How to implement natural logarithm with continued

问题:

Here I have a little problem. Create something from this formula:

This is what I have, but it doesn't work. Franky, I really don't understand how it should work.. I tried to code it with some bad instructions. N is number of iteration and parts of fraction. I think it leads somehow to recursion but don't know how.

Thanks for any help.

double contFragLog(double z, int n)
{
    double cf = 2 * z;
    double a, b;
    for(int i = n; i >= 1; i--)
    {
        a = sq(i - 2) * sq(z);
        b = i + i - 2;
        cf = a / (b - cf);

    }
    return (1 + cf) / (1 - cf);
}

回答1:

The central loop is messed. Reworked. Recursion not needed either. Just compute the deepest term first and work your way out.

double contFragLog(double z, int n) {
  double zz = z*z;
  double cf = 1.0;  // Important this is not 0
  for (int i = n; i >= 1; i--) {
    cf = (2*i -1) - i*i*zz/cf;
  }
  return 2*z/cf;
}

void testln(double z) {
  double y = log((1+z)/(1-z));
  double y2 = contFragLog(z, 8);
  printf("%e %e %e\n", z, y, y2);
}

int main() {
  testln(0.2);
  testln(0.5);
  testln(0.8);
  return 0;
}

Output

2.000000e-01 4.054651e-01 4.054651e-01
5.000000e-01 1.098612e+00 1.098612e+00
8.000000e-01 2.197225e+00 2.196987e+00

---

[Edit]

As prompted by @MicroVirus, I found double cf = 1.88*n - 0.95; to work better than double cf = 1.0;. As more terms are used, the value used makes less difference, yet a good initial cf requires fewer terms for a good answer, especially for |z| near 0.5. More work could be done here as I studied 0 < z <= 0.5. @MicroVirus suggestion of 2*n+1 may be close to my suggestion due to an off-by-one of what n is.

This is based on reverse computing and noting the value of CF[n] as n increased. I was surprised the "seed" value did not appear to be some nice integer equation.

回答2:

Here's a solution to the problem that does use recursion (if anyone is interested):

#include <math.h>
#include <stdio.h>

/* `i` is the iteration of the recursion and `n` is
   just for testing when we should end. 'zz' is z^2 */
double recursion (double zz, int i, int n) {
  if (!n)
    return 1;

  return 2 * i - 1 - i * i * zz / recursion (zz, i + 1, --n);
}

double contFragLog (double z, int n) {
  return 2 * z / recursion (z * z, 1, n);
}

void testln(double z) {
  double y = log((1+z)/(1-z));
  double y2 = contFragLog(z, 8);
  printf("%e %e %e\n", z, y, y2);
}

int main() {
  testln(0.2);
  testln(0.5);
  testln(0.8);
  return 0;
}

The output is identical to the solution above:

2.000000e-01 4.054651e-01 4.054651e-01
5.000000e-01 1.098612e+00 1.098612e+00
8.000000e-01 2.197225e+00 2.196987e+00