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

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