Function: sumnumlagrange
Section: sums
C-Name: sumnumlagrange0
Prototype: V=GEDGp
Help: sumnumlagrange(n=a,f,{tab}): numerical summation of f(n) from
 n = a to +infinity using Lagrange summation.
 a must be an integer, and tab, if given, is the output of sumnumlagrangeinit.
Wrapper: (,Gp)
Description:
  (gen,gen,?gen):gen:prec sumnumlagrange(${2 cookie}, ${2 wrapper}, $1, $3, $prec)
Doc: Numerical summation of $f(n)$ from $n=a$ to $+\infty$ using Lagrange
 summation; $a$ must be an integer, and the optional argument \kbd{tab} is
 the output of \kbd{sumnumlagrangeinit}. By default, the program assumes that
 the $N$th remainder has an asymptotic expansion in integral powers of $1/N$.
 If not, initialize \kbd{tab} using \kbd{sumnumlagrangeinit(al)}, where
 the asymptotic expansion of the remainder is integral powers of $1/N^{al}$;
 $al$ can be equal to $1$ (default), $1/2$, $1/3$, or $1/4$, and also
 equal to $2$, but in this latter case it is the $N$th remainder minus one
 half of the last summand which has an asymptotic expansion in integral
 powers of $1/N^{2}$.
 \bprog
 ? \p1000
 ? z3 = zeta(3);
 ? sumpos(n = 1, n^-3) - z3
 time = 4,440 ms.
 %2 = -2.08[...] E-1001
 ? sumnumlagrange(n = 1, n^-3) - z3 \\ much faster than sumpos
 time = 25 ms.
 %3 = 0.E-1001
 ? tab = sumnumlagrangeinit();
 time = 21 ms.
 ? sumnumlagrange(n = 1, n^-3, tab) - z3
 time = 2 ms. /* even faster */
 %5 = 0.E-1001

 ? \p115
 ? tab = sumnumlagrangeinit([1/3,1/3]);
 time = 316 ms.
 ? sumnumlagrange(n = 1, n^-(7/3), tab) - zeta(7/3)
 time = 24 ms.
 %7 = 0.E-115
 ? sumnumlagrange(n = 1, n^(-2/3) - 3*(n^(1/3)-(n-1)^(1/3)), tab) - zeta(2/3)
 time = 32 ms.
 %8 = 1.0151767349262596893 E-115
 @eprog

 \misctitle{Complexity}
 The function $f$ is evaluated at $O(D)$ integer arguments,
 where $D \approx \kbd{realprecision} \cdot \log(10)$.

 \synt{sumnumlagrange}{(void *E, GEN (*eval)(void*, GEN), GEN a, GEN tab, long prec)}
 where an omitted \var{tab} is coded as \kbd{NULL}.
