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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.

` = 1 + x/1! + /2! + /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 `` ` `// 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

 ``
Output:
`54.597883`

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