Program for sum of cosh(x) series upto Nth term

0 0 3 minutes read

Given two numbers x and N, the task is to find the value of cosh(x) from the series upto N terms.
The expansion of cosh(x) is given below:

cosh(x) = 1 + x²/2! + x⁴/4! + …………

Examples:

Input: x = 1, N = 5
Output: 1.54308035714

Input: x = 1, N = 10
Output: 1.54308063497

Approach:
The above series can be easily implemented using a factorial function and loops.
The nth term of the series is:

Below is the implementation of the above approach:

C++

// C++ program for
// the sum of cosh(x) series
 
#include <bits/stdc++.h>
using namespace std;
 
// function to return the factorial of a number
int fact(int n)
{
 
    int i = 1, fac = 1;
    for (i = 1; i <= n; i++)
        fac = fac * i;
 
    return fac;
}
 
// function to return the sum of the series
double log_Expansion(double x, int n)
{
 
    double sum = 0;
    int i = 0;
 
    for (i = 0; i < n; i++) {
 
        sum = sum
              + pow(x, 2 * i)
                    / fact(2 * i);
    }
 
    return sum;
}
 
// Driver code
int main()
{
    double x = 1;
    int n = 10;
    cout << setprecision(12)
         << log_Expansion(x, n)
         << endl;
 
    return 0;
}

Java

// Java program for the sum of
// cosh(x) series
import java.util.*;
 
class GFG
{
 
// function to return the factorial of a number
static int fact(int n)
{
    int i = 1, fac = 1;
    for (i = 1; i <= n; i++)
        fac = fac * i;
 
    return fac;
}
 
// function to return the sum of the series
static double log_Expansion(double x, int n)
{
    double sum = 0;
    int i = 0;
 
    for (i = 0; i < n; i++) 
    {
        sum = sum + Math.pow(x, 2 * i) / 
                           fact(2 * i);
    }
 
    return sum;
}
 
// Driver code
public static void main(String[] args) 
{
    double x = 1;
    int n = 10;
    System.out.println(log_Expansion(x, n));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program for the Sum of cosh(x) series
 
# function to return the factorial of a number
def fact(n):
 
    i, fac = 1, 1
    for i in range(1, n + 1):
        fac = fac * i
 
    return fac
 
# function to return the Sum of the series
def log_Expansion(x, n):
 
    Sum = 0
    i = 0
 
    for i in range(n):
 
        Sum = Sum + pow(x, 2 * i) / fact(2 * i)
 
    return Sum
 
# Driver code
x = 1
n = 10
print(log_Expansion(x, n))
 
# This code is contributed by Mohit Kumar

C#

// C# program for the sum of
// cosh(x) series
using System;
 
class GFG
{
 
// function to return the 
// factorial of a number
static int fact(int n)
{
    int i = 1, fac = 1;
    for (i = 1; i <= n; i++)
        fac = fac * i;
 
    return fac;
}
 
// function to return the sum of the series
static double log_Expansion(double x, int n)
{
    double sum = 0;
    int i = 0;
 
    for (i = 0; i < n; i++) 
    {
        sum = sum + Math.Pow(x, 2 * i) / 
                        fact(2 * i);
    }
 
    return sum;
}
 
// Driver code
public static void Main(String[] args) 
{
    double x = 1;
    int n = 10;
    Console.WriteLine(log_Expansion(x, n));
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

<script>
 
// Javascript program for the sum of
// cosh(x) series
 
    // function to return the factorial of a number
    function fact( n) {
        let i = 1, fac = 1;
        for (i = 1; i <= n; i++)
            fac = fac * i;
 
        return fac;
    }
 
    // function to return the sum of the series
    function log_Expansion( x , n) {
        let sum = 0;
        let i = 0;
 
        for (i = 0; i < n; i++) {
            sum = sum + Math.pow(x, 2 * i) / fact(2*i);
        }
 
        return sum;
    }
 
    // Driver code
      
        let x = 1;
        let n = 10;
        document.write(log_Expansion(x, n).toFixed(11));
 
// This code is contributed by shikhasingrajput
 
</script>

Output:

1.54308063497

Time Complexity: O(n²), where n represents the value of the given integer.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!