Exercise 1.15
Posted on Thursday, September 13th, 2007
a. The procedure p is applied 5 times.
(sine 12.15))
(p (sine 4.05))) (p (p (sine 1.35)))) (p (p (p (sine .45))))) (p (p (p (p (sine .15)))))) (p (p (p (p (p (sine .05)))))))
b. So I’m really unclear about a lot of this but I will try to express what I can based on my research and what other people answered. It appears that the order of growth in space and number of steps are related to the division of the angle by 3. I am learning then that this means that it is logarithmic. A logarithmic unit is and abstraction technique used in mathematics to express any quantity that is proportional to the value of a logarithm function. A logarithm function is written like this: logb(x). It is used to represent y = logb(x) and is equivalent to x = by. So we are expressing that something (b but the b or base is usually irrelevant according to footnote 37 on pate 46 in SICP) is proportionate to some power (y) is equal to x.
So in our case, we are dividing a by 3 until we arrive to something less than 0.1. How many times we have to do that is what determines the number of steps our procedure has to take. So we would express that as θlog3(a) to say that the number or steps required is whatever it takes to multiply θ * 3 * 3 . . . * 3 to equal our angle, a. I guess in our case θ would be .1. And since the base of 3 is irrelevant our final answer is θlog(a). The 3 is irrelevant because we are really just expressing the idea of something being proportionate to something else.
Anyway, if you’re reading this try not to laugh too much :) My last math class was Calculus in my freshman year of college which was 7 years ago. And as I recall, I wasn’t too interested in math or programming at the time. Now I am seeing the importance of math and wish I could go back.
Please feel free to comment and correct my thinking.