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What is the mathematical justification for the convergence of the Babylonian method sequence?

The mathematical justification relies on the Monotone Convergence Theorem. The sequence xn+1=12(xn+axn)x_{n+1} = \frac{1}{2}(x_n + \frac{a}{x_n}) is shown to be bounded below by a\sqrt{a} (using the AM-GM inequality) and monotone decreasing for n≥1n \ge 1 (assuming x1>ax_1 > \sqrt{a}). A bounded monotone sequence must converge to a limit LL. By taking the limit on both sides of the recurrence relation, we get L=12(L+aL)L = \frac{1}{2}(L + \frac{a}{L}), which solves to L2=aL^2 = a, hence L=aL = \sqrt{a} (since L>0L>0). This proves both the existence of the limit and its value.

Conditions

  • The constant a>0a > 0.
  • The initial guess x0>0x_0 > 0.
  • The sequence is defined by xn+1=12(xn+axn)x_{n+1} = \frac{1}{2}(x_n + \frac{a}{x_n}).

Reasoning, step by step

  1. Apply the AM-GM inequality to show xn+1≥ax_{n+1} \ge \sqrt{a} for all n≥0n \ge 0.
  2. Show that if xn≥ax_n \ge \sqrt{a}, then xn+1≤xnx_{n+1} \le x_n, establishing monotonicity for n≥1n \ge 1.
  3. Invoke the Monotone Convergence Theorem to conclude that lim⁡n→∞xn\lim_{n \to \infty} x_n exists. Let this limit be LL.
  4. Take the limit of the recurrence relation: L=12(L+aL)L = \frac{1}{2}(L + \frac{a}{L}).
  5. Solve for LL: 2L=L+aL  ⟹  L=aL  ⟹  L2=a2L = L + \frac{a}{L} \implies L = \frac{a}{L} \implies L^2 = a.
  6. Since xn>0x_n > 0, L=aL = \sqrt{a}.
  7. Conclude that the sequence converges to a\sqrt{a}.

Example

The video states: 'The solution strategy involves proving monotonicity and boundedness.' It further notes that 'Since the sequence is bounded below by a\sqrt{a} (for n≥1n \ge 1) and decreasing (after the first step), it must converge. Solving the limit equation confirms the value is exactly the square root.'

Common misconceptions

  • Believing that boundedness alone is sufficient for convergence.
  • Thinking that the limit could be −a-\sqrt{a}.
  • Confusing the geometric intuition with the rigorous analytical proof.

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