From Osher Doctorow Ph.D. mdoctorow@comcast.net

A Lemma is a "sub-theorem" or intermediate stage Theorem on the way to (hopefully) more important results. Here's one based on the previous ideas with a proof. LEMMA 1. In the differential equation model: 1) dy/dt = A1(exp(t) - y) + y^2 exp(-t), y(1) = 1, the condition (dy/dt)(1) = 1, that is to say dy/dt is 1 at time t = 1, holds if and only if: 2) A1(1) = 1/e which is to say A1(t) at time t = 1 is given by 1/e, which is the boundary of Exponential Explosions. PROOF. Set dy/dt = 1 in (1), in which case exp(1) is e^1 = e and y is y(t) = y(1) = 1, and exp(-t) = exp(-1), so we get: 3) 1 = A1(e - 1) + 1/e and solving for A1: 4) A1 = (1 - 1/e)/(e - 1) = [(e - 1)/e]/[e - 1] = 1/e. Q.E.D. REMARK. We can thus regard A1(t) as a function or curve which takes on the Exponential Explosion boundary value 1/e when t = 1. It is a sort of Generalized Exponential Explosion Boundary. LEMMA 2. In equation (1) except that y(1) = 1 is not assumed, if the solution is y(t) = (1/3)exp(t) (so that y(1) = e/3), then A(t) = 1/3. In that case, the Universe is accelerating at least in reference to variable y with respect to time t. PROOF. Substituting y(t) = (1/3)exp(t) into (1), we get: 5) (1/3)exp(t) = A(t)[(2/3)exp(t)] + (1/9)exp(2t)exp(-t) = exp(t)[(2/3)A(t) + 1/9] and therefore (dividing through by exp(t)) A(t) = 1/3. Then dy/dt = (1/3)exp(t) so the acceleration is d(dy/dt)/dt = d[(1/3)exp(t)]/dt and since d(c f(t))/dt = cdf(t)/dt for a constant c, this is (1/3)exp(t) which is always positive for all real t. Q.E.D. REMARK. In Lemma 2, no discontinuity at t = 1 is assumed. We could still define A(t) separately in [0, .05) without changing the Lemma basically. The relationship of y(1) and (dy/dt)(1) to 1/e is now proportional to the inverse of 1/e which is e. Osher Doctorow