Author's dialogue. Alexander Smirnov. - page 34

[Candid]

Mathemat:
Through the muwings? :)

Hmm, am I not the one asking the questions here? :) The wrong RMS will be through the muwings, you need an RMS for the current LR.

2 Korey: And why should M-qWMA be called via iCustom? It is more logical to make 2 cycles of history in one indicator.

P.S. Or maybe I got your hint wrong? (it's to Mathemat)

[Sceptic Philozoff]

Oops, didn't see that one, I don't know, Candid. The RMS is a non-linear price function, you can't do without muwings, obviously. Are you trying to build a channel or something?

About the two cycles: yes, they increase that much. But who will invent the recurrence formula to calculate QWMA?

[Candid]

Mathemat:
Oops, didn't see that one, I don't know, Candid. The RMS is a non-linear price function, you can't do without muwings, obviously. Are you trying to build a channel or something?

I usually build a channel, yes, but there are other applications.

[Sceptic Philozoff]

In principle RMS^2 = M[X^2] - (M[X])^2. From here you can try to construct something recurrent.

[Candid]

Mathemat:

But who will come up with the recurrence formula to calculate QWMA?

And the following idea would not help: P*i*i - P*(i-1)*(i-1) = P*(2*i-1)?

[Sceptic Philozoff]

Yes, of course, that's what I'm dancing around.

[Candid]

Mathemat:
Yes, of course, that's where I'm dancing from.

Now if we do it again, that is P*(2*i-1)-P*(2*(i-1)-1)=2* P, we get the sum independent of i . Now let's calculate the sum P*(2*i-1) together with the sum P*i*i in the first loop ; then let's recurrenceally calculate the price sum and use it as P*(2*i-1), and
P*i*i. Do you get it?

Mathemat:

In principle RMS^2 = M[X^2] - (M[X])^2. You can try to build something recurrenceally from here.

I have to think about it.

[Aleksandr Pak]

lna01:

2 Korey: Why should M-qWMA be called via iCustom? It's an extra overhead, it's more logical to make 2 cycles of history in one indicator.

P.S. Or I misunderstood your hint? (it's to Mathemat)

You are welcome!

[Sceptic Philozoff]

Well yes, it should work, Candid. The second differences from the quadratic function are a constant. I wonder how many steps it takes to go from known QWMA[i] to QWMA[i-1]?

[Candid]

Mathemat:
Well yes, it should work, Candid. The second differences from the quadratic function are a constant. I wonder how many steps it takes to go from known QWMA[i] to QWMA[i-1]?

Yes, something a bit cumbersome is going on. For small periods, head-on calculation may be preferable