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Part BCSIR NET December 2024the-other-term-has-already-spent-the-budget

The other term has already spent the budget

Let f:R→Rf : \mathbb{R} \to \mathbb{R} be such that sup(x≠y) |f(x) − f(y)|/|x − y| = L, where 1<L<∞1 < L < \infty. Let h:R→Rh : \mathbb{R} \to \mathbb{R} be a differentiable function satisfying |h′(x)| ≤ 3/4 for all x∈Rx \in \mathbb{R}. For α>0\alpha > 0, define g(x)=αf(x)+h(x)g(x) = \alpha{}f(x) + h(x) for x∈Rx \in \mathbb{R}. Consider the sequence {xkx_{k}}(k≥0) defined by x(k+1)=g(xk),k=0,1,…x_(k+1) = g(x_{k}), k = 0, 1, \dots, where x0∈Rx_{0} \in \mathbb{R}. The sequence {xkx_{k}}(k≥0) converges to the solution of the equation x = g(x) if

  1. A.α<2/(3L)\alpha < 2/(3L)
  2. B.α<3/(2L)\alpha < 3/(2L)
  3. C.α<4L\alpha < 4L
  4. D.α<1/(4L)\alpha < 1/(4L)✓ Correct

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

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