The provided code is an implementation of calculating the **power**...

The provided code is an implementation of calculating the power of a number, ( x^n ), using recursion. Let's break the code down:

Detailed Explanation:

This function myPow(double x, int n) computes ( x^n ) (x raised to the power of n). It uses recursion with multiple conditions to optimize the calculation:

1. Base Cases:

   • If ( x == 1 ) or ( n == 0 ): [ x^n = 1 ] This is because ( x^0 = 1 ) for any non-zero ( x ), and ( 1^n = 1 ) for any ( n ).
   • If ( n == 1 ): [ x^n = x ] Simply return ( x ).

2. Negative Powers:

   • If ( n < 0 ): ( x^{-n} = \frac{1}{x^n} ). The function handles this by calling itself with ( -n ), effectively flipping the sign of ( n ).
     • There is a special check for ( n = INT_MIN ) to prevent integer overflow, as the negative counterpart ( -INT_MIN ) would exceed the range of integers. The formula used for ( INT_MIN ) is: [ x^{INT_MIN} = \frac{1}{x^{(INT_MAX)} \cdot x} ]

3. Recursive Optimization:

   • The recursive solution leverages the property of exponents: [ x^n = x^{n/2} \cdot x^{n/2} \quad \text{(when $n$ is even)} , \text{or} , x^n = x \cdot x^{n-1} \quad \text{(when $n$ is odd)}. ]
   • To reduce the number of recursive calls, the function divides ( n ) by 2 if ( n % 2 == 0 ) (even case), further optimizing the calculation.

Specific Problem:

Given ( x = 2.00000 ) and ( n = 10 ), the function calculates:

[ 2^{10} = 1024 ]

However, the output shown (4.00000) is incorrect because the provided implementation is missing a step when ( n % 2 == 0 ): it should multiply the result of the recursive call instead of just returning it unprocessed. Here's the corrected logic for that case:

if (n % 2 == 0) {
    double half = myPow(x, n / 2);
    return half * half;  // Return the square of half to compute x^(n/2) * x^(n/2).
}

As written, the provided code incorrectly implements ( x^n ) for even powers, leading to the output mismatch.

Conclusion:

• The code calculates ( x^n ) using recursion but has an implementation bug in the even exponent case. Fixing it by squaring the result as shown above would correct the issue.
• The expected output of ( 2^{10} ) should indeed be 1024.00000, not 4.00000.