Brain-training tasks related to trading in one way or another. Theorist, game theory, etc. - page 20

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Mathemat:
It's not linear... it's not even polynomial. In short, it's non-linear.

I see. I'm googling it now... I'm kind of stumped myself... :-)
 
new-rena:

Maybe you missed it...

I.e. we cautiously give a geometric progression of lot increase. And you don't get a graph of change of calculation result, with the condition that we take only minimum lot and above? And about and - the picture is not inserted:

i.e. bx = N and log ( ab ) = log a + log b, i.e.log a + log b = log( ab )

Using these formulas we seem to get something

And this:

log ( b k ) = k - log b .

this refers to the properties of logarithms

https://ru.wikipedia.org/wiki/Логарифм

 
new-rena: Well, yes. Exactly. It turns out the "stretch" on the expected pullback is the action of a stretched rubber band. And the money should be enough, because the lot does not grow that much further ))) This is the mathematical model of forex, basically

OK, I'll show you how the lot changes further (x=0.5):

0.01^(0.5^0) = 0.01,

0.01^(0.5^1) = 0.1,

0.01^(0.5^2) = 0.316.

0.01^(0.5^3) = 0.562,

0.01^(0.5^4) = 0.750,

0.01^(0.5^5) = 0.866.

0.01^(0.5^6) = 0.931,

0.01^(0.5^7) = 0.965,

0.01^(0.5^8) = 0.982.

In short, each next term is a square root of the previous one (it is at x=0.5), and lot tends to 1.

If we take the same x=0.5, but initial lot is 1, then the lot will always be the same (1).

And if the initial lot is larger than 1 (say, 2), then the lot will gradually decrease to 1.

In short, no matter how you spin it, still in the limit the lot will be 1, regardless of the initial lot.

Everything is as you planned?

 
avtomat:

it refers to the properties of logarithms


Logarithm

reply

I see. Can I check the results of my calculations on any of the pairs?

 
Mathemat:

OK, I'll show you how the lot changes further (x=0.5):

0.01^(0.5^0) = 0.01,

0.01^(0.5^1) = 0.1,

0.01^(0.5^2) = 0.316.

0.01^(0.5^3) = 0.562,

0.01^(0.5^4) = 0.750,

0.01^(0.5^5) = 0.866.

0.01^(0.5^6) = 0.931,

0.01^(0.5^7) = 0.965,

0.01^(0.5^8) = 0.982.

In short, each next term is the square root of the previous one (it's at x=0.5).

If we take the same x=0.5, but initial lot is 1, the lot will always be the same (1).

And if the initial lot is larger than 1 (say, 2), then the lot will gradually fall to 1.

In short, no matter how you spin it, still in the limit the lot will be 1, regardless of the initial lot.

Is everything just as you planned it?

Yes, that's right... And Logarithm will give infinity
 
new-rena:

I see. Can I check my calculations?


er... and here I find myself in a stupor! :)))

What was counting? How was it calculated? I wish I had a hint. ....

 
avtomat:

er... and here I found myself in a stupor! :)))

What was counting? How was it counted? I wish I had a hint....


MiniLot^(x^0)+MiniLot^(x^1)+MiniLot^(x^2) ... + MiniLot^(x^(N-1))=VolMax,

where N-maximum estimated number of orders, (_MaxOtders)

VolMax-maximum possible total volume of all N orders (_MaxLots)

so far by simple brute force finding x
Maybe someone knows solution to this equation where only x (_Stepen) is unknown?

 

how do I know what's in the table... spreads, points, degrees, amounts, spreads... What are we talking about?

Give me the specific input data and you'll get your answer.

 
avtomat:

how do I know what's in the table... spreads, points, degrees, amounts, spreads... What are we talking about?

Give me the specific raw data and you'll get your answer.

0,01^(0.5587^0)+ 0,01 ^(0.5587^1)+ 0,01 ^(0.5587^2) ... + 0.01 ^(0.5587^76)=5.96 - Is this correct?, 

for ( k=0; k<=Pars; k++ )//начало цикла перебора инструментов
         {
            Instr=s[k];
            for(int s=1; s<10000;s++)
               {
                  double Lot=MinLot,Lot_0=0,Stepn;Sum=0;int Lts=0;
                  
                  for ( double a=0; a<=NormalizeDouble(GlobalVariableGet(Instr+"_Razmah")/GlobalVariableGet(Instr+"_Pips"),0); a++ )//начало цикла перебора инструментов
                     {
                        Stepn=s*0.0001;
                        Lot_0=MathPow(MinLot,Stepn);
                        Lot=Lot+Lot_0;
                        if(a>=GlobalVariableGet(Instr+"_Razmah")/(GlobalVariableGet(Instr+"_Pips")+2*GlobalVariableGet(Instr+"_Spread")) && Lot>=GlobalVariableGet(Instr+"_MaxLots") && Lot<=NormalizeDouble(GlobalVariableGet(Instr+"_Razmah")/GlobalVariableGet(Instr+"_Pips"),0) && Lot_0>MinLot)
                        {
                           GlobalVariableSet(Instr+"_Stepen",Stepn);
                           GlobalVariableSet(Instr+"_MaxOtders",a);
                           break;
                        }
                     }
                }               
         }


 
new-rena:
0,01^(0.5587^0)+ 0,01 ^(0.5587^1)+ 0,01 ^(0.5587^2) ... + 0.01 ^(0.5587^(76))=5.96 - Is this correct?

Right would be like this:

.

.

and if x=0.5587

1...1314151617181920212223
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