In the special case that P is a constant and Q = 0, this represents the very important equation for exponential growth or decay (such as radioactive decay) whose solution is y = ke−Px, where e is the base of the natural logarithm.
The Bayesian argument is from the very easily proved equation which says that if h is a hypothesis, e the evidence, and k the relevant background information, then the probability of h given e & k is equal to the probability of e given h & k divided by the probability of e given k.
Why does the very simple Hill equation match so well to the range of observed input-output relations?
The following very similar equation describes the model with 13 fields: logit = −11.5+22.8(honey bee density) +0.38 area of Bt cotton within 750 m) +10.7 adventitious Bt plants) – 1.1 area of Bt cotton within 750 m)(adventitious Bt plants).
I know that because I have received an avalanche of insider stories - Watergate it isn't - including one from an academic in psychology who was offered money by Porter Novelli PR agency to put his name to the very same Sky Travel equation story that Arnall sold his to.
So we obtain that (uin W^{1, s}(Omega, Cell_{n})) is the very weak solution to equation (4), the theorem follows.
By substitution in (1), we have: Similarly, by interchanging 'HI' and 'T', we also have: If we then divide (2) by (3) we arrive at the following very useful equation: So far, this is simply a matter of probability theory.
The Navier-Stokes equation has been recognized as the basic equation and the very starting point of all problems in fluid physics (see [1]).
Most of these problems could be translated into solutions for systems of linear equations: the very topic of numerical linear algebra.
Indeed, in the latter case the very similarity of Coulomb's equation to Newton's was taken to be in its favour.
