Romberg Integration from a Table: 04-BS-5, May 2017, Q5

Question 5: Romberg Integration from a Table (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 equally spaced ordinates of an unknown curve, spacing $h=1.0$ on $a=-4$ to $b=4$:

$x$	−4.0	−3.0	−2.0	−1.0	0	1.0	2.0	3.0	4.0
$f(x)$	15.0	13.0	12.0	12.0	14.0	17.0	19.0	20.0	17.0

Find. $\displaystyle\int_{-4}^{4}f(x)\,dx$ by Romberg’s algorithm, carried as far as the tabulated data allow.

The nine tabulated ordinates and the region whose area is required. Nine points support four halvings of the interval, hence a four-row Romberg array terminating at $R(4,4)$.

Approach. Build the first Romberg column as successively halved trapezoidal rules (each reusing the previous estimate plus only the new midpoints), then apply Richardson extrapolation across the rows to cancel successive powers of $H^2$. Nine ordinates permit $H=8,4,2,1$, so the array is $4\times4$ and terminates at $R(4,4)$.

First column, $k=1$ — the crude trapezoid. With $H_1=(b-a)/2^0=8$, $$R(1,1)=\dfrac{H_1}{2}\big[f(-4)+f(4)\big]=\dfrac{8}{2}\big[15.0+17.0\big]=128.000000$$
Refine to $k=2$ and $k=3$. Each step halves $H$ and adds only the newly sampled midpoints, weighted by the previous spacing. With $H_2=4$ the single new abscissa is $a+H_2=0$: $$R(2,1)=\tfrac12\big[128.000000+8\cdot f(0)\big]=\tfrac12\big[128+8(14.0)\big]=120.000000$$ With $H_3=2$ the two new abscissae are $a+H_3=-2$ and $a+3H_3=2$: $$R(3,1)=\tfrac12\big[120.000000+4\big(f(-2)+f(2)\big)\big]=\tfrac12\big[120+4(12.0+19.0)\big]=122.000000$$
Refine to $k=4$. With $H_4=1$ the four new abscissae are $-3,-1,1,3$: $$R(4,1)=\tfrac12\big[122.000000+2\big(13.0+12.0+17.0+20.0\big)\big]=\tfrac12\big[122+2(62.0)\big]=123.000000$$ This exhausts the table — every one of the nine ordinates has now been used, so no fifth row is possible.
Richardson extrapolation across the rows. The second column removes the leading $O(H^2)$ error using $j=2$, for which $4^{j-1}-1=3$: $$R(2,2)=120+\dfrac{120-128}{3}=117.333333,\qquad R(3,2)=122+\dfrac{122-120}{3}=122.666667$$ $$R(4,2)=123+\dfrac{123-122}{3}=123.333333$$ (These second-column entries are exactly the composite Simpson estimates; $R(4,2)=123.333333$ agrees with Simpson’s rule applied directly at $h=1$, a useful independent check.)
Third and fourth columns. With $j=3$ the divisor is $4^2-1=15$, and with $j=4$ it is $4^3-1=63$: $$R(3,3)=122.666667+\dfrac{122.666667-117.333333}{15}=123.022222$$ $$R(4,3)=123.333333+\dfrac{123.333333-122.666667}{15}=123.377778$$ $$R(4,4)=123.377778+\dfrac{123.377778-123.022222}{63}=123.383422$$
Assemble the array and state the result. The completed triangular array is

$k$ $H_k$ $R(k,1)$ $R(k,2)$ $R(k,3)$ $R(k,4)$

1 8 128.000000

2 4 120.000000 117.333333

3 2 122.000000 122.666667 123.022222

4 1 123.000000 123.333333 123.377778 123.383422

The best estimate is the last entry of the leading diagonal, $$\boxed{\int_{-4}^{4}f(x)\,dx\approx R(4,4)=123.3834\approx123.38}$$ The rapid settling of the final column — $123.02$, $123.38$, $123.383$ — indicates roughly four significant figures of accuracy, which is as much as nine data points can justify for a curve of unknown form.

$k$	$H_k$	$R(k,1)$	$R(k,2)$	$R(k,3)$	$R(k,4)$
1	8	128.000000
2	4	120.000000	117.333333
3	2	122.000000	122.666667	123.022222
4	1	123.000000	123.333333	123.377778	123.383422

Quantity	Result
$R(1,1),\ R(2,1),\ R(3,1),\ R(4,1)$	$128.000000,\ 120.000000,\ 122.000000,\ 123.000000$
$R(2,2),\ R(3,2),\ R(4,2)$	$117.333333,\ 122.666667,\ 123.333333$
$R(3,3),\ R(4,3)$	$123.022222,\ 123.377778$
$R(4,4)$ — best estimate	$\boxed{123.3834}$
Area sought	$\approx123.38$ square units

Topic: Romberg integration — recursive trapezoidal refinement combined with Richardson extrapolation.

Key relations: $H_k=(b-a)/2^{k-1}$; recursive trapezoid $R(k,1)=\frac12\big[R(k-1,1)+H_{k-1}\sum f(\text{new midpoints})\big]$; extrapolation $R(k,j)=R(k,j-1)+\dfrac{R(k,j-1)-R(k-1,j-1)}{4^{\,j-1}-1}$.

Why this works: the composite trapezoidal rule has an error expansion in even powers of $H$ only, $I-R(k,1)=c_1H^2+c_2H^4+\cdots$. Halving $H$ divides the leading term by exactly four, so the specific combination in the second column annihilates it and leaves an $O(H^4)$ estimate; repeating with divisors $4^2-1$ and $4^3-1$ kills the $H^4$ and $H^6$ terms in turn. That is why accuracy climbs so steeply across a row while the first column is still creeping. The recursive form of the first column matters practically too: each refinement reuses every ordinate already evaluated and touches only the new midpoints, so the whole array costs no more function evaluations than the finest trapezoid alone.

Common pitfall: using $H_k$ instead of $H_{k-1}$ as the multiplier on the new-midpoint sum (this paper’s stated form puts the factor $\frac12$ outside the bracket, so the effective weight is $H_{k-1}/2=H_k$ — getting this wrong halves or doubles the correction). Also common is running the array past the data: with nine ordinates the finest possible spacing is $H=1$, so any fifth row would require values that were never measured.

Source: Kreyszig, Advanced Engineering Mathematics, 10th ed., Ch. 19 §19.5 (Numeric Integration and Differentiation); Chapra & Canale, Numerical Methods for Engineers, 7th ed., Ch. 22 (Romberg Integration).

Question 5 of 7: Romberg Integration from a Table

Question 5: Romberg Integration from a Table (20 marks)