Continued fraction natural logarithm(number of iterations needed to calculate right logarithm)

Question

I have problem with my continued fraction algorithm for natural logarithm. I need to calculate natural logarithm for example ln(0.31) with accuracy on 1e-6 in 6 iterations, my algorithm will do it in 8.

This is my implementation:

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

double c_frac_log(double x, unsigned int n)
{
    double z=(x-1)/(x+1);
    double zz = z*z,res=0;
    double cf = 1;
    for (int i = n; i >= 1; i--) 
    {
        cf = (2*i-1) - i*i*zz/cf;
    }
    res=2*z/cf;
    return res;
}

int c_frac_eps(double x,double eps)
{
    int a=0;
    double loga=log(x),fraclog=c_frac_log(x,a); 
    double roz=(loga-fraclog);
    roz=fabs(roz);
    for(a=0;roz >= eps;a++)
    {   
        fraclog=c_frac_log(x,a);    
        roz=(loga-fraclog);
        roz=fabs(roz);
    }
    return a-1;
}

int main()
{
    double x=0.31,eps=0.000001;
    printf("c_frac_log (%0.4f) =%0.12f    \n",x,c_frac_log(x,c_frac_eps(x,eps)));
    printf("math.h - log(%0.4f)=%0.12f\n",x,log(x));
    printf("minimum of iterations with accuracy %f is %d\n",eps,c_frac_eps(x,eps));

    return 0;
}

Do any of you have some idea how to refine my code?

Answer 1

The initial cf affects the end result - a little bit.

In a previous comment, found that cf = 1.88*n-0.95; yielded a better result than double cf = 1;

This was found by reversing the algorithm and finding a correlation with n. YMMV.

Original results

c_frac_log (0.3100) =-1.171182812362    
math.h - log(0.3100)=-1.171182981503
minimum of iterations with accuracy 0.000001 is 8

With this change

c_frac_log (0.3100) =-1.171183069158    
math.h - log(0.3100)=-1.171182981503
minimum of iterations with accuracy 0.000001 is 6

6 beats 8 and meets OP's "in 6 iterations".

---

Note: type consistency:

double c_frac_log(double x, unsigned int n) { 
  ...
  // for (int i = n; i >= 1; i--) {
  for (unsigned i = n; i >= 1; i--) {