编写函数fun，功能是计算如下多项式：s=1+x+x**2/2!+x**3/3!+•••+x**n/n!

Sure! 下面是函数fun的代码，可以计算给定n和x下的多项式值：

import math

def fun(n, x):
    s = 0
    for i in range(1, n+1):
        s += (x ** i) / math.factorial(i)
    return s

希望能够帮到您！如果您还有其他问题，欢迎随时提出。

相关问题

编写函数fun，计算下列多项式：s=1+x+x**2/2!+x**3/3!+•••+x**n/n!

Sure, I can help you with that! Here is the function:

def fun(x, n):
    s = 1
    fact = 1
    for i in range(1, n+1):
        fact *= i
        s += (x ** i) / fact
    return s

Simply pass in the value of `x` and the number of terms `n` you want to use in the approximation. The function will return the value of the sum of the series.

Hope this helps!

/*请编写函数fun,其功能是计算并输出如下多项式的值: Sn=l+1/1!+1/2!+1/3!+1/4!+…+1/n!

*/#include <stdio.h>

double fun(int n)
{
double result = 0, factorial = 1;
int i;

for (i = 1; i <= n; i++) {
factorial *= i;
result += 1 / factorial;
}

return result;
}

int main()
{
int n;

printf("请输入n的值：");
scanf("%d", &n);
printf("Sn的值为：%lf\n", fun(n));

return 0;
}