# SICP - Solution: Exercise 1.36

## Oct 28, 2018 04:03 · 243 words · 2 minute read

Exercise 1.36: Modify fixed-point so that it prints the sequence of approximations it generates, using the newline and display primitives shown in Exercise 1.22. Then find a solution to ${x^x=1000}$ by finding a fixed point of $x\mapsto{\log(1000)/\log(x)}$. (Use Scheme’s primitive log procedure, which computes natural logarithms.) Compare the number of steps this takes with and without average damping. (Note that you cannot start fixed-point with a guess of 1, as this would cause division by ${\log(1)=0}$.)

Solution

We can update fixed-point to show the sequence of approximation:

(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))


Without damping:

(display (fixed-point (lambda (x) (/ (log 100) (log x))) 2)) (newline)
2
6.643856189774725
2.4318468141875065
5.18220349492049
2.799103592609581
4.474083040767624
3.0735939466718745
4.101331665508161
3.263042550471523
3.89390861350767
3.3876156118518117
3.774338878643513
3.467160585111255
3.703882334279048
3.5170574911778987
3.6618000473016026
3.548020342802485
3.636455531432125
3.5671080319002773
3.621113728089196
3.5788280227423996
3.611797964445781
3.58600665606369
3.606130550334198
3.590397119344493
3.602678677221703
3.593079898951771
3.6005747438550824
3.594718296719416
3.5992918324007075
3.5957185433616616
3.598509347733154
3.5963290711231792
3.5980320110257193
3.596701676536737
3.597740794209986
3.5969290603825685
3.5975631160407135
3.597067815732927
3.597454706485773
3.5971524853260233
3.597388559396976
3.5972041504698735
3.5973481985922007
3.597235676058036
3.5973235715358247
3.5972549125527054
3.597308544705186
3.597266650359501
3.5972993756703846
3.597273812567488
3.5972937809205825
3.5972781828172766
3.5972903671194523
3.5972808494622672


With damping, the evaluation becomes:

(display (fixed-point (lambda (x) (average x (/ (log 100) (log x)))) 2)) (newline)
2
4.321928094887363
3.7340886925505887
3.6147319789883383
3.599236305662878
3.597499081067613
3.597308454173251
3.597287587624866
3.5972853041284294


It takes 55 steps to converge without damping, but only 9 steps with the damping method. It is clear that damping makes the convergence faster in this case.