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Find all polynomials of the form $$ P(x)=x^n-nx^{n-1}+\left(\frac{n^2-n}{2}\right)x^{n-2}+a_{n-3}x^{n-3} + \ldots+a_1x+a_0$$

where $n>2$ and all roots of $P(x)$ are real.

Please check my work :

Let $r_1, r_2, \ldots, r_n $ be roots of $P(x)$.

By Vieta formula, $\displaystyle\sum^{n}_{i=1}r_i=n$, $\displaystyle\sum_{1\leq i<j\leq n}r_ir_j=\frac{n(n-1)}{2}$

so $\displaystyle\sum^{n}_{i=1}r^2_i=\left(\displaystyle\sum^{n}_{i=1}r_i\right)^2-2\left(\displaystyle\sum_{1\leq i<j\leq n}r_ir_j\right)=n^2-n(n-1)=n$

By Power Mean and Triangle inequalities,

We have $1 = \sqrt{\frac{\displaystyle\sum^{n}_{i=1}r^2_i}{n}} \geq \frac{\displaystyle\sum^{n}_{i=1}|r_i|}{n}\geq \left|\displaystyle\sum^{n}_{i=1}\frac{r_i}{n}\right|=1$

then $\displaystyle\sum^{n}_{i=1}|r_i|=n$

Since $\displaystyle\sum^{n}_{i=1}r_i=n$, $\displaystyle\sum^{n}_{i=1}|r_i|=n$ and $\frac{\displaystyle\sum^{n}_{i=1}|r_i|}{n}\geq \left|\displaystyle\sum^{n}_{i=1}\frac{r_i}{n}\right|$,

the triangle inequality will have equality hold when $r_1= r_2= \ldots= r_n $

then $r_1= r_2= \ldots= r_n =1$

Therefore, $P(x)=(x-1)^n $

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  • 1
    $\begingroup$ Assuming $a_k$ are binomial coefficients, why not just look up proofs of the binomial theorem? $\endgroup$ – Simply Beautiful Art Aug 27 '17 at 14:25
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    $\begingroup$ It's not clear what "$\dotsb$" means in this context. $\endgroup$ – Chappers Aug 27 '17 at 14:30
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    $\begingroup$ The triangle equality holds for the much weaker condition $r_i \geq 0$ so that last step won't work. But consider $\sum(r_i - 1)^2$ instead. $\endgroup$ – WimC Aug 27 '17 at 17:16
  • $\begingroup$ @Winther. Sorry for the misunderstanding. Only $a_n=1$ and $a_{n−1}=−n$ and $a_{n−2}= {n\choose 2}$ fixed and the rest free. $\endgroup$ – carat Aug 27 '17 at 17:16
  • $\begingroup$ your last assertion is very false. It only means $a_k>0$ and their sum is $n$. $\endgroup$ – dezdichado Aug 27 '17 at 17:19
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I think your solution is true.

I like the following reasoning.

Since $$\sum_{1\leq i<j\leq n}(r_i-r_j)^2=(n-1)\left(\sum_{i=1}^nr_i\right)^2-2n\sum_{1\leq i<j\leq n}r_ir_j=(n-1)n^2-2n\cdot\frac{n^2-n}{2}=0,$$ we obtain $r_i=1$ and $P(x)=(x-1)^n$.

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  • $\begingroup$ Your solution is elegant. I'm grateful for your kind help. $\endgroup$ – carat Aug 28 '17 at 15:25
  • $\begingroup$ @carat Thank you! Good luck! $\endgroup$ – Michael Rozenberg Aug 28 '17 at 15:27
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It only suffices to notice that $$\sqrt{\dfrac{{\sum_{i=1}^n}r_i^2}{n}} = \dfrac{\sum_{i=1}^nr_i}{n}.$$

This ensures $r_1=r_2=...=r_n$ because that is the only condition where the power-mean inequality is an equality. You can finish the problem here.

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  • $\begingroup$ Thank you, dezdichado. $\endgroup$ – carat Aug 28 '17 at 15:23

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