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For a given value of $n$, how can we find integers $a_0, ..., a_n$ so that $\sum\limits_{k=0}^n a_k x^k$ has $n$ distinct real roots and $S=\sum\limits_{k=0}^n |a_k|$ is minimized?

For the first few smallest values of $n$, I have tried to find the coefficients by trial and error. These polynomials have $n$ distinct real roots, but I'm not sure if $S$ is minimized.

$n=1: x, S=\color{red}{1}$
$n=2: x^2-1$ or $x^2-x, S=\color{red}{2}$
$n=3: x^3-x, S=\color{red}{2}$
$n=4: x^4-2x^2+x, S=\color{red}{4}$
$n=5: x^5-3x^3+x, S=\color{red}{5}$
$n=6: x^6-4x^4+3x^2-x, S=\color{red}{9}$
$n=7: x^7-4x^5+4x^3-x, S=\color{red}{10}$

UPDATE

From the comments:

@Xiangxiang Xu confirmed my conjectured results up to $n=7$ using a Python program.

@Peter Košinár confirmed that for $n=8$, $x^8-3x^7-x^6+7x^5-5x^3+x, S=\color{red}{18}$ is best.

@Martin R confirmed that for $n=9$, $x^9-7x^7+11x^5-6x^3+x, S=\color{red}{26}$ is best.

UPDATE $2$

@Tom Sirgedas at Puzzling found that for $n=10$, $x^{10}-4x^{9}-2x^{8}+14x^{7}+x^{6}-16x^{5}+7x^{3}-x, S=\color{red}{46}$, is best.

Dan
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  • Are complex solutions allowed? Or only reals or rationals? – emacs drives me nuts Mar 19 '23 at 11:36
  • @emacsdrivesmenuts The polynomial must have $n$ distinct real roots. – Dan Mar 19 '23 at 12:00
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    Just an observation: When $\phi$ is the Golden Ratio, then we have that your polynomials satisfy $$p_5(x)=x(x-\phi)(x+\phi)(x-\phi^{-1})(x+\phi^{-1}) $$ $$p_7(x)=(x-1)(x+1)p_5(x) $$ and $p_6$ is divisible by $(x+\phi)(x-\phi^{-1})=x^2+x-1$. – emacs drives me nuts Mar 19 '23 at 16:59
  • Did you find those example with a computer program? If yes, can you share the source? – Martin R Mar 28 '23 at 06:49
  • @MartinR I found the examples by drawing graphs on desmos. – Dan Mar 28 '23 at 07:11
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    @MartinR I wrote a python program to search for the solution by pure brute force, which verifies the above results up to $n = 7$: https://colab.research.google.com/drive/1Jddu_y2m50xjno77CimBgWrMjz1MH1lc#scrollTo=nXgHum8CiyoI – Xiangxiang Xu Mar 31 '23 at 16:49
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    It seems that $n=8$ admits a better solution: $p_8(x)=x^8 - 3x^7 - x^6 + 7x^5 - 5x^3 + x$, with sum of $18$. – Peter Košinár Mar 31 '23 at 22:03
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    $n = 9$ also admits a better solution $x^9 - 7x^7 + 11 x^5 - 6 x^3 + x$, with $S = 26$. – Xiangxiang Xu Apr 01 '23 at 13:38
  • @XiangxiangXu: I have modified your program to show all “minimal polynomials” for a given degree. It seems that they are unique if $n$ is odd, but not if $n$ is even. – Martin R Apr 01 '23 at 15:50
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    @MartinR That program can be too slow for large $n$. Have you obtained a solution for $n=9$ better than $S = 26$? Also, all the non-unique results I obtained so far can be retrieved from the properties I mentioned in the answer. So up to this equivalence (flipping the order or sign of coefficients), it seems the solution is still unique . – Xiangxiang Xu Apr 01 '23 at 16:39
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    @XiangxiangXu: I have started writing a PARI/GP program to find the polynomials. PARI/GP works exactly with arbitrarily large integers, and also has methods to determine the number of real distinct roots with Sturm chains, so that there are no precision problems. I can confirm that $p_8(x)=x^8 - 3x^7 - x^6 + 7x^5 - 5x^3 + x$ is best possible, but my program is still too slow for $n=9$. – Martin R Apr 02 '23 at 16:53
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    @MartinR I also tried to implement the Sturm theorem and added pruning: while certain sign change doesn't happen, it would not compute the proceeding polynomials. But it seems slower than brute-force root evaluation. You can check the colab link for more details. – Xiangxiang Xu Apr 02 '23 at 19:30
  • Based on the reported optimal polynomials (in the comments here, and the comments after one of the answers) for $n=1$ to $9$, I would expect that for $n=8$ there should be a (non-unique) optimal polynomial of the form $x^8−a_6x^6+a_4x^4−a_2x^2+x$ with $a_6,a_4,a_2$ positive and less than $11$, and $S=18$. I searched systematically (by drawing graphs on desmos) for such a polynomial, but did not find one. Strange. – Dan Apr 04 '23 at 23:50
  • Related question on MO: Given $n$, how to find integers $a_0,...,a_n$ so that $\sum_{k=0}^n a_k x^k$ has $n$ distinct real roots, and $\max{(|a_0|,...,|a_n|)}$ is minimized? – Dan Apr 05 '23 at 04:02
  • Results up to $n=10$ are here on puzzling – Ross Millikan Oct 03 '25 at 14:03

2 Answers2

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Not an answer, but a list of partial observations and conjectures:

Properties (can be easily proved)

  1. If $\displaystyle p_n(x) = \sum_{k = 0}^n a_k x^k$ is optimal, then $a_0^2 + a_1^2 > 0$.
  2. If $\displaystyle p_n(x) = \sum_{k = 0}^n a_k x^k$ is optimal, then $\displaystyle p_n(-x) = \sum_{k = 0}^n (-1)^ka_k x^k$ is also optimal.
  3. If $\displaystyle p_n(x) = \sum_{k = 0}^n a_k x^k$ is optimal and $a_0 = 0$, then $\displaystyle p_n(x^{-1})x^{n-1} = \sum_{k = 1}^n a_{n+1-k} x^k$ is also optimal for degree $n$.

Conjectures for $n > 2$

  • The optimal polynomial $\displaystyle p_n(x) = \sum_{k = 0}^n a_k x^k$ satisfies $a_0 = 0$.
  • If $n$ is odd, then we have $p_n(x) = x \cdot q_{\frac{n-1}{2}}(x^2)$, where $q_k(x)$ is an interger polynomial of degree $k$ that has $k$ distinct positive roots with the minimum $\ell_1$-norm.
  • If $n$ is even, then we have $p_n(x) = x \cdot r_{n-1}(x)$, where $r_k(x)$ is an integer polynomial of degree $k$ that has $k$ distinct non-zero real roots with the minimum $\ell_1$-norm.

where we have defined the $\ell_1$-norm of $\displaystyle\sum_{k = 0}^n a_k x^k$ as $\displaystyle S = \sum_{k = 0}^n |a_k|$.

Xiangxiang Xu
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  • It is clear (as you say) that $a_0$ and $a_1$ can not both be zero. I wonder if one can show that no two consecutive coefficients can be zero. That would reduce the number of combinations to test. – Martin R Apr 03 '23 at 06:43
  • Except for the middle two? For $p_8$ the middle two are both non-zero. – Xiangxiang Xu Apr 03 '23 at 11:44
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    $p_8(x)=x^8 - 3x^7 - x^6 + 7x^5 - 5x^3 + x$ has $a_4=a_2 = a_0 = 0$. What I meant is that the optimal polynomial does not have $a_j = a_{j+1} = 0$ for some index $j$. – Martin R Apr 03 '23 at 11:48
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    I can now confirm that $S=26$ is best possible for $n=9$. Apart from the obvious transformations there are two optimal polynomials $x^9 - 7 x^7 + 11 x^5 - 6 x^3 + x$ and $x^9 - 6 x^7 - x^6 + 10 x^5 + x^4 - 6 x^3 + x$. – Martin R Apr 03 '23 at 19:57
  • Very nice example. So the optimal solution may not be unique, and my conjecture cannot capture all solutions. – Xiangxiang Xu Apr 04 '23 at 00:20
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    If one restricts to odd polynomials in degree $n=11$, then $f(x)=xh(-x^2)$ with $h(x)=x^5 + 8x^4 + 20x^3 + 20x^2 + 8x + 1$ is optimal an unique with $S=58$. – Peter Mueller Apr 07 '23 at 13:03
  • @PeterMueller So we shall have $h(x) = q_5(-x)$ defined above, right? – Xiangxiang Xu Apr 07 '23 at 16:00
  • @XiangxiangXu Yes, that's right. I missed that the $q_k$'s were introduced above. – Peter Mueller Apr 07 '23 at 16:37
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Just an observation: For $n = 1 \mod 4$, the solutions you have listed are of the form: if $+r$ is a root then $-r$ is also a root. $0$ is a root. if $r$ is a root, then $1/r$ is also a root.

So try out: $$f(x) = x \prod_{i=1}^{n-1} (x-r_i) (x+r_i) (x+1/r_i) (x-1/r_i)$$.