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

• Last Updated : 25 May, 2022

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 using namespace std; // function to return the factorial of a numberint 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 seriesdouble 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 codeint 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) seriesimport java.util.*; class GFG{ // function to return the factorial of a numberstatic 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 seriesstatic 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 codepublic 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 numberdef 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 seriesdef 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 codex = 1n = 10print(log_Expansion(x, n)) # This code is contributed by Mohit Kumar

## C#

 // C# program for the sum of// cosh(x) seriesusing System; class GFG{ // function to return the// factorial of a numberstatic 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 seriesstatic 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 codepublic static void Main(String[] args){    double x = 1;    int n = 10;    Console.WriteLine(log_Expansion(x, n));}} // This code is contributed by PrinciRaj1992

## Javascript



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.

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