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Simpson's Rule for Cubics


Date: 10/23/2000 at 02:30:23
From: George Stratis
Subject: Simpson's rule with cubic polynomials

I don't understand why Simpson's rule gives the exact answer for cubic 
polynomials. Can you explain that using a simple example, like x^3?


Date: 10/23/2000 at 06:18:32
From: Doctor Mitteldorf
Subject: Re: Simpson's rule with cubic polynomials

There's no simple verbal explanation that I know of. However, it's an 
interesting and instructive exercise to just do the calculation: Take 
the function y = x^3 between any two points - you may want to do a and 
b for generality, but I'll do 0 and 2 for illustration:

The Simpson weightings are 1, 4, 1. So the Simpson sum is

     (1*0^3 + 4*1^3 + 1*2^3) = 12

To get the Simpson integral estimate, divide by 1+4+1 = 6 (this is 
part of the averaging process) and then multiply by the length of the 
interval, 2.

So the Simpson integral for this case is 12/6 * 2 = 4. Of course, the 
exact value of the integral is 2^4/4, which is exactly 4.

Once you've proven this is true for any two numbers a...b, not just 
for 0...2, you can combine several intervals together to get Simpson 
sums with the usual weightings 1, 4, 2, 4, 2, 4, ...2, 4, 1. If it 
works exactly on each of the intervals separately, then it works on 
the entire interval.

- Doctor Mitteldorf, The Math Forum
  http://mathforum.org/dr.math/   


Date: 10/23/2000 at 07:43:40
From: Doctor Anthony
Subject: Re: Simpson's rule with cubic polynomials

Assume that y = f(x) is a CUBIC polynomial of the form

     y = ax^3 + bx^2 + cx + d

We consider two intervals each of width h, and take the origin at the 
middle ordinate. So the following three points are on the curve: 
(-h,y0), (0,y1), (h,y2).

The TRUE area is given by 

     INT[(ax^3 + bx^2 + cx + d) dx] from -h to h

Integrating, we get [ax^4/4 + bx^3/3 + cx^2/2 + dx] from -h to h.

       ah^4/4 + bh^3/3 + ch^2/2 + dh
     -[ah^4/4 - bh^3/3 + ch^2/2 - dh]
     ----------------------------------
               2bh^3/3 +         2dh                  

So the area is 

     A = 2bh^3/3 + 2dh

Substituting our three points into the equation of the curve, we have

     y0 = -ah^3 + bh^2 - ch + d
     y1 =   0   +  0   +  0 + d
     y2 =  ah^3 + bh^2 + ch + d  

       y0+y2  = 2bh^2   + 2d
     h(y0+y2) = 2bh^3   + 2dh
 (h/3)(y0+y2) = 2bh^3/3 + 2dh/3
              = 2bh^3/3 + 2dh - 4dh/3
              =       A       - 4(y1)(h/3)

So the exact area of the two slices is 

     A = (h/3)(y0 + 4y1 + y2)

Adding a succession of PAIRS of slices, we obtain

     A = (h/3)[y0 + 4y1 + y2 + y2 + 4y3 + y4 + ... + yn]
       = (h/3)[y0 + yn + 4(y1 + y3 + ... + y(n-1))  
                       + 2(y2 + y4 + ... + y(n-2))]

For the formula to work there must be an EVEN number of slices, that 
is, n must be an even number. In words the formula is

     A = (h/3)[(first + last) + 4*(sum of odd) + 2*(sum of even)]

and note that the formula is exact for polynomials of degree 3.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Calculus
High School Calculus

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