Programming Challenge: a classic trigonometry problem

[Albert Chan]

(03-12-2023 05:18 AM)Albert Chan Wrote:  edge(1/v) = 1/c ± 1/a = (1 ± c/a) / c

There are many ways to look at effect at the edges.
edge(v/c) = (1 ± c/a)^-1, with -1 power, gives minimum guess for q

Here is another way.

edge(v/c) = edge(1+1/q) = (1 ± b/a), should gives maximum guess for q

(1+q) = ±√((a²-b²)/c² + (v/c)²), with both ways for edge(v/c), we have:

lua> S = require2'solver'
lua> Q = fn'q: (1+q)^2*(1-q^-2) - k'
lua> a, b, c = 40, 30, 15
lua> k = (a+b)*(a-b)/(c*c)
lua> eps = -1e-9 -- swap worse point as first

lua> S.secant(Q, -1+sqrt(k+(1-c/a)^-2), -1+sqrt(k+(1+b/a)^2), eps, true)
1.3814094799322336
1.4846752526459288
1.4445475888872654
1.4443992821167904
1.4443995665205573
1.4443995665182763
1.4443995665182763

lua> S.secant(Q, -1-sqrt(k+(1+c/a)^-2), -1-sqrt(k+(1-b/a)^2), eps, true)
-2.907888028932993
-2.781463193869329
-- points swapped
-2.8799914833475135
-2.8807791881861102
-2.8807853504286443
-2.8807853490169144
-2.880785349016917
-2.880785349016917

The 2 guesses are asymptote of extreme cases.
Minimum q guess is better when k is huge
Maximum q guess is better when k is tiny.

For middle ground, we may use geometric mean of 2 edge(v/c):

q ≈ -1 ± sqrt(k + (±a+b)/(±a-c))