## Approximating Pi Using Numerical Summation

### Explanation
This method approximates the value of \( \pi \) using a numerical summation technique. The approach is based on the concept of inscribing and circumscribing rectangles under a quarter-circle to estimate the area. By summing the areas of these rectangles, we can approximate the area of the quarter-circle, which in turn helps us estimate \( \pi \).

### Mathematical Representation
The equation used is:

![Pi Equation](.../.././base.png)

Where:
- \( a \) and \( b \) are summations that approximate the quarter-circle area.
- The sum iterates over a set of discrete steps to refine the approximation.

More specifically, \( a \) and \( b \) are defined as:

![A Equation](.../.././a.png)

![B Equation](.../.././b.png)

### Python Equivalent

```python file.py
import math

def approximate_pi(n=46325):
    lange = n - 2
    a, b = 0, 0

    # Compute summation for 'a'
    for k in range(1, lange + 1):  # Changed to include lange
        a += math.sqrt(n**2 - k**2)

    # Compute summation for 'b'
    for j in range(lange + 1):  # Changed to include lange
        b += math.sqrt(n**2 - j**2)

    # Normalize the results
    b = 4 / (n**2) * b
    a = 4 / (n**2) * a
    result = (a + b) / 2

    return result

# Run the function
print(approximate_pi())
```

### How It Works
1. **Initialization**: The variable `lange` is set to \( n - 2 \), and two summation variables `a` and `b` are initialized to zero.
2. **Summation for 'a'**: Iterates from 1 to `lange`, calculating the square root of \( n^2 - k^2 \) for each \( k \).
3. **Summation for 'b'**: Iterates from 0 to `lange`, calculating the square root of \( n^2 - j^2 \) for each \( j \).
4. **Normalization**: Both summations are normalized by multiplying by \( \frac{4}{n^2} \).
5. **Result Calculation**: The final approximation of \( \pi \) is obtained by averaging the normalized values of `a` and `b`.

### Conclusion
This method provides an approximation of \( \pi \) using numerical summation. The accuracy increases as \( n \) grows larger, effectively refining the result. By adjusting the value of \( n \), you can control the trade-off between computational effort and precision.

### Notes
- The choice of `n = 46325` is arbitrary and can be adjusted for better performance or higher precision.
- This method is computationally intensive and may require optimization for very large values of \( n \).