Hi Lars,
Now, consider f(t) = a*log(b*t+1), then the derivate is a*b/(b*t+1) and
second derivate - a * b^2 / (b*t + 1)^2.
Forming first f'(t) and second f"(t) derivate estimates from data is
trivial. Given that we can estimate a and b using
a = - f('t)^2 / f"(t)
b = - f'(t) / (f'(t) * t - a)
= - f"(t) / (f("t) * t - f'(t))
A bit of paper and pen work or you get Maxima to do some work for you.
I haven't seen how any real estimator of this drift function is
implemented, but I wanted to provide some notes from note-book of stuff
being unfinished.
Cheers,
Magnus
On 11/18/2016 07:26 PM, Lars Walenius wrote:
Bob wrote:
As mentioned earlier in this thread. The function that has been used in several
posts
isn’t the right log function. The proper fit is to ln(bt+1)
You are absolutely right. It was my mistake to use the ln(t) in the graph. As
that was what I know in Excel and I don´t have Stable32 or MatLab. In Excel I
actually double checked that (a*ln(bt+1)) with b 5 to 1000 gave about the same
as (a*ln(t)) for my data set (only the offset was largely different).
Hopefully someone can find the correct a and b for a*ln(bt+1) with stable32 or
matlab for this data set:
Days ppb
2 2
4 3.5
7 4.65
8 5.05
9 5.22
12 6.11
13 6.19
25 7.26
32 7.92
It would also be interesting if I could get the drift after 10 years to see if
it is about 6E-13/day as with the ln(t).
Peter wrote:
I'm not very good with Excel, but this curve-fitting function sounds very
useful. Could you please tell me how it's done?
In the graph I only right-clicked the curve and selected ”add trendline” here I
checked the logarithmic and show equation.
Lars
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