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++ 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 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 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
# 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# 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 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.
?>
<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.