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I have already managed to recall the Mendeleev-Clapeyron equation, isochoric, isothermal and isobaric processes, but as I remembered the equation, I immediately crossed myself:)
I propose not to voice any objections to the topicstarter until 28 February. Just two days :)
Any thoughts in support of the topic are welcome.
I don't know about Alsu, but I'm already waiting :)
It's about time, it's 1 a.m.
I propose not to voice any objections to the topicstarter until 28 February. That's all, two days :)
Any thoughts in support of the topic are welcome.
Sustained. We'll have time to bitch.
I second that. We'll have time to bitch about it.
Good day, colleagues.
The idea of the algorithm itself is not difficult for me. The issue is purely procedural - automated selection of a single coefficient. There are some difficulties of my own. I will solve them in the next couple of days. In order to post it here today I have picked up the required coefficient in the algorithm manually. Therefore there will be small inaccuracies on the charts, in small fractions of a pip.
So, here we go.
In order not to clutter up the idea with unnecessary bars, I suggest taking the last 12 hours from the ends of these files. That is 144 bars (M5).
We will take the dollar-yen as a cross of these two. The triangle, so to speak, is closed.
Now pay attention to the trick.
The assertion is this: if we put D=1, E=ED, Y=1/DY in a bar 12 hours in the past from the end, and try to plot E, D, and Y separately to these initial values in this bar 12 hours in the past from the end, then the changes in E, D, and Y in these 12 hours (144 bars) were so:
In order that all who wish can construct for themselves and be convinced of the following, I give the data columns of these curves:
Proved simply: their relations are equal to the known relations ED, EY, DY: