Renter - page 20
You are missing trading opportunities:
- Free trading apps
- Over 8,000 signals for copying
- Economic news for exploring financial markets
Registration
Log in
You agree to website policy and terms of use
If you do not have an account, please register
.
that's more convenient.
You can reproduce it in Matcadet in no time at all --- you'll be able to adjust the parameters quickly.
A little later. You need to move in.
just in case:
eps=ln(1+q)
I don't see an obvious expression for the alpha, Oleg. You said it was already there. But here you have a function equal to the derivative on k, which Neutron already calculated long ago.
2 Neutron: well, now everything is even simpler. Fix t=50 and look for experimental dependence alpha = k/q on q. It is smooth and straightforward, easy to approximate. We use the approximation as a first approximation for... tangent method and stop at the first step. Let's try it.
I don't see an explicit expression for alpha, Oleg. You said it was already there. But what you have here is a function equal to the derivative on k, which Neutron calculated long ago.
and there is no explicit expression for alpha,
.
In engineering practice, various kinds of nomograms are used in such cases.
.
uh... am I missing somewhere... Where is that correct formula, long ago calculated?
https://www.mql5.com/ru/forum/131914/page2 - Neutron's second post on the page. It's the same in the form of an equation to be solved with respect to k. And your exponent, Oleg, is just a stepped function in disguise, since xi is a logarithm...
Anyway, a little later, hopefully, I will post analytical solution for t=50 at q=0.1...0.3.
https://www.mql5.com/ru/forum/131914/page2 - Neutron's second post on the page. It's the same as an equation to be solved with respect to k. And your exponent, Oleg, is just a stepped function in disguise, since xi is a logarithm...
Anyway, later hopefully I will post analytical solution for t=50 at q=0.1...0.3.
do you think it's the same thing...?
.
mmm... except maybe with a little bit of a stretch....
Oleg, you got it mixed up. Neutron's second post on the page. Here is the equation:
Substitute t=36 and q=0.3 into the function on the left and build it as a function of k.
Oleg, you got it mixed up. Neutron's second post on the page. Here is the equation:
Yes, it is quite a normal "solution", but a bit contrived and overloaded with terms from ATS. Besides it is already obtained on the 2nd page of this thread.
This is not a solution, but just a function whose zero needs to be found. We have been trying to find this zero for a long time based on this function.
I thought you said that conversion of diphunts to class of algebraic equations can help to solve this problem. But so far you've only got the same function whose zero we're looking for. If you think you got the same result "more reasonably" than Neutron- justify why.
In the "simple" derivation of the function we didn't need to translate the "lattice" functions into continuous time, as we simply had no such problems. But you did! But what is the advantage of your "solution"?