\[f(\alpha , \beta , \gamma ,...)\]
is a polynomimal such that if the arguments \[\alpha , \beta , \gamma , \...\]
are raeearanged, then the value of \[f\]
remains the same, because rearranges the arguments only changes the order of the terms.Example:
\[f(\alpha , \beta , \gamma )= \alpha^2 + \beta^2 + \gamma^2\]
is a symmetric polynomial because \[f(\alpha , \beta , \gamma )=f(\alpha , \beta , \gamma )=f(\alpha , \beta , \gamma )= \alpha^2 + \beta^2 + \gamma^2\]
.Suppose
\[f(\alpha , \beta )= \alpha - \beta\]
.\[f(\alpha , \beta )\]
is not a symmetric polynomial because \[f(\alpha , \beta )= \alpha - \beta \neq f(\beta , \alpha )= \beta - \alpha\]
.We can however create a symmetric polynomial from
\[f(\alpha , \beta , \gamma )\]
by squaring it.\[(f(\alpha , \beta ))=(\alpha - \beta )^2 = \alpha^2 - 2 \alpha \beta \beta^2 = \beta^2 - 2 \beta \alpha + \alpha^2 = (f(\beta , \alpha ))^2 \]
.Can we then find the value of
\[f(\alpha , \beta ) = \alpha - \beta\]
by finding the square root of \[(f(\alpha , \beta))^2\]
?Suppose the root of the equation
\[z^2- 3z-5\]
are \[\alpha , \: \beta \]
.Since the roots are
\[\alpha , \: \beta\]
, \[(z- \alpha )(z - \beta )= z^2 - (\alpha + \beta ) z + \alpha \beta =0\]
.Comparing with the equation above,
\[\alpha + \beta = 3, \: \alpha \beta = -5\]
.
\[(f(\alpha , \beta ))^2 = z^2 - 2 \alpha \beta + \beta ^2 = (\alpha + \beta )^2 - 4 \alpha \beta =3^2-4 \times -5 =29\]
.Then
\[f(\alpha , \beta )= \pm \sqrt{29}\]
.This will only work if the function
\[f\]
has terms such that powers of roots of symmetric, and the only things that stops symmetry are the signs of some terms.The value of
\[\alpha - 3 \beta\]
cannot be found, and neither can the value of \[\alpha + 3 \beta\]