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<title>GNU Scientific Library – Reference Manual: Example of accelerating a series</title>
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<a name="Example-of-accelerating-a-series"></a>
<div class="header">
<p>
Next: <a href="Series-Acceleration-References.html#Series-Acceleration-References" accesskey="n" rel="next">Series Acceleration References</a>, Previous: <a href="Acceleration-functions-without-error-estimation.html#Acceleration-functions-without-error-estimation" accesskey="p" rel="previous">Acceleration functions without error estimation</a>, Up: <a href="Series-Acceleration.html#Series-Acceleration" accesskey="u" rel="up">Series Acceleration</a> [<a href="Function-Index.html#Function-Index" title="Index" rel="index">Index</a>]</p>
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<hr>
<a name="Examples-22"></a>
<h3 class="section">30.3 Examples</h3>
<p>The following code calculates an estimate of <em>\zeta(2) = \pi^2 / 6</em>
using the series,
After <var>N</var> terms the error in the sum is <em>O(1/N)</em>, making direct
summation of the series converge slowly.
</p>
<div class="example">
<pre class="verbatim">#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;
}
</pre></div>
<p>The output below shows that the Levin <em>u</em>-transform is able to obtain an
estimate of the sum to 1 part in
<em>10^10</em> 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.
</p>
<div class="example">
<pre class="example">$ ./a.out
</pre><pre class="verbatim">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
</pre></div>
<p>Note that a direct summation of this series would require
<em>10^10</em> terms to achieve the same precision as the accelerated
sum does in 13 terms.
</p>
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