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Prove that $\displaystyle\sum_{i=1}^n\frac{(1-a_i)^n}{a_i\prod_{j\neq i}(a_j-a_i)}=\frac{1}{a_1\cdots a_n}-1$ for distinct $a_1,\cdots,a_n\in\mathbb{R}\backslash\{0\}$.

I noticed that it is equivalent to $\displaystyle\sum_{i=1}^nx_i^n\prod_{j\neq i}\frac{1-x_j}{x_i-x_j}=1-\prod_{k=1}(1-x_k)$ where $x_i=1-a_i$, then the LHS can be regarded as Lagrange interpolation formula, the value at $x=1$ of the polynomial of degree $n-1$ going through $(x_1,x_1^n),\cdots,(x_n,x_n^n)$.

But I don't know if it is useful for the proof.

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    $\begingroup$ You can edit a question instead of reposting it. $\endgroup$
    – Aig
    Commented Mar 20 at 17:33
  • $\begingroup$ Oh I see, thank you! $\endgroup$ Commented Mar 21 at 2:41

2 Answers 2

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You are almost done using your approach. You have an $n-1$ degree polynomial $p(x)$ such that $p(a_i) = a_i^n$. The $a_i$ are the roots of $x^n - p(x)$ and since this is monic, it follows that $x^n - p(x) = (x-a_1)...(x-a_n)$. Then, plugging in 1 gives $p(1) = 1 - (1-a_1)...(1-a_n)$ which is what you wanted.

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  • $\begingroup$ Got it, thank you very much! $\endgroup$ Commented Mar 21 at 2:38
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Another approach, using partial fractions

Solve by partial fractions:

$$\frac{(x-1)^n}{(x-a_1)(x-a_2)\cdots (x-a_n)}=1+\sum_{i=1}^n \frac{b_i}{x-a_i}\tag1$$

Multiplying by $x-a_i$ and evaluating both sides at $x=a_i,$ you get:

$$b_i=\frac{(a_i-1)^n}{\prod_{j\neq i} (a_i-a_j)}=-\frac{(1-a_i)^n}{\prod_{j\neq i}(a_j-a_i)} $$

Evaluating $(1)$ at $x=0,$ you get:

$$\frac{1}{a_1\dots a_n}=1+\sum_{i=1}^n \frac{(1-a_i)^n}{a_i\prod_{j\neq i}(a_j-a_i)}$$

Which easily gives your result.


More generally, if $p(x)$ is a polynomial of degree less than or equal to $n$ with coefficient $c$ at $x^n,$ we get:

$$\frac{(-1)^np(0)}{a_1\dots a_n}=c-\sum_i\frac{p(a_i)}{a_i\prod_{j\neq i}(a_i-a_j)}$$

If $p(x)=(x-k)^n,$ then you get:

$$\frac{k^n}{a_1\dots a_n}-1=\sum_{i=1}^n\frac{(k-a_i)^n}{a_i\prod_j(a_j-a_i)}$$

If $p(x)=(x-k)^m$ for $m<n$ then $c=0$ and:

$$\frac{k^m}{\prod a_i}=\sum_{i=1}^n \frac{(k-a_i)^m}{a_i\prod(a_j-a_i)}$$

If $1\leq m\leq n$ then you get, by subtracting $k^m/\prod a_i$ from $k\cdot k^{m-1}/\prod a_i$

$$\sum_{i=1}^n\frac{(k-a_i)^{m-1}}{\prod_j(a_j-a_i)}=\delta_{mn}=\begin{cases}1&m=n\\0&m\neq n\end{cases}.$$

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  • $\begingroup$ Got it, thank you very much! $\endgroup$ Commented Mar 21 at 2:39
  • $\begingroup$ My answer isn't really that different from the Lagrange interpolation view - there is a strong relationship between partial fractions and Legrange interpolation of polynomials. I just think partial fractions first when I see such equalities. $\endgroup$ Commented Mar 21 at 14:18

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