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I'm having some issues with proving that the following sum is $0$ for any value of $n \geq 2$:

$$ \sum_{j=1}^{n} \frac{1}{\prod_{i=1,i\neq j}^{n}(a_{j}-a_{i})} $$

where the $a_i$ are non-zero and positive. Just wondering if anyone had any pointers, the first few cases are easy to verify, i.e $n=2,3$.

So for $n=2$ the sum reads:

$$ \frac{1}{a_2 -a_1}+\frac{1}{a_1-a_2} = 0 $$

and for $n=3$ it would be $$ \frac{1}{(a_{3}-a_{1})(a_{2}-a_{1})}+\frac{1}{(a_1-a_2)(a_3-a_2)}+\frac{1}{(a_1-a_3)(a_2-a_3)} = 0, $$ it's just the general case I'm struggling to find an argument for. Thanks!

Joshua
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2 Answers2

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The degree of the polynomial

$$q(x)=\sum_{j=1}^n \prod_{\substack{i=1 \\ i \neq j}}^{n}\frac{x-a_i}{a_{j}-a_{i}}-1$$ is $n-1$ and the $a_j$ for $1 \le j \le n$ are $n$ roots. Therefore $q$ is the always vanishing polynomial and $q$ leading coefficient which is $$\sum_{j=1}^n \frac{1}{\prod_{\substack{i=1 \\ i \neq j}}^{n}(a_{j}-a_{i})}$$ is equal to zero.

Note: all this is closely related to Lagrange polynomials.

mathcounterexamples.net
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  • In the formula for $q$, $X$ should be $x$. – Gary Feb 01 '21 at 16:16
  • Very nice solution, I thought by the appearance of the sum that there would be some sort of link to Lagrange polynomials, but I didn't see it! Thanks for this! – Joshua Feb 02 '21 at 11:13
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Not an answer... but maybe useful

Let $p$ be the polynomial $p(x) = (x-a_1) \dots (x-a_n)$. Then for $1 \le j \le n$

$$p^\prime(a_j)= \prod_{\substack{i=1 \\ i \neq j}}^{n}(a_{j}-a_{i})$$ and the result to be proven is equivalent to

$$\sum_{j=1}^n \frac{1}{p^\prime(a_j)} = 0.$$ However, I don't know how to prove that for the time being...

One can also notice that

$$\lim\limits_{x \to a_j} \frac{x-a_j}{p(x)} = \frac{1}{p^\prime(a_j)}.$$

mathcounterexamples.net
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  • Thank you for this insight. I had tried a similar thing myself as the problem originally comes from me studying the polynomial $\frac{1}{p(x)}$ as you have written it, and I first noted that $1/p = \frac{1}{\prod_{i \neq 1}^{m}(a_{1}-a_{i})}\frac{1}{x-a_{1}} + \frac{1}{\prod_{i \neq 2}^{m}(a_{2}-a_{i})}\frac{1}{x-a_{2}} + ... + \frac{1}{\prod_{i \neq m}^{m}(a_{m}-a_{i})}\frac{1}{x-a_{m}}$ – Joshua Feb 01 '21 at 15:34
  • I wrote another answer which is now complete. – mathcounterexamples.net Feb 01 '21 at 16:13