Sum of N terms in the expansion of Arcsin(x)

Given two integers N and X, the task is to find the value of Arcsin(x) using expansion upto N terms.

Examples:

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

Approach: The expansion of arcsin(x) is given by :
 arcsin(x) = x + (1/2).(x^3)/3 + ((1.3)/(2.4)).(x^5)/5 + .....

Note: |x| < 1

The above expansion is solved by using two variables maintaining the numerator and the denominator.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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;
}

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Java

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//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

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Python3

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# 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

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C#

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// 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

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Output:

-0.5233948501


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