Absolute courses - page 68

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didn't get it
 
Dima.A.:
didn't get it

There is a system, reversal, there are signals such as 1 and -1, so when the signal 1 opens buy, pose is held until the reverse signal. This is to say that the maximum profit from the signal (especially in the reversal system) is much earlier than the reverse signal. That is why I say the way out of this situation is either by TP or by equity.
 
grell:

There is a system, a reversal, there are signals like 1 and -1, so when the signal 1 opens a buy, the pose is held until the reverse signal appears. This is to say that the maximum profit from the signal (especially in the reversal system) is much earlier than the reverse signal. That is why I say the way out of the situation is either by TP or equity.

Not so much from real trading experience but from the TS testing experience: as soon as permanent TakeProfit is introduced in a system, the system degrades its characteristics.
 
Dima.A.:

Not so much from real trading experience, but from the experience of testing the TS: as soon as a permanent TakeProfit is introduced into the system - the characteristics of this system deteriorate.

Then the second option, but certainly not closing on the opposite signal. The trawl will not help either.
 
so... Based on the top post here https://www.mql5.com/en/forum/ru/47342/page5 select a number, the power function of which will be equal to the product of the relative magnitudes of the increments of the currency pairs, but in the calculation make all pairs co-directional, that is, at this stage do not take into account the sign of removal from the point, and take into account the size of the removal.Then we can do the same with the whole tangle of pairs of all currencies, also choose such a number, and then consider the difference between the chosen number and the chosen number in each cluster of currencies separately.
 

came across this 5^x-3^x=2, F(X)= 5^x-3^x

Such equations are not usually solved analytically.
In this case, the trick is to find the root easily by eye.
The next step is to prove that it is unique:
for x<= 0 there are no solutions because ...
For x>0 F(x)=2 can occur only once, since ...
it follows that there are no other roots.

 
Joperniiteatr:

came across this 5^x-3^x=2


Next it remains to prove that it is singular:


To do this, it suffices to show the monotonicity of the function 5^x-3^x, which is done by the usual differentiation followed by convincing the submitter that the derivative is always positive (or negative, at the wizard's choice).
 
I just wanted to write about monotony, but didn't have time.)
 
alsu:
To do this, just show the monotonicity of the function 5^x-3^x
It is not monotonous )
 

I regret to inform you that the correlated absolute rate solutions discussed at the beginning of the thread allow for an infinite number of solutions.

Here is an example for the last 24 hours (288 bars M5):

this solution is possible:

and possibly this:

and it is possible that this:

and many other forms are possible, including many more exotic ones.

For example, like this:

1...616263646566676869707172737475...113
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