All integer polynomials of the form $x^{d_t}-x^{d_1-1}-x^{d_2-1}…x^{d_t-d_t}$ with the same maximal (real)...












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Let $d=(d_1,dots,d_{d})$ be a vector of integer exponents of a polynomials such that $d_1leq d_2leq dots leq d_{|d|}$. Let us define the polynomial $p_d(x)=x^{d_{|d|}}-sum_i{x^{d_{|d|}-d_i}}$ (such that $1-sum_i{x^{-d_i}}=0$ iff $p_d(x)=0$). These are all polynomials where only the power of the largest exponent has a positive coefficient.



I am interested in every polynomial $p_{d'}(x)$ (or vector of integers $d'$) of the same form such that the largest real root $r$ of $p_{d'}(x)$ and $p_{d}(x)$ is the same. For example $x^{1}-2cdot 1$, $x^{2}-4cdot 1$, $x^{3}-8cdot 1,,dots$, with the root $x=2$, or $x^{2}-x^{1}-1,x^{4}-x^{1}-x^{2}-x^{2}-1,dots$ with the (real) roots $frac{1pm sqrt{5}}{2}$.



I already know that we can get another vector $d'$ of exponents from a known one: Just add all combinations of the exponents: From $(2,1)$ we can get $(4,3,3,2)$. The real roots of the related polynomial should be the same then. However, it is not clear for me, if a) there can be multiple vectors of the same length, and b) if there are other vectors possible that cannot be produced by this method.






functions polynomials







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  • $begingroup$
    For example, all the polynomials $p(x) = x^m - x^k$ for $k < m$ have largest real root $1$.
    $endgroup$
    – Robert Israel
    Jan 10 at 16:11










  • $begingroup$
    That's right, but unfortunately these polynomials are not what I mean. There should be $−x^{d_{|d|}−d_{|d|}}=x^0$ in the term.
    $endgroup$
    – ptid91
    Jan 11 at 9:51


















0












$begingroup$


Let $d=(d_1,dots,d_{d})$ be a vector of integer exponents of a polynomials such that $d_1leq d_2leq dots leq d_{|d|}$. Let us define the polynomial $p_d(x)=x^{d_{|d|}}-sum_i{x^{d_{|d|}-d_i}}$ (such that $1-sum_i{x^{-d_i}}=0$ iff $p_d(x)=0$). These are all polynomials where only the power of the largest exponent has a positive coefficient.



I am interested in every polynomial $p_{d'}(x)$ (or vector of integers $d'$) of the same form such that the largest real root $r$ of $p_{d'}(x)$ and $p_{d}(x)$ is the same. For example $x^{1}-2cdot 1$, $x^{2}-4cdot 1$, $x^{3}-8cdot 1,,dots$, with the root $x=2$, or $x^{2}-x^{1}-1,x^{4}-x^{1}-x^{2}-x^{2}-1,dots$ with the (real) roots $frac{1pm sqrt{5}}{2}$.



I already know that we can get another vector $d'$ of exponents from a known one: Just add all combinations of the exponents: From $(2,1)$ we can get $(4,3,3,2)$. The real roots of the related polynomial should be the same then. However, it is not clear for me, if a) there can be multiple vectors of the same length, and b) if there are other vectors possible that cannot be produced by this method.






functions polynomials







share|cite|improve this question











$endgroup$












  • $begingroup$
    For example, all the polynomials $p(x) = x^m - x^k$ for $k < m$ have largest real root $1$.
    $endgroup$
    – Robert Israel
    Jan 10 at 16:11










  • $begingroup$
    That's right, but unfortunately these polynomials are not what I mean. There should be $−x^{d_{|d|}−d_{|d|}}=x^0$ in the term.
    $endgroup$
    – ptid91
    Jan 11 at 9:51
















0












0








0





$begingroup$


Let $d=(d_1,dots,d_{d})$ be a vector of integer exponents of a polynomials such that $d_1leq d_2leq dots leq d_{|d|}$. Let us define the polynomial $p_d(x)=x^{d_{|d|}}-sum_i{x^{d_{|d|}-d_i}}$ (such that $1-sum_i{x^{-d_i}}=0$ iff $p_d(x)=0$). These are all polynomials where only the power of the largest exponent has a positive coefficient.



I am interested in every polynomial $p_{d'}(x)$ (or vector of integers $d'$) of the same form such that the largest real root $r$ of $p_{d'}(x)$ and $p_{d}(x)$ is the same. For example $x^{1}-2cdot 1$, $x^{2}-4cdot 1$, $x^{3}-8cdot 1,,dots$, with the root $x=2$, or $x^{2}-x^{1}-1,x^{4}-x^{1}-x^{2}-x^{2}-1,dots$ with the (real) roots $frac{1pm sqrt{5}}{2}$.



I already know that we can get another vector $d'$ of exponents from a known one: Just add all combinations of the exponents: From $(2,1)$ we can get $(4,3,3,2)$. The real roots of the related polynomial should be the same then. However, it is not clear for me, if a) there can be multiple vectors of the same length, and b) if there are other vectors possible that cannot be produced by this method.






functions polynomials







share|cite|improve this question











$endgroup$




Let $d=(d_1,dots,d_{d})$ be a vector of integer exponents of a polynomials such that $d_1leq d_2leq dots leq d_{|d|}$. Let us define the polynomial $p_d(x)=x^{d_{|d|}}-sum_i{x^{d_{|d|}-d_i}}$ (such that $1-sum_i{x^{-d_i}}=0$ iff $p_d(x)=0$). These are all polynomials where only the power of the largest exponent has a positive coefficient.



I am interested in every polynomial $p_{d'}(x)$ (or vector of integers $d'$) of the same form such that the largest real root $r$ of $p_{d'}(x)$ and $p_{d}(x)$ is the same. For example $x^{1}-2cdot 1$, $x^{2}-4cdot 1$, $x^{3}-8cdot 1,,dots$, with the root $x=2$, or $x^{2}-x^{1}-1,x^{4}-x^{1}-x^{2}-x^{2}-1,dots$ with the (real) roots $frac{1pm sqrt{5}}{2}$.



I already know that we can get another vector $d'$ of exponents from a known one: Just add all combinations of the exponents: From $(2,1)$ we can get $(4,3,3,2)$. The real roots of the related polynomial should be the same then. However, it is not clear for me, if a) there can be multiple vectors of the same length, and b) if there are other vectors possible that cannot be produced by this method.






functions polynomials




functions polynomials






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edited Jan 11 at 9:49







ptid91

















asked Jan 10 at 15:22









ptid91ptid91

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1












  • $begingroup$
    For example, all the polynomials $p(x) = x^m - x^k$ for $k < m$ have largest real root $1$.
    $endgroup$
    – Robert Israel
    Jan 10 at 16:11










  • $begingroup$
    That's right, but unfortunately these polynomials are not what I mean. There should be $−x^{d_{|d|}−d_{|d|}}=x^0$ in the term.
    $endgroup$
    – ptid91
    Jan 11 at 9:51




















  • $begingroup$
    For example, all the polynomials $p(x) = x^m - x^k$ for $k < m$ have largest real root $1$.
    $endgroup$
    – Robert Israel
    Jan 10 at 16:11










  • $begingroup$
    That's right, but unfortunately these polynomials are not what I mean. There should be $−x^{d_{|d|}−d_{|d|}}=x^0$ in the term.
    $endgroup$
    – ptid91
    Jan 11 at 9:51


















$begingroup$
For example, all the polynomials $p(x) = x^m - x^k$ for $k < m$ have largest real root $1$.
$endgroup$
– Robert Israel
Jan 10 at 16:11




$begingroup$
For example, all the polynomials $p(x) = x^m - x^k$ for $k < m$ have largest real root $1$.
$endgroup$
– Robert Israel
Jan 10 at 16:11












$begingroup$
That's right, but unfortunately these polynomials are not what I mean. There should be $−x^{d_{|d|}−d_{|d|}}=x^0$ in the term.
$endgroup$
– ptid91
Jan 11 at 9:51






$begingroup$
That's right, but unfortunately these polynomials are not what I mean. There should be $−x^{d_{|d|}−d_{|d|}}=x^0$ in the term.
$endgroup$
– ptid91
Jan 11 at 9:51












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$x^7-x^5-x^4-x^2-1$ and $x^7-x^6-x^2-x-1$ both have the same largest real root, namely the real root of $x^3-x^2-1$ which is a factor of both.






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    $begingroup$

    $x^7-x^5-x^4-x^2-1$ and $x^7-x^6-x^2-x-1$ both have the same largest real root, namely the real root of $x^3-x^2-1$ which is a factor of both.






    share|cite|improve this answer









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      $begingroup$

      $x^7-x^5-x^4-x^2-1$ and $x^7-x^6-x^2-x-1$ both have the same largest real root, namely the real root of $x^3-x^2-1$ which is a factor of both.






      share|cite|improve this answer









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        0












        0








        0





        $begingroup$

        $x^7-x^5-x^4-x^2-1$ and $x^7-x^6-x^2-x-1$ both have the same largest real root, namely the real root of $x^3-x^2-1$ which is a factor of both.






        share|cite|improve this answer









        $endgroup$



        $x^7-x^5-x^4-x^2-1$ and $x^7-x^6-x^2-x-1$ both have the same largest real root, namely the real root of $x^3-x^2-1$ which is a factor of both.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 11 at 13:37









        Robert IsraelRobert Israel

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