A numerical method produces estimates A(h) = I + c₁h + c₂h² + ··· You compute A(h) and A(h/2). What does the Richardson extrapolation formula 2A(h/2) − A(h) produce, and why?
AAn average of the two estimates, which reduces random numerical noise
BAn estimate with error starting at O(h²) instead of O(h), because the O(h) terms cancel exactly
CThe exact value I, because combining two estimates eliminates all error
DAn estimate with error starting at O(h/2), because we used the smaller step size
Richardson extrapolation cancels the leading error term algebraically. A(h/2) = I + (c₁/2)h + (c₂/4)h² + ···. Multiply by 2: 2A(h/2) = 2I + c₁h + (c₂/2)h² + ···. Subtract A(h) = I + c₁h + c₂h² + ···: result = I + 0·h + (c₂/2 − c₂)h² + ··· = I − (c₂/2)h² + ···. The O(h) term cancels exactly; accuracy improves by a full order. This is not averaging — different coefficients are required, and averaging (weighting both estimates by 1/2) would not cancel the linear term.
Question 2 Multiple Choice
For Richardson extrapolation to dramatically improve accuracy, what must be known about the method being used?
AThe exact value of the true answer I, so that errors can be measured and corrected
BThe structure of the error expansion — specifically, that it takes the form c₁h^p + c₂h^q + ··· with known exponents
CThe values of the error coefficients c₁, c₂, … so the cancellation formula can be derived
DThat the step size h is already below machine epsilon, ensuring floating-point arithmetic is exact
You need to know the *form* of the error expansion — the powers of h that appear — not the coefficients themselves. For a method with error c₁h + c₂h² + ···, the cancellation weights (2 and −1 in 2A(h/2) − A(h)) are determined entirely by the exponent p = 1. For centered differences with error c₂h² + c₄h⁴ + ··· (only even powers), different weights cancel the O(h²) term. If the error expansion is unknown or irregular (e.g., due to singularities), Richardson extrapolation cannot be applied reliably — the unknown powers mean you don't know which linear combination to form.
Question 3 True / False
Richardson extrapolation improves accuracy by using results already computed at two step sizes, requiring no additional function evaluations beyond those two estimates.
TTrue
FFalse
Answer: True
This is the key practical advantage: Richardson extrapolation is 'free' in the sense that once you have A(h) and A(h/2), the improved estimate 2A(h/2) − A(h) requires only arithmetic, no new function evaluations. Compare this to the naive approach of simply using a smaller step size h/2 — that achieves the same O(h²) error but at the cost of recomputing the estimate from scratch. Richardson extrapolation recycles existing work and extracts higher accuracy from it.
Question 4 True / False
Richardson extrapolation works by averaging two numerical estimates, giving each estimate equal weight of 1/2.
TTrue
FFalse
Answer: False
Richardson extrapolation uses deliberately unequal weights chosen to cancel the leading error term, not to average. For a method with O(h) error, the formula is 2A(h/2) − 1·A(h): weights +2 and −1. Simple averaging would give (A(h) + A(h/2))/2 = I + (3c₁/4)h + ···, which does not cancel the O(h) term at all. The specific weights are derived algebraically from the requirement that the coefficient of h in the combined estimate equals zero. Different error structures require different weights.
Question 5 Short Answer
Why does Richardson extrapolation fail or behave unpredictably when applied to a method whose error expansion contains a logarithmic term (e.g., error ~ c₁h ln h + c₂h²) rather than a pure power series in h?
Think about your answer, then reveal below.
Model answer: Richardson extrapolation is designed to cancel specific powers of h. The standard formula 2A(h/2) − A(h) is derived assuming error terms proportional to h^p for integer p. If the error contains h ln h, replacing h with h/2 gives (h/2)ln(h/2) = (h/2)(ln h − ln 2), which does not scale simply as a half-integer power of h. The resulting combination 2A(h/2) − A(h) no longer cancels the leading error — the logarithmic term survives in modified form. Richardson extrapolation requires a clean polynomial error expansion in h to work; irregular or logarithmic error structures break the algebraic cancellation.
The deeper point: Richardson extrapolation exploits the *structure* of errors, not just their magnitude. When you know errors are pure powers of h, you can engineer their cancellation. When error structure is complicated by logs, fractional powers, or singularities, the method's assumptions fail and you may actually worsen accuracy by combining estimates with the wrong weights.