计算的总和1+（1/2！）+ ...（1 / N！）n个C语言 [英] Calculate sum of 1+(1/2!)+…+(1/n!) n number in C language

问题描述

就像标题说，我是如何计算的形式正数的总和：（！1/2）1+ +⋯（1 / N！）？我已经得到了code为调和级数：

Like the title say, how I calculate the sum of n number of the form: 1+(1/2!)+⋯(1/n!)? I already got the code for the harmonic series:

#include <stdio.h>

int main( void )
{
    int v=0,i,ch;
    double x=0.;

    printf("Introduce un número paracalcular la suma: ");
    while(scanf("%d",&v)==0 || v<=0)
    {
        printf("Favor de introducir numeros reales positivos: ");
        while((ch=getchar())!='\n')
            if(ch==EOF)
                return 1;
    }
    for (i=v; i>=1; i--)
        x+=1./i;

    printf("EL valor de la serie es %f\n", x);
    getch();
    return 0;
}

这里的问题是..我已经得到的总和的比例，但如何使变量i阶乘？

The question here is.. I already got the sum as the fraction, but how make the variable "i" factorial?

注：I'm编程语言C，与DEV - C ++ 4.9.9.2

Note: I´m programming in language C, with DEV -C++ 4.9.9.2

推荐答案

您得到的谐波求和1 / I + 1 ./（I-1）... 1/1稍微更准确的答案。建议你留在这个顺序。

You got a slightly more accurate answer for the harmonic summing 1./i + 1./(i-1) ... 1./1. Suggest you stay with that order.

改写：感谢@ pablo197您指出我的方式错误

[edit] Rewrite: Thanks to @pablo197 for pointing out the error of my ways.

要计算谐波的和的1+（1/2！）+ ... +（1 / N！），继续总结了至少显著术语以及第一因为这有助于最大限度地减少precision亏损。用最少的显著期限开始 1 / N 为之，的那笔和n-1项是：和=（1 +总和）/（N-1）等。 （见下文）

To calculate harmonic and 1+(1/2!)+…+(1/n!), continue summing the least significant terms together first as that helps to minimize precision loss. Starting with the least significant term 1/n as sum, sum of that and the n-1 term is : sum = (1 + sum)/(n-1) and so on. (See below)

double x = 0.0;
double one_over_factorial_series = 0.0;
for (i = v; i >= 1; i--) {
  x += 1.0/i;
  one_over_factorial_series = (one_over_factorial_series + 1)/i;
}
printf("harmonic:%le\n", x); 
// 2.828968e+00
printf("one_over_factorial:%.10le\n", one_over_factorial_series); 
// 1.7182815256e+00

添加 1.0 或 1/0！到 one_over_factorial_series ，结果约 E = 2.7182818284 ...

详细说明如何直接N！避免了计算。

Detail showing how direct n! calculation is avoided.

1 + (1/2!) + … + (1/n!) =  
1/n!  +  1/((n-1)!)   +  1/((n-2)!)  +  1/((n-3)!)  + ... + 1 =  
(1/n + 1)/((n-1)!)    +  1/((n-2)!)  +  1/((n-3)!)  + ... + 1 =  
((1/n + 1)/(n-1) + 1)/((n-2)!)       +  1/((n-3)!)  + ... + 1 =  
...
((((1/n + 1)/(n-1) + 1)/(n-2) + 1)/(n-3) + 1)/(n-4) + ... =  

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