Lecture 19: Roots: Eurekas from Mathematics

1. #|
       find-root: (N -> N) N -> N u #f
       (define (find-root f i) ...)
   
       Purpose: 
        j in [0,i] such that (f i) = 0
        otherwise #f
   |#
   
   (define (find-root f i)
     (cond
       ((zero? i) (cond
   		 ((= (f i) 0) i)
   		 (else #f)))
       (else (cond
   	    ((= (f i) 0) i)
   	    (else (find-root f (sub1 i)))))))
   

   Here is an improvement, based on the laws of cond:

   (define (find-root f i)
     (cond
       ((= (f i) 0) i)
       ((zero? i) #f)
       (else (find-root f (sub1 i)))))
   

   Does this tell us that find-root recurs on N? (NOPE)
   
   
2. let's find the root of a continuous function g

   #|
       find-root: (R -> R) R -> R
       (define (find-root f a) ...)
   
       Purpose: 
        x in [0,a] such that (abs (f a)) <= 0
   |#
   

   How do we know that f has a root? We don't. So let's assume a bit more: f(0) < 0 and f(a) > 0. Here is the graph:

   The Eureka: Mean Value Theorem. If f(b) is below 0 and f(a) is above 0, then f(x) = 0 for some x between b and a. -- So we split the interval into two parts and focus on the one that again satisfies the condition. When the interval is small enough, we stop.

   #|
      find-root : (number -> number) number number -> number
      the result $r$ is the left boundary of an interval [r,r+TOLERANCE]
      in which f has a root (\ie, (f x) = 0 for some x in the interval
      
      ASSUMPTION: f is continuous and monotone
   |#
   (define (find-root f left right)
     (cond
       ((<= (- right left) TOLERANCE) left)
       (else 
         (local ((define mid (/ (+ left right) 2)))
   	(cond
   	  ((<= (f left) 0 (f mid)) 
              (find-root left mid))
   	  (else 
              (find-root mid right)))))))
   

   
   
   
3. Now let's return to the function on N. What do we have to change? TOLERANCE must become 1 or 0? What else needs to change? The division. #| find-root-N : (N -> N) N N -> N Purpose: see above |# (define (find-root-N f left right) (cond ((<= (- right left) 1) left) (else (local ((define mid (quotient (+ left right) 2))) (cond ((<= (f left) 0 (f mid)) (find-root left mid)) (else (find-root mid right)))))))

   We will see another application of this idea somewhere down the road.
   
   

   General idea: consider the continuous counterpart, use knowledge from analysis, and go back to discrete world.