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 (define (average x y) 
   (/ (+ x y) 2.0)) 
 (define (average-damp f) 
   (lambda (x) (average x (f x)))) 
 (define tolerance 0.00001) 
 (define (fixed-point f first-guess) 
   (define (close-enough? v1 v2) 
     (< (abs (- v1 v2)) tolerance)) 
   (define (try guess) 
     (let ((next (f guess))) 
       (if (close-enough? guess next) 
           (try next)))) 
   (try first-guess)) 
 (define (repeated f n) 
   (if (= n 1) 
       (lambda (x) (f ((repeated f (- n 1)) x))))) 
 (define (get-max-pow n) 
   (define (iter p r) 
     (if (< (- n r) 0) 
         (- p 1) 
         (iter (+ p 1) (* r 2)))) 
   (iter 1 2)) 
 (define (pow b p) 
   (define (even? x) 
     (= (remainder x 2) 0)) 
   (define (sqr x) 
     (* x x)) 
   (define (iter res a n) 
     (if (= n 0) 
         (if (even? n) 
             (iter res (sqr a) (/ n 2)) 
             (iter (* res a) a (- n 1))))) 
   (iter 1 b p)) 
 (define (nth-root n x) 
   (fixed-point ((repeated average-damp (get-max-pow n)) 
                 (lambda (y) (/ x (pow y (- n 1))))) 

Example: (nth-root 5 32)


The number of times to repeat average-damp can also be calculated using floor and log to the base 2 as follows

 (define (log2 x) (/ (log x) (log 2))) 
 (define (nth-root n x) 
 (fixed-point ((repeated average-damp (floor (log2 n)))  
               (lambda (y) (/ x (pow y (- n 1))))) 

The above solution of applying average-damp (floor (log2 n)) times damps more than is necessary, this overdamping looks to be an error in various other solutions on the net as well. For example it would appear one only needs to damp 5 times for the (2^17)th root to converge. At these high nth roots and high damping application amounts, we may get into a realm where the floating point error can not be ignored.