Secant method

	1. The method:


	This method  requires guessing the two first values close to the root x0 
	and x1 for the solution of the equation f(x) = 0. The secant from the point 
	(x0, f(x0)) to the point (x1, f(x1)) cross 
	the x axis at the point (x2, 0)). From this x-intercept, we get the point 
	(x2, f(x2)). 
	The second iteration consists of drawing the secant from the point (x1, f(x1)) 
	to the point (x2, f(x2)). The related x-intercept of this secant  is x3. 
	From this, we get the point (x3, f(x3)). The third iteration consists of 
	drawing the secant from the point (x2, f(x2)) to the point 
	(x3, f(x3)). The related x-intercept of this secant is x4, 
	... and so on until we get an x-intercept closer to the root of the function f(x); that is 
	a sufficiently high level of precision corresponding to a small difference between xn and xn-1.

	The secant related to the segment 0 and x1, can be found as follow:

	
	y ax + b 

	f(x0) = a x0 + b 	   (1)
	f(x1) = a x1 + b 	   (2)

	Subtracting (1) from  (2), yields:
	f(x1) -  f(x0) = a (x1 - x0), 
	then: a = (f(x1) -  f(x0))/(x1 - x0)

	Subtracting (1) from  2 x (2), yields:

	2 f(x1) -  f(x0) = a (2 x1 - x0) + b
	then:
	b =   2 f(x1) -  f(x0) - a (2 x1 - x0) 

	Substituting the expression of "a" yields:
	b =  2 f(x1) -  f(x0) - (f(x1) -  f(x0)) 
	- x1 (f(x1) -  f(x0))/(x1 - x0) 

	=  f(x1) - x1 (f(x1) -  f(x0))/(x1 - x0) 

	Then:

	y ax + b  = x (f(x1) -  f(x0))/(x1 - x0) + 
	f(x1) - x1 (f(x1) -  f(x0))/(x1 - x0) ) = 


	=   (x - x1) (f(x1) -  f(x0))/(x1 - x0) + 
	f(x1) 

	For y = 0, we have x2:

	x2= -  f(x1) (x1 - x0)/ (f(x1) -  f(x0)) + x1


	The next will have the following expression:

	x3= -  f(x2) (x2 - x1)/ (f(x2) -  f(x1)) + x2

	...

	The nth x-intercept is:
	
	xn=  xn-1 -  f(xn-1) (xn-1 - xn-2)/ (f(xn-1) -  f(xn-2)) 
	


	2. Example in clanguage
	

	#include 
	#include 
	#include 

	double g = 9.81;
	float y = 500, to, t1, z = 0.05;


	// First function --------
	float f(float t)
	{
	return 	y - (g/2)*t*t;
	}

	typedef float (*PFUN) (float);

	float secant (PFUN f, float to, float t1){
	float ss = 0;
	float c[21];
	c[0] = to ;
	c[1] = t1 ;

	int i=1,n = 20;
	for (i =1; i<=n; i++){
	c[i+1] = c[i] - (c[i] - c[i-1])*f(c[i])/(f(c[i]) - f(c[i-1]));
	printf("%f \n", c[i+1]);
	if (abs((c[i] - c[i-1])/c[i-1]) <= z/100.0 )
	{
	printf("\nThe solution is t =  %f seconds.\n", c[i+1]);
	ss= c[i];
	i = n+1;
	}
	}
	return ss;
	}

	// ---------------------The main -------------

	int main()
	{
	printf("\n Input values for t0 and t1 (in seconds) : --> ");
	scanf("%f%f", &to, &t1);
	secant (&f,to,t1);
	return 0;
	}
	


	The execution gives:
	-------------------
	secant_free_fall
	10 11

	Input values for t0 and t1 (in seconds) : --> 10.092229 

	The solution is t =  10.092229 seconds.
	>Exit code: 0