a trading strategy based on Elliott Wave Theory - page 101
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In principle, yes. Although it's not clear what to output with the indicator. Just so you don't waste time waiting for an idea :)
I made Murray a long time ago for history. Solved the buffer shortage problem by cloning, one Odd, one Even :). But it turns out to be a double calculation. An option - to write "superfluous" levels in global variables and read them from there by a special simple indicator - I will do it that way, when I feel like it.
I did not notice anything special in this place on the original.
It is clear that if we do selection by RMS for parabolas, the samples of the best channels by the criterion of 1<2/3 will not coincide. Although it's not a problem to do a search for the best parabolas now.
ZZY has posted 1000 posts, you can say it's a kind of anniversary. congratulations to all!
Congrats to all. I can't understand how to test EAs (like Mak's Graphic Expert Advisor) running on channels (like Shi_Channell) and you have solved the problem. And this despite the fact that Shi_Channell already has this indicator buffer:)
Saw a strange jump on the RMS of a parabola.
If it refers to the same parabola, then the first thought is a bug. If it is the RMS of the "current" parabola, then I have encountered that the approximation parameters change by a jump, for linear regression this is more of a rule. I haven't tracked the RMS of parabolas, so I can't say anything.
Увидел странный скачок на СКО парабол.
If it applies to the same parabola, then the first thought is bug. If it is the RMS of the "current" parabola, then I have encountered that the approximation parameters change by leaps and bounds, for linear regression this is more of a rule. I haven't tracked the RMS of a parabola, so I can't say anything.
This means that in the 215-bar sample the RMS of the parabola is 0.005744, in the 216-bar sample it is 0.00648, and in the 217-bar sample it drops again to 0.006199 .
The signal energy calculation is guaranteed to be correct. With channel potential energy calculation, it's too early to look particularly closely, I think there are still errors. I mean, I'm even willing to bet there are some. Negative values of potential energy appear to be due to the choice of initial "reference point". It is the same for all channels. On my arrival, I will look into it more specifically. (:о)
To make it clearer, I decided to describe how channels are sampled.
Zone is a so called "dead zone" - the prices in it are taken into account for the calculated channels, but the channels smaller than that are not calculated. It is set manually for the time being.
On the chart: Calculated from the zero bar.
JR_SERIES_U potential energy of channels
JR_SERIES_ENG signal energy (DSP)
If by eye, as the sampling boundary moves to the left, the parabola is simply bound to flip at some point. That is, change the sign of A (if A*x^2+...). The RMS at the transition point should flicker. I think this is the same as what I observed for linear regression.
This seems to be clear, but then what gives us a reason to talk about the potentiality of the field if we say that earnings and work are different things? How then does it follow that the closed-loop work in our case will be 0?
A field is potential if the work on the closed loop equals 0. That is what they like to say in physics.
A curvilinear integral is independent of the form of the integration curve if dQ(x,y)/dy=dP(x,y)/dx. This is how it sounds in matanalysis. This equality is always true if Q(x,y)=dU(x,y)/dx and P(x,y)=dU(x,y)/dy.
Keeping in mind that Q(x,y) and P(x,y) are Cartesian components of the vector F(x,y), we obtain that F(x,y)=grad(U(x,y))
Translated into Russian, it sounds like this. If in a field of a scalar potential a force is equal to the gradient of that potential, such a field is potential and the work to move along a closed loop in this field is 0.
It follows that you can represent potential by ANY scalar function and this field will be potential. It is clear that the work in this case will also have a different value, depending on the type of potential. And earnings, it is clearly proportional to the price difference.
You can, of course, postulate that your earnings are equal to your work. But in that case it would immediately lead to a particular kind of potential and I personally don't like that kind. :-)