A numerical analyst has computed composite trapezoidal estimates T(h), T(h/2), T(h/4), and T(h/8). When they form column 1 of the Romberg table by combining T(h) and T(h/2), what has been accomplished?
AThe estimate is now equivalent to using the trapezoidal rule with step size h/4
BThe O(h²) error term has been cancelled, leaving an estimate with O(h⁴) error
CThe result is simply the average of T(h) and T(h/2), providing a better mean
DAll error terms have been eliminated, giving the exact integral
Richardson extrapolation combines two estimates with known error structure to cancel the leading error term. The trapezoidal rule has error proportional to h², so combining T(h) and T(h/2) eliminates the O(h²) term, leaving an estimate with O(h⁴) error. This is not simply averaging — it is the weighted combination (4·T(h/2) − T(h))/3, which equals Simpson's rule. Further columns of the Romberg table cancel successively higher-order terms.
Question 2 Multiple Choice
Why does Romberg integration require relatively few new function evaluations when adding a new row to the table by halving the step size?
AIt uses a Monte Carlo sampling scheme that does not require a fine grid
BThe new, finer grid contains all the points of the previous grid as a subset
CIt only evaluates the integrand at the endpoints, regardless of step size
DRomberg avoids evaluating the integrand entirely, relying on symbolic integration
When you halve the step size from h to h/2, the new grid at h/2 includes all the points that were already in the grid at h, plus new midpoints between them. Only the new midpoints require fresh function evaluations. This means each new row of the Romberg table costs only about half as many new evaluations as the total evaluations in that row, making the method remarkably efficient for smooth functions.
Question 3 True / False
The entries in column 1 of the Romberg table are equivalent to Simpson's rule approximations.
TTrue
FFalse
Answer: True
Column 0 of the Romberg table holds composite trapezoidal approximations. Column 1 applies Richardson extrapolation to cancel the O(h²) error, producing estimates equivalent to Simpson's rule (which achieves O(h⁴) accuracy). Column 2 cancels the O(h⁴) error, equivalent to Boole's rule. Each successive column corresponds to a higher-order quadrature rule, with extrapolation systematically eliminating the next error term.
Question 4 True / False
Romberg integration converges faster than the composite trapezoidal rule primarily because it uses more function evaluations per step.
TTrue
FFalse
Answer: False
Romberg integration's speedup comes from Richardson extrapolation eliminating error terms algebraically, not from using more function evaluations. It achieves high accuracy by reusing existing evaluations and combining them cleverly, rather than simply taking more or smaller steps. A key structural fact enabling this is that the trapezoidal rule's error consists only of even powers of h, which allows successive Richardson steps to cancel each term precisely.
Question 5 Short Answer
Explain why the composite trapezoidal rule's error consisting exclusively of even powers of h is crucial to Romberg integration's effectiveness.
Think about your answer, then reveal below.
Model answer: The Euler–Maclaurin formula shows that the trapezoidal error is T(h) = I + c₁h² + c₂h⁴ + ..., with no odd-power terms. Halving h to get T(h/2) scales each term by powers of 1/4. This predictable structure means you can algebraically combine T(h) and T(h/2) to cancel the c₁h² term exactly, leaving an error of order h⁴. If the error had odd-power terms, the cancellation would not be clean and the extrapolation would not eliminate the leading term so precisely.
The absence of odd-power terms follows from the trapezoidal rule's symmetry — errors from symmetric intervals cancel at even powers. Without this regularity, Romberg's systematic table construction would not work: you could not predict which combination of T(h) and T(h/2) eliminates the leading error, because the error structure would be less rigid.