Ser

Function: Ser
Section: conversions
C-Name: Ser0
Prototype: GDnDGDP
Help: Ser(s,{v='x},{d=seriesprecision}): convert s into a power series with
 variable v and precision d, starting with the constant coefficient.
Doc: transforms the object $s$ into a power series with main variable $v$
 ($x$ by default) and precision (number of significant terms) equal to
 $d \geq 0$ ($d = \kbd{seriesprecision}$ by default). If $s$ is a
 scalar, this gives a constant power series in $v$ with precision \kbd{d}.
 If $s$ is a polynomial, the polynomial is truncated to $d$ terms if needed
 \bprog
 ? \ps
   seriesprecision = 16 significant terms
 ? Ser(1)  \\ 16 terms by default
 %1 = 1 + O(x^16)
 ? Ser(1, 'y, 5)
 %2 = 1 + O(y^5)
 ? Ser(x^2,, 5)
 %3 = x^2 + O(x^7)
 ? T = polcyclo(100)
 %4 = x^40 - x^30 + x^20 - x^10 + 1
 ? Ser(T, 'x, 11)
 %5 = 1 - x^10 + O(x^11)
 @eprog\noindent The function is more or less equivalent with multiplication by
 $1 + O(v^d)$ in theses cases, only faster.

 For the remaining types, vectors and power series, we first explain what
 occurs if $d$ is omitted. In this case, the function uses exactly the amount
 of information given in the input:

 \item If $s$ is already a power series in $v$, we return it verbatim;

 \item If $s$ is a vector, the coefficients of the vector are
 understood to be the coefficients of the power series starting from the
 constant term (as in \tet{Polrev}$(x)$); in other words we convert
 \typ{VEC} / \typ{COL} to the power series whose significant terms are exactly
 given by the vector entries.

 On the other hand, if $d$ is explicitly given, we abide by its value
 and return a series, truncated or extended with zeros as needed, with
 $d$ significant terms.

 \bprog
 ? v = [1,2,3];
 ? Ser(v, t) \\ 3 terms: seriesprecision is ignored!
 %7 = 1 + 2*t + 3*t^2 + O(t^3)
 ? Ser(v, t, 7)  \\ 7 terms as explicitly requested
 %8 = 1 + 2*t + 3*t^2 + O(t^7)
 ? s = 1+x+O(x^2);
 ? Ser(s)
 %10 = 1 + x + O(x^2)  \\ 2 terms: seriesprecision is ignored
 ? Ser(s, x, 7)  \\ extend to 7 terms
 %11 = 1 + x + O(x^7)
 ? Ser(s, x, 1)  \\ truncate to 1 term
 %12 = 1 + O(x)
 @eprog\noindent
 The warning given for \kbd{Pol} also applies here: this is not a substitution
 function.