Return a vector with the probabilities.
The flexit equation is published in:
Abreu-Grobois, F.A., Morales-Mérida, B.A., Hart, C.E., Guillon, J.-M., Godfrey, M.H.,
Navarro, E. & Girondot, M. (2020) Recent advances on the estimation of the thermal
reaction norm for sex ratios. PeerJ, 8, e8451.
If dose < P then \((1 + (2^K1 - 1) * exp(4 * S1 * (P - x)))^(-1/K1)\)
If dose > P then \(1-((1 + (2^K2 - 1) * exp(4 * S2 * (x - P)))^(-1/K2)\)
with:
$$S1 = (2^(K1 - 1) * S * K1)/(2^K1 - 1)$$
$$S2 = (2^(K2 - 1) * S * K2)/(2^K2 - 1)$$
If \(2^K1\) is too large to be estimated, the approximation \(S1 = S*K1/2\) is used.
Demonstration:
$$S1 = (2^(K1 - 1) * S * K1)/(2^K1 - 1)$$
$$S1 = exp(log((2^(K1 - 1) * S * K1)/(2^K1 - 1)))$$
$$S1 = exp(log(2^(K1 - 1)) + log(S * K1) - log(2^K1 - 1))$$
When \(K1\) is very large, \(2^K1 - 1 = 2^K1\) then
$$S1 = exp((K1 - 1) * log(2) + log(S * K1) - K1 * log(2))$$
$$S1 = exp((K1 * log(2) - log(2) + log(S * K1) - K1 * log(2))$$
$$S1 = exp(log(S * K1)- log(2))$$
$$S1 = S * K1 / 2$$
If \(2^K2\) is too large to be estimated, the approximation \(S2 = S*K2/2\) is used.
If \((1 + (2^K1 - 1) * exp(4 * S1 * (P - x)))^(-1/K1)\) is not finite,
the following approximation is used:
$$exp((-1/K1)*(K1*log(2)+(4*S1*(P-x))))$$
If \(1-((1 + (2^K2 - 1) * exp(4 * S2 * (x - P)))^(-1/K2)\) is not finite,
the following approximation is used:
$$1 - exp((-1/K2)*(K2*log(2)+(4*S2*(x - P))))$$