# C++ Program to Estimate value of 'e' and e^x

This is a simple code for the program to estimate value of e and e^x (e power x) using factorials. It is coded by building on three different programs.

- The first part finds the factorial of a given number.
- The second one incorporates this factorial code to estimate the value of 'e'.
- The third program incorporates the second one with some editions to estimate the value of e^x.

## C++ Program to find Factorial of a given Number

#include <iostream> using namespace std; int main () { int A, B, D=1; cout<<"Enter The Integer\n"; cin>>A; for (B=1; B<=A ; B++ ) { D = D * B; } cout<<"Factorial is : "<<D<<endl; system("pause"); }

Logic:We have to multiply that given number whose factorial is going to be computed by all the numbers less then that number. To do this we will be using a for loop and its condition will be set a that given number.

**Line 6:** All necessary variables are initialized, it is necessary to set D as 1 initially.

**Line 9:** Asks us to input that number whose factorial is to be found. It is stored in variable A.

**Line 11:** Initiates a for loop with A as its condition.

**Line 13:** This is the most important step. Due to the for loop B will have all the values from 1 till A and its value will increase by 1 each time the loop runs. D was already initialized as 1.

**Now Understand this bit**, first time the loop runs B is 2 it is multiplied by D which was 1 and the answer 2 is stored again in D. When the loop runs next time B is 3 and D is 2, they both are multiplied and result 6 is stored in D. It goes on like this until loop terminates and at that times D gives the value of the factorial.

## C++ Program to estimate the value of 'e'

#include <iostream> #include <iomanip> using namespace std; int main () { int A, B; double E, D; for (A=1; A<=15; A++ ) { D=1; for (B=1; B<=A ; B++ ) { D = D * B; } E = E + 1 / D; } cout<<setprecision(10)<<"Estimated Value of e is "<< E + 1 <<endl; system("pause");

We will use the formula **1 + 1/1! + 1/2! + 1/3!...** to estimate value of 'e' so see you have to use the factorial is this computation and hence the C++ code for finding factorial given above will be incorporated into it.

Logic:Note that the formula uses factorials of numbers which are increasing by a factor of 1 so just simply put the factorial code you wrote earlier into another for loop, the rest is just simple maths and the job is done.

**Line 2:** Includes the <iomanip> library file which includes the 'set precision' function.

**Line 8:** Now it is important to initialize our variables as 'doubles' because now calculation will involve decimal values.

**Line 10:** Initiates a for loop to calculate factorials from 1 to 15.

**Line 12-16:** The code for finding factorial is incorporated here. It becomes a nested for statement.

**Line 17:** Performs other calculation according to the formula.

**Line 19:** Sets the precision of the answer to 10 decimal places and prints the value of e which was stored in variable E. (we did E + 1 according to the formula in the last)

## C++ Program to estimate value of e^x

#include <iostream> #include <iomanip> #include <cmath> using namespace std; int main () { int A, B; double E, D; double X, Y; cout<<"Enter The Value of X to estimate e^X : "; cin>>X; for (A=1; A<=100; A++ ) { D=1; for (B=1; B<=A ; B++ ) { D = D * B; } Y = pow (X,A); E = E + Y / D; } cout<<setprecision(10)<<"Estimated Value of e^X is "<< E + 1 <<endl; system("pause"); }

We will use the formula **e^x = 1+x/1! + (x^2)/2! + (x^3)/3!....** to estimate value of e^x (e power x) so the C++ code used to estimate value of e will be incorporated into it.

Logic:Again we can achieve this by doing a few additions to our previous code. We only need to replace one which was being divided by the factorial with a variable X. We will enter this into program. Power of X can be calculated by the function pow(base,power). Where A is going to be the power which will increase by a factor of one due to the for loop.

**Line 3:** Includes the <cmath> library which contains the pow() function.

**Line 14:** Prompts for value of X.

**Line 16-22:** Code for estimating value of 'e' is incorporated here.

**Line 23:** Variable Y is assigned the value of pow(X,A) where A increases by a factor of 1 each time the loop runs.

**Line 24:** Calculation according to the given formula.

**Line 26:** Prints the estimated value of e^x after adding one to it like we did in our previous code.

**© 2014 StormsHalted**

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## Comments 2 comments

though I am a software engineer and my platform was in MFC and then core C++.Useful hub I must say..Good Job.

thank you so much! it helps a lot~ good job