SICP - Solution: Exercise 1.38

Oct 29, 2018 05:03 · 381 words · 2 minute read

Exercise 1.38: In 1737, the Swiss mathematician Leonhard Euler published a memoir De Fractionibus Continuis, which included a continued fraction expansion for $e−2$, where $e$ is the base of the natural logarithms. In this fraction, the $N_i$ are all 1, and the $D_i$ are successively 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, …. Write a program that uses your cont-frac procedure from Exercise 1.37 to approximate $e$, based on Euler’s expansion.

Solution

The key to this exercice is writing a function that will return successively 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, …. Let’s build this function step by step. In order to facilitate our work, we can write a couple of functions to display the first n elements of any serie and use a dummy implementation of d-euler:

(define (d-euler i)
  0)

(define (display-serie f n)
  (define (rec i)
    (display (f i)) (display ", ")
    (if (= i n)
        (newline)
        (rec (add1 i))))
  (rec 1))

(display-serie d-euler 12)

which evaluates to:

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,

The regular grouping of 3 indicates a modulo 3 going on. Let’s try to add it:

(define (d-euler i)
  (if (= (modulo i 3) 2)
      9
      1))

Which evaluates to:

1, 9, 1, 1, 9, 1, 1, 9, 1, 1, 9, 1,

Now we need to compute the proper number instead of 9. It seems to be twice the number of the triplet in the series:

(define (d-euler i)
  (if (= (modulo i 3) 2)
      (* 2(/ (+ i 1) 3))
      1))

and we evaluate it:

1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1,

Finally, we can put all together and have an approximation for $e$:

(define (cont-frac-iter n d k)
  (define (iter i result)
    (if (= 0 i)
        result
        (iter (sub1 i) (/ (n i) (+ result (d i))))))
  (iter (sub1 k) (/ (n k) (d k))))

(define (d-euler i)
  (if (= (modulo i 3) 2)
      (* 2(/ (+ i 1) 3))
      1))

(define (display-serie f n)
  (define (rec i)
    (display (f i)) (display ", ")
    (if (= i n)
        (newline)
        (rec (add1 i))))
  (rec 1))

(display-serie d-euler 12)

;  exact result is 0.718281828459
(cont-frac-iter (lambda (i) 1.0)
                 d-euler
                 1)  (newline)

Which gives:

0.7182818284590452