Question:medium

Consider the sequence \( \{ f_n \} \) of continuous functions on \( [0, 1] \) defined by \[ f_1(x) = \frac{x}{2}, \quad f_{n+1}(x) = f_n(x) - \frac{1}{2} \left( (f_n(x))^2 - x \right), \quad n = 1, 2, 3, \dots \] Then, which of the following is/are TRUE?

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For recursive function sequences, check the behavior of the differences between successive terms to determine uniform convergence. Additionally, ensure that the functions remain bounded by the limit function, which in this case is \( \sqrt{x} \).
Updated On: Feb 2, 2026
  • The sequence \( \{ f_n \} \) converges pointwise but not uniformly on \( [0, 1] \)
  • The sequence \( \{ f_n \} \) converges uniformly on \( [0, 1] \)
  • \( \sqrt{x} - f_n(x)>\frac{2\sqrt{x}}{2 + n\sqrt{x}} \quad {for all} \, x \in [0, 1] \, {and} \, n = 1, 2, 3, \dots \)
  • \( 0 \leq f_n(x) \leq \sqrt{x} \quad {for all} \, x \in [0, 1] \, {and} \, n = 1, 2, 3, \dots \)
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The Correct Option is B, D

Solution and Explanation

We are given a sequence of continuous functions \( \{ f_n \} \) on the interval \([0, 1]\), where the functions are defined as follows:

  • \( f_1(x) = \frac{x}{2} \)
  • \( f_{n+1}(x) = f_n(x) - \frac{1}{2} \left( (f_n(x))^2 - x \right) \)

We need to determine which of the provided statements about the sequence is/are true.

Analysis of the Sequence: 

The recurrence relation given for \( f_{n+1}(x) \) can be interpreted as a form of Newton's method for finding the square root of \( x \), with \( f_n(x) \approx \sqrt{x} \).

Step-by-Step Verification:

  1. Convergence Analysis:

Notice that the recurrence relation:

  1. \(f_{n+1}(x) = f_n(x) - \frac{1}{2} ((f_n(x))^2 - x)\)

is a form of iterative method intended to converge to the square root of \( x \). Particularly, given the form, this resembles the Newton-Raphson method for solving \((f_n(x))^2 = x\). It is known that Newton's method converges quadratically under suitable conditions.

Hence, \( f_n(x) \to \sqrt{x} \) pointwise on \([0, 1]\).

  1. Uniform Convergence:

To determine if the convergence is uniform, consider the form:

  1. \(|f_{n+1}(x) - \sqrt{x}| = \left| f_n(x) - \sqrt{x} - \frac{1}{2} ((f_n(x))^2 - x) \right|\)

This eventually reduces the error in each step, indicating uniform convergence over compact intervals like \([0, 1]\). Thus, the sequence converges uniformly on \([0, 1]\).

Therefore, the option claiming uniform convergence is correct.

  1. Bound Analysis:

We need to show that the functions are bound by:

  1. \(0 \leq f_n(x) \leq \sqrt{x}\)

We begin with \( f_1(x) = \frac{x}{2} \) and observe:

  • Clearly, \( 0 \leq \frac{x}{2} \leq x \leq \sqrt{x} \) for \( 0 \leq x \leq 1 \) (since \( x \leq \sqrt{x} \) for \( x \) in this range).
  • Inductive Hypothesis: Assume \( 0 \leq f_n(x) \leq \sqrt{x} \).
  • Then \( f_{n+1}(x) \) is calculated to remain within these bounds by the structure of the iterative formula.
  1. Analyzing the other options:

The claim \( \sqrt{x} - f_n(x) > \frac{2\sqrt{x}}{2 + n\sqrt{x}} \) is incorrect. With uniform convergence, the error diminishes rapidly, contradicting this lower bound.

Therefore, the correct answers are:

  • The sequence \( \{ f_n \} \) converges uniformly on \([0, 1]\).
  • \( 0 \leq f_n(x) \leq \sqrt{x} \) for all \( x \in [0, 1] \) and \( n = 1, 2, 3, \ldots \).
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