Well, I recently proved a formula (at least, I think) to the sum of the inverse of the roots $x_{1}, x_{2}, x_{3},\ldots, x_{n} \in \mathbb{C}$, and $\neq 0$. It starts:

Let a polynomial $P(x) = a_nx^n+a_{n-1}x^{n-1}+\cdot\cdot\cdot+a_1x + a_0$ of roots $x_{1}, x_{2}, x_{3},\ldots,x_{n} \in \mathbb{C^*}$ and $a_n \neq 0$. So, $$\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3}+\cdots+ \frac{1}{x_n} = -\frac{a_1}{a_0}, \qquad a_0 \neq 0.$$

Now, I am trying to prove another formula, the sum of the square of the roots, but I think it's getting pretty difficult to me.

Let $x_{1}^{2} + x_{2}^{2} + x_{3}^{2}+\cdots+x_{n}^{2}=u.$ So, if $x_1\cdot x_2\cdot x_3 \cdots x_n = (-1)^n\frac{a_0}{a_n}$, then $$x_1 =(-1)^n\frac{a_0}{(x_2\cdot x_3 \cdots x_n)a_n},$$ and $$x_1^2 =(-1)^nx_1\frac{a_0}{(x_2\cdot x_3 \cdots x_n)a_n}.$$


$$\begin{align*} x_{1}^{2} &+ x_{2}^{2} +\cdots+x_{n}^{2}=u\\ &=(-1)^nx_1\frac{a_0}{(x_2\cdot x_3 \cdots x_n)a_n} + (-1)^nx_2\frac{a_0}{(x_1\cdot x_3 \cdots x_n)a_n} \\ &\qquad +\cdots+(-1)^nx_n\frac{a_0}{(x_1\cdot x_2 \cdots x_{n-1})a_n}. \end{align*}$$

It can be written as $$(-1)^n\cdot \frac{a_0}{a_n}\left(\frac{x_1}{x_2\cdot x_3 \cdots x_n} + \frac{x_2}{x_1\cdot x_3 \cdots x_n} +\cdots+ \frac{x_n}{x_1\cdot x_2 \cdots x_{n-1}}\right) = u,$$ And I'm stuck here. That's my question. How can I write $u$ in function of the coefficients (or its impossibility)? Any help will be very appreciated. I'm young, and I do not have experience with proving things. That's all.

Thank you.

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    $\begingroup$ $\LaTeX$ tip: use \cdots instead of \cdot\cdot\cdot; and use \ldots when they should be at baseline level. $\endgroup$ – Arturo Magidin Dec 12 '11 at 20:59
  • $\begingroup$ @ArturoMagidin thanks, I was looking for that. $\endgroup$ – Ian Mateus Dec 12 '11 at 21:01

If you have a polynomial $$P(x) = a_nx^n + a_{n-1}x^{n-1}+\cdots + a_1x + a_0$$ and $a_n\neq 0$, $a_0\neq 0$, then consider the "reversal polynomial" $Q(x)$, $$Q(x) = a_0x^n + a_{1}x^{n-1} + \cdots + a_{n-1}x + a_n.$$

Then $r$ is a root of $P(x)$ if and only if $\frac{1}{r}$ is a root of $Q(x)$. Indeed, if $P(r)=0$, then $$\begin{align*} r^nQ\left(\frac{1}{r}\right) &= r^n\left(a_0\left(\frac{1}{r}\right)^n + \cdots + a_{n-1}\left(\frac{1}{r}\right) + a_n\right)\\ &= a_0 + a_1r + \cdots + a_{n-1}r^{n-1}+ a_nr^n\\ &= P(r) = 0. \end{align*}$$ Since $r\neq 0$, it follows that $Q(\frac{1}{r})=0$. Since $P$ is the reversal of $Q$, the symmetric argument establishes the converse implication.

So what you are trying to do is essentially equivalent to finding the sums of powers of the roots of a polynomial.

These can be obtained from the coefficients by using Newton's identities, which express them in terms of the elementary symmetric polynomials, which are in turn equal to $\frac{a_i}{a_n}$ by the factor theorem.

  • $\begingroup$ So, it means that a formula is not impossible. Is there a way to continue what I have started? I have never seen those Newton's identities, so I will read these articles. But I realized that $\frac{1}{x_n} = x_n^{-1}$. $\endgroup$ – Ian Mateus Dec 12 '11 at 21:00
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    $\begingroup$ (I'm going to use $r_i$ instead of $x_i$, because it's too easy to confuse $x_n$ with $x^n$). The key observation is that $$r_1^2+\cdots+r_n^2 = (r_1+\cdots+r_n)^2 - 2(r_1r_2+r_1r_3+\cdots+r_1r_n+r_2r_3+\cdots+r_{n-1}r_n).$$ Now note that $r_1+\cdots+r_n$ is $-\frac{a_{n-1}}{a_n}$, and $r_1r_2+\cdots+r_1r_n+r_2r_3+\cdots+r_{n-1}r_n$ is $\frac{a_{n-2}}{a_n}$, and that gives you the desired formula. Now apply this formula to $Q$ in order to get the value of $r_1^{-2}+\cdots+r_n^{-2}$. $\endgroup$ – Arturo Magidin Dec 12 '11 at 21:09
  • $\begingroup$ Now, I really understand and I'm going to find the formula. And I think it can be extended for $x_{n}^3, x_{n}^4, x_{n}^n$, sure?. Thanks. $\endgroup$ – Ian Mateus Dec 12 '11 at 21:17
  • $\begingroup$ @Ian: Yes, it can be extended to the sum of the $k$th powers of the roots, again, this was done by Newton and the identities that express the sum of the $k$th powers of the roots in terms of the coefficients (equivalently, of the elementary symmetric polynomials in the roots) are known as Newton's Identities. $\endgroup$ – Arturo Magidin Dec 12 '11 at 21:19
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    $\begingroup$ Can this be "extended" to the square root of the roots? i.e. summing $\sqrt{x_1}+\sqrt{x_2}+\ldots$? $\endgroup$ – Ziofil Aug 7 '14 at 20:21

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