Title | Derivation of Simpson\'s Rule |
---|---|
Author | Hi Hi |
Course | Math |
Institution | Singapore University of Technology and Design |
Pages | 2 |
File Size | 78.7 KB |
File Type | |
Total Downloads | 34 |
Total Views | 154 |
Download Derivation of Simpson's Rule PDF
Simpson’s Rule Simpson’s rule is a numerical method that approximates the value of a definite integral by using quadratic polynomials. Let’s first derive a formula for the area under a parabola of equation y = ax2 + bx + c passing through the three points: (−h, y0 ), (0, y1 ), (h, y2 ). y
y = ax2 + bx + c
(0, y1 )
(−h, y0 )
(h, y2 )
y1
y2
y0 ∆x 0
−h
A= =
Z
h
x
h
(ax2 + bx + c) dx −h
ax3 bx2 + + cx 3 2
h
−h
2ah3 = + 2ch 3 h = 2ah2 + 6c 3
Since the points (−h, y0 ), (0, y1 ), (h, y2 ) are on the parabola, they satisfy y = ax2 + bx + c. Therefore, y0 = ah2 − bh + c y1 = c y2 = ah2 + bh + c Observe that y0 + 4y1 + y2 = (ah2 − bh + c) + 4c + (ah2 + bh + c) = 2ah2 + 6c. Therefore, the area under the parabola is A=
∆x h (y0 + 4y1 + y2 ) = (y0 + 4y1 + y2 ) . 3 3
Gilles Cazelais. Typeset with LATEX on April 23, 2008.
We consider the definite integral Z
b
f (x) dx.
a
We assume that f (x) is continuous on [a, b] and we divide [a, b] into an even number n of subintervals of equal length b−a ∆x = n using the n + 1 points x0 = a,
x1 = a + ∆x,
x2 = a + 2∆x,
...,
xn = a + n∆x = b.
We can compute the value of f (x) at these points. y0 = f (x0 ),
y1 = f (x1 ),
y2 = f (x2 ),
...,
yn = f (xn ).
y
y4
y0 y1
a = x0 x1
y2
x2
yn
y3
x3
yn−2
x4
···
∆x
···
yn−1
xn−2 xn−1 xn = b
x
We can estimate the integral by adding the areas under the parabolic arcs through three successive points. Z b ∆x ∆x ∆x (y2 + 4y3 + y4 ) + · · · + (y0 + 4y1 + y2 ) + (yn−2 + 4yn−1 + yn ) f (x) dx ≈ 3 3 3 a By simplifying, we obtain Simpson’s rule formula. Z
b a
f (x) dx ≈
∆x (y0 + 4y1 + 2y2 + 4y3 + 2y4 + · · · + 4yn−1 + yn ) 3
Example. Use Simpson’s rule with n = 6 to estimate Z
1
Solution. For n = 6, we have ∆x = x √ y = 1 + x3
4−1 6
1 √ 2
4p
1 + x3 dx.
= 0.5. We compute the values of y0 , y1 , y2 , . . . , y6 . 1.5 √ 4.375
2 3
2.5 √ 16.625
3 √ 28
3.5 √ 43.875
4 √ 65
Therefore, Z
1
4p
1 + x3 dx ≈
√ √ √ √ √ 0.5 √ 2 + 4 4.375 + 2(3) + 4 16.625 + 2 28 + 4 43.875 + 65 3
≈ 12.871...