# Using the solution of AsymptoticSolve into another function

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 $$J$$.

FD1[(d - 1)/t, η] - FD1[(d - 1)/t, η - vd]/.AsymptoticSolve[eqn, {ηs, 2^(-(t/d)) nd^(t/d)}, {vd, 0, 2}]

• This may not work as written, because AsymptoticSolve yields a conditional expression in this case. Mar 22 '20 at 0:42