Lines Matching refs:sumnum

1 Function: sumnum
5 Help: sumnum(n=a,f,{tab}): numerical summation of f(n) from
11   (gen,gen,?gen):gen:prec sumnum(${2 cookie}, ${2 wrapper}, $1, $3, $prec)
24  ? sumnum(n = 1, n^-3) - z3 \\ here slower than sumpos
48  ? sumnum(n = 1, 1/n^3, tab) - z3 \\ now much faster than sumpos
60  ? sumnum(n = 1, lngamma(1+1/n)/n, tab);
68  By default, \kbd{sumnum} assumes that \var{expr} decreases slowly at infinity,
80  ? sumnum(n=1,2^-n)
81   ***   at top-level: sumnum(n=1,2^-n)
84  ? tab = sumnuminit([+oo,log(2)]); sumnum(n=1,2^-n, tab)
90    sumnum(n = [a, asymp], f)
94    sumnum(n = a, f, tab)
100  ? sumnum(n = 1, n^(-2)) - zeta(2) \\ accurate, fast
109  ? sumnum(n=1,n^(-4/3)) - zeta(4/3)  \\ fast but inaccurate
112  ? sumnum(n=[1,-4/3],n^(-4/3)) - zeta(4/3) \\ with decrease rate, now accurate
118  ? sumnum(n=1, n^(-4/3), tab) - zeta(4/3) \\ faster with precomputations
121  ? sumnum(n=1,-log(n)*n^(-4/3), tab) - zeta'(4/3)
129  \kbd{sumnum} is given for completeness and comparison purposes only: one
132  ? sumnum(n=[1, 1], 2^-n) \\ pretend we decrease as exp(-n)
147  ? sumnum(n=1, f(n)) - z
150  @eprog\noindent As \kbd{sumnum(n=1, print(n))} shows, we evaluate $f(n)$ for
158  ? sumnum(n=1, if(n > 1e7, subst(F,x,1/n), f(n))) - z
162  \synt{sumnum}{(void *E, GEN (*eval)(void*, GEN), GEN a, GEN tab, long prec)}