Re: Acceleration of the Universe as Acceleration of Probable Influence Derivative 16 Remarkable Approximation of Dyt Operator



>>From Osher Doctorow mdoctorow@xxxxxxxxxxx

To begin proving the Theorem (but note a modification below of equation
(3) of the Theorem), let's write B(t) as B for brevity, and the
generalized Logistic Differential equation in standardized form is:

1) dy/dt = yB(1 - y), 0 < = y < = 1, B > 0

So we have:

2) dydt = By - By^2

Hence Dy(dy/dt), which I also write as Dyt(y), is:

3) Dy(dy/dt) = B - 2By

which is just the partial derivative with respect to y of the right
hand side of (2) with y and t (and hence B(t)) assumed independent.
Therefore:

4) Dy(dy/dt) = B - 2By > 0 iff B(1 - 2y) > 0 iff 2y < 1 iff y < 1/2

On the other hand, Dtt(y) is the second derivative of y with respect to
t assuming that y is a function of t, that is y = y(t), and
differentiating both sides of (2) with respect to t yields:

5) Dtt(y) = yB' + By' - y^2 B' - 2Byy' = yB' (1 - y) + By' (1 - 2y)

Let's examine where Dtt(y) = 0:

6) Dtt(y) = 0 iff yB' (1 - y) + By' (1 - 2y) = 0 iff B' (1 - y) =
-By'(1 - 2y)

Solving for y' yields:

7) Dtt(y) = 0 iff y' = -B' (1 - y)/[B(1 - 2y)]

But y' is By(1 - y) so substituting for y' in (7) yields:

8) Dtt(y) = 0 iff By(1 - y) = -B' (1 - y)/[B(1 - 2y)]

from which:

9) Dtt(y) = 0 iff B' /B^2 = (2y - 1)

So instead of equation (3) of the Theorem, which stated that Dtt(y) > 0
iff the absolute value of B' /B^2 << 1, the correct requirement is:

10) Dtt(y) > 0 iff B' /B^2 > 2y - 1

I apologize for this error.

The way in which the "wrong" condition that the absolute value of B'
/B^2 << 1 was obtained is to see when Dyt(y) and Dtt(y) are
approximately equal. Notice that if we solve (9) for y we get:

11) 2y = 1 + B' /B^2

so:

12) y = (1/2) + B' /(2B^2)

which is approximately (1/2) when B' /(2B^2) << 1. However, this
condition cannot be assumed - it has to be examined in each case, and
the correct equation (3) of the Theorem stated in the last few posts is
(10) above.

Osher Doctorow

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