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a.

```
(define (product term a next b)
(if (> a b) 1
(* (term a) (product term (next a) next b))))
```

factorial function, in terms of product function above, can be written as below.

(define (identity x) x) (define (next x) (+ x 1)) (define (factorial n) (product identity 1 next n))

approximations to pi using john wallis' formula, can be found by defining a new term to be used by product function as below

```
(define (pi-term n)
(if (even? n)
(/ (+ n 2) (+ n 1))
(/ (+ n 1) (+ n 2))))
```

And it can be used as follows.

(* (product pi-term 1 next 6) 4) ;;= 3.3436734693877552 (* (product pi-term 1 next 100) 4) ;;= 3.1570301764551676

b.

```
(define (product term a next b)
(define (iter a res)
(if (> a b) res
(iter (next a) (* (term a) res))))
(iter a 1))
```

Another approach would be to recognize the series as 4*4*6*6*.../3*3*5*5*... The leading factor of 2 in the numerator must be dealt with.

```
(define (pi n)
(define (term x) (* x x))
(define (next x) (+ x 2))
; since we are increasing the numbers by two on every iteration
(define limit (* n 2))
; upper term: - 2 always goes first, start building product from 4
; - as 2 numbers are skipped, the limit must respect that, too
; - since we are squaring one time too often at the end,
; we have to divide that back out of the result
; lower term: start with 3, which is 1 more than the upper term
; -> so increase limit by 1
;
(* 4 (/ (/ (* 2 (product term 4 next (+ limit 2)))
(+ limit 2))
(product term 3 next (+ limit 1)))))
```

Yielding:

(pi 6) ;;= 3.255721745 (pi 100) ;;= 3.149378473

n defines steps in terms of pairs of factors. So where the above solution does 6 steps when n=6, this solution does 12 steps when n=6:

(pi 3) ;;= 3,343673469 (pi 50) ;;= 3,157030176

A third way of defining the pi-term function, using integer arithmetic instead of the even? predicate:

```
(define (pi n)
(define (pi-term n)
(/ (* 2 (ceiling (/ (+ 1 n) 2)))
(+ 1 (* 2 (ceiling (/ n 2))))))
(product pi-term 1 inc n))
```

A fourth way of doing it is to recognize pi = 4 * (2 * (1/3) *4 * (1/5) ...) * ((1/3) * 4 * (1/5) ...)

Arithmetic progression formula was used to generate the nth number of the sequence

namely a_k = a_1 + (k-1)d

```
(define (pi-prod n)
(define (inv k)
(if (even? k)
k
(/ 1 k)))
(* 4.0 (* (product 2 (+ 2 (- n 1)) inc inv )
(product 3 (+ 3 (- n 1)) inc inv))))
```

Another way is to recognise that the formula for pi is composed of two separate products:

pi = 4 * ((2/3 * 4/5 * 6/7 ... * n/(n+1)) * (4/3 * 6/5 * ... * n/(n-1))

This gives:

```
(define (approx-pi n)
(* 4
(product
(lambda (n) (/ n (+ 1 n)))
2
(lambda (n) (+ n 2))
n)
(product
(lambda (n) (/ n (- n 1)))
4
(lambda (n) (+ n 2))
n)))
```

Here's how I visualize the problem: If you recognize that the denominator follows the pattern 3,3,5,5,7,7 and that for any given pair of n's, the numerators will be n-1 and n+2, then there is no need to take into account the "stray" 2 at the start of the numerator sequence.

(define (product-iter f a next b) (define (iter a result) (if (> a b) result (iter (next a) (* (f a) result)))) (iter a 1)) (define (pi-term n) (* (/ (- n 1.0) n) (/ (+ n 1.0) n))) (define (next-pi n) (+ n 2.0)) (define (pi accuracy) (* 4 (product-iter pi-term 3.0 next-pi accuracy)))

My method is the same as the above. A clarification though, this method works by splitting the formula into:

(2⋅4/3⋅3) ⋅ (4⋅6/5⋅5) ⋅ (6⋅8/7⋅7)

We can then see the general expression for each pair of terms:

((n-1)⋅(n+1))/n^2 --> (n^2 - 1)/n^2 --> n^2/n^2 - 1/n^2 --> 1 - 1/n^2

This translates into code as:

(define (pi-term a) (- 1 (/ 1 (square a)))) (define (pi-next a) (+ a 2)) (define (pi accuracy) (* 4.0 (product-iter pi-term 3 pi-next (+ (* accuracy 2) 1))))