Methods for ruling out rational roots of polynomials
$begingroup$
Given a polynomial $P(x)$ of degree $m>1$: $$P(x)=a_m x^m +...+ a_k x^k +...+ a_2 x^2 + a_1 x - alpha$$
Where $lvert a_m rvert >...> lvert a_k rvert > ... > lvert a_1 rvert = 1$ and $alpha$ are integer coefficients, such that $a_k <0$ $forall k neq m$. What methods are there to rule out a specific rational root $ frac {1}{beta} >0 $ ? (It is supposed that this case that $beta$ divides $a_m$, so the rational root theorem applies, and thus this root $ frac {1}{beta} $ cannot be ruled out by contradiction with the theorem).
According to Sturm's Theorem, one can assert that there is a root in the interval $[0,F]$, where F is a positive number. It is known that $P(x)$ has no irrational roots.
polynomials
asked Jan 15 at 22:56
Federico OmarFederico Omar
63
$endgroup$
|
show 1 more comment
$begingroup$
Given a polynomial $P(x)$ of degree $m>1$: $$P(x)=a_m x^m +...+ a_k x^k +...+ a_2 x^2 + a_1 x - alpha$$
Where $lvert a_m rvert >...> lvert a_k rvert > ... > lvert a_1 rvert = 1$ and $alpha$ are integer coefficients, such that $a_k <0$ $forall k neq m$. What methods are there to rule out a specific rational root $ frac {1}{beta} >0 $ ? (It is supposed that this case that $beta$ divides $a_m$, so the rational root theorem applies, and thus this root $ frac {1}{beta} $ cannot be ruled out by contradiction with the theorem).
According to Sturm's Theorem, one can assert that there is a root in the interval $[0,F]$, where F is a positive number. It is known that $P(x)$ has no irrational roots.
polynomials
asked Jan 15 at 22:56
Federico OmarFederico Omar
63
$endgroup$
2
$begingroup$
Plugging it in?
$endgroup$
– Severin Schraven
Jan 15 at 23:05
$begingroup$
Won't work... I am looking for insights on how to tackle this problem with the given conditions, and evaluating the polynomial on the given value is impossible, because the degree of the polynomial isn't defined. i.e., the polynomial has arbitrary degree m.
$endgroup$
– Federico Omar
Jan 15 at 23:08
$begingroup$
Well, we have to know what "is defined and what not" in the question. For my taste, i would take the reciprocal and search for possibilities to avoid some integer root. It is unclear from the post if $beta$ is positive or general. What kind of answer is expected? What is fixed and what is moving? The $a_m$ is fixed, then $beta$ but $m$ and all other coefficients not?
$endgroup$
– dan_fulea
Jan 15 at 23:46
$begingroup$
@dan_fulea $beta$ is a positive value, and $frac{1}{beta}$ is a candidate for rational root. All roots of P(x) are rational. deg(P(x)) is non trivial ( can be greater than 4). All the coefficients of P(x) are negative, except $a_m$.
$endgroup$
– Federico Omar
Jan 15 at 23:53
$begingroup$
That $beta$ is $>0$ is a new information. The fact that $P$ completely splits over $Bbb Q$ is also new. Please rewrite the question starting with the given data. Best, introduce the objects one by one, and take care for the usual $forall$ and $exists$ quantifiers. Please do not introduce a polynomial, and after that the degree and the coefficients, if some $forall$ and/or $exists$ apply only to parts of them. Please insert all details in the posted answer, not in the comments.
$endgroup$
– dan_fulea
Jan 16 at 0:01
|
show 1 more comment
$begingroup$
Given a polynomial $P(x)$ of degree $m>1$: $$P(x)=a_m x^m +...+ a_k x^k +...+ a_2 x^2 + a_1 x - alpha$$
Where $lvert a_m rvert >...> lvert a_k rvert > ... > lvert a_1 rvert = 1$ and $alpha$ are integer coefficients, such that $a_k <0$ $forall k neq m$. What methods are there to rule out a specific rational root $ frac {1}{beta} >0 $ ? (It is supposed that this case that $beta$ divides $a_m$, so the rational root theorem applies, and thus this root $ frac {1}{beta} $ cannot be ruled out by contradiction with the theorem).
According to Sturm's Theorem, one can assert that there is a root in the interval $[0,F]$, where F is a positive number. It is known that $P(x)$ has no irrational roots.
polynomials
asked Jan 15 at 22:56
Federico OmarFederico Omar
63
$endgroup$
Given a polynomial $P(x)$ of degree $m>1$: $$P(x)=a_m x^m +...+ a_k x^k +...+ a_2 x^2 + a_1 x - alpha$$
Where $lvert a_m rvert >...> lvert a_k rvert > ... > lvert a_1 rvert = 1$ and $alpha$ are integer coefficients, such that $a_k <0$ $forall k neq m$. What methods are there to rule out a specific rational root $ frac {1}{beta} >0 $ ? (It is supposed that this case that $beta$ divides $a_m$, so the rational root theorem applies, and thus this root $ frac {1}{beta} $ cannot be ruled out by contradiction with the theorem).
According to Sturm's Theorem, one can assert that there is a root in the interval $[0,F]$, where F is a positive number. It is known that $P(x)$ has no irrational roots.
polynomials
polynomials
asked Jan 15 at 22:56
Federico OmarFederico Omar
63
asked Jan 15 at 22:56
Federico OmarFederico Omar
63
edited Jan 16 at 0:13
Federico Omar
asked Jan 15 at 22:56
Federico OmarFederico Omar
63
asked Jan 15 at 22:56
Federico OmarFederico Omar
63
asked Jan 15 at 22:56
Federico OmarFederico Omar
63
63
2
$begingroup$
Plugging it in?
$endgroup$
– Severin Schraven
Jan 15 at 23:05
$begingroup$
Won't work... I am looking for insights on how to tackle this problem with the given conditions, and evaluating the polynomial on the given value is impossible, because the degree of the polynomial isn't defined. i.e., the polynomial has arbitrary degree m.
$endgroup$
– Federico Omar
Jan 15 at 23:08
$begingroup$
Well, we have to know what "is defined and what not" in the question. For my taste, i would take the reciprocal and search for possibilities to avoid some integer root. It is unclear from the post if $beta$ is positive or general. What kind of answer is expected? What is fixed and what is moving? The $a_m$ is fixed, then $beta$ but $m$ and all other coefficients not?
$endgroup$
– dan_fulea
Jan 15 at 23:46
$begingroup$
@dan_fulea $beta$ is a positive value, and $frac{1}{beta}$ is a candidate for rational root. All roots of P(x) are rational. deg(P(x)) is non trivial ( can be greater than 4). All the coefficients of P(x) are negative, except $a_m$.
$endgroup$
– Federico Omar
Jan 15 at 23:53
$begingroup$
That $beta$ is $>0$ is a new information. The fact that $P$ completely splits over $Bbb Q$ is also new. Please rewrite the question starting with the given data. Best, introduce the objects one by one, and take care for the usual $forall$ and $exists$ quantifiers. Please do not introduce a polynomial, and after that the degree and the coefficients, if some $forall$ and/or $exists$ apply only to parts of them. Please insert all details in the posted answer, not in the comments.
$endgroup$
– dan_fulea
Jan 16 at 0:01
|
show 1 more comment
2
$begingroup$
Plugging it in?
$endgroup$
– Severin Schraven
Jan 15 at 23:05
$begingroup$
Won't work... I am looking for insights on how to tackle this problem with the given conditions, and evaluating the polynomial on the given value is impossible, because the degree of the polynomial isn't defined. i.e., the polynomial has arbitrary degree m.
$endgroup$
– Federico Omar
Jan 15 at 23:08
$begingroup$
Well, we have to know what "is defined and what not" in the question. For my taste, i would take the reciprocal and search for possibilities to avoid some integer root. It is unclear from the post if $beta$ is positive or general. What kind of answer is expected? What is fixed and what is moving? The $a_m$ is fixed, then $beta$ but $m$ and all other coefficients not?
$endgroup$
– dan_fulea
Jan 15 at 23:46
$begingroup$
@dan_fulea $beta$ is a positive value, and $frac{1}{beta}$ is a candidate for rational root. All roots of P(x) are rational. deg(P(x)) is non trivial ( can be greater than 4). All the coefficients of P(x) are negative, except $a_m$.
$endgroup$
– Federico Omar
Jan 15 at 23:53
$begingroup$
That $beta$ is $>0$ is a new information. The fact that $P$ completely splits over $Bbb Q$ is also new. Please rewrite the question starting with the given data. Best, introduce the objects one by one, and take care for the usual $forall$ and $exists$ quantifiers. Please do not introduce a polynomial, and after that the degree and the coefficients, if some $forall$ and/or $exists$ apply only to parts of them. Please insert all details in the posted answer, not in the comments.
$endgroup$
– dan_fulea
Jan 16 at 0:01
2
2
$begingroup$
Plugging it in?
$endgroup$
– Severin Schraven
Jan 15 at 23:05
$begingroup$
Plugging it in?
$endgroup$
– Severin Schraven
Jan 15 at 23:05
$begingroup$
Won't work... I am looking for insights on how to tackle this problem with the given conditions, and evaluating the polynomial on the given value is impossible, because the degree of the polynomial isn't defined. i.e., the polynomial has arbitrary degree m.
$endgroup$
– Federico Omar
Jan 15 at 23:08
$begingroup$
Won't work... I am looking for insights on how to tackle this problem with the given conditions, and evaluating the polynomial on the given value is impossible, because the degree of the polynomial isn't defined. i.e., the polynomial has arbitrary degree m.
$endgroup$
– Federico Omar
Jan 15 at 23:08
$begingroup$
Well, we have to know what "is defined and what not" in the question. For my taste, i would take the reciprocal and search for possibilities to avoid some integer root. It is unclear from the post if $beta$ is positive or general. What kind of answer is expected? What is fixed and what is moving? The $a_m$ is fixed, then $beta$ but $m$ and all other coefficients not?
$endgroup$
– dan_fulea
Jan 15 at 23:46
$begingroup$
Well, we have to know what "is defined and what not" in the question. For my taste, i would take the reciprocal and search for possibilities to avoid some integer root. It is unclear from the post if $beta$ is positive or general. What kind of answer is expected? What is fixed and what is moving? The $a_m$ is fixed, then $beta$ but $m$ and all other coefficients not?
$endgroup$
– dan_fulea
Jan 15 at 23:46
$begingroup$
@dan_fulea $beta$ is a positive value, and $frac{1}{beta}$ is a candidate for rational root. All roots of P(x) are rational. deg(P(x)) is non trivial ( can be greater than 4). All the coefficients of P(x) are negative, except $a_m$.
$endgroup$
– Federico Omar
Jan 15 at 23:53
$begingroup$
@dan_fulea $beta$ is a positive value, and $frac{1}{beta}$ is a candidate for rational root. All roots of P(x) are rational. deg(P(x)) is non trivial ( can be greater than 4). All the coefficients of P(x) are negative, except $a_m$.
$endgroup$
– Federico Omar
Jan 15 at 23:53
$begingroup$
That $beta$ is $>0$ is a new information. The fact that $P$ completely splits over $Bbb Q$ is also new. Please rewrite the question starting with the given data. Best, introduce the objects one by one, and take care for the usual $forall$ and $exists$ quantifiers. Please do not introduce a polynomial, and after that the degree and the coefficients, if some $forall$ and/or $exists$ apply only to parts of them. Please insert all details in the posted answer, not in the comments.
$endgroup$
– dan_fulea
Jan 16 at 0:01
$begingroup$
That $beta$ is $>0$ is a new information. The fact that $P$ completely splits over $Bbb Q$ is also new. Please rewrite the question starting with the given data. Best, introduce the objects one by one, and take care for the usual $forall$ and $exists$ quantifiers. Please do not introduce a polynomial, and after that the degree and the coefficients, if some $forall$ and/or $exists$ apply only to parts of them. Please insert all details in the posted answer, not in the comments.
$endgroup$
– dan_fulea
Jan 16 at 0:01
|
show 1 more comment
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2
$begingroup$
Plugging it in?
$endgroup$
– Severin Schraven
Jan 15 at 23:05
$begingroup$
Won't work... I am looking for insights on how to tackle this problem with the given conditions, and evaluating the polynomial on the given value is impossible, because the degree of the polynomial isn't defined. i.e., the polynomial has arbitrary degree m.
$endgroup$
– Federico Omar
Jan 15 at 23:08
$begingroup$
Well, we have to know what "is defined and what not" in the question. For my taste, i would take the reciprocal and search for possibilities to avoid some integer root. It is unclear from the post if $beta$ is positive or general. What kind of answer is expected? What is fixed and what is moving? The $a_m$ is fixed, then $beta$ but $m$ and all other coefficients not?
$endgroup$
– dan_fulea
Jan 15 at 23:46
$begingroup$
@dan_fulea $beta$ is a positive value, and $frac{1}{beta}$ is a candidate for rational root. All roots of P(x) are rational. deg(P(x)) is non trivial ( can be greater than 4). All the coefficients of P(x) are negative, except $a_m$.
$endgroup$
– Federico Omar
Jan 15 at 23:53
$begingroup$
That $beta$ is $>0$ is a new information. The fact that $P$ completely splits over $Bbb Q$ is also new. Please rewrite the question starting with the given data. Best, introduce the objects one by one, and take care for the usual $forall$ and $exists$ quantifiers. Please do not introduce a polynomial, and after that the degree and the coefficients, if some $forall$ and/or $exists$ apply only to parts of them. Please insert all details in the posted answer, not in the comments.
$endgroup$
– dan_fulea
Jan 16 at 0:01