[amibroker] Re: DT's Ct prediction for the Ti3

["DIMITRIS TSOKAKIS" <TSOKAKIS@xxxxxxxxx>]

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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