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Let $P(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0$ be a polynomial of degree $n \ge 3$, Knowing that $a_{n-1}=- {n \choose 1}$ and $a_{n-2}={n \choose 2}$, and that all roots are real, find the remaining coefficients.

$n$ is obviously even. Now the product of its roots is $a_0$ and the sum is $n$.I cannot do anything else. Please help me.

Please don't use Calculus.

Sufaid Saleel
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  • the product of its roots is 1 The question doesn't mention that. and the sum is −n No, the sum is $,+n,$. But it may help that $,P^{(n-2)}(x)=\frac{n!}{2}(x-1)^2,$ has a double root at $,1,$. – dxiv Oct 02 '18 at 02:56
  • My guess is that $P(x)=\sum_{r=0}^{n}a_{r}x^{r}$ where $a_{r}=(-1)^{r}{n \choose r}$ when $n$ is even. If $n$is odd, then the formula will be the same except for $a_{r}=(-1)^{r+1}{n \choose r}$. But that's just a guess. – clathratus Oct 02 '18 at 02:59
  • @above I also guessed that but we have to proof it! – Sufaid Saleel Oct 02 '18 at 03:20
  • What is the context of this problem? I'm surprised that the rest of the coefficients may be determined uniquely – TomGrubb Oct 02 '18 at 03:33
  • @SufaidSaleel n is obviously even Why? the product of its roots is a_0 No, it is $,(-1)^n a_0,$. – dxiv Oct 02 '18 at 03:39
  • Offfff!! This problem is disturbing me. Please solve it! – Sufaid Saleel Oct 02 '18 at 03:41

2 Answers2

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Alt. hint:   for a purely algebraic solution, consider the equality case of the RMS-AM inequality, which must hold since all roots $\,x_k\,$ are real:

$$ 1 = \frac{\sum_k x_k}{n} \le \sqrt{\frac{\sum_k x_k^2}{n}} = \sqrt{\frac{\left(\sum_k x_k\right)^2 - 2 \sum_{i \lt j} x_i x_j}{n}} = \sqrt{\frac{n^2 - n(n-1)}{n}} = 1 $$

dxiv
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Hint:   if $\,P(x)\,$ has all roots real, then the $\,k^{th}\,$ derivatives $\,P^{(k)}(x)\,$ have all roots real for $1 \le k \lt n\,$.

In the case here $\,P^{(n-2)}(x) = \frac{n!}{2!}(x-1)^2\,$, and therefore $\,P^{(n-3)}(x)=\frac{n!}{3!}(x-1)^3 + C\,$ for some constant $\,C \in \Bbb R\,$. The roots of $\,P^{(n-3)}(x)=0\,$ are $\,x=1 + \frac{3!}{n!} \cdot \sqrt[3]{-C}\cdot \omega^k \mid k = 0,1,2\,$ where $\,\omega\,$ is a primitive cube root of unity. For all the roots to be real, it is necessary and sufficient that $\,C=0\,$, so $\,P^{(n-3)}(x)=\frac{n!}{3!}(x-1)^3\,$.

Repeat the same argument to show that in the end $\,P(x)=\frac{n!}{n!}(x-1)^n=(x-1)^n\,$.

dxiv
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