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Newton's Relations

Let $s_i$ be the sum of the products of distinct Roots $r_j$ of the Polynomial equation of degree $n$

\begin{displaymath}
a_n x^n + a_{n-1} x^{n-1} + \ldots+a_1x+a_0 = 0,
\end{displaymath} (1)

where the roots are taken $i$ at a time (i.e., $s_i$ is defined as the Elementary Symmetric Function $\Pi_i(r_1,
\ldots, r_n)$) $s_i$ is defined for $i=1$, ..., $n$. For example, the first few values of $s_i$ are
$\displaystyle s_1$ $\textstyle =$ $\displaystyle r_1+r_2+r_3+r_4+\ldots$ (2)
$\displaystyle s_2$ $\textstyle =$ $\displaystyle r_1r_2+r_1r_3+r_1r_4+r_2r_3+\ldots$ (3)
$\displaystyle s_3$ $\textstyle =$ $\displaystyle r_1r_2r_3+r_1r_2r_4+r_2r_3r_4+\ldots,$ (4)

and so on. Then
\begin{displaymath}
s_i=(-1)^i {a_{n-i}\over a_n}.
\end{displaymath} (5)

This can be seen for a second Degree Polynomial by multiplying out,


\begin{displaymath}
a_2x^2+a_1x+a_0=a_2(x-r_1)(x-r_2)=a_2[x^2-(r_1+r_2)x+r_1r_2],
\end{displaymath} (6)

so
$\displaystyle s_1$ $\textstyle =$ $\displaystyle \sum_{i=1}^2 r_i=r_1+r_2=-{a_1\over a_2}$ (7)
$\displaystyle s_2$ $\textstyle =$ $\displaystyle \sum_{\scriptstyle i,j=1\atop\scriptstyle i\not=j}^2 r_ir_j=r_1r_2={a_0\over a_2},$ (8)

and for a third Degree Polynomial,

$a_3x^3+a_2x^2+a_1x+a_0=a_3(x-r_1)(x-r_2)(x-r_3)$
$ = a_3[x^3-(r_1+r_2+r_3)x^2+(r_1r_2+r_1r_3+r_2r_3)x-r_1r_2r_3],\quad$ (9)
so

$\displaystyle s_1$ $\textstyle =$ $\displaystyle \sum_{i=1}^3 r_i = -{a_2\over a_3}$ (10)
$\displaystyle s_2$ $\textstyle =$ $\displaystyle \sum_{\scriptstyle i,j\atop\scriptstyle i\not=j}^3 r_ir_j = r_1r_2+r_1r_3+r_2r_3={a_1\over a_3}$ (11)
$\displaystyle s_3$ $\textstyle =$ $\displaystyle \sum_{\scriptstyle i,j,k\atop\scriptstyle i\not=j\not=k}^3 r_ir_jr_k = r_1r_2r_3 =-{a_0\over a_3}.$ (12)

See also Elementary Symmetric Function


References

Coolidge, J. L. A Treatise on Algebraic Plane Curves. New York: Dover, pp. 1-2, 1959.



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© 1996-9 Eric W. Weisstein
1999-05-25