[JAVA程式語言]例題2-7 已知方程式 e^x + x^-2 + 2 cosx -6 利用正割法 找出f(x)=0的根=? TOL = 0.00001

[JAVA程式語言]例題2-7 例題2-7 已知方程式 e^x + x^-2 + 2 cosx -6 利用正割法 找出f(x)=0的根=?

The method[edit]

The first two iterations of the secant method. The red curve shows the function f, and the blue lines are the secants. For this particular case, the secant method will not converge to the visible root.

The secant method is defined by the recurrence relation

${\displaystyle x_{n}=x_{n-1}-f(x_{n-1}){\frac {x_{n-1}-x_{n-2}}{f(x_{n-1})-f(x_{n-2})}}={\frac {x_{n-2}f(x_{n-1})-x_{n-1}f(x_{n-2})}{f(x_{n-1})-f(x_{n-2})}}.}$

As can be seen from the recurrence relation, the secant method requires two initial values, x₀ and x₁, which should ideally be chosen to lie close to the root.

Derivation of the method[edit]

Starting with initial values x₀ and x₁, we construct a line through the points (x₀, f(x₀)) and (x₁, f(x₁)), as shown in the picture above. In slope–intercept form, the equation of this line is

${\displaystyle y={\frac {f(x_{1})-f(x_{0})}{x_{1}-x_{0}}}(x-x_{1})+f(x_{1}).}$

The root of this linear function, that is the value of x such that y = 0 is

${\displaystyle x=x_{1}-f(x_{1}){\frac {x_{1}-x_{0}}{f(x_{1})-f(x_{0})}}.}$

We then use this new value of x as x₂ and repeat the process, using x₁ and x₂ instead of x₀ and x₁. We continue this process, solving for x₃, x₄, etc., until we reach a sufficiently high level of precision (a sufficiently small difference between x_n and x_n−1):

${\displaystyle {\begin{aligned}x_{2}&=x_{1}-f(x_{1}){\frac {x_{1}-x_{0}}{f(x_{1})-f(x_{0})}},\\[6pt]x_{3}&=x_{2}-f(x_{2}){\frac {x_{2}-x_{1}}{f(x_{2})-f(x_{1})}},\\[6pt]&\,\,\,\vdots \\[6pt]x_{n}&=x_{n-1}-f(x_{n-1}){\frac {x_{n-1}-x_{n-2}}{f(x_{n-1})-f(x_{n-2})}}.\end{aligned}}}$

程式

/* ex2-7.java Secant Method is similar to Newton-Raphson

 *  Method used for find solutions to f(x)=0 given

 *  initial approximations x0 and x1.

例題2-7 例題2-7 已知方程式 e^x + x^-2 + 2 cosx -6 利用正割法 找出f(x)=0的根=?

*/

public class Main {

    double fx(double x1) {

    return (Math.exp(x1)+1/Math.pow(2,x1)+2*Math.cos(x1)-6);

}

public static void main(String args[]){

    Main fun = new Main();

    final int MAX = 50;  /* maximum iterations */

    final double TOL = 0.00001;/* maximum iterations */

        int i=2;

        double x0,x1,x,q0,q1;

        x0=1.8;

        x1=2.0;

        q0=fun.fx(x0);

        q1=fun.fx(x1);

        System.out.printf("i       xi           f(x)\n");

        System.out.printf("%-2d   %10.6f   %10.6f\n",0,x0,q0);

        System.out.printf("%-2d   %10.6f   %10.6f\n",1,x1,q1);

        

        while(i<=MAX) {

            x=x1-q1*(x1-x0)/(q1-q0);

            System.out.printf("%-2d   %10.6f   %10.6f\n",i,x,fun.fx(x));

            if(Math.abs(x-x1) < TOL) {

                System.out.printf("The Root=%10.6f f(%10.6f)=%10.6f\n",x,x,fun.fx(x));

                break;

        }

            else {

            i=i+1;

            x0=x1;

            q0=q1;

            x1=x;

            q1=fun.fx(x);

            }

        }

        if(i>MAX){

            System.out.printf("Secant Method faileds!!!\n");

        }    

}

}

    輸出畫面

i       xi           f(x)

0      1.800000    -0.117582

1      2.000000     0.806762

2      1.825441    -0.016116

3      1.828860    -0.002147

4      1.829385     0.000007

5      1.829384    -0.000000

The Root=  1.829384 f(  1.829384)= -0.000000