Arcsin(x)展开式中N个项的总和

给定两个整数N和X ，任务是使用扩展到N个项来找到Arcsin(x)的值。
例子：

Input: N = 4, X = 0.5 
Output: 0.5233863467 
Sum of first 4 terms in the expansion of Arcsin(x) for 
x = 0.5 is 0.5233863467.
Input: N = 8, X = -0.5 
Output: -0.5233948501 
 

方法： arcsin(x)的扩展由下式给出：
$arcsin(x) = x + (1/2).(x^3)/3 + ((1.3)/(2.4)).(x^5)/5 + .....$
注意： | x | <1
通过使用两个保留分子和分母的变量来解决上述扩展问题。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to find the arcsin(x)
void find_Solution(double x, int n)
{
    double sum = x, e = 2, o = 1, p = 1;
    for (int i = 2; i <= n; i++) {
 
        // The power to which 'x' is raised
        p += 2;
 
        sum += (double)(o / e) * (double)(pow(x, p) / p);
 
        // Numerator value
        o = o * (o + 2);
 
        // Denominator value
        e = e * (e + 2);
    }
    cout << setprecision(10) << sum;
}
 
// Driver code
int main()
{
    double x = -0.5;
 
    if (abs(x) >= 1) {
        cout << "Invalid Input\n";
        return 0;
    }
 
    int n = 8;
    find_Solution(x, n);
    return 0;
}

Java

//Java implementation of the approach
import java.io.*;
 
class GFG
{
 
// Function to find the arcsin(x)
static void find_Solution(double x, int n)
{
    double sum = x, e = 2, o = 1, p = 1;
    for (int i = 2; i <= n; i++)
    {
 
        // The power to which 'x' is raised
        p += 2;
 
        sum += (double)(o / e) *
               (double)(Math.pow(x, p) / p);
 
        // Numerator value
        o = o * (o + 2);
 
        // Denominator value
        e = e * (e + 2);
    }
    System.out.println (sum);
}
 
// Driver code
public static void main (String[] args)
{
    double x = -0.5;
 
    if (Math.abs(x) >= 1)
    {
        System.out.println ("Invalid Input");
    }
     
    int n = 8;
    find_Solution(x, n);
}
}
 
// This code is contributed by ajit

Python3

# Python3 implementation of the approach
 
# Function to find the arcsin(x)
def find_Solution(x, n):
    Sum = x
    e = 2
    o = 1
    p = 1
    for i in range(2, n + 1):
 
        # The power to which 'x' is raised
        p += 2
 
        Sum += (o / e) * (pow(x, p) / p)
 
        # Numerator value
        o = o * (o + 2)
 
        # Denominator value
        e = e * (e + 2)
    print(round(Sum, 10))
 
# Driver code
x = -0.5
 
if (abs(x) >= 1):
    print("Invalid Input\n")
 
n = 8
find_Solution(x, n)
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;
class GFG
{
 
// Function to find the arcsin(x)
static void find_Solution(double x, int n)
{
    double sum = x, e = 2, o = 1, p = 1;
    for (int i = 2; i <= n; i++)
    {
 
        // The power to which 'x' is raised
        p += 2;
 
        sum += (double)(o / e) *
               (double)(Math.Pow(x, p) / p);
 
        // Numerator value
        o = o * (o + 2);
 
        // Denominator value
        e = e * (e + 2);
    }
    Console.WriteLine(sum);
}
 
// Driver code
public static void Main (String[] args)
{
    double x = -0.5;
 
    if (Math.Abs(x) >= 1)
    {
        Console.WriteLine("Invalid Input");
    }
     
    int n = 8;
    find_Solution(x, n);
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

 
// JavaScript implementation of the approach
 
// Function to find the arcsin(x)
function find_Solution(x, n)
{
    let sum = x, e = 2, o = 1, p = 1;
    for (let i = 2; i <= n; i++) {
 
        // The power to which 'x' is raised
        p += 2;
 
        sum += (o / e) * (Math.pow(x, p) / p);
 
        // Numerator value
        o = o * (o + 2);
 
        // Denominator value
        e = e * (e + 2);
    }
    document.write(sum.toFixed(10));
}
 
// Driver code
 
    let x = -0.5;
 
    if (Math.abs(x) >= 1) {
        document.write("Invalid Input<br>");
    }
         
 
    let n = 8;
    find_Solution(x, n);
     
// This code is contributed by Surbhi Tyagi.
 

输出：

-0.5233948501