Summation with polynomial roots
by fungarwai, May 26, 2020, 12:52 PM
Reciprocal type
![$\sum_{i=1}^m \frac{-1}{(x-x_i)^2}=\frac{f''(x)f(x)-[f'(x)]^2}{[f(x)]^2}$](//latex.artofproblemsolving.com/a/b/3/ab3a6c2f444a2265b208b177fea06bfc3e7196d8.png)
![$\sum_{i=1}^m \frac{2}{(x-x_i)^3}=\frac{f'''(x)[f(x)]^2-3f''(x)f'(x)f(x)+2[f'(x)]^3}{[f(x)]^3}$](//latex.artofproblemsolving.com/9/d/a/9da9b44ce18202b3021555318f2c8ebe471836e7.png)
Identity of a conditional symmetric polynomial
Let
Lagrange polynomial type
where
,
![$\sum_{i=1}^m \frac{-1}{(x-x_i)^2}=\frac{f''(x)f(x)-[f'(x)]^2}{[f(x)]^2}$](http://latex.artofproblemsolving.com/a/b/3/ab3a6c2f444a2265b208b177fea06bfc3e7196d8.png)
![$\sum_{i=1}^m \frac{2}{(x-x_i)^3}=\frac{f'''(x)[f(x)]^2-3f''(x)f'(x)f(x)+2[f'(x)]^3}{[f(x)]^3}$](http://latex.artofproblemsolving.com/9/d/a/9da9b44ce18202b3021555318f2c8ebe471836e7.png)
Identity of a conditional symmetric polynomial
Let
balls into 
boxes
has 
balls
be the event of box 
contains at least one ball
,
as example
is true for 
Lagrange polynomial type
,
,
![$\displaystyle =\frac{1}{b-c}\left[\left(\sum_{k=1}^{n}{a^{n-k}b^{k-1}}\right)-\left(\sum_{j=1}^{n}{a^{n-j}c^{j-1}}\right)\right]$](http://latex.artofproblemsolving.com/9/e/4/9e407f635285bfda0a141f2c19e6863c4a346292.png)
![$\displaystyle =\frac{1}{b-c}\left[\sum_{i=1}^{n}a^{n-i}\left(b^{i-1}-c^{i-1}\right)\right]$](http://latex.artofproblemsolving.com/1/f/8/1f86840d023c731e7706b09c7d0de7293707888d.png)
![$\displaystyle =\sum_{s=2}^{n}\left[a^{n-s}\frac{\left(b^{s-1}-c^{s-1}\right)}{b-c}\right]$](http://latex.artofproblemsolving.com/5/e/9/5e97fe5d25462f9546fabfe70e306d8710f99071.png)
![$\displaystyle =\sum_{s=2}^{n}\left[a^{n-s}\sum_{t=0}^{s-1}{b^{s-t-2}c^t}\right]$](http://latex.artofproblemsolving.com/8/6/8/868f452e9542a4aa9d498d91df7856bddffabc46.png)
should be 0![$[x^k]P(x)=0,~\forall m\le k\le n$](http://latex.artofproblemsolving.com/e/e/c/eec149e97f1674238203b8de81e4b91781a0214a.png)
![$[x^{m-1}]P(x)=\sum_{i=1}^m x_i^n \prod_{j=1\atop i\neq j}^m\frac{1}{x_i-x_j}$](http://latex.artofproblemsolving.com/9/3/3/933e37fd232bc27983ac056623d279f3460a8f12.png)
![$\begin{cases}
[x^n]P(x)=1-1=0\\
[x^{n-1}]P(x)=-\left(q_1-\sigma_{1,m}\right)=0\\
[x^{n-2}]P(x)=-\left(q_2-\sigma_{1,m} q_1+(-1)^{K_2}\sigma_{K_2,m}\right)=0\\
\vdots\\
\displaystyle [x^m]P(x)=-\left(q_{n-m}+\sum_{k=1}^{K_{n-m}-1} (-1)^k \sigma_{k,m} q_{n-m-k}+(-1)^{K_{n-m}}\sigma_{K_{n-m},m}\right)=0\\
\displaystyle [x^{m-1}]P(x)=-\left(\sum_{k=1}^{K_{n-m+1}-1} (-1)^k \sigma_{k,m} q_{n-m+1-k}+(-1)^{K_{n-m+1}}\sigma_{K_{n-m+1},m}\right)\end{cases}$](http://latex.artofproblemsolving.com/e/5/f/e5f9aa5011209a8abcb849c61cae2067a13303b9.png)
![$[x^{m-1}]P(x)=-\left(-S_{n-m+1,m}\right)=S_{n-m+1,m}$](http://latex.artofproblemsolving.com/5/4/9/549b85f0b17c81e85eb9826d6fa829e68d66ec4f.png)
The desired expression is the coefficient of 
of 
.
(and cyclically) so
But 
and 
should be 
.
So
The coefficient of 
.
where
,
,
so 
This post has been edited 23 times. Last edited by fungarwai, May 19, 2023, 12:07 PM
March 2024
February 2023