0
$\begingroup$

If $a_1,a_2,\cdots,a_n\ (n\geq2)$ are real and $(n-1)a_1^2-2na_2<0$, then prove that at least two roots of the equation, $$x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots+a_n=0$$ are imaginary.

Let $\{\alpha_i\}_{i=1}^n$ be the $n$ roots of the given equation. Then,$$\sum_i\alpha_i=-a_1;$$ also $$\sum_{i \ne j} \alpha_i\alpha_j = a_2.$$ It follows that \begin{align}(n-1)a_1^2-2na_2&= (n-1)\bigg(\sum_i \alpha_i\bigg)^2-2n\sum_{i \ne j} \alpha_i\alpha_j \\ &=n\bigg[{\bigg(\sum_i \alpha_i\bigg)^2-2\sum_{i \ne j} \alpha_i \alpha_j}\bigg]-\bigg(\sum_i \alpha_i\bigg)^2&\\&=n\sum_i \alpha_i^2-\bigg(\sum_i \alpha_i\bigg)^2 < 0.\end{align}

How to carry it further from here?

$\endgroup$
  • $\begingroup$ When you say "imaginary", do you actually mean "complex and not real"? $\endgroup$ – Arthur Jul 18 '18 at 6:34
  • $\begingroup$ @Arthur yes complex $\endgroup$ – mnulb Jul 18 '18 at 12:42
3
$\begingroup$

If $\alpha_i$ 's are all real then Cauchy - Schwarz inequality gives $(\sum \alpha_i)^{2} \leq n \sum (\alpha_i )^{2}$ which contradicts the inequality you have derived. Hence at least one root must be non-real. Since complex roots appear in conjugate pairs there must be two non-real roots. [Cauchy - Schwarz inequality gives $(\sum \alpha_i \beta_i )^{2} \leq \sum \alpha_i ^{2} \sum \beta_i ^{2}$. Put $\beta_i =1$ for all $i$.]

$\endgroup$
2
$\begingroup$

CS is a good way. Another way is to note that between every two real roots of a polynomial, its derivative needs to have a real root. Hence if $P(x) = x^n+a_1x^{n-1}+a_2x^{n-2}+\cdots+a_n$ has all real roots, then the quadratic $P^{(n-2)}(x)$ must have real roots. As $$P^{(n-2)}(x) = (n-2)!\binom{n}{n-2}x^2 + (n-2)!\binom{n-1}{n-2} a_1x+(n-2)!\binom{n-2}{n-2}a_2$$ the quadratic discriminant condition $\implies (n-1)a_1^2 - 2na_2\geqslant 0$, a contradiction.

$\endgroup$
0
$\begingroup$

The claim is false. If the coefficients $f(x):=x^n+a_1x^{n-1}+\ldots +a_n$ match the inequality, then so do the coefficients of $g(x):=f(x+\epsilon)$ provided $|\epsilon|$ is small. But it cannot be the case that $g$ has imaginary roots for all small $\epsilon$ because that would mean $f$ has infinitely many roots.

$\endgroup$
0
$\begingroup$

$$n\sum \alpha_i^2=\begin{bmatrix}\alpha_1&\alpha_2&\cdots&\alpha_n\end{bmatrix}\begin{bmatrix}n&0&\cdots&0\\0&n&\cdots&0\\.\\.\\.\\0&0&\cdots&n\end{bmatrix}\begin{bmatrix}\alpha_1\\\alpha_2\\.\\.\\.\\\alpha_n\end{bmatrix}$$and $$(\alpha_1+\alpha_2+\cdots+\alpha_n)^2=\begin{bmatrix}\alpha_1&\alpha_2&\cdots&\alpha_n\end{bmatrix}\begin{bmatrix}1&1&\cdots&1\\1&1&\cdots&1\\.\\.\\.\\1&1&\cdots&1\end{bmatrix}\begin{bmatrix}\alpha_1\\\alpha_2\\.\\.\\.\\\alpha_n\end{bmatrix}$$if all the $a_i$s are real, the inequality imposes that $$\begin{bmatrix}n&0&\cdots&0\\0&n&\cdots&0\\.\\.\\.\\0&0&\cdots&n\end{bmatrix}-\begin{bmatrix}1&1&\cdots&1\\1&1&\cdots&1\\.\\.\\.\\1&1&\cdots&1\end{bmatrix}=\begin{bmatrix}n-1&-1&\cdots&-1\\-1&n-1&\cdots&-1\\.\\.\\.\\-1&-1&\cdots&n-1\end{bmatrix}$$must be negative definite while this is a contradiction since the set of eigenvalues is $$\{0,n\}$$ therefore at least two of $\alpha_i$'s are conjugate unreals since all the coefficients are real.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.