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Given $1<\alpha_{1}<\alpha_{2}<\alpha_{3}\cdots<\alpha_{N}<2$

I need to construct a degree-$L$ (with $L>N$) real polynomial $f(x)=x^{L}+\sum\limits_{i=1}^{L}b_ix^{L-i}$ which satisfies

1) $f(\alpha_{i})=0$, $1\leq i\leq N$

and

2) $\sum\limits_{i=1}^{L}|b_i|\leq 2$

user1551
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Tao
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  • What is the relation between $L$ and $N$? For example, what you asks for is often impossible if $L<N$. Also, are the $b_i$s real numbers? – user1551 Mar 13 '13 at 15:51
  • I've changed the title to better reflect the content of the question. If you think it's inappropriate, please feel free to roll back. – user1551 Mar 13 '13 at 15:58
  • From condition 1, L is bigger than N, $b_i$ are real. anyway, I just want to a polynomial satisfies 2 conditions. – Tao Mar 13 '13 at 16:09

1 Answers1

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Just a hint but maybe helpful: If $L=N$ the polynomial is uniquely determined (since your leading monomial is by assumption equal to 1) by $f(x)=\prod_{i=1}^{N} (x-\alpha_{i})$. Now you can think about how the coefficients $b_{i}$ depend on $\{\alpha_{i}\}$. If I am not wrong every coefficient has the property $|b_{i}|>1$, so it seems to be impossibly for $L\geq 2$. Maybe if you choose a larger polynomial degree?

Alex
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  • I have considered the relationship between $b_i$ and $\alpha_i$, but it cannot work, the coefficients are not necessary to have $|b_i|>1$. – Tao Mar 13 '13 at 16:15
  • You are right, I was a bit too fast and wrong :). Nevertheless the leading coefficient $b_1$ should be something like $b_{1}=-\sum_{i=1}^{L}\alpha_{i}$ which is as absolute value certainly larger than 2. Also the absolute value of the constant $b_{L}=(-1)^{L}\prod_{i=1}^{L}\alpha_{i}$ is by assumption larger than one, but as I see now you want $L>N$ and then the question is open again :). If you add for a higher degree a further root lets say $\alpha$ than it is still impossible by the formula for $b_{1}$. – Alex Mar 13 '13 at 16:34
  • and $\alpha>0$. – Alex Mar 13 '13 at 16:42
  • The desirable polynomial can have complex roots!!! – Tao Mar 13 '13 at 16:54
  • yes, if you say so :), I can't know it :). Let us first assume that $\alpha\in \mathbb{R}$ and $L=N+1$. Then we have again $b_{1}=-\alpha-\sum_{i=1}^{L-1}\alpha_{i}$ and $b_{L}=(-1)^{L}\alpha\prod_{i=1}^{L-1}}\alpha_{i}$. By assumption we can conclude that $|\alpha|<2$ but also $\alpha<-N+1$, a contradiction for $N>2$. – Alex Mar 13 '13 at 19:18
  • If we have a complex root, we have also as root its complex conjugate since you postulated that f is a real polynomial. So we have at least a polynomial of degree $N+2$ which factorizes into the real roots $\alpha_{i}$ and in a quadratic polynomial $x^2-2xa+a^2+b^2$, where the complex roots are $a+ib$ and $a-ib$. Maybe this is helpful... – Alex Mar 13 '13 at 19:30