snowplow problem

[Albert Chan]

(03-19-2021 04:46 PM)Albert Chan Wrote:  Let z = x/(x+1), 0 < z < 1

z = (2-z)^(-k)

f(z) = z - (2-z)^(-k)
f'(z) = 1 - k*(2-z)^(-k-1)

With heavy snowstorm, k is big.
Guess of z=0 is good, apply Newton's method on it:

z = 0 - f(0)/f'(0) = 2^(-k)/(1-k*2^(-k-1)) = 1/(2^k-k/2)
x = z/(1-z) = 1/(2^k-k/2-1)

With almost no snow (k≈1) , we have x → inf
Or, matching above form: x = 1/(2^k-2 - c*(k-1))

We already shown k=2 ⇒ x = φ-1 ≈ 0.618
This suggested c = 2-1/x = 2-1/(φ-1) = 2-φ ≈ 0.382

The fit is pretty good

Note: table is minutes between snow started and snowplow started, for different k's

lua> fmt = function(...) print(('%.3g\t%.3g\t%.3g'):format(...)) end
lua> for k=1.25,5,0.25 do fmt(k, 60*snowplow(k), 60/(2^k-2-0.382*(k-1))) end

1.25   212    212
1.5    94.1   94.1
1.75   55.7   55.7
2      37.1   37.1
2.25   26.3   26.3
2.5    19.5   19.5
2.75   14.8   14.8
3      11.5   11.5
3.25   9.04   9.02
3.5    7.2    7.18
3.75   5.79   5.77
4      4.69   4.67
4.25   3.82   3.8
4.5    3.12   3.11
4.75   2.57   2.56
5      2.12   2.11