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I am trying to solve a set of equations in Mathematica. My input is

Solve[y*x - 1/x - 1/x^2 == 1 && z*x - 1/x^2 + 1/x == 2, {y, z}]

and output is

{{y -> (1 + x + x^2)/x^3, z -> (1 - x + 2 x^2)/x^3}}.

Now I want the answer only upto the order of 1/x, neglecting all higher orders(i.e neglecting 1/x^2, 1/x^3 and so on). If I have a single equation, I am able to get the desired answer (only keeeping 1/x and neglecting higher orders)using AsymptoticSolve.

AsymptoticSolve[y*x-(1/x)-(1/x^2)==1,y,x->inf]

and output,

 y->1/x. 

But for more than 1 equations, AsymptoticSolve is not working. Can anyone tell how can I use AsymptoticSolve to get solution of simultaneous equations keeping only 1/x order terms in answer ?

My input is

AsymptoticSolve[y*x-(1/x)-(1/x^2)==1 && z*x-(1/x^2)+(1/x)==2,{y,z},x->inf]

here, as I want to keep upto 1/x terms in answer , answer should be y-> 1/x, z-> 2/x, but I am getting answer {}. Can anyone help me with this ?

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  • $\begingroup$ inf should be written as Infinity (or one has to define it so), though this doesn't change the result {} of AsymptoticSolve with multiple variables. $\endgroup$
    – tueda
    Dec 25, 2021 at 10:17

1 Answer 1

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As I understand it, Mathematica has a problem with the order of the series expansion at infinity. The following works in 13.0.0 and produces the required result.

AsymptoticSolve[{y*x - (1/x) - (1/x^2) == 1, 
z*x - (1/x^2) + (1/x) == 2} /. x -> 1/t, {{y, z}, {0, 0}}, {t, 0,1}] /. t -> 1/x

{{y -> 1/x, z -> 2/x}}

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  • $\begingroup$ Here, {t,0,1} means we are exapanding t around 0 upto order 1, right ? What does {{y,z},{0,0}} mean ? $\endgroup$
    – apk
    Dec 25, 2021 at 13:36
  • $\begingroup$ @apk: 1. Yes. 2. A solution near the point {0,0}. $\endgroup$
    – user64494
    Dec 25, 2021 at 14:07

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