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Exactly! And in your picture it is quantised. Look for the clamp.
It's a matcad, I presume? I can't say, I don't have it.
Read carefully, the difference cannot be unquantised. I.e. it would be the opposite strange if there were no quantization.
Please explain why the quantum for R(B)-R(A) should be different from the quantum for R(A) ? It seems to me that in both cases it should correspond to Point.
If we write the equality ln(Price + i * Point) = ln(Price) + k[i], then obviously the value of k[i] is not proportional to i.
If we write the equality ln(Price + i * Point) = ln(Price) + k[i], then obviously the value of k[i] is not proportional to i.
ln(Price + Point ) - ln(Price) = ln(Price) + ln(1 + Point/ Price ) - ln(Price) ≈ Point/ Price.
That is, the quantum of both R(A) and R(B) is equal to Point / Price. And for their differences for some reason visually it is an order of magnitude bigger.
ln(Price + Point) - ln(Price) = ln(Price) + ln(1 + Point/ Price ) - ln(Price) ≈ Point/ Price.
That is, the quantum of both R(A) and R(B) is equal to Point / Price. And for their differences for some reason visually it is an order of magnitude bigger.
In principle, the paradox is solved by taking each stroke as a single point. Especially since then we also get a quantum of the order of 0.0001, which is just the order of Point/Price.
The conversion to strokes is due to different Price values for different R(A). But for the corresponding R(B) Price is about the same, so there is no vertical blurring of a point into a stroke.
In short the last posts should be transferred here:).