29.3 Examples

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=1}^{\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.

     $ ./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.