## Solution to SICP Exercise 1.37 Solution to Exercise 1.37:

```; part a (recursive process) (define (cont-frac n d k) (define (sub-expr i) (if (= i k) (/ (n i) (d i)) (/ (n i) (+ (d i) (sub-expr (+ i 1)))))) (sub-expr 1))```

``` (define (golden-ratio k) (cont-frac (lambda (i) 1.0) (lambda (i) 1.0) k)) (define (iterations-required) (define (iter i previous) (let ((current (golden-ratio i))) (if (= (round (* current 10000)) (round (* previous 10000))) (- i 1) (iter (+ i 1) current)))) (iter 1 0.0)) ; > (iterations-required) ; 12 ```

```; part b (iterative process) (define (cont-frac-i n d k) (define (iter i sub-expr) (if (= i 0) sub-expr (iter (- i 1) (/ (n i) (+ (d i) sub-expr))))) (iter (- k 1) (/ (n k) (d k)))) ```

## Solution to SICP Exercise 1.36 Solution to Exercise 1.36:

```(define tolerance 0.00001) (define (fixed-point f first-guess) (define (close-enough? v1 v2) (< (abs (- v1 v2)) tolerance)) (define (try guess) (display guess) (newline) (let ((next (f guess))) (if (close-enough? guess next) next (try next)))) (try first-guess))```

``` (define (average a b) (/ (+ a b) 2)) (define result-with-damping (fixed-point (lambda (x) (average x (/ (log 1000) (log x)))) 2.0)) (display "result-with-damping: ") (display result-with-damping) (newline) (newline) (define result-without-damping (fixed-point (lambda (x) (/ (log 1000) (log x))) 2.0)) ```

```(display "result-without-damping: ") (display result-without-damping) ```

Output:

```2.0 5.9828921423310435 4.922168721308343 4.628224318195455 4.568346513136242 4.5577305909237005 4.555909809045131 4.555599411610624 4.5555465521473675 result-with-damping: 4.555537551999825```

``` ```

```2.0 9.965784284662087 3.004472209841214 6.279195757507157 3.759850702401539 5.215843784925895 4.182207192401397 4.8277650983445906 4.387593384662677 4.671250085763899 4.481403616895052 4.6053657460929 4.5230849678718865 4.577114682047341 4.541382480151454 4.564903245230833 4.549372679303342 4.559606491913287 4.552853875788271 4.557305529748263 4.554369064436181 4.556305311532999 4.555028263573554 4.555870396702851 4.555315001192079 4.5556812635433275 4.555439715736846 4.555599009998291 4.555493957531389 4.555563237292884 4.555517548417651 4.555547679306398 4.555527808516254 4.555540912917957 result-without-damping: 4.555532270803653 ```

## Solution to SICP Exercise 1.35 Solution to Exercise 1.35:

```(define golden-ratio (fixed-point (lambda (x) (+ 1 (/ 1 x))) 1.0))```

``` ```

```> golden-ratio 1.6180327868852458 ```

## Solution to SICP Exercise 1.34 Solution to Exercise 1.34:

DrScheme prints out an error message that reads:

`procedure application: expected procedure, given: 2; arguments were: 2`.

This is hardly surprising since `(f f)` evaluates to `(f 2)`, which evaluates to `(2 2)`, which is nonsense. The number `2` is not a procedure.

## Solution to SICP Exercise 1.33 Solution to Exercise 1.33:

```(define (filtered-accumulate predicate combiner null-value term a next b) (define (iter a result) (if (> a b) result (iter (next a) (if (predicate a) (combiner (term a) result) result)))) (iter a null-value))```

``` ; part a (define (square x) (* x x)) (define (sum-of-squared-primes a b) (define (square-and-add i result) (+ (square i) result)) (filtered-accumulate prime? square-and-add 0 identity a inc b)) ```

```; part b (define (product-of-positives-relatively-prime-to n) (define (relatively-prime? i) (= 1 (gcd i n))) (filtered-accumulate relatively-prime? * 1 identity 1 inc (- n 1))) ```

## Solution to SICP Exercise 1.32 Solution to Exercise 1.32:

`; part a (recursive process)`

``` (define (accumulate-r combiner null-value term a next b) (if (> a b) null-value (combiner (term a) (accumulate-r combiner null-value term (next a) next b)))) (define (sum-r term a next b) (accumulate-r + 0 term a next b)) (define (product-r term a next b) (accumulate-r * 1 term a next b)) ; part b (iterative process) ```

```(define (accumulate-i combiner null-value term a next b) (define (iter a result) (if (> a b) result (iter (next a) (combiner result (term a))))) (iter a null-value)) ```

## Solution to SICP Exercise 1.31 Solution to Exercise 1.31:

```; part a (recursive process) (define (product-r term a next b) (if (> a b) 1 (* (term a) (product-r term (next a) next b))))```

``` (define (identity i) i) (define (inc i) (+ i 1)) (define (factorial n) (product-r identity 1 inc n)) (define (pi-approximator terms) (define (numerator i) (+ i (if (odd? i) 1 2))) (define (denominator i) (+ i (if (odd? i) 2 1))) (define (pi-term i) (/ (numerator i) (denominator i))) (* 4.0 (product-r pi-term 1 inc terms))) ```

```; part b (iterative process) (define (product-i term a next b) (define (iter a result) (if (> a b) result (iter (next a) (* result (term a))))) (iter a 1)) ```

## Solution to SICP Exercise 1.30 Solution to Exercise 1.30:

```(define (sum term a next b) (define (iter a result) (if (> a b) result (iter (next a) (+ result (term a))))) (iter a 0)) ```

## Solution to SICP Exercise 1.29 Solution to Exercise 1.29:

```(define (sum term a next b) (if (> a b) 0 (+ (term a) (sum term (next a) next b))))```

``` (define (inc n) (+ n 1)) ```

```(define (simpsons-integral f a b n) (define (do-it h) (define (y k) (f (+ a (* k h)))) (define (simpson-term k) (* (y k) (cond ((or (= k 0) (= k n)) 1) ((odd? k) 4) (else 2)))) (* (/ h 3) (sum simpson-term 0 inc n))) (do-it (/ (- b a) n))) ```