type/puiseuxseries
¶$\operatorname{type}(f, {\mathit puiseuxseries})$ $\operatorname{type}(f, {\mathit puiseuxseries}(K))$ $\operatorname{type}(f, {\mathit puiseuxseries}(K, v))$
libname := libname, FileTools:-JoinPath(["maple","lib","dcfun.mla"],base=homedir):
with(dcfun):
A series whose exponents are half-integers is not a Puiseux series, in the sense of dcfun.
f := series(sin(x)^(1/2)/x^5, x);
$$\frac{1}{x^{\frac{9}{2}}}-\frac{1}{12 x^{\frac{5}{2}}}+\frac{1}{1440 \sqrt{x}}+O\! \left(x^{\frac{3}{2}}\right)$$
type(f, puiseuxseries);
$$\mathit{false}$$
Let us remove the ramification by changing x into x^2.
f := simplify(eval(f, x = x^2), power, symbolic);
$$\frac{1}{x^{9}}-\frac{1}{12 x^{5}}+\frac{1}{1440 x}+O\! \left(x^{3}\right)$$
f := series(f, x);
$$x^{-9}-\frac{1}{12} x^{-5}+\frac{1}{1440} x^{-1}+\mathrm{O}\! \left(x^{3}\right)$$
We can define a Puiseux series in the sense of dcfun using the PuiseuxSeries procedure.
f := PuiseuxSeries(2, f);
$$\frac{1}{x^{\frac{9}{2}}}-\frac{1}{12 x^{\frac{5}{2}}}+\frac{1}{1440 \sqrt{x}}+O\! \left(x^{\frac{3}{2}}\right)$$
Now we have a Puiseux series in the sense of dcfun.
type(f, puiseuxseries(anything,x));
$$\mathit{true}$$
type(f, puiseuxseries(rational, x));
$$\mathit{true}$$
type(f, puiseuxseries(polynom(anything, y), x));
$$\mathit{true}$$