_Market description - page 18

 
LProgrammer писал(а) >>

You don't seem to have tried it, would you accept the argument that this is how encryption works? You don't have a phase at all. If you don't understand what in the context of this problem is a granular condition, you probably are just reasoning about something you haven't tried. Once again, take a formula, you can even drop the phase from there (I don't seem to remember it now - it all happened long ago) and get your feckin' spectrum and generate a continuation beyond the boundary conditions... By the way... haha ... better not throw away the phase... the disappointment will be quicker, I remember...

Prival, what are you trying to say anyway - that you're cool and I'm shit... OK, but I already told you... You're right, I'm not... How else can I make it clear to you? I leaked (in troll slang) ... And you'll probably get a maths diploma soon. Goodbye. Let's talk about the market.)

Apparently you haven't tried to solve it.

I don't have to solve it, it was solved long ago

Жаном Батистом Жозефом Фурье https://ru.wikipedia.org/wiki/%D0%9F%D1%80%D0%B5%D0%BE%D0%B1%D1%80%D0%B0%D0%B7%D0%BE%D0%B2%D0%B0%D0%BD%D0%B8%D0%B5_%D0%A4%D1%83%D1%80%D1%8C%D0%B5

Will you accept the argument that encryption is based on such a principle?

We can talk about encryption too, but I think you'll get lost there too. Encryption and spectrum are different notions, respectively methods of processing and algorithms. You will be satisfied with the counter-argument that any information can be encrypted and decrypted, even without knowing what the spectrum is, and the crypto-proofness will be great.

You don't have a phase at all.

Easy, here's a phase made by random at all.

If you do not understand what in the context of this zadchy grand. condition. then apparently you are just reasoning that you have not tried.

Yes did not do (not tried), oh the horror of how I get it in 5 minutes programming and show you that you are wrong, I put pictures. Maybe you're not paying attention, it's not Photoshop, it's Matkad.

Fig. Wonder of miracles, again all works, though the initial phase is random ))))

Once again, take the formula, you can even drop the phase from there ( I don't seem to remember, it was a long time ago ) and get your fucking spectrum and generate a sequel with boundary conditions ... By the way haha ... better not throw away the spectrum ... the disappointment will be quicker, I remember...

Why throw it out, you can do it with the phase (see picture above). Very good that, something started to come to mind, can you enlighten me. Second time I ask you, what are the boundary conditions according to you?

It is this "Boundary conditions are conditions connecting values of intensities and inductances of magnetic and electric fields on different sides of surfaces, characterized by certain surface electric charge and/or electric current density".

Or what else? I could add boundary conditions ))))

>> Or really, let's talk about what you know and how to apply it to forex.

 
Prival писал(а) >>

Apparently you haven't tried to solve it.

I don't have to solve it, it was solved a long time ago.

Жаном Батистом Жозефом Фурье https://ru.wikipedia.org/wiki/%D0%9F%D1%80%D0%B5%D0%BE%D0%B1%D1%80%D0%B0%D0%B7%D0%BE%D0%B2%D0%B0%D0%BD%D0%B8%D0%B5_%D0%A4%D1%83%D1%80%D1%8C%D0%B5

Will you accept the argument that encryption is based on this principle?

We can talk about encryption too, but I think you'll get lost there too. Encryption and spectrum are different notions, respectively methods of processing and algorithms. You will be satisfied with the counter-argument that any information can be encrypted and decrypted, even without knowing what spectrum is, and the crypto-proofness will be great.

You don't have a phase at all.

Easy, here's a phase made by random at all.

If you do not understand what in the context of this zadchy grand. condition. then apparently you are just reasoning that you have not tried.

Yes did not do (did not try), oh the horror of how I get it in 5 minutes programming and show you that you are wrong, I put pictures. Maybe you're not paying attention, it's not Photoshop, it's Matkad.

Fig. Wonder of miracles, again all works, though the initial phase is random ))))

Once again, take the formula, you can even drop the phase from there ( I don't seem to remember, it was a long time ago ) and get your fucking spectrum and generate a sequel with boundary conditions ... By the way haha ... better not throw away the spectrum ... the disappointment will be quicker, I remember...

Why throw it out, you can do it with the phase (see picture above). Very good that, something started to come to mind, can you enlighten me. Second time I ask you, what are the boundary conditions according to you?

It is this "Boundary conditions are conditions connecting values of intensities and inductances of magnetic and electric fields on different sides of surfaces, characterized by certain surface electric charge and/or electric current density".

Or what else? I could add boundary conditions ))))

Z.U. Or really, let's talk about what you know and how to apply it to forex.

Halt, this is some nonsense :)). You are talking nonsense...

I've told you everything before about Fourier and his decomposition... Why are you giving me more references? You're just some kind of frankly radio amateur... :))

I was also saying that the goal is not to determine the fuckin' sperkt. :)) Which you're just wiping yourself with... And of which you can not gather anything beyond the range on which you have decomposed the signal ... Did I or did I not tell you about the perfect filter? Do you know what it is? It's a tough job dragging the behemoth out of the swamp... Start with the basics... :)

Next, while talking to you, note respectfully, for I hope that you are sincerely mistaken, in those matters in which you should rather understand the mathematics ... So, next - to explain... At kindergarten level...

there is a signal - you decomposed it into some sinusoids (don't know that it was made out of them! :) ) getting these sinusoids you got nothing... It's a zero result... Do you hear me? These sine waves just draw a signal... the same shape... They're just repeating, but ONLY in the time range that you're drawing it in. Just because you found a way to draw the signal you've found a way to continue.... I don't even know how to explain it to you - I'm afraid I'm starting to get frustrated... shit... In a word, here's the problem, take a line segment, from zero to 10... The function is something like kx + c... And break it down into a Fourier series... And continue it beyond those grunts. By the way, just so you know - grand conditions apply to diff. equations as well... :)) Well so as a result you will receive, not at all direct going further...

Now for the filter phase... Oh, Prival, how you don't know how much you're a bit of a swimmer when it comes to filters... :) Well, practically none of the drying filters gives the right phase... They all distort it... They lag...

In a word, all of them... I'm tired of trying to explain - what do you want from me? :) I told you, you're a genius... I'm a shit, I'm no forex either... :) leave me alone... That's it, you win, just remember to invite me to the award ceremony...

God, I guess I was wrong about you the most. :) That's all I can tell you. You're a genius. Stop. I'm shit. You're right, I'm wrong. End of story?

 

25 again.

I will repeat to you the three conditions I mentioned before. Not your mythical diff. equations, but the ones that anyone who has worked with spectra knows about.

Prival писал(а) >>

Amplitude, frequency and phase are not constant but are functions of time. They change. But this is not a problem. This can be handled. The other two conditions are worse.

2. The sampling frequency is not constant, this is the problem which ruins everything.

3. Kotelnikov's theorem often fails (gaps...).

1. So in your example k*x+b , Kotelnikov's theorem is not satisfied.

2. In the formula you gave, Y=P0+K0*sin(T0+F0)+P1+K1*sin(T1+F1) .... all the coefficients are constant, i.e. do not change with time, hence construction of an optimal filter is possible and since nothing changes, it can be easily built beyond the time interval of analysis.

Take the file I posted earlier and study + model + think + read a proper book. There is a simple example for students. Given y(x)=A*sin(2*pi*f+fi) - by means of PF the parameters A, f and fi are determined.

And now, on my fingers, I'll explain what you're trying to prove here, that it's kind of impossible. To build a sinusoid, you need to know only three parameters amplitude, frequency and phase, if over time they (the parameters) do not change, then by determining at some point in time these parameters, its sinusoid can safely build from - infinity, to + infinity.

LProgrammer to school to learn the basics, not to manage someone what you want, and learn, learn and learn again. Especially this concerns the presentation of your thoughts. So I have not heard from you about boundary conditions, very interesting, three I gave, see above. Bring the 4th one different from the first three.

 
Prival >> :

25 again.

I will repeat to you the three conditions I mentioned before. Not your mythical ones from diff. equations, but the ones that anyone who has worked with spectra knows about.

1. So in your example k*x+b the Kotelnikov theorem is not satisfied.

2. In the formula you gave Y=P0+K0*sin(T0+F0)+P1+K1*sin(T1+F1) .... all the coefficients are constant, i.e. do not change with time, hence construction of an optimal filter is possible and since nothing changes, it can be easily built beyond the time interval of analysis.

Take the file I posted earlier and study + model + think + read a proper book. There is a simple example for a student. Given y(x)=A*sin(2*pi*f+fi) - by means of PF the parameters A, f and fi are determined.

And now, on my fingers, I'll explain what you're trying to prove here, that it's kind of impossible. To build a sinusoid, you need to know only three parameters amplitude, frequency and phase, if over time they (the parameters) do not change, then by determining at some point in time these parameters, its sinusoid can safely build from - infinity, to + infinity.

LProgrammer to school to learn the basics, not to manage someone what you want, and learn, learn and learn again. Especially this concerns the presentation of your thoughts. So I have not heard from you about boundary conditions, very interesting, three I gave, see above. Bring the 4th one different from the first three.

Do you know how to identify targets with the help of this tool? I mean where to place the take, if it's common sense. If not, with all its complexity of work, what for was it needed...

 
Prival >> :


1. amplitude, frequency and phase are not constant but are functions of time. They change. But this is no problem. This can be handled. The other two conditions are worse.

2. The frequency of dicretization is not constant, this is the problem that makes everything fall apart.

3. often Kotelnikov's theorem is not fulfilled (gaps).


It is possible to transform the quotes - by shift of some amplitude,

e.g. by 50 pips, and interpret each such shift as

time step, in this case sampling frequency will be automatically constant.

and the gaps will fill. But in my experience it does not give much - the fractal

The structure of the resulting series differs little from the initial one.

 
Prival писал(а) >>

25 again.

I will repeat to you the three conditions I mentioned before. Not your mythical ones from diff. equations, but the ones that anyone who has worked with spectra knows about.

1. So in your example k*x+b the Kotelnikov theorem is not satisfied.

2. In the formula you gave Y=P0+K0*sin(T0+F0)+P1+K1*sin(T1+F1) .... all the coefficients are constant, i.e. do not change with time, hence construction of an optimal filter is possible and since nothing changes, it can be easily built beyond the time interval of analysis.

Take the file I posted earlier and study + model + think + read a proper book. There is a simple example for students. Given y(x)=A*sin(2*pi*f+fi) - by means of PF the parameters A, f and fi are determined.

And now, on my fingers, I'll explain what you're trying to prove here, that it's kind of impossible. To build a sinusoid, you need to know only three parameters amplitude, frequency and phase, if over time they (the parameters) do not change, then by determining at some point in time these parameters, its sinusoid can safely build from - infinity, to + infinity.

LProgrammer to school to learn the basics, not to manage someone what you want, and learn, learn and learn again. Especially this concerns the presentation of your thoughts. So I have not heard from you about boundary conditions, very interesting, three I gave, see above. Bring the 4th one different from the first three.

:)

Quietly, by 1) :) If you don't understand what I'm talking about by this example then you simply have absolutely no idea of the real (not abstract, theoretical, mathematical ) conditions of application of "this" mathematics.

2) Here are the conditions for your trouble, and I hope it will clear up for you, why you spend so much time "screwing" with these sinusoids :))) and can't get anything in the end... Although I do not even really want, to be honest, to open your eyes to your own delusion ... So, take 10 sinusoids with these parameters as constant components :) (I hope you understand what "it" is) - 10000, 9000, 8000, 0.1, 0.0001, 0.00001, 0.00001, -200, -10000 : and the periods... attention not frequencies as you say here... :)) periods - 1 000 000 000, 2 000 000, 500 000, 50000, 15000, 7000, 150, 110, 60, 13 ; coefficients - 100, 200, 50, 0.1,0.001,0.001,0.00001,-10,-0.1,0.1; by phase shifts, then coefficients at period: 0.5, 0.8, 0.1, 0.33,0.97,-0.3,-0.88,0,-0.77,0.1; . Take and build the resultant function... :) say, on a range of t ( or x as you like) from 0 and up to 1440*365*3 ... The resulting curve, decompose it into your Fourier series or whatever you want... And using your series values, extend the curve further by 15000... And subtract the value from the curve function and display the difference in the range from 0 to the end of the forecast...

:) So, zy, I'm sorry you wasted so much time on this bullshit, I'll be honest, I realized there's nothing in this "super idea" after about a couple of weeks of research. I've already cited a certain Mr. Berg's dissertation on a certain forum, do you know who that is? By the way... Well, if it were as fine as you say, this unhappy man would not have suffered over it, and he would have been carried in his time.

By the way, a question, I'm already curious, maybe you've dug something here, name a filter that has minimal phase distortion?

About the lead, Prival, where did you get that nonsense from? :) I'm sick and tired of being in charge for probably 7 years now... :)

Good luck.

 
Yourmindmy писал(а) >>

Do you know how to identify targets with this tool? Well, I mean where to put the stake, if you want to put it in a common way. If not, then with all the complexity of the work done, what was it for?

He won't be able to tell anything from it except what was at the "0" point :) Too bad for the poor guy...

that Prival, by the way... Did you at least see how you added a phase to the spectrum, and the harmonics... No, and that's exactly what it is... :))

 
By the way, Prival... Do you even understand the essence of the theorem Найквиста-Шеннона применительно к фильтрам и следовательно к тому о чем тут высупаете... ? :) Похоже что нет.

 
LProgrammer писал(а) >>

By the way, Prival... Do you even understand the essence of Nyquist-Shannon theorem in relation to filters and consequently to what you are talking about here... ? :) Apparently not.

You should at least read the last name in there first. Wipe your eyes and read it, I've been telling you about it for 3 pages and you're finally starting to get it. Congratulations, you're beginning to understand something.

Remains to understand the other two limitations, although I think you do not fully understand it, because to understand it you need to work with this.

>> I'm sick of you doing this.

 
Prival >> :

By the way, Prival... Do you even understand the essence of Nyquist-Shannon theorem in relation to filters and consequently to what you are talking about here... ? :) Apparently not.

You should at least read the last name in there first. Wipe your eyes and read it, I've been telling you about it for 3 pages and you're finally starting to get it. Congratulations, you're beginning to understand something.

Remains to understand the other two limitations, although I think you do not fully understand it, because to understand it you need to work with this.

Z.I. I'm tired of your enlightenment.

So don't bother with your lectures and don't get on your and LP's nerves :). And LP is now thinking about how to translate that paradox about the same point into the task at hand, I'm sure of it.