Sure—but can you offhand fit an exponential curve and calculate its summation? I’m sure it’s doable with the specified endpoints and # of periods (just steal a simple interest formula), but it’s more work than halving and multiplying.
Well… integral from t0 to t1 of exp(at+b) dt = (exp(at1+b)-exp(a*t2+b))/a i.e. the difference between the endpoints times the time needed to increase by a factor of e… a 6-million-fold increase is about 22.5 doublings (knowing 2^20 = 1 million), hence about 15 factors of e (knowing that ln 2 = 0.7) i.e. about one in 150… hence the total number of tourists is about 1 billion (about six times less than Rhwawn’s estimate—my eyeballs had told me it would be about one third… close enough!)
The first approximation which springs in my mind would be an exponential growth rather than a linear one.
Sure—but can you offhand fit an exponential curve and calculate its summation? I’m sure it’s doable with the specified endpoints and # of periods (just steal a simple interest formula), but it’s more work than halving and multiplying.
Well… integral from t0 to t1 of exp(at+b) dt = (exp(at1+b)-exp(a*t2+b))/a i.e. the difference between the endpoints times the time needed to increase by a factor of e… a 6-million-fold increase is about 22.5 doublings (knowing 2^20 = 1 million), hence about 15 factors of e (knowing that ln 2 = 0.7) i.e. about one in 150… hence the total number of tourists is about 1 billion (about six times less than Rhwawn’s estimate—my eyeballs had told me it would be about one third… close enough!)
I’m actually a little surprised that his such gross approximation puts it off by only 6x. For a Fermi estimate that’s perfectly acceptable.