Summation with polynomial roots

by fungarwai, May 26, 2020, 12:52 PM

Reciprocal type

$\sum_{i=1}^m \frac{1}{x-x_i}=\frac{f'(x)}{f(x)},~f(x)=\prod_{i=1}^m (x-x_i)$

Credit to the idea posted by ccmmjj in mathchina forum

Proof

$\sum_{i=1}^m \frac{-1}{(x-x_i)^2}=\frac{f''(x)f(x)-[f'(x)]^2}{[f(x)]^2}$

$\sum_{i=1}^m \frac{2}{(x-x_i)^3}=\frac{f'''(x)[f(x)]^2-3f''(x)f'(x)f(x)+2[f'(x)]^3}{[f(x)]^3}$

Identity of a conditional symmetric polynomial

Let $\displaystyle S_{n,m}=\sum_{r_1+r_2+\cdots+r_m=n} x_1^{r_1}x_2^{r_2}\cdots x_m^{r_m},~\sigma_{k,m} =\sum_{1\le j_1 < j_2 < \cdots < j_k \le m} x_{j_1} \dots x_{j_k}$
$\displaystyle \boxed{S_{j,m}=\sum_{k=1}^{m-1} (-1)^{k-1} \sigma_{k,m} S_{j-k,m}+(-1)^{m-1}\sigma_{m,m} S_{j-m,m}}$
$\displaystyle \text{or}~ \boxed{S_{j,m}=\sum_{k=1}^{j-1} (-1)^{k-1} \sigma_{k,m} S_{j-k,m}+(-1)^{j-1}\sigma_{j,m}}$

Proof

Example

$\displaystyle S_{j,K}=\sum_{k=1}^{M-1} (-1)^{k-1}
\sigma_{k,M} S_{j-k,K}+(-1)^{M-1}\sigma_{M,M}S_{j-M,K},~\forall K\le M\le m$

Proof

Example

Lagrange polynomial type

$S_{n,m}=\sum_{i=1}^m x_i^n \prod_{j=1\atop i\neq j}^m\frac{1}{x_i-x_j}
=\begin{cases}0 & 0\le n\le m-2\\1 & n=m-1\end{cases}$

Proof

$S_{n,m}=\sum_{i=1}^m x_i^n \prod_{j=1\atop i\neq j}^m\frac{1}{x_i-x_j}
=\begin{cases}0 & 0\le n\le m-2\\1 & n=m-1\\
\displaystyle
\boxed{\sum_{r_1+r_2+\cdots+r_m=n-m+1} x_1^{r_1}x_2^{r_2}\cdots x_m^{r_m}}
 & n\ge m
\end{cases}$

$S_{m,m}=\sum a$
$S_{m+1,m}=\sum a^2+\sum ab$
$S_{m+2,m}=\sum a^3+\sum a^2 b+\sum abc$
$S_{m+3,m}=\sum a^4+\sum a^3 b+\sum a^2 b^2+\sum a^2 bc+\sum abcd$

Best Proof Credit to the idea with four variables posted by natmath at #2

Proof Credit to the idea with three variables posted by gasgous at #10

Other Proof

$S_{n,m}=\sum_{i=1}^m x_i^n \prod_{j=1\atop i\neq j}^m\frac{1}{x_i-x_j}=\begin{cases}0 & 0\le n\le m-2\\1 & n=m-1\\
\displaystyle
\boxed{\sum_{r_1 + 2r_2 + \cdots + mr_m = s \atop r_1\ge 0, \ldots, r_m\ge 0} (-1)^n \frac{(r_1 + r_2 + \cdots + r_m)!}{r_1!r_2! \cdots r_m!} \prod_{i=1}^m (-\sigma_{i,m})^{r_i}} & n=s+m-1\ge m
\end{cases}$

where $\sigma_{k,m} =\sigma_k (x_1 , \ldots , x_m )=\sum_{1\le  j_1 < j_2 < \cdots < j_k \le m} x_{j_1} \dots x_{j_k}$

$S_{m,m}=\sigma_{1,m}$
$S_{m+1,m}=\sigma_{1,m}^2-\sigma_{2,m}$
$S_{m+2,m}=\sigma_{1,m}^3-2\sigma_{1,m}\sigma_{2,m}+\sigma_{3,m}$
$S_{m+3,m}=\sigma_{1,m}^4-3\sigma_{1,m}^2\sigma_{2,m}+\sigma_{2,m}^2-2\sigma_{1,m}\sigma_{3,m}+\sigma_{4,m}$

Proof

Example
This post has been edited 23 times. Last edited by fungarwai, May 19, 2023, 12:07 PM

Comment

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woah after checking your blog and my posts from 2020 again i remember you used the lagrange polynomial to solve my problem and that example with $S_{4,3}$ was indeed my problem. really nice identities and proof! :)

by MelonGirl, Sep 11, 2021, 2:51 AM

Notable algebra methods with proofs and examples

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    by jasperE3, Dec 3, 2021, 10:01 PM

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    by fungarwai, Aug 28, 2021, 4:54 AM

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    by mufree, May 26, 2019, 6:40 AM

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