# Continued Fraction Expansion

The continued fraction expansion gives us the answer to life's most persistent questions, like what is sin(x) or even tanh(x). This surprisingly short code allows you to estimate SIN, COS, TAN, EXP, SINH, COSH and TANH fairly accurately. Careful, this code is not for the usual TV crowd! You should at least know what trigonometry is, and have a mild interest in mathematics.

``````// Continued Fraction Expansion yields seven transcendental
// values from one function.
// Original Turbo C, modified and tested with Pelles C  vegaseat  8oct2004

#include <stdio.h>   // in/out functions
#include <math.h>   // math functions

#define SIN   1
#define COS   2
#define TAN   3
#define EXP   4
#define SINH  5
#define COSH  6
#define TANH  7
#define PI    3.1415926536  // or use atan(1)*4.0

double trigfunc(int p, int n, double x);

//
// test trigfunc() against compiler functions
//
int main()
{
double x;

x = 30 * PI/180;  // convert 30 degrees to radians
puts("Continued Fraction Expansion can yield seven transcendental values ... \n");
printf("x = 30 degrees = %11.10f radians \n\n",x);
printf("compiler sin(x)    = %11.10f \n",sin(x));
printf("trigfunc(SIN,2,x)  = %11.10f \n",trigfunc(SIN,2,x));
printf("trigfunc(SIN,3,x)  = %11.10f \n\n",trigfunc(SIN,3,x));
printf("compiler cos(x)    = %11.10f \n",cos(x));
printf("trigfunc(COS,3,x)  = %11.10f \n\n",trigfunc(COS,3,x));
printf("compiler tan(x)    = %11.10f \n",tan(x));
printf("trigfunc(TAN,3,x)  = %11.10f \n\n",trigfunc(TAN,3,x));
printf("compiler exp(x)    = %11.10f \n",exp(x));
printf("trigfunc(EXP,3,x)  = %11.10f \n\n",trigfunc(EXP,3,x));
printf("compiler sinh(x)   = %11.10f \n",sinh(x));
printf("trigfunc(SINH,3,x) = %11.10f \n\n",trigfunc(SINH,3,x));
printf("compiler cosh(x)   = %11.10f \n",cosh(x));
printf("trigfunc(COSH,3,x) = %11.10f \n\n",trigfunc(COSH,3,x));
printf("compiler tanh(x)   = %11.10f \n",tanh(x));
printf("trigfunc(TANH,3,x) = %11.10f \n\n",trigfunc(TANH,3,x));
getchar();
return 0;
}

//
// generate 7 transcendental values with one function
// p = function selector,  #define the following:
// SIN 1   COS 2   TAN 3   EXP 4   SINH 5   COSH 6   TANH 7
// n = convergence cutoff, n = 3 is typical,
// higher n gives higher precision, but slower execution
//
double trigfunc(int p, int n, double x)
{
int     k;
double  r, s, t;

if (p <= 3)
{
r = - x * x;    // trig
}
else
{
r = x * x;      // hyperbolic
}
s = 4 * n + 2;
for (k = n; k > 0; k--)
{
s = 4 * k - 2 + r/s;
}
switch (p % 4)
{
case 0 : t = (s + x)/(s - x);         // exp
break;
case 1 : t = 2 * x * s/(s * s - r);   // sin, sinh
break;
case 2 : t = (s * s + r)/(s * s - r); // cos, cosh
break;
case 3 : t = 2 * x * s/(s * s + r);   // tan, tanh
break;
}
return (t);
}``````
bumsfeld 413

Even my professor of mathematics couldn't believe this easy algorithm.

Be a part of the DaniWeb community

We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.