the factorial of non-negative integer n is written n! and
is defined as follows:
n!=n*(n-1)*(n-2)........1(for values of n greater than or
equal to 1 and
n!=1(for n=0)
Perform the following
1.write a c program that reads a non-negative integer and
computes and prints its factorial.
2. write a C program that estimates the value of the
mathematical constant e by using the formula:
e=1+1/!+1/2!+1/3!+....
3. write a c program the computes the value ex by using the
formula
ex=1+x/1!+xsquare/2!+xcube/3!+....
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Answer / shanthimathi
The factorial of non-negative integer n is written n! and is
defined as follows:?
n! = n * (n – 1) * (n – 2) . …. .1 (for values of n greater
than or equal to 1).
and
n! = 1 (for n =0).
Perform the following:
a) Write a C program that reads a non-negative integer and
computes and prints its factorial.
b) Write a C program that estimates the value of the
mathematical constant e by using the formula:
e = 1 + 1/1! + 1/2! + 1/3! + …..
c) Write a C program that computes the value ex by using the
formula
ex= 1 + x/1! + x2/2! + x3/3! + …..
Answer:
a) int main()
{
int num;
long unsigned int factorial=1;
printf("Enter a number to compute factorial : ");
scanf("%d",&num);
for(int i=0;i<num;num--){
factorial *= num;
}
printf("The factorial is %d\n",factorial);
return 0;
}
//use upto num = 10 or 12 I guess
(b) For this
int main(){
int num = 3;
float e_value=1;
float int factorial;
for(int i=0;i<num;num--){
factorial = 1;
for(int j=num;j>0;j--){
factorial *= j;
}
e_value += 1/factorial;
}
printf("The e value is %f\n",e_value);
return 0;
}
(c) for this the one you can figure out now I hope
int main(){
int num=3;
float e_value = 1;
float factorial;
int x;
printf("enter value of x : ");
scanf("%d",&x);
for(int i=0;i<num;num--){
factorial = 1;
for(int j=num;j>0;j--){
factorial *= j;
}
e_value += (x*num)/factorial;
}
printf("The value of ex is : %f",e_value);
return 0;
}
Is This Answer Correct ? | 35 Yes | 11 No |
Answer / dally
#include<stdio.h>
int main()
{
int i,n,sum =0;
printf("Enter value for n\n");
scanf("%d",&n);
sum = sum+1/fact(i);
printf("sum of Total result %d",sum);
}
int fact(int j)
{
int k =1;
if(k <= j)
fact = k*fact(--j);
return fact;
}
Is This Answer Correct ? | 16 Yes | 10 No |
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