如何在Java中使用牛顿多项式从给定的x和f（x）点生成插值函数？

这是我的密码：

Double[] x = {1.0, 2.0, 3.0};
Double[] fx = {1.0, 8.0, 27.0};
s = x.length;

Double[][] newton = new Double[(2*s)-1][s+1];
    for(int i=0, z=0;i<s;i++,z+=2){
        newton[z][0]=x[i];
        newton[z][1]=fx[i];
    }

int i=1, ii=2, j=2, ss=s;
for(int z=0;z<s-1;z++,j++,ss-=1,ii++){
        for(int y=0;y<ss-1;y++,i+=2){
            newton[i][j]=(newton[i+1][j-1]-newton[i-1][j-1])/(newton[i+(ii-1)][0]-newton[i-(ii-1)][0]);
        }
        i=ii;
    }
}

这是一个分差表

---

然后，我想生成它的插值多项式函数。因此，在上面的例子中，使用牛顿多项式规则，输出函数应该是1+7（x-1）+6（x-1）（x-2）=6x^2-11x+6。我真的被困在这个问题上了，有人能帮我生成这样的输出吗？

根据我的理解，从你的微分和x-位置，你想要牛顿多项式的系数，对吗？ 下面是一个小技巧：

public class PolynomProduct {

    public static void main(String[] args) {
        double[] values = {1.0, 2.0, 3.0};
        double[] diffs = {1.0, 7.0, 6.0};

        // Initialize result array
        double[] result = new double[values.length];
        for (int i = 0; i < values.length; ++i) {
            result[i] = 0.0;
        }

        for (int i = 0; i < values.length; ++i) {
            // 'poly' has a degree 'i'. We use 'i - 1' because only terms
            // from 0 to 'i - 1' are used
            double[] poly = getPoly(values, i - 1);

            // Now add to result, do not forget to multiply by the divided
            // difference !
            for (int j = 0; j < poly.length; ++j) {
                result[j] += poly[j] * diffs[i];
            }
        }

        for (int i = 0; i < result.length; ++i) {
            System.out.println("Coef for x^" + i + " is: " + result[i]);
        }
    }

    public static double[] getPoly(double[] values, int i) {
        // Start poly: 1.0, neutral value for multiplication
        double[] coefs = {1.0};

        // Accumulate values of products
        for (int j = 0; j <= i; ++j) {
            // 'coefsLocal' represent polynom of 1st degree (x - values[j])
            double[] coefsLocal = {-values[j], 1.0};
            coefs = getPolyProduct(coefs, coefsLocal);
        }

        return coefs;
    }

    public static double[] getPolyProduct(double[] coefs1, double[] coefs2) {
        // Get lengths and degree
        int s1 = coefs1.length - 1;
        int s2 = coefs2.length - 1;
        int degree = s1 + s2;

        // Initialize polynom resulting from product, with null values
        double[] coefsProduct = new double[degree + 1];
        for (int k = 0; k <= degree; ++k) {
            coefsProduct[k] = 0.0;
        }

        // Compute products
        for (int i = 0; i <= s1; ++i)   {
            for (int j = 0; j <= s2; ++j)   {
                coefsProduct[i + j] += coefs1[i] * coefs2[j];
            }
        }
        return coefsProduct;
    }
}

---

这就是你期望的多项式

6-11x+6x^2

。

多项式插值的关键点是你永远不会计算多项式的系数。每次需要值时，都必须计算除法差分算法。原因是1）其复杂性非常小（O（d^2），d为插值次数，通常小于10），2）多项式的系数通常较大且具有交替符号，因此对其进行评估会导致灾难性的对消错误。如果你想一劳永逸地计算高次多项式，看看切比雪夫插值。即使性能不是问题，我建议你在这里使用

double

而不是

double

。我也建议你考虑如何使代码更清晰，而不是使用一个字母变量。@亚历山大：谢谢，但我想要的是使用牛顿的方法。这就是为什么我把它作为一个评论。如果你在做家庭作业，那么好吧，如果你打算插入一些真实的东西，不要这样做。

public class PolynomProduct {

    public static void main(String[] args) {
        double[] values = {1.0, 2.0, 3.0};
        double[] diffs = {1.0, 7.0, 6.0};

        // Initialize result array
        double[] result = new double[values.length];
        for (int i = 0; i < values.length; ++i) {
            result[i] = 0.0;
        }

        for (int i = 0; i < values.length; ++i) {
            // 'poly' has a degree 'i'. We use 'i - 1' because only terms
            // from 0 to 'i - 1' are used
            double[] poly = getPoly(values, i - 1);

            // Now add to result, do not forget to multiply by the divided
            // difference !
            for (int j = 0; j < poly.length; ++j) {
                result[j] += poly[j] * diffs[i];
            }
        }

        for (int i = 0; i < result.length; ++i) {
            System.out.println("Coef for x^" + i + " is: " + result[i]);
        }
    }

    public static double[] getPoly(double[] values, int i) {
        // Start poly: 1.0, neutral value for multiplication
        double[] coefs = {1.0};

        // Accumulate values of products
        for (int j = 0; j <= i; ++j) {
            // 'coefsLocal' represent polynom of 1st degree (x - values[j])
            double[] coefsLocal = {-values[j], 1.0};
            coefs = getPolyProduct(coefs, coefsLocal);
        }

        return coefs;
    }

    public static double[] getPolyProduct(double[] coefs1, double[] coefs2) {
        // Get lengths and degree
        int s1 = coefs1.length - 1;
        int s2 = coefs2.length - 1;
        int degree = s1 + s2;

        // Initialize polynom resulting from product, with null values
        double[] coefsProduct = new double[degree + 1];
        for (int k = 0; k <= degree; ++k) {
            coefsProduct[k] = 0.0;
        }

        // Compute products
        for (int i = 0; i <= s1; ++i)   {
            for (int j = 0; j <= s2; ++j)   {
                coefsProduct[i + j] += coefs1[i] * coefs2[j];
            }
        }
        return coefsProduct;
    }
}

Coef for x^0 is: 6.0
Coef for x^1 is: -11.0
Coef for x^2 is: 6.0