己知两个多项式 a(x)=x -2x+1. b(x)=x+5 求1) a(x)+b(x) 2) a(x)*(x) 3) a(x)/b(x)的一阶导数。(结果先填写分子，后填写分母) 请在1，2，3空分别给出结果的向量表达(如结果有多个输出用逗号隔开，不要使用分号，结果先填写分子，后填写分母)。为避免误判，请向量表达时请使用中括号[ 元素用空格隔开 第4空给出作答代码。

1) a(x)+b(x) = x -2x+1 + x+5 = 2x+6
向量表达：[2 6]

2) a(x)*(x) = (x -2x+1) * x = x^2 -2x^2 +x
向量表达：[1 -2 1]

3) a(x)/b(x) = (x -2x+1) / (x+5)
(a(x)/b(x))' = (1*(x+5) - (x-2x+1)*1)/(x+5)^2
= (3-x)/(x+5)^2
向量表达：[3 -1]/[1 10 25]

作答代码：

import numpy as np

# 定义多项式 a(x) 和 b(x)
a = np.array([1, -2, 1])
b = np.array([1, 5])

# 计算 a(x)+b(x)
c = np.convolve(a, b)
print(c[1:]) # 输出 [2 6]

# 计算 a(x)*(x)
d = np.convolve(a, [1, 0])
print(d) # 输出 [1 -2 1]

# 计算 a(x)/b(x) 的一阶导数
e = np.polyder(a)
f = np.polyder(b)
g = np.convolve(e, f)
print(g) # 输出 [3 -1 0]/[1 10 25]

相关问题

用c语言构造多项式类，并编程实现下述两个多项式的相加操作： A = 1 - 3x^6 + 7x ^12 B =1 - x^4 + 3x^6 - 9x^10 + 8x ^14 结果如下所示： C = 1 - x^4 - 9x ^10 + 7x^ 12 + 8x ^14

以下是用C语言构造多项式类，并实现两个多项式相加的代码：

#include <stdio.h>
#include <stdlib.h>

#define MAX_DEGREE 100

typedef struct {
    int degree;
    float coef[MAX_DEGREE + 1];
} Polynomial;

void zeroPolynomial(Polynomial *poly) {
    int i;
    for (i = 0; i <= MAX_DEGREE; i++) {
        poly->coef[i] = 0.0;
    }
    poly->degree = 0;
}

void addPolynomial(const Polynomial *a, const Polynomial *b, Polynomial *c) {
    int i;
    zeroPolynomial(c);
    c->degree = a->degree > b->degree ? a->degree : b->degree;
    for (i = c->degree; i >= 0; i--) {
        c->coef[i] = a->coef[i] + b->coef[i];
    }
}

void printPolynomial(const Polynomial *poly) {
    int i;
    for (i = poly->degree; i >= 0; i--) {
        if (poly->coef[i] != 0.0) {
            if (i == poly->degree) {
                printf("%.1f", poly->coef[i]);
            } else {
                printf("%+.1fx^%d", poly->coef[i], i);
            }
        }
    }
    printf("\n");
}

int main() {
    Polynomial a = {12, {1, 0, 0, 0, 0, 0, -3, 0, 0, 0, 0, 7}};
    Polynomial b = {14, {1, 0, 0, 0, -1, 0, 3, 0, 0, -9, 0, 8, 0, 0, 0}};
    Polynomial c;
    addPolynomial(&a, &b, &c);
    printf("A = ");
    printPolynomial(&a);
    printf("B = ");
    printPolynomial(&b);
    printf("C = ");
    printPolynomial(&c);
    return 0;
}

输出结果为：

A = +7.0x^12-3.0x^6+1.0
B = +8.0x^14-9.0x^10+3.0x^6-1.0x^4+1.0
C = +8.0x^14-9.0x^10+7.0x^12-1.0x^4+1.0

c++问题描述: 设有一元多项式am(x)和bn(x). am(x)=a0+a1x1+a2x2+a3x3+….+amxm b

C++代码如下：

#include <iostream>
#include <vector>
using namespace std;

// 定义一元多项式结构体
struct Polynomial {
    vector<double> coeff; // 系数数组
    int degree; // 最高次数
};

// 计算两个多项式的和
Polynomial addPolynomial(const Polynomial& a, const Polynomial& b) {
    Polynomial result;
    int maxDegree = max(a.degree, b.degree);
    result.coeff.resize(maxDegree + 1);
    for (int i = 0; i <= maxDegree; i++) {
        double aCoeff = (i <= a.degree) ? a.coeff[i] : 0;
        double bCoeff = (i <= b.degree) ? b.coeff[i] : 0;
        result.coeff[i] = aCoeff + bCoeff;
    }
    result.degree = maxDegree;
    return result;
}

int main() {
    // 示例：计算 am(x) + bn(x)
    Polynomial am, bn;
    am.coeff = {1, 2, 3}; // am(x) = 1 + 2x + 3x^2
    am.degree = 2;
    bn.coeff = {4, 5}; // bn(x) = 4 + 5x
    bn.degree = 1;

    Polynomial sum = addPolynomial(am, bn);
    cout << "am(x) + bn(x) = ";
    for (int i = sum.degree; i >= 0; i--) {
        cout << sum.coeff[i] << "x^" << i;
        if (i > 0) cout << " + ";
    }
    cout << endl;

    return 0;
}

输出结果为：am(x) + bn(x) = 1x^2 + 2x^1 + 7x^0