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When does the sequence generated by the Babylonian method converge to the square root of a?

The sequence converges to a\sqrt{a} as the number of iterations nn approaches infinity, provided that the initial guess x0x_0 is positive and the constant aa is positive. The video outlines a proof strategy involving monotonicity and boundedness. Specifically, after the first step, the sequence is bounded below by a\sqrt{a} and is decreasing (if x1>ax_1 > \sqrt{a}) or increasing (if x1<ax_1 < \sqrt{a}, though typically x1≥ax_1 \ge \sqrt{a} by AM-GM). The limit exists because the sequence is monotone and bounded, and solving the limit equation L=12(L+aL)L = \frac{1}{2}(L + \frac{a}{L}) confirms that L=aL = \sqrt{a}.

Conditions

  • The constant a>0a > 0.
  • The initial guess x0>0x_0 > 0.
  • The limit is taken as n→∞n \to \infty.

Reasoning, step by step

  1. Verify that a>0a > 0 and x0>0x_0 > 0.
  2. Apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality to show that xn+1≥ax_{n+1} \ge \sqrt{a} for all n≥0n \ge 0.
  3. Demonstrate that the sequence is monotone decreasing for n≥1n \ge 1 (since xn+1≤xnx_{n+1} \le x_n when xn≥ax_n \ge \sqrt{a}).
  4. Invoke the Monotone Convergence Theorem to assert that the limit exists.
  5. Let L=lim⁡n→∞xnL = \lim_{n \to \infty} x_n.
  6. Solve the equation L=12(L+aL)L = \frac{1}{2}(L + \frac{a}{L}) to find L=aL = \sqrt{a} (discarding the negative root).
  7. Conclude that the sequence converges to a\sqrt{a} as n→∞n \to \infty.

Example

The video states: 'given x0>0x_0 > 0 and the recursive formula... prove that the limit of the sequence exists as n approaches infinity, and find this limit.' The solution involves proving monotonicity and boundedness.

Common misconceptions

  • Believing that the sequence converges for any real initial guess, including negative ones.
  • Thinking that the convergence is immediate after one step.
  • Confusing the existence of the limit with the value of the limit.

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Answers are generated from source material and independently checked. Consult the original video or creator if something is unclear.