通过递归计算e ^ x的程序

指数函数的值可以使用泰勒级数计算。

$e^x$ = 1 + x / 1！ + $x^2$ / 2！ + $x^3$ / 3！ + ……

为了使用递归来找到它的值，我们将使用静态变量。对于x的幂，我们将使用p，对于阶乘，我们将使用f作为静态变量。
如下所示的函数用于增加x的幂。

p = p*x 

下面的函数用于查找阶乘。

f = f*n

下面的函数用于计算序列的总和。

r+p/f

其中r是对函数的递归调用。

以下是上述想法的实现。

C++

// C++ implementation of the approach
#include 
  
// 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

输出：

54.597883