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Program to Calculate e^x by Recursion
  • Difficulty Level : Medium
  • Last Updated : 23 Apr, 2021

The value of the Exponential function can be calculated using Taylor Series. 
 

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

To find its value using recursion, we will use static variables. For the power of x, we will use p and for factorials, we will use f as static variables. 
The function shown below is used to increase the power of x. 
 

p = p*x 

The function below is used to find factorials. 
 

f = f*n

The function below is used to calculate the summation of the series. 
 

r+p/f

Where r is the recursive call to the function.
Below is the implementation of the above idea. 
 



C++




// C++ implementation of the approach
#include <stdio.h>
  
// Recursive Function with static
// variables p and f
double e(int x, int n)
{
    static double p = 1, f = 1;
    double r;
  
    // Termination condition
    if (n == 0)
        return 1;
  
    // Recursive call
    r = e(x, n - 1);
  
    // Update the power of x
    p = p * x;
  
    // Factorial
    f = f * n;
  
    return (r + p / f);
}
  
// Driver code
int main()
{
    int x = 4, n = 15;
    printf("%lf \n", e(x, n));
  
    return 0;
}

Java




// Java implementation of the approach
import java.text.*;
  
class GFG
{
      
// Recursive Function with static
// variables p and f
static double p = 1, f = 1;
static double e(int x, int n)
{
    double r;
  
    // Termination condition
    if (n == 0)
        return 1;
  
    // Recursive call
    r = e(x, n - 1);
  
    // Update the power of x
    p = p * x;
  
    // Factorial
    f = f * n;
  
    return (r + p / f);
}
  
// Driver code
public static void main (String[] args) 
{
    int x = 4, n = 15;
    DecimalFormat df = new DecimalFormat("0.######");
    System.out.println(df.format(e(x, n)));
  
}
}
  
// This code is contributed by mits

Python3




# Python implementation of the approach
  
# Recursive Function 
# global variables p and f
p = 1.0
f = 1.0
  
def e(x, n) :
  
    global p, f
      
    # Termination condition
    if (n == 0) :
        return 1
      
    # Recursive call
    r = e(x, n - 1)
      
    # Update the power of x
    p = p * x
      
    # Factorial
    f = f * n
      
    return (r + p / f)
  
# Driver code
  
x = 4
n = 15
print(e(x, n))
  
# This contributed by ihritik

C#




// C# implementation of the approach
using System;
  
class GFG
{
      
// Recursive Function with static
// variables p and f
static double p = 1, f = 1;
static double e(int x, int n)
{
    double r;
  
    // Termination condition
    if (n == 0)
        return 1;
  
    // Recursive call
    r = e(x, n - 1);
  
    // Update the power of x
    p = p * x;
  
    // Factorial
    f = f * n;
  
    return (r + p / f);
}
  
// Driver code
static void Main()
{
    int x = 4, n = 15;
    Console.WriteLine(Math.Round(e(x, n),6));
  
}
}
  
// This code is contributed by mits

Javascript




<script>
  
// Javascript implementation of the approach
  
// Recursive Function with static
// variables p and f
p = 1, f = 1;
function e(x, n)
{
    var r;
  
    // Termination condition
    if (n == 0)
        return 1;
  
    // Recursive call
    r = e(x, n - 1);
  
    // Update the power of x
    p = p * x;
  
    // Factorial
    f = f * n;
  
    return (r + p / f);
}
  
// Driver Code
var x = 4, n = 15;
var res = e(x, n); 
  
document.write(res.toFixed(6));
  
// This code is contributed by kirti
  
</script>
Output: 
54.597883

 

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