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Let $x_1,...,x_n$ be $n$ distinct real numbers.

I'd like to prove that $$\sum_{j=1}^{n} { \frac {1} {\prod_{i=1 \land i \neq j}^{n}{(x_j-x_i)}} } = 0.$$

This makes me think of Cauchy determinants but I don't see any direct link. I tried to reduce to the same denominator but it led to nothing good. Any suggestion?

NB : Note that this sum equals $\sum_{k=1}^{n}{\frac{1}{Q'(x_k)}}$ where $Q = \prod_{k=1}^{n}{(X-x_k)}$.

Thank you!

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More generally, the identity holds also when $x_1,...,x_n$ are $n$ distinct complex numbers.

Consider the polynomial of degree at most $n-1$, given by $$P(z):=\sum_{j=1}^{n} { \frac {{\prod_{i=1 \land i \neq j}^{n}{(z-x_i)}}} {\prod_{i=1 \land i \neq j}^{n}{(x_j-x_i)}} }-1.$$ It is easy to verify that $P(x_i)=0$ for $i=1,\dots,n$. Therefore, since $x_1,...,x_n$ are $n$ distinct numbers, the polynomial $P$ is identically zero. It follows that the coefficient of $z^{n-1}$ of $P$, which is just $$\sum_{j=1}^{n}\frac {1} {\prod_{i=1 \land i \neq j}^{n}{(x_j-x_i)}}, $$ has to be zero.

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  • $\begingroup$ Very clever, thank you sir! $\endgroup$ – Sylve Feb 18 '18 at 13:25
  • $\begingroup$ Ah, what an elegant proof! $\endgroup$ – Rellek Feb 18 '18 at 13:30
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Using the hint, we may rewrite $$\sum_{k=1}^n \frac{1}{Q' (x_k)} = \lim_{h \to 0} \sum_{k=1}^n \frac{h}{Q(x_k + h)}$$ Where the above uses that $Q(x_k) = 0$. Then, getting a common denominator on the inside, $$\lim_{h \to 0} \sum_{k=1}^n \frac{h}{Q(x_k + h)} = \lim_{h \to 0} \frac{h}{\prod_{i=1}^n Q (x_i + h)} \sum_{k=1}^n \prod_{i \neq k} Q(x_i+h)$$ Now, note that the outside limit is simply $$\frac{1}{\prod_{i=1}^n Q'(x_i)}$$ And $\prod_{i=1}^n Q'(x_i)$ is precisely the discriminant of $Q$ which is nonzero since $Q$ has distinct roots (up to a factor of $(-1)^{n(n-1)/2}$, but this doesn't change anything). Thus, we see: $$\lim_{h \to 0} \frac{h}{\prod_{i=1}^n Q (x_i + h)} \sum_{k=1}^n \prod_{i \neq k} Q(x_i+h) =\frac{1}{\prod_{i=1}^n Q'(x_i)} \lim_{h \to 0} \sum_{k=1}^n \prod_{i \neq k} Q(x_i+h)$$ $$ = \frac{1}{\prod_{i=1}^n Q'(x_i)} \sum_{k=1}^n \prod_{i \neq k} Q(x_i) = 0$$ Where we've used that each $x_i$ is a root of $Q$ in the above.

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  • $\begingroup$ That actually wasn't the hint but what I wanted to prove in the first place! :-) Thank you very much, that's nice! I didn't know about the determinant of a polynomial, that is the occasion to learn about it. $\endgroup$ – Sylve Feb 18 '18 at 13:25

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