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+; <procedure +>
square; <procedure square>

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def square(x): return x*x
square # <function __main__.square(x)>
# `+` is syntax in Python — it cannot be referenced as a value (SyntaxError)

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(define abs
  (lambda (x) (cond [(< x 0) (- x)] [(= x 0) 0] [(> x 0) x])))

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(assert (= (abs -20) 20))

(assert (= (abs 0) 0))

(assert (= (abs 20) 20))

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(define abs2
  (lambda (x)
    (if (< x 0) (- x) x)))

(assert (= (abs2 -5) 5))

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(define (sqrt x)
  (define avg
    (lambda (x y)
      (/ (+ x y) 2)))
  (define (square x)
    (* x x))
  (define abs
    (lambda (x)
      (cond
        [(< x 0) (- x)]

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; avg/square/abs/improve are scoped inside sqrt — verified values:
; (avg 0 100)=>50  (square 3)=>9  (abs -1)=>1  (improve 50 100)=>26
(assert (< (abs (- (sqrt 9) 3)) 0.001)) ; ✅ approx check — sqrt returns 65537/21845, not exactly 3
(sqrt 9) ; => 65537/21845 ≈ 3.0

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TO FIND AN APPROXIMATION TO SQUARE ROOT OF X (9):
• MAKE A GUESS G (1)
• IMPROVE THE GUESS BY AVERAGING G AND X/G (1+(9/1))/2 = 5; 👇
• KEEP IMPROVING THE GUESS UNTIL IT IS GOOD ENOUGH (5+(9/5))/2 = 3,4; (3,4+(9/3/4))/2 = 3.02 ✅

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# GENERAL
METHOD FOR FINDING A FIXED POINT OF A FUNCTION F
(THAT IS, A VALUE Y SUCH THAT F(Y) = Y):

• START WITH A GUESS FOR Y
• KEEP APPLYING F OVER AND OVER UNTIL THE RESULT DOESN'T CHANGE VERY MUCH.

# APPLIED
TO COMPUTE A SQUARE ROOT OF X, FIND A >>FIXED POINT<<
OF THE FUNCTION F, WHICH TAKES Y OVER THE AVERAGE OF Y AND X/Y

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TO FIND AN APPROXIMATION TO SQUARE ROOT OF X (e.g. 9):
  • MAKE A GUESS G  (start: 1)
  • IMPROVE: average G and X/G  → (1 + 9/1) / 2 = 5
  • KEEP IMPROVING UNTIL GOOD ENOUGH:
      (5 + 9/5)   / 2 = 3.4
      (3.4 + 9/3.4) / 2 ≈ 3.02  ✅