ITPEC FE Subject B April 2025 Question 14

Maclaurin series cosine approximation: iterative term recurrence

The program computes cos(y) using the Maclaurin series:
cos(y) = 1 − y²/2! + y⁴/4! − y⁶/6! + ...

Each successive term is derived from the previous one by multiplying by (-1) × y² / (2n × (2n − 1)), where n is the iteration number starting from 1.

• •Blank A = `term`: The while loop checks whether the current term is still significant. term represents the latest addition to the sum; once its absolute value drops below the threshold, convergence is reached.
• •Blank B = `squared` (i.e., `y²`): The recurrence relation between consecutive terms of the cosine series involves multiplying by y² (not a higher power). Verification:
• •n=1: 1 × (−y²) / (2 × 1) = −y²/2 ✓
• •n=2: (−y²/2) × (−y²) / (4 × 3) = y⁴/24 ✓
• •n=3: (y⁴/24) × (−y²) / (6 × 5) = −y⁶/720 ✓

Why not others:
- a–c (A = cosy): Checking convergence against the accumulated sum is incorrect — cosy does not approach zero
- d (B = y^(2n)): Would multiply by increasing powers each iteration, producing wrong factorial growth
- e (B = y⁴): Fixed power of 4 breaks the recurrence after the first correct step

Key rule: In iterative series approximation, the convergence check uses the current term, and the recurrence multiplier uses the constant ratio between consecutive terms — for cosine, that ratio involves y².

AI-generated — may contain errors