8
$\begingroup$

With distinct numbers $a_1, a_2, \ldots, a_n$, let's denote the products of the differences of each of these numbers with the each of the rest of them by the following principle:

\begin{align} (a_1-a_2)(a_1-a_3)(a_1-a_4)\dots(a_1-a_n) &= \prod_{1\lt j\leqslant n}(a_1-a_j) = \mu_1, \\ (a_2-a_1)(a_2-a_3)(a_2-a_4)\dots(a_2-a_n) &= \prod_{\substack{1\leqslant j\leqslant n \\ j\neq 2}}(a_2-a_j) = \mu_2, \\ (a_3-a_1)(a_3-a_2)(a_3-a_4)\dots(a_3-a_n) &= \prod_{\substack{1\leqslant j\leqslant n \\ j\neq 3}}(a_3-a_j) = \mu_3, \\ &\ \ \vdots \\ (a_n-a_1)(a_n-a_2)(a_n-a_3)\dots(a_n-a_{n-1}) &= \prod_{1\leqslant j\lt n}(a_2-a_j) = \mu_n \end{align}

There is a theorem that the sum of reciprocals of that resulting sequence will always be equal to $0$:

\begin{align} \sum_{i=1}^n\dfrac{1}{\mu_i}=\dfrac{1}{\mu_1}+\dfrac{1}{\mu_2}+\ldots+\dfrac{1}{\mu_n}=0 \end{align}

There also is a more general claim that for any integer $\xi$, $0\leqslant\xi\lt n-1$ (in other words, $\xi$ is a nonnegative integer which is less than the number of factors in each $\mu$, which happens to be exactly $n-1$), the result is just the same:

\begin{align} \sum_{i=1}^n\dfrac{a_i^{\xi}}{\mu_i}=\dfrac{a_1^{\xi}}{\mu_1}+\dfrac{a_2^{\xi}}{\mu_2}+\ldots+\dfrac{a_n^{\xi}}{\mu_n}=0 \end{align}

Leonhard Euler provides a very similar theorem in his Institutionum Calculi Integralis, Vol. II, §1169 (available in English here, see page 26 of this document), and the way he proves it seems quite confusing to me. I have a couple of questions related to this theorem:

  1. Does this theorem have a special name, or is it just a generic fact?

  2. In Corollary 1 (§1170), Euler claims that in case if $\xi\geqslant n-1$ the sum no longer vanishes (i.e. is no longer equal to zero because in this case $\xi$ is out of bounds mentioned in the theorem), but is equal to $\dfrac{1}{N}$ instead. But, due to $N$ not being defined directly in the mentioned corollary, how to understand what $N$ is equal to in $\dfrac{1}{N}$? It is known to me that $N=1$ in case $\xi = n-1$, and that $\dfrac{1}{N}=\sum_{i=1}^na_i$ in case $\xi=n$. What happens to $N$ in case $\xi\gt n$?

    After more thorough reading, I believe that $N$ might be the real polynomial function in the denominator of a proper rational function $\dfrac{1}{N}$, according to Euler's Introductione ad Analysin Infinitorum, Ch. II, §40 (available in English here, see page 41 of this document), which Euler himself seems to be referencing.

  3. Can you explain Euler's proof of the theorem in question rigorously, so it's clearly seen what goes from where?


Edit: in my search for something to improve the question, I've found some useful references that can help.

  • Case $\xi = 0$ had been reviewed in this question. The currently accepted answer makes use of Lagrange polynomial, while Euler seems to have a much more basic proof through partial fraction decomposition, which makes the essence of my question different.

  • Case $\xi = n-1$ had been mentioned in this question. The currently accepted answer is based on a polynomial analysis, thus it's also different from what I'm looking for.

  • Finally, $\xi = n$ is mentioned in my other question. I've got an elegant answer based on Vandermonde determinant and polynomial analysis, and it's not too much of what I am looking for.

$\endgroup$

0

You must log in to answer this question.