Absolute courses - page 70

 

As I said before "watch your hands". We take the last 288 M5 bars from the posted data files. Plot the following EURUSD, EURJPY, USDJPY rates:

see file ED_EY_DY.txt

Files:
ed_ey_dy.txt  8 kb
 
Dr.F.:

As I said before "watch your hands". We take the last 288 M5 bars from the posted data files. Plot the following EURUSD, EURJPY, USDJPY rates:

see file ED_EY_DY.txt


So? Where's the calculation method? Why are we digging through thousands of variations of this nonsense? There was a method. But it could be developed further, which is probably what you are doing, not the triangles but the whole list of currencies.
 
Joperniiteatr:

0.998683^x + 1.00216908^x+ 1.002040888^x+ 0.998182^x+ 1.003999^x=1

And this x=?


alsu, so which solutions are analytical?)
 

Now let's construct one "decent" and one "ripped" case as an example.

Here's the "tattered" one first:

You want the numbers? Here's the data file:

Everybody can see that correlation of curves in the file is real 0,999999+ and their correlation coincides with initial EURUSD, EURJPY, USDJPY correlation.

Files:
 
(a^x)'=a^xln a.
 

And here is (one of the infinitely many) "decent" case of the solution. I purposely made it based on a blunt straight line up to show that the solution is determined by the arbitrariness of the solver and nothing else.

Here is the corresponding data file with curves E, D, Y.

Everyone may make sure that correlation of curves given in the file is real 0,9999+ and their relations coincide with initial relations of EURUSD, EURJPY, USDJPY. There is an infinite number of curves in this way. I don't care if they vary sine-wave.

Files:
 
Joperniiteatr:
(a^x)'=a^xln a.


It is known that the derivative of any function is the product of the function itself by the derivative of its natural logarithm. So your notation is wrong.

 
Shit, doctor, calm down, no one is interested in this..... I'm personally interested in other things here...
 
(x^n) = nx^(n-1)
 
Joperniiteatr:
(x^n) = nx^(n-1)

Exactly. On the primes of power functions it is good (convenient) to see the validity of the formula "

It is known that the derivative of any function is the product of the function itself by the derivative of its natural logarithm."

In the case of the exponential it means exactly that(a^x)'=a^xln a, I'm sorry, you wrote it down correctly. Didn't look carefully right away.