0

Let $f(x)=(x-a_{1})(x-a_{2})...(x-a_{n})$, where $a_{1},a_{2},...,a_{n}$ are pairwise distinct and neither of them is zero. Prove that

$ \sum_{j=1}^{n} \frac{a_{j}^{n-1}}{f'(a_{j})}=1$

I've proved that using Cauchy residue theorem, but I'm looking for solution requiring less knowledge. Any help will be greatly appreciated.

  • For the same rational function that you used, do partial fraction decomposition. The residue theorem can be replaced by the uniqueness of the partial fraction decomposition. – plop Oct 04 '22 at 18:30
  • You can just use the partial fraction expansion of $x^{n-1}/f(x)=\sum_{j=1}^n\frac {a_j^{n-1}}{f'(a_j)(x-a_j)} $ and then multiply by $f(x) $ and compare coefficients of $x^{n-1}$. – Paramanand Singh Oct 07 '22 at 03:50

2 Answers2

0

A partial fractions argument similar to the one described in this answer: https://math.stackexchange.com/a/891102/154826 would go as follows.

First, we have that $f'(a_j) = \prod_{i \neq j}(a_j - a_i)$, so that we're dealing with the quantity \begin{align*} \sum_{j=1}^n \frac{a_j^{n-1}}{f'(a_j)} = \sum_{j=1}^n \frac{a_j^{n-1}}{\prod_{i\neq j}(a_j - a_i)}. \end{align*}

Now, consider the rational functionwith corresponding partial fraction decomposition \begin{equation} R(x) = \frac{x^{n-1}}{\prod_{i=1}^n(x - a_i)} = \sum_{i=1}^n \frac{b_i}{x - a_i},\tag{1} \end{equation} where \begin{align*} b_j = \lim_{x \rightarrow a_j}(x - a_j)R(x) = \frac{a_j^{n-1}}{\prod_{i\neq j}(a_j - a_i)}. \end{align*}

But now, multiplying both sides of (1) by $x$, we get \begin{align*} \frac{x^n}{\prod_{j=1}^n (x - a_j)} &= \sum_{j=1}^n \frac{x}{x - a_j} b_j. \end{align*}

Taking $x\rightarrow \infty$ on both sides then yields \begin{align*} 1 = \sum_{j=1}^n b_j = \sum_{j=1}^n \frac{a_j^{n-1}}{\prod_{i\neq j}(a_j - a_i)}. \end{align*}

Damian Pavlyshyn
  • 5,428
0

This problem can be easily shown with Lagrange polynomials. Since $f(x)=\prod_{i=1}^n\left(x-a_i\right)$ we obtain \begin{align*} f^{\prime}(x)&=\sum_{j=1}^n\prod_{{i=1}\atop{i\ne j}}^n\left(x-a_i\right) \qquad\quad f^{\prime}(a_k)=\prod_{{i=1}\atop{i\ne k}}\left(a_k-a_i\right)\quad 1\leq k\leq n \end{align*} and the claim can be written as \begin{align*} \color{blue}{\sum_{j=1}^n\prod_{{i=1}\atop{i\ne j}}^n\frac{a_j}{a_j-a_i}=1}\tag{1} \end{align*}

We consider the Lagrange polynomial $L(x)$ \begin{align*} L(x)=\sum_{j=1}^n\prod_{{i=1}\atop{i\ne j}}^n\frac{a_j-x}{a_j-a_i} \qquad\text{with}\qquad L(a_k)=1\quad 1\leq k\leq n \end{align*} Since $L(x)-1$ is a polynomial of degree $n-1$ with $n$ zeros $a_k,1\leq k\leq n$ it follows \begin{align*} \color{blue}{L(x)-1=0} \end{align*} and the claim (1) \begin{align*} \color{blue}{L(0)=\sum_{j=1}^n\prod_{{i=1}\atop{i\ne j}}^n\frac{a_j}{a_j-a_i}=1} \end{align*} follows.

Markus Scheuer
  • 112,413