# Find inverse of a complicated function

I am looking for an analytic inverse function for

u0[alpha_] := (p0 (1/2 (((m1 - m2) (m1 + m2))/p0 + p0) + ((1 -
2 alpha)^2 m1^2 (((m1 - m2) (m1 + m2))/p0 + p0))/(
2 (p0^2 - p0 (((m1 - m2) (m1 + m2))/p0 + p0) +
1/4 (1 - (1 - 2 alpha)^2) (((m1 - m2) (m1 + m2))/p0 +
p0)^2)) - \[Sqrt](((1 -
2 alpha)^4 m1^4 (((m1 - m2) (m1 + m2))/p0 + p0)^2)/(
4 (p0^2 - p0 (((m1 - m2) (m1 + m2))/p0 + p0) +
1/4 (1 - (1 - 2 alpha)^2) (((m1 - m2) (m1 + m2))/p0 +
p0)^2)^2) + ((1 - 2 alpha)^2 m1^4 +
m1^2 (1/2 (-(((m1 - m2) (m1 + m2))/p0) - p0) + p0)^2)/(
p0^2 - p0 (((m1 - m2) (m1 + m2))/p0 + p0) +
1/4 (1 - (1 - 2 alpha)^2) (((m1 - m2) (m1 + m2))/p0 +
p0)^2))))/(p0 + ((1 -
2 alpha)^2 m1^2 (((m1 - m2) (m1 + m2))/p0 + p0))/(
2 (p0^2 - p0 (((m1 - m2) (m1 + m2))/p0 + p0) +
1/4 (1 - (1 - 2 alpha)^2) (((m1 - m2) (m1 + m2))/p0 +
p0)^2)) - \[Sqrt](((1 -
2 alpha)^4 m1^4 (((m1 - m2) (m1 + m2))/p0 + p0)^2)/(
4 (p0^2 - p0 (((m1 - m2) (m1 + m2))/p0 + p0) +
1/4 (1 - (1 - 2 alpha)^2) (((m1 - m2) (m1 + m2))/p0 +
p0)^2)^2) + ((1 - 2 alpha)^2 m1^4 +
m1^2 (1/2 (-(((m1 - m2) (m1 + m2))/p0) - p0) + p0)^2)/(
p0^2 - p0 (((m1 - m2) (m1 + m2))/p0 + p0) +
1/4 (1 - (1 - 2 alpha)^2) (((m1 - m2) (m1 + m2))/p0 +
p0)^2)) + (1 -
2 alpha) \[Sqrt](-m1^2 + (-(((1 -
2 alpha)^2 m1^2 (((m1 - m2) (m1 + m2))/p0 + p0))/(
2 (p0^2 - p0 (((m1 - m2) (m1 + m2))/p0 + p0) +
1/4 (1 - (1 - 2 alpha)^2) (((m1 - m2) (m1 + m2))/p0 +
p0)^2))) + \[Sqrt](((1 -
2 alpha)^4 m1^4 (((m1 - m2) (m1 + m2))/p0 + p0)^2)/(
4 (p0^2 - p0 (((m1 - m2) (m1 + m2))/p0 + p0) +
1/4 (1 - (1 - 2 alpha)^2) (((m1 - m2) (m1 + m2))/p0 +
p0)^2)^2) + ((1 - 2 alpha)^2 m1^4 +
m1^2 (1/2 (-(((m1 - m2) (m1 + m2))/p0) - p0) + p0)^2)/(
p0^2 - p0 (((m1 - m2) (m1 + m2))/p0 + p0) +
1/4 (1 - (1 - 2 alpha)^2) (((m1 - m2) (m1 + m2))/p0 +
p0)^2)))^2));


With p0 > 0, 0 < m1 < p0/2 , 0 < m2 < p0/2. I have already tried to invert this function in full generality with the help of Gröbner basis, with InverseFunction ad with Solve, but nothing succeeded so far. For InverseFunction, i used

Assuming[p0 > 0 && 0 < m1 < 1/2 && 0 < m2 < 1/2 && 0 <= xi <= 1 &&
0 <= # <= 1,
Simplify[InverseFunction[(p0 (1/
2 (((m1 - m2) (m1 + m2))/p0 + p0) + ((1 -
2 #)^2 m1^2 (((m1 - m2) (m1 + m2))/p0 + p0))/(
2 (p0^2 - p0 (((m1 - m2) (m1 + m2))/p0 + p0) +
1/4 (1 - (1 - 2 #)^2) (((m1 - m2) (m1 + m2))/p0 +
p0)^2)) - \[Sqrt](((1 -
2 #)^4 m1^4 (((m1 - m2) (m1 + m2))/p0 + p0)^2)/(
4 (p0^2 - p0 (((m1 - m2) (m1 + m2))/p0 + p0) +
1/4 (1 - (1 - 2 #)^2) (((m1 - m2) (m1 + m2))/p0 +
p0)^2)^2) + ((1 - 2 #)^2 m1^4 +
m1^2 (1/2 (-(((m1 - m2) (m1 + m2))/p0) - p0) + p0)^2)/(
p0^2 - p0 (((m1 - m2) (m1 + m2))/p0 + p0) +
1/4 (1 - (1 - 2 #)^2) (((m1 - m2) (m1 + m2))/p0 +
p0)^2))))/(p0 + ((1 -
2 #)^2 m1^2 (((m1 - m2) (m1 + m2))/p0 + p0))/(
2 (p0^2 - p0 (((m1 - m2) (m1 + m2))/p0 + p0) +
1/4 (1 - (1 - 2 #)^2) (((m1 - m2) (m1 + m2))/p0 +
p0)^2)) - \[Sqrt](((1 -
2 #)^4 m1^4 (((m1 - m2) (m1 + m2))/p0 + p0)^2)/(
4 (p0^2 - p0 (((m1 - m2) (m1 + m2))/p0 + p0) +

1/4 (1 - (1 - 2 #)^2) (((m1 - m2) (m1 + m2))/p0 +
p0)^2)^2) + ((1 - 2 #)^2 m1^4 +
m1^2 (1/2 (-(((m1 - m2) (m1 + m2))/p0) - p0) + p0)^2)/(
p0^2 - p0 (((m1 - m2) (m1 + m2))/p0 + p0) +
1/4 (1 - (1 - 2 #)^2) (((m1 - m2) (m1 + m2))/p0 +
p0)^2)) + (1 -
2 #)*\[Sqrt](-m1^2 + (-(((1 -
2 #)^2 m1^2 (((m1 - m2) (m1 + m2))/p0 + p0))/(
2 (p0^2 - p0 (((m1 - m2) (m1 + m2))/p0 + p0) +
1/4 (1 - (1 - 2 #)^2) (((m1 - m2) (m1 + m2))/p0 +
p0)^2))) + \[Sqrt](((1 -
2 #)^4 m1^4 (((m1 - m2) (m1 + m2))/p0 + p0)^2)/(
4 (p0^2 - p0 (((m1 - m2) (m1 + m2))/p0 + p0) +
1/4 (1 - (1 - 2 #)^2) (((m1 - m2) (m1 + m2))/p0 +
p0)^2)^2) + ((1 - 2 #)^2 m1^4 +
m1^2 (1/2 (-(((m1 - m2) (m1 + m2))/p0) - p0) +
p0)^2)/(
p0^2 - p0 (((m1 - m2) (m1 + m2))/p0 + p0) +
1/4 (1 - (1 - 2 #)^2) (((m1 - m2) (m1 + m2))/p0 +
p0)^2)))^2)) &][u0]]]


But this does not seem to produce any results at least on my machine. Do you have any suggestions how to tackle this issue? Please keep in mind that i am explicitly looking for an analytic function alpha[u0] in the end.

• Would a series approximation suffice? – rhermans Jul 18 '19 at 11:18
• I am pretty sure that an approximation is not even needed. You can calculate the inverse function for a given set of constants like p0 = 1 , m1 = 1/10, m2 = 1/10 and you will see that this expression is very simple, just a square root appears. You can generalize this even further by assuming p0 =1 and m1 = m2 and the result will still be very simple. – RealDestructor Jul 18 '19 at 11:21
• Not invertible: m1 = 1; m2 = 2; p0 = 3; u0[0] == u0[1] returns True. – Henrik Schumacher Jul 18 '19 at 12:09
• This case is forbidden by physics, check out the assumptions. In my restricted area, this function should be bijective. – RealDestructor Jul 18 '19 at 12:53