Exercise 2.34 of SICP

Exercise 2.34: Evaluating a polynomial in x at a given value of x can be formulated as an accumulation. We evaluate the polynomial $$p(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x+ a_0$$ using a well-known algorithm called Horner’s rule, which structures the computation as $$(\cdots (a_n x + a_{n-1})x + \cdots + a_1)x + a_0$$ In other words, we start with $$a_n$$, multiply by $$x$$, add $$a_{n-1}$$, multiply by $$x$$, and so on, until we reach $$a_0$$. Fill in the following template to produce a procedure that evaluates a polynomial using Horner’s rule. Assume that the coefficients of the polynomial are arranged in a sequence, from $$a_0$$ through $$a_n$$.

(define (horner-eval x coefficient-sequence)
(accumulate (lambda (this-coeff higher-terms) <??>)
0
coefficient-sequence))


For example, to compute $$1 + 3x + 5x^3 + x^5$$ at $$x = 2$$ you would evaluate

(horner-eval 2 (list 1 3 0 5 0 1))


I included a non-abstracted version of Horner’s method for clarity.

(define (accumulate fn init-value items)
(if (null? items)
init-value
(fn (car items)
(accumulate fn init-value (cdr items)))))

(define (he x coeff-seq)
(if (null? coeff-seq)
0
(+ (car coeff-seq)
(* x (he x (cdr coeff-seq))))))

(define (horner-eval x coeff-seq)
(accumulate (lambda (this-coeff higher-terms)
(+ this-coeff (* x higher-terms)))
0
coeff-seq))

> (horner-eval 2 (list 1 3 0 5 0 1))
79
> (horner-eval 2 (list 1))
1
> (horner-eval 2 (list ))
0
> (horner-eval 2 (list 1 1))
3