Efficient program to calculate e^x

The value of Exponential Function e^x can be expressed using following Taylor Series.

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

How to efficiently calculate the sum of above series?
The series can be re-written as 

e^x = 1 + (x/1) (1 + (x/2) (1 + (x/3) (........) ) ) 

Let the sum needs to be calculated for n terms, we can calculate sum using following loop.

for (i = n - 1, sum = 1; i > 0; --i )
    sum = 1 + x * sum / i; 

Following is implementation of the above idea. 

C++

// C++ Efficient program to calculate
// e raise to the power x
#include <bits/stdc++.h>
using namespace std;

// Returns approximate value of e^x
// using sum of first n terms of Taylor Series
float exponential(int n, float x)
{
    float sum = 1.0f; // initialize sum of series

    for (int i = n - 1; i > 0; --i )
        sum = 1 + x * sum / i;

    return sum;
}

// Driver code
int main()
{
    int n = 10;
    float x = 1.0f;
    cout << "e^x = " << fixed << setprecision(5) << exponential(n, x);
    return 0;
}

// This code is contributed by rathbhupendra

C

// C Efficient program to calculate
// e raise to the power x
#include <stdio.h>

// Returns approximate value of e^x
// using sum of first n terms of Taylor Series
float exponential(int n, float x)
{
    float sum = 1.0f; // initialize sum of series

    for (int i = n - 1; i > 0; --i )
        sum = 1 + x * sum / i;

    return sum;
}

// Driver program to test above function
int main()
{
    int n = 10;
    float x = 1.0f;
    printf("e^x = %f", exponential(n, x));
    return 0;
}

Java

// Java efficient program to calculate
// e raise to the power x
import java.io.*;

class GFG
{
    // Function returns approximate value of e^x
    // using sum of first n terms of Taylor Series
    static float exponential(int n, float x)
    {
        // initialize sum of series
        float sum = 1;

        for (int i = n - 1; i > 0; --i )
            sum = 1 + x * sum / i;

        return sum;
    }

    // driver program
    public static void main (String[] args)
    {
        int n = 10;
        float x = 1;
        System.out.println("e^x = "+exponential(n,x));
    }
}

// Contributed by Pramod Kumar

Python3

# Python program to calculate
# e raise to the power x

# Function to calculate value
# using sum of first n terms of
# Taylor Series
def exponential(n, x):

    # initialize sum of series
    sum = 1.0
    for i in range(n, 0, -1):
        sum = 1 + x * sum / i
    print ("e^x =", sum)

# Driver program to test above function
n = 10
x = 1.0
exponential(n, x)

# This code is contributed by Danish Raza

C#

// C# efficient program to calculate
// e raise to the power x
using System;

class GFG
{
    // Function returns approximate value of e^x
    // using sum of first n terms of Taylor Series
    static float exponential(int n, float x)
    {
        // initialize sum of series
        float sum = 1;

        for (int i = n - 1; i > 0; --i )
            sum = 1 + x * sum / i;

        return sum;
    }

    // driver program
    public static void Main ()
    {
        int n = 10;
        float x = 1;
        Console.Write("e^x = " + exponential(n, x));
    }
}

// This code is contributed by nitin mittal.

PHP

<?php
// PHP Efficient program to calculate
// e raise to the power x

// Returns approximate value of e^x
// using sum of first n terms
// of Taylor Series
function exponential($n, $x)
{
    // initialize sum of series
    $sum = 1.0;

    for ($i = $n - 1; $i > 0; --$i )
        $sum = 1 + $x * $sum / $i;

    return $sum;
}

// Driver Code
$n = 10;
$x = 1.0;
echo("e^x = " . exponential($n, $x));

// This code is contributed by Ajit.
?>

Javascript

<script>
// javascript efficient program to calculate
// e raise to the power x

    // Function returns approximate value of e^x
    // using sum of first n terms of Taylor Series
    function exponential(n , x) {
        // initialize sum of series
        var sum = 1;

        for (i = n - 1; i > 0; --i)
            sum = 1 + x * sum / i;

        return sum;
    }

    // driver program

        var n = 10;
        var x = 1;
        document.write("e^x = " + exponential(n, x).toFixed(6));

// This code contributed by Rajput-Ji

</script>

Output: 

e^x = 2.718282

Time Complexity: O(n)

Auxiliary Space: O(1), since no extra space has been taken.

This article is compiled by Rahul and reviewed by GeeksforGeeks team.