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> Hopefully, you'll find my math below not too bad?
Well, I'm OK on the first part. But here you lost me:
> FactorZ*((C*(FactorX-FactorY) + (1-FactorX)*xavgX[1] -
> (1-FactorY)*xavgY[1])) + (1-FactorZ)*macdZ[1]
> = P*(FactorX-FactorY) + (1-FactorX)*xavgX - (1-FactorY)*xavgY
It looks as though you're saying the smoothed MACD (the part before
the =) is equal to the unsmoothed MACD with C replaced by P !?
I'd approach it like so, and again, I am whipping this off the top of
my head so if you want to use it, it's your job to check it!! :-)
the smoothed MACD is
> FactorZ*((C*(FactorX-FactorY) + (1-FactorX)*xavgX[1] -
> (1-FactorY)*xavgY[1])) + (1-FactorZ)*macdZ[1] + (1-FactorY)*xavgY -
> (1-FactorX)*xavgX
... as you said. (You have an extra set of ()'s but that doesn't
hurt anything.) You want to solve for the value of C that causes it
to equal zero on the next bar. So: (using FX, FY, FZ instead of
FactorX/Y/Z)
0 = FZ*C*(FX-FY) + FZ*(1-FX)*xavgX - FZ*(1-FY)*xavgY + (1-FZ)*macdZ
FZ*C*(FX-FY) = -FZ(1-FX)*avgX + FZ*(1-FY)*avgY - (1-FZ)*macdZ
C = -FZ(1-FX)*avgX + FZ*(1-FY)*avgY - (1-FZ)*macdZ
-----------------------------------------------
FZ*(FX-FY)
which could be simplified as
C = -(1-FX)*avgX + (1-FY)*avgY - (1-FZ)*macdZ/FZ
--------------------------------------------
FX-FY
Gary
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