I have used Series & Normal commands a lot to ignore higher order terms in final results. However, I have no idea how to solve a system of equations ignoring higher order terms. For example, consider a system of equations

0, (b)x^2+(4a)y^2+(a^2)xy+(a)x+(b)y+(b-a)F == 0}, {x, y}]. 

Can we solve this system of equations for x and y, ignoring any second or higher order terms in a and b? Of course I can erase such higher order terms in the first place like

0, (b)x^2+(4a)y^2+(a)x+(b)y+(b-a)F == 0}, {x, y}],

but even in that case the solution x and y might have higher order terms in a and b. What I want is to ignore higher order terms in solving a system of equations and get the final results linear in a and b

I would really appreciate if you help me out with this...! (But please don't care about the above specific example too much since I have just made it up randomly.)

  • 1
    $\begingroup$ Please checkand correct your equations: xy should be x y? ab should be a b?fshould be F! If the order O[a] equals O[b]you can solve your problem by substituting {a -> eps a, b -> eps b} and seriesexpansion eps->0 $\endgroup$ Commented Dec 16, 2018 at 12:38
  • $\begingroup$ Are you saying a and b are small, so a a or a b are really small? What is the justification for ignoring your HOTs? It is possible to have a system equations be 'ill conditioned' such that the HOTs are important to the result. One must be careful. $\endgroup$
    – MikeY
    Commented Dec 16, 2018 at 18:10
  • $\begingroup$ 1) Oh, sorry for simple mistakes... Yes, xy and ab should be x y and a b, and F should be f. I will try! 2) Yes, I am trying to make any second or higher order terms consisting of a and b small enough, since the full system of equation is too complicated even for mathematica to be solved. I will try and see if something weird thing happens as you said! $\endgroup$
    – StudyHard
    Commented Dec 17, 2018 at 13:11
  • $\begingroup$ To Neumann. In fact, my original system of equations is so complicated, so Mathematica can't solve it exactly. So I want Mathematica solve this perturbatively with respect to small variables (for example, with respect to a and b above), within Solve command. Can we do that...? $\endgroup$
    – StudyHard
    Commented Dec 18, 2018 at 13:48


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