Showing that $\sum_{i = 1}^m \frac{1}{\prod_{j = 1, j \neq i}^m (a_j - a_i)}$ is zero

Question

Question: How can I prove, for $m \geq 2$ and reals $a_1 < a_2 < \dots < a_m$ that $$\sum_{i = 1}^m \frac{1}{\prod_{j = 1, j \neq i}^m (a_j - a_i)} = 0?$$

Context: In Gamelin's text on Complex Analysis, exercise VII.6.4 asks to prove that $$\text{PV}\int_{-\infty}^\infty \frac{1}{\prod_{k = 1}^m (x - a_k)} dx = 0,$$ which can be done using a contour integral around a half disk $\partial D$ (of radius $R$) in the upper half-plane, with small semicircular indents (of radius $\varepsilon$) above the singularities $a_1, a_2, \dots, a_m$ on the real axis. The method is straightforward, but when applying the fractional residue theorem to the semicircle indents $\gamma_b$, the sum of their contributions becomes $$\sum_{b = 1}^m \lim_{\varepsilon \rightarrow 0} \int_{\gamma_b} \frac{1}{\prod_{k = 1}^m (x - a_k)} dx = \sum_{b = 1}^m \frac{-\pi i}{\prod_{j = 1, j \neq b}^m (a_j - a_b)},$$ and it is easy to show (using the ML-estimate) that the contribution of the integral over the semicircumference is negligible as $R \to \infty$. Thus, by Cauchy's Theorem, we have that $$\lim_{\varepsilon \to 0, R \to \infty}\int_{\partial D} \frac{1}{\prod_{k = 1}^m (x - a_k)} dx = \sum_{b = 1}^m\frac{-\pi i}{\prod_{j = 1, j \neq b}^m (a_j - a_b)} + \text{PV} \int_{-\infty}^\infty \frac{1}{\prod_{k = 1}^m (x - a_k)} dx = 0,$$ which gives the result the question wants, if the identity I am trying to prove is true.

I convinced myself the identity holds by trying for small values of $m$, but I have yet to come up with any rigorous proof. I have attempted an induction argument, but I am having trouble constructing the inductive step. Any hints/advice would be greatly appreciated.

Answer 1

Let $\displaystyle\;P(\lambda) = (\lambda - a_1)(\lambda - a_2) \cdots (\lambda - a_m) = \prod_{i=1}^m (\lambda - a_i)$.
By product rule, we have $$P'(\lambda) = {\small\begin{align} & (\lambda - a_1)'(\lambda - a_2)\cdots(\lambda - a_m)\\ + & (\lambda - a_1)(\lambda - a_2)' \cdots (\lambda - a_m ) \\ + & \cdots\\ + & (\lambda - a_1)(\lambda - a_2) \cdots (\lambda - a_m)' \end{align}} = \sum_{i=1}^m (\lambda - a_i)'\prod_{j=1,\ne i}^m (\lambda - a_j) = \sum_{i=1}^m \prod_{j=1,\ne i}^m (\lambda - a_j) $$ RHS is a sum of $m$ terms and for each $i$, the factor $\lambda - a_i$ appear in $m-1$ terms (i.e. all terms excluding the $i^{th}$ term ). At $\lambda = a_i$, they won't contribute. As a result $$P'(a_i) = \prod_{j=1,\ne i}^m (a_i - a_j)$$ This allow us to rewrite the sum at hand as $$\mathcal{S}\stackrel{def}{=}\sum_{i=1}^m \frac{1}{\prod_{j=1,\ne i}^m(a_j - a_i)} = (-1)^{m-1} \sum_{i=1}^m \frac{1}{\prod_{j=1,\ne i}^m(a_i - a_j)} = (-1)^{m-1}\sum_{i=1}^m\frac{1}{P'(a_i)}$$

Since the roots of $P(\lambda)$ are distinct, the partial fraction decomposition for $\displaystyle\;\frac{1}{P(\lambda)}$ equals to

$$\frac{1}{P(\lambda)} = \sum_{i=1}^m \frac{1}{P'(a_i)(\lambda - a_i)}\tag{*1}$$

As a consequence, $$\mathcal{S} = (-1)^{m-1}\lim_{\lambda\to\infty} \sum_{i=1}^m\frac{\lambda}{P'(a_i)(\lambda-a_i)} = (-1)^{m-1}\lim_{\lambda\to\infty} \frac{\lambda}{P(\lambda)} = 0 $$

In case one need to justify $(*1)$, multiply RHS$(*1)$ by $P(\lambda)$, one obtain

$$Q(\lambda) \stackrel{def}{=}{\rm RHS}(*1) P(\lambda) = \sum_{i=1}^m \frac{1}{P'(a_i)(\lambda - a_i)}\prod_{j=1}^m(\lambda - a_j) =\sum_{i=1}^m \frac{1}{P'(a_i)} \prod_{j=1,\ne i}^m (\lambda - a_j) $$ This is a sum of $m$ polynomials in $\lambda$ with degree $m-1$. This means $Q(\lambda)$ is also a polynomial in $\lambda$ with $\deg Q \le m-1$. Once again, for each $i$, the factor $\lambda - a_i$ appear in all but the $i^{th}$ polynomials. At $\lambda = a_i$, only the $i^{th}$ polynomial contributes and

$$Q(a_i) = \frac{1}{P'(a_i)} \prod_{j=1,\ne i}^m (a_i - a_j) = \frac{P'(a_i)}{P'(a_i)} = 1$$

Since $Q(\lambda) = 1$ at $m > \deg Q$ values of $\lambda$, $Q(\lambda)$ equals to $1$ identically. This establishes $(*1)$.

Answer 2

Given $n$ distinct abscissas $a_i$ and arbitrary ordinates $y_i$, the Lagrange interpolating polynomial is

$$P(x)=\sum_{i=1}^n y_i\dfrac{\prod\limits_{j\ne i} (x-a_j)}{\prod\limits_{j\ne i} (a_i-a_j)}$$

We can choose the $y_i$, and build the corresponding interpolating polynomial $P$. For some $x_0$ such that $x_0\ne a_i$ for all $i$, let

$$\dfrac 1M=\prod\limits_{i=1}^n (x_0-a_i)$$

and

$$y_i=M(x_0-a_i)=\dfrac{1}{\prod\limits_{j\ne i} (x_0-a_j)}$$

We have, on one hand,

$$P(x_0)=\sum_{i=1}^n y_i\dfrac{\prod\limits_{j\ne i} (x_0-a_j)}{\prod\limits_{j\ne i} (a_i-a_j)}=\sum_{i=1}^n \dfrac{1}{\prod\limits_{j\ne i} (a_i-a_j)}$$

On the other hand, since $P$ is an interpolating polynomial, for all $i$,

$$P(a_i)=y_i=M(x_0-a_i)$$

But since the interpolating polynomial is unique among polynomials of degree at most $n-1$ (and $n-1\ge 1$ here, as we assume $n\ge2$), that means that for all $x$,

$$P(x)=M(x_0-x)$$

Then

$$P(x_0)=\sum_{i=1}^n \dfrac{1}{\prod\limits_{j\ne i} (a_i-a_j)}=0$$

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Note: the interpolation polynomial is unique because if $P$ and $Q$ are two interpolating polynomials of degree at most $n-1$ for the same abscissas and ordinates, the polynomial $P-Q$, which is also of degree at most $n-1$, has $n$ roots (the $a_i$), hence it's the null polynomial.

Answer 3

Recall that the Lagrange interpolation formula says that if we have $P(x_i) = y_i$ with $ \deg P < m$, then $P$ is the unique polynomial $ \sum y_i \times \prod_{i\neq j} \frac{ x-x_j } { x_i - x_j } $.

Let's consider the polynomial $P$ where $P(x_i) = 1 \, \forall i$, with $\deg P < m$.
Clearly, $ P = 1$.

Consider the coefficient of $ x^{m-1}$ in $P$, with $ m - 1 \geq 1$.
On the LHS, this is $ \sum_i \prod_{i\neq j} \frac{1}{a_i - a_j}$, which is $(-1)^{m-1}$ of what we want.
On the RHS, this is 0.

Note: This also shows where the condition $m \geq 2$ is used, and why for $m = 1$, we get the value of 1 instead.