泰勒展开&牛顿迭代的例子

KataMathTaylorFormula.java

/**
 *
 * 1. 通过泰勒展开的方式求e、sin95
 * 2. 牛顿迭代法求平方根 (https://netedu.xauat.edu.cn/jpkc/netedu/jpkc2009/jsff/content/dzja/3.4.htm)
 */
public class KataMathTaylorFormula {


    public static void main(String[] args) {

        System.out.println(exp(1.5));
        System.out.println(Math.exp(1.5));

        System.out.println(sin95());
        System.out.println(Math.sin(Math.PI * 95.0 / 180));

        // sqrt:2.0, loopCount:5
        // sqrt:21.0, loopCount:7
        // sqrt:211.0, loopCount:9
        // sqrt:2111.0, loopCount:11
        // sqrt:21111.0, loopCount:12
        // sqrt:121111.0, loopCount:14
        // sqrt:2.121110001E9, loopCount:21
        // 收敛的速度还是蛮快的
        checkSqrt(2);
        checkSqrt(21);
        checkSqrt(211);
        checkSqrt(2111);
        checkSqrt(21111);
        checkSqrt(121111);
        checkSqrt(2121110001);
    }


    public static void checkSqrt(double n) {
        double mySqrt = sqrt(n);
        double mathSqrt = Math.sqrt(n);
        if (Math.abs(mathSqrt - mySqrt) > 0.0001) {
            System.out.println(mySqrt + ", mathSqrt:" + mathSqrt + ", diff:" + (mathSqrt - mySqrt));
        }
    }



    // e^x 在0展开，求e
    public static double exp(double x) {

        double exp = 1;
        long factorial = 1;
        double xn = 1;
        for (int i = 1; i < 10; ++i) {
            factorial *= i;
            xn *= x;
            exp += xn / factorial;
        }
        return exp;
    }


    // sinx 在 pi/2 展开，就sin(95度)
    public static double sin95() {

        double delta = Math.PI * 5.0 / 180;
        double result = 1;
        long factorial = 1;
        double deltaN = 1;
        double dsin90N = -1;
        for (int i = 1; i < 10; i++) {
            factorial *= i;
            deltaN *= delta;
            if (i % 2 == 0) {
                result += dsin90N * deltaN / factorial;
                dsin90N = -1 * dsin90N;
            }
        }
        return result;
    }



    // 通过牛顿迭代法求开平方根
    public static double sqrt(double n) {

        // f(x) = x^2-n
        // 做泰勒在x0展开，只考虑前前两项
        // f(x) = x0^2 -n  + 2*x0 (x-x0)
        // f(x) = 0，得x = x0 -  (x0*x0 -n) / (2*x0)
        // 迭代即可
        int loopCount = 1;
        double xn = n;
        while (Math.abs(xn * xn - n) > 0.000001 && loopCount++ < 100) {
            xn = xn - (xn * xn - n) / (2 * xn);
        }
        if (loopCount == 100) {
            throw new RuntimeException("不收敛:" + loopCount);
        }
        System.out.println("sqrt:" + n + ", loopCount:" + loopCount);
        return xn;
    }

}