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FD1[k_, η_] := η^(k + 1) ;  

eqn = FD1[(d - t)/t, ηs] + FD1[(d - t)/t, ηs - vd] == nd;
(* I want to solve eqn for ηs  where d,t,vd are constants in the limit of very small vd*)

Solve[FD1[(d - t)/t, ηs] + FD1[(d - t)/t, ηs] == nd, {ηs}]
(*First I get an estimate of ηs by solving eqn for vd=0 *)

AsymptoticSolve[eqn, {ηs, 2^(-(t/d)) nd^(t/d)}, {vd, 0, 2}]
(*Then I use AsymptoticSolve to get a solution for ηs in Taylor series of vd, I have only started with 3 terms in the series but I also want to see higher terms dependence*)

(*Now I want to put this value of ηs in another expression J *)
J[η_] := FD1[(d - 1)/t, η] - FD1[(d - 1)/t, η - vd];

(*In the above expression for J, I want to evaluate the J[ηS] which I evaluated from eqn*)

How do I use the expression for $\eta s$ from eqn and susbtitue in the expression for $J$. Even when I do it manually, the expression is very huge, and I cannot simplify it. How can Mathematica simplify the expression. I basically finally want to get the dependence of $J$ on $nd$ once I substitute $\eta s$ from eqn into $J$.

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The normal way is simply

FD1[(d - 1)/t, η] - FD1[(d - 1)/t, η - vd]/.AsymptoticSolve[eqn, {ηs, 2^(-(t/d)) nd^(t/d)}, {vd, 0, 2}]
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  • $\begingroup$ This may not work as written, because AsymptoticSolve yields a conditional expression in this case. $\endgroup$
    – bbgodfrey
    Mar 22 '20 at 0:42
  • $\begingroup$ So you need to try and anticipate the effect of the conditions. Try to simplify the problem until it can be explained. $\endgroup$ Mar 22 '20 at 10:48

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