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Thank you DT, I tried this code but it requires that pb is greater than pa
and also it uses C in both Ti3s. I am looking for a sulution where pa==pb
and we use different price arrays.
best regards,
herman
-----Original Message-----
From: DIMITRIS TSOKAKIS [mailto:TSOKAKIS@xxxxxxxxx]
Sent: Monday, October 25, 2004 1:57 AM
To: amibroker@xxxxxxxxxxxxxxx
Subject: [amibroker] Re: DT's Ct prediction for the Ti3
--- In amibroker@xxxxxxxxxxxxxxx, "Herman van den Bergen"
<psytek@xxxx> wrote:
> 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.
Herman,
this specific question is already in
http://finance.groups.yahoo.com/group/amibroker-ts/files/A%20Ti3%
20application/
"Cross(Ti3a,Ti3b) predictions.txt"
Dimitris
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@x...]
> 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);
>
>
>
>
>
> Check AmiBroker web page at:
> http://www.amibroker.com/
>
> Check group FAQ at:
> http://groups.yahoo.com/group/amibroker/files/groupfaq.html
>
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