Lagrange interpolating polynomial

最新推荐文章于 2024-06-24 16:37:34 发布

原创最新推荐文章于 2024-06-24 16:37:34 发布 · 951 阅读

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CC 4.0 BY-SA版权

本文通过一个具体的例子展示了拉格朗日插值法的实现过程。使用了三个已知的数据点来估算未知点的函数值，并给出了详细的计算步骤及最终结果。

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#include "interpolating.h"
#include <iostream>

Interpolating::Interpolating()
{
  int n = 2;
  float* x = new float[n+1];
  x[0] = 1.0f;
  x[1] = 4.0f;
  x[2] = 6.0f;
  
  float* y = new float[n+1];
  y[0] = 0;
  y[1] = 1.386294f;
  y[2] = 1.791760f;

  float xx = 2.0f;

  float interpolatingSum = Lagrng(x, y, n, xx);
  std::cout << "f(2)="<< interpolatingSum <<std::endl;
}

Interpolating::~Interpolating()
{
}

float Interpolating:: Lagrng(float* x, float* y, int n, float xx)
{
  float sum = 0.0f;

  for(int i = 0; i <= n; i++) {
    float product = y[i];

    std::cout << "x[" << i << "]=" << x[i]
	      << ", y[" << i << "]=" << y[i] << std::endl;
    
    for(int j = 0; j <= n; j++) {
      if (i != j) {
	product = product * (xx - x[j]) / (x[i] - x[j]);
      }
    }

    sum = sum + product;
  }

  return sum;
}

results:

x[0]=1, y[0]=0
x[1]=4, y[1]=1.38629
x[2]=6, y[2]=1.79176
f(2)=0.565844

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