From Osher Doctorow Ph.D.
Here's a slightly more general Lemma. LEMMA 3. In the equation: 1) dy/dt = A1(exp(t) - y) + exp(-t) y^2 if y = k exp(t), then A1(t) = k. PROOF. dy/dt = k d(exp(t))/dt = k exp(t) from calculus, and y^2 = k^2 (e^t)^2 = k^2 e^(2t), so we get: 2) ke^t = A1(e^t - ke^t) + e(-t)k^2 e^(2t) which simplifies to: 3) ke^t = A1e^t(1 - k) + k^2 e^t since e(2t)e^(-t) = e^(2t - t) = e^t. Divide through by e^t: 4) k = A1(1 - k) + k^2 and solve for A1(t): 5) A1(t) = (k - k^2)/(1 - k) = k(1 - k)/(1 - k) = k. Q.E.D. REMARK. It follows that if y = (1/e)exp(t) (which is exp(t - 1)), then A1(t) = 1/e. LEMMA 4. If (dy/dt)(1) = 1/e, then there is a real positive value of A1(1) which satisfies equation (1) and such that 0 < = A(1) < = 2. PROOF. From (1) for (dy/dt)(1) = 1/e, we get a quadratic equation in y(1)^2 with A(1) in two of the coefficients, and from the square root in the quadratic equation this is always real if and only if 0 < = A(1) < = 2. Q.E.D. REMARK. It is interesting that the values of A(1) and y(1) are linked, especially when constrained by (dy/dt)(1) = 1/e. The linkage is quadratic, that is to say an algebraic quadratic equation in y(1) with A1(1) in two of the coefficients. The equation is in fact: 6) y(1)^2 - A1(1)ey(1) + (A1(1)e^2 - 1) = 0 This is a parabola in the A1(1)-y(1) plane across different pairs of values of (A1(1), y(1)). It is also interesting that "exponential Information," or as I refer to it in Rare Event Theory "Knowledge", of the form y(t) = kexp(t), is an inverse of the Shannon Information-Entropy kernel (in fact, the Hartley Information) log(t) to base 2 up to a constant. Osher Doctorow