AsymptoticSolve for the Inverse

How can I find an asymptotic expansion for the inverse of the function $$f[x]=x(1+x^{1/4})$$ near $$0$$?

I tried substituting $$z=x^{1/4}$$ and using AsymptoticSolve to solve $$y = z^4+z^5$$ for $$z[y]$$ and then do $$x[y] = z[y]^4$$.

(z /. AsymptoticSolve[y == z^(4) + z^(5), {z, 0}, {y, 0, 1}][[1]])^(4)
(* output x[y] = (-y^(1/4) - Sqrt[y]/4 - (7 y^(3/4))/32 - y/4)^4 *)


If this is indeed the inverse, the difference $$x[f[x]]-x$$ should be small. When I do a series expansion for this difference, the leading term is $$2 x^{5/4}$$, but I suspect I am not doing things correctly since if I just try $$x[y] = y(1-y^{1/4})$$ as the inverse of $$f$$, I get a better leading term for $$x[f[x]]-x$$. I get something of the order $$x^{3/2}$$.

• Have you already seen InverseSeries[]? May 21 '20 at 15:23
• Thanks! Something like: InverseSeries[x + x^(4/3) + O[x]^2] ? May 21 '20 at 15:25
• Any help?: z /. AsymptoticSolve[y == z^(4) + z^(5), {z, 0}, {y, 0, 1}][[1]] /. First@Solve[y == z^(4) + z^(5), y] // Series[#, {z, 0, 3}, Assumptions -> z > 0] & -- Note which branch was chosen, though. May 21 '20 at 15:31
• The same as you I guess, but the branch is for the solution to y == x - x^(5/4). Since the 4th root has four branches, you have to pay attention to which was chosen by AsymptoticSolve. What you want is asol = AsymptoticSolve[y == z^(4) + z^(5), {z, 0}, {y, 0, 2}][[4]], the last branch returned instead of the first. May 21 '20 at 16:21
• @user64494 You have to work over the complex numbers. Plot does not do that. May 21 '20 at 16:30

You question is somewhat unclearly formulated. If I correctly understand it, the following does the job.

Series[InverseFunction[# (1 + #^(1/4)) &][x], {x, 0, 2}]


$$x-x^{5/4}+\frac{5 x^{3/2}}{4}-\frac{55 x^{7/4}}{32}+\frac{5 x^2}{2}+O\left(x^{9/4}\right)$$

• Thanks, this answers my question. (what was unclear? perhaps I can edit the question) May 21 '20 at 15:48
• @Cantor: You denote by $x$ both argument of $f(x)$ and argument of $f^{-1}(x)$. May 21 '20 at 15:57
• This solution doesn't seem to work if I change 1/4 to 1/3. May 21 '20 at 18:14

As indicated in one of my comments, $$y = z^4 + z^5$$ has four branches in the neighborhood of $$z=0$$. AsymptoticSolve returns asymptotic series for all four. The last one is the one corresponding to $$y = x+x^{5/4}$$:

ClearAll[asol];
asol[n_] :=  (* n = order of series sought *)
AsymptoticSolve[y == z^(4) + z^(5), {z, 0}, {y, 0, n}][[4]];

z^4 /. asol[2];
Series[%, {y, 0, 2}]
(*
y - y^(5/4) + (5 y^(3/2))/4 - (55 y^(7/4))/32 + (5 y^2)/2 -
(7735 y^(9/4))/2048 + (3003 y^(5/2))/512 -
(609615 y^(11/4))/65536 + O[y]^3
*)


The error that the OP was interested in ($$x(f(x)) - x$$):

z^4 /. asol[1] /. y -> (x + x^(5/4));
Series[% - x, {x, 0, 2}]
(*
-((663 x^2)/512) + (381 x^(9/4))/2048 - (1113 x^(5/2))/8192 +
(3485 x^(11/4))/32768 + O[x]^3
*)

z^4 /. asol[2] /. y -> (x + x^(5/4));
Series[% - x, {x, 0, 3}]
(*
-((13042315 x^3)/2097152) - (34610147 x^(13/4))/8388608 +
(46787 x^(7/2))/4194304 + (284843 x^(15/4))/16777216 + O[x]^4
*)

• As far as I understand it, the inverse function of $y=z^4+z^5$ has five branches in the neighborhood of $y=0$. May 21 '20 at 16:53
• In addition to my previous comment: the result ofTable[Root[-y + #1^4 + #1^5 &, k] /. y -> 0, {k, 1, 5}] is {-1, 0, 0, 0, 0}. May 21 '20 at 17:04
• @user64494 OK, but how does that relate to what I said? May 22 '20 at 13:19