如何使用拉格朗日插值计算多项式系数

Question

我需要使用拉格朗日插值多项式计算多项式的系数,作为我的作业,我决定在Javascript中执行此操作.

这里是拉格朗日多项式(L(x))的定义

拉格朗日基多项式定义如下

计算特定X(W(x)函数)的y值很简单,但我需要计算多项式的系数([a0,a1,...,an]的数组)我需要这样做n <= 10但它将有任意n很好,然后我可以将该函数放入horner函数并绘制该多项式.

我有在第一个等式中计算分母的函数

function denominator(i, points) {
   var result = 1;
   var x_i = points[i].x;
   for (var j=points.length; j--;) {
      if (i != j) {
        result *= x_i - points[j].x;
      }
   }
   return result;
}

和使用horner方法返回y的函数(我也有使用canvas的绘图功能)

function horner(array, x_scale, y_scale) {
   function recur(x, i, array) {
      if (i == 0) {
         return x*array[0];
      } else {
         return array[i] + x*recur(x, --i, array);
      }
   }
   return function(x) {
      return recur(x*x_scale, array.length-1, array)*y_scale;
   };
}

任何人都知道这样做的算法,或者想知道如何计算这些系数

Answer 1

好吧,你可以用天真的方式做到这一点.通过其系数数组(阵列)表示多项式

[a_0,a_1,...,a_n]

对应于a_0 + a_1*X + ... + a_n*X^n.我对JavaScript不好,所以伪代码必须这样做:

interpolation_polynomial(i,points)
    coefficients = [1/denominator(i,points)]
    for k = 0 to points.length-1
        if k == i
            next k
        new_coefficients = [0,0,...,0] // length k+2 if k < i, k+1 if k > i
        if k < i
            m = k
        else
            m = k-1
        for j = m downto 0
            new_coefficients[j+1] += coefficients[j]
            new_coefficients[j] -= points[k]*coefficients[j]
        coefficients = new_coefficients
    return coefficients

从常数多项式开始1/((x_1-x_0)* ... *(x_i-x_{i-1})*(x_i-x_{i+1})*...*(x_i-x_n))并与X - x_k所有相乘k != i.因此,给出L _i的系数,然后你只需将它们与y _i相乘(你可以通过初始化coefficients为y_i/denominator(i,points)y ,如果你将y值作为参数传递)并最终将所有系数加在一起.

polynomial = [0,0,...,0] // points.length entries
for i = 0 to points.length-1
    coefficients = interpolation_polynomial(i,points)
    for k = 0 to points.length-1
        polynomial[k] += y[i]*coefficients[k]

计算每个L _i是O(n²),因此总计算为O(n³).

更新:在你的jsFiddle中,你在多项式乘法循环中有一个错误,除了(现在已经纠正的)错误和我做的起始索引,它应该是

for (var j= (k < i) ? (k+1) : k; j--;) {
     new_coefficients[j+1] += coefficients[j];
     new_coefficients[j] -= points[k].x*coefficients[j];
}

由于您j在测试时会减少,因此需要更高一些.

这还没有产生正确的插值,但它至少比以前更明智.

另外,在你的horner功能中,

function horner(array, x_scale, y_scale) {
   function recur(x, i, array) {
      if (i == 0) {
         return x*array[0];
      } else {
         return array[i] + x*recur(x, --i, array);
      }
   }
   return function(x) {
      return recur(x*x_scale, array.length-1, array)*y_scale;
   };
}

你将最高系数乘以两次x,它应该是

if (i == 0) {
    return array[0];
}

代替.尽管如此,仍然没有好结果.

Update2:最终错字修复,以下工作:

function horner(array, x_scale, y_scale) {
   function recur(x, i, array) {
      if (i == 0) {
         return array[0];
      } else {
         return array[i] + x*recur(x, --i, array);
      }
   }
   return function(x) {
      return recur(x*x_scale, array.length-1, array)*y_scale;
   };
}

// initialize array
function zeros(n) {
   var array = new Array(n);
   for (var i=n; i--;) {
     array[i] = 0;
   }
   return array;
}

function denominator(i, points) {
   var result = 1;
   var x_i = points[i].x;
   for (var j=points.length; j--;) {
      if (i != j) {
        result *= x_i - points[j].x;
      }
   }
    console.log(result);
   return result;
}

// calculate coefficients for Li polynomial
function interpolation_polynomial(i, points) {
   var coefficients = zeros(points.length);
    // alert("Denominator " + i + ": " + denominator(i,points));
   coefficients[0] = 1/denominator(i,points);
    console.log(coefficients[0]);
    //new Array(points.length);
   /*for (var s=points.length; s--;) {
      coefficients[s] = 1/denominator(i,points);
   }*/
   var new_coefficients;

   for (var k = 0; k<points.length; k++) {
      if (k == i) {
        continue;
      }
      new_coefficients = zeros(points.length);
       for (var j= (k < i) ? k+1 : k; j--;) {
         new_coefficients[j+1] += coefficients[j];
         new_coefficients[j] -= points[k].x*coefficients[j];
      }   
      coefficients = new_coefficients;
   }
   console.log(coefficients);
   return coefficients;
}

// calculate coefficients of polynomial
function Lagrange(points) {
   var polynomial = zeros(points.length);
   var coefficients;
   for (var i=0; i<points.length; ++i) {
     coefficients = interpolation_polynomial(i, points);
     //console.log(coefficients);
     for (var k=0; k<points.length; ++k) {
       // console.log(points[k].y*coefficients[k]);
        polynomial[k] += points[i].y*coefficients[k];
     }
   }
   return polynomial;
}