c语言编程题:要求用泰勒级数求e的近似值。直到最后一项小于1e—6为止,并且将文件写入到磁盘中去。 C语言的程序设计:用泰勒级数求e的近似值,直到最后一项小于1...

c\u8bed\u8a00\uff0c\u7528\u6cf0\u52d2\u7ea7\u6570\u6c42e\u7684\u8fd1\u4f3c\u503c\u3002\u76f4\u5230\u6700\u540e\u4e00\u9879\u5c0f\u4e8e1e-6\u4e3a\u6b62\u3002

#include
"stdio.h"
void
main\uff08\uff09
{
float
e=1.0,t=1.0\uff1b
int
n=1;
while(1/t>(1e-6))
//(1/n!)<(1e-6)\u662f\u7ed3\u675f\uff0c\u4f60\u4e66\u5199\u6709\u9519\uff0ci\u5e94\u662f1
{
t=t*n;
//\u5f97\u5230
n!
e=e+1/t;
//\u7d2f\u52a0(1/n!)\uff0c\u5373\u8ba1\u7b97\u6b64\u884c\u524d
e
=
1+(1/1!)+(1/2!)+....+(1/(n-1)!)\u518d\u52a0\u4e0a(1\u4fef\u653b\u7898\u7eaa\u9122\u614c\u7889\u82c7\u51ac\u4fe947;n!)
n++;
//n
:
2
3
4
...
}
printf("e=%8.6f\n",e);
}

#include "stdio.h"
void main()
{double sum=1,k=1,t=1;
int i=1;
while(k>=1e-6)
{ t=t*i;
k=1.0*i/t;
sum+=k;
i++;
}
printf("e=%lf\n",sum);
}

#include<stdio.h>
int main()
{
    float i=1,e=0;
    int p = 1;
    e+=1/i;
    do
    {
        i*=(p++);
        e+=1/i;
    }while((1/i)>1e-9);
    printf("%f",e);
    FILE* fp = fopen("result.txt","w");
    fprintf(fp,"%f",e);
    fclose(fp);
}

会输出到result.txt磁盘文件中



#include<stdio.h>

void main()
{
int i = 1;
int t = 1;
double e = 1;
do
{
t *= i++;
e += 1.0 / t;
}
while(t < 10e6);
printf("%.6f
",e);
FILE * fp = fopen("e.txt", "w");
if(fp)
{
fprintf(fp, "%.6f
", e);
fclose(fp);
}
else printf("File open error!
");
}


#include<stdio,h>
main()
{
float i=1,e=0;
e+=1/i;
do
{
i=i*(i+1)
e+=1/i;
}while((1/i)<10e-6)
printf("%f",e)
}

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