MACHINE period with minus value - page 32
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strange, I thought 2/3 was more than 1/3
well yeah))) it's d1 and d2 - the increments of X1 + 2/3*d1 + 1/3*d2
well yeah))) these are d1 and d2 - increments of X1 + 2/3*d1 + 1/3*d2
i.e. the last increment (d1) has more weight than the older one (d2)?
i.e. does the most recent increment (d1) have more weight than the older one (d2)?
no d1 is the earlier increment
You forgot about the shift function in your code on the previous page, it's missing...(( add it to the code by someone
Freeloader.
Caesar, either do it yourself or go here, or get others interested. Stop messing around and making a fool of yourself who doesn't understand anything.
I'm just a moderator and I'm sick of watching this circus. You can't see any creativity from you, you only demand something from others.
I'm not going to ban you yet. But I might just shut it down as inadequate.
P.S. Topicstarter banned for a week for insulting.
:)))) The only way to look into the future is to look into the past.
The problem is solvable.
Draw the MA with period x. We draw MA with period x and with a shift minus x. Then we need to complete the necessary tail in x bars of the shifted MA. The task is reduced to a pattern recognition problem: we look for on the whole (or a given) set of history data a maximally similar current segment of the last N bars (we only consider its slope (increment)) so that it lags behind the current one by more than x bars. Its continuation in x bars is the required tail. It will fit perfectly as the tangent line will maximally coincide. The only disadvantage is that this tail will "bounce" (change) all the time when the current situation changes (like ZigZag).
I will definitely implement it and post it in the indicators.
The only way, is in the past, in the future... In the tester easily..... I just popped in, I swear I wasn't going to... I didn't mean to... Got my results from a hundred quid to a few million in a few months.
no d1 is an earlier increment
It makes no difference, you can write it backwards following the same principle and you will get /in the same notation/: (X1 + X2 +X3)/3 = (X3 - d1 -d2 + X3 - d2 +X3)/3 = X3 - 2/3*d2 - 1/3 * d1 (you have X1 + 2/3*d1 + 1/3*d2), that is, following your logic about greater or lesser consideration of increments, exactly the opposite. And all because the formula mixes signal values and incremental values, no conclusions can be drawn from such expressions.
In general, the input signal for a filter is usually not incremental, but directly the value of the input. If you want to filter the incremental values, you must first use a differentiating element.