a(z)=z^2-2-2left(frac{z^2-2+1}{z-1}right)=z^2+z-1$$
has one of its roots $z_1:=frac{-1+sqrt{5}}{2}$ inside $mathbb{U}$, the other one $z_1:=frac{-1+sqrt{5}}{2}$ being outside $mathbb{U}$ (have you recognized the golden ratio ?).
answered Jan 19 at 14:50
Jean MarieJean Marie
30.2k42051
$endgroup$
$begingroup$
Thank you for this, I think it is a useful example. However, I don't think it counts as a counter-example since $a$ is assumed to be real.
$endgroup$
– mzp
Jan 21 at 14:43
$begingroup$
I failed to notice this restriction...
$endgroup$
– Jean Marie
Jan 21 at 22:03
$begingroup$
Can p have complex coefficients ?
$endgroup$
– Jean Marie
Jan 21 at 23:01
1
$begingroup$
I should have specified this in the question, sorry. I meant to have $p$ be real valued.
$endgroup$
– mzp
Jan 21 at 23:04
1
$begingroup$
Thank you very much! I really appreciate your help!
$endgroup$
– mzp
Jan 22 at 9:14
|
show 1 more comment
$begingroup$
Let :
$$Q_a(z)=frac{zP(z)-aP(a)}{z-a}=P(z)+aleft(frac{P(z)-P(a)}{z-a}right)tag{1}$$
Let $mathbb{U}$ denote the unit disk.
Let $P(z)=z+b$, then $Q_a(z)=z+a+b$.
It suffices then to take $a$ and $b$ in such a way that :
$$|b|>1 text{and} |a+b|<1$$
knowing that this can be done in many different ways !
(see a similar answer, though less general, in Can $frac{zR(z)-aR(a)}{z-a}$ have an inside root if $R$ does not? ).
We could find similar counter-examples for higher degree polynomials. For example here is one for a quadratic polynomial and $a=1$ : take $P(z)=z^2-2$, whose roots are $z=pmsqrt{2}$, thus outside $mathbb{U}$, whereas, according to (1) :
$$Q_a(z)=z^2-2-2left(frac{z^2-2+1}{z-1}right)=z^2+z-1$$
has one of its roots $z_1:=frac{-1+sqrt{5}}{2}$ inside $mathbb{U}$, the other one $z_1:=frac{-1+sqrt{5}}{2}$ being outside $mathbb{U}$ (have you recognized the golden ratio ?).
answered Jan 19 at 14:50
Jean MarieJean Marie
30.2k42051
$endgroup$
Let :
$$Q_a(z)=frac{zP(z)-aP(a)}{z-a}=P(z)+aleft(frac{P(z)-P(a)}{z-a}right)tag{1}$$
Let $mathbb{U}$ denote the unit disk.
Let $P(z)=z+b$, then $Q_a(z)=z+a+b$.
It suffices then to take $a$ and $b$ in such a way that :
$$|b|>1 text{and} |a+b|<1$$
knowing that this can be done in many different ways !
(see a similar answer, though less general, in Can $frac{zR(z)-aR(a)}{z-a}$ have an inside root if $R$ does not? ).
We could find similar counter-examples for higher degree polynomials. For example here is one for a quadratic polynomial and $a=1$ : take $P(z)=z^2-2$, whose roots are $z=pmsqrt{2}$, thus outside $mathbb{U}$, whereas, according to (1) :
$$Q_a(z)=z^2-2-2left(frac{z^2-2+1}{z-1}right)=z^2+z-1$$
has one of its roots $z_1:=frac{-1+sqrt{5}}{2}$ inside $mathbb{U}$, the other one $z_1:=frac{-1+sqrt{5}}{2}$ being outside $mathbb{U}$ (have you recognized the golden ratio ?).
answered Jan 19 at 14:50
Jean MarieJean Marie
30.2k42051
edited Jan 22 at 1:04
answered Jan 19 at 14:50
Jean MarieJean Marie
30.2k42051
answered Jan 19 at 14:50
Jean MarieJean Marie
30.2k42051
answered Jan 19 at 14:50
Jean MarieJean Marie
30.2k42051
30.2k42051
$begingroup$
Thank you for this, I think it is a useful example. However, I don't think it counts as a counter-example since $a$ is assumed to be real.
$endgroup$
– mzp
Jan 21 at 14:43
$begingroup$
I failed to notice this restriction...
$endgroup$
– Jean Marie
Jan 21 at 22:03
$begingroup$
Can p have complex coefficients ?
$endgroup$
– Jean Marie
Jan 21 at 23:01
1
$begingroup$
I should have specified this in the question, sorry. I meant to have $p$ be real valued.
$endgroup$
– mzp
Jan 21 at 23:04
1
$begingroup$
Thank you very much! I really appreciate your help!
$endgroup$
– mzp
Jan 22 at 9:14
|
show 1 more comment
$begingroup$
Thank you for this, I think it is a useful example. However, I don't think it counts as a counter-example since $a$ is assumed to be real.
$endgroup$
– mzp
Jan 21 at 14:43
$begingroup$
I failed to notice this restriction...
$endgroup$
– Jean Marie
Jan 21 at 22:03
$begingroup$
Can p have complex coefficients ?
$endgroup$
– Jean Marie
Jan 21 at 23:01
1
$begingroup$
I should have specified this in the question, sorry. I meant to have $p$ be real valued.
$endgroup$
– mzp
Jan 21 at 23:04
1
$begingroup$
Thank you very much! I really appreciate your help!
$endgroup$
– mzp
Jan 22 at 9:14
$begingroup$
Thank you for this, I think it is a useful example. However, I don't think it counts as a counter-example since $a$ is assumed to be real.
$endgroup$
– mzp
Jan 21 at 14:43
$begingroup$
Thank you for this, I think it is a useful example. However, I don't think it counts as a counter-example since $a$ is assumed to be real.
$endgroup$
– mzp
Jan 21 at 14:43
$begingroup$
I failed to notice this restriction...
$endgroup$
– Jean Marie
Jan 21 at 22:03
$begingroup$
I failed to notice this restriction...
$endgroup$
– Jean Marie
Jan 21 at 22:03
$begingroup$
Can p have complex coefficients ?
$endgroup$
– Jean Marie
Jan 21 at 23:01
$begingroup$
Can p have complex coefficients ?
$endgroup$
– Jean Marie
Jan 21 at 23:01
1
1
$begingroup$
I should have specified this in the question, sorry. I meant to have $p$ be real valued.
$endgroup$
– mzp
Jan 21 at 23:04
$begingroup$
I should have specified this in the question, sorry. I meant to have $p$ be real valued.
$endgroup$
– mzp
Jan 21 at 23:04
1
1
$begingroup$
Thank you very much! I really appreciate your help!
$endgroup$
– mzp
Jan 22 at 9:14
$begingroup$
Thank you very much! I really appreciate your help!
$endgroup$
– mzp
Jan 22 at 9:14
|
show 1 more comment
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1
$begingroup$
A transformation that somewhat simplifies the issue : writing the numerator under the form $(x-a)p(x)+a (p(x)-p(a))$ gives a quotient of the form : $p(x)+a frac{p(x)-p(a)}{x-a}=p(x)+a(p'(a)+frac12(x-a)p''(a)+...)$ (the last expression coming from Taylor's expansion of $p$ at point $a$).
$endgroup$
– Jean Marie
Jan 16 at 16:21
1
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@JeanMarie I follow what you have written so far, but I don't see how it helps to figure out whether or not there is an inside root. Is it obvious?
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– mzp
Jan 16 at 16:33
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Had you known complex function theory, I would have advise you to try to use the so-called "Rouché's theorem" en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem because it has the form p(x)+q(x)$, as my transformation evidences it,
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– Jean Marie
Jan 16 at 16:34
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@JeanMarie I see. Thank you! Then the problem becomes that of showing that $|p(x)|>|a(p'(a)+frac{1}{2}(x-a)p''(a)+cdots)|$ for $x$ inside the unit circle, right?
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– mzp
Jan 16 at 16:39
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I don't say I have a solution : I just indicated this transformation as a way to simplify your issue. But I think it can be a path towards the proof...
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– Jean Marie
Jan 16 at 16:42