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Let $n$ real roots of the equation $\displaystyle y^n-2ny^{n-1}+2n(n-1)y^{n-2}+ay^{n-3}+by^{n-4}+\cdots +c=0$ has roots $y_1,y_2,y_3,\cdots ,y_n$. Then $\displaystyle \sum^n_{k=1}(-1)^{k-1}y_k=$

What I try : $\displaystyle y_1+y_2+y_3+\cdots y_n=2n$

And $\displaystyle y_1y_2+y_2y_3+\cdots +y_{n-1}y_{n}=2n(n-1)$

Also $\displaystyle y_1\cdot y_2 \cdots y_n=(-1)^{n}c$

But I did not understand How can I find value of

$\displaystyle y_1-y_2+y_3-y_4+\cdots +(-1)^{n-1}y_n$

Please have a look , Thanks

Leox
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jacky
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  • Note: Not $\displaystyle y_1y_2+y_2y_3+\cdots +y_{n-1}y_{n}=2n(n-1)$ but $\displaystyle y_1y_2+y_1y_3+\cdots +y_{n-1}y_{n}=2n(n-1)$. It's $n(n-1)/2$ products on the left hand side, not just $n-1$ products. – darij grinberg Mar 25 '24 at 20:41

1 Answers1

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Unless all the roots are the same, it's impossible to calculate $\sum_{k=1}^n (-1)^{k-1}y_k$ because the roots are indistinguishable, but the value of the expression can be changed by permutation of roots. So the question boils down to whether $y_1 = y_2 = .. =y_n$. Consider the following expression

$$\sum_{1\leq i< j \leq n}(y_i-y_j)^2=(n-1)\sum_{i=1}^n{y^2_i}-2\sum_{1\leq i< j\leq n}y_iy_j$$

Since $\sum_{0\leq i< j \leq n}(y_i-y_j)^2 \geq 0$, therefore $\sum_{i=1}^n{y^2_i}\geq \dfrac{2}{n-1}\sum_{1\leq i< j \leq n}y_iy_j $.

We have $$\begin{align} (\sum_{i=1}^n{y_i})^2&=\sum_{i=1}^n{y^2_i}\;+2\sum_{1\leq i< j\leq n}y_iy_j\\ &\geq \frac{2n}{n-1} \sum_{1\leq i< j\leq n}y_iy_j \end{align} $$

On the other hand, $(\sum_{i=1}^n{y_i})^2 = 4n^2 = 2n(2n(n-1))=\dfrac{2n}{n-1} \sum_{1\leq i< j\leq n}y_iy_j$. Therefore $\sum_{1\leq i< j \leq n}(y_i-y_j)^2 = 0$ hence $y_i =y_j$ for $1\leq i< j \leq n$. In other words $y_1=y_2=..=y_n$

It follows straightforwardly that $y_i=2$ and $\sum_{k=1}^n(-1)^{k-1}y_k = 2(1-(-1)^n)$

ioveri
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  • From $$\left(\sum_{i=1}^n{y_i}\right)^2 =\dfrac{2n}{n-1} \sum_{1\leq i< j\leq n}y_iy_j$$ it does not follow that $$\sum_{1\leq i< j \leq n}(y_i-y_j)^2 = 0$$ – Leox Mar 26 '24 at 12:10
  • @Leox please read about the inequality that I mentioned above. If $\sum_{1\leq i < j \leq n}(y_i - y_j)^2 > 0$ then we won't have the equality, hence contradiction. – ioveri Mar 26 '24 at 12:17
  • Let $n=3$ and consider the equation $y^3-6y^2+12y-16=0$. Then, for its roots, the following holds: $y_1+y_2+y_3=6=2n$ and $y_1y_2+y_1y_3+y_2y_3=12=2n(n-1)$ but the $y_i$ are not equal to each other and are not equal to $2$. – Leox Mar 26 '24 at 16:39
  • @Leox your polynomial has complex roots :) Remember we're talking about a polynomial with $n$ *real roots* – ioveri Mar 26 '24 at 17:05