How to reverse irreversible function in Mathematica?

Question

How to reverse formula

$y(x)=x (\frac{1}{sin \frac{\pi x}{2}})^\alpha$

i.e. express it as

$x = x(y)$

in Mathematica?

I did this way

Clear[α, x, y]
Solve[y == x (1/Sin[Pi x/2])^α, x]

and it answered

During evaluation of In[39]:= Solve::nsmet: This system cannot be solved with the methods available to Solve. >>

Out[40]= Solve[y == x Csc[(π x)/2]^α, x]

How to know, what prevent equation from solvation?

I have plotted formula for some values

Plot[Table[
  x (1/Sin[Pi x/2])^α, {α, {-1, -0.5, 0, 0.5, 1}}], {x, 
  0, 1}]

and found nothing criminal in my expected domain

I have tried to add domain to the formula, but it didn't gave me answer anyway

Solve[y == x (1/Sin[Pi x/2])^α, x, 
 x >= 0 && x <= 1 && α >= -1 && α <= 1]

During evaluation of In[41]:= Solve::bdomv: Warning: x>=0&&x<=1&&α>=-1&&α<=1 is not a valid domain specification. Mathematica is assuming it is a variable to eliminate. >>

During evaluation of In[41]:= Solve::ivar: x>=0&&x<=1&&α>=-1&&α<=1 is not a valid variable. >>

Out[41]= Solve[y == x Csc[(π x)/2]^α, x, 
 x >= 0 && x <= 1 && α >= -1 && α <= 1]

How to set conditions correctly?

How to force to give computable result like as series?

UPDATE

Using InverseFunction I wrote:

MyFun2[x_, α_] := x*(1/Sin[Pi x/2])^α
MyFunInverse2 = InverseFunction[MyFun2, 1, 2]

which gave apparently good result:

Plot[Table[MyFunInverse2[y, α], {α, -1, 1, 0.25}], {y, 
  0, 1}]

but trying to produce computable expression failed:

Series[MyFunInverse2[y, α], {y, 0, 5}]

Out[73]= SeriesData[y, 0, {
InverseFunction[MyFun2, 1, 2][0, α], 
Derivative[1, 0][
InverseFunction[MyFun2, 1, 2]][0, α], 
  Rational[1, 2] Derivative[2, 0][
InverseFunction[MyFun2, 1, 2]][0, α], 
  Rational[1, 6] Derivative[3, 0][
InverseFunction[MyFun2, 1, 2]][0, α], 
  Rational[1, 24] Derivative[4, 0][
InverseFunction[MyFun2, 1, 2]][0, α], 
  Rational[1, 120] Derivative[5, 0][
InverseFunction[MyFun2, 1, 2]][0, α]}, 0, 6, 1]

UPDATE 2

I need to port function to C-like programming language, so final result should consist of "computable" operations. Is it possible?

How does Mathematica itself plots the result? Does just perform optimization algorithm inside? Is it really impossible to do otherwise, for example do with multivariate series?

UPDATE 3

If I take 3 terms of a series

Fwd[x_, \[Alpha]_] := x (1/Sin[Pi x/2])^\[Alpha]
In[115]:= Normal[Series[Fwd[x, \[Alpha]], {x, 0, 3}]]
Fwd1 = Function[{x, \[Alpha]}, %]

I see that result is quite applicable for me

nevertheless I can't reverse this (now much simpler) formula with neither Solve, InverseFunction and InverseSeries.

Isn't it really possible to get some short numeric series with 2-3-4 terms in the case???

Answer 1

If all you're interested in is the inverse power series, then don't calculate the InverseFunction; instead, use the InverseSeries function:

fseries[α_] := Series[x (1/Sin[Pi x/2])^α, {x, 0, 5}]
Table[InverseSeries[fseries[α], y], {α, -1, 1, 1/2}]

Note that $\alpha$ has to be a rational number for this to work. In particular, if you replace the 1/2 with 0.5 in the Table command above, it doesn't work:

fseries[α_] := Series[x (1/Sin[Pi x/2])^α, {x, 0, 5}]
Table[InverseSeries[fseries[α], y], {α, -1, 1, 0.5}]

Answer 2

You can get very close to the solution in three iterations of Newton's method:

f[x_, a_] := x/Sin[Pi/2 x]^a

fInvNewton[y_?NumericQ, a_?NumericQ, iter_: 3] :=
 Nest[# - (f[#, a] - y)/Derivative[1, 0][f][#, a] &, 
  If[y > 2/Pi, y, ((Pi/2)^a y)^(1/(1 - a))], iter]

Example:

Plot[Evaluate@{fInvNewton[y, 0.5, 3], 
   InverseFunction[f, 1, 2][y, 0.5]}, {y, 0, 1}]

LogPlot[Evaluate@
  Abs[fInvNewton[y, 0.5, 3] - InverseFunction[f, 1, 2][y, 0.5]], {y, 
  0, 1}]

Here's some pseudocode for the method:

def fInverseNewton(y, a):
  # choose initial value
  if y > math.pi/2
    x = y
  else:
    x = ((math.pi/2)**a * y)**(1/(1-a))
  # three iterations of netwton's method
  for i in range(3):
    p = math.pi/2*x
    c = math.cos(p)
    s = math.sin(p)
    x = (a*x*p*c - y*s**(a+1)) / (a*p*c - s)
  return x