Q.59 Let a0 = 0 and define an =½(1 + an−1) for all positive integers n ≥ 1.
The least value of n for which |1 − an| < 1 / 210 is ______.
(Answer in integer)
Least n for |1 − an| < 1/210 in Newton Sequence
The sequence defined by
a0 = 0 and
an = ½(1 + an−1) for n ≥ 1
converges to 1. The error decreases geometrically, halving at each iteration.
The least value of n such that
|1 − an| < 1/210 is:
Correct Answer
n = 11
Sequence Convergence
The recurrence relation
an = ½(1 + an−1)
is equivalent to the Newton–Raphson iteration for solving
x² − x = 0, starting from a0 = 0.
The iteration converges to the root x = 1.
Unrolling the recurrence gives the closed-form solution:
an = 1 − 2−n
Therefore, the error is exactly:
|1 − an| = 2−n
This shows that the error halves at each step:
|1 − an| = ½ |1 − an−1|
Solving the Inequality
We require:
2−n < 2−10
Taking powers of 2 on both sides:
n > 10
Checking Values
- n = 10:
|1 − a10| = 2−10 = 1/1024 (equal, not strictly less ❌) - n = 11:
|1 − a11| = 2−11 = 1/2048 < 1/1024 ✔
Hence, the least positive integer satisfying the condition is n = 11.
Option Analysis (Exam Perspective)
- n = 9: Error = 1/512 > 1/1024 (too large)
- n = 10: Error = 1/1024 (fails strict inequality)
- n = 11: First value satisfying the condition ✔
- n > 11: Valid but not the least
