Question:
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?
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?
Show Hint
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 \( [0, 1] \)
- \( \sqrt{x} - f_n(x)>\frac{2\sqrt{x}}{2 + n\sqrt{x}} \quad {for all} \, x \in [0, 1] \, {and} \, n = 1, 2, 3, \dots \)
- \( 0 \leq f_n(x) \leq \sqrt{x} \quad {for all} \, x \in [0, 1] \, {and} \, n = 1, 2, 3, \dots \)
Show Solution
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The Correct Option is B, D
Solution and Explanation
Step 1: Analyzing the Sequence of Functions
The given sequence of functions is defined recursively. We can observe that the functions \( f_n(x) \) are continuous and are getting closer to the function \( \sqrt{x} \) as \( n \) increases. To establish uniform convergence, we analyze the difference between successive terms and observe that the convergence becomes uniform across the interval \( [0, 1] \).
Step 2: Uniform Convergence
Since the sequence \( f_n(x) \) converges to \( \sqrt{x} \) and the rate of convergence does not depend on \( x \), the sequence converges uniformly on \( [0, 1] \). This justifies Option (B).
Step 3: Verifying Option (D)
We also know that \( 0 \leq f_n(x) \leq \sqrt{x} \) for all \( x \in [0, 1] \) because the function sequence is designed to converge to \( \sqrt{x} \), and each \( f_n(x) \) is bounded by \( \sqrt{x} \). This justifies Option (D).
Step 4: Conclusion
Thus, the correct answers are:
Final Answer
\[ \boxed{B} \quad \text{The sequence } \{ f_n \} \text{ converges uniformly on } [0, 1] \]
\[ \boxed{D} \quad 0 \leq f_n(x) \leq \sqrt{x} \quad \text{for all } x \in [0, 1] \text{ and } n = 1, 2, 3, \dots \]
The given sequence of functions is defined recursively. We can observe that the functions \( f_n(x) \) are continuous and are getting closer to the function \( \sqrt{x} \) as \( n \) increases. To establish uniform convergence, we analyze the difference between successive terms and observe that the convergence becomes uniform across the interval \( [0, 1] \).
Step 2: Uniform Convergence
Since the sequence \( f_n(x) \) converges to \( \sqrt{x} \) and the rate of convergence does not depend on \( x \), the sequence converges uniformly on \( [0, 1] \). This justifies Option (B).
Step 3: Verifying Option (D)
We also know that \( 0 \leq f_n(x) \leq \sqrt{x} \) for all \( x \in [0, 1] \) because the function sequence is designed to converge to \( \sqrt{x} \), and each \( f_n(x) \) is bounded by \( \sqrt{x} \). This justifies Option (D).
Step 4: Conclusion
Thus, the correct answers are:
Final Answer
\[ \boxed{B} \quad \text{The sequence } \{ f_n \} \text{ converges uniformly on } [0, 1] \]
\[ \boxed{D} \quad 0 \leq f_n(x) \leq \sqrt{x} \quad \text{for all } x \in [0, 1] \text{ and } n = 1, 2, 3, \dots \]
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