billp wrote:
Quote:
You are confusing production deceleration with production rate.
You may be right.
Please check if my
differentiatons of the normal distribution are correct or not.
My senior citizen math buddies not of much help. What's his name's disease?
bestAnd, what do you think is going to happen?
http://www.prosefights.org/nmlegal/dcvo ... m#gabreski
From your website:
dy/dx = y' = 1/(σ(2π)(1/2) )e- ½((x -µ)/σ)2d(- ½((x -µ)/σ)2)/dx
= 1/(σ(2π)(1/2) )e- ½((x -µ)/σ)2-((x -µ)/σ2)
= -y ((x -µ)/σ2)
I think that is correct.
I think I found your mistake.
Your website says "The second derivative of the normal distribution is zero. This occurs at µ - σ. "
It should say "The first derivative of the normal distribution is zero. This occurs when x = µ " . You can check by substituting x with µ below
dy/dx = -y ((x - µ)/σ2)
The time "x = µ" is the peak oil time. Remember that x is (time) and y is (production per day).
Better still, replace x with t and y with P
And if you believe that peak oil occurs at 2006 then replace µ with 2006