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 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
<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.
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