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Chapter 6: Techniques of Integration

Section 6.7: Numeric Methods

Example 6.7.9

Sketch a derivation of Simpson's rule.

Solution

Mathematical Solution

•	Figure 6.7.8(a) shows a black curve with five points corresponding to $x = x_{k}, k = 0, \dots, 4$ , with the first three points interpolated with a quadratic whose graph is shown in red along the curve.

•	The rightmost of these three points, and the next two are interpolated by a second quadratic whose graph is shown in green along the curve.

•	These five nodes determine two contiguous "double panels." Simpson's rule amounts to approximating the area under a curve with the area under such approximating quadratics.

•	The line along which the two double panels in the figure touch deliberatelyshows both red and green.

use plots in
module()
local g,a,b,c,P1,P2,p1,p2,p3,p4,p5,p6,p7;
g := x-> 2+sin(x);
a := .5; b := -1.;c:=.1;
P1:=.2347601860*x^2+1.076231171*x+2.;
P2:=-.2347601860*x^2+1.076231171*x+2.;
p1:=plot([[x,P1,x =-1..0],[x,P2,x=0..1],[x,g(x),x=-1..1],[x,0,x=-1..1]],color=[red,green,black,black], thickness=[8,8,1,1],transparency=[.7,.7,0,0],scaling=constrained,axes=none,view=[-1..1,-.2..3]);
p2:=textplot({[-1,-.1,x[k]],[0,-.1,x[k+2]],[1,-.1,x[k+4]],[-.5,-.1,x[k+1]],[.5,-.1,x[k+3]]});
p3:=plot([[[-1,g(-1)],[-1,0]],[[-.5,g(-.5)],[-.5,0]],[[0,0],[0,g(0)]]],color=[red$3],thickness=[1,1,4]);
p4:=plot([[[.01,0],[.01,g(.01)]],[[.5,0],[.5,g(.5)]],[[1,0],[1,g(1)]]],color=[green$3],thickness=[2,1,1]);
p5:=plot([[-1,g(-1)],[-.5,g(-.5)],[0,g(0)],[.5,g(.5)],[1,g(1)]],style=point,symbol=solidcircle,symbolsize=15,color=black):
p6:=textplot({[-1,g(-1)+c,typeset(``(x[k],f[k]))],[-.5,g(-.5)+c,typeset(``(x[k+1],f[k+1]))],[0,g(0)+c,typeset(``(x[k+2],f[k+2]))],[.5,g(.5)+c,typeset(``(x[k+3],f[k+3]))],[1,g(1)+c,typeset(``(x[k+4],f[k+4]))]},align=left);
p7:=display(p1,p2,p3,p4,p5,p6);
print(p7);
end module:
end use:

Figure 6.7.8(a) Two interpolating quadratics

•

The expression for a quadratic interpolating the three points in Table 6.7.8(a) is cumbersome, and integrating it between $x = x_{k} = a + k h$ and $x = x_{k + 2} = a + (k + 2) h$ is an even more complex task. Surprisingly, though, this definite integral simplifies to $\frac{h}{3} (f_{k} + 4 f_{k + 1} + f_{k + 2})$ , so that the sum of a succession of such expressions collapses to the compact form of Simpson's rule.

$(x_{k}, f_{k})$	$= (a + k h, f_{k})$
$(x_{k + 1}, f_{k + 1})$	$= (a + (k + 1) h, f_{k + 1})$
$(x_{k + 2}, f_{k + 2})$	$= (a + (k + 2) h, f_{k + 2})$
Table 6.7.8(a) Points interpolated by quadratic

•	The actual calculations are so cumbersome that they are carried out only in Maple.

Maple Solution

Interpolate $(x_{k}, f_{k}), (x_{k + 1}, f_{k + 1}), (x_{k + 2}, f_{k + 2})$ with a polynomial quadratic in $x$

1.	Enter $[a + (k + n) h, f_{k + n}]$ and press the Enter key.

2.	Context Panel: Sequence≻ $n = 0, 2$

3.	Context Panel: Conversions≻To List

4.	Context Panel: Curve Fitting≻Polynomial Interpolation≻x

$[a + (k + n) h, f_{k + n}]$

$\overset{sequence w.r.t. n}{\to}$

$[h k + a, f_{k}], [a + (k + 1) h, f_{k + 1}], [a + (k + 2) h, f_{k + 2}]$

$\overset{to list}{\to}$

$[[h k + a, f_{k}], [a + (k + 1) h, f_{k + 1}], [a + (k + 2) h, f_{k + 2}]]$

$\overset{polynomial curve fit}{\to}$

$\frac{1}{2} \frac{(f_{k + 2} - 2 f_{k + 1} + f_{k}) x^{2}}{h^{2}} + \frac{1}{2} \frac{(- 2 h k f_{k} + 4 h k f_{k + 1} - 2 h k f_{k + 2} - 2 a f_{k} + 4 a f_{k + 1} - 2 a f_{k + 2} - 3 h f_{k} + 4 h f_{k + 1} - h f_{k + 2}) x}{h^{2}} + \frac{1}{2} \frac{h^{2} k^{2} f_{k} - 2 h^{2} k^{2} f_{k + 1} + h^{2} k^{2} f_{k + 2} + 2 a h k f_{k} - 4 a h k f_{k + 1} + 2 a h k f_{k + 2} + 3 h^{2} k f_{k} - 4 h^{2} k f_{k + 1} + h^{2} k f_{k + 2} + a^{2} f_{k} - 2 a^{2} f_{k + 1} + a^{2} f_{k + 2} + 3 a h f_{k} - 4 a h f_{k + 1} + a h f_{k + 2} + 2 h^{2} f_{k}}{h^{2}}$

(1)

Find the area under the interpolating polynomial

•	Context Panel: Definite Integral template Integrate between $a + k h$ and $a + (k + 2) h$ then press the Enter key.

•	Context Panel: Simplify≻Simplify

$\int_{a + k h}^{a + (k + 2) h} &DifferentialD; x$

$\frac{1}{6} \frac{(f_{k + 2} - 2 f_{k + 1} + f_{k}) ({(a + (k + 2) h)}^{3} - {(h k + a)}^{3})}{h^{2}} + \frac{1}{4} \frac{(- 2 h k f_{k} + 4 h k f_{k + 1} - 2 h k f_{k + 2} - 2 a f_{k} + 4 a f_{k + 1} - 2 a f_{k + 2} - 3 h f_{k} + 4 h f_{k + 1} - h f_{k + 2}) ({(a + (k + 2) h)}^{2} - {(h k + a)}^{2})}{h^{2}} + \frac{1}{2} \frac{(h^{2} k^{2} f_{k} - 2 h^{2} k^{2} f_{k + 1} + h^{2} k^{2} f_{k + 2} + 2 a h k f_{k} - 4 a h k f_{k + 1} + 2 a h k f_{k + 2} + 3 h^{2} k f_{k} - 4 h^{2} k f_{k + 1} + h^{2} k f_{k + 2} + a^{2} f_{k} - 2 a^{2} f_{k + 1} + a^{2} f_{k + 2} + 3 a h f_{k} - 4 a h f_{k + 1} + a h f_{k + 2} + 2 h^{2} f_{k}) ((k + 2) h - k h)}{h^{2}}$

$\overset{simplify}{=}$

$\frac{1}{3} h (4 f_{k + 1} + f_{k + 2} + f_{k})$

(2)

Add the areas under three contiguous interpolating parabolas

•	Expression palette: Evaluation template Set $k = 0, 2, 4$ and add the results.

•	Context Panel: Factor

$\binom{}{} | \binom{}{k = 0} + \binom{}{} | \binom{}{k = 2} + \binom{}{} | \binom{}{k = 4}$

$\frac{1}{3} h (4 f_{1} + f_{2} + f_{0}) + \frac{1}{3} h (4 f_{3} + f_{4} + f_{2}) + \frac{1}{3} h (4 f_{5} + f_{6} + f_{4})$

$\overset{factor}{=}$

$\frac{1}{3} h (4 f_{1} + 2 f_{2} + f_{0} + 4 f_{3} + 2 f_{4} + 4 f_{5} + f_{6})$

Generalize to the case of $n / 2$ panels and $n + 1$ nodes, $x_{0}, x_{1}, \dots, x_{n}$ .

•

If $a = x_{0}$ and $b = x_{n}$ , then there are $n + 1$ equispaced nodes in the interval $[a, b]$ . If $n$ is even, then the interval can be subdivided into $n / 2$ contiguous "double panels" spanning three nodes. These three nodes can be interpolated by a quadratic polynomial (whose graph is a parabola), and the area under the generic parabola so formed is $A_{k} = \frac{h}{3} (f_{k} + 4 f_{k + 1} + f_{k + 2})$ . The sum of $n / 2$ areas of this type will then be

$\sum_{k = 0}^{n / 2} A_{k} = \frac{h}{3} (y_{0} + 4 \sum_{k = 1}^{n / 2} f_{2 k - 1} + 2 \sum_{k = 1}^{n / 2 - 1} f_{2 k} + f_{n})$

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