[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 547

 
alsu:
You assholes.
I think we're doing you a favour. At 11 years old, you should be enjoying your childhood, not surfing the forums and teaching grown-up guys and aunts.
 
alsu:
You assholes. The man may have brightened up and wanted to make everyone happy, but you're in his face in the mud(((


Yeah, that's not good... he's offended and he's gone...

And about enlightenment - that's 100% out of the question....

Otherwise I'd have been on Skype a long time ago.)

 
jelizavettka:


And about enlightenment - that's 100% out of the question....

Otherwise I'd have been on Skype a long time ago)

I'm a little confused as to what's the cause and what's the effect? )))
 
alsu:

I have a problem too))

solve a system of 4 equations to the 4th degree with 4 variables))

a0*b0+a1*b1 = A1

a0^2+4a0*a1*b0*b1+a1^2 = A2

a0*a1*(a1*b0+a0*b1) = A3

a0^2*a1^2 = A4

Ai - parameters, values from the corresponding acceptable area.

Something nothing comes to mind, maybe someone will see a simple solution... And by the way I don't recall how such systems are solved numerically?


I forgot

ai >0

-1 <= bi <= 1


... and most likely even 0 < bi <= 1

I solved it numerically. The result, I must say, is a bit stunning (as it is suspiciously exactly converged to the theoretical prediction), but more about this later.

The question - Newton's method converges well to admissible values for bi, but for ai something goes into negative. Who knows how to take into account constraints on the range of admissible values during iterations?

 
alsu:

Solved it numerically. The result, I must say, is a bit staggering (because it suspiciously exactly converged to the theoretical prediction), but about this later.

The question - Newton's method converges well to admissible values for bi, but for ai something goes into negative. Who knows how to take into account constraints on the range of admissible values during iterations?


Numerically I do not know, but there is an analytical solution. Attached is a short summary of the key third-degree equation with one variable. Unless, of course, I'm wrong.
Files:
alsu_1.zip  5 kb
 

It's bad enough that the equations are heterogeneous. The second equation spoils the whole thing. But there are some symmetry properties.

If (a0,a1,b0,b1) is the solution, then (a1,a0,b1,b0) is too.

Or changing all signs to minus at the same time also gives a solution.

 
jelizavettka:


Yeah, that didn't go well... he got offended and left...

Well, it's kind of boring on the forum now, it was so lively before...
 
sand:
Well, it's kind of boring on the forum now, and it was so lively before...

It's fine. It's calm and familiar :)

Reshetov's been revitalised.

 
TheXpert:

It's fine. Peaceful and familiar :)

Reshetov has stepped up.

Oh, come on! :)

The Solano is making me laugh, and I wish you the same.

 
but from Albert's simplicity? physbook.ru/index.php/Kvant._Inertia_ of the body ;)