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epi,
Nice question !
Let Ti3 the today´s Tillson smoother, Ct the next bar Close and Ti3t
the next bar Ti3.
For any e1=EMA(C,p); the next bar EMA is e1t=f*Ct+(1-f)*e1;
The repeated use of this fundamental relation will finally give Ti3t
as a function of Ct.
The solution of the
Cond=Ct==Ti3t;
will give the Ct.
Will you do it ?
[Else I will do it later]
Dimitris Tsokakis
For the verification, paste in IB the
p=10;s=0.7;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;
Plot(Ti3,"Ti3",colorblack,1);
Plot(Ref(Ti3,1),"nextTi3",colorred,8);//the next bar Ti3
// Ti3t is the next bar Ti3
Ct=Ref(C,1);
e1t=f*Ct+(1-f)*e1;
e2t=f*e1t+(1-f)*e2;
e3t=f*e2t+(1-f)*e3;
e4t=f*e3t+(1-f)*e4;
e5t=f*e4t+(1-f)*e5;
e6t=f*e5t+(1-f)*e6;
Ti3t=c1*e6t+c2*e5t+c3*e4t+c4*e3t;// relation I
//by cascade replacements
e1t=f*Ct+(1-f)*e1;
e2t=f*(f*Ct+(1-f)*e1)+(1-f)*e2;
e3t=f*(f*(f*Ct+(1-f)*e1)+(1-f)*e2)+(1-f)*e3;
e4t=f*(f*(f*(f*Ct+(1-f)*e1)+(1-f)*e2)+(1-f)*e3)+(1-f)*e4;
e5t=f*(f*(f*(f*(f*Ct+(1-f)*e1)+(1-f)*e2)+(1-f)*e3)+(1-f)*e4)+(1-f)*e5;
e6t=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;
// Ti3t as a function of 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);// relation II
Plot(Ti3t,"Ti3t",colorwhite,1);//the analytic next bar Ti3
Cond=Ct==Ti3t;
As you see, the red and the white lines match point by point.
--- In amibroker@xxxxxxxxxxxxxxx, "epintoem" <epintoem@xxxx> wrote:
> Is there any way this can be accomplished...
>
> the question is...What price must the security close at for a
> crossover the next day?
>
> TIA
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