Inverting series with symbolic coefficients?

Question

I am trying to invert the series symbolically. Is this possible in Mathematica?

Example 1 -

Let $p = u + au^2 + bu^3$, where $a,b$ are symbolic variables. I am trying to invert the series around $u=0$ say up to $4$ th order. That is I am trying to get inverse series in terms of $p$ with coefficients in terms of $a,b$.

How do we manage this once we have two dimensional case?

Example 2 -

Suppose we have $$p = u + au^2 + b uv + cv^2$$ $$q = v + dv^2 + euv+ fv^2$$

I am trying to invert the series say upto $4$th order. That is I am trying to get the inverse series in terms of $p,q$. Like $$u = ()p + ()p^2 + ()q + ()q^2 + ()pq + \ldots$$ $$v = ()p + ()p^2 + ()q + ()pq + ()q^2 + \ldots$$, where the coefficients are in terms of $a,b,c,d,e,f$.

I was trying Inverse Series $[u + a u^2 + b u^3 , p]$ but seems like it is not working.

I tried defining a function like $f[u] = u + au^2 + bu^3$ and then using the inverse series Inverse Series$[f[u],p]$. Seems this is also not working.

Any help?

EDIT - Trying Carl Woll suggestions, still I am making mistake somewhere -

asymptoticSolve[args__] := CloudEvaluate[System`AsymptoticSolve[args]]
asymptoticSolve[{p == u + a u^2 + b u v + c v^2, 
  q == v + d u^2 + e u v + f v^2}, {{u, v}, {0, 0}}, {{p, q}, {0, 0}, 
  3}]

results into output

CloudEvaluate[
 AsymptoticSolve[{p == {u + u^2 + u v + c v^2, 
     u + 2 u^2 + 2 u v + c v^2, u + 3 u^2 + 3 u v + c v^2}, 
   q == d u^2 + v + e u v + v^2 Function[x, 4 x (1 - x)]}, {{u, 
    v}, {0, 0}}, {{p, q}, {0, 0}, 3}]]

Not sure why?

Answer 1

You can use the new in M12 function AsymptoticSolve for this:

AsymptoticSolve[
    {
    p == u + a u^2 + b u v + c v^2,
    q == v + d v^2 + e u v + f v^2
    }, 
    {{u, v}, {0, 0}},
    {{p, q}, {0, 0}, 3}
]

{{u -> p - a p^2 + 2 a^2 p^3 - b p q + (3 a b + b e) p^2 q - c q^2 + (b^2 + 2 a c + b d + 2 c e + b f) p q^2 + (b c + 2 c d + 2 c f) q^3, v -> q - e p q + (a e + e^2) p^2 q + (-d - f) q^2 + (b e + 3 d e + 3 e f) p q^2 + (2 d^2 + c e + 4 d f + 2 f^2) q^3}}

Answer 2

Example 1

Let's define series like this:

p[u_] := u + a u^2 + b u^3 + O[u]^4

Inversing series p gives:

q[x_] := Evaluate[InverseSeries[p[x], x]];
q[p]

$p-a p^2+p^3 \left(2 a^2-b\right)+O\left(p^4\right)$

Check that q is an inverse series for p:

q[p[u]]

$u+O\left(u^4\right)$

Example 2

I believe the solution that you are looking for is documented here.

Answer 3

Let us consider your example 1 (I think example 2 can be done in future versions of Mathematica only.). There are several cases depending on parameters and four branches of u as a function of p up to the result of

s = Reduce[p == u + a*u^2 + b*u^3, u] // ToRadicals

(b != 0 && (u == -(a/(3 b)) - (2^(1/3) (-a^2 + 3 b))/( 3 b (-2 a^3 + 9 a b + 27 b^2 p + Sqrt[ 4 (-a^2 + 3 b)^3 + (-2 a^3 + 9 a b + 27 b^2 p)^2])^( 1/3)) + (-2 a^3 + 9 a b + 27 b^2 p + Sqrt[ 4 (-a^2 + 3 b)^3 + (-2 a^3 + 9 a b + 27 b^2 p)^2])^(1/3)/( 3 2^(1/3) b) || u == -(a/(3 b)) + ((1 + I Sqrt[3]) (-a^2 + 3 b))/( 3 2^(2/3) b (-2 a^3 + 9 a b + 27 b^2 p + Sqrt[ 4 (-a^2 + 3 b)^3 + (-2 a^3 + 9 a b + 27 b^2 p)^2])^( 1/3)) - ((1 - I Sqrt[3]) (-2 a^3 + 9 a b + 27 b^2 p + Sqrt[ 4 (-a^2 + 3 b)^3 + (-2 a^3 + 9 a b + 27 b^2 p)^2])^(1/3))/( 6 2^(1/3) b) || u == -(a/(3 b)) + ((1 - I Sqrt[3]) (-a^2 + 3 b))/( 3 2^(2/3) b (-2 a^3 + 9 a b + 27 b^2 p + Sqrt[ 4 (-a^2 + 3 b)^3 + (-2 a^3 + 9 a b + 27 b^2 p)^2])^( 1/3)) - ((1 + I Sqrt[3]) (-2 a^3 + 9 a b + 27 b^2 p + Sqrt[ 4 (-a^2 + 3 b)^3 + (-2 a^3 + 9 a b + 27 b^2 p)^2])^(1/3))/( 6 2^(1/3) b))) || (b == 0 && a == 0 && u == p) || (b == 0 && a != 0 && (u == (-1 - Sqrt[1 + 4 a p])/(2 a) || u == (-1 + Sqrt[1 + 4 a p])/(2 a)))

Let us consider the first case and the first branch. Then

Series[Part[s[[1, 2, 1]], 2], {p, 0, 4}]

does the job (A long output is omitted.).