Romberg Integration of Tabulated Data: 04-BS-5, May 2013, 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 data
$x$	-1.00	-0.75	-0.50	-0.25	0	0.25	0.50	0.75	1.00
$y$	3.50	12.50	14.00	17.50	16.00	13.50	12.00	12.50	18.50

Given. 9 tabulated $(x,y)$ pairs spanning $[-1,1]$ in steps of $h=0.25$ (table above).

Find. A Romberg estimate of $\displaystyle\int_{-1}^{1}y\,dx$.

Approach. Build the trapezoidal row $R(k,1)$ at $h=2,1,0.5,0.25$ (using the nested subsets of 2, 3, 5, 9 tabulated points), then Richardson-extrapolate the triangular array using $R(k,j)=R(k,j-1)+\dfrac{R(k,j-1)-R(k-1,j-1)}{4^{j-1}-1}$.

Trapezoidal row $R(k,1)$ for $h=2,1,0.5,0.25$. $$R(1,1)=\tfrac{2}{2}\big[f(-1)+f(1)\big]=1(3.50+18.50)=\boxed{22.00}$$ $$R(2,1)=\tfrac{1}{2}\big[f(-1)+2f(0)+f(1)\big]=0.5(3.50+32.00+18.50)=\boxed{27.00}$$ $$R(3,1)=\tfrac{0.5}{2}\Big[f(-1)+f(1)+2\!\!\sum_{-0.5,0,0.5}\!\!f\Big]=0.25\big(22.00+2(42.00)\big)=\boxed{26.50}$$ $$R(4,1)=\tfrac{0.25}{2}\Big[f(-1)+f(1)+2\!\!\!\sum_{\text{7 interior pts}}\!\!\!f\Big]=0.125\big(22.00+2(98.00)\big)=\boxed{27.25}$$
First Richardson column, $j=2$ (divide by $4^1-1=3$). $$R(2,2)=27.00+\tfrac{27.00-22.00}{3}=\boxed{28.667},\qquad R(3,2)=26.50+\tfrac{26.50-27.00}{3}=\boxed{26.333}$$ $$R(4,2)=27.25+\tfrac{27.25-26.50}{3}=\boxed{27.500}$$
Second column, $j=3$ (divide by $4^2-1=15$), and third column, $j=4$ (divide by $4^3-1=63$). $$R(3,3)=26.333+\tfrac{26.333-28.667}{15}=\boxed{26.178},\qquad R(4,3)=27.500+\tfrac{27.500-26.333}{15}=\boxed{27.578}$$ $$R(4,4)=27.578+\tfrac{27.578-26.178}{63}=\boxed{27.600}$$

$k$	$R(k,1)$	$R(k,2)$	$R(k,3)$	$R(k,4)$
1	22.000
2	27.000	28.667
3	26.500	26.333	26.178
4	27.250	27.500	27.578	27.600

Concept & Theory

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

Key relations. $R(k,1)$ = trapezoid rule with $H_k=(b-a)/2^{k-1}$; $R(k,j)=R(k,j-1)+\dfrac{R(k,j-1)-R(k-1,j-1)}{4^{j-1}-1}$.

Why this works. The trapezoidal rule's error is an even power series in $h$ ($h^2,h^4,h^6,\dots$); each Richardson step algebraically cancels the leading error term, so $R(4,4)$ is effectively an 8th-order-accurate estimate versus $R(4,1)$'s 2nd order — visible here in how the estimate settles from $22.00$ to $27.60$ as the table fills in.

Common pitfall. Using the wrong divisor $4^{j-1}-1$ (not $4^j-1$) when extrapolating column $j$, and re-sampling the function for each coarser row instead of reusing exactly the tabulated subset already used at the finer level.

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 19 (§19.3 Numerical Integration, Romberg Integration).

Question 5 of 7: Romberg Integration of Tabulated Data

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