The factorial of a nonnegative integer n is written n ! (pronounced "n factorial") and is defined as follows: n ! = n · (n - 1)

Question

13 days ago

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The factorial of a nonnegative integer n is written n ! (pronounced "n factorial") 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). For example, 5! = 5 · 4 · 3 · 2 · 1, which is 120.
a) Write a program that reads a nonnegative integer and computes and prints its factorial.
b) Write a 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 program that computes the value of by using the formula
e^x = 1 + x/1! + x^2/2! + x^3/3! +.......

Computers and Technology

1 answer:

Harlamova29_29 [932] · Answer 1 · 2026-03-23T02:42:40+03:00

Here are the programs. I have written C++ and Python scripts:

a)

C++

#include<iostream>

using namespace std;

int factorial(int num) {

if (num == 0)

return 1;

return num * factorial(num - 1); }

int main() {

int integer;

cout<<"Enter a non negative integer: ";

cin>>integer;

cout<< "Factorial of "<< integer<<" is "<< factorial(integer)<< endl; }

Python:

def factorial(num):

if num == 0:

return 1

return num * factorial(num-1)

integer = int(input("Enter a non negative integer: "))

print("Factorial of", integer, "is", factorial(integer))

b)

C++

#include <iostream>

using namespace std;

double factorial(int number) {

if (number == 0)

return 1;

return number * factorial(number - 1); }

double estimate_e(int num){

double e = 1;

for(int i = 1; i < num; i++)

e = e + 1/factorial(i);

cout<<"e: "<< e; }

int main(){

int term;

cout<<"Enter a term to evaluate: ";

cin>>term;

estimate_e(term);}

Python:

def factorial(number):

if number == 0:

return 1

return number * factorial(number-1)

def estimate_e(term):

if not term:

return 0

else:

return (1 / factorial(term-1)) + estimate_e(term-1)

number = int(input("Enter how many terms to evaluate "))

print("e: ", estimate_e(number))

c)

C++

#include <iostream>

using namespace std;

int main(){

float terms, sumSeries, series;

int i, number;

cout << " Input the value of x: ";

cin >> number;

cout << " Input number of terms: ";

cin >> terms;

sumSeries = 1;

series = 1;

for (i = 1; i < terms; i++) {

series = series * number / (float)i;

sumSeries = sumSeries + series; }

cout << " The sum is: " << sumSeries << endl; }

Python

def ePowerx(number,terms):

sumSeries = 1

series =1

for x in range(1,terms):

series = series * number / x;

sumSeries = sumSeries + series;

return sumSeries

num = int(input("Enter a number: "))

term=int(input("Enter a number: "))

print("e^x: ",ePowerx(num,term))

Explanation:

a)

The program includes a factorial method that takes a number as an argument and calculates its factorial using recursion. For instance, if number = 3

The base case occurs at if (number == 0)

and the recursion is handled with return number * factorial(number - 1);

With number = 3 not equaling zero, the function calls itself recursively to get the factorial of 3

return 3* factorial(3- 1);

3 * factorial(2)

3* [2* factorial(2- 1) ]

3 * 2* [ factorial(1)]

3 * 2 * [1* factorial(1- 1) ]

3 * 2 * 1* [factorial(0)]

At this point at factorial(0), the base condition is satisfied as number==0, so factorial(0) returns 1

The resulting output is:

3 * 2 * 1* 1

yielding 6

So, the final program output will be

Factorial of 3 is 6

b)

The estimate_e method takes a number, termed as num, which signifies the term to estimate the mathematical constant e

The for loop extends through each term. For example, if num is set to 3

Then the core statement:

e = e + 1/factorial(i);

The preceding calculation works as:

e = 1 + 1/1! +1/2!

Since the term count is 3

Initially, e is set to 1

i is initialized at 1

Inserting this into the calculation gives us:

e = 1 + 1/factorial(1)

The factorial function computes and returns 1, as the factorial of 1 is 1. Thus,

e = 1 + 1/1

This results in e = 2

Proceeding to the next iteration, where i = 2 and e = 2, we calculate e = 2 + 1/factorial(2)

Thus, e = 2 + 1/2 results in e = 2.5

Following to the next iteration with i = 3, we have e = 3 + 1/factorial(3)

This yields e = 3 + 1/6 resulting in approximately e = 3.16666

Therefore, the output is:

e: 3.16666

c)

This program calculates the sum of a series based on the formula:

e^x = 1 + x/1! + x^2/2! + x^3/3! +...

The for loop iterates according to the number set for terms. Assuming x is 2, and the number of terms is set to 3, the series would read:

e^2 = 1 + 2/1! + 2^2/2!

In this setup: number = 2 and terms = 3

Initial values for series and sumSeries are both 1

Starting with i equal to 1, the update statement series = series * number / (float)i; applies as follows:series = 1 * 2 /1 results in series = 2

Then, for sumSeries, we have sumSeries = sumSeries + series; Outputs sumSeries as 1 + 2, yielding 3

Continuing to the next iteration: i=2, with series = 2 and sumSeries = 3, we recalculate as series = 2 * 2/2 imposing series = 2 again. Thus, we find: sumSeries = 3 + 2 giving a final sumSeries value of 5

After the loop concludes, the result shows the value of sumSeries, leading finally to the output value of 5