Romberg Integration from Tabulated Data: 04-BS-5, December 2019, Q5

Question 5: Romberg Integration from 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. Nine tabulated points of an unknown $y=f(x)$ on $[-2,2]$ at spacing $0.5$ (below), $a=-2.0$, $b=2.0$.

$x$	$-2.0$	$-1.5$	$-1.0$	$-0.5$	$0$	$0.5$	$1.0$	$1.5$	$2.0$
$y$	$123$	$126$	$128$	$129$	$134$	$149$	$174$	$208$	$249$

Fig. 3 — the nine tabulated points on $[-2,2]$; the shaded region under this curve (down to the x-axis) is the area Romberg's algorithm estimates.

Find. The area $\displaystyle\int_{-2}^{2}y\,dx$ via Romberg's algorithm, using the full triangular array up to $R(4,4)$.

Approach. Compute the composite-trapezoidal estimate $R(k,1)$ at four successively halved step sizes ($H_1=4,H_2=2,H_3=1,H_4=0.5$, using $1,2,4,8$ trapezoids respectively) directly from the tabulated $y$-values, then apply Richardson extrapolation column by column.

Trapezoidal column, $R(k,1)$. With $H_1=4$: $R(1,1)=\frac{4}{2}[123+249]=744$. With $H_2=2$ (points $-2,0,2$): $R(2,1)=\frac{2}{2}[123+2(134)+249]=640$. With $H_3=1$ (points $-2,-1,0,1,2$): $R(3,1)=\frac{1}{2}[123+2(128+134+174)+249]=622$. With $H_4=0.5$ (all nine points): $R(4,1)=\frac{0.5}{2}[123+2(126+128+129+134+149+174+208)+249]=617$.
First Richardson column, $R(k,2)=R(k,1)+\dfrac{R(k,1)-R(k-1,1)}{3}$. $$R(2,2)=640+\tfrac{640-744}{3}=605.333333,\quad R(3,2)=622+\tfrac{622-640}{3}=616.000000,\quad R(4,2)=617+\tfrac{617-622}{3}=615.333333$$
Second column, $R(k,3)=R(k,2)+\dfrac{R(k,2)-R(k-1,2)}{15}$. $$R(3,3)=616+\tfrac{616-605.333333}{15}=616.711111,\qquad R(4,3)=615.333333+\tfrac{615.333333-616}{15}=615.288889$$
Third column, $R(4,4)=R(4,3)+\dfrac{R(4,3)-R(3,3)}{63}$. $$R(4,4)=615.288889+\frac{615.288889-616.711111}{63}=\boxed{615.266314}$$

$k$	$R(k,1)$	$R(k,2)$	$R(k,3)$	$R(k,4)$
1	$744.000000$
2	$640.000000$	$605.333333$
3	$622.000000$	$616.000000$	$616.711111$
4	$617.000000$	$615.333333$	$615.288889$	$615.266314$

Quantity	Result
Trapezoid estimates $R(1,1)\dots R(4,1)$	$744,\ 640,\ 622,\ 617$
Best Romberg estimate	$R(4,4)=615.266314$

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

Key relations: composite trapezoidal rule $R(k,1)=\frac{H_k}{2}\big[f(a)+2\sum f(\text{interior})+f(b)\big]$; Richardson extrapolation $R(k,j)=R(k,j-1)+\dfrac{R(k,j-1)-R(k-1,j-1)}{4^{j-1}-1}$, which cancels the leading error term of order $H^{2(j-1)}$ at each step.

Why this works: the trapezoidal rule's error expansion contains only even powers of $H$ (Euler–Maclaurin), so successively combining two step-halved estimates with weight $4^{j-1}$ eliminates one more order of error each column — turning four crude $O(H^2)$ estimates into one estimate accurate to $O(H^8)$, without ever needing the function itself (only the tabulated values).

Common pitfall: re-using a whole trapezoidal sum's interior points as "new" points at the next level instead of only the odd-indexed halving points; or applying the wrong denominator ($4^{j-1}-1$, not $4^j-1$) when extrapolating column $j$.

Source: Kreyszig, Advanced Engineering Mathematics, 10th ed., Ch. 19.4 (Romberg Integration); Chapra & Canale, Numerical Methods for Engineers, 7th ed., Ch. 22.

Question 5 of 7: Romberg Integration from Tabulated Data

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