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Part CCSIR NET December 2024the-rearrangement-decides-which-root-attracts

The rearrangement decides which root attracts

Consider the following statements: S1S_{1}: There exists an ε>0\varepsilon > 0 such that ∀x0∈(2−ε,2+ε)\forall{}x_{0} \in (2 - \varepsilon, 2 + \varepsilon), the iterative sequence defined by x(n+1)=6/(5−xn),n=0,1,2,3,…x(n+1) = 6/(5 - x_{n}), n = 0, 1, 2, 3, \dots, converges to 2.S22. S_{2}: There exists an ε>0\varepsilon > 0 such that ∀x0∈(3−ε,3+ε)\forall{}x_{0} \in (3 - \varepsilon, 3 + \varepsilon), the iterative sequence defined by x(n+1)=(xn2+6)/5,n=0,1,2,3,…x(n+1) = (x_{n}^{2} + 6)/5, n = 0, 1, 2, 3, \dots, converges to 3.S33. S_{3}: There exists an ε>0\varepsilon > 0 such that ∀x0∈(3−ε,3+ε)\forall{}x_{0} \in (3 - \varepsilon, 3 + \varepsilon), the iterative sequence defined by x(n+1)=(5xn−6)/xn,n=0,1,2,3,…x(n+1) = (5x_{n} - 6)/x_{n}, n = 0, 1, 2, 3, \dots, converges to 3. Then which of the following statements are true?

  1. A.S1S_{1} and S2S_{2} are true but NOT S3S_{3}
  2. B.S2S_{2} and S3S_{3} are true but NOT S1S_{1}
  3. C.S1S_{1} and S3S_{3} are true but NOT S2S_{2}✓ Correct
  4. D.Each of S1,S2S_{1}, S_{2}, and S3S_{3} is true

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 numerical convergence.

See pricing

50 are analysed free — try those first.

The trap it tests

Numerical convergence

A scheme assumed to converge, or to converge at its advertised rate.

Drill statements like this

Related counterexample: Newton's method converges quadratically to any root

More on this topic

The chapter behind this: Root finding and orders of convergence — free to read

From Numerical Analysis › Root finding: bisection, Newton–Raphson, fixed point, order of convergence

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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