**Exercise 1.37:**

a. An infinite *continued fraction* is an expression of the form

$$ f = \cfrac{N_1}{D_1 + \cfrac{N_2}{D_2 + \cfrac{N_3}{D_3 +\cdots}}} $$

As an example, one can show that the infinite continued fraction expansion with the Ni
and the Di all equal to 1 produces \( \frac{1}{\phi} \), where \( \phi \)
is the golden ratio (described in section 1.2.2). One way to approximate an infinite continued
fraction is to truncate the expansion after a given number of terms. Such a truncation –
a so-called *k-term finite continued fraction* – has the form

$$ f = \cfrac{N_1}{D_1 + \cfrac{N_2}{{\ddots,} + \cfrac{N_k}{D_k}}} $$

Suppose that `n`

and `d`

are procedures of one argument (the term index i) that return the Ni and
Di of the terms of the continued fraction. Define a procedure `cont-frac`

such that evaluating
`(cont-frac n d k)`

computes the value of the k-term finite continued fraction. Check your procedure
by approximating \( \frac{1}{\phi} \) using

```
(cont-frac (lambda (i) 1.0)
(lambda (i) 1.0)
k)
```

for successive values of `k`

. How large must you make `k`

in order to get an approximation that is accurate to
4 decimal places?

```
(define (cont-frac n d k)
(define (cf i)
(if (= i (+ k 1))
0
(/ (n i)
(+ (d i) (cf (+ i 1))))))
(if (not (> k 0))
0
(cf 1)))
```

```
> (cont-frac (lambda (i) 1.0)
(lambda (i) 1.0)
11)
.6180555555555556
```

It takes 11 iterations to get to 4 decimal places within the reciprocal of golden ratio.

$$ \frac{1}{\phi} = 0.618033988 $$

It’s amusing that the LaTeX looks almost identical to the above definition in scheme:

$$ \cfrac{N_1}{D_1 + \cfrac{N_2}{D_2 + \cfrac{N_3}{D_3 +\cdots}}} $$

b. If your `cont-frac`

procedure generates a recursive process, write one that generates an iterative process.
If it generates an iterative process, write one that generates a recursive process.

```
(define (cont-frac-iter n d k)
(define (iter k result)
(if (= k 0)
result
(iter (- k 1) (/ (n k) (+ (d k) result)))))
(iter k 0))
```

```
> (cont-frac-iter (lambda (i) 1.0)
(lambda (i) 1.0)
11)
.6180555555555556
```