Python牛顿插值法的解决办法

这篇文章主要为大家详细介绍了Python牛顿插值法的简单示例，具有一定的参考价值，可以用来参考一下。

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一、牛顿多项式

拉格朗日多项式的公式不具备递推性，每个多项式需要单独构造。但很多时候我们需要从若干个逼近多项式选择一个。这个时候我们就需要一个具有递推关系的方法来构造——牛顿多项式

这里的的a0，a1…等可以通过逐一带入点的值来求得。但是当项数多起来时，会发现式子变得很大，这个时候我们便要引入差商的概念（利用差分思想）具体见式子能更好理解

这里在编程实现中我们可以推出相应的差商推导方程

d(k,0)=y(k)d(k,j)=(d(k,j-1)-d(k-1,j-1)) / (x(k)-x(k-j))

二、例题

【问题描述】考虑[0,3]内的函数y=f(x)=cos(x)。利用多个（最多为6个）节点构造牛顿插值多项式。【输入形式】在屏幕上依次输入在区间[0,3]内的一个值x*，构造插值多项式后求其P(x*)值，和多个节点的x坐标。【输出形式】输出牛顿插值多项式系数向量，差商矩阵，P(x*)值（保留6位有效数字），和与真实值的绝对误差（使用科学计数法，保留小数点后6位有数字）。【样例1输入】0.80 0.5 1【样例1输出】-0.429726-0.029972111 0 00.877583 -0.244835 00.540302 -0.674561 -0.4297260.7009984.291237e-03【样例1说明】输入：x为0.8，3个节点为(k, cos(k)),其中k = 0, 0.5, 1。输出：牛顿插值多项式系数向量，表示P2(x)=-0.429726x^2 - 0.0299721x + 1；3行3列的差商矩阵；当x为0.8时，P2(0.8)值为0.700998与真实值的绝对误差为：4.291237*10^(-3)【评分标准】根据输入得到的输出准确

三、ACcode:

C++(后面还有python代码）

代码如下：

/*
 * @Author: csc
 * @Date: 2021-04-30 08:52:45
 * @LastEditTime: 2021-04-30 11:57:46
 * @LastEditors: Please set LastEditors
 * @Description: In User Settings Edit
 * @FilePath: \code_formal\course\cal\newton_quo.cpp
 */
#include <bits/stdc++.h>
#define pr printf
#define sc scanf
#define for0(i, n) for (i = 0; i < n; i++)
#define for1n(i, n) for (i = 1; i <= n; i++)
#define forab(i, a, b) for (i = a; i <= b; i++)
#define forba(i, a, b) for (i = b; i >= a; i--)
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define int long long
#define pii pair<int, int>
#define vi vector<int>
#define vii vector<vector<int>>
#define vt3 vector<tuple<int, int, int>>
#define mem(ara, n) memset(ara, n, sizeof(ara))
#define memb(ara) memset(ara, false, sizeof(ara))
#define all(x) (x).begin(), (x).end()
#define sq(x) ((x) * (x))
#define sz(x) x.size()
const int N = 2e5 + 100;
const int mod = 1e9 + 7;
namespace often
{
    inline void input(int &res)
    {
        char c = getchar();
        res = 0;
        int f = 1;
        while (!isdigit(c))
        {
            f ^= c == '-';
            c = getchar();
        }
        while (isdigit(c))
        {
            res = (res << 3) + (res << 1) + (c ^ 48);
            c = getchar();
        }
        res = f ? res : -res;
    }
    inline int qpow(int a, int b)
    {
        int ans = 1, base = a;
        while (b)
        {
            if (b & 1)
                ans = (ans * base % mod + mod) % mod;
            base = (base * base % mod + mod) % mod;
            b >>= 1;
        }
        return ans;
    }
    int fact(int n)
    {
        int res = 1;
        for (int i = 1; i <= n; i++)
            res = res * 1ll * i % mod;
        return res;
    }
    int C(int n, int k)
    {
        return fact(n) * 1ll * qpow(fact(k), mod - 2) % mod * 1ll * qpow(fact(n - k), mod - 2) % mod;
    }
    int exgcd(int a, int b, int &x, int &y)
    {
        if (b == 0)
        {
            x = 1, y = 0;
            return a;
        }
        int res = exgcd(b, a % b, x, y);
        int t = y;
        y = x - (a / b) * y;
        x = t;
        return res;
    }
    int invmod(int a, int mod)
    {
        int x, y;
        exgcd(a, mod, x, y);
        x %= mod;
        if (x < 0)
            x += mod;
        return x;
    }
}
using namespace often;
using namespace std;

int n;

signed main()
{
    auto polymul = [&](vector<double> &v, double er) {
        v.emplace_back(0);
        vector<double> _ = v;
        int m = sz(v);
        for (int i = 1; i < m; i++)
            v[i] += er * _[i - 1];
    };
    auto polyval = [&](vector<double> const &c, double const &_x) -> double {
        double res = 0.0;
        int m = sz(c);
        for (int ii = 0; ii < m; ii++)
            res += c[ii] * pow(_x, (double)(m - ii - 1));
        return res;
    };

    int __ = 1;
    //input(_);
    while (__--)
    {
        double _x, temp;
        cin >> _x;
        vector<double> x, y;
        while (cin >> temp)
            x.emplace_back(temp), y.emplace_back(cos(temp));
        n = x.size();
        vector<vector<double>> a(n, vector<double>(n));
        int i, j;
        for0(i, n) a[i][0] = y[i];
        forab(j, 1, n - 1) forab(i, j, n - 1) a[i][j] = (a[i][j - 1] - a[i - 1][j - 1]) / (x[i] - x[i - j]);
        vector<double> v;
        v.emplace_back(a[n - 1][n - 1]);
        forba(i, 0, n - 2)
        {
            polymul(v, -x[i]);
            int l = sz(v);
            v[l - 1] += a[i][i];
        }

        for0(i, n)
            pr("%g\n", v[i]);
        for0(i, n)
        {
            for0(j, n)
                pr("%g ", a[i][j]);
            puts("");
        }
        double _y =  polyval(v, _x);
        pr("%g\n", _y);
        pr("%.6e",fabs(_y-cos(_x)));
    }

    return 0;
}

分析Python牛顿插值法

python代码

代码如下：

'''
Author: csc
Date: 2021-04-29 23:00:57
LastEditTime: 2021-04-30 09:58:07
LastEditors: Please set LastEditors
Description: In User Settings Edit
FilePath: \code_py\newton_.py
'''
import numpy as np


def difference_quotient(x, y):
    n = len(x)
    a = np.zeros([n, n], dtype=float)
    for i in range(n):
        a[i][0] = y[i]
    for j in range(1, n):
        for i in range(j, n):
            a[i][j] = (a[i][j-1]-a[i-1][j-1])/(x[i]-x[i-j])
    return a


def newton(x, y, _x):
    a = difference_quotient(x, y)
    n = len(x)
    s = a[n-1][n-1]
    j = n-2
    while j >= 0:
        s = np.polyadd(np.polymul(s, np.poly1d(
            [x[j]], True)), np.poly1d([a[j][j]]))
        j -= 1
    for i in range(n):
        print('%g' % s[n-1-i])
    for i in range(n):
        for j in range(n):
            print('%g' % a[i][j], end=' ')
        print()
    _y = np.polyval(s, _x)
    print('%g' % _y)
    # re_err
    real_y = np.cos(_x)
    err = abs(_y-real_y)
    print('%.6e' % err)


def main():
    _x = float(input())
    x = list(map(float, input().split()))
    y = np.cos(x)
    newton(x, y, _x)


if __name__ == '__main__':
    main()

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