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In any case, the last few bars are redrawn, which means the value of the code = 0.
On the "four digits"?
yes, on the 4
if more than that, the signal
no
You are either wrong or deceiving.
agree
In the construction of the last visible bar the next bar is involved, which is absent.
the next bar is missing.
I suspect that in the quote that we see it is exactly what is missing
there is only one last bar which is not longer than 10 minutes hidden //the version, not the fact
I started to calculate it, but so far I haven't had the brains to do it.agree
the construction of the last visible bar involves the next bar, which is absent
the next bar is missing
I suspect that in the quote we see it is exactly what is missing
only one last bar is hidden, the duration of which is no more than 10 minutes
I started to calculate it, but so far I haven't had the brains to do it.Here is, for example, a redrawable Fourier approximation:
And here is an ordinary non-drawn SMA with a period of 4, but shifted left by 2:
These are all toys and self-deception
Here is a redrawable Fourier approximation, for example:
And here is a regular SMA with a period of 4 and a shift to the left by 2:
It's all toys and self-delusion.
Fourier approximation is even steeper than mine.
What is redrawn there?
Fourier is even cooler than mine.
What is redrawn there?
Everything is redrawn. I dumped the code of this indicator here
animated GIF:
Everything redraws. I' ve dumped the code for this indicator here
Animated GIF:
It's funny and cool on the other hand
//have to learn animation too
really hard to redraw
and if the starting point of the calculation is tied to a certain time stamp?really hard to redraw
What if the starting point of calculation is attached to a certain time stamp?To see the dynamics.
The same Expert Advisor on M15.