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DT, I am not sure i understand the Ti3t... what i am trying to find out is
what tomorrows closing price would be to cause the crossing of two Ti3
functions. For example:
F1 = T3( Array1, 3, 0.8 );
F2 = T3( ref(Array2,-1), 3, 0.8 );
Array1 and Array2 are two different arrays and could be any type of
indicator array. For development you can plug in any of the OHLC arrays, but
you cannot use the same price array for both function calles.
The Periods and Sensitivities are the same for both T3 function calls.
F2 is based on yesterdays values, F1 is based on todays values.
I would like to calculate tomorrows value for Array1 that would cause the
two functions to cross: cross(F1, F2).
Do you think this is possible?
thanks for you help DT!
herman
-----Original Message-----
From: DIMITRIS TSOKAKIS [mailto:TSOKAKIS@xxxxxxxxx]
Sent: Saturday, October 23, 2004 5:34 AM
To: amibroker@xxxxxxxxxxxxxxx
Subject: [amibroker] Re: DT's Ct prediction for the Ti3
Herman,
Ti3t (the next bar Ti3) is a direct function of the next bar Close Ct.
Ti3t=
c1*(f*(f*(f*(f*(f*(f*Ct+(1-f)*e1)+
(1-f)*e2)+(1-f)*e3)+(1-f)*e4)+(1-f)*e5)+(1-f)*e6)+
c2*(f*(f*(f*(f*(f*Ct+(1-f)*e1)+(1-f)*e2)+(1-f)*e3)+
(1-f)*e4)+(1-f)*e5)+
c3*(f*(f*(f*(f*Ct+(1-f)*e1)+(1-f)*e2)+(1-f)*e3)+(1-f)*e4)+
c4*(f*(f*(f*Ct+(1-f)*e1)+(1-f)*e2)+(1-f)*e3);
If I understand your question, you want to solve Ct for any given
Ti3t.
Let me know and I will do it.
Dimitris
--- In amibroker@xxxxxxxxxxxxxxx, "Herman van den Bergen"
<psytek@xxxx> wrote:
> Hello,
>
> 'been looking at DT's Ct formula (nice work!) to predict where
tomorrow's
> Close will touch the Ti3 - see code below. Can anybody see a way to
use this
> formula to predict the Close of tomorrow needed to have the Ti3
touch any
> arbitrary point? For example a point on another indicator.
>
> My math is not up to this, any help would be appreciated!
>
> herman.
>
> p=3;s=0.84;f=2/(p+1);
> // Ti3
> e1=EMA(C,p);
> e2=EMA(e1,p);
> e3=EMA(e2,p);
> e4=EMA(e3,p);
> e5=EMA(e4,p);
> e6=EMA(e5,p);
> c1=-s*s*s;
> c2=3*s*s+3*s*s*s;
> c3=-6*s*s-3*s-3*s*s*s;
> c4=1+3*s+s*s*s+3*s*s;
> Ti3=c1*e6+c2*e5+c3*e4+c4*e3;
> //The value of tomorrow´s Close Ct that touches tomorrow´s Ti3 is
> Ct=
> (1-f)*((c1*f^5+c2*f^4+c3*f^3+c4*f^2)*e1+(c1*f^4+c2*f^3+c3*f^2+c4*f)
*e2+(c1*f
> ^3+c2*f^2+c3*f+c4*1)*e3+(c1*f^2+c2*f+c3*1)*e4+(c1*f+c2*1)*e5+c1*e6)/
(1-(C1*f
> ^6+C2*f^5+C3*f^4+C4*f^3)) ;//relation III
> Plot(C,"Close",1,128);
> Plot(Ti3,"Ti3",4,1);
> Plot(Ct,"Ct",2,1);
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