Newton's method has quadratic convergence and bisection has linear convergence on a problem. Starting from the same initial guess x₀, which method is guaranteed to have a smaller error after the first iteration?
ANewton's method, because quadratic convergence always beats linear convergence
BBisection, because it guarantees halving the interval every step
CNeither — convergence order is asymptotic and says nothing about early iterations far from the root
DThey will have the same error after the first iteration
Order of convergence is an asymptotic property: it describes how errors shrink when x_n is already close to the solution x*. Far from the root, Newton's method can behave poorly — it may overshoot, oscillate, or even diverge. Bisection is the one that guarantees halving the error every step regardless of starting point. The practical implication: when far from the root, lower-order but globally convergent methods like bisection may be preferable; switch to Newton's method once you are close enough that quadratic convergence kicks in.
Question 2 Multiple Choice
A method with order 2 (Newton's) requires evaluating both f and f′ per iteration, while a method with order 1.618 (secant) requires only one function evaluation per iteration. If evaluating f′ is very expensive, which method might deliver more accuracy per unit of computational cost?
ANewton's method, because quadratic convergence always dominates in total iterations needed
BThe secant method, because fewer expensive evaluations per step may compensate for slightly lower convergence order
CThey are equivalent in practice because the secant method's order is close enough to 2
DNewton's method, because it requires fewer total iterations to reach any given tolerance
Order of convergence counts accuracy per iteration, not accuracy per function evaluation. If evaluating f′ costs as much as 5 evaluations of f, then Newton's method may be 5× more expensive per step despite needing fewer steps. The secant method needs only one new function evaluation per step (it reuses the previous one), making it highly competitive. The right comparison is accuracy per unit of computational work, not accuracy per iteration — a distinction that matters greatly when f is expensive to evaluate.
Question 3 True / False
For a method with quadratic convergence, once the error is around 10⁻⁴, the next iteration's error will be approximately 10⁻⁸.
TTrue
FFalse
Answer: True
Quadratic convergence means |e_{n+1}| ≈ C·|e_n|². With e_n ≈ 10⁻⁴ and C ≈ 1, we get e_{n+1} ≈ (10⁻⁴)² = 10⁻⁸. The number of correct decimal digits roughly doubles each step — from 4 to 8 in this case. This doubling behavior is what makes quadratic convergence so powerful in the final iterations: a few steps deliver enormous accuracy gains that linear convergence would take hundreds of steps to achieve.
Question 4 True / False
If a numerical method has quadratic convergence, it will converge to the solution faster than a linear method starting from any initial guess.
TTrue
FFalse
Answer: False
Order of convergence is asymptotic — it only describes behavior in the tail of the iteration sequence, when x_n is already very close to x*. Starting from an arbitrary initial guess, quadratic methods like Newton's are not guaranteed to outperform linear methods and may even diverge. Bisection, despite its linear (order 1) convergence, converges reliably from any starting interval. The comparison 'quadratic beats linear' only applies once you are close enough that the asymptotic regime has kicked in.
Question 5 Short Answer
Why is 'order of convergence' not the same as 'efficiency' of a numerical method? Give an example.
Think about your answer, then reveal below.
Model answer: Order of convergence measures how quickly errors shrink per iteration, not per unit of computational work. A method with lower order may be more efficient if each iteration is cheaper. For example, Newton's method (order 2) requires both f and f′ per step; the secant method (order ≈ 1.618) requires only one function evaluation. If f′ is expensive, the secant method may produce more accuracy per computation even though it needs slightly more iterations.
The distinction between per-iteration and per-evaluation efficiency is crucial in practice. In scientific computing, function evaluations may involve solving PDEs or running simulations, costing minutes each. In such settings, the secant or even Brent's method (which avoids derivatives entirely) often outperforms Newton's despite technically lower order. Always measure computational cost, not just iteration count.