Find all polynomials with coefficients from set $-1,1$ and which have all their roots real.Polynomials with integer coefficients, with value close to $0$, in the interval $[-1,1]$Bounding the roots of the sum of two monic polynomials with real coefficients.Proving that polynomials with rational coefficients have integer rootsReal roots of a polynomial of real co-efficients , with the co-efficients of $x^2 , x$ and the constant term all $1$Polynomials and their (real) rootsA polynomial with real coefficients expressible as sum of squares of two polynomials will have not all roots as real.Roots of polynomials with bounded integer coefficientsContinuity in roots of polynomials as functions of Real Coefficients for multidimensional polynomialsConvex combination of monic polynomials of real coefficients with roots lying on the left half planeCan we prove that all roots of those polynomials are real negative?
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Find all polynomials with coefficients from set $-1,1$ and which have all their roots real.
Polynomials with integer coefficients, with value close to $0$, in the interval $[-1,1]$Bounding the roots of the sum of two monic polynomials with real coefficients.Proving that polynomials with rational coefficients have integer rootsReal roots of a polynomial of real co-efficients , with the co-efficients of $x^2 , x$ and the constant term all $1$Polynomials and their (real) rootsA polynomial with real coefficients expressible as sum of squares of two polynomials will have not all roots as real.Roots of polynomials with bounded integer coefficientsContinuity in roots of polynomials as functions of Real Coefficients for multidimensional polynomialsConvex combination of monic polynomials of real coefficients with roots lying on the left half planeCan we prove that all roots of those polynomials are real negative?
$begingroup$
Find all polynomials with coefficients from
set $-1,1$ and which have all their roots real.
What I have tried:
Assume polynomial is $a_nx^n+a_n-1x^n-1+a_n-2x^n-2+cdots +a_0=0$ and $a_iin -1,1$ for $i=0,1,2,3,4,cdots ,n$. Let its roots be $x=r_i$ for $i=1,2,3,cdots n$. Then
$beginalignedsum r_i=-fraca_n-1a_n=pm 1\\
mathopsum_i<jr_ir_j=+fraca_n-2a_n=pm 1\\
mathopsum_i<j<kr_ir_jr_k=+fraca_n-2a_n=pm 1\\
prod r_i=(-1)^nfraca_0a_nendaligned$
How do I solve it?
polynomials roots
edited Mar 16 at 8:55
TheSimpliFire
12.9k62462
asked Mar 16 at 8:47
jackyjacky
1,264816
$endgroup$
add a comment |
$begingroup$
Find all polynomials with coefficients from
set $-1,1$ and which have all their roots real.
What I have tried:
Assume polynomial is $a_nx^n+a_n-1x^n-1+a_n-2x^n-2+cdots +a_0=0$ and $a_iin -1,1$ for $i=0,1,2,3,4,cdots ,n$. Let its roots be $x=r_i$ for $i=1,2,3,cdots n$. Then
$beginalignedsum r_i=-fraca_n-1a_n=pm 1\\
mathopsum_i<jr_ir_j=+fraca_n-2a_n=pm 1\\
mathopsum_i<j<kr_ir_jr_k=+fraca_n-2a_n=pm 1\\
prod r_i=(-1)^nfraca_0a_nendaligned$
How do I solve it?
polynomials roots
edited Mar 16 at 8:55
TheSimpliFire
12.9k62462
asked Mar 16 at 8:47
jackyjacky
1,264816
$endgroup$
$begingroup$
Are you actually required to use that form? It'd be a lot easier - to the point of triviality - to use the roots form $aproduct_k(x-x_k)$
$endgroup$
– Dan Uznanski
Mar 16 at 9:00
$begingroup$
Note if $a_n = -1$, you can multiply the equation by $-1$ and still get the same roots. Thus, WLOG, you can set $a_n = 1$, with this at least simplifying your equations slightly. Also, your triple product formula should have $-fraca_n-3a_n$ in the middle part.
$endgroup$
– John Omielan
Mar 16 at 9:09
2
$begingroup$
Have you tried solving the problems explicitly for low degrees and looking for patterns that you can try to prove?
$endgroup$
– Henning Makholm
Mar 16 at 9:48
$begingroup$
I haven't solved it, but for $n ge 2$, I have determined so far that, if you let $a_n = 1$, then $a_n-2 = -1$, plus that $sum r_i^2 = 3$. Are you able to see why this is so?
$endgroup$
– John Omielan
Mar 16 at 10:30
add a comment |
$begingroup$
Find all polynomials with coefficients from
set $-1,1$ and which have all their roots real.
What I have tried:
Assume polynomial is $a_nx^n+a_n-1x^n-1+a_n-2x^n-2+cdots +a_0=0$ and $a_iin -1,1$ for $i=0,1,2,3,4,cdots ,n$. Let its roots be $x=r_i$ for $i=1,2,3,cdots n$. Then
$beginalignedsum r_i=-fraca_n-1a_n=pm 1\\
mathopsum_i<jr_ir_j=+fraca_n-2a_n=pm 1\\
mathopsum_i<j<kr_ir_jr_k=+fraca_n-2a_n=pm 1\\
prod r_i=(-1)^nfraca_0a_nendaligned$
How do I solve it?
polynomials roots
edited Mar 16 at 8:55
TheSimpliFire
12.9k62462
asked Mar 16 at 8:47
jackyjacky
1,264816
$endgroup$
Find all polynomials with coefficients from
set $-1,1$ and which have all their roots real.
What I have tried:
Assume polynomial is $a_nx^n+a_n-1x^n-1+a_n-2x^n-2+cdots +a_0=0$ and $a_iin -1,1$ for $i=0,1,2,3,4,cdots ,n$. Let its roots be $x=r_i$ for $i=1,2,3,cdots n$. Then
$beginalignedsum r_i=-fraca_n-1a_n=pm 1\\
mathopsum_i<jr_ir_j=+fraca_n-2a_n=pm 1\\
mathopsum_i<j<kr_ir_jr_k=+fraca_n-2a_n=pm 1\\
prod r_i=(-1)^nfraca_0a_nendaligned$
How do I solve it?
polynomials roots
polynomials roots
edited Mar 16 at 8:55
TheSimpliFire
12.9k62462
asked Mar 16 at 8:47
jackyjacky
1,264816
edited Mar 16 at 8:55
TheSimpliFire
12.9k62462
asked Mar 16 at 8:47
jackyjacky
1,264816
edited Mar 16 at 8:55
TheSimpliFire
12.9k62462
edited Mar 16 at 8:55
TheSimpliFire
12.9k62462
edited Mar 16 at 8:55
TheSimpliFire
12.9k62462
12.9k62462
asked Mar 16 at 8:47
jackyjacky
1,264816
asked Mar 16 at 8:47
jackyjacky
1,264816
asked Mar 16 at 8:47
jackyjacky
1,264816
1,264816
$begingroup$
Are you actually required to use that form? It'd be a lot easier - to the point of triviality - to use the roots form $aproduct_k(x-x_k)$
$endgroup$
– Dan Uznanski
Mar 16 at 9:00
$begingroup$
Note if $a_n = -1$, you can multiply the equation by $-1$ and still get the same roots. Thus, WLOG, you can set $a_n = 1$, with this at least simplifying your equations slightly. Also, your triple product formula should have $-fraca_n-3a_n$ in the middle part.
$endgroup$
– John Omielan
Mar 16 at 9:09
2
$begingroup$
Have you tried solving the problems explicitly for low degrees and looking for patterns that you can try to prove?
$endgroup$
– Henning Makholm
Mar 16 at 9:48
$begingroup$
I haven't solved it, but for $n ge 2$, I have determined so far that, if you let $a_n = 1$, then $a_n-2 = -1$, plus that $sum r_i^2 = 3$. Are you able to see why this is so?
$endgroup$
– John Omielan
Mar 16 at 10:30
add a comment |
$begingroup$
Are you actually required to use that form? It'd be a lot easier - to the point of triviality - to use the roots form $aproduct_k(x-x_k)$
$endgroup$
– Dan Uznanski
Mar 16 at 9:00
$begingroup$
Note if $a_n = -1$, you can multiply the equation by $-1$ and still get the same roots. Thus, WLOG, you can set $a_n = 1$, with this at least simplifying your equations slightly. Also, your triple product formula should have $-fraca_n-3a_n$ in the middle part.
$endgroup$
– John Omielan
Mar 16 at 9:09
2
$begingroup$
Have you tried solving the problems explicitly for low degrees and looking for patterns that you can try to prove?
$endgroup$
– Henning Makholm
Mar 16 at 9:48
$begingroup$
I haven't solved it, but for $n ge 2$, I have determined so far that, if you let $a_n = 1$, then $a_n-2 = -1$, plus that $sum r_i^2 = 3$. Are you able to see why this is so?
$endgroup$
– John Omielan
Mar 16 at 10:30
$begingroup$
Are you actually required to use that form? It'd be a lot easier - to the point of triviality - to use the roots form $aproduct_k(x-x_k)$
$endgroup$
– Dan Uznanski
Mar 16 at 9:00
$begingroup$
Are you actually required to use that form? It'd be a lot easier - to the point of triviality - to use the roots form $aproduct_k(x-x_k)$
$endgroup$
– Dan Uznanski
Mar 16 at 9:00
$begingroup$
Note if $a_n = -1$, you can multiply the equation by $-1$ and still get the same roots. Thus, WLOG, you can set $a_n = 1$, with this at least simplifying your equations slightly. Also, your triple product formula should have $-fraca_n-3a_n$ in the middle part.
$endgroup$
– John Omielan
Mar 16 at 9:09
$begingroup$
Note if $a_n = -1$, you can multiply the equation by $-1$ and still get the same roots. Thus, WLOG, you can set $a_n = 1$, with this at least simplifying your equations slightly. Also, your triple product formula should have $-fraca_n-3a_n$ in the middle part.
$endgroup$
– John Omielan
Mar 16 at 9:09
2
2
$begingroup$
Have you tried solving the problems explicitly for low degrees and looking for patterns that you can try to prove?
$endgroup$
– Henning Makholm
Mar 16 at 9:48
$begingroup$
Have you tried solving the problems explicitly for low degrees and looking for patterns that you can try to prove?
$endgroup$
– Henning Makholm
Mar 16 at 9:48
$begingroup$
I haven't solved it, but for $n ge 2$, I have determined so far that, if you let $a_n = 1$, then $a_n-2 = -1$, plus that $sum r_i^2 = 3$. Are you able to see why this is so?
$endgroup$
– John Omielan
Mar 16 at 10:30
$begingroup$
I haven't solved it, but for $n ge 2$, I have determined so far that, if you let $a_n = 1$, then $a_n-2 = -1$, plus that $sum r_i^2 = 3$. Are you able to see why this is so?
$endgroup$
– John Omielan
Mar 16 at 10:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
WLOG, let $a_n = 1$. Also, let $r_1,ldots,r_n$ be the roots of the polynomial.
As you noted, $displaystylesum_k = 1^nr_k = -a_n-1$, $displaystylesum_1 le k < ell le nr_kr_ell = a_n-2$, and $displaystyleprod_k = 1^nr_k = (-1)^na_0$. Hence,
$$displaystylesum_k = 1^nr_k^2 = left(sum_k = 1^nr_kright)^2-2left(sum_1 le k < ell le nr_kr_ellright) = a_n-1^2-2a_n-2 = 1-2a_n-2 = begincases3 & textif a_n-2 = -1 \ -1 & textif a_n-2 = 1 endcases.$$
Since all the roots are real, $displaystylesum_k = 1^nr_k^2 ge 0$. Hence, we must have $a_n-2 = -1$ and thus $displaystylesum_k = 1^nr_k^2 = 3$.
Trivially, we then have $displaystylesum_k = 1^n|r_k|^2 = sum_k = 1^nr_k^2= 3$, as well as $displaystyleprod_k = 1^n|r_k| = left|prod_k = 1^nr_kright| = left|(-1)^na_0right| = 1$.
So by the RMS-GM inequality, we have $$displaystyle 1 = left(prod_k = 1^n|r_k|right)^1/n le left(dfrac1nsum_k = 1^n|r_k|^2right)^1/2 = sqrtdfrac3n.$$
Therefore, we must have $n le 3$. It is easy to check that the only solutions are:
$x-1$, $x+1$, $x^2-x-1$, $x^2+x-1$, $x^3-x^2-x+1$, $x^3+x^2-x-1$, and their negatives.
answered Mar 17 at 9:27
JimmyK4542JimmyK4542
41.3k245107
$endgroup$
add a comment |
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$begingroup$
WLOG, let $a_n = 1$. Also, let $r_1,ldots,r_n$ be the roots of the polynomial.
As you noted, $displaystylesum_k = 1^nr_k = -a_n-1$, $displaystylesum_1 le k < ell le nr_kr_ell = a_n-2$, and $displaystyleprod_k = 1^nr_k = (-1)^na_0$. Hence,
$$displaystylesum_k = 1^nr_k^2 = left(sum_k = 1^nr_kright)^2-2left(sum_1 le k < ell le nr_kr_ellright) = a_n-1^2-2a_n-2 = 1-2a_n-2 = begincases3 & textif a_n-2 = -1 \ -1 & textif a_n-2 = 1 endcases.$$
Since all the roots are real, $displaystylesum_k = 1^nr_k^2 ge 0$. Hence, we must have $a_n-2 = -1$ and thus $displaystylesum_k = 1^nr_k^2 = 3$.
Trivially, we then have $displaystylesum_k = 1^n|r_k|^2 = sum_k = 1^nr_k^2= 3$, as well as $displaystyleprod_k = 1^n|r_k| = left|prod_k = 1^nr_kright| = left|(-1)^na_0right| = 1$.
So by the RMS-GM inequality, we have $$displaystyle 1 = left(prod_k = 1^n|r_k|right)^1/n le left(dfrac1nsum_k = 1^n|r_k|^2right)^1/2 = sqrtdfrac3n.$$
Therefore, we must have $n le 3$. It is easy to check that the only solutions are:
$x-1$, $x+1$, $x^2-x-1$, $x^2+x-1$, $x^3-x^2-x+1$, $x^3+x^2-x-1$, and their negatives.
answered Mar 17 at 9:27
JimmyK4542JimmyK4542
41.3k245107
$endgroup$
add a comment |
$begingroup$
WLOG, let $a_n = 1$. Also, let $r_1,ldots,r_n$ be the roots of the polynomial.
As you noted, $displaystylesum_k = 1^nr_k = -a_n-1$, $displaystylesum_1 le k < ell le nr_kr_ell = a_n-2$, and $displaystyleprod_k = 1^nr_k = (-1)^na_0$. Hence,
$$displaystylesum_k = 1^nr_k^2 = left(sum_k = 1^nr_kright)^2-2left(sum_1 le k < ell le nr_kr_ellright) = a_n-1^2-2a_n-2 = 1-2a_n-2 = begincases3 & textif a_n-2 = -1 \ -1 & textif a_n-2 = 1 endcases.$$
Since all the roots are real, $displaystylesum_k = 1^nr_k^2 ge 0$. Hence, we must have $a_n-2 = -1$ and thus $displaystylesum_k = 1^nr_k^2 = 3$.
Trivially, we then have $displaystylesum_k = 1^n|r_k|^2 = sum_k = 1^nr_k^2= 3$, as well as $displaystyleprod_k = 1^n|r_k| = left|prod_k = 1^nr_kright| = left|(-1)^na_0right| = 1$.
So by the RMS-GM inequality, we have $$displaystyle 1 = left(prod_k = 1^n|r_k|right)^1/n le left(dfrac1nsum_k = 1^n|r_k|^2right)^1/2 = sqrtdfrac3n.$$
Therefore, we must have $n le 3$. It is easy to check that the only solutions are:
$x-1$, $x+1$, $x^2-x-1$, $x^2+x-1$, $x^3-x^2-x+1$, $x^3+x^2-x-1$, and their negatives.
answered Mar 17 at 9:27
JimmyK4542JimmyK4542
41.3k245107
$endgroup$
add a comment |
$begingroup$
WLOG, let $a_n = 1$. Also, let $r_1,ldots,r_n$ be the roots of the polynomial.
As you noted, $displaystylesum_k = 1^nr_k = -a_n-1$, $displaystylesum_1 le k < ell le nr_kr_ell = a_n-2$, and $displaystyleprod_k = 1^nr_k = (-1)^na_0$. Hence,
$$displaystylesum_k = 1^nr_k^2 = left(sum_k = 1^nr_kright)^2-2left(sum_1 le k < ell le nr_kr_ellright) = a_n-1^2-2a_n-2 = 1-2a_n-2 = begincases3 & textif a_n-2 = -1 \ -1 & textif a_n-2 = 1 endcases.$$
Since all the roots are real, $displaystylesum_k = 1^nr_k^2 ge 0$. Hence, we must have $a_n-2 = -1$ and thus $displaystylesum_k = 1^nr_k^2 = 3$.
Trivially, we then have $displaystylesum_k = 1^n|r_k|^2 = sum_k = 1^nr_k^2= 3$, as well as $displaystyleprod_k = 1^n|r_k| = left|prod_k = 1^nr_kright| = left|(-1)^na_0right| = 1$.
So by the RMS-GM inequality, we have $$displaystyle 1 = left(prod_k = 1^n|r_k|right)^1/n le left(dfrac1nsum_k = 1^n|r_k|^2right)^1/2 = sqrtdfrac3n.$$
Therefore, we must have $n le 3$. It is easy to check that the only solutions are:
$x-1$, $x+1$, $x^2-x-1$, $x^2+x-1$, $x^3-x^2-x+1$, $x^3+x^2-x-1$, and their negatives.
answered Mar 17 at 9:27
JimmyK4542JimmyK4542
41.3k245107
$endgroup$
WLOG, let $a_n = 1$. Also, let $r_1,ldots,r_n$ be the roots of the polynomial.
As you noted, $displaystylesum_k = 1^nr_k = -a_n-1$, $displaystylesum_1 le k < ell le nr_kr_ell = a_n-2$, and $displaystyleprod_k = 1^nr_k = (-1)^na_0$. Hence,
$$displaystylesum_k = 1^nr_k^2 = left(sum_k = 1^nr_kright)^2-2left(sum_1 le k < ell le nr_kr_ellright) = a_n-1^2-2a_n-2 = 1-2a_n-2 = begincases3 & textif a_n-2 = -1 \ -1 & textif a_n-2 = 1 endcases.$$
Since all the roots are real, $displaystylesum_k = 1^nr_k^2 ge 0$. Hence, we must have $a_n-2 = -1$ and thus $displaystylesum_k = 1^nr_k^2 = 3$.
Trivially, we then have $displaystylesum_k = 1^n|r_k|^2 = sum_k = 1^nr_k^2= 3$, as well as $displaystyleprod_k = 1^n|r_k| = left|prod_k = 1^nr_kright| = left|(-1)^na_0right| = 1$.
So by the RMS-GM inequality, we have $$displaystyle 1 = left(prod_k = 1^n|r_k|right)^1/n le left(dfrac1nsum_k = 1^n|r_k|^2right)^1/2 = sqrtdfrac3n.$$
Therefore, we must have $n le 3$. It is easy to check that the only solutions are:
$x-1$, $x+1$, $x^2-x-1$, $x^2+x-1$, $x^3-x^2-x+1$, $x^3+x^2-x-1$, and their negatives.
answered Mar 17 at 9:27
JimmyK4542JimmyK4542
41.3k245107
answered Mar 17 at 9:27
JimmyK4542JimmyK4542
41.3k245107
answered Mar 17 at 9:27
JimmyK4542JimmyK4542
41.3k245107
answered Mar 17 at 9:27
JimmyK4542JimmyK4542
41.3k245107
41.3k245107
add a comment |
add a comment |
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$begingroup$
Are you actually required to use that form? It'd be a lot easier - to the point of triviality - to use the roots form $aproduct_k(x-x_k)$
$endgroup$
– Dan Uznanski
Mar 16 at 9:00
$begingroup$
Note if $a_n = -1$, you can multiply the equation by $-1$ and still get the same roots. Thus, WLOG, you can set $a_n = 1$, with this at least simplifying your equations slightly. Also, your triple product formula should have $-fraca_n-3a_n$ in the middle part.
$endgroup$
– John Omielan
Mar 16 at 9:09
2
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Have you tried solving the problems explicitly for low degrees and looking for patterns that you can try to prove?
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– Henning Makholm
Mar 16 at 9:48
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I haven't solved it, but for $n ge 2$, I have determined so far that, if you let $a_n = 1$, then $a_n-2 = -1$, plus that $sum r_i^2 = 3$. Are you able to see why this is so?
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– John Omielan
Mar 16 at 10:30