bisection_method.cpp File Reference

Solve the equation \(f(x)=0\) using bisection method More...

#include <cmath>
#include <iostream>
#include <limits>

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Macros
#define	EPSILON    1e-6
#define	MAX_ITERATIONS   50000
	Maximum number of iterations to check.

Functions
static double	eq (double i)
template<typename T >
int	sgn (T val)
int	main ()

Detailed Description

Solve the equation \(f(x)=0\) using bisection method

Given two points \(a\) and \(b\) such that \(f(a)<0\) and \(f(b)>0\), then the \((i+1)^\text{th}\) approximation is given by:

\[ x_{i+1} = \frac{a_i+b_i}{2} \]

For the next iteration, the interval is selected as: \([a,x]\) if \(x>0\) or \([x,b]\) if \(x<0\). The Process is continued till a close enough approximation is achieved.

See also

newton_raphson_method.cpp, false_position.cpp, secant_method.cpp

Function Documentation

eq()

static double eq ( double  i )

static

define \(f(x)\) to find root for

                           {
    return (std::pow(i, 3) - (4 * i) - 9);  // original equation
}

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main()

int main

(

void

)

main function

           {
    double a = -1, b = 1, x, z;
    int i;
 
    // loop to find initial intervals a, b
    for (int i = 0; i < MAX_ITERATIONS; i++) {
        z = eq(a);
        x = eq(b);
        if (sgn(z) == sgn(x)) {  // same signs, increase interval
            b++;
            a--;
        } else {  // if opposite signs, we got our interval
            break;
        }
    }
 
    std::cout << "\nFirst initial: " << a;
    std::cout << "\nSecond initial: " << b;
 
    // start iterations
    for (i = 0; i < MAX_ITERATIONS; i++) {
        x = (a + b) / 2;
        z = eq(x);
        std::cout << "\n\nz: " << z << "\t[" << a << " , " << b
                  << " | Bisect: " << x << "]";
 
        if (z < 0) {
            a = x;
        } else {
            b = x;
        }
 
        if (std::abs(z) < EPSILON)  // stoping criteria
            break;
    }
 
    std::cout << "\n\nRoot: " << x << "\t\tSteps: " << i << std::endl;
    return 0;
}

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sgn()

template<typename T >

int sgn

(

T

val

)

get the sign of any given number

               {
    return (T(0) < val) - (val < T(0));
}