Program \"Euler\'s number\" The Eulers number e is a mathematical constant appro

Question

Program "Euler's number" The Eulers number e is a mathematical constant approximated at 2. 718. The formula for computing it is as follows: e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + . where n! = nx(n-l)x(n-2)x. . xl, e. g. , 4!=4x3x2xl Write a program euler. c that approximates the value of e by using the following equation: e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + . + 1/n! where n is a number entered by the user. Your program will first compute e as 1, then as 1 + 1/1!, then as 1 + 1/1! + 1/2! and so on. In each step, print the step number and the value of e, as follows: 2. 000000 2. 500000 2. 666666

Explanation / Answer

#include <stdio.h>

int main()
{
int i,j,n;
float product=1,e=1;
printf("enter the n value ");
scanf("%d",&n);
for(i=1;i<=n;i++)
{
product = 1;
for(j=1;j<=i;j++)
{
product*= j;
}
e+=(1/product);
}
printf("e = %f ",e);
return 0;
}

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