Euler's Continued Fraction in Lisp

In 1748, Leonhard Euler published a formula describing an identity that connected and generalized an infinite series and infinite continued fraction. The continued fraction function can be defined as follows. Let n represent a numerator, d a denominator, and k a number of iterations. Upon each loop, our function will decrement its count and recurse on itself until reaching its iteration limit:

$$\begin{array}{r}\text{cont-frac}(n,d,k)=\{\begin{array}{ll}\text{result}& \text{if }i=0\\ \text{iter}(i-1,\frac{n(i)}{d(i)+\text{result}})& \text{otherwise}\end{array}\end{array}$$ .. where the iter function is defined as follows: $$\begin{array}{r}\text{iter}(i,\text{result})=\{\begin{array}{ll}\text{result}& \text{if }i=0\\ \text{iter}(i-1,\frac{n(i)}{d(i)+\text{result}})& \text{otherwise}\end{array}\end{array}$$

Altogether, courtesy of Wikipedia:

$$\begin{array}{r}{\displaystyle a_0+a_0{a}_1+a_0{a}_1{a}_2+\cdots +a_0{a}_1{a}_2\cdots a_n=\frac{a_0}{1-\frac{a_1}{1+a_1-\frac{a_2}{1+a_2-\frac{\ddots }{\ddots \frac{a_{n-1}}{1+a_{n-1}-\frac{a_n}{1+a_n}}}}}}}\end{array}$$ To calculate the continued fraction using higher-order functions, such that our function accepts another function as input, we can use lambda functions, like so: $$\begin{array}{r}\text{cont-frac}(\lambda (i)1.0,\lambda (i)1.0,10)\end{array}$$

Putting it together, we can write this recursive function concisely in Lisp, with k defining the iterations or approximation accurary. We'll set it to 10.

(defun cont-frac (n d k)
  (labels ((iter (i result)
             (if (= i 0)
                 result
                 (iter (- i 1) (/ (funcall n i) (+ (funcall d i) result))))))
    (iter k 0)))

(format t "~a" (cont-frac (lambda (i) 1.0) (lambda (i) 1.0) 10))

Upon execution, the Lisp interpreter iterates ten times, then prints our result, 0.6179775.