C program for secant method to find roots of polynomial - c

Question

Here is the question:
Write a program to find the real roots of the following equation using the Secant method:
f(x) = 23x^4 -13x^3 + 3x^2 - 5x + 38.
Let h = r(i) - r(i-1), where r(i) is the root computed in iteration i of your program. Your program should continue to refine its answer until h < 10 - 3. This value is known as the convergence criterion. Your program should print the final value of the root and the number of iterations required to compute it.
Here is my program:
#include <stdio.h>
double function ( double i );
int main ()
{
double x_1, x_2, h_x, temp, a, b;
int count = 0;
printf("Enter first approximation: ");
scanf("%lf", &x_1);
printf("Enter second approximation: ");
scanf("%lf", &x_2);
a = function ( x_1 );
b = function ( x_2 );
h_x = x_2 - x_1;
if ( h_x < 0 )
{
h_x = ( h_x < 0 )? - h_x: h_x;
}
while ( h_x >= ( 1.E-3 ) && count <= 999999999 )
{
a = function ( x_1 );
b = function ( x_2 );
temp = ( ( x_1 * b ) - ( x_2 * a ) / ( b - a ) );
x_1 = x_2;
x_2 = temp;
printf("%lf\n", x_1);
printf("%lf\n", x_2);
count += 1;
h_x = x_2 - x_1;
if ( h_x < 0 )
{
h_x = ( h_x < 0 )? - h_x: h_x;
}
}
printf("Final value of root is %lf\n", x_2);
printf("%d iterations were required to compute it\n", count);
return 0;
}
double function ( double i )
{
double result;
result = ( 23 * i * i * i * i ) - ( 13 * i * i * i ) + ( 3 * i * i ) - ( 5 * i ) + 38;
return result;
}
The problem with my code is that it doesn't work for any other input that is different from 0 and 1.
I don't see what's the problem with my code though, I have added a code for absolute error, and the formula looks right to me. It isn't logical to me that with different initial guesses, the root returned is different, and just loops forever.
I would just like to know if it is a coding mistake, or a math error. Or is it that the roots oscillate? I just don't know what's wrong with my program. Can anyone help me out? Thanks.

Answer 1

Your midpoint formula is off. It seems to be a mix of two variants of the formula. The formula close to the derivation as root of the secant line
y = f(x_2)+(x-x2) * (f(x2)-f(x1))/(x2-x1)
gives
x3 = x2 - f(x2)*(x2-x1)/(f(x2)-f(x1)).
Bringing everything into one fraction gives the equivalent
x3 = (x1*f(x2) -x2*f(x1)) / (f(x2)-f(x1)).
You get a shorter formula for the polynomial using the Newton-Horner trick,
f(x) = ((( 23 * x - 13 ) * x + 3 ) * x - 5 ) * x + 38.
You should avoid the repeated computation of the same quantity, even if it does not have any noticeable effect here. Declare variables for the values f(x1) and f(x2).
Use the absolute value function to compute absolute values. hx = fabs(x2-x1). Or use hx=(hx<0)?-hx:hx; to avoid including math.h.