The mathematical constant \"e\" can be expressed as an infinite series: e = 1 +

Question

The mathematical constant "e" can be expressed as an infinite series: e = 1 + 1/1! + 1/2! + 1/3! + ... Write a program that will print an approximate value of "e" by computing the series out to 1/n!, where n is an integer entered by the user. In other words, calculate: e = 1 + 1/1! + 1/2! + ... + 1/n! The Fibonacci numbers are 0, 1, 1, 2, 3, 5, 8, 13,... where each number (after the first two values) is the sum of the previous two numbers. Write a program that declares an integer array called fib_numbers with a length of 25 elements. The program then fills the array with the first 25 Fibonacci numbers using a loop. The program then prints the values to the screen using a separate loop. Write a function that determines the average of the values in an array. In the main() function declare a floating point array with a length of 8 elements, and have the user enter the value of each element using a loop. The program should then call and pass the array to the function, then display the average.

Explanation / Answer

2). Code :

#include <stdio.h>

double sumfunction(double);

main()

{

    double num,sum;

    printf(" Enter the value: ");

    scanf("%lf", &num);

    sum = sumfunction(num);

    printf(" Sum of the above series = %lf ", sum);

}

double sumfunction(double n)

{

    double result = 1, k= 1, i;

    for (i = 1; i <= n; i++)

    {

        k = k * i;

        result = result +(1 / k);

    }

    return(result);

}

output:

Enter the value: 4
Sum of the above series = 2.708333

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.