mohanrajendran

10
; 10

(+ 5 3 4)
; 12

(- 9 1)
; 8

(/ 6 2)

mohanrajendran

(/ (+ 5 4 (- 2 (- 3 (+ 6 (/ 4 5))))) 
   (* 3 (- 6 2) (- 2 7)))
; -37/150

mohanrajendran

(define (sq x) (* x x))
; sq

(define (sumsq x y) (+ (sq x) (sq y)))
; sumsq

(define (largest2sq a b c)
        (cond ((and (<= a b) (<= a c)) (sumsq b c))
              ((and (<= b a) (<= b c)) (sumsq a c))
              (else (sumsq a b))))

mohanrajendran

(define (a-plus-abs-b a b)
  ((if (> b 0) + -) a b))

When the if function is evaluated, based on the condition, the operator + or - is applied on the operands a and b. If b > 0, + operator is used, evaluating to a+b. Otherwise, - operator is issued which yields a-b. Ultimately, we evaluate a+|b|.

mohanrajendran

(define (p) (p))

(define (test x y)
  (if (= x 0)
      0
      y))

(test 0 (p))

mohanrajendran

(define (new-if predicate then-clause else-clause)
  (cond (predicate then-clause)
        (else else-clause)))

(define (sqrt-iter guess x)
  (new-if (good-enough? guess x)
          guess
          (sqrt-iter (improve guess x)
 x)))

mohanrajendran

The good-enough? function is not so effective when we use small numbers as arguments in the function.

For example, when we want to find the square root of 0.0001 which is 0.01,

(sqrt 0.0001)

;.03230844833048122

mohanrajendran

(define (cbrt-iter guess prev-guess x)
  (if (good-enough? guess prev-guess)
      guess
      (sqrt-iter (improve guess x) 
      		 guess
                 x)))
; cbrt-iter

(define (improve guess x)
  (/ (+ (/ x (square guess)) (* 2 guess)) 3))

mohanrajendran

Expansion of (+ 4 5) using first definition

(define (+ a b)
  (if (= a 0)
      b
      (inc (+ (dec a) b))))

(+ 4 5)
(inc (+ (dec 4) 5))

mohanrajendran

The following are the results from defining and evaluating the Ackermann's function on various arguments.

(define (A x y)
  (cond ((= y 0) 0)
  	((= x 0) (* 2 y))
	((= y 1) 2)
	(else (A (- x 1)
	      	 (A x (- y 1))))))
; A