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Part CCSIR NET June 2023recursive-sequence-fixed-point

Recursive sequence fixed point

Let f:R→Rf : \mathbb{R} \to \mathbb{R} be defined as f(x)=1/4+x−x2f(x) = 1/4 + x - x^{2}. Given a∈Ra \in \mathbb{R}, define the sequence {xnx_{n}} by x0=ax_{0} = a and xn=f(xnx_{n} = f(x_{n}₋1)_{1}) for n ≥ 1. Which of the following statements are true?

  1. A.If a = 0, then the sequence {xnx_{n}} converges to 1/2.✓ Correct
  2. B.If a = 0, then the sequence {xnx_{n}} converges to −1/2.
  3. C.The sequence {xnx_{n}} converges for every a ∈ (−1/2, 3/2), and it converges to 1/2.✓ Correct
  4. D.If a = 0, then the sequence {xnx_{n}} does not converge.

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests limit assumed to exist.

See pricing

50 are analysed free — try those first.

The trap it tests

Limit assumed to exist

Reasoning with a limit before establishing there is one. limsup is not lim.

Drill statements like this

Related counterexample: aₙ^{1/n} → L ⇒ aₙ₊₁/aₙ → L

More on this topic

The chapter behind this: Sequences in ℝ — the implication map — free to read

From The Real Line › Sequences: convergence, monotone, Bolzano–Weierstrass, Cauchy

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