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algcurves

 puiseux
 determines the Puiseux expansions of an algebraic function

 Calling Sequence puiseux(a, x=p, n) puiseux(f, x=p, y, n) puiseux(f, x=p, y, n, T)

Parameters

 a - algebraic function in RootOf form x=p - gives the point around which the expansions are computed n - number specifying the desired accuracy of the expansions y - variable f - square-free polynomial in x and y T - (optional) variable, used for obtaining a different output syntax

Description

 • A square-free polynomial f of degree $N$ in the variable y with coefficients in a field $L\left(x\right)$ has $N$ roots in the algebraic closure of the field $L\left(\left(x-p\right)\right)$. These roots are called the Puiseux expansions of f at x=p. Each Puiseux expansion is of the form ${\sum }_{i=-N}^{\mathrm{\infty }}{a}_{i}{\left(x-p\right)}^{\frac{i}{r}}$ for some integer $r$ ($r$ is called the ramification index of the Puiseux expansion), and some integer $N$ and elements ${a}_{i}$ in the algebraic closure of $L$.
 • The polynomial f must be square-free, otherwise the puiseux procedure does not work.
 • If f is irreducible then f gives an algebraic extension $L\left(x\right)\left[y\right]/\left(f\right)$ of $L\left(x\right)$. Instead of giving f, this algebraic extension can also be specified with a RootOf a of f. This a can be viewed as a multivalued function in x. The Puiseux expansions give the local expansions of this multivalued function.
 • The procedure puiseux determines the field $L$ from the input. The groundfield $L$ of the computation is the smallest field such that f and p are in $L\left(x\right)\left[y\right]$ .
 • The Puiseux expansions are only computed up to conjugation over $L\left(\left(x-p\right)\right)$. So if a number of expansions are algebraically conjugated over $L\left(\left(x-p\right)\right)$ then only one of these expansions is given.
 • The Puiseux expansions are computed modulo ${x}^{n}$. So if for instance n=10, then the term ${x}^{49}{5}}$ would be computed, but not the term ${x}^{10}$.
 • If n=0 then the expansions are not computed modulo ${x}^{0}$, but in this case the number of terms that is computed is precisely the number that is needed to distinguish the expansions from the other expansions. If n='minimal' then the same output is given.
 • To avoid an output containing fractional powers of x one can specify a fifth argument T. Then the Puiseux expansions in the output are represented in a different way, namely as

$[x=\left(a\mathrm{constant}\right)\left(a\mathrm{power}\mathrm{of}T\right),y=a\mathrm{power series}\mathrm{in}T]$

 So then x and y are expressed in terms of a local parameter.
 Note: The Maple "alias" function does not recognize an alias in terms of another alias. Therefore, you must not use nested aliases for algebraic numbers because then the puiseux algorithm is not able to construct the field L over which the algebraic function is defined.

Examples

 > $\mathrm{with}\left(\mathrm{algcurves}\right):$
 > $\mathrm{\alpha }≔\mathrm{RootOf}\left({x}^{3}+7\right)$
 ${\mathrm{\alpha }}{≔}{\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{3}}{+}{7}\right)$ (1)
 > $f≔{y}^{8}+x{y}^{5}+{x}^{4}-{x}^{6}\mathrm{\alpha }$
 ${f}{≔}{{y}}^{{8}}{-}{\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{3}}{+}{7}\right){}{{x}}^{{6}}{+}{x}{}{{y}}^{{5}}{+}{{x}}^{{4}}$ (2)
 > $\mathrm{puiseux}\left(f,x=0,y,0\right)$
 $\left\{{\left({-}{x}\right)}^{{1}}{{3}}}{,}{-}{{x}}^{{3}}{{5}}}\right\}$ (3)
 > $\mathrm{puiseux}\left(f,x=0,y,5\right)$
 $\left\{{-}\frac{{170}{}{\left({-}{x}\right)}^{{13}}{{3}}}}{{81}}{+}\frac{{\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{3}}{+}{7}\right){}{\left({-}{x}\right)}^{{11}}{{3}}}}{{3}}{+}\frac{{2}{}{{x}}^{{3}}}{{3}}{-}\frac{{\left({-}{x}\right)}^{{5}}{{3}}}}{{3}}{+}{\left({-}{x}\right)}^{{1}}{{3}}}{,}\left({-}\frac{{25389}}{{15625}}{+}\frac{{2}{}{{\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{3}}{+}{7}\right)}^{{2}}}{{25}}\right){}{{x}}^{{23}}{{5}}}{+}\frac{{42}{}{\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{3}}{+}{7}\right){}{{x}}^{{21}}{{5}}}}{{125}}{-}\frac{{483}{}{{x}}^{{19}}{{5}}}}{{625}}{+}\frac{{4}{}{\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{3}}{+}{7}\right){}{{x}}^{{17}}{{5}}}}{{25}}{-}\frac{{2}{}{{x}}^{{3}}}{{5}}{+}\frac{{\mathrm{RootOf}}{}\left({{\mathrm{_Z}}}^{{3}}{+}{7}\right){}{{x}}^{{13}}{{5}}}}{{5}}{-}\frac{{6}{}{{x}}^{{11}}{{5}}}}{{25}}{-}\frac{{{x}}^{{7}}{{5}}}}{{5}}{-}{{x}}^{{3}}{{5}}}\right\}$ (4)

To get all 8 Puiseux expansions substitute all conjugates over $L\left(\left(x\right)\right)$ of ${x}^{1}{5}}$ and ${\left(-x\right)}^{1}{3}}$ in this set.