Romberg Integration of Tabulated Data: 04-BS-5, December 2017, Q5

Question 5: Romberg Integration of Tabulated Data (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $y=f(x)$ tabulated at nine equally spaced points, $x=1$ to $x=5$ in steps of $h=0.5$.

Given data for Romberg integration
$x$	$1$	$1.5$	$2$	$2.5$	$3$	$3.5$	$4$	$4.5$	$5$
$y$	$215$	$345$	$444$	$537$	$600$	$763$	$856$	$955$	$1085$

Find. A best Romberg estimate of $\displaystyle\int_1^5f(x)\,dx$, presented as the full triangular array $R(k,j)$, $1\le j\le k\le4$.

Approach. Compute composite-trapezoid estimates $R(k,1)$ at successively halved step sizes $H=4,2,1,0.5$ (using $2^{k-1}+1$ of the nine tabulated points each time), then Richardson-extrapolate across the table using the supplied $R(k,j)$ recurrence.

Row 1 — coarsest trapezoid, $H_1=4$ (endpoints only, $x=1,5$). $$R(1,1)=\dfrac{4}{2}\big[f(1)+f(5)\big]=2\big[215+1085\big]=\boxed{2600.0}$$
Row 2 — $H_2=2$ (add $x=3$). $$R(2,1)=\dfrac{2}{2}\big[f(1)+2f(3)+f(5)\big]=1\big[215+1200+1085\big]=\boxed{2500.0}$$ Extrapolating with $j=2$ ($4^1-1=3$ in the denominator): $$R(2,2)=R(2,1)+\dfrac{R(2,1)-R(1,1)}{3}=2500+\dfrac{-100}{3}=\boxed{2466.667}$$
Row 3 — $H_3=1$ (add $x=2,4$). $$R(3,1)=\dfrac{1}{2}\big[215+2(444+600+856)+1085\big]=\dfrac{1}{2}\big[215+3800+1085\big]=\boxed{2550.0}$$ $$R(3,2)=2550+\dfrac{2550-2500}{3}=\boxed{2566.667},\qquad R(3,3)=2566.667+\dfrac{2566.667-2466.667}{15}=\boxed{2573.333}$$
Row 4 — $H_4=0.5$ (all nine points). $$R(4,1)=\dfrac{0.5}{2}\big[215+2(345+444+537+600+763+856+955)+1085\big]=\dfrac{0.25}{1}\big[215+9000+1085\big]=\boxed{2575.0}$$ $$R(4,2)=2575+\dfrac{2575-2550}{3}=2583.333,\quad R(4,3)=2583.333+\dfrac{2583.333-2566.667}{15}=2584.444$$ $$R(4,4)=2584.444+\dfrac{2584.444-2573.333}{63}=\boxed{2584.621}$$

Figure: the tabulated curve $y=f(x)$ over $[1,5]$ used to build each row of the Romberg table.

$k\backslash j$	1	2	3	4
1	2600.000
2	2500.000	2466.667
3	2550.000	2566.667	2573.333
4	2575.000	2583.333	2584.444	2584.621

Topic: Romberg integration — Richardson extrapolation of the composite trapezoidal rule.

Key relations: composite trapezoid rule $R(k,1)$; Richardson extrapolation $R(k,j)=R(k,j-1)+\dfrac{R(k,j-1)-R(k-1,j-1)}{4^{j-1}-1}$; Euler–Maclaurin expansion showing the trapezoid error is a pure power series in $h^2$.

Why this works: because the trapezoidal error expands as $c_1h^2+c_2h^4+\cdots$, combining two estimates at step sizes $h$ and $h/2$ with the weight $4^{j-1}$ (matched to the leading power of $h$ being eliminated) cancels that leading term exactly, so each successive column of the array is asymptotically far more accurate than the raw trapezoid estimates feeding it.

Common pitfall: using the wrong denominator ($4^{j}-1$ instead of $4^{j-1}-1$), or extrapolating $R(k,j-1)$ against the wrong "previous row" entry $R(k-1,j-1)$ — the recurrence always looks one row up at the same column, not diagonally.

Source: Kreyszig, Advanced Engineering Mathematics, 10th ed., Ch. 19.5 (Numerical Integration and Differentiation: Romberg).

Question 5 of 7: Romberg Integration of Tabulated Data

Question 5: Romberg Integration of Tabulated Data (20 marks)