(Alternating) Binomial Sums via Rmpfr
Compute (alternating) binomial sums via high-precision arithmetic.
If sBn(f, n) :=sumBinomMpfr(n, f), (default
alternating is true, and n0 = 0),
sBn(f,n) = sum(k = n0:n ; (-1)^(n-k) choose(n,k) * f(k)) = Δ^n f,
see Details for the n-th forward difference operator
Δ^n f.  If alternating is false, the
(-1)^(n-k) factor is dropped (or replaced by 1) above.
Such sums appear in different contexts and are typically challenging, i.e., currently impossible, to evaluate reliably as soon as n is larger than around 50--70.
sumBinomMpfr(n, f, n0 = 0, alternating = TRUE, precBits = 256,
             f.k = f(mpfr(k, precBits=precBits)))| n | upper summation index (integer). | 
| f | 
 | 
| n0 | lower summation index, typically  | 
| alternating | logical indicating if the sum is alternating, see below. | 
| precBits | the number of bits for MPFR precision, see
 | 
| f.k | can be specified instead of  | 
The alternating binomial sum sB(f,n) := sumBinom(n, f, n0=0) is equal to the n-th forward difference operator Δ^n f,
sB(f,n) = Δ^n f,
where
Delta^n f = sum(k = n0:n ; (-1)^(n-k) choose(n,k) * f(k)),
is the n-fold iterated forward difference Δ f(x) = f(x+1) - f(x) (for x = 0).
The current implementation might be improved in the future, notably
for the case where
sB(f,n)=sumBinomMpfr(n, f, *) is to be computed for a whole sequence
n = 1,...,N.
an mpfr number of precision precBits.
s. If alternating is true (as per default),
s = sum(k = n0:n ; (-1)^k choose(n,k) * f(k)),
if alternating is false, the (-1)^k factor is dropped (or
replaced by 1) above.
Martin Maechler, after conversations with Christophe Dutang.
Wikipedia (2012) The N\"orlund-Rice integral, https://en.wikipedia.org/wiki/Rice_integral
Flajolet, P. and Sedgewick, R. (1995) Mellin Transforms and Asymptotics: Finite Differences and Rice's Integrals, Theoretical Computer Science 144, 101–124.
chooseMpfr, chooseZ from package gmp.
## "naive" R implementation:
sumBinom <- function(n, f, n0=0, ...) {
  k <- n0:n
  sum( choose(n, k) * (-1)^(n-k) * f(k, ...))
}
## compute  sumBinomMpfr(.) for a whole set of 'n' values:
sumBin.all <- function(n, f, n0=0, precBits = 256, ...)
{
  N <- length(n)
  precBits <- rep(precBits, length = N)
  ll <- lapply(seq_len(N), function(i)
           sumBinomMpfr(n[i], f, n0=n0, precBits=precBits[i], ...))
  sapply(ll, as, "double")
}
sumBin.all.R <- function(n, f, n0=0, ...)
   sapply(n, sumBinom, f=f, n0=n0, ...)
n.set <- 5:80
system.time(res.R   <- sumBin.all.R(n.set, f = sqrt)) ## instantaneous..
system.time(resMpfr <- sumBin.all  (n.set, f = sqrt)) ## ~ 0.6 seconds
matplot(n.set, cbind(res.R, resMpfr), type = "l", lty=1,
        ylim = extendrange(resMpfr, f = 0.25), xlab = "n",
        main = "sumBinomMpfr(n, f = sqrt)  vs.  R double precision")
legend("topleft", leg=c("double prec.", "mpfr"), lty=1, col=1:2, bty = "n")Please choose more modern alternatives, such as Google Chrome or Mozilla Firefox.