Question

给定这种计算x的余弦的方法：

public static double myCos(double x){
        double alteSumme, neueSumme, summand;
        int j = 0;
        neueSumme = 1.0;
        summand = 1.0;
        do
        {
            j++;
            summand *= -x * x / j;
            j++;
            summand /= j;
            alteSumme = neueSumme;
            neueSumme += summand;
        } while ( neueSumme != alteSumme );
        return alteSumme;
    }

我得到大x的不准确结果：

我必须将域名从实数转换为区间 [0，PI / 2] 以获得准确的结果。

所以，我写了这个方法：

public static double transform(double x){
        x = x%(2*Math.PI);
        if(!(0<=x&&x<Math.PI)){
            x = -(x+Math.PI);
        }
        if(!(0<=x&&x<Math.PI/2)){
            x = -(x-2*Math.abs(x-Math.PI/2));
        }
        return x;
    }

它应分三步完成。

将域从reals转换为[0,2PI）。
将域名从[0,2PI]转换为[0，PI）。
将域名从[0，PI）转换为[0，PI / 2]

但不知何故，它会产生错误的结果。

修改

public class Main {

    public static double transform(double x) {
        x = x % (2 * Math.PI);
        if (!(0 <= x && x < Math.PI)) {
            x = -(x + Math.PI);
        }
        if (!(0 <= x && x < Math.PI / 2)) {
            x = -(x - 2 * Math.abs(x - Math.PI / 2));
        }
        return x;
    }

    public static double myCos(double x) {

        x = transform(x);

        double alteSumme, neueSumme, summand;
        int j = 0;
        neueSumme = 1.0;
        summand = 1.0;
        do {
            j++;
            summand *= -x * x / j;
            j++;
            summand /= j;
            alteSumme = neueSumme;
            neueSumme += summand;
        } while (neueSumme != alteSumme);
        return alteSumme;
    }

    public static void main(String[] args) {
        int n = 0;
        int k =50;
        for (double y = k * Math.PI; y <= (k + 2) * Math.PI; y += Math.PI/4) {

            System.out.println(n + ":" + y + ": " + myCos(y) + " " + Math.cos(y));
            n++;
        }
    }
}

Answer 1

<强>更新以下是transform仅适用于cos的实现。

/**
 * @param x any value
 * @return x within [0..2PI)
public static double transform(double x) {
    x = Math.abs(x);   // We can do this because Cosine is symmetric around the y axis.
    double y = Math.floor(x / (Math.PI * 2));
    return x - y * Math.PI * 2;
}

根据要求，这是我的测试类：

package cosine;

public class Main {

    public static double transform(double x) {
        x = Math.abs(x);
        double y = Math.floor(x / (Math.PI * 2));
        return x - y * Math.PI * 2;
    }

    public static double myCos(double x) {

        x = transform(x);

        double alteSumme, neueSumme, summand;
        int j = 0;
        neueSumme = 1.0;
        summand = 1.0;
        do {
            j++;
            summand *= -x * x / j;
            j++;
            summand /= j;
            alteSumme = neueSumme;
            neueSumme += summand;
        } while (neueSumme != alteSumme);
        return alteSumme;
    }

    public static void main(String[] args) {
        int n = 0;
        for (double y = -20 * Math.PI; y < 20 * Math.PI; y += Math.PI / 3) {
            double x = Math.PI / 3 + y;
            double tmpa, tmpb;
            System.out.println(
                    n + ":" + x + ": " + (tmpa = myCos(x)) + " " + (tmpb = Math.cos(x)) + "  DIFF: " + (tmpa - tmpb));
            n++;
        }
    }
}

<强>更新

仅通过计算余弦超过0..π：

，这是一个小小的改进

public static double myCos(double x) {
    // Cosine is symmetric around the Y axis: get rid of the sign.
    x = Math.abs(x);

    // Calculate the number of times 2*PI fits in x
    double y = Math.floor(x / (Math.PI * 2));

    // and subtract that many 2*PI
    x -= y * Math.PI * 2;
    // x is now within 0 and 2*PI.

    // The PI..2PI range is the negated version of 0..PI.
    double sign = 1;
    if ( x > Math.PI ) {
       sign = -1;
       // mirror x in the line x=Math.PI:
       x =  - x + Math.PI; // or: Math.PI * 2 - x
    }

    /* cosine approximation ... */

    return alteSumme * sign;
}

提高余弦精度

1 个答案: