
Re: My comments on "traditional" algorithms
Posted:
Apr 6, 2000 1:05 PM


>I came unstuck eventually when someone cleverly asked for 1.5245^20 (or >something like that). I argued that this was unfair because it was really 19 >calculations but was howled down.
I'm coming to the discussion a bit late, but if you have your log estimates memorized, you can do some reasonable approximations here, too... but that just postpones the crash by one level of complexity. Also the errors do multiply pretty quickly.
Let me see.... the only ones I remember off the top of my head is ln(10) is pretty close to 2.3 also ln(3) is 1.099 and logbase10 of 2 is .30103 (I like the palindrome, so I remembered it. Also, I know 2^10 is a bit more than 10^3, which nails the relationship down) ln(2) is a bit less than .7, but I don't remember the exact amount)
can I use those? 1.5245 is around 3/2, so ln(1.5245) must be a bit MORE than .399 times 20 must be about 8.. but what on earth is e^8? ln(10^3) is 6.9 ln(10^4) is 9.2, so it's somewhere in there... I need something with a ln of 8  6.9... well, that sounds like 3 (quite close!)... so 3*1000 = 3000 is my estimate.
Phooey, I undershot by about 50%. Heck, most of the error comes from replacing 1.5245 by 1.5, once you compound the error to the 20th power... as an estimate for 1.5^20 I'm only off by around 10%.

