2
$\begingroup$

Let $Q(z)=(z-\alpha_1)\cdots(z-\alpha_n)$ be a polynomial of degree $>1$ with distinct roots outside the real line.

We have

$$\sum_{j=1}^n \frac{1}{Q'(\alpha_j)}=0.$$

I know an interesting but indirect proof using the continuity of the Fourier transform, but I want to know whether there is a proof relying on more rudimentary techniques.

  • 2
    you can apply residue theorem to $1/Q(z)$ (though you probably want something even more elementary) – 2017-01-21
  • 0
    @user8268 That actually the same idea I used. Thanks. – 2017-01-21

2 Answers 2

2

Let $n \geq 1$ and $\alpha_1, \cdots, \alpha_n \in \Bbb{C}$ be distinct. If we put $Q(z) = (z-\alpha_1)\cdots(z-\alpha_n)$, then from the partial fraction decomposition, it follows that

$$ \frac{1}{Q(z)} = \sum_{j=1}^{n} \frac{1}{Q'(\alpha_j)(z - \alpha_j)}. $$

From this, we have

$$ \sum_{j=1}^{n} \frac{1}{Q'(\alpha_j)} = \lim_{|z|\to\infty} \sum_{j=1}^{n} \frac{z}{Q'(\alpha_j)(z - \alpha_j)} = \lim_{|z|\to\infty} \frac{z}{Q(z)} = \begin{cases} 1, & \text{if } n = 1 \\ 0, & \text{if } n > 1 \end{cases}. $$

  • 0
    This is indeed cool!! I didn't thought of it in this way. Thanks. – 2017-01-21
  • 1
    @HyJu Thank you! I am also curious about the proof using Fourier transform. Is it related to the Poisson kernel on the upper-half plane? – 2017-01-21
  • 1
    Well, it actually relies on the continuity statement with simple residue calculus. Because we have $$\int_{-\infty}^{\infty}\frac{e^{-2\pi i x \xi}}{Q(x)}dx = \left\{ \begin{align} 2\pi i \sum_{j=1}^{r} \frac{e^{-2\pi i \alpha_j \xi}}{Q'(\alpha_j)},&&\xi<0 \\ -2\pi i \sum_{j=r+1}^{n} \frac{e^{-2\pi i \alpha_j \xi}}{Q'(\alpha_j)},&& \xi>0 \end{align} \right.. $$ By continuity of the Fourier transform, we find the identity as $\xi \to 0$. Here $\alpha_j$ is above the real line if $1 \leq j \leq r$ and below if $r+1 \leq j \leq n$. – 2017-01-21
  • 0
    @HyJu, Aha, that makes sense. – 2017-01-21
  • 0
    But I didn't wanted to use a continuity argument again to check the identity still holds even if a root lies on the line. Thanks! – 2017-01-21
2

$f(z)=\frac{1}{Q(z)}$ is a meromorphic function with simple poles at $\alpha_1,\ldots,\alpha_n$.
Since $f(z)=O\left(\frac{1}{\|z\|}\right)$ as $\|z\|\to +\infty$, by integrating $f(z)$ along a big circle
centered at the origin we have:

$$ 2\pi i\sum_{k=1}^{n}\text{Res}\left(f(z),z=\alpha_k\right) = \lim_{R\to +\infty}\oint_{\|z\|=R}f(z)\,dz = 0.$$ On the other hand, $$ \text{Res}\left(f(z),z=\alpha_k\right)=\lim_{z\to \alpha_k}\frac{z-\alpha_k}{Q(z)} \stackrel{dH}{=}\frac{1}{Q'(\alpha_k)} $$ by de l'Hospital rule, so $\sum_{k=1}^{n}\frac{1}{Q'(\alpha_k)}=0$ as wanted.