If the option 'makeindependent' = true is given, ConvertRootOf will attempt to rewrite all algebraic objects, including RootOfs that are present in f, in terms of independent RootOfs, using the command evala/Algfield. Note that this computation can be very expensive. If 'makeindependent' = false is given, then ConvertRootOf will not perform any independence checking, and the RootOfs in the output may not be independent. By default, 'makeindependent' = FAIL is assumed, and a rewrite in terms of independent RootOfs will only be attempted if there are at most $4$ distinct algebraic objects in f.

If the option 'substitutions' = true is given (or 'substitutions' for short), then ConvertRootOf does not actually perform the conversion but instead returns an expression sequence F, B, S, where

F is a set of forward substitutions, such that the result of subs(F, f) is equal to what ConvertRootOf returns without option substitutions,

B is a set of backward substitutions, essentially the inverse of the substitution induced by F, and

S, of type boolean, indicates whether the RootOfs in subs(F, f) are independent.

The ConvertRootOf command changes all occurrences of algebraic objects (typically radicals) in f to indexed RootOf notation.

Usually, the radical $a^{\frac{p}{m}}$, for integers $p<m$, is transformed into the equivalent expression ${\mathrm{RootOf}\left({\mathrm{\_Z}}^m-a\&comma;\mathrm{index}=1\right)}^p$.

The imaginary unit $\mathrm{I}$ is replaced by $\mathrm{RootOf}\left({\mathrm{\_Z}}^2+1\&comma;\mathrm{index}=1\right)$, unless it occurs in the second or third argument of a RootOf.

This command descends recursively into subexpressions, including tables. In particular, nested radicals are converted.

By default, this command performs a limited amount of independence checking, for efficiency reasons, unless the option 'makeindependent' is given (see below).

If the input is a single algebraic object, then ConvertRootOf always returns an irreducible RootOf.

$\mathrm{Algebraic}\left[\mathrm{ConvertRootOf}\right]\left(\mathrm{sqrt}\left(5\right)\right)$

$\mathrm{RootOf}\left({\mathrm{\_Z}}^2-5\&comma;\mathrm{index}=1\right)$

$\mathrm{Algebraic}\left[\mathrm{ConvertRootOf}\right]\left(\mathrm{I}+\mathrm{sqrt}\left(5\right)\right)$

$\mathrm{RootOf}\left({\mathrm{\_Z}}^2+1\&comma;\mathrm{index}=1\right)+\mathrm{RootOf}\left({\mathrm{\_Z}}^2-5\&comma;\mathrm{index}=1\right)$

$\mathrm{Algebraic}\left[\mathrm{ConvertRootOf}\right]\left(\mathrm{I}+\mathrm{sqrt}\left(5\right)\&comma;'\mathrm{substitutions}'\right)$

$\left\{\mathrm{I}=\mathrm{RootOf}\left({\mathrm{\_Z}}^2+1\&comma;\mathrm{index}=1\right)\&comma;\sqrt{5}=\mathrm{RootOf}\left({\mathrm{\_Z}}^2-5\&comma;\mathrm{index}=1\right)\right\},\left\{\mathrm{RootOf}\left({\mathrm{\_Z}}^2-5\&comma;\mathrm{index}=1\right)=\sqrt{5}\&comma;\mathrm{RootOf}\left({\mathrm{\_Z}}^2+1\&comma;\mathrm{index}=1\right)=\mathrm{I}\right\},\mathrm{true}$

$\mathrm{Algebraic}\left[\mathrm{ConvertRootOf}\right]\left(\mathrm{RootOf}\left({\mathrm{\_Z}}^2-\mathrm{sqrt}\left(2\right)\mathrm{\_Z}+1\&comma;\mathrm{index}=2\right)\right)$

$\mathrm{RootOf}\left({\mathrm{\_Z}}^2-\mathrm{RootOf}\left({\mathrm{\_Z}}^2-2\&comma;\mathrm{index}=1\right)\mathrm{\_Z}+1\&comma;\mathrm{index}=2\right)$

$\mathrm{Algebraic}\left[\mathrm{ConvertRootOf}\right]\left(x+\mathrm{I}+\mathrm{RootOf}\left({\mathrm{\_Z}}^3-\mathrm{I}\&comma;0..1+\mathrm{I}\right)\right)$

$x+\mathrm{RootOf}\left({\mathrm{\_Z}}^2+1\&comma;\mathrm{index}=1\right)+\mathrm{RootOf}\left({\mathrm{\_Z}}^2-\mathrm{RootOf}\left({\mathrm{\_Z}}^2+1\&comma;\mathrm{index}=1\right)\mathrm{\_Z}-1\&comma;\mathrm{index}=1\right)$

$\mathrm{Algebraic}\left[\mathrm{ConvertRootOf}\right]\left(\left[\mathrm{sqrt}\left(2\right)\&comma;\mathrm{sqrt}\left(3\right)\&comma;\mathrm{sqrt}\left(6\right)\right]\right)$

$\left[\mathrm{RootOf}\left({\mathrm{\_Z}}^2-2\&comma;\mathrm{index}=1\right)\&comma;\mathrm{RootOf}\left({\mathrm{\_Z}}^2-3\&comma;\mathrm{index}=1\right)\&comma;\mathrm{RootOf}\left({\mathrm{\_Z}}^2-3\&comma;\mathrm{index}=1\right)\mathrm{RootOf}\left({\mathrm{\_Z}}^2-2\&comma;\mathrm{index}=1\right)\right]$

$\mathrm{Algebraic}\left[\mathrm{ConvertRootOf}\right]\left(\left[\mathrm{sqrt}\left(2\right)\&comma;\mathrm{sqrt}\left(3\right)\&comma;\mathrm{sqrt}\left(6\right)\right]\&comma;'\mathrm{substitutions}'\right)$

$\left\{\sqrt{2}=\mathrm{RootOf}\left({\mathrm{\_Z}}^2-2\&comma;\mathrm{index}=1\right)\&comma;\sqrt{3}=\mathrm{RootOf}\left({\mathrm{\_Z}}^2-3\&comma;\mathrm{index}=1\right)\&comma;\sqrt{6}=\mathrm{RootOf}\left({\mathrm{\_Z}}^2-3\&comma;\mathrm{index}=1\right)\mathrm{RootOf}\left({\mathrm{\_Z}}^2-2\&comma;\mathrm{index}=1\right)\right\},\left\{\mathrm{RootOf}\left({\mathrm{\_Z}}^2-3\&comma;\mathrm{index}=1\right)=\sqrt{3}\&comma;\mathrm{RootOf}\left({\mathrm{\_Z}}^2-2\&comma;\mathrm{index}=1\right)=\sqrt{2}\right\},\mathrm{true}$

$\mathrm{Algebraic}\left[\mathrm{ConvertRootOf}\right]\left(\left[\mathrm{I}\&comma;\mathrm{sqrt}\left(2\right)\&comma;\mathrm{sqrt}\left(3\right)\&comma;\mathrm{sqrt}\left(5\right)\&comma;\mathrm{sqrt}\left(6\right)\right]\right)$

$\left[\mathrm{RootOf}\left({\mathrm{\_Z}}^2+1\&comma;\mathrm{index}=1\right)\&comma;\mathrm{RootOf}\left({\mathrm{\_Z}}^2-2\&comma;\mathrm{index}=1\right)\&comma;\mathrm{RootOf}\left({\mathrm{\_Z}}^2-3\&comma;\mathrm{index}=1\right)\&comma;\mathrm{RootOf}\left({\mathrm{\_Z}}^2-5\&comma;\mathrm{index}=1\right)\&comma;\mathrm{RootOf}\left({\mathrm{\_Z}}^2-6\&comma;\mathrm{index}=1\right)\right]$

$\mathrm{Algebraic}\left[\mathrm{ConvertRootOf}\right]\left(\left[\mathrm{I}\&comma;\mathrm{sqrt}\left(2\right)\&comma;\mathrm{sqrt}\left(3\right)\&comma;\mathrm{sqrt}\left(5\right)\&comma;\mathrm{sqrt}\left(6\right)\right]\&comma;'\mathrm{makeindependent}'\right)$

$\left[\mathrm{RootOf}\left({\mathrm{\_Z}}^2+1\&comma;\mathrm{index}=1\right)\&comma;\mathrm{RootOf}\left({\mathrm{\_Z}}^2-2\&comma;\mathrm{index}=1\right)\&comma;\mathrm{RootOf}\left({\mathrm{\_Z}}^2-3\&comma;\mathrm{index}=1\right)\&comma;\mathrm{RootOf}\left({\mathrm{\_Z}}^2-5\&comma;\mathrm{index}=1\right)\&comma;\mathrm{RootOf}\left({\mathrm{\_Z}}^2-3\&comma;\mathrm{index}=1\right)\mathrm{RootOf}\left({\mathrm{\_Z}}^2-2\&comma;\mathrm{index}=1\right)\right]$

$\mathrm{Algebraic}\left[\mathrm{ConvertRootOf}\right]\left(\sin \left(\frac{\mathrm{\pi }}{5}\right)\right)$

$\frac{\mathrm{RootOf}\left({\mathrm{\_Z}}^4-5{\mathrm{\_Z}}^2+5\&comma;\mathrm{index}=1\right)}{2}$

$\mathrm{Algebraic}\left[\mathrm{ConvertRootOf}\right]\left(\sin \left(\frac{\mathrm{\pi }}{5}\right)+\mathrm{I}\right)$

$\frac{\mathrm{RootOf}\left({\mathrm{\_Z}}^2-2\&comma;\mathrm{index}=1\right)\mathrm{RootOf}\left({\mathrm{\_Z}}^2-5+\mathrm{RootOf}\left({\mathrm{\_Z}}^2-5\&comma;\mathrm{index}=1\right)\&comma;\mathrm{index}=1\right)}{4}+\mathrm{RootOf}\left({\mathrm{\_Z}}^2+1\&comma;\mathrm{index}=1\right)$