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