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Fri Mar 15 13:09:45 EDT 2013

Exponentials: computing the log of |p| ~= 1

Hmm... still something is missing.

Given that P ~= log(p) and p=c+si with |p| = 1 so re(P)=0, what
happens if this is plugged into the Taylor expansion?

log(1 + x) = \sum_n=1 1/n x^n


      P ~= (c-1 + s i) 
         - 1/2 (c-1 + s i) ^2  
         + 1/3 (c-1 + s i) ^3

Taking only the imaginary coefficients:

        = s i
        - (c-1) s i
        + (c-1)^2 s i - 1/3 (s^3) 

From this it seems that computing log(1 + x) for x small is not such a
problem.  How to use this?

It could be used to compute log (p1/p0), and use that to update the
estimate of P1 = log(p1).

I'm still missing a way to make all the relations concrete:
- Using P1 directly avoids drift, but is imprecise
- Using (P1-P0)/N has drift, but is more precise.

These are two estimators.  How to combine two estimators into one in
an attempt to avoid drift?


The thing is, there is an estimate already.  To use this information,
an update method should be used.  Using N-R, log(P) is a solution of
f(x) = 0 with:

   f(x) = exp(x) - P

The N-R update step for this is  

        u(x) = x - f'(x)/f(x)

             = x - exp(x) / ( exp(x) - P )

Maybe it pays to combine these, i.e. take exp(x) and x to be the
estimates we have...

To be continued..



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