Exercise 1.29 of SICP
Exercise 1.29: Simpson's Rule is a more accurate method of numerical integration than the method illustrated above. Using Simpson's Rule, the integral of a function f between a and b is approximated as
![\frac{h}{3}[y_0+4y_1+2y_2+4y_3+2y_4+ \cdots + 2y_{n-2} + 4y_{n-1} + y_n] \frac{h}{3}[y_0+4y_1+2y_2+4y_3+2y_4+ \cdots + 2y_{n-2} + 4y_{n-1} + y_n]](http://danboykis.com/wp-content/latex/b4d/b4dd6ba08ee1e0f8641525dced561c20-ffffff-000000-0.png)
where
, for some even integer n, and
. (Increasing n increases the accuracy of the approximation.) Define a procedure that takes as arguments f, a, b, and n and returns the value of the integral, computed using Simpson's Rule. Use your procedure to integrate cube between 0 and 1 (with n = 100 and n = 1000), and compare the results to those of the integral procedure shown above.
(define (cube x) (* x x x)) (define (simpson-integral f a b n) (define h (/ (- b a) n)) (define (next k) (+ k 1)) (define (coeff k) (cond ((or (= k 0) (= k n)) 1) ((even? k) 2) (else 4))) (define (term k) (* (coeff k) (f (+ a (* k h))))) (* (/ h 3.0) (sum term 0 next n))) (define (sum term a next b) (if (> a b) 0 (+ (term a) (sum term (next a) next b))))
There are several things I want to note here. All the auxiliary functions to represent the leading coefficients as well as h make life a lot easier. It's also important to note that the variables a and b in sum are not performing the same role as in the integral example earlier. In this case they are just counting indices from a = 0 to b = n. It might not be immediately obvious: When the function term is called from sum a (the start of the integrating interval) is bound to it from the original call to simpson-integral . Term only receives the index which it promptly evaluates.