How to prove that the sum of the squares of the roots of the nth Hermite polynomial is $\frac{n(n-1)}{2}$?

Filipinacws
2022-08-13
Answered

How to prove that the sum of the squares of the roots of the nth Hermite polynomial is $\frac{n(n-1)}{2}$?

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Lisa Acevedo

Answered 2022-08-14
Author has **18** answers

Let us write

$\begin{array}{rl}{H}_{n}(x)& ={A}_{n}(x-{x}_{1})\dots (x-{x}_{n})=\\ \text{(1)}& & ={A}_{n}({x}^{n}-{e}_{1}({x}_{1},\dots ,{x}_{n}){x}^{n-1}+{e}_{2}({x}_{1},\dots ,{x}_{n}){x}^{n-2}+{\mathrm{p}\mathrm{o}\mathrm{l}\mathrm{y}}_{n-3}(x)),\end{array}$

where ${e}_{k}({x}_{1},\dots ,{x}_{n})$ denote elementary symmetric polynomials:

$\begin{array}{rl}& {e}_{1}({x}_{1},\dots ,{x}_{n})=\sum _{k=1}^{n}{x}_{k},\\ & {e}_{2}({x}_{1},\dots ,{x}_{n})=\sum _{1\le i<j\le n}^{n}{x}_{i}{x}_{j}.\end{array}$

We want to find

$\begin{array}{}\text{(2)}& \sum _{k=1}^{n}{x}_{k}^{2}={e}_{1}^{2}({x}_{1},\dots ,{x}_{n})-2{e}_{2}({x}_{1},\dots ,{x}_{n}),\end{array}$

and therefore it will suffice to know the coefficients of ${x}^{n}$, ${x}^{n-1}$ and ${x}^{n-2}$ in ${H}_{n}(x)$. But they can be determined from the series representations of Hermite polynomials:

$\begin{array}{}\text{(3)}& {H}_{n}(x)={2}^{n}({x}^{n}-\frac{n(n-2)}{4}{x}^{n-2}+{\mathrm{p}\mathrm{o}\mathrm{l}\mathrm{y}}_{n-4}(x)).\end{array}$

Together with (1) and (2), this gives the result:

$\sum _{k=1}^{n}{x}_{k}^{2}=\frac{n(n-1)}{2}.$

$\begin{array}{rl}{H}_{n}(x)& ={A}_{n}(x-{x}_{1})\dots (x-{x}_{n})=\\ \text{(1)}& & ={A}_{n}({x}^{n}-{e}_{1}({x}_{1},\dots ,{x}_{n}){x}^{n-1}+{e}_{2}({x}_{1},\dots ,{x}_{n}){x}^{n-2}+{\mathrm{p}\mathrm{o}\mathrm{l}\mathrm{y}}_{n-3}(x)),\end{array}$

where ${e}_{k}({x}_{1},\dots ,{x}_{n})$ denote elementary symmetric polynomials:

$\begin{array}{rl}& {e}_{1}({x}_{1},\dots ,{x}_{n})=\sum _{k=1}^{n}{x}_{k},\\ & {e}_{2}({x}_{1},\dots ,{x}_{n})=\sum _{1\le i<j\le n}^{n}{x}_{i}{x}_{j}.\end{array}$

We want to find

$\begin{array}{}\text{(2)}& \sum _{k=1}^{n}{x}_{k}^{2}={e}_{1}^{2}({x}_{1},\dots ,{x}_{n})-2{e}_{2}({x}_{1},\dots ,{x}_{n}),\end{array}$

and therefore it will suffice to know the coefficients of ${x}^{n}$, ${x}^{n-1}$ and ${x}^{n-2}$ in ${H}_{n}(x)$. But they can be determined from the series representations of Hermite polynomials:

$\begin{array}{}\text{(3)}& {H}_{n}(x)={2}^{n}({x}^{n}-\frac{n(n-2)}{4}{x}^{n-2}+{\mathrm{p}\mathrm{o}\mathrm{l}\mathrm{y}}_{n-4}(x)).\end{array}$

Together with (1) and (2), this gives the result:

$\sum _{k=1}^{n}{x}_{k}^{2}=\frac{n(n-1)}{2}.$

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