2025-09-12

1. The error of values \(x_k\) in a bisection iteration is given by \(e_k \approx \left( b - a \right) \left(\frac{1}{2} \right)^k\) for the initial interval \(\left[ a, b \right]\). This method is

  1. divergent

  2. linearly convergent

  3. \(q\)-linearly convergent

  4. quadratically convergent

Answerc

2. The irrational number \(e\) can be approximated by applying Newton’s method to solve the nonlinear equation \(f \left( x \right) = ln \left( x \right) - 1 = 0\). What is the Newton iteration formula? (Note: \(\frac{d}{dx} ln \left( x \right) = \frac{1}{x}\))

  1. \(x_{k + 1} = ln \left( x_k \right)\)

  2. \(x_{k + 1} = x_k - ln \left( x_k \right)\)

  3. \(x_{k + 1} = x_k \left( ln \left( x_k \right) - 1 \right)\)

  4. \(x_{k + 1} = 2 x_k - x_k ln \left( x_k \right)\)

Answerd

3. Sketch the graph of a continuous function \(f \left( x \right)\) that has a root \(f \left( 1 \right) = 0\) that cannot be located using bisection. Explain why this root cannot be located with bisection.

AnswerAny graph of a continuous function with a zero at \(f \left( 1 \right) = 0\) and all positive values or all negative values will work, as bisection requires an interval in which the function has a positive output at one endpoint and a negative output at the other endpoint.