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I have a strange case here! The story begins with this equation

t22b=(r - rh)*(-2*rh*\[Kappa]g - (24*G^2*\[Eta]*\[Kappa]g^2)/rh^2 + 
  (Q^2*(3 - 8*b^2*G*Sqrt[Pi]*(-3*Sqrt[Pi] + 
        Sqrt[4*Pi + Q^2/(b^2*rh^4)])*rh^2 - 
      4*b*G*Sqrt[Pi]*Sqrt[Q^2 + 4*b^2*Pi*rh^4] - 3*rh^2*\[CapitalLambda]) + 
    12*b^2*Pi*rh^4*(1 + 8*b^2*G*Pi*rh^2 - 
      4*b*G*Sqrt[Pi]*Sqrt[Q^2 + 4*b^2*Pi*rh^4] - rh^2*\[CapitalLambda]))/
   (3*(Q^2 + 4*b^2*Pi*rh^4)))

Firstly, one can expand t22b as follows

(Series[t22b, {b, Infinity, 0}] // Normal // PowerExpand // 
    FullSimplify // Expand) // FullSimplify

The result will be

-(((r - rh) (G (Q^2 + 24 G \[Eta] \[Kappa]g^2) + 
    rh^2 (-1 + 2 rh \[Kappa]g + rh^2 \[CapitalLambda])))/rh^2)

Now, solving above term equal to zero for \[Kappa]g we have (note that, this answer is true for my problem)

\[Kappa]g -> (-2 rh^3 + Sqrt[
  4 rh^6 - 96 G^2 \[Eta] (G Q^2 - rh^2 + rh^4 \[CapitalLambda])])/(
 48 G^2 \[Eta]).

In the second method, one can find \[Kappa]g from the eq22b=0 and then expand the answer.

sol1 = Solve[(t22b) == 0, \[Kappa]g] // PowerExpand // FullSimplify
\[Kappa]a = \[Kappa]g /. sol1[[1]];
\[Kappa]b = \[Kappa]g /. sol1[[2]];

The expanded result is

Series[\[Kappa]a, {b, Infinity, 0}] // PowerExpand // FullSimplify
Series[\[Kappa]b, {b, Infinity, 0}] // PowerExpand // FullSimplify

Non of these expanded answers reproduce the \[Kappa]g of previous procedure. Why?! wHy?! WhY?!!! It is driving me crazy :)) Can you help?

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  • $\begingroup$ I do not understand why do you need all these PowerExpand and FullSimplify. Assuming[b > 0 && rh > 0, Series[t22b, {b, Infinity, 0}]]does the job. Notice extra term. $\endgroup$
    – yarchik
    Commented Oct 24, 2019 at 9:12
  • $\begingroup$ @yarchik : yes it is true. However, doesn't matter so much. $\endgroup$ Commented Oct 24, 2019 at 10:37

1 Answer 1

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I get identical answers. Let sA be the result of the series expansion and solution, and let sB be the result of solution followed by the series expansions. There are 2 roots. They are identical modulo permutation.

sA=Assuming[b>0&&rh>0,κg/.Solve[(Series[t22b,{b,Infinity,0}]//Normal)==0,κg]];
sB=Assuming[b>0&&rh>0,Series[κg/.Solve[t22b==0,κg],{b,Infinity,0}]//Normal];
FullSimplify[sA[[1]]-sB[[2]]]
FullSimplify[sA[[2]]-sB[[1]]]
Out[1]= 0
Out[2]= 0
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  • $\begingroup$ Thanks. The problems are missing the Normal command and also conditions on rh and b. Now it is true :) $\endgroup$ Commented Oct 24, 2019 at 16:02

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