28.3 Example of accelerating a series

The following code calculates an estimate of \zeta(2) = \pi^2 / 6 using the series,

\zeta(2) = 1 + 1/2^2 + 1/3^2 + 1/4^2 + ...

After N terms the error in the sum is O(1/N), making direct summation of the series converge slowly.

#include <stdio.h> #include <gsl/gsl_math.h> #include <gsl/gsl_sum.h> #define N 20 int main (void) { double t[N]; double sum_accel, err; double sum = 0; int n; gsl_sum_levin_u_workspace * w = gsl_sum_levin_u_alloc (N); const double zeta_2 = M_PI * M_PI / 6.0; /* terms for zeta(2) = \sum_{n=0}^{\infty} 1/n^2 */ for (n = 0; n < N; n++) { double np1 = n + 1.0; t[n] = 1.0 / (np1 * np1); sum += t[n]; } gsl_sum_levin_u_accel (t, N, w, &sum_accel, &err); printf("term-by-term sum = % .16f using %d terms\n", sum, N); printf("term-by-term sum = % .16f using %d terms\n", w->sum_plain, w->terms_used); printf("exact value = % .16f\n", zeta_2); printf("accelerated sum = % .16f using %d terms\n", sum_accel, w->terms_used); printf("estimated error = % .16f\n", err); printf("actual error = % .16f\n", sum_accel - zeta_2); gsl_sum_levin_u_free (w); return 0; }

The output below shows that the Levin u-transform is able to obtain an estimate of the sum to 1 part in 10^10 using the first eleven terms of the series. The error estimate returned by the function is also accurate, giving the correct number of significant digits.

bash$ ./a.out term-by-term sum = 1.5961632439130233 using 20 terms term-by-term sum = 1.5759958390005426 using 13 terms exact value = 1.6449340668482264 accelerated sum = 1.6449340668166479 using 13 terms estimated error = 0.0000000000508580 actual error = -0.0000000000315785

Note that a direct summation of this series would require 10^10 terms to achieve the same precision as the accelerated sum does in 13 terms.