a trading strategy based on Elliott Wave Theory - page 98

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No... You're trying to lead us astray :)
 
No... you decided to give us the wrong trail :)

Yeah, looks like you got the wrong trail, too :) . Alas, it's still not clear where it leads.
 
I am putting up another idea for public judgment :).

Looking at my channels with parabolas, I thought, but at the top of the parabola potential energy is maximal. Maybe we should select channels with the top of parabola found by OLS (or any other way, Vladislav said that he is not looking for parabola) lying outside the sample.
 
There are likely to be many such parabolas, I think there is no escaping the problem of choice.
 
Looking at my channels with parabolas, I thought, but at the top of the parabola, the potential energy is maximal.

It's not clear where that comes from. It doesn't even matter where the parabola has its horns. Just try to explain where it follows that "at the top of the parabola the potential energy is maximal".
 
I thought it was obvious, throw a rock upwards. It will fly along a parabola at some point its vertical component of velocity will be 0, that's the point where its kinetic energy is fully converted to potential energy (and actually it has the highest potential there) that will be the top of the parabola it's flying along. I think that's it.
 
Jhonny:
I have a feeling you drew channels 2/3 of a sample, you probably forgot at the last moment to substitute the full length in the channel drawing function. :)

In general, the observation is very interesting. By definition, every vertex is the beginning of descent. Well, it looks like we can see in the picture that as soon as the algorithm stops finding the best representations (i.e. a channel reaches perfection in the process), a breakdown follows.
 
... which ones represent the same object ....


I wonder how you set this up?
 
I thought it was obvious, throw a rock upwards. It will fly along a parabola at some point its vertical component of velocity will be 0, that's the point where its kinetic energy is fully converted to potential energy (and actually it has the highest potential there) that will be the top of the parabola it's flying along. I think so.

So you implicitly assume that the top of the graph is the same as our top. That is, that the greater the rate (not the price), the greater the potential energy. Then it is natural to assume that the lower the exchange rate, the lower the potential energy. And, as we know, every system tends to have a minimum potential. You and I, for example, walk on the ground rather than jumping on rooftops. :-)
So why is the course not falling and, moreover, its graph does not give the impression that the minimum of potential is at the bottom ?
 
So you are implicitly suggesting that the top of the graph is the same as our top.


I didn't say that, I gave a comparison to the Earth's potential field in which the source of the field is the Earth itself, here I already said the difficult question of what is the source, but the superposition I think there is, and this source can be both top and bottom.

that the bigger the rate (not the price),


ZS I am ashamed, but I do not know the difference between the rate and the price, please explain.
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