给定两个数x和N的任务是找到COSH(X)从高达系列的N项的值。
cosh(x)的扩展如下:
cosh(x) = 1 + x2/2! + x4/4! + …………
例子:
Input: x = 1, N = 5
Output: 1.54308035714
Input: x = 1, N = 10
Output: 1.54308063497
方法:
上面的系列可以使用阶乘函数和循环轻松实现。
该系列的第n个术语是:
下面是上述方法的实现:
C++
// C++ program for
// the sum of cosh(x) series
#include
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
输出:
1.54308063497